Subject: Re: Red Shift {Was: Center of Universe?}
From: [email-masked] (john baez)
Date: 23 Jan 1996 13:03:24 -0800
Newsgroups: sci.physics

In article <[email-masked]> [email-masked] (Bronis Vidugiris) writes:
>In article <[email-masked]>,
>Oz <[email-masked]> wrote:

>)Now of course the first thing (trying to keep 4D in mind) is
>)what a tangent in this context would be. In particular a
>)tangent to what? Now I never really needed to do anything
>)with tensors in anger, or even at all. However I did read a
>)little about them many years ago and I vaguely remember
>)deciding they were basically vectors with position.

I haven't seen the original of this post by Oz yet so I'll respond to
this quoted bit.  Everything Vidugiris says is true and good to know,
but let me just say some other stuff that's also true and good to know.

To really understand geometry, hence to understand GR, you gotta
understand tangent vectors.  Tangent to what, you ask?  Tangent to
a given point in spacetime!  What does that mean?  Well, this is easier
to visualize if we consider not 4d curved spacetime, but a 2d curved
space, like the surface of a pumpkin.  (Yes, the pumpkin again.)  Now
the surface of a pumpkin is a "curved 2-dimensional Riemannian manifold", but
it sits conveniently in (more or less) flat 3-dimensional Euclidean
space, so we can think of a tangent vector to it as being an arrow whose
base is at one point of the pumpkin, and which sticks out tangent to the
pumpkin.  We say it's a "tangent vector at a point" of the pumpkin.

Now we have to abstract things a bit!  First, remove the 3d ambient
Euclidean space and think only of the surface of the pumpkin!  We can
still define a "tangent vector"... the actual definition being rather
mathematical... but one way to visualize it is as a teeny-weeny
itsy-bitsy little arrow drawn on the surface of the pumpkin, with its
base at the specified point.  We make it small --- in fact,
infinitesimal --- just in order to avoid worrying about the fact that
the pumpkin is curved.  After all, if we had an ambient 3d space as
before, we could ignore the difference between a vector tangent to the
pumpkin, and a vector actually drawn *on* the pumpkin, in the limit
where the arrow became very small.  

This is the sort of thing mathematicians can make precise; don't worry
about it too much for now.

Now what's a tensor?  Well, there are a million ways to think of it, but
a good way is to think of it as a machine that eats a list of say, 3
tangent vectors, and spits out a number, for example, or maybe a tangent
vector.  (This isn't quite the most general sort of tensor but it's good
enough for starters.)  We require that the output depend in a linear way
on each of the inputs.  

So for example a while back I discussed the Riemann tensor R^a_{bcd}.  
This was a thing that ate 3 tangent vectors and spit out a tangent
vector... which is why there are 3 subscripts and one superscript.  

Namely, if we parallel translate a tangent vector u around the little
parallelogram of size epsilon whose edges point in the directions of the
tangent vectors v and w, it changes by a little bit.  Namely, it changes
by the tangent vector whose component in the a direction is

- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3

Here we sum over b, c, and d.  The thing "v^b" is the component of the
vector v in the b direction... in whatever the hell coordinate system we
happen to be using.  And remember, indices like a,b,c,d range
from 0 to 3 if we are working in 4d spacetime.  

From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 23 Jan 1996 13:13:32 -0800
Organization: University of California, Riverside
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>

In article <[email-masked]> [email-masked] (Bronis Vidugiris) writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:

>)Well, I started by giving a description of curvature: a space (or
>)spacetime) is "curved" if when we take an arrow and carry it around a
>)loop, doing our best to keep it the same length and pointing in the same
>)direction, it may come back rotated.  I gave the example of carrying
>)a horizontally-pointing javelin from the north pole to the equator along
>)a line of constant longitude, then around the equator a bit, then back
>)to the north pole.   It's very good to work this out in detail.  

>Unfortunately, from this description I imagine a gyroscope (allowing
>arrow to pivot vertically so it is always parallel to surface) attached
>to the arrow, and the thing winding up pointing in the original
>direction :-(.

Why you imagine a gyroscope when I say "javelin" is beyond me.  :-)

>Also I imagine being at/near the north pole, going down to the 
>equator, back up, and being very confused about the forth leg!
>(I can only go south from the north pole - what do I do _now_?).

There ain't no fourth leg.  Where'd I say there was a fourth leg?
True, in my later description of the Riemann tensor I talked about
carrying a tangent vector around a little square.  But here we are
carrying it around a big "triangle".  In fact, one can do this parallel
translation game for any sort of loop that ends where it started, or
even any path. 

>Can you (while retaining this degree of informality) also achieve
>the precision to avoid the above (presumably unintentional) interpretation
>of what you actually meant by this?

Well, I really meant just what I said.  Say you were a Roman gladiator
up at the north pole (historical accuracy not being my strong point) and
you were handed a javelin.  "Hold it horizontally, pointing that way,"
says your commander, pointing at a lump of ice at the horizon.  You do
so smartly, an exemplar of military precision.  It points off to your
left.  "Now march forwards!  Go straight ahead, and never rotate the
javelin in the least, under penalty of death!  Stop when you reach the
equator!"  And so on.  You march along, never letting the javelin sway
or rotate in the least.

(If you buy Gauge Fields, Knots and Gravity by Baez and Muniain, you
will see the result on pp. 232-233, although this loop goes from the
north pole all the way to the south pole along a line of longitude, and
then back up along another one.)  

From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 23 Jan 1996 12:43:09 -0800
Organization: University of California, Riverside
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>

In article <[email-masked]> [email-masked] writes:

>...plonking down a set of
>co-ordinates is also somewhat meaningless if they are
>conventional ones since these are usually 'distance'
>co-ordinates (taking time as a distance).

I don't know what "distance coordinates" are.  But I think I vaguely
understand your discomfiture.  You are getting used to the fact that the
world doesn't come with a grid of lines painted on it.  Of course, even
in the old Newtonian days folks knew that any rotated or translated
Cartesian coordinate system was "just as good" as any other.  So it's
not as if they thought there was a particular coordinate system handed
down by God that was the "best" one.  Instead, one had a manageable
family of "good" coordinates, any one of which was related to any other
by an easily understandable sort of transformation: rotation or
translation.  

The same applied back in the days of special relativity.  It may freak
some people out that in addition to rotations and translations one has
"Lorentz transformations" in which the new t' coordinate depends on the
old t and x coordinates (say), but there are still a manageable set of
"best" coordinates, corresponding physically to the inertial frames.

General relativity takes a completely different approach to spacetime.
In general relativity, spacetime is wiggly in a fairly arbitrary way
(though it must satisfy Einstein's equation), so there is in general no
manageable set of "best" coordinates.

Think in terms of curved 2d space if you have trouble visualizing curved
4d spacetime.  (In the long run you should learn how to visualize curved
4d spacetime, but visualizing curved 2d space is the best way I know to
get to that point.)  Say someone hands you a flat piece of paper.  Then
you can draw a Cartesian coordinate system on it in which every line is
straight and all intersecting lines meet at right angles.  There are
different ways to do it, but they all differ only by rotations and
translations.  (And reflections, if one wants to nitpick.)

Now say someone hands you a pumpkin.  It's wiggly and bumpy... so
there's no obvious best way to coordinatize its surface, not even any
obvious manageable set of better-than-average ways.   Say you try to
draw a straight line on it.  The best you can do is to draw a
"geodesic": a curve that goes "locally as straight as possible".  This
is the curve a very tiny ant might follow if it was doing its best
to follow its nose and walk as straight as possible.  

(Subconsiously, this should remind you of the fact that free fall in
curved spacetime is motion along a geodesic, "locally as straight as
possible", i.e. feeling no acceleration.  I'm secretly indoctrinating
you in a new worldview: physics as geometry.)

Okay, so you don't have "straight lines" but you do have "geodesics" on
a pumpkin.  So say you try to draw a grid on the pumpkin such that each
curve in it is a geodesic and whenever two geodesics intersect, they do
so at right angles.  If you could do that, it would be a good stab at
Cartesian coordinates.  But you can't.  That's because the surface of
the pumpkin is a "curved 2-dimensional Riemannian manifold" --- with the
emphasis here on *curved*.  

So what do we do in general relativity, where spacetime is as bumpy as
the surface of a pumpkin.  We give up all attempts to pick "best" or
"good" coordinates ---- except in working on certain very special
problems with lots of symmetry! ---- and decide to do things in such a
way that ANY choice of coordinates will work as smoothly as ANY OTHER.  

We don't exactly abandon coordinates; we just relegate them to the
status of completely arbitrary tools.  

>I wonder if a lot of my confusion is in the intermix of
>cosmology and GR. Too much to follow all in one shot, too
>many new ways to look at things at once.

Certainly this is a big part of it.  Personally I can't teach you
cosmology without teaching you GR, any more than I could teach you
celestial mechanics without teaching you F = ma.  There might be some
way to study cosmology without GR, but to me that seems to miss the whole
grandeur and strangeness of the subject.  

For example, it's true that one could not bother learning GR and only
try to understand one solution of Einstein's equation, the
Robertson-Friedman-Walker metric which describes the standard big bang
cosmology.  This might allow one a certain tempting conceptual
sloppiness: one could write this metric down in the "standard
coordinates" which take advantage of the symmetry of that solution, and
ignore my remarks above about the arbitrariness of coordinates.  But one
would really be missing a lot of the fun!  For example, you've already
seen that in the Milne cosmology, different coordinates can give one
very different pictures of what's going on.  This mental flexibility is
crucial if one gets into questions like "why is there a redshift: is it
really due to the expansion of space, or is it a gravitational effect?"
Without understanding physics as geometry and the arbitrary nature of
ones choice of coordinates, these riddles can really throw one for a
loop.  *With* this understanding one can, so to speak, deconstruct the
question and figure out the *right* question and the right answer.

>Perhaps a simple GR example might help. I have assumed that
>the moon orbits the earth in a GR explanation because it
>simply travels in a straight line, but space (3D) is curved
>(ie orbital sized curving). However I am getting the
>impression that the GR curvature is very very tiny. 

Yes indeed. 

>I also
>remember Baez commenting that in spacetime the geodesic of a
>stone tossed up into the air is very very long and only very
>slightly curved since the time distance is ct. Presumably
>then the spacetime curvature that bends the moons orbit
>round the earth is similarly only curved in a minuscule and
>almost undetectable amount, but because of the 20 lightday
>distance in the time direction we see it as taking a very
>curved path in our *almost flat 3D slice* of spacetime. I
>hope this has a small element of correctness in it because I
>am finding it difficult piecing all the various questions
>and answers together to form some coherent model that
>doesn't fall foul of your devastatingly correct objections.

Yes, this is right.  Visualize the x and y directions as lying in the
horizontal plane and the t direction as pointing "up".  Then the geodesic
path traced out by the moon is a very very stretched-out helix which
goes "up" in time 28 lightdays each time it goes around, while its
radius is less than a measly lightminute.  This is very close to what a
geodesic would be in flat spacetime (namely, a straight line).  That's
as expected, because the curvature is so small.

Article 95270 (3700 more) in sci.physics:
From: Michael Weiss
(SAME) Subject: Re: Feeling stressed out?
Date: 22 Jan 1996 22:20:37 GMT
Organization: OSF Research Institute
Lines: 22
NNTP-Posting-Host: pleides.osf.org
In-reply-to: [email-masked]'s message of 21 Jan 1996 21:15:09 -0500
cc: [email-masked]

You raise a bunch of interesting questions; I won't have time (or
space!) now to make even a decent stab at an explanation.  I think you
might enjoy looking at Eddington's discussion in "The Nature of the
Physical World"; there are always the usual book recommendations
(e.g., Taylor and Wheeler, "Spacetime Physics").

The short answer is yes, not all distinctions between time and space
are erased in relativity.  We can say that the event "the pitcher
throws the ball" occurs *before* the event "the batter hits that ball,
on that pitch", and this notion of "before" is absolute.

If you're athletic enough, you can pick any spatial direction for your
x-axis, but you can't interchange the x-axis with the t-axis.

This distinction can be expressed mathematically in a number of ways.
For example, in the Minkowski formula for the spacetime "interval":

     ds^2 = dx^2 + dy^2 + dz^2 - dt^2

the dt^2 is the only term that gets a minus sign.  Something called
Sylvester's Theorem on Signatures implies that we'll get the same
pattern of +'s and -'s no matter what frame of reference we use.

Article 95116 (3694 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Feeling stressed out?
Date: 22 Jan 1996 12:40:51 -0800
Organization: University of California, Riverside
Lines: 93
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:

>Suppose somebody plopped us down at rest in a strange coordinate system.
>Now,  none of us would have any trouble identifying which of the four
>dimesions in this frame was X_0, aka time, unless we had been chewing a few
>too many peyote buttons. 

Your claim here ambiguous because you don't say what if anything being
"at rest" has to do with the coordinate system.  Is one of the
coordinate directions the direction your worldline is pointing along ---
in which case we'd be tempted to call that direction "time" --- or
is there no relation between the coordinates and your worldline?

Rather than attempting to grapple with this ambiguity let me just state
some facts in order of diminishing obviousness.

1) Given a point in spactime, there is no such thing as "the time
direction" at that point.  I.e., there are lots of tangent vectors at
that point, pointing in all sorts of directions, and a bunch of them are
timelike, a bunch are spacelike, and a bunch are lightlike.  But there's
no way (given the data provided: just the point in spacetime) to pick
out one timelike vector and say it is "the" time direction.  This is
already true in special relativity: if we Lorentz boost any timelike
vector we get another equally good timelike vector.

2) If you are made of ordinary matter, the tangent vector to your
worldline is timelike.  I.e., you can't go fast than light.  Thus at any
given point along your worldline we can erect a (nonunique, local) coordinate
system about that point, say (t,x,y,z), such that the tangent vector to
your worldline points in the t direction, and all vectors pointing in
the t direction are timelike.  If we did this for else moving relative
to you, we'd get a different coordinate system (see 1 about Lorentz
boosts).  

3) If someone just hands you a random coordinate system on some patch of
space, there is no good way to pick out which of the four coordinates to
call "time".  Consider for example the usual coordinates on Minkowski
space, (t,x,y,z), and then define new coordinates (T,X,Y,Z) with

T = t + x
X = t + y
Y = t + z
Z = y - z

Despite the alluring names T, X, Y, and Z, you would be hard pressed to
say any one of these coordinates "was time".  Note that (in units with c
= 1, of course) the coordinate directions T, X, and Y are lightlike,
while Z is spacelike.

>I had hoped somebody with an abstract mathematical bent would say something
>like this:  "Ok,  while we loosely call both the objects relating to
>crystals in three space and the objects living in four dimensional
>spacetime "tensors",  one is actually a more general concept.  In ordinary
>three space all the coordinates have the same flavor,  but in spacetime one
>of them has a different flavor,  and no matter how we mix them,  one of
>them always comes out tasting more of time than the others." 

Remark 3 certainly shows that no one coordinate need be more like time
than the others.  However, what's true is that in spacetime (or more
technically, a "Lorentzian manifold") some directions are timelike, some
are spacelike, and some are lightlike, and they have very different
flavors.

>"We capture this mathematically by saying that instead of living on a
>single n-space, real or complex,  these tensors actually live on n
>"timelike" dimensions,  n "spacelike" dimesions, and possibly, k "k-like"
>dimensions, l "l-like" dimensions,  and so forth.  When we transform
>coordinates they are all mixed together,  but it always turns out we can
>identify the same number of coordinates of the same flavors we started
>with.  These tensors are really defined on direct sum spaces of the form (
>n | m | l | k | ...).  Their theory is due to [fill in eastern European
>sounding name]... and they are a treated in the branch of mathematics call
>[fill in any suitable sounding name]... :-) " 

Well, what we really say is that there's a theory of n-dimensional
semi-Riemannian manifolds of signature (p,q), where p + q = n.  This
means that at any point we can find an orthonormal basis of tangent
vectors, p of which are spacelike and q of which are timelike.  (Or the
other way round, depending on your conventions!)  This has been
intensively studied, but the most deeply studied cases are the case
where q = 0 --- the "Riemannian" case, which one might think of
physically as the geometry of "space" --- and the case where q = 1 ---
the "Lorentzian" case, which one might think of physically as the
geometry of "spacetime".  One can study worlds with 2 timelike and 2
spacelike directions... and in fact people do; the symmetry makes things
very interesting... but from the viewpoint of physics this is rather
decadent.  

To get started, read

Semi-Riemannian geometry : with applications to relativity / Barrett
   O'Neill.  New York : Academic Press, 1983.

Article 95097 (3693 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Feeling stressed out?
Date: 22 Jan 1996 18:39:48 GMT
Organization: University of California, Riverside
Lines: 85
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>Yet more. 

>I left my house today with a warm glow of understanding.  Now I knew what
>the stress-energy tensor is.  Then,  on the NYC subway,  a dark thought hit
>me:   John Baez was pulling my leg! 

No, I wasn't.

>Yes. 

No.

>He claims that this object is a "tensor".  Now,  what little I know
>about tensors leads me to believe they represent some linearization of a
>local property of space.  Their transformation properties are a guarantee
>that they express a physical property of space,  not an artifact of
>coordinates. 

It's a guarantee that they express a physical property of space(time) or
*stuff in it*.  Here by stuff I mean fields and that sort of thing.

>So what's so special about the top row??? 
>In three space the top row of a two index object may be called "x",  but
>conceptually it's not any different from "y".  There is no special "x-ness"
>about this position.  So how can we say the top row of T keeps tract of
>flow wrt *time*.  Sounds to me like we are trying to shoe-horn time into
>some calculational scheme,  and it doesn't quite fit.  Even if different
>observers have different t's,  there is still something intrinsically
>"timelike" about this row sticking out like a sore thumb. 

We can work with tensors using whatever coordinates we want, and their
transformation properties are a guarantee that if we watch our step, our
results will not be a mere artifact of a coordinate choice.  Pick any
local coordinates x^0, x^1, x^2, x^3 on spacetime whatsoever.  Use
these to define the notion of a tangent vector in the ith direction (i =
0,1,2,3).  Use them also to define the notion of momentum in the jth
direction (0,1,2,3).  The entry T_{ij} in the ith row and jth column of
the tensor T then keeps track of the flow in the ith direction of the
jth component of momentum.  

Nothing special about the top row of T_{ij} here!

It is merely a matter of *convention* that, if we have a metric on
spacetime, and thus a way of determining whether a given vector is
timelike or spacelike, that we often use coordinates such that the
tangent vector in the 0th direction is timelike, while those in the 1st,
2nd, and 3rd directions are spacelike.  

If we follow this convention, it is then common to indulge in a quaint
anachronism and call momentum in the 0th direction "energy".  It is also
common to call flow in the 0th direction "density".  

That's what I was doing.

But I'm not sure that's what you're worrying about.  Maybe you're
worrying about the fact that in spacetime, as opposed to space, not all
directions were created equal?  Some are spacelike and some are timelike!
A tangent vector v is spacelike if its dot product with itself --- which
we write as g(v,v) --- is positive, and timelike if g(v,v) is negative.

(The gadget g is called the metric; we'll probably talk about that a lot
more later if I wind up teaching a sci.physics course on general
relativity.  Often people use the opposite sign conventions from those
above.)  

Now if you ask me why one can find a basis of orthogonal tangent vectors
with 3 spacelike ones and 1 timelike one --- i.e., why there is only one
dimension's worth of space and 3 of time --- I have to say I don't know.
Nobody knows.  Not yet, anyway.

But don't mistake that puzzle for anything funny about *coordinates*.  
It's a completely coordinate-independent fact --- unlike the fact that
folks like to use coordinates in which the x^0 coordinate corresponds to
a time-like direction.  The statement "there exists a basis of
orthogonal tangent vectors with 3 spacelike ones and 1 timelike one" is
a completely coordinate-independent statement.

Article 95066 (3692 more) in sci.physics:
From: Edward Green
(SAME) Subject: Re: Feeling stressed out?
Date: 21 Jan 1996 21:15:09 -0500
Organization: The Pipeline
Lines: 53
NNTP-Posting-Host: pipe11.nyc.pipeline.com
X-PipeUser: egreen
X-PipeHub: nyc.pipeline.com
X-PipeGCOS: (Edward Green)
X-Newsreader: The Pipeline v3.4.0

[email-masked] (Michael Weiss)' wrote: 
> 
>4.  So what's so special about the top row???  
> 
>Nothing.  No more is there anything special about the 0-component of 
>the energy-momentum 4-vector 
 
Thank you for your answers!   I still feel there must be something special
about the top-row,  though where it falls between the mathematics and the
physics I couldn't say.  I gave this my best shot last time,  but let me
try one more analogy. 
 
Suppose somebody plopped us down at rest in a strange coordinate system.
Now,  none of us would have any trouble identifying which of the four
dimesions in this frame was X_0, aka time, unless we had been chewing a few
too many peyote buttons.  But if somebody dropped a set of orthogonal
spacial axes in our lap, and asked us "which one is X_1",  we could rightly
answer "how the hell should I know,  jack?  which ever one you want it to
be,  ok?"  (I guess we are a little testy from being plopped in a strange
coordinate frame).  The point is,  even though our transformations mix time
with space,  when we stop and consider a particular one there is always a
special dimension sticking out,  a dimension physically different from the
others,  which we call "time". 
 
I had hoped somebody with an abstract mathematical bent would say something
like this:  "Ok,  while we loosely call both the objects relating to
crystals in three space and the objects living in four dimensional
spacetime "tensors",  one is actually a more general concept.  In ordinary
three space all the coordinates have the same flavor,  but in spacetime one
of them has a different flavor,  and no matter how we mix them,  one of
them always comes out tasting more of time than the others." 
 
He continues... 
 
"We capture this mathematically by saying that instead of living on a
single n-space, real or complex,  these tensors actually live on n
"timelike" dimensions,  n "spacelike" dimesions, and possibly, k "k-like"
dimensions, l "l-like" dimensions,  and so forth.  When we transform
coordinates they are all mixed together,  but it always turns out we can
identify the same number of coordinates of the same flavors we started
with.  These tensors are really defined on direct sum spaces of the form (
n | m | l | k | ...).  Their theory is due to [fill in eastern European
sounding name]... and they are a treated in the branch of mathematics call
[fill in any suitable sounding name]... :-) " 
 
I hope this doesn't seem too sophmoric or argumentative coming from someone
who started with so many questions!  Call it mathematical fiction if you
like.  Does life imitate art? 
  
-- 
 
Ed Green 
[email-masked]
Article 95294 (3690 more) in sci.physics:
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
From: Oz
Date: Mon, 22 Jan 1996 08:45:19 GMT
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
Lines: 61

>In article <[email-masked]> [email-masked] writes:
>>Cosmologists often talk about the curvature of *space* (meaning
>>ordinary three-dimensional space) rather than spacetime.  This is
>>perhaps unfortunate, since it confuses people like Oz, but that's the
>>way it is. 

>[email-masked] (john baez) wrote:
>Oh, so that's what the confusion is about.  Sorry.  Yes indeed, if you
>choose to slice up spacetime into slices called "space", the curvature
>of the slices depends on how you do the slicing.  Indeed, the slices can
>be curved even if the spacetime itself is flat.  Here's an example that
>starts with 3d space rather than 4d spacetime: take 3-dimensional
>Euclidean space --- nice and flat --- and slice it up into a bunch of
>concentric spheres --- which are curved.

I am a little loath to ask a question, since I seem to have
irritated the supervisors (apologies). However I am left in
a rather unsatisfactory situation. Obviously I need a new
paradigm, my quasi-classical one having been comprehensively
demolished.

For example.
I can see than 'distance' is not a description of anything,
cosmologically. As a result, plonking down a set of
co-ordinates is also somewhat meaningless if they are
conventional ones since these are usually 'distance'
co-ordinates (taking time as a distance). Another measure or
mechanism is needed. Even the initially non-physical
co-moving co-ordinate system now looks to be more attractive
although I am beginning to suspect that it sort of
linearises the time direction only, dunno.

I can also see why Baez insists that we can only know what
is observed at a point. This cannot be ambiguous.

I wonder if a lot of my confusion is in the intermix of
cosmology and GR. Too much to follow all in one shot, too
many new ways to look at things at once.

Perhaps a simple GR example might help. I have assumed that
the moon orbits the earth in a GR explanation because it
simply travels in a straight line, but space (3D) is curved
(ie orbital sized curving). However I am getting the
impression that the GR curvature is very very tiny. I also
remember Baez commenting that in spacetime the geodesic of a
stone tossed up into the air is very very long and only very
slightly curved since the time distance is ct. Presumably
then the spacetime curvature that bends the moons orbit
round the earth is similarly only curved in a minuscule and
almost undetectable amount, but because of the 20 lightday
distance in the time direction we see it as taking a very
curved path in our *almost flat 3D slice* of spacetime. I
hope this has a small element of correctness in it because I
am finding it difficult piecing all the various questions
and answers together to form some coherent model that
doesn't fall foul of your devastatingly correct objections.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 95018 (3683 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 21 Jan 1996 13:09:30 -0800
Organization: University of California, Riverside
Lines: 37
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:
>[email-masked] (john baez)' wrote: 

>>Think about this.  Say you start at the north pole holding a javelin 
>>that points horizontally in some direction, and you carry the javelin to 
>>the equator, always keeping the javelin pointing "in as same a direction 
>>as possible", subject to the constraint that it point horizontally, 
>>i.e., tangent to the earth.  (The idea is that we're taking "space" to 
>>be the 2-dimensional surface of the earth, and the javelin is the 
>>"little arrow" or "tangent vector", which must remain tangent to 
>>"space".)  After marching down to the equator, march 90 degrees around 
>>the equator, and then march back up to the north pole, always keeping  
>>the javelin pointing horizontally and "in as same a direction as 
>>possible".    

>>By the time you get back to the north pole, the javelin is pointing a 
>>different direction! 

>>That's because the surface of the earth is curved.  

>I'm sorry,  but I have a really obvious question here,  so obvious you
>didn't address it.  What do we do with the javelin at the corners?  Do we
>try to hold it "pointing in the same direction" as near as possible,  while
>we turn,  or do do we rotate it with us by the angle we turn through? 

Hold the javelin pointing in the same direction even at the corners!
The idea is to do your damnedest to never rotate that sucker, even at
corners.  We want to study how, even when we do our best not to rotate a
tangent vector as we carry it around, it can come back rotated due to
the curvature of spacetime (or space, or the earth).

Also, since corners are just a limiting case of sharp turns without
corners, it would be a bad rule to do something different at corners
than at non-corners.  

Article 95016 (3682 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 21 Jan 1996 13:03:04 -0800
Organization: University of California, Riverside
Lines: 73
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:
>>No.  Curvature is a thing that has meaning independent of coordinates;
>>no coordinate change can turn a curved space -- or spacetime -- into a
>>flat one, or vice versa.  

>Cosmologists often talk about the curvature of *space* (meaning
>ordinary three-dimensional space) rather than spacetime.  This is
>perhaps unfortunate, since it confuses people like Oz, but that's the
>way it is. 

Oh, so that's what the confusion is about.  Sorry.  Yes indeed, if you
choose to slice up spacetime into slices called "space", the curvature
of the slices depends on how you do the slicing.  Indeed, the slices can
be curved even if the spacetime itself is flat.  Here's an example that
starts with 3d space rather than 4d spacetime: take 3-dimensional
Euclidean space --- nice and flat --- and slice it up into a bunch of
concentric spheres --- which are curved.

Whenever you do this slicing stuff, say with spacetime, there are nice
relationships between: 

1. the curvature of the spacetime

2. the curvature of the slices themselves --- their so-called "intrinsic
curvature" 

and

3. the curvedness of how the slices sit inside the spacetime --- their
so-called "extrinsic curvature".

These are called the Gauss-Codazzi relations, and they are very
important whenever we want to take spacetime, slice it, and study
general relativity that way.  I won't go in to them here, but for
starters it's worthwhile trying to understand that 1, 2, and 3 are
different concepts.

> When someone asks whether space is curved, what they mean
>is this.  Choose a particular instant of time, and consider the slice
>through four-dimensional spacetime that represents space at that one
>instant.  This is a perfectly good three-dimensional manifold, and
>it's perfectly reasonable to ask whether or not this manifold is
>curved.  

Yes, this is the so-called "intrinsic curvature" of the slice we're
calling "space".

>Worse, the
>question of whether space is curved depends on what coordinates you
>use.  Specifically, choosing a particular "instant of time" everywhere
>in space is a coordinate-dependent act, because there's no
>coordinate-independent way of deciding whether two distant events are
>at the same time.  So your answer to the question of whether space (as
>opposed to spacetime) is curved will in general depend on how you
>choose to slice spacetime up.

I'd prefer to say, not that the curvature of the slice depends on what
coordinates you use, but that it depends on the slice!

In other words, I like the last sentence in the above bit more than the
first one.  Of course, if you define your slice by taking a "time
coordinate" t and letting your slice be described by the equation t =
constant, the slice, hence its curvature, depends on the coordinate t.  

This indeed comes up a lot, since slicing spacetime into slices of
"space" is a very practical tool.  Still, it's worth keeping in mind
Eddington's remark, that if we're trying to study the anatomy of a pig
it can be rather confusing to study bacon.

Article 95340 (3681 more) in sci.physics:
From: john baez
Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 23 Jan 1996 13:03:24 -0800
Organization: University of California, Riverside
Lines: 66
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Br
onis Vidugiris) writes:
>In article <[email-masked]>,
>Oz <[email-masked]> wrote:

>)Now of course the first thing (trying to keep 4D in mind) is
>)what a tangent in this context would be. In particular a
>)tangent to what? Now I never really needed to do anything
>)with tensors in anger, or even at all. However I did read a
>)little about them many years ago and I vaguely remember
>)deciding they were basically vectors with position.

I haven't seen the original of this post by Oz yet so I'll respond to
this quoted bit.  Everything Vidugiris says is true and good to know,
but let me just say some other stuff that's also true and good to know.

To really understand geometry, hence to understand GR, you gotta
understand tangent vectors.  Tangent to what, you ask?  Tangent to
a given point in spacetime!  What does that mean?  Well, this is easier
to visualize if we consider not 4d curved spacetime, but a 2d curved
space, like the surface of a pumpkin.  (Yes, the pumpkin again.)  Now
the surface of a pumpkin is a "curved 2-dimensional Riemannian manifold", but
it sits conveniently in (more or less) flat 3-dimensional Euclidean
space, so we can think of a tangent vector to it as being an arrow whose
base is at one point of the pumpkin, and which sticks out tangent to the
pumpkin.  We say it's a "tangent vector at a point" of the pumpkin.

Now we have to abstract things a bit!  First, remove the 3d ambient
Euclidean space and think only of the surface of the pumpkin!  We can
still define a "tangent vector"... the actual definition being rather
mathematical... but one way to visualize it is as a teeny-weeny
itsy-bitsy little arrow drawn on the surface of the pumpkin, with its
base at the specified point.  We make it small --- in fact,
infinitesimal --- just in order to avoid worrying about the fact that
the pumpkin is curved.  After all, if we had an ambient 3d space as
before, we could ignore the difference between a vector tangent to the
pumpkin, and a vector actually drawn *on* the pumpkin, in the limit
where the arrow became very small.  

This is the sort of thing mathematicians can make precise; don't worry
about it too much for now.

Now what's a tensor?  Well, there are a million ways to think of it, but
a good way is to think of it as a machine that eats a list of say, 3
tangent vectors, and spits out a number, for example, or maybe a tangent
vector.  (This isn't quite the most general sort of tensor but it's good
enough for starters.)  We require that the output depend in a linear way
on each of the inputs.  

So for example a while back I discussed the Riemann tensor R^a_{bcd}.  
This was a thing that ate 3 tangent vectors and spit out a tangent
vector... which is why there are 3 subscripts and one superscript.  

Namely, if we parallel translate a tangent vector u around the little
parallelogram of size epsilon whose edges point in the directions of the
tangent vectors v and w, it changes by a little bit.  Namely, it changes
by the tangent vector whose component in the a direction is

- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3

Here we sum over b, c, and d.  The thing "v^b" is the component of the
vector v in the b direction... in whatever the hell coordinate system we
happen to be using.  And remember, indices like a,b,c,d range
from 0 to 3 if we are working in 4d spacetime.  

Article 95253 (1 more) in sci.physics:
From: Bruce Scott TOK 
(SAME) Subject: Re: Feeling stressed out?
Date: 22 Jan 1996 19:19:27 GMT
Organization: Rechenzentrum der Max-Planck-Gesellschaft in Garching
Lines: 61
NNTP-Posting-Host: s4bds.aug.ipp-garching.mpg.de
X-Newsreader: TIN [version 1.2 PL2]

Edward Green ([email-masked]) wrote:

: Thank you for your answers!   I still feel there must be something special
: about the top-row,  though where it falls between the mathematics and the
: physics I couldn't say.  I gave this my best shot last time,  but let me
: try one more analogy. 

[...]

The way you make the top row unique is to look at it in the rest frame
of the thing whose stress tensor you're studying.  For example, a
dissipationless fluid has a stress tensor of

    T^ab = n e u^a u^b + p (g^ab + u^a u^b)

(n is the particle density in the fluid's rest frame, e is the thermal
energy per particle, and p is the pressure), which looks a bit
complicated.  In fact, there is _nothing_ special about the top row of
this if g^ab has time-space components (example: Kruskal coordinates for
the black hole problem) and u is relativistic (especially if it has
significant variation in any of the four coordinates).  Note that the
only physical requirement on u is that it must be a timelike _unit_
vector:  u_a u^a = -1 (in units with c = 1).

But if you contract this with the fluid's velocity you get
a peek at the fluid's rest frame:

    - T^ab u_b = n e u^a.

Now, that looks a lot more like an energy flux, doesn't it?  It is in
fact a four-vector.  But what about the pressure?  You can find the
fluid's rest frame by transforming such that u^a = (1,0,0,0).  Note you
have to do a local Lorentz transformation: one that is defined for the
small neighborhood about a single point.  This is because u for a fluid
is itself a field variable.  But in the local rest frame you find

    T^ab = diag ( ne, p, p, p ),

where "diag" denotes a diagonal matrix.  The 00-component is the thermal
energy density and the ii-components are the pressure (one assumes an
isotropic pressure for a ideal fluid).

Note that "thermal energy" here includes rest energy.

Your question was actually aimed at distinguishing among the
ii-components.  This is, as you surmise, not possible without
arbitrariness, unless there is a physical phenomenon (such as a local
magnetic field; gee, isn't plasma physics nice :-] ) to show you the
way. 

There is a popular book I saw once where the question of explaining left
and right to an extraterrestrial was discussed, to show the
arbitrariness in our convention.  A complicated set of mental gymnastics
was offered by which one could use symmetry violation in kaon decays to
do this.  No, I don't remember the details :-)

--
Mach's gut!
Bruce Scott                                The deadliest bullshit is
Max-Planck-Institut fuer Plasmaphysik       odorless and transparent
[email-masked]                               -- W Gibson

From: Michael Weiss
Organization: OSF Research Institute
Subject: Re: Feeling stressed out?
Date: 23 Jan 1996 20:01:12 GMT
Organization: OSF Research Institute
Lines: 41
NNTP-Posting-Host: pleides.osf.org
In-reply-to: [email-masked]'s message of 22 Jan 1996 22:20:37 GMT
cc: [email-masked]

Between John Baez's beautiful job on the difference between
"time-like" separations and the t-axis, and Bruce Scott's nice
explanation of the stress-energy tensor for the perfect fluid, I don't
know that there's much to add.  However, I do remember one thing about
T_ij that puzzled me when I first encountered it, and the "aha" mental
lightbulb that cleared things up.

Here's the puzzle: if T_ij measures the amount of i-momentum being
transferred in the j-direction, then how come T_ij isn't identically
zero in the rest frame of the fluid, when the momentum vanishes?

Let's polish off a couple of points quickly.  As Bruce Scott points
out, in general a fluid doesn't have a global rest frame.  We can say
we're just talking about an infinitesimal neighborhood of a point, or
the special case of a motionless fluid.

Next, T_00 is a special case.  In the rest frame of the fluid, the
velocity 4-vector of a fluid element is (E,0,0,0) -- you can make the
v_i components vanish for i=1,2,3 by picking the right frame of
reference, but not the v_0 component, aka the energy.  (I'm being
sloppy about the difference between tensors and tensor-densities, but
let's ignore that.  What's a determinant among friends?)  So it's not
surprising that T_00 doesn't vanish, but why doesn't T_ii vanish for
i=1,2,3?  (Once we've plopped ourselves down in the rest frame of the
fluid, that is.)

Answer: think microscopically, and remember that T depends
*quadratically* on v.  Consider the yz-plane, for yucks.  Particles of
fluid constantly zip through this plane in all directions.  If a
particle with 4-vector (E, v_x, v_y, v_z) passes across the plane,
headed in the +x direction (i.e., v_x>0), then it transfers a bit of
v_x momentum in that direction, and likewise v_y momentum and v_z
momentum.

OK, now here's the kicker.  If we look at the *average* transfer of
v_y and v_z across the yz-plane, we get zero.  That's because v_x is
uncorrelated with v_y and v_z, so the average of (v_x.v_y) is zero,
ditto (v_x.v_z).  (If we weren't sitting in the rest-frame of the
fluid--- if we felt a breeze--- then this wouldn't necessarily be
true.)  But the average of (v_x.v_x) is positive, of course; it
doesn't take much imagination to see why this represents the pressure.

From [email-masked] Wed Jan 24 10:54 PST 1996
From: Oz <[email-masked]>
To: john baez <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: Wed, 24 Jan 1996 13:32:13 GMT
Reply-To: [email-masked]
Content-Length: 3093
Status: OR

Maybe this one?

On 19 Jan 1996 13:00:31 -0800, you wrote:

>In article <[email-masked]> [email-masked] (Edward Green) writes:
>
>>If you do succeed in setting up this tutorial,  I would like to sit very
>>quietly in the corner,  and ask the odd question.  With your permission, 
>>of course. 
>
>Well, I started by giving a description of curvature: a space (or
>spacetime) is "curved" if when we take an arrow and carry it around a
>loop, doing our best to keep it the same length and pointing in the same
>direction, it may come back rotated.  I gave the example of carrying
>a horizontally-pointing javelin from the north pole to the equator along
>a line of constant longitude, then around the equator a bit, then back
>to the north pole.   It's very good to work this out in detail.  
>
>Okay, now for some more jargon: by "arrow" we really mean "tangent
>vector", which we can think of roughly as an "infinitesimal arrow" based
>some point of space.  And this process of carrying a tangent vector
>around while doing our best not to rotate or stretch it is called
>"parallel translation".  
>
>Now for the definition of the Riemann curvature tensor R^a_{bcd}.  This
>gadget knows all about the curvature of space and we will use it to cook
>up the Einstein tensor G_{ab} that sits on the left hand side of
>Einstein's equation
>
>G_{ab} = T_{ab}.
>
>Here I will use letters at the beginning of the alphabet to denote the
>numbers 0,1,2,3.  The 0,1,2 and 3 components of something are the t,x,y,
>and z components of something... where we use *completely arbitrary*
>coordinates t,x,y, and z.  (In setting up general relativity, everything
>should work nicely NO MATTER WHAT COORDINATES we use.)  I have already
>said what T_{ab} is: it's the flow in the a direction of momentum in the
>b direction.  (E.g., T_{00} is the flow in the time direction of
>momentum in the time direction, i.e. the density of energy.)  So now I
>gotta say what G_{ab} is.... but the crucial thing is to define the
>Riemann curvature tensor R^a_{bcd}.
>
>Remember, the letters a,b,c and d stand for anything like 0,1,2 or 3, 
>or in other words, t,x,y and z.
>
>Say we take a tangent vector pointing in the d direction and carry it
>around a little square in the b-c plane.  We go in the b direction until
>the b coordinate has changed by epsilon, then we go in the c direction
>until the c coordinate has changed by epsilon, then we go back in the b
>direction until the b coordinate is what it started out as, and then we
>go back in the c direction until the c coordinate is what it started out
>as.  Our tangent vector may have rotated a little bit since space is
>curved.  Its component in the a direction has changed a bit, say
>
>-epsilon^2 R^a_{bcd}
>
>plus terms of order epsilon^3.  
>
>This defines R^a_{bcd}.  The minus sign is just an annoying convention.
>
>That's the glorious Riemann curvature tensor.
>
>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 95429 (10 more) in sci.physics:
From: john baez
Subject: Getting tenser?
Date: 24 Jan 1996 10:58:54 -0800
Organization: University of California, Riverside
Lines: 76
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked]
writes, concerning my divagations on tangent vectors at a point of
spacetime: 

>Now of course the first thing (trying to keep 4D in mind) is
>what a tangent in this context would be. In particular a
>tangent to what? 

Tangent to spacetime itself!  I hope my pumpkin metaphor made that
clear... but let me repeat: if we think of a manifold, like the
surface of a pumpkin, as embedded in some higher-dimensional Euclidean
space: it's easy to visualize what we mean by a vector *tangent* to a
point of that manifold.  But in fact, even if we do not think of our spacetime
as embedded in some higher-dimensional space, one may define the notion
of a "tangent vector" at a point of spacetime.  One could just as well
drop the "tangent" business and call it a "vector", but hotshot
physicists use so many different kinds of vectors they like to keep
track of things this way, and besides, the "tangent vector" terminology
encourages the useful kind of geometrical thinking that I was trying to
cultivate in folks with that pumpkin metaphor.  

Think of a tangent vector at a point of spacetime, if you like, as a wee
arrow whose tail is pinned to that point.  

>Now I never really needed to do anything
>with tensors in anger, or even at all. However I did read a
>little about them many years ago and I vaguely remember
>deciding they were basically vectors with position. 

Hmm, that's not what tensors are... "vectors with position" is actually
a pretty decent way of saying what *tangent vectors* are.

>So I
>would imagine them defining a 4D vector at every point in
>spacetime. Presumably one does some sort of path integral
>incorporating one's 'original' direction and the Riemann
>curvature tensor. The Riemann curvature tensor can
>presumably be expressed as some function over all points on
>the path you are interested in. It sounds the sort of
>operation you hope you only have to do with conveniently
>tractable functions, say 4D conic sections or whatever.

Hmm.  I can't understand a word of this, and I'm afraid that if I keep
on trying I will.  

>Ah-ha. I presume your '4-tangents' are related to geodesics.
>So I stuff a 'rectangular' co-ordinate system over space (I
>hope that's allowed) then fire off some object from some
>point in spacetime. This Riemann curvature tensor will tell
>me the change in direction at every point and in this way I
>can plot it's path wrt the chosen co-ordinate system.

Hmm.  There is a certain vague truth to this, but I would 
be reluctant to say it's right.  There are many things we have to get to
before we can fully understand how the following things are related:

1. the metric
2. parallel translation
3. the Riemann curvature tensor 
4. geodesics
5. the Einstein tensor
6. the stress-energy tensor

I've described how to compute 3 in terms of 2.  (Did someone out there
keep a copy of my definition of the Riemann tensor in terms of carrying
a vector around a little square?  I want to save a copy!  Oz and Ed
Green have not coughed one up.)  I've also said that Einstein's equation
says 5 and 6 are proportional.  I've also said that freely falling
objects move along 4.  But I need to say what 4 has to do with 2.  And,
to really understand GR, I need to tell you how you compute 5 in terms
of 3 --- that's pretty easy --- and how you compute 2 in terms of 1 ---
that's a bit hard.  Then everything will be all hooked together.

From [email-masked] Wed Jan 24 16:23 PST 1996
From: Oz <[email-masked]>
To: john baez <[email-masked]>
Newsgroups: sci.physics,sci.astro,sci.philosophy.meta
Subject: Re: Stressed out?
Date: Wed, 24 Jan 1996 13:32:22 GMT
Reply-To: [email-masked]
Content-Length: 1998
Status: OR

Ahh, is this it? Oz

[email-masked] (john baez) wrote:

>In article <[email-masked]> [email-masked] (Edward Green) writes:
>
>>What stress tensor? 
>
>The stress-energy tensor, aka energy-momentum tensor, T_{ij}, where
>i,j go from 0 to 3.  This gadget is the thing that appears on the right
>side of Einstein's equation for general relativity:
>
>G_{ij} = T_{ij}
>
>(in nice units).  The thing on the left is the Einstein tensor, that
>summarizes some information about spacetime curvature.
>
>>  What top row? 
>
>The top row, T_{0j}, of this 4x4 matrix, keeps track of the *density* of 
>energy --- that's T_{00} --- and the density of momentum in the x,y, and
>z directions --- that's T_{01}, T_{02}, and T_{03} respectively.
>
>This should make sense if you remember that i=0,1,2,3 correspond to 
>t,x,y, and z respectively.  And that energy is just the same as momentum
>in the time direction.
>
>The other entries of the stress-energy keep track of the *flow* of
>energy and momentum in various spatial directions.
>
>In brief, T_{ij} is the flow in the i direction of momentum in the j
>direction. 
>
>This gadget tells you everything about what energy and momentum are
>doing at your given point of spacetime.
>
>>What language are you talking?   
>
>Physics, ca 20th century.
>
>(I sure can be snide.  Sorry.)
>
>>You wouldn't care to give me a crash descriptive course in GR, would you? 
>>(That is the language you are speaking, isn't it?). 
>
>Well, the stress-energy tensor is a basic gadget throughout physics,
>since we all want to know where energy and momentum are going and how
>much there is sitting around, right?  But it's only in general
>relativity where the stress-energy tensor is sitting proudly on the
>right side of an equation, telling spacetime how to curve.
>
>
>
>
>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From [email-masked] Fri Jan 26 12:30 PST 1996
From: "Ken Muldrew" <[email-masked]>
Organization:  McCaig Centre
To: [email-masked]
Date:          Fri, 26 Jan 1996 13:37:35 MDT
Subject:       parallel transport around a parallellogram
Reply-To: [email-masked]
Priority: normal
Content-Length: 11093
Status: OR

========
Newsgroups: sci.physics
Subject: Re: Red Shift  {Was: Center of Universe?}
From: [email-masked] (Bronis Vidugiris)
Date: Mon, 22 Jan 1996 21:31:39 GMT

In article <[email-masked]>,
Oz <[email-masked]> wrote:

)Now of course the first thing (trying to keep 4D in mind) is
)what a tangent in this context would be. In particular a
)tangent to what? Now I never really needed to do anything
)with tensors in anger, or even at all. However I did read a
)little about them many years ago and I vaguely remember
)deciding they were basically vectors with position.

Umm - well, vectors with position (a vector field) *could* be a 
tensor, but that's not a general definition by any means.

Tensors are sort of a generalization of matrices, with different
notation.
                 a
If you think of T as being a column vector, Ta as being a row
vector, you won't go too far wrong.  (T is a tensor, a 1-element
tensor in this case).

A standard matrix multiplication is then

Y = AX (matrix notation)

 b    b     a
Y  = A a   X  (tensor notation).

Because the 'a' is repeated, it us understood that one sums over it -
one sums over any repeated indeex.

In this case, one sums over the row of A (row, because it's a subscript) of a
and the column (column because it's a superscript) of X.

The convention is that one always sums over one row and one column,
never a row and a row or a column or a column.

The difference between row and column vectors becomes very important and
is called "tangent and cotangent spaces" or "covariant and contravariant"
components.  The distinction doesn't matter when one has a nice, uniform,
Cartesian metric, (i.e. s= x^2+y^2+z^2) but becomes important when one has
to deal with other metrics, such as with either non-cartesian co-ordinate
systems or the x^2+y^2+z^2 -t^2 metric of GR.  Actually the difference
isn't fundamentally important as long as one keeps tract of what's what. 
If one misplaces an index, one is off by the value of the metric, which is
no longer the identity matrix, which it used to be. in the simplest
case.   So one has to keep tract of it, but once one is aware of the need
to keep tract of it one relegates this to an unpleasant but necessary task
that one has to do to get the right answers).

Dunno if the metric stuff made any sense - do you recall having
seen at least some stuff about matrices and "quadratic forms"?
If you have, the metric stuff should make more sense - the
quadratic form for x^2+y^2+z^2 is the identity matrix, which is
why it's possible to not to worry about covariance vs. contravariance.
This isn't possible when the quadratic form of the metric isn't
equivalent to an identity metric.

Anyway, a matrix is just a two-element tensor, one can think of a matrix
as either generating a column vector from a column vector (the usual way),
or one can consider it as taking in a column vector, and a row vector (aka
- one form), and generating a scalar result.

A three element tensor can be thought of as taking in two row/column
vectors and producing a r/c vector, or it can be thought of as taking in
three r/c vectors and producing a scalar.  The last definition is what
most mathematicians use, a general tensor of degree m+n takes in m row
vectors and n column vectors and spits out a scalar.  (This may
seem odd, but it's an easy way to generalize).

As I recall, the Rienmann tensor is a 4-element tensor - it takes
in 3 things and spits out  another.  This is what defines
the "curvature" of space in GR.

One other point I should make explicitly - it's all linear.

Hope this isn't two confused - that's my take on the topic, anyway,
I'm not as familiar with all this yet as I'd really like myself.

ps - "Gravitation" does have an intro to this stuff, but it moves
quickly, like it's supposed to be a review (IMO).

========
Newsgroups: sci.physics
Subject: Getting tenser?
From: [email-masked] (john baez)
Date: 24 Jan 1996 10:58:54 -0800

In article <[email-masked]> [email-masked]
writes, concerning my divagations on tangent vectors at a point of
spacetime: 

>Now of course the first thing (trying to keep 4D in mind) is
>what a tangent in this context would be. In particular a
>tangent to what? 

Tangent to spacetime itself!  I hope my pumpkin metaphor made that
clear... but let me repeat: if we think of a manifold, like the
surface of a pumpkin, as embedded in some higher-dimensional Euclidean
space: it's easy to visualize what we mean by a vector *tangent* to a
point of that manifold.  But in fact, even if we do not think of our spacetime
as embedded in some higher-dimensional space, one may define the notion
of a "tangent vector" at a point of spacetime.  One could just as well
drop the "tangent" business and call it a "vector", but hotshot
physicists use so many different kinds of vectors they like to keep
track of things this way, and besides, the "tangent vector" terminology
encourages the useful kind of geometrical thinking that I was trying to
cultivate in folks with that pumpkin metaphor.  

Think of a tangent vector at a point of spacetime, if you like, as a wee
arrow whose tail is pinned to that point.  

>Now I never really needed to do anything
>with tensors in anger, or even at all. However I did read a
>little about them many years ago and I vaguely remember
>deciding they were basically vectors with position. 

Hmm, that's not what tensors are... "vectors with position" is actually
a pretty decent way of saying what *tangent vectors* are.

>So I
>would imagine them defining a 4D vector at every point in
>spacetime. Presumably one does some sort of path integral
>incorporating one's 'original' direction and the Riemann
>curvature tensor. The Riemann curvature tensor can
>presumably be expressed as some function over all points on
>the path you are interested in. It sounds the sort of
>operation you hope you only have to do with conveniently
>tractable functions, say 4D conic sections or whatever.

Hmm.  I can't understand a word of this, and I'm afraid that if I keep
on trying I will.  

>Ah-ha. I presume your '4-tangents' are related to geodesics.
>So I stuff a 'rectangular' co-ordinate system over space (I
>hope that's allowed) then fire off some object from some
>point in spacetime. This Riemann curvature tensor will tell
>me the change in direction at every point and in this way I
>can plot it's path wrt the chosen co-ordinate system.

Hmm.  There is a certain vague truth to this, but I would 
be reluctant to say it's right.  There are many things we have to get to
before we can fully understand how the following things are related:

1. the metric
2. parallel translation
3. the Riemann curvature tensor 
4. geodesics
5. the Einstein tensor
6. the stress-energy tensor

I've described how to compute 3 in terms of 2.  (Did someone out there
keep a copy of my definition of the Riemann tensor in terms of carrying
a vector around a little square?  I want to save a copy!  Oz and Ed
Green have not coughed one up.)  I've also said that Einstein's equation
says 5 and 6 are proportional.  I've also said that freely falling
objects move along 4.  But I need to say what 4 has to do with 2.  And,
to really understand GR, I need to tell you how you compute 5 in terms
of 3 --- that's pretty easy --- and how you compute 2 in terms of 1 ---
that's a bit hard.  Then everything will be all hooked together.

Article 95557 (379 more) in sci.physics:
From: Oz
Subject: Re: Red Shift  {Was: Center of Universe?}
Date: Wed, 24 Jan 1996 13:31:55 GMT
Lines: 133
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
X-Newsreader: Forte Agent .99c/16.141

[email-masked] (john baez) wrote:

>  You are getting used to the fact that the
>world doesn't come with a grid of lines painted on it.  Of course, even
>in the old Newtonian days folks knew that any rotated or translated
>Cartesian coordinate system was "just as good" as any other.  So it's
>not as if they thought there was a particular coordinate system handed
>down by God that was the "best" one.  Instead, one had a manageable
>family of "good" coordinates, any one of which was related to any other
>by an easily understandable sort of transformation: rotation or
>translation.  

Yes, got it in one. I was imagining that it was reasonable
to transform between a rectangular (or whatever) co-ordinate
system and 'another view', or at least be able to chose one
that suited the problem and could be induced to give
'sensible' answers. Here we seem to be in the situation that
the problem, which is non trivial, is to ask a clearly
defined question first. Indeed many reasonable questions
seem to have become undefined, this (luckily for you)
seriously inhibits question asking unless you know the
subject very well indeed at a deep level. For this (dammit)
you need to have studied it in it's technical depth. A
somewhat depressing conclusion. However I appreciate the
small scent of the subject I have gleaned. Not really
satisfactory, but I suppose that's life.

>The same applied back in the days of special relativity.  It may freak
>some people out that in addition to rotations and translations one has
>"Lorentz transformations" in which the new t' coordinate depends on the
>old t and x coordinates (say), but there are still a manageable set of
>"best" coordinates, corresponding physically to the inertial frames.

Yes, this mislead me for a very long while.

>General relativity takes a completely different approach to spacetime.
>In general relativity, spacetime is wiggly in a fairly arbitrary way
>(though it must satisfy Einstein's equation), so there is in general no
>manageable set of "best" coordinates.

<smip for brevity>

>Okay, so you don't have "straight lines" but you do have "geodesics" on
>a pumpkin.  So say you try to draw a grid on the pumpkin such that each
>curve in it is a geodesic and whenever two geodesics intersect, they do
>so at right angles.  If you could do that, it would be a good stab at
>Cartesian coordinates.  But you can't.  That's because the surface of
>the pumpkin is a "curved 2-dimensional Riemannian manifold" --- with the
>emphasis here on *curved*.  
>
>So what do we do in general relativity, where spacetime is as bumpy as
>the surface of a pumpkin.  We give up all attempts to pick "best" or
>"good" coordinates ---- except in working on certain very special
>problems with lots of symmetry! ---- and decide to do things in such a
>way that ANY choice of coordinates will work as smoothly as ANY OTHER. 

Comment:

Most of the structures that seem to be bandied about seem to
be exactly these highly symmetrical structures. In fact
precious little else. Should I deduce that in general these
are the only ones that we (I mean 'you') can properly handle
in an unambiguous enough way to make a reasonable and
general statement of both the problem and it's solution?

>We don't exactly abandon coordinates; we just relegate them to the
>status of completely arbitrary tools. 

Hmmm. Some esoteric mathematical construct that takes a near
genius several years to begin to master. Gloom.

>>I wonder if a lot of my confusion is in the intermix of
>>cosmology and GR. Too much to follow all in one shot, too
>>many new ways to look at things at once.
>
>Certainly this is a big part of it.  Personally I can't teach you
>cosmology without teaching you GR, any more than I could teach you
>celestial mechanics without teaching you F = ma.  There might be some
>way to study cosmology without GR, but to me that seems to miss the whole
>grandeur and strangeness of the subject.  

This is, of course, true. However I was rather hoping that a
basic understanding, but below the level of being able to
properly manipulate real problems, might be enough to follow
the reasoning well enough. For example understanding F=ma
will not allow you to calculate most dynamic problems, but
will allow you to follow someone elses explanation. Anyway I
still live in some small hope.

>For example, it's true that one could not bother learning GR and only
>try to understand one solution of Einstein's equation, the
>Robertson-Friedman-Walker metric which describes the standard big bang
>cosmology.  This might allow one a certain tempting conceptual
>sloppiness: one could write this metric down in the "standard
>coordinates" which take advantage of the symmetry of that solution, and
>ignore my remarks above about the arbitrariness of coordinates.  But one
>would really be missing a lot of the fun!  For example, you've already
>seen that in the Milne cosmology, different coordinates can give one
>very different pictures of what's going on.  This mental flexibility is
>crucial if one gets into questions like "why is there a redshift: is it
>really due to the expansion of space, or is it a gravitational effect?"
>Without understanding physics as geometry and the arbitrary nature of
>ones choice of coordinates, these riddles can really throw one for a
>loop.  *With* this understanding one can, so to speak, deconstruct the
>question and figure out the *right* question and the right answer.

Exactly, of course, why it interests me. I rather like the
unintuitive behaviour of the cosmos, it stretches the brain.
I do not have the time, or enough brain cells any more, to
take a prostgrad (UK postgrad that is) course on the
subject. I must make do with crumbs from the table. <sigh>

>>Perhaps a simple GR example might help. I have assumed that
>>the moon orbits the earth in a GR explanation because it
>>simply travels in a straight line, but space (3D) is curved
>>(ie orbital sized curving). However I am getting the
>>impression that the GR curvature is very very tiny. 
 ...........
<snip>

Well, it took me long enough to glean *that* one out from
the crumbs that fell off the table! Worth the effort
although I must resist the temptation to see it in
understandable maths. Oh, I don't know. It's the only way to
get more of a qualitative feel. OK hit me if you fancy it.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,

Article 95762 (377 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 25 Jan 1996 13:19:37 -0800
Organization: University of California, Riverside
Lines: 162
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>[email-masked] (john baez) wrote:

>>Now we have to abstract things a bit!  First, remove the 3d ambient
>>Euclidean space and think only of the surface of the pumpkin!  We can
>>still define a "tangent vector"... the actual definition being rather
>>mathematical... but one way to visualize it is as a teeny-weeny
>>itsy-bitsy little arrow drawn on the surface of the pumpkin, with its
>>base at the specified point.  We make it small --- in fact,
>>infinitesimal --- just in order to avoid worrying about the fact that
>>the pumpkin is curved.  After all, if we had an ambient 3d space as
>>before, we could ignore the difference between a vector tangent to the
>>pumpkin, and a vector actually drawn *on* the pumpkin, in the limit
>>where the arrow became very small.  

>OK, perhaps, just perhaps, I get the idea. 'Normal' geometry
>is obsessed with orthogonality in some way. We abandon this
>and go for tangents. Now it's tempting to say that any small
>bit of pumpkin is flat, but that loses the curvaceousness of
>it. On the other hand we can't really define a big long
>tangent line on a 2D pumpkin since it cruises off into
>another non-existant dimension. We can however, always
>define an infinitesimally short one, possibly in a
>similarish way to Newton, at least in concept.

That's the basic idea.  Mathematicians have had many years to make
Newton's "infinitesimals" precise... in quite a variety of different 
ways which are pretty much equivalent for practical purposes.  Thus
we may unabashedly imagine a tangent vector to a pumpkin as an
vector tangent to the pumpkin, but infinitesimal, so that it doesn't
cruise off into the 3d space which is, after, quite nonexistent to the
Pumpkin People, the 2-dimensional beings who inhabit the surface of the
pumpkin.  

>Incidentally I presume your warty old pumpkin is properly
>2D, ie it's really completely smooth but has some strange
>spacial properties where different paths give different
>distances in "whatever co-ordinate system we define". Since
>we find it hard to visualise this we bend it up into a
>non-existent 3rd dimension.

Precisely.  For us, of course, the 3rd dimension is real and the surface
of the pumpkin is a mere "submanifold", but for the Pumpkin People the
pumpkin is all of space and the 3rd dimension would simply be a
mathematical fiction.  Luckily, there is no need to argue.
Mathematicians have mastered both "extrinsic geometry," in which a
curved space is treated as a "submanifold" of some other, perhaps flat,
space, and "intrinsic geometry", where you treat curved space in its own
right and don't imagine it sitting in some higher-dimensional space.

The intrinsic viewpoint, developed by Gauss and Riemann in the late
1800s, is harder to get good at but it's often better.  Why bother with
extra dimensions you never really see or use anyway, if you don't need
to?

>>a good way is to think of it as a machine that eats a list of say, 3
>>tangent vectors, and spits out a number, for example, or maybe a tangent
>>vector.  (This isn't quite the most general sort of tensor but it's good
>>enough for starters.)  We require that the output depend in a linear way
>>on each of the inputs. 

>OK, I bet in reality it's not quite as simple as this
>however. 

Well, that was a pretty precise definition a large class of tensors, but
not of the most general kind. 

Here it is again, more formally so you will feel the suffering normally
associated with education:

A tensor of "rank (0,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a number.  The output must depend linearly on each input.   

A tensor of "rank (1,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a tangent vector at the point p.  The output must depend linearly
on each input.    

I am avoiding defining tensors of rank (n,k) for other values of n,
because there is actually a fair amount of physics I can do without
dragging them in.  Eventually I would need to explain them, and you'd
see they weren't much worse.

>I take it that "linear way" is a fundamental
>property of a Tensor.

Indeed!!!!!!!!!!  It's a branch of Linear Algebra.  

>>So for example a while back I discussed the Riemann tensor R^a_{bcd}.  
>>This was a thing that ate 3 tangent vectors and spit out a tangent
>>vector... which is why there are 3 subscripts and one superscript.  
>>
>>Namely, if we parallel translate a tangent vector u around the little
>>parallelogram of size epsilon whose edges point in the directions of the
>>tangent vectors v and w, it changes by a little bit.  Namely, it changes
>>by the tangent vector whose component in the a direction is
>>
>>- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3
>>
>>Here we sum over b, c, and d.  The thing "v^b" is the component of the
>>vector v in the b direction... in whatever the hell coordinate system we
>>happen to be using.  And remember, indices like a,b,c,d range
>>from 0 to 3 if we are working in 4d spacetime.  

>1) In general I assume there are an infinity of possible
>tangent vector directions like v,w above defining some
>parallelogram of size epsilon (which some nasty person will
>presumably tend to zero). I presume this is operational for
>a well behaved space over which tangent vectors are
>definable.

Sure, there are infinitely many tangent vectors at a point.
This is not so bad.  For example, note that the output 

R^a_{bcd} v^b w^c u^d 

(let's ignore that epsilon junk and the niggly minus sign)
depends linearly on the inputs u, v, and w, so we don't need to know it
for *infinitely* many choices of u, v, and w to figure out what it will
be for all possible choices.  Linearity keeps life simple.

>2) Somebody has sneakily brought in some co-ordinates whilst
>nobody was looking (a,b,c ....). Now telling me they are
>local just won't do and nor will telling me they are
>'whatever co-ordinates you desire'. 

Well, certainly they are local coordinates, because you would be hard
pressed to flatten out a whole pumpkin and impose *global* coordinates
on it, rendering it a mere plane.  And certainly the coordinates above are
indeed "whatever coordinates you desire".  But you are starting to sound
like a mathematician --- high praise in my book --- with your complaint
about the unpleasant appearance of coordinates.  So to reward you, I
will explain how it works without coordinates.  You'll see it's much
simpler.

The Riemann tensor is a tensor of rank (1,3) at each point of spacetime.
Thus it takes three tangent vectors, say u, v, and w as inputs, and
outputs 1 tangent vector, say R(u,v,w).  As usual, the output depends
linearly on each input.  The Riemann tensor is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges are the vectors  epsilon u  and  epsilon v , where epsilon
is a tiny number longing to approach zero.  The vector w comes back a
bit changed by its journey; it is now a new vector w'.  We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Note: I am now saying just what I said before, but without those yucky
coordinates!  If you insist on using coordinates, I will say "Go ahead!
Pick any that you like!  I don't care which!"  The only effect will be
to turn the above elegant equation into the grungier but sometimes more
practical one:  

w'^a - w^a = - epsilon^2 R^a_{bcd} v^b w^c u^d + terms of order epsilon^3

>I have had (in GR
>context) been bludgeoned into realising that I need to
>reconsider the concept of 'co-ordinates' altogether.

Good.  There's nothing like a good bludgeoning now and then.

Article 95805 (376 more) in sci.physics:
From: Keith Ramsay
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 25 Jan 1996 04:43:34 GMT
Organization: Boston College
Lines: 89
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, [email-masked] (john baez) wrote:
[Tangent vectors on pumpkins....]
|This is the sort of thing mathematicians can make precise; don't worry
|about it too much for now.

It's not very hard, anyway.

There are two sort of obvious constructions you can make on the space-time 
pumpkin. You can draw parametrized curves on it. You can have functions
varying from point to point on it. Assume we're dealing in both cases
with curves and functions which are differentiable (no "corners").

Given a parametrized curve passing through a point, and a function, you
can consider the rate of change of the function as you follow the point.
If you like:

    d f(v(t))
    _________
       dt

where v(t) parametrizes the curve, and f is the function varying on the
pumpkin.

So if, as I tell my students, the parametrized curve describes the path
of a fly, and f describes the temperature at each point, then we're
talking about how fast the fly is getting hotter when it is passing
through the point.

Now various parametrized curves passing through the same point will be
"equivalent" in the sense that the rate of increase is the same for any
given f. Likewise, various functions will be "equivalent" in the sense
that for any given parametrized curve (did I say it had to be differentiable?)
passing through the point, the functions are both increasing at the same
rate.

The set of all the curves which are equivalent in this way to a given one
are said to define a "tangent vector" at that point. (If it were in space, 
it would be a "velocity".) It's the tangent vector for those curves.

The set of all functions which are equivalent in this respect to a given
one are said to define a "cotangent vector" at the point. It is essentially
the gradient vector of the function.

|Now what's a tensor?  Well, there are a million ways to think of it, but
|a good way is to think of it as a machine that eats a list of say, 3
|tangent vectors, and spits out a number, for example, or maybe a tangent
|vector.  (This isn't quite the most general sort of tensor but it's good
|enough for starters.)  We require that the output depend in a linear way
|on each of the inputs.  

What John is skirting around here is simply the bit about cotangent
vectors. Some tensors you have to think of as taking in cotangent
vectors, if you want to think of them this way.

Taking in a tangent vector is a lot like spitting out a cotangent
vector, and vice versa.

|So for example a while back I discussed the Riemann tensor R^a_{bcd}.  
|This was a thing that ate 3 tangent vectors and spit out a tangent
|vector... which is why there are 3 subscripts and one superscript.  

It's equivalent to think of it as taking in three tangent vectors
and a cotangent vector (at the same point), and spitting out a number 
in a multi-linear way. (Let Baez's gizmo take the three vectors and
give back one. Then combine it with the given cotangent vector and
output the result.)

Up to a point, people are happy to blur the distinction between
tangent vectors and cotangent vectors. They both work out as
coordinate vectors if you have a fixed coordinate system. But to
identify them with each other, you need a metric. Concretely,
you associate a tangent vector with a covector by imagining that
your fly is going directly toward the hotter air, and moves at a
speed in proportion to how quickly it is warming up with distance.

This requires, however, having a notion of "distance".

That's typically okay. In GR, though, the metric is something
you have to figure out. It's not given. So one is more careful
when converting vectors to covectors and vice-versa.

In terms of notation, it's called "raising and lowering indices".
Given a metric, different types of tensors of the same rank can
be put in correspondence with the others.

I guess I should stop.

Keith Ramsay

Article 95871 (375 more) in sci.physics:
From: Keith Ramsay
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 25 Jan 1996 18:32:58 GMT
Organization: Boston College
Lines: 105
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, [email-masked] wrote:
[email-masked] (john baez) wrote:
[...]
|>Now what's a tensor?  Well, there are a million ways to think of it, but
|>a good way is to think of it as a machine that eats a list of say, 3
|>tangent vectors, and spits out a number, for example, or maybe a tangent
|>vector.  (This isn't quite the most general sort of tensor but it's good
|>enough for starters.)  We require that the output depend in a linear way
|>on each of the inputs. 
|
|OK, I bet in reality it's not quite as simple as this
|however. 

It very nearly is. The "most general sort" of tensor on a manifold 
might not correspond to such a machine if it only takes tangent
vectors as inputs. But if you include also such "machines" which
possibly take a given number of cotangent vectors at the same point
as well, then you have all of them. Moreover, the kinds of tensors 
which you can think of as taking as inputs just tangent vectors at 
the given point are sufficient for a lot of the GR story, so you 
needn't worry.

|I take it that "linear way" is a fundamental
|property of a Tensor.

Yes. If you fix all but one of the inputs, the result depends linearly
upon the remaining one.

You get a "tensor field" just like a "vector field": attach a tensor
(of one homogeneous kind) to each point. Whatever operations you 
perform on a tensor at just one point can be applied en masse to
tensor fields.

|>Namely, it changes
|>by the tangent vector whose component in the a direction is
|>
|>- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3
|>
|>Here we sum over b, c, and d.  The thing "v^b" is the component of the
|>vector v in the b direction... in whatever the hell coordinate system we
|>happen to be using.  And remember, indices like a,b,c,d range
|>from 0 to 3 if we are working in 4d spacetime.  
[...]
|2) Somebody has sneakily brought in some co-ordinates whilst
|nobody was looking (a,b,c ....). Now telling me they are
|local just won't do and nor will telling me they are
|'whatever co-ordinates you desire'.

The point is that there are some things which are hard to describe
without using coordinates, but can be described in a coordinate-
invariant way. If you change the coordinates, you have to change
the numbers (R^0_{000}, R^1_{230}, v^2 etc.) appearing in the formula.
But the result you get will be the same tangent vector, but expressed
in the new coordinates.

Here's a simple example. T_a would be a tensor which takes a
tangent vector and gives back a number (i.e., a cotangent vector). 
v^a would be a tangent vector.

If we have coordinates at a point, then the tangent vector v there
has coordinates (v^0, v^1, v^2, v^3). The coordinates of T are
(T_0, T_1, T_2, T_3). They are defined as the answers T gives when
fed the vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), and (0,0,0,1)
respectively.

By the linearity of T, the answer you get when you plug in v is
v^0 times T applied to (1,0,0,0) + v^1 times T applied to (0,1,0,0)
+ etc., i.e. T_0 v^0 + T_1 v^1 + T_2 v^2 + T_3 v^3. Now, to keep
things managable we have summation notation: sum{a} T_a v^a. Then,
to make it even simpler to write, we have the "Einstein summation
convention": if an index appears as a subscript and as a superscript,
then sum over it. So we just have to write T_a v^a. So you see, it
doesn't depend on which coordinate system you have.

I like the notation Wald uses in his book, General Relativity. For
formulas which are like the ones we have here, in which
any coordinate system will do, but it's helpful to write it out
suggesting how we would calculate the result if we had a particular 
one in mind, one has an "abstract index notation".

One way to think of it is as a way of labelling the inputs and outputs
of your tensors. In some products, wires are labelled red, green, blue
for which socket matches which plug. In calculations with tensors, it's
convenient sometimes to label which "slots" in tensors match up with
which other ones. So the formula above is just a way to describe plugging
the vectors v,w,u into R and getting a tangent vector out.

Here's an odd thing one can do. (Those with dirty minds may be excused.)
Take a tensor T^a_b, the kind which takes a tangent vector and gives
back one. It turns out T^a_a (plugging one end into the other) makes
sense as a number, independent of coordinate system. The components 
T^i_j of T in a given coordinate system can be thought of as the entries
of a matrix giving the linear transformation on tangent vectors which T
is. T^a_a represents the sum T^0_0+T^1_1+T^2_2+T^3_3 of the diagonal
entries of T's matrix. This is called the "trace" of the matrix.
Interestingly enough, if you change the coordinate system, the matrix
in the new coordinate system has the same trace.

Since also sometimes you want to calculate with the components of
tensors in a given coordinate system, the notation allows for that.
Switch the index letters to Greek! This is just a way to say, "No,
this formula is not guaranteed to be correct for all coordinate
systems."

Keith Ramsay

Article 96015 (374 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Red Shift  {Was: Center of Universe?}
Date: 26 Jan 1996 13:27:13 -0800
Organization: University of California, Riverside
Lines: 113
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>Yes, got it in one. I was imagining that it was reasonable
>to transform between a rectangular (or whatever) co-ordinate
>system and 'another view', or at least be able to chose one
>that suited the problem and could be induced to give
>'sensible' answers. Here we seem to be in the situation that
>the problem, which is non trivial, is to ask a clearly
>defined question first. Indeed many reasonable questions
                                    ^seemingly           
>seem to have become undefined, this (luckily for you)
>seriously inhibits question asking unless you know the
>subject very well indeed at a deep level.

For some reason your asking of questions seems not to have been
inhibited one whit.  

>For this (dammit)
>you need to have studied it in it's technical depth. 

Perhaps.  Or else you need to learn the art of asking operational
questions --- questions that come along with an experimental procedure
for determining their answer.  Those are the two ways to stay reasonable
in modern physics: either learn the mathematical formalism so you can
ask questions about that, or stick with operational questions. 

>>So what do we do in general relativity, where spacetime is as bumpy as
>>the surface of a pumpkin.  We give up all attempts to pick "best" or
>>"good" coordinates ---- except in working on certain very special
>>problems with lots of symmetry! ---- and decide to do things in such a
>>way that ANY choice of coordinates will work as smoothly as ANY OTHER. 

>Most of the structures that seem to be bandied about seem to
>be exactly these highly symmetrical structures. In fact
>precious little else. Should I deduce that in general these
>are the only ones that we (I mean 'you') can properly handle
>in an unambiguous enough way to make a reasonable and
>general statement of both the problem and it's solution?

It's true that things get trickier when there are no symmetries around,
hence no "specially nice" coordinates.  There are two aspects to this
trickiness.  The first and simpler one is that in the absence of
symmetries, the math gets tough: there are very few "exactly solvable"
problems in GR without symmetry.  In these cases one typically needs a
computer to solve things numerically.  So, for example, the problem of
two in-spiralling black holes is one of the National Science
Foundation's "grand challenge problems", and lots of people are writing
code to solve it.  

The second one is the deeper issue of what questions make sense.  For a
while now I have been attempting to force you to only ask questions that
make sense in the *general case*, i.e. that make sense when spacetime is
wiggly and bumpy in an arbitrary way.  While you continue to kick and
scream and ask questions that don't make sense, I think you are getting
the idea of why they don't make sense.  But I hope you also get an idea
of what questions DO make sense!  There are lots of them out there... as
there had damn well better be!  Basically, IF YOU TELL ME EXACTLY HOW TO
PERFORM AN EXPERIMENT, I CAN TELL YOU WHAT WOULD HAPPEN.  I write this
in capital letters for three reasons: one, because it's the fundamental
criterion for a theory to be complete, two, because there's never been
any reason to expect any *more* from a theory, and three, because only a
crackpot would say that they personally could meet *any* challenge of
this kind: some physics problems like this may be too hard in practice,
but *in principle* the theory should tell us what would happen.

Of course this is a bit sneaky, because when you start describing an
experiment and say "measure the distance from A to B", I will say "how?"
And you will have to tell me a procedure.  And if that procedure
involves coordinates, I will say "which coordinates?"  And so on.  This
question-and-answer game may take a while to play, but if you play it
well, it *does* end with you giving a exact specification of an
experiment, which I can then in principle describe the results of.

(Of course, in quantum mechanics the results would be probabilistic in
nature.  But we're not talking about that.)

>>We don't exactly abandon coordinates; we just relegate them to the
>>status of completely arbitrary tools. 

>Hmmm. Some esoteric mathematical construct that takes a near
>genius several years to begin to master. Gloom.

Huh?  It's not some "esoteric mathematical construct" you need to
master, it's an *attitude*.  You have already mastered the esoteric
mathematical construct known as coordinates.  Now you just need to
master the attitude that the coordinates are arbitrary: that any
calculation you want to do, you can do in any coordinates whatsoever.
(Sometimes it's easier in some coordinates than others, of course.) 
This means that you don't get attached to any particular choice of
coordinates; you don't attribute too much "physical reality" (whatever
that is) to it.

>This is, of course, true. However I was rather hoping that a
>basic understanding, but below the level of being able to
>properly manipulate real problems, might be enough to follow
>the reasoning well enough. For example understanding F=ma
>will not allow you to calculate most dynamic problems, but
>will allow you to follow someone elses explanation. Anyway I
>still live in some small hope.

Oh, certainly you can get this basic understanding without being able to
do any calculations.  That's what I'm shooting for.  You need to learn
about tensors, metrics, parallel transport, and curvature, but you don't
need to know how to calculate your way out of a paper bag.  Luckily,
they are all geometrical concepts so they are very easy to understand
without any long and complicated formulas.  

Article 96159 (373 more) in sci.physics:
From: Oz
Subject: Re: Red Shift  {Was: Center of Universe?}
Date: Fri, 26 Jan 1996 13:43:08 GMT
Lines: 48
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
X-Newsreader: Forte Agent .99c/16.141

[email-masked] (Keith Ramsay) wrote:

>In article <[email-masked]>, [email-masked] wrote:
>[email-masked] (john baez) wrote:
>[...]
>|>Now what's a tensor?  Well, there are a million ways to think of it, but
>|>a good way is to think of it as a machine that eats a list of say, 3
>|>tangent vectors, and spits out a number, for example, or maybe a tangent
>|>vector.  (This isn't quite the most general sort of tensor but it's good
>|>enough for starters.)  We require that the output depend in a linear way
>|>on each of the inputs. 
>|
>|OK, I bet in reality it's not quite as simple as this
>|however. 
>
>It very nearly is. The "most general sort" of tensor on a manifold 
>might not correspond to such a machine if it only takes tangent
>vectors as inputs. But if you include also such "machines" which
......

Thank you very much for your posts. May they continue to
come. However to avoid disappointment do not expect me to
get more than a superficial understanding. This is sad, but
unfortunately realistic. I suspect this has upset John in
the past because I haven't properly understood him at the
level he wants, although I sometimes get close after a few
months. Do not underestimate the value of a full blown
lecture course, supervision, and the interaction with other
students that is impossible on the net.

I wouldn't want you to stop mathematical descriptions at the
sort of level you think is appropriate. Even if I only get a
superficial idea I can see (often) roughly where you are
going and get some idea of the structures and manipulation
involved. This does help the visualisation and the physics
even though quantitative analysis will (sigh) forever be
beyond me. Short of taking an Open University course (say)
that I haven't got time to do.

So providing you can put up with all sorts of
misinterpretations and frustration at this 'nit on the net',
I am prepared to work at understanding your posts! 

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 96017 (365 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 26 Jan 1996 13:40:29 -0800
Organization: University of California, Riverside
Lines: 33
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:
>[email-masked] (Bruce Scott TOK )' wrote: 

>>The 
>>definition of R is given in terms of the failure of two of these 
>>derivatives to commute, so if you change the orientation of the path 
>>around the little quadrilateral, you change the sign of R. 

>Now I find this a provocative characterization,  on the tip of
>understanding...   

>So rich in resonances! 

You betcha.  If we ever get around to being detailed and technical about
the definition of the Riemann curvature, we'll see it has a lot to do
with noncommutativity.  I.e., if we have one of those teeny
parallelograms of size epsilon that I keep talking about, we can
parallel translate a vector first along the u side and then along the v
side, or first along the v side and then along the u side, and the
results will differ when there's curvature.  In fact, we can define
the Riemann curvature to be the difference of the two results, divided
by epsilon^2, in the limit as epsilon -> 0.  (Perhaps with a minus signs
thrown in for spice.)  

Now if you really want to get mystical you may reflect on the fact that
quantum mechanics also has to do with the failure of certain things to
commute.  So both general relativity and quantum theory have to do with
situations where things that we used to expect to commute, don't.  

Nobody has made too much use out of this observation, though Shahn Majid
has tried quite hard.

Article 96049 (364 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 26 Jan 1996 15:26:07 -0800
Organization: University of California, Riverside
Lines: 64
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>Uh-huh. So as expected there is some 'definition' of
>parallel transportation. Is it a straightforward one or do I
>not want to know?

In what I've said so far in this mini-course on general relativity, I
have taken parallel translation as a (fairly) easy-to-grasp starting
point.  It is simply a function that, given a point p, a curve from p to
q, and a tangent vector v at p, spits out a tangent vector at q.  
One can do this in the nitty-gritty mathematical theory, as
well.  One then requires this function to satisfy a few obvious axioms:
e.g., if you have a curve from p to q, and another curve from q to r, 
you can glom them together to get a curve from p to r, and then we can
parallel translate a tangent vector at p over to r either in two stages
or in one fell swoop, and we had better get the same answer.  (There are
a few more axioms.)  

But what we need to get to eventually is the marvelous fact that the
*metric* on spacetime determines a "best" recipe for parallel
translation.   Remember, the "metric" g is a tensor such that if you
feed in two tangent vectors v and w, g(v,w) is the "dot product" or
"inner product" of v and w.  This is the gadget that lets us calculate
angles between vectors, lengths of curves, and all that.  

Our spacetime has a metric on it, a "Lorentzian" metric g (meaning
*roughly* that the dot product of a vector with itself can be
negative... for details see the stuff I posted in response to Ed).  
By a certain amount hard work we can get from

the metric g_{ab}

to 

parallel translation

to            <--------------------- I already described this step!

the Riemann curvature tensor R^a_{bcd}

to

the Einstein tensor G_{ab}

(I stuck in indices just so you know how many inputs there
are to each of the tensors listed.  Parallel translation is not a
tensor because it involves two points and a curve between them, not just
one point.)

Once we have done this, when we look at Einstein's equation

G_{ab} = T_{ab}

we will know how the left hand side is cooked up from the metric, and
what it has to do with the curvature of spacetime.  We already know
about the right hand side, the stress-energy (or energy-momentum)
tensor.  So we will understand how the presence of energy and momentum
curve spacetime!

I hope this course outline inspires you and reassures you that there's
not a vast indefinite number of things you need to learn.  

Article 96392 (33 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 28 Jan 1996 11:46:40 -0800
Organization: University of California, Riverside
Lines: 119
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>[email-masked] (john baez) wrote:

>>Think of a tangent vector at a point of spacetime, if you like, as a wee
>>arrow whose tail is pinned to that point. 

>Woah, woah. Slow down. I am as impatient as you to get to
>the really good stuff, but I have found that understanding,
>and I mean understanding, the basics is time well spent.

Indeed.  Getting the geometry figured out is 99% of the battle as far as
general relativity is concerned.  And understanding tangent vectors is
really crucial.  (It's also good to understand cotangent vectors,
so you see what all the "covariant/contravariant" stuff is about, but
I'm planning on postponing that for as long as possible!)

>Tangents and tangent vectors are superficially obvious. The
>sort of thing that you don't like asking the Prof about
>because he might think you are stupid. Since you already
>know that, I have no qualms. As far as I can see a tangent
>vector is basically a little thingy that points in a
>direction *from a point*. In fact any direction at all as
>*all* directions are tangents. It is just a *little*
>direction thingy from here to there. So it's a rather
>general and cares not for any co-ordinates since it points
>from here to there, and there can be as close as we want and
>we could (probably) co-ordinatise here and there if we
>wanted to in any co-ordinate system we chose.
>Am I in the ball park? Same orbit?

Yes, this is a good informal summary of what tangent vectors are like.
In particular, while we can describe them using any coordinates we like,
we don't need any coordinates to get our hands on them.  This is true
both at the intuitive level --- what does a teeny infinitesimal arrow
care about coordinates?? --- and at the mathematically rigorous level.

>Riemann curvature tensor derivation. This little
>parallelogram that results from 'parallel transport' really
>does need to be looked at a teeny bit more deeply. It's
>John's own fault really, he has battered my preconceived
>ideas into abject submission, so I am hypercautious. In fact
>I am trying to cultivate an outlook that makes Ted look
>positively reckless.

A praiseworthy goal, though you'd be hard put to make Ted look reckless.

>A) Our loop round local space.
>1) Am I right in assuming that we always arrive back at our
>starting point after our little local excursion?

Well, we can parallel transport a tangent vector at the point P along
any path from P to the point Q and get a tangent vector at Q.  E.g., our
Roman can carry his javelin from the north pole to the equator and leave
it there.  This is important.  But for the present purposes, let's
concentrate on what happens when we parallel translate a tangent vector
around a teeny loop that ends where it started.

>2) For n dimensions I presume we have n tangent vectors to
>follow so we enclose an n dimensional volume. 

Yikes!!!!!!!!!  No, I like to say what I mean, and you must respect that
quirk of mine.  I gave the definition of the Riemann tensor in any
number of dimensions a while back; I'll quote a post of mine where I did
it without coordinates:

The Riemann tensor takes three tangent vectors, say u, v, and w, as
inputs, and outputs one tangent vector, say R(u,v,w).  It's defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges go in the directions epsilon u and epsilon v , where
epsilon is a tiny number longing to approach zero.  The vector w comes
back a bit changed by its journey; it is now a new vector w'.  We then
have

w' - w = - epsilon^2 R(u,v,w) + terms of order epsilon^3

The point is this.  The Riemann tensor is supposed to tell us in a
"local" or "infinitesimal" way how space (or spacetime) is curved.  
What does that mean?  Well, space being curved means that when you
parallel transport a tangent vector around a loop, it comes back
changed.  The idea of the Riemann tensor is that we can take any big
loop, span it with a surface, and then chop up that surface into tons of
wee parallelograms, thus reducing the problem of "parallel translation
around a big loop" to lots of problems of "parallel translation around a
wee parallelogram".   This works in any dimension.  So we never need to
worry about "parallel translating around a hyperquadrilateral."  

(Note: if our space has a funky topology, there may be no surface
spanning a given big loop!  Let's not worry about that now, though there
is a whole branch of math devoted to it.  There are always lots of loops
that *can* be spanned by a surface.)

>Same ball-park? Same solar system?

Well, you gotta let that parallelogram thing sink in.  Get rid of the
hyperquadrilaterals.  

>Typical prof. Hang the carrot out and the donkies just gotta
>follow for the payoff. Hmmm, looks really nice and juicy, if
>only I could reach it!

Sure you can; if it was that hard to understand general relativity there
wouldn't be thousands of people who understood it.  Especially since you
just want a rough idea of it, it's really just a matter of putting some
time into thinking about what I say, and keeping on asking questions.

>Due to the unreal time element of usenet there are active
>threads all over on, or associated, with this subject from
>the same posters. In order to allow John the chance to be
>coherent I propose that we all post to one thread, and this
>one is as good as any. 

I will try to start collecting the various threads along these lines
and renaming them

Re: General relativity tutorial

which is what one of them is already called.

Article 96402 (25 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 28 Jan 1996 12:45:21 -0800
Organization: University of California, Riverside
Lines: 89
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>This particular path to the equator and
>back,  walking the outline of an imaginary segment of a chocolate orange, 
>contains the gratuitous symmetry of constant compass direction. 

Compass direction!?  As you know, any attempt to take coordinates
seriously will only confuse us here when studying general relativity, 
so clearly parallel translation can have nothing to do with something
like "constant compass direction".  Parallel translation is something
you can do on any curved space or spacetime whatsoever, with no
reference to any compass or map.   

So, for example, if you want to avoid symmetries, you can parallel
translate a vector around the coastline of Eurasia.  It's just a bit
hard to talk about that example using ASCII.

>A reasonable person may have some doubt about the well-defined nature of
>"trying very hard not to rotate the arrow".  The point of the exercise is
>after all,  try as hard as we like and the damn thing will have rotated
>when we get back!   But just how "hard" is "very hard"... 

The key thing is the *local* nature of parallel transport.  I think I
noted this in my first explanation: at *each step of the way* you do
your best not to rotate the vector.  It's cheating to know your route
ahead of time and sneakily diddle with your vector so that it comes
out pointing the same way at the end of the journey.  

I think you know this and are just trying to cause trouble.  

Let's go back to the example I originally gave.

Say our Roman starts at the north pole with his javelin.  Say his
javelin points directly in front of him as he begins his journey down
the meridian to the equator.  Assuming the earth is a perfect sphere and
he doesn't gratuitously swing his javelin from side to side, he will end
up at the equator with the javelin pointing due south.  Now he turns 90
degrees --- NOT rotating the javelin, of course!!! --- and walks along
the equator due west.  The javelin continues to point due south, to his
left.  He goes 90 degrees along the equator and stops.  He does NOT
sneakily rotate his vector because he guesses what's coming.  He's a
Roman soldier, after all, trained to follow orders.  He turns 90 degrees
again until he's facing north.  He does NOT rotate the javelin --- jeez,
how many times do I need to say this: he doesn't EVER rotate that
javelin --- so it is still facing due south, directly behind him.  He
now marches up the meridian to the north pole, carrying the javelin
pointing directly behind him, not rotating it even a teeny weeny little
bit.  When he returns to the north pole the javelin is pointing a
different direction than when he started.  In fact, it's rotated by an
angle of 90 degrees from his initial position.

For fun, notice that if we think of the earth as a unit sphere, it's
area is 4 pi, and our Roman has just travelled around a region of land
having area one eighth of that, hence pi/2.  His javelin has rotated by
an angle of pi/2!  This is no coincidence: on the unit sphere, whenever you
go around a simple closed curve enclosing an area A, parallel
translation gives a rotation of angle A.

>It's like SR...  when we speak of length contraction,   that means we have
>tried "very hard" to measure the correct length... but not too hard!   If
>we tried "hard enough",  ie,  allowing for the Lorentz transfomation,  we
>would measure the correct rest length!   And if we tried hard enough here, 
>maybe doing a bit of local surveying,   perhaps we *could* triumphantly
>come back to the same orientation,  even in curved space! 

Of course you can always cleverly correct for things, but that's not the
point.  DON'T cleverly correct for things.  In the case of Lorentz
contractions, just read what the damn ruler says.  In the case of
parallel translations, just follow orders like a Roman soldier and don't
ever swing that javelin around.  

>[Other perverse objections deleted.]

>In fact,  let me take a wild educated guess:   If we get around to stating
>some STOKES like theorem,  to the effect the total discrepency accumulated
>in walking a path is equal to some surface integral,  it will turn out that
>we first state it for a path made of segments of geodesics,  and then
>"arbitrarily well approximate" our arbitrary path by little pieces of
>geodesic,  in an appropriate sense,  of course.  

Well, you can see in the paragraph beginning "for fun" that there is a
Stokes-like theorem operating here.  But there's no need to prove it
first for piecewise geodesic paths; in fact it applies just as well to
situations where parallel translation is well-defined but geodesics are
not --- situations that are irrelevant to general relativity, but very
important in discussing the strong, weak, and electromagnetic forces,
which are ALSO all about parallel translation.

Article 96411 (21 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 28 Jan 1996 13:28:44 -0800
Organization: University of California, Riverside
Lines: 154
NNTP-Posting-Host: guitar.ucr.edu

To keep this course rolling, let me just state where we've gotten so
far, and rapidly finish explaining general relativity!!  Well, I won't
really explain all of general relativity here.  But if you read
this and understand it somewhat, you will know what Einstein's equation,
the basic equation of general relativity, says.

1.  A TANGENT VECTOR at the point p of spacetime may be visualized as an
infinitesimal arrow with tail at the point p.  

2.  A TENSOR of "rank (0,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a number.  The output must depend linearly on each input.   

A TENSOR of "rank (1,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a tangent vector at the point p.  The output must depend linearly
on each input.    

3.  The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
v,w and spits out a number g(v,w), which we think of as the "dot
product" or "inner product" of the vectors v and w.  This lets us
compute the length of any tangent vector, or the angle between two
tangent vectors.  Since we are talking about spacetime, the metric need
not satisfy g(v,v) > 0 for all nonzero v.  A vector v is SPACELIKE if 
g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

4.  PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  

I don't think I want to go into the math of "torsion" just yet.  Let's
see the overall picture first. 

5.  The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point
of spacetime.  Thus it takes three tangent vectors, say u, v, and w as
inputs, and outputs one tangent vector, say R(u,v,w).  The Riemann tensor
is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number.  The vector w comes back a bit changed by its
journey; it is now a new vector w'.  We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Thus the Riemann tensor keeps track of how much parallel translation
around a wee parallelogram changes the vector w.             

6.  Introducing COORDINATES.  Now say we choose coordinates on some
patch of spacetime near the point p.  Call these coordinates x^a (where
a = 0,1,2,3).  Then given any tangent vector v at p, we may speak of its
components v^a in this basis.  The inner product g(v,w) of two tangent
vectors is given by

        g(v,w) = g_{ab} v^a w^b

for some matrix of numbers g_{ab}, where as usual we sum over the 
repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION.
Another way to think of it is that our coordinates give us a basis of
tangent vectors at p, and g_{ab} is the inner product of the
basis vector pointing in the x^a direction and the basis vector pointing
in the x^b direction.

Similarly, the vector R(u,v,w) has components 

        R(u,v,w)^a = R^a_{bcd} u^b v^c w^d

where we sum over the indices b,c,d.  

7.  The EINSTEIN TENSOR.  The matrix g_{ab} is invertible
and we write its inverse as g^{ab}.  We use this to cook up some tensors
starting from the Riemann curvature tensor and leading to the Einstein
tensor, which appears on the left side of Einstein's marvelous equation
for general relativity.  We will do this using coordinates to save
time... though later we should do this over again without coordinates.
This part is the only profound and mysterious part, at least to me.

Okay, starting from the Riemann tensor, which has components R^a_{bcd},
we now define the RICCI TENSOR to have components

         R_{bd} = R^c_{bcd}

where as usual we sum over the repeated index c.  Then we "raise an
index" and define

         R^a_d = g^{ab} R_{bd},

and then we define the RICCI SCALAR by

         R = R^a_a

The Riemann tensor knows everything about spacetime curvature, but these
gadgets distill certain aspects of that information which turn out to be
important in physics.  Finally, we define the Einstein tensor by
         G_{ab} = R_{ab} - (1/2)R g_{ab}.

You should not feel you understand why I am defining it this way!!
Don't worry!  That will take quite a bit longer to explain.  But we are
almost at Einstein's equation; all we need is 

8.  The STRESS-ENERGY TENSOR.  The stress-energy is what appears on the
right side of Einstein's equation.  It is a tensor of rank (0,2), and it
defined as follows: given any two tangent vectors u and v at a point p,
the number T(u,v) says how much momentum-in-the-v-direction is flowing
through the point p in the u direction.  Writing it out in terms of
components in any coordinates, we have

         T(u,v) = T_{ab} u^a v^b

In coordinates where x^0 is the time direction t while x^1, x^2, x^3 are
the space directions (x,y,z), we have the following physical
interpretation of the components T_{ab}:

The top row of this 4x4 matrix, keeps track of the density of energy ---
that's T_{00} --- and the density of momentum in the x,y, and z
directions --- those are T_{01}, T_{02}, and T_{03} respectively.  This
should make sense if you remember that "density" is the same as "flow in
the time direction" and "energy" is the same as "momentum in the time
direction".  The other components of the stress-energy tensor keep track
of the flow of energy and momentum in various spatial directions.

9.  EINSTEIN'S EQUATION: This is what general relativity is based on.
It says that

          G = T

or if you like coordinates and more standard units,

          G_{ab} = 8 pi k T_{ab}

where k is Newton's gravitational constant.  So it says how the flow of
energy and momentum through a given point of spacetime affect the
curvature of spacetime there.

That's it!

Article 96587 (29 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 29 Jan 1996 12:57:52 -0800
Organization: University of California, Riverside
Lines: 243
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:

>I see there already *is* a tiny analogue to Stokes theorem floating around
>here,  since the discrepency here depends on the *area* of the path, 
>epsilon^2 .  The Riemann curvature tensor is thus some analogue
>(generalization?)  of the curl of a vector field.   

Very, very, very very smart observation.

This observation is at the basis of all our present-day theories of
forces: Maxwell's equations for electromagnetism, Einstein's theory of
general relativity, and the Yang-Mills equations for the electroweak and
strong forces.  We say they are all "gauge theories".  What this means
is that the basic field involved in any of these forces is a
"connection" which describes what happens to particles when you move
them along a path.  Various internal degrees of freedom get "parallel
transported".  When you take them around a loop they don't come back as
they were.  When you study this infinitesimally, you get a "curvature
tensor" describing parallel translation around infinitesimal
parallelograms --- of which the Riemann curvature is an example.  There
is a formula for this curvature tensor as a kind of "curl" of the
connection.  In the case of magnetostatics, this takes the simple form:

                       B = curl A

That is, the magnetic field is the curl of the vector potential.  
The vector potential is the connection in this case, and the magnetic
field is the curvature.  

In the case of electromagnetism in 4d spacetime we have the same sort of
thing:

                       F = dA

where the electromagnetic field F is the curvature and the (4d) vector
potential A is the connection.  Here d is a 4d analog of the curl,
called the "exterior derivative".  

In Yang-Mills theory and gravity we have the same sort of thing, only
somewhat fancier.  Not too surprising, in a way, since Einstein, Yang
and Mills were all deliberately trying to copy the shining example of
Maxwell's equations.  

Now you might ask: in general relativity, when a parallel translate a
tangent vector around a loop, its actual direction changes.  But in
electromagnetism, if I carry a charged particle around a loop, what
changes? 

It's phase!  (Here quantum theory rears its ugly head.)

Article 96602 (18 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 29 Jan 1996 14:01:27 -0800
Organization: University of California, Riverside
Lines: 71
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>[email-masked] (john baez) wrote:

>Just an odd notation I used. Honest. I forgot the 'proper'
>ones. Ow!

It wasn't a notational error you made, I think.  It was a conceptual
error, but an easy one to make, especially if you haven't been following
my notation.  

>Actually notation is a problem here, I think. OK, could be
>memory too. Anyway could some kind soul take a simple
>concrete example and run it through so I can see properly
>what all these curly brackets and so on actually are.

When I write something like F_{abc} this is just short for

                       F
                        abc

This notation has become standard among folks trying to talk about math
on usenet.  Here curly brackets just tell you what stuff is part of the
subscript.  Just as _ means subscript, ^ means superscript.  So I would
write the Riemann tensor R^a_{bcd} as

                         a
                        R
                         bcd

if I had the whole damn day to type this stuff.

>Usually these things are simple and straightforward once you
>know what people mean, but can seem very ambiguous if you
>don't.

Yup.

>>Someone else may have put it a bit clearly than I did: we want F to be
>>linear in each argument *when we hold the others fixed*.  What's an
>>example of this sort of thing?  Well, a good example is
>>
>>F(u,v,w) = F_{abc} u^a v^b w^c
>>
>>Here I have introduced some coordinates, and u^a, v^b, and w^c stand for
>>the components of u, v and w in these coordinates.  As usual we sum over
>>the indices a,b,c (say from 0 to 3 if we're in good old four-dimensional
>>spacetime).  What's F_{abc}, you ask?   Any old batch of numbers!!!

So typographically speaking, the above equation just means

                               a  b  c
              F(u,v,w) = F    u  v  w
                          abc

If you're wondering what THAT means, reread the above explanation and
then ask question... or maybe someone will be so kind as to go over
an example of how this tensor stuff works.  I gotta go teach class.

>>It just works out that way: the amount the vector w changes when you
>>carry it around a little parallelogram with sides epsilon u and epsilon
>>v is roughly proportional to epsilon^2.  

>Hmmm, sounds plausible. One might expect the amount of
>curvature to relate to area.

Yup.

Article 96646 (58 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 29 Jan 1996 17:23:09 -0800
Organization: University of California, Riverside
Lines: 115
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:

>Parallel transport is well defined but geodesics are not:   Just what kind
>of mathematical structure allows this? 

Well, I alluded to it in my gleeful reply to your post where you noticed
the relationship between the Riemann curvature tensor and the curl of a
vector field... the answer is: "gauge theories".

In general relativity, you parallel transport tangent vectors around.
In other gauge theories, you parallel transport vectors living in more
abstract vector spaces.  You could think of these other theories as insane
mathematical generalizations of general relativity, if it weren't for
the little fact that all the forces in nature are described by gauge
theories.  
 
>I think you mentioned parallel transport can be well defined in terms of a
>metric?  Surely this is sufficient for geodesics also. 

Let me outline the dependencies.  If I say something you know, don't
assume I think you didn't know it.  

1.  If you have a metric, one can define parallel transport of tangent
vectors in terms of it.  See the end for more details on how this works:
this is very important in general relativity.

2. If you have a notion of parallel transport of tangent vectors, one
can define the notion of geodesics.  This is also very important in
general relativity since things in free fall trace out geodesics in
spacetime.  A geodesic is simply a curve whose own tangent vector is
parallel transported along it.  Note: the "tangent vector to a curve" at
the point p is a special "tangent vector" at the point p; the term
"tangent vector" has two meanings here!  

There are branches of geometry where you have parallel transport of
tangent vectors, but not a metric.  These are irrelevant for general
relativity, though they have been considered by physicists trying to
soup up general relativity to handle other forces.

3.  However, if we are parallel transporting vectors that aren't tangent
vectors, a metric is often quite irrelevant.  That's what I was alluding
to.  In this context there might well be no metric, hence no notion of
geodesics.  This comes up in fancy-ass mathematical physics like
"topological quantum field theories".  

>A topological space
>with no metric?  Do such things actually have physical applications? 

Well, we don't need to consider topological spaces with no metric, we
can simply consider smooth manifolds with no metric, but with some other
geometrical structure, to get situations where parallel transport of
some *other* sort of vectors (not tangent vectors) might be
well-defined, but not geodesics.

It's sort of amusing: to the outsider these possibilities must seem
quite bizarre, but in fact they have been under intense scrutiny ever
since the early 20th century, and are "standard material" that every
mathematician or theoretical physicist should be acquainted with.  They
call it simply "geometry" or "differential geometry".

Just to reminisce slightly... when I learned general relativity in
college, I realized I needed to learn more geometry to really understand
what was going on.  The assigned text was Weinberg's Gravitation and
Cosmology, but in a notorious passage in that book Weinberg writes `the
passage of time has taught us not to expect that the strong, weak, and
electromagnetic interactions can be understood in geometrical terms, and
too great an emphasis on geometry can only obscure the deep connections
between gravitation and the rest of physics.'  Many theoretical
physicists these days would laugh their heads off if someone said that
now.  Anyway, I naturally turned to Misner Thorne and Wheeler's
Gravitation, which *does* emphasize the geometry, and I liked this much
better, but I realized I needed a good solid mathematical treatment of
geometry, so I eventually bumped into 

Analysis, Manifolds, and Physics, by Yvonne Choquet-Bruhat, Cecile
DeWitt-Morette, and Margaret Dillard-Bleick, North Holland, New York,
1982.

This covers a lot of the geometry and other math physicists need, and I
fell in love with it; I just kept reading and rereading it 'til I knew
all that stuff.  Later in grad school I studied differential geometry
with Guillemin (famous for his work on symplectic geometry, which is the
geometry of *phase space* rather than physical space or spacetime), but
I didn't go too deep in the subject.  When I started working in earnest
on quantum gravity, I found I needed to dig deeper into geometry.  So
I've been learning a bit more about it.  Not a whole lot more, mind you:
it's a vast and deep field, and if I got too carried away with it I'd
never get to all the other stuff I need to know, and do, in studying
quantum gravity.  

Okay, here's a reminder of how you get from the metric to parallel
transport: 

PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  

Article 96675 (57 more) in sci.physics:
From: john baez
(SAME) Subject: Re: general relativity tutorial
Date: 29 Jan 1996 19:16:01 -0800
Organization: University of California, Riverside
Lines: 60
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Bro
nis Vidugiris) writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:
>)In article <[email-masked]> [email-masked] writes:

>)>Anyway, it would seem that what you are saying is
>)>that (oh dear, I feel a Baez bashing coming) if I took a
>)>scalar (say density) and wanted to know how it was varying
>)>from point p in the 2u+v+7w local direction, I could plug it
>)>into my tensor of rank (0,3) at p and get out an answer.

>)You seem to think that the function
>)
>)F(u,v,w) = 2u + v + 7w
>)
>)is linear in each argument.  It's not. 

>Hmmm - well, my impression was that Oz was thinking of u (not to mention v and
>w) as vectors - in which case he was on the right track.  His notation
>is fairly standard, except that he didn't boldface them (difficult
>to do in Ascii!). 

I had lots of trouble understanding what Oz was actually thinking of, so
I just took a stab at it.  Rather than trying to guess further, it
probably makes more sense to think about various precisely described
functions and get good at deciding whether they are tensors or not.

So: 

>But onto the other example 
>
>But isn't F(u,v,w) as you have written it above a one-form?  I always
>thought those wore considered linear (?).  Or are they called
>anti-linear?  Or what?

Let's see.  I was thinking of u, v, and w as three different vectors.
In my example, I considered the function which eats three vectors u, v,
and w, and outputs the vector 2u + v + 7w.  You might naively think this
is a tensor of rank (1,3), since there are 3 vector inputs and 1 vector
output, and it looks sorta linear.  My point was that it's not, since
it's not separately linear in each argument.

But you seem to be thinking of u, v, and w as the 3 components of a
single vector!  Perfectly understandable, since I didn't make myself
terribly clear!  So let's think about this.  Now we are saying F is a
function which eats one vector (u,v,w) and spits out a number 2u + v +
7w.  Now F turns out to be a tensor of rank (0,1), since there is one
vector input and one number output, and it now *is* linear in its one
input.  You are right; this kind of tensor is also called a one-form or
cotangent vector.

So with sufficiently ambiguous notation it's completely impossible to
tell whether something is a tensor or not.  :-)

I'll try to be clearer about notation in the future.

Article 96864 (58 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 30 Jan 1996 11:25:16 -0800
Organization: University of California, Riverside
Lines: 162
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Bron
is Vidugiris) writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:
>)In article <[email-masked]> [email-masked] (Bronis V
idugiris) writes:

>)>Anwyay, just to clarify what should be obvious but is for some
>)>reason giving me problems:

>)>I keep the javelin at a constant angle relative to my path, as long
>)>as my path stays constant. 

>)Your "path stays constant"?  What does THAT mean? 

>It's something that better be defined to use my idea of "don't
>change the angle" :-)

Your honesty is disarming.

>If a constant path is a great circle, that's one thing, and hopefully
>it's all then co-ordinate invariant.

Yes, "geodesic" is a coordinate-invariant notion on any Riemannian
manifold.  To move along a geodesic is, so to speak, simply to "follow
your nose" while keeping your nose pointing straight forwards.  (By the
latter I secretly mean that your nose is being parallel transported.)
Thus the geodesics on the sphere may be defined in a
coordinate-independent way, and they work out to be simply the great
circles.    

>OTOH if a constant path was a line of constant
>lattitude, it becomes something totally different (probably co-ordinate
>dependent, too).

"Lines of latitude" are, of course, coordinate-dependent.  The only line
of latitude that's a geodesic is the equator.  If you go on any other
line of latitude, you are constantly "swerving" slightly.  This is
easiest to understand if you consider the line of latitude which is a
circle one foot in diameter going around the north pole!  

So it's bad to think of lines of latitude as having any special
privileged role as "constant paths".  Parallel translation along lines
of latitude is complicated.  If you parallel transport a vector all
around a line of latitude, it does not come back to where it started,
unless the line of latitude happened to be the equator.

(For smart alecks, I suppose I should add that the north pole and south
pole could be considered degenerate "lines of latitude", and parallel
transport along these... i.e., just sitting there... doesn't change a
tangent vector.)

>)But if you know what it means for the tangent vector to your path not to
>)rotate, how come you don't know what it means for the javelin ---
>)another vector --- not to rotate???
>
>I can accept as a primitive operation "move in the dirction of the
>tangent vector", but I have difficulty accepting "don't rotate"
>as a primitive operation - it just doesn't click for me.

Well, believe it or not, it's well-defined in a coordinate-independent
way.  It is, however, rather subtle!  Recall what I said:

PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  

The notion of "torsion-free" parallel transport is what you are worrying
about.  I am beginning to fear that I am going to have to give you the
mathematical definition to make you happy.  :-)  Luckily I know a way
to do this without writing down any big equations.  But I won't do it
just yet.  

You see, I had hoped that the notion of "carrying a vector along while
not rotating it" would be geometrically obvious in a rough intuitive
sort of way.  Certainly, for example, if two guys in a column of
soldiers are marching along carrying javelins (not necessarily pointing
forwards), and one is not rotating it, while the other is sassily
swinging it back and forth, you can tell who is following orders! 

You don't need to have coordinates around to see who is rotating
something and who is not.  You might think this is only true in flat
space, but actually it's still true in curved space.  That's the point
of "torsion-freeness".

>)I think the more primitive concept is that of parallel translation ---
>)"carrying a vector along while never rotating or stretching it".  The
>)concept of geodesic may then be defined as a path whose tangent
>)vector is parallel translated along that path.  In this approach, one
>)doesn't want to go back and define parallel translation using the notion
>)of geodesic.  However, if the notion of geodesic seems more basic to
>)you, then your approach is okay.  At least it suffices to define
>)parallel translation along curves that are piecewise geodesic:

>Hmmm - well, I saw some stuff in "Gravitation" that may give me a clue
>as to your motivations the other day more or less by chance
>(you probably want to avoid having to define a metric), but
>it seems kind of strange and backwards to me as of yet.

I defined a metric a few times by now...

The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
v,w and spits out a number g(v,w), which we think of as the "dot
product" or "inner product" of the vectors v and w.  This lets us
compute the length of any tangent vector, or the angle between two
tangent vectors.  Since we are talking about spacetime, the metric need
not satisfy g(v,v) > 0 for all nonzero v.  A vector v is SPACELIKE if 
g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

If my pedagogy seems perverse, it's because I'm secretly trying to 
make clear the following facts:

1. A metric determines a unique notion of parallel transport satisfying
a-c as above.

2. Not every notion of parallel transport comes from a metric in this
way!  It's often good to think of parallel transport as a fundamental
notion.  

3. Given a notion of parallel transport we may define geodesics.

4. Just given geodesics we *cannot* easily define parallel transport.
Two different notions of parallel transport may happen to give the same
geodesics.  

It's true that one may take a shortcut and define geodesics directly
from the metric, by saying:

A geodesic is a path that's locally a path of minimum (or maximum)
length.

But I wasn't wanting to take this shortcut, because the notion of
parallel transport is needed for all sorts of things in GR, like
defining the Riemann curvature, so we must face up to it.

As you can see, there is a fairly intricate web of concepts here, and I
was attempting to cut my way through it with maximum efficiency, but now
you have gone and made me explain more of the mathematics.  :-)

Article 96867 (57 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 30 Jan 1996 11:55:21 -0800
Organization: University of California, Riverside
Lines: 147
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>It would seem to me that (despite all the contortions to the
>contrary) the legionary is in fact *locally* in a flat
>spacetime. We can make it as arbitrarily as flat as we like
>by considering a small enough local area (<<epsilon). In
>this case there is no problem with local angles, the
>legionary can head off and maintain at any local angle he
>likes, and hold the javelin (as long as it's not too long!)
>at any angle to his path as is required, quite sensibly.
>Everything he does locally is quite flat, valid and
>uncomplicated. He has no idea that "globally" his little
>local area is being rotated by his path though the larger
>geometry.

This is correct and is a good way to see why there's no problem
with defining "parallel transport with no rotation or stretching"
even in curved space (or spacetime).  Parallel transport is a locally
defined thing, meaning that at each itsy-bitsy step of the way the
legionary carries his javelin, "not rotating or stretching it",
according to a recipe that depends only on what his path is like and
what space is like in the immediate vicinity.  If we make the vicinity
immediate enough, we can get away with pretending it's flat, for these
purposes. 

There is, however, a certain care one must exercise in this way of
thinking, which is why I haven't brought it up sooner.  It has to do
with whether one is keeping track of terms of order epsilon, or
epsilon^2, or what.  We can pretend space is flat locally to a good enough
degree to define parallel transport, but it's still true that if we
parallel transport a vector around a rectangle whose edges have length
epsilon, it changes by an amount proportional to epsilon^2... so it's
not as if space really *is* flat.  

>Baez intends to discard terms in epsilon^3 in the time-honoured Newtonian
>way. 

Something like that.  If this all seems confusing, remember the freshman
calculus mistake: "if I change x a teeny bit epsilon, then f(x) changes
by a teeny bit proportional to epsilon.  But then as I take the limit
epsilon -> 0 that means f doesn't change at all.  So the derivative of f
is zero."  

Or for example

(d/dx) x^2 |            = (d/dx) 2^2 =  (d/dx) 4 = 0.
            x = 2

>I take it that Baez is going to show us the next steps with minimal
maths ..... ? 

Sure, I am not going to get into these epsilons and deltas more than
absolutely necessary.  I'll prefer to send them to zero and say to hell
with them.  

I have laid out a road map that'll take us to Einstein's equation, and
now I will wait for people to ask me questions about it.  To repeat it,
leaving out the boring stuff we have already been over thoroughly (??),
it goes like this: 

1.  A TANGENT VECTOR is....

2.  A TENSOR of "rank (0,k)" is....

    A TENSOR of "rank (1,k)" is...

3.  The METRIC g is...

4.  PARALLEL TRANSLATION is...

5.  The RIEMANN CURVATURE TENSOR is...

6.  Introducing COORDINATES.  Now say we choose coordinates on some
patch of spacetime near the point p.  Call these coordinates x^a (where
a = 0,1,2,3).  Then given any tangent vector v at p, we may speak of its
components v^a in this basis.  The inner product g(v,w) of two tangent
vectors is given by

        g(v,w) = g_{ab} v^a v^b

for some matrix of numbers g_{ab}, where as usual we sum over the 
repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION.
Another way to think of it is that our coordinates give us a basis of
tangent vectors at p, and g_{ab} is the inner product of the
basis vector pointing in the x^a direction and the basis vector pointing
in the x^b direction.

Similarly, if R is the Riemann tensor, the vector R(u,v,w) has components 

        R(u,v,w)^a = R^a_{bcd} u^b v^c w^d

where we sum over the indices b,c,d.  

7.  The EINSTEIN TENSOR.  The matrix g_{ab} is invertible
and we write its inverse as g^{ab}.  We use this to cook up some tensors
starting from the Riemann curvature tensor and leading to the Einstein
tensor, which appears on the left side of Einstein's marvelous equation
for general relativity.  We will do this using coordinates to save
time... though later we should do this over again without coordinates.
This part is the only profound and mysterious part, at least to me.

Okay, starting from the Riemann tensor, which has components R^a_{bcd},
we now define the RICCI TENSOR to have components

         R_{bd} = R^c_{bcd}

where as usual we sum over the repeated index c.  Then we "raise an
index" and define

         R^a_d = g^{ab} R_{bd},

and then we define the RICCI SCALAR by

         R = R^a_a

The Riemann tensor knows everything about spacetime curvature, but these
gadgets distill certain aspects of that information which turn out to be
important in physics.  Finally, we define the Einstein tensor by
         G_{ab} = R_{ab} - (1/2)R g_{ab}.

You should not feel you understand why I am defining it this way!!
Don't worry!  That will take quite a bit longer to explain.  But we are
almost at Einstein's equation; all we need is 

8.  The STRESS-ENERGY TENSOR is...

9.  EINSTEIN'S EQUATION: This is what general relativity is based on.
It says that

          G = T

or if you like coordinates and more standard units,

          G_{ab} = 8 pi k T_{ab}

where k is Newton's gravitational constant.  So it says how the flow of
energy and momentum through a given point of spacetime affect the
curvature of spacetime there.

That's it!

Article 96869 (56 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 30 Jan 1996 12:02:56 -0800
Organization: University of California, Riverside
Lines: 32
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> Geoffrey A. Landis <geoffrey.landi
[email-masked]> writes:

>Topics you might consider mentioning:
>
>What is the geometrical meaning of the Ricci tensor?  What is the
>geometrical meaning of the contracted Ricci tensor?  When we contract the
>Riemann tensor to make the Ricci tensor, what information do we discard?

Ah.  I wish I understood the answers to these questions better.
I am hoping, in fact, that in the process of explaining this stuff on
sci.physics, I will be forced to learn more about this.  It should start
happening soon, since I have laid out a road map to Einstein's equation,
and most gnarly part of it is getting from the Riemann tensor to the
Einstein tensor, via the Ricci tensor and Ricci scalar.

>Since the metric has only 4x4 components, why do we need the Riemann
>tensor with 4x4x4x4 components to describe curvature?

That at least is clear: curvature tells us, when we take any
little vector and parallel transport it around any little parallelogram,
how much it changes.  By (multi)linearity, we need to know this for
vectors in each of the 4 directions (t, x, y, and z), for each of 4x4
possible parallelograms in the tx, xy, xz, etc. planes, and we need to
know how the vector changes in each of the 4 directions.   That's
4x4x4x4.  However there are lots of relationships.  For example, if we
know what happens when we go around a parallelogram in the xt plane we
know it when go around a parallelogram in the tx plane!!  If we discard
all the redundancies, the Riemann tensor has only 20 independent
components in 4 dimensions.  

Article 96871 (55 more) in sci.physics:
From: Matthew P Wiener
Subject: Re: General relativity tutorial
Date: 30 Jan 1996 18:03:06 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 60
Distribution: world
NNTP-Posting-Host: sagi.wistar.upenn.edu
In-reply-to: Geoffrey A. Landis <[email-masked]>

In article <[email-masked]>, Geoffrey A. Landis <geoffrey.land
is@lerc writes:
>What is the geometrical meaning of the Ricci tensor?  What is the
>geometrical meaning of the contracted Ricci tensor?  When we contract the
>Riemann tensor to make the Ricci tensor, what information do we discard?

>Since the metric has only 4x4 components, why do we need the Riemann
>tensor with 4x4x4x4 components to describe curvature?

I posted the following before.  Perhaps it will make sense to you.
------------------------------------------------------------------------
It is not too hard to get a good geometric feel for the curvature of
curves and surfaces in 3-space.  Perhaps the best introduction is a
neat new book, John McCleary GEOMETRY FROM A DIFFERENTIABLE VIEWPOINT,
Cambridge, 1994.  In particular, he explains well that it shows up
rather naturally via lim_{r->0} (2.pi.r - circum(radius r))/r^3: the
second derivative, so to speak, of the variation from Euclidean geometry.

At any point on a smooth manifold, you can pick local coordinates whose
origin is the point in question, and which has zero first derivatives
of each coordinate with respect to each other.  In general, that is the
best you can do.  Curvature shows up via non-zero second derivatives.
This is vague, but it's enough to motivate what happens next.

Given a tangent vector at the starting point, one can interpolate a
geodesic whose starting velocity is that tangent vector.  Now push off
like this for a plane P's worth of tangent vectors in the tangent space.
Near the point in question, these describe a 2-dimensional submanifold
P* of the original manifold.  Let u,v be an orthonormal basis for the
tangent plane P.  Then Riemann(u,v,u,v)=Gaussian curvature(P*).

The full tensor can be recovered from this by a simple polarization.

This explains R_ijkl.  One would also to understand R^i_jkl just as
concretely.  That's where holonomy comes in.

Given P,u,v as above, one can push off on these geodesics in an epsilon
rhombus as follows.  First to u.eps, then u.eps+v.eps, then backwards to
v.eps, then back to 0.  You don't quite hit home again, but the error
shrinks like eps^2, so ignore it.  Now, in addition to running around,
carry an arbitrary vector w with you along for the ride.  Carry it in
a parallel way the whole ride.  When you get home, the vector, now w',
could be pointing anywhere, thanks to curvature.  What's very interesting
is that (w'-w)/eps^2 approaches a limit as eps->0, and this limit is linear
in w.  So for each u,v, we have R(u,v):w|->e, a kind of second derivative
that measures holonomic deviation.  What's very very very interesting is
that R itself is linear in u and v!

And so we have R(u_j,u_k)u_l=Sum R^i_jkl.u_i, the Riemann curvature tensor.

The Ricci tensor, summing over i and l, can be thought of as an "average"
measurement of R(u,v)'s absolute scaling without twisting.  That is, cut
to just the positive orthant, and ignore any holonomic deviation in other
directions by projecting back on to the original vector each time, and
see how big it is.  That's the part that contributes to gravity.  (Which
makes a bit of rough sense, even: no sidewise gravity!)

(Note: signs and factors have been gleefully ignored.  For one thing,
there are different conventions out there.)
-- 
-Matthew P Wiener ([email-masked])

Article 97058 (41 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 31 Jan 1996 11:06:56 -0800
Organization: University of California, Riverside
Lines: 84
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>[email-masked] (john baez)' wrote: 
>(Bronis Vidugiris wrote) 

>>>I can accept as a primitive operation "move in the dirction of the 
>>>tangent vector", but I have difficulty accepting "don't rotate" 
>>>as a primitive operation - it just doesn't click for me. 

>Look, sir,  I'm with Bronis!   It doesn't really click for me,  either. 

If you don't watch out I'm going to have explain the concept of
"torsion".  Do you really want that? 

Seriously, I was hoping to gloss over torsion for now so I could get to
Einstein's equation.  If you insist, I'll talk about it, but you have to
decide what it is you really want to be spending time on.

For now, let me just stage a little rhetorical holding action.  

>I think the problem is,  we are both just sophisticated enough to think
>there may be some problem with the idea  "just walk along and don't rotate
>the javelin",  and I think we are right to be worried!   I know you would
>not want us to just accept this on faith.   :-) 

Yes, but the problem is that you are not sophisticated to also think
that there may be exactly the same problem with the idea "just walk
along a geodesic".  

To walk along a geodesic seems as simple as "following your nose", and
it is, but only if you know how not to *turn your nose* in the process.
In other words, you can only "keep walking as straight as you can" ---
which is what following a geodesic means --- if you know how not to
turn!  This is what I mean by saying that one of the most widely-used
definitions of geodesic *relies upon* the concept of "parallel transport
without rotation".  Folks define a geodesic as a path whose tangent
vector is parallel translated along that very path.  

So if you are scared that you don't know how to carry a javelin along
without rotating it, you should be scared that you don't know how to
walk along in a straight line.  Personally I have no trouble doing
either.  

>This is why I tried to take refuge in the idea of geodesics,  like
>Bronis.  They make some sense to me as a primitive notion....

One is allowed to take whatever notions one wants as primitive and then
try to develop others on that basis.  It's all a matter of elegance.
You guys like geodesics, so you might prefer an approach based on those.
But see below for a problem with trying to develop an approach like
that.  

>...and in that
>case the idea of holding the javelin at a constant angle w.r.t. the
>path also makes sense  (I assume I have that much right... that if we
>*do* walk a geodesic, holding a constant angle w.r.t. it is the right
>thing to do).  

Let me just point out one problem which I failed to notice before.  Say
you know what a geodesic is.  Then in TWO dimensions you are right: when
we are walking along a geodesic, you can parallel translate a vector by
holding it at a constant angle from the tangent vector to the geodesic.
But in three or more dimensions that's not good enough: even though the
vector remained at a constant angle from the tangent vector to the
geodesic, it could "swing around".  It's only parallel transported when
it doesn't rotate at all!

So there is trouble seeing how to reconstruct the concept of
geodesics from the concept of parallel transport.  And indeed, now that
I think about it, this is obvious; two different connections can have
the same geodesics, and to pick out the "good" one, the one to use in
Einstein's equation, we need to pick the one with no "torsion".  

>Too late.  Look,  say I am strolling along the surface of a sphere with my
>javelin (let's keep this simple):  can you give me an *operational*
>definition of how to keep the javelin pointing in the same direction as I
>move around? And please don't tell me "just don't rotate it".  I am
>rotationally challenged,  and that would be discriminating against me.  I
>would like you to tell me how to do it in terms of surveying instruments, 
>or rulers and protractors,  and other such crutchs for the differently
>torsioned. 

Okay, I'll do this after I come back from class.

Article 97064 (40 more) in sci.physics:
From: Edward Green
(SAME) Subject: Re: General relativity tutorial
Date: 31 Jan 1996 07:33:10 -0500
Organization: The Pipeline
Lines: 47
NNTP-Posting-Host: pipe11.nyc.pipeline.com
X-PipeUser: egreen
X-PipeHub: nyc.pipeline.com
X-PipeGCOS: (Edward Green)
X-Newsreader: The Pipeline v3.4.0

[email-masked] (Oz)' wrote: 
 
>It would seem to me that (despite all the contortions to the 
>contrary) the legionary is in fact *locally* in a flat 
>spacetime. We can make it as arbitrarily as flat as we like 
>by considering a small enough local area (<<epsilon). In 
>this case there is no problem with local angles, the 
>legionary can head off and maintain at any local angle he 
>likes, and hold the javelin (as long as it's not too long!) 
>at any angle to his path as is required, quite sensibly. 
>Everything he does locally is quite flat, valid and 
>uncomplicated. He has no idea that "globally" his little 
>local area is being rotated by his path though the larger 
>geometry. 
 
Thank you,  Oz!   After letting that thought simmer overnight,  that really
seems to clear up my CD (cognitive dyspepsia) here.  I think the *two*
infinitesimal distance scales are really the key to thinking about this
sensibly. 
 
On the one hand,  we let epsilon get small enough that *something* is
effectively constant over the dimensions of our little path,  that
something presumably being the "curvature" or the Riemann tensor.  On the
other hand,  we then let distances shrink still further, an "infinitesimal
of an infinitesimal",  in considering the legionary's path,  so that space
is effectively flat right in front of his face.  Then presumably there is
no ambiguity,  even for drunken wanderings on this scale,  in what we mean
by "holding the javelin at a constant angle".   We piece together all these
epsilon-tisimo bits of path to form the path of dimensions or order
epsilon.   
On the scale of epsilon,  the legionary is a microbe!   
 
I am not sure this view is totally coherent,  but it has the strong enough
flavor for me of something on its way from speculation into rigor that I
think I am about ready to accept it and move on.   I assume this insight
could be made rigorous. 
 
And I am beginning to think you understood this all along... 
 
By the way,  did legionaries carry javelins?   That's the real CD problem
here...  he was pointing with his short sword.   Now,  all is clear  :-) . 

-- 
 
Ed Green 
[email-masked]

Article 97130 (17 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 31 Jan 1996 15:28:57 -0800
Organization: University of California, Riverside
Lines: 78
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>[email-masked] (Oz)' wrote: 

>>It would seem to me that (despite all the contortions to the 
>>contrary) the legionary is in fact *locally* in a flat 
>>spacetime. We can make it as arbitrarily as flat as we like 
>>by considering a small enough local area (<<epsilon). In 
>>this case there is no problem with local angles, the 
>>legionary can head off and maintain at any local angle he 
>>likes, and hold the javelin (as long as it's not too long!) 
>>at any angle to his path as is required, quite sensibly. 
>>Everything he does locally is quite flat, valid and 
>>uncomplicated. He has no idea that "globally" his little 
>>local area is being rotated by his path though the larger 
>>geometry. 

>Thank you,  Oz!   After letting that thought simmer overnight,  that really
>seems to clear up my CD (cognitive dyspepsia) here. 

Whew.  Thanks, Oz!  Now if Bronis also buys this description of parallel
transport, maybe I won't have to launch into my heavy-handed spiel about
torsion just yet.

>I think the *two*
>infinitesimal distance scales are really the key to thinking about this
>sensibly. 
> 
>On the one hand,  we let epsilon get small enough that *something* is
>effectively constant over the dimensions of our little path,  that
>something presumably being the "curvature" or the Riemann tensor.  On the
>other hand,  we then let distances shrink still further, an "infinitesimal
>of an infinitesimal",  in considering the legionary's path,  so that space
>is effectively flat right in front of his face.  Then presumably there is
>no ambiguity,  even for drunken wanderings on this scale,  in what we mean
>by "holding the javelin at a constant angle".   We piece together all these
>epsilon-tisimo bits of path to form the path of dimensions or order
>epsilon.   

>I am not sure this view is totally coherent, but it has the strong enough
>flavor for me of something on its way from speculation into rigor that I
>think I am about ready to accept it and move on.   I assume this insight
>could be made rigorous. 

Yes, it can.  Let me try to restate it a little more mathematically, just for
fun.  Let's take any old n-dimensional space with a Riemannian metric on it.
Say we want to parallel transport a vector along a curve.  Since this is
a "local" process, it suffices to parallel transport it from point P along an
itsy-bitsy stretch of the curve, of length epsilon, to the nearby point
Q.  

So: consider a little patch of spacetime, whose length, width, etc. are
all about epsilon, and which contains the curve from P to Q.  It's not flat,
but it's almost so.  Flatten it out, i.e., take a flat metric that
closely approximates our original metric in the patch.  Now that we're
in flat space, parallel transport is unambiguous!  So do parallel
transport that way.

But wait --- this "flattening out" process was ill-defined: there might
be different flat metrics that approximate our original metric pretty
well in the patch.  That's true, but if two people do this flattening
out in different but still reasonable ways, it will only affect the
answer by an amount of order at most espilon^2.  

This means that if we take a long curve of length L and chop it into
L/epsilon pieces of length epsilon, and do the above trick repeatedly,
the accumulated error will be at most of order L epsilon, when epsilon
is sufficiently small.  

So chopping ever finer we can get as good an approximation as we want.

[If any of you ask how the "flattening out" business works I'll challenge you
to a duel with javelins at 50 paces.  If your manifold is embedded in a
metric-preserving way in some higher-dimensional space, this is easy:
just project onto the plane tangent to the manifold.]

Article 97164 (69 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 31 Jan 1996 18:35:24 -0800
Organization: University of California, Riverside
Lines: 89
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward Gr
een) writes:
>[email-masked] (Edward Green)' wrote: 

>>...because there are 4x4 little squares we could 
>>translate around ...  We notice that it is 
>>antisymmetric w.r.t. interchanging the indices for the little epsilon 
>>paths,   

>That didn't go far enough.  Clearly the antisymmetry w.r.t. these indices
>extends to the diagonal,  which is zero.  So there are at most  
> 
>       (4 choose 2) = 6   
> 
>independent components generated by permuting these indices while holding
>the others fixed,  or 
> 
>       4x4x4x4 -> 4x4x6   
> 
>independent components in all,  at most.   And that's only the first
>symmetry we found... 

Right.  Remember that R^a_{bcd} describes how much more the vector in the d
direction points in the a direction after touring around the parallelogram
that goes first in the b direction and then the c direction.  You are
noting that

R^a_{bcd} = -R^a_{cbd}.

Here are the other symmetries.  Note that parallel translation around 
the little parallelogram *rotates* a little bit.  So if we consider the
a and d indices only, what we have is an "infinitesimal rotation".  This
implies

R_{abcd} = -R^{dbca}.

Get it?  Here I did a trick called "lowering indices" to push down the
first one; don't worry about that.  Here's the point: a rotation in 2d
is a matrix like 

cos theta     -sin theta
sin theta      cos theta

If you differentiate this with respect to theta and set theta to zero
you get the "infinitesimal rotation"

0 -1
1  0

This matrix T has T_{ab} = - T_{ba}, and that's always true for
infinitesimal rotations, in any number of dimensions of space.  

There is one other symmetry, which is a bit more profound.  
It's called the "algebraic Bianchi identity" --- to be distinguished
from the still more profound "differential Bianchi identity" that we'll
get to later.  It's

R^a_{bcd} + R^a_{cdb} + R^a_{dbc} = 0.

To derive this one you gotta use the fact that the connection is
"torsion-free" --- whereas the other two identities didn't need that.

Naturally, real experts in general relativity know these identities by
heart and constantly use them to pull all sorts of fiendish tricks.
Don't bother remembering them, since they won't be on the final.  But
some of you folks might enjoy seeing how you can use them to boil the
Riemann tensor down to 20 independent components when spacetime has
dimension 4.

Here is something cool.  In 1 dimension, these identities mean
that the Riemann tensor has NO nonzero components --- there ain't no
intrinsic curvature in a 1-dimensional universe.  In 2 dimensions,
these identities mean it has ONE linearly independent component --- the
Gaussian curvature of a surface.  In 3 dimensions, there are 6
independent components, and in 4 dimensions, there are 20.  

The Einstein tensor G, which appears in Einstein's equation G = T, has
exactly as many independent components as the Riemann tensor in
dimensions 1, 2, and 3, but not in 4 or more!  This means that in the
vacuum, where G = 0, the Riemann tensor must vanish in dimensions 1, 2,
and 3.  So empty space is flat unless you have at least 4 dimensions.
In 4 dimensions it needn't be --- which is why gravitational waves
exist.

Article 97321 (32 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 1 Feb 1996 13:33:21 -0800
Organization: University of California, Riverside
Lines: 33
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>[email-masked] (john baez)' wrote: 

>>In 1 dimension, these identities mean 
>>that the Riemann tensor has NO nonzero components --- there ain't no 
>>intrinsic curvature in a 1-dimensional universe. 

>Interesting.   I see this,  and yet, if we embed a 1-dimensional universe
>in higher dimensional space,  we *can* have some kind of curvature,  which
>is evidently not captured by the Riemann tensor. 

>Oh,  maybe that is why you said "intrinisic" curvature... 

Yes, the curve in space only has extrinsic curvature.  I mentioned this
intrinsic/extrinsic curvaturre distinction before.  Poetically speaking,
the little 1-dimensionsal Linelanders can't do any experiments crawling
back forth along their line that enables them to detect any curvature.
Their universe is flat as far as they are concerned.  Similarly, in GR
we don't care about any sort of extrinsic curvature our universe might
have if embedded in some 47-dimensional spacetime, even if the folks in
47 dimensions are snickering at us as we crawl about.  

Here's a good example.  The metric on a flat sheet of paper is "flat" in
the technical sense: parallel translation around a little parallelogram
has no effect.  Now take the sheet of paper and roll it into a cylinder.
It has extrinsic curvature now but still no intrinsic curvature.  It's
metric is still the same (if you haven't stretched or ripped the paper),
so it's still flat.

Ponder that and ye shall become enlightened.

Article 97363 (85 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 1 Feb 1996 17:35:25 -0800
Organization: University of California, Riverside
Lines: 43
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:

>Anyway,  I persist, perversely, in thinking that even from the "follow your
>nose" point of view,  the geodesic (the path obtained by following your
>nose) is more primitive than following an *arbitrary* path and holding the
>javelin "without rotating it".  

Well, perhaps what your intuition is related to the fact that there are
lots of recipe for parallel translation that preserves lengths and
angles, and all of these have the same geodesics, even though they have
different "torsion".  Requiring the torsion to be zero picks out a
unique "best" recipe for parallel translation (given the metric), but
this torsion stuff is a bit subtler.

>Seriously,  sir,  could we have a short definition of torsion?  I promise
>not to argue. 

I will give a definition that depends on a choice of local coordinates.
You promised not to argue, so this is okay.  

Say we have coordinates x^a.  Take a little vector of size epsilon
pointing in the a direction, and a little vector of size epsilon
pointing in the b direction.  Parallel translate the vector pointing in
the a direction by an amount epsilon in the b direction.  Similarly,
parallel translate the vector pointing in the b direction by an amount
epsilon in the a direction.  (Draw the resulting two vectors.)  If the
tips touch, up to terms of epsilon^3, there's no torsion!  Otherwise
take the difference of the tips and divide by epsilon^2.  Taking the
limit as epsilon -> 0 we get the torsion t_{ab}.

(Since we are working in coordinates, we are allowed to subtract two
points by subtracting their coordinates, just like in the good old days.
We also can speak of the "size" of a vector pointing in the a or b
direction, meaning not its length (no metric is needed here) but how far
it goes in the given coordinates.  Ditto for the term "amount".  Of
course these tricks introduce a seemingly dangerous coordinate-
dependence, so I should really give you the coordinate-independent
definition of torsion to make you feel happier, but it's a bit more
technical.  Take my word for it that t_{ab} is actually a tensor which
could be defined in a coordinate-independent way.)

From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 1 Feb 1996 17:39:29 -0800
Organization: University of California, Riverside
Lines: 21
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Daryl McCullough) writ
es:

>Anyway, in Misner, Thorne, and
>Wheeler's _Gravitation_ (the big, fat, black book), they define parallel
>transport geometrically using the sorts of constructions one might
>use in high-school geometry, except that geodesics replace straight-edges.

That's nice; I didn't know this construction.

>[construction deleted]

>This construction parallel-transports vector AB to the point C. Of
>course, when I say "draw a line between points X and Y", I mean,
>construct a geodesic, so this construction requires geodesics. It
>also requires some notion of the distance along a geodesic (otherwise,
>we couldn't take half to get a midpoint).

I should add that we are supposed to think of A and B as being
"infinitesimally far apart", so that AB is really a tangent vector.

Article 97524 (27 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 2 Feb 1996 10:32:51 -0800
Organization: University of California, Riverside
Lines: 33
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:

>>When we contract the Riemann tensor to make the Ricci tensor, what 
>>information do we discard? 

I'll try in a while to say in intuitive terms what information we
*retain*.  As I said a while back, I've never really understood the
geometrical significance of the Ricci tensor as well as I wanted.  Matt
Wiener's post helps... but to be really happy I am going to have learn
Raychaudhuri's theorem.  This relates the Ricci tensor to the focussing
(or defocussing) of geodesics, and it's a key ingredient in Hawking and
Penrose's proof that black holes (or other singularities) must form in
certain circumstances.  Anyone who wants to beat me to it can check out
p. 218 in Wald.

>I don't know.  But going back to my crystal physics book to check on the
>stiffness tensor,  I see that symmetry reduces its 81 components to no more
>than 36 independent entries.  So I wonder how much information we do
>discard here in contracting the Riemann tensor? 

I don't know if you saw it yet, but in one of my posts I noted that the
Riemann tensor has 20 independent entries (in 4d spacetime), while the
Ricci satisfies

R_{ab} = R_{ba},

so it has 10.  So we lose 10 juicy facts about the curvature of spacetime
when we contract the Riemann to get the Ricci.

Article 97727 (109 more) in sci.physics:
From: Keith Ramsay
Subject: Re: general relativity tutorial
Date: 3 Feb 1996 01:49:25 GMT
Organization: Boston College
Lines: 98
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, 
[email-masked] (Edward Green) wrote:
|Maybe YOU would like to explain to me the difference between "covariant"
|and "contravariant" then sir.   :-)    Poor JB has invested enough time
|here, and I have a feeling that is what you are getting at. 

I'm not sure which of these means "tangent vector" and which means
"cotangent vector", but they mean the same thing in some order.

A tangent vector is what John Baez has been explaining. One way to
represent a tangent vector is by giving a (smooth) parametrized
curve through the point. Two such curves yield the same tangent 
vector if they are tangent to each other, and if increments in their
respective parameters move them along at the same rate at the point
in question.

In coordinates, one has a curve (x(s),y(s),z(s),t(s)), and its
tangent vector has coordinates (x'(0),y'(0),z'(0),t'(0)) at the
point (x(0),y(0),z(0),t(0)). Of course, we don't want to be stuck
with just one coordinate system, but if two curves yield the same
tangent vector in one coordinate system, they do in all.

To tell our Roman a tangent vector, we put a dagger in his hand, aim
it, and say, "That direction, but think of it as 5 stadia in magnitude."

A cotangent vector can be thought of as a gradient. I sometimes remind
my students that these tend to be in different units. A gradient is in
units *per* distance.

To tell our Roman a cotangent vector, spray a cloud of perfume 
near him, in such a way that if he moves in certain directions
the smell gets stronger. :-) Assuming, I suppose, that one has
an agreed-upon scale for intensity of smell.

One can present a cotangent vector by giving a function of which
it is the gradient at the point in question. In coordinates, the
gradient of f has coordinates (@f/@x,@f/@y,@f/@z,@f/@t). Again,
we need not get stuck on any one coordinate system; if two functions
have the same gradient in one coordinate system, they do in all.

They are often treated in elementary math courses as being essentially
the same type of thing ("vector"). Part of the reason is that if one 
has a *metric* (q.v.) one can identify the two. You can associate to
the cotangent vector the tangent vector which suggests moving in the
direction of fastest increase of the function, and whose length is
the rate of increase. This only makes sense, however, because we can
compare lengths in different directions. Taking one step to the south,
say, increases the smell more than one step in any other direction.
Without such a measure of distance as "steps", though, there's no
direct comparison between the rate at which the function increases
in different directions.

Also, tangent and cotangent vectors transform differently when you
change coordinates. I mentioned already a difference in how they 
thansform under a change of units. If you multiply the coordinates
of all the points by 10, then the coordinates of a tangent vector
also get multiplied by 10, but the coordinates of a cotangent vector
are reduced by a factor of 10: the amount by which the function 
increases per "unit" change in a coordinate is less, not greater.

If you like, a more mathematical example. Let u=x+y and v=y be new
coordinates for the x-y plane. The parametrized line x=t, y=0
defines a tangent vectorat the origin; call it a. The parametrized 
line x=0, y=t defines a tangent vector; call it b. In the original 
(x,y) coordinates these are (1,0) and (0,1). In the (u,v) coordinates
these lines become u=t,v=0 and u=t,v=t. Thus in the new coordinates
they are (1,0) and (1,1).

Now consider the cotangent vectors. The gradient of the function
x is often denoted by dx. The gradient of y is dy. They are
of course represented by coordinates (1,0) and (0,1). Now consider
how they are represented in (u,v) coordinates. We have to take
partial derivatives with respect to u and v. We carefully avoid
the pitfall of supposing that taking the partial derivative
with respect to v and with respect to y are the same, because
v=y. In the one case, one is leaving u the same, in the other x.
Since x=u-v and y=v, dx=du-dv and dy=dv. Hence in the new 
coordinates, dx=(1,-1) and dy=(0,1).

Thus the matrix for transforming tangent vectors is

(1 1)
(0 1)

but for cotangent vectors it is the inverse,

(1 -1)
(0  1).

The reason for being extra careful in relativity theory is that
the metric is part of what one is trying to figure out, and even
with a known metric, the correspondence between tangent and 
cotangent vectors requires calculation as a function of the 
metric. They call it "raising and lowering indices", because one
uses upper indices for one aspect and lower for the other.

Keith Ramsay

Article 97714 (29 more) in sci.physics:
From: Keith Ramsay
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 03:04:24 GMT
Organization: Boston College
Lines: 78
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, 
[email-masked] (Bronis Vidugiris) wrote:

Regarding parallel transport:
|1) a constant angle, i.e. a constant dot product, which requires
|a metric.
|
|Ummm - well this works for this example, but may have problems
|in higher dimensional spaces.  A constant dot product implies
|a constant angle in 2-space, but not in 3-space.  I think
|this may be where I'm going slightly ........ wrong :-(.
|
|2) a notion of "moving in the direction of the tangent vector".
|(There may be some problems with this notion, but if so I don't
|quite get it, though some of the stuff I've read indicates there
|may be some issues with this seemingly simple idea).

Well, it's possible to set out in the same direction with two
paths that start to diverge, and you need something besides
just a manifold and a tangent direction to tell you which one
is "straighter".

John has avoided defining geodesics independently of parallel
transport, but I think there may be some merit in stating how
one can do so. Geodesics are locally extremes of length. In 
ordinary geometry, paths which take the shortest route between
points which are close together on the path. Take a strong cord
and stretch it tight.... In general relativity, it is sometimes
a maximum of elapsed time. Feynman has a cute illustration in
a book of his. Suppose you want to arrive back where you are
now in one hour of local time, but with a maximum of time having 
elapsed for you. Note that going uphill takes you to a place
where, informally speaking, time goes faster. But moving fast
causes time to go "slower" (informally speaking). What is the
tradeoff between the two which leads to an optimum of wasted
time?

Geodesics in space-time are the *free-fall* paths of objects.
So the right thing to do is to shoot yourself out of a cannon
so that in free fall, you return to the same spot on the ground.

If you transport the "forward" direction on a geodesic along
the geodesic, it really ought to remain the "forward" direction.
If our Roman turns out to be a clutz, we make him follow a taut 
cord, and point his spear along it to keep it from wobbling.
The more difficult problem is to deal with carrying vectors
which are pointing off to one side.

One does also require that parallel transport preserve the
metric. So the spear shouldn't lengthen, say, as one goes.
And the angles between different vectors should remain the
same.

As you've pointed out, on a two-dimensional manifold, this
is enough. But in 3 or more, spears pointed to the side
can wobble around while keeping the same angle to the 
taut cord.

Let the Roman take not just a spear, but a set of three
which are at right angles. One is pointing along the cord;
the others remain at right angles to it and to each other.
So far, our instructions don't prevent the spears pointed 
perpendicular to the path from turning around in the plane 
perpendicular to the path. "Torsion", as it were.

Someone should correct me if I'm mistaken, but I believe that
this can be corrected if we require that the tips of the spears
take the shortest path compatible with our prior requirements.
Spinning them around makes them take a longer path. This is
less than absolutely precise, since in fact we want to consider
the spears infinitesimal arrows, and for the distance to be
an issue, they have to have some small finite length. But I
believe that if we make them very small, in the limit the way
to carry them which moves them the least will be to parallel
transport them as tangent vectors. And this is enough to show
how to transport any other tangent vector.

Keith Ramsay

Article 97745 (28 more) in sci.physics:
From: Keith Ramsay
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 02:36:08 GMT
Organization: Boston College
Lines: 100
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, [email-masked] wrote:
|I feel that the concept of 'metric' has been rather glossed
|over. I mean, it has a proper name that almost means
|something. Something to do with 'measurement'. Could you
|possibly expand a little on the above, please?

Yes, that's the derivation of the word.

The metric on space-time consists of the information describing 
lengths and times. (For those who have time for a book, Geroch's
"Relativity from A to B" gives what seems a good explanation of
the physical significance of the metric.)

In Minkowski space, one has a metric dt^2-dx^2-dy^2-dz^2. 
This takes two vectors (x1,y1,z1,t1) and (x2,y2,z2,t2) and gives
back the number, t1t2-x1x2-y1y2-z1z2. Here I'm setting the units
so that the speed c of light is 1. I'm also adopting the "+---"
convention. The "-+++" convention (change all the signs) is also
used.

Suppose you are in Minkowski space-time, and that your path can
be described by the parametrized curve (x(s),y(s),z(s),t(s)).
The elapsed time on your clock between s=0 and whenever you 
are at a given value of s is a function of s. Let's call it f(s).
f'(s) is the rate at which time is passing per unit of s (whatever
s may be).

From special relativity, one knows that in this model, the rate
of elapsed time for you, relative to t, is sqrt(1-v^2/c^2).
Since I'm setting c=1, this is just sqrt(1-v^2). To find the
rate of elapsed time w.r.t. s, multiply by dt/ds:
sqrt(1-v^2)dt/ds. Then of course v^2=(dx/dt)^2+(dy/dt)^2+(dz/dt)^2,
so one gets

   sqrt((dt/ds)^2 (1-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2))
  =sqrt((dt/ds)^2 - (dx/ds)^2 - (dy/ds)^2 - (dz/ds)^2)

using the chain rule from calculus. Now the square of this

   (dt/ds)^2 - (dx/ds)^2 - (dy/ds)^2 - (dz/ds)^2

is what you get by applying the metric, above, to the vector
(dx/ds,dy/ds,dz/ds,dt/ds), the tangent vector to your parametrization
of your path (world-line) through space-time.

This is directly analogous to the way that one can compute the
length of a parametrized curve (x(s),y(s),z(s)) in Euclidean 
space by integrating

   sqrt((dx/ds)^2+(dy/ds)^2+(dz/ds)^2)

over the portion of the curve in question. The metric in space-time
"measures" elapsed time, roughly the way the metric in Euclidean
space "measures" distance.

It also measures distances, although the interpretation seems a little
less direct. If you have a tiny rigid rod moving along a path in
space-time, the points in space-time occupied by its ends, which are
simultaneous in the locally inertial frame for the rod, are a distance
apart equal to the length of the rod. Here, you can compute distance
using the metric with the sign changed:

   sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2-(t1-t2)^2),

where (x1,y1,z1,t1) and (x2,y2,z2,t2) are the coordinates of the
points of space-time in question.

Now, all that was special relativity (!). In general relativity, one
deals with other space-times. The metric, however, will do the same
for you. Some astronaut has had an amazing career in which he's flown
close to event horizons of black holes and things like that. Once he's
retired and died of old age, it's up to you to write an obituary. One
thing you want to work out: how old was he, after all that? Sure he was
born in 2025 and died only in 2211, but what with all the time-dilation
he certainly wasn't 186 years old. So draw his world-line in space-time,
and parametrize it somehow. Then use the metric to determine the rate
of elapsed personal time for him, per unit of parameter. (It's sometimes
nice to make elapsed time itself the parameter.) Apply the metric tensor
to the tangent vector v at a point and v. (It's a 2-tensor, which takes 
two tangent vectors and gives back a number. Put the same vector in both 
arguments.) Take the square root of the result. That's the rate of elapsed 
time per unit of parameter.

Angles come out of the metric in a way similar to the way angles can
be figured out from distances in ordinary geometry. The metric is
directly analogous to the "dot product", v.w=|v||w|cos(angle).

The metric is assumed to be one which at any one point can be transformed
by a change of coordinates into a Minkowski metric. So there is always
an implicit "+---" (or "-+++") "signature". This makes the geometry used
in general relativity different from much of metric differential geometry,
where the signature is positive. One manifestation of this feature is that
in some tangent directions to space-time, the metric is 0. These are
"null" directions, and massless particles (such as light, presumedly)
travel along them. A common tactic is to employ some mathematical 
tool to convert the problem into one which involves a positive metric
(using "imaginary time").

Keith Ramsay

Article 97823 (12 more) in sci.physics:
From: john baez
Subject: Re: general relativity tutorial
Date: 3 Feb 1996 13:01:39 -0800
Organization: University of California, Riverside
Lines: 64
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward Gr
een) writes:
>[email-masked] (Daryl McCullough)' wrote: 

>>According to General 
>>Relativity, the clock on the mountain will run *faster* than the 
>>clock on the ground. General Relativity says that the clock with 
>>the lowest gravitational potential will run the slowest. (It isn't 
>>the gravitational field that is important, but the gravitational 
>>potential.  

>This assertion is very, very, very strange to me.  Maybe now would be a
>good time to hash this out a bit. 

>Surely the "gravitational potential" is something that is globally
>determined,  not locally,  while on the other hand the rate at which a
>clock runs is a local property of space.  How in the cosmos could a clock
>which saw the same local conditions "know" that it was in a higher
>gravitational potential,  and hence run faster? 

While everything Daryl McCullough writes is correct, Ed Green is right
to be very suspicious.  In general, there's no such thing as the
"gravitational potential" in GR.  This is a mathematical tool that crops
up in a very special situation: the spherically symmetric static
solution of GR.  (Maybe in some other situations I don't know about too,
but still very special ones.)  There is also in general not much sense
to saying that time runs slower or faster at one or another location.
You need to worry about whether you are making a coordinate-dependent
statement here, or an operational one; if it's an operational one, you
should make it operational by making it precise. 

Let's do that in this example.  There are lots of different ways, but
let's take a simple one.  Say you have two clocks next to each other in
your house, at spacetime point A.  Then you move one up to a mountain
for a while, and then move it back down and compare the two clocks at
spacetime point B.  You see the one you took up the mountain is ahead of
the other. 

Note that since I am only comparing clocks when they are right next to
each other, I don't need to worry about how the information was passed
from one to the other, and whether extra time lags were introduced in
the process.  

Note also that I don't need any coordinate system!  I just need two clocks.

In terms of the mathematics of GR, what's going on?  Well, we have two
paths from spacetime point A to spacetime point B, and the "length" (or
more precisely, "proper time") along one path is more than the other.
That's all. 

The gravitational potential can be a useful tool in *certain* problems,
but I prefer this other way of thinking of it, since it tells you
something relevant to quite general GR problems.  Also, it fits with the
idea that the *metric on spacetime* is what really matters: that's what
you'd use to compute the proper time along a curve.  Also, it seems
pretty simple and unmysterious.

Article 97828 (9 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 13:27:40 -0800
Organization: University of California, Riverside
Lines: 43
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Bron
is Vidugiris) writes:

>Ouch!

Sorry, I was being a bit sharp-tongued there.  I gotta mellow out a bit
here... 

>It's fine to prove that the stress energy tensor is a contraction
>of the Rienmann tensor, but this seems incomplete unless
>either it's possible to recover the full Rienmann tensor from the
>contraction, *or* if it's possible to calculate what we really want to
>know given only the contraction and not the full tensor.

It's definitely NOT true that you can recover the Riemann tensor from
the stress-energy tensor in 4 dimensions.  You can do it dimensions 1,
2, or 3.  This is why there are gravitational waves in 4 dimensions but
not in lower dimensions.  I sorta said this a while back.  To expand:

In 4 dimensions, the Riemann tensor R^a_{bcd} has 20 independent
components.  The Ricci tensor R_{ab} --- or for that matter the Einstein
tensor G_{ab} --- has 10.  So if you know the stress-energy T_{ab},
Einstein's equation 

G_{ab} = T_{ab} 

gives you the Einstein tensor.  But you can't get the Riemann tensor
back... even using whatever sneaky tricks you might attempt to dream up.
Indeed, there are lots of different solutions of Einstein's equation
with T_{ab} = 0 --- so-called "vacuum solutions", since there's no
energy or momentum anywhere.

There are lots of such solution with R^a_{bcd} nonzero, in addition to
the obvious one with R^a_{bcd} = 0, namely flat Minkowski spacetime.
One of them is the Schwarzschild solution.  I.e., in the solar system,
spacetime is curved even in the vacuum, due to the gravity of the sun.
Another sort is the gravitational wave solution, in which spacetime has
ripples of curvature moving through the vacuum.

So the business of actually solving Einstein's equation proceeds a bit
differently than you might have suspected.

Article 97841 (7 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 14:56:44 -0800
Organization: University of California, Riverside
Lines: 258
NNTP-Posting-Host: guitar.ucr.edu

Okay, on today's episode of GENERAL RELATIVITY TUTORIAL we are going to
step outside of the studio and see how a typical fan out there is doing 
at learning general relativity:

In article <[email-masked]> [email-masked] writes:
>[email-masked] (john baez) wrote:

>>1.  A TANGENT VECTOR at the point p of spacetime may be visualized as an
>>infinitesimal arrow with tail at the point p. 

>I think we've got this one, but could I just double check
>that it is nothing more than an 'infinitesimal vector
>attached to a point'. Any point and a vector pointing in
>*any* direction whatever within the 'spacetime'. Like any
>vector it has magnitude.

Yes!  Quite right!  You've definitely got that down.  

>>1.  A TENSOR of "rank (0,k)" at a point p of spacetime is a function that
>>takes as input a list of k tangent vectors at the point p and returns as
>>output a number.  The output must depend linearly on each input.   

>>A TENSOR of "rank (1,k)" at a point p of spacetime is a function that
>>takes as input a list of k tangent vectors at the point p and returns as
>>output a tangent vector at the point p.  The output must depend linearly
>>on each input.    

>I'll just consider then as a machine that does as specified
>for now. So there doesn't seem any intrinsic problem here at
>this level.

There really isn't anything more to them than what I said, so if you
feel fine, we're okay.  Of coure there are all sorts of sneaky things
you can do WITH tensors, and in a minute we'll start doing a few,
but that's different than the simplicity of the basic concept.

Also, I haven't yet said what a general tensor of rank (j,k) is.
Luckily we haven't been forced to deal with those too much just yet.

>>2.  The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
>>v,w and spits out a number g(v,w), which we think of as the "dot
>>product" or "inner product" of the vectors v and w.  This lets us
>>compute the length of any tangent vector, or the angle between two
>>tangent vectors.  Since we are talking about spacetime, the metric need
>>not satisfy g(v,v) > 0 for all nonzero v.  A vector v is SPACELIKE if 
>>g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

>I feel that the concept of 'metric' has been rather glossed
>over. I mean, it has a proper name that almost means
>something. Something to do with 'measurement'. Could you
>possibly expand a little on the above, please? For example,
>given the generality of the stuff you are doing, I would
>expect two scalars to be produced simultaneously, one
>magnitude and one angle.

You can get magnitudes and angles from the dot product.

The "metric" is no more than the generalization from space to spacetime
of the dot product you know and love.  It "measures" lengths and angles.
So in the following little blurb, let me not use the scary symbol

g(v,w).

Instead let me use the hopefully more familiar symbol

v.w

(where unfortunately the dot has fallen on the floor instead of hovering
directly between the v and the w).  We use the metric to measure
the length of a tangent vector v as follows:

|v| = sqrt(v.v)

and we use it to measure the angle between vectors v and w as follows:

arccos(v.w/|v| |w|)

Now actually, if we are in spacetime rather than space, the dot product
v.v can be negative.  (See the above stuff I wrote earlier.)  So we have
to be a bit flexible in our use of the terms "length" and "angles",
unless the vectors we're dealing with are spacelike.  But it's no big
deal to those whose brain arteries aren't completely calcified. 

The idea is that if you can measure the lengths and angles between
tangent vectors, you know EVERYTHING about the geometry of spacetime.
That may seem shocking, but really, that's what the point of all this
Riemanninan geometry is.... the metric gives you parallel translation,
curvature, the whole works!

>>3.  PARALLEL TRANSLATION is an operation which, given a curve from p to
>>q and a tangent vector v at p, spits out a tangent vector v' at q. 

>I think I have this grasped.

Good!

>>4.  The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point
>>of spacetime.  

>OK, provisionally.

Okay, good!

>>5.  Introducing COORDINATES.  Now say we choose coordinates on some
>>patch of spacetime near the point p.  Call these coordinates x^a (where
>>a = 0,1,2,3).  Then given any tangent vector v at p, we may speak of its
>>components v^a in this basis.  The inner product g(v,w) of two tangent
>>vectors is given by
>>
>>        g(v,w) = g_{ab} v^a v^b
>
>Terminology: Whilst there is no particular reason why
>'metric', 'inner product' and 'g(v,w)' should be synonymous,
>they do seem to be pretty similar. I wonder if you could
>elucidate on these a little. 

They are the exact same thing.  

>I also note that 'b' has
>appeared unexplained, just a terminology thing I expect, but
>where did it come from? 

It is the next letter after 'a'.  It was invented by ancient
Phoenicians, I think.    

>(NB co-ordinates were defined as 'a' where 'a'=0,1,2,3)

Oh, I see!  Sorry.  We feel free to use, not just a, but all letters,
particularly those near the beginning of the alphabet, to stand for any
of the numbers 0,1,2,3.  So when I write g_{ab}, I am quickly referring
to

g_{00}   g_{01}   g_{02}   g_{03}
g_{10}   g_{11}   g_{12}   g_{13}
g_{20}   g_{21}   g_{22}   g_{23}
g_{30}   g_{31}   g_{32}   g_{33}

You can see why I prefer to say g_{ab}.  Indeed, if we were in 26
dimensions, the way some nuts believe, the indices a,b,c, etc. would go
from 0 to 25.  It's much better just to say g_{ab}.

>>for some matrix of numbers g_{ab}, where as usual we 

>'as usual we'? What a compliment. Unfortunately ......

>>sum over the 
>>repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION.

Well, you'll soon get used to it.  The way you tell the difference
between a physicist and a mathematician these days, by the way, is that
physicists instinctively know to sum over repeated indices.  So I was
just flaunting the fact that, while a mathematician, I can play the 
physicist.  You can too.  

>>6.  The EINSTEIN TENSOR.  The matrix g_{ab} is invertible
>
>Another g(,). Presumably another tensor of rank (0,2)

It's not another one, it's the same bloody one every time.  The metric.  

>>7.  The STRESS-ENERGY TENSOR.  The stress-energy is what appears on the
>>right side of Einstein's equation.  It is a tensor of rank (0,2), and it
>>defined as follows: given any two tangent vectors u and v at a point p,
>>the number T(u,v) says how much momentum-in-the-u-direction is flowing
>>through the point p in the v direction.
>
>At last, something that even *sounds* like a physical
>system. Hmmm. Momentum, time as a dimension. OhMyGod shades
>of VV, yuk! What does momentum in the time dimension mean?
>Oh, do a vaguely remember Baez calling it energy. No I am
>probably confused. This needs sorting out. Heeeeeellllpppp!

Oh-oh, our ratings our dropping!!  What's the problem, dear viewer?
Energy is momentum in the time direction.  Consider a rock "just sitting
there".  It has lots of velocity in the time direction --- one second
per second --- hence lots of momentum in the time direction.  That's
what we call it's "rest energy".  You know, E = mc^2 and all that stuff
they taught you in grade school....

>>The top row of this 4x4 matrix, keeps track of the density of energy ---
>>that's T_{00} --- and the density of momentum in the x,y, and z
>>directions --- those are T_{01}, T_{02}, and T_{03} respectively.  This
>>should make sense if you remember that "density" is the same as "flow in
>>the time direction" and "energy" is the same as "momentum in the time
>>direction".  The other components of the stress-energy tensor keep track
>>of the flow of energy and momentum in various spatial directions.

>Oh. Is it just a generalised expression of the conservation
>of momentum in spacetime?

We haven't said anything about *conservation* of energy-momentum here.
But perhaps you are peeking at the next chapter, where I explain that
the law

D^a T_{ab} = 0

expresses local conservation of energy and momentum, and that it's an
automatic consequence of Einstein's equation

T_{ab} = G_{ab}.

>Hmmm, quite a philosophical point
>worthy of some deep consideration. I guess you have all
>considered this, I wonder what sort of conclusions/observations you came to?

Well, it all boils down to this: local energy-momentum conservation is
really just an expression of the fact that the laws of physics are the
same in any coordinate system.   Cool, huh?  It's a special case of
"Noether's theorem" relating symmetries and conservation laws.  But
really, this is quite a digression.

>>8.  EINSTEIN'S EQUATION: This is what general relativity is based on.
>>It says that
>>
>>          G = T
>>
>>or if you like coordinates and more standard units,
>>
>>          G_{ab} = 8 pi k T_{ab}
>>
>>where k is Newton's gravitational constant.  So it says how the flow of
>>energy and momentum through a given point of spacetime affect the
>>curvature of spacetime there.

>[Cryptic allusions deleted.]

>So, this says [...] that the flow of
>energy and momentum through a given point of spacetime
>affect the curvature of spacetime there. 

Yes.  Precisely.

>(That can't be wrong). 

What's more, it's precisely right!

>Hang on a tick. Does this refer to one tiny point in
>spacetime?

In fact, each and every tiny little point!

>Surely not, what if that point is in a vacuum?.

Then T_{ab} = 0, of course, 'cause there's no energy and momentum there.
What's the big deal????  It just means G_{ab} = 0.

>BANG! THUD!

Hmm, well, our viewer has died.  Or possibly just collapsed in a faint?
Perhaps he needs a little review course in index juggling.  It seems
that he started losing it right around when we formed the Einstein
tensor by contracting some indices on the Riemann curvature tensor.
Or maybe a more geometric explanation of the Ricci and Einstein tensors
would help.

We'll be back next week!

Article 97842 (6 more) in sci.physics:
From: Marco de Innocentis <[email-masked]>
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 10:53:20 GMT
Organization: The Mathematical Institute, Oxford
Lines: 61
NNTP-Posting-Host: yukon.maths.ox.ac.uk
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: 7bit
X-Mailer: Mozilla 1.1N (X11; I; SunOS 4.1.3_U1 sun4c)
X-URL: news:[email-masked]

[email-masked] (john baez) wrote:
>In article <[email-masked]> [email-masked] (Edward 
Green) writes:
>
>>>When we contract the Riemann tensor to make the Ricci tensor, what 
>>>information do we discard? 
>
>I'll try in a while to say in intuitive terms what information we
>*retain*.  As I said a while back, I've never really understood the
>geometrical significance of the Ricci tensor as well as I wanted.  Matt
>Wiener's post helps... but to be really happy I am going to have learn
>Raychaudhuri's theorem.  This relates the Ricci tensor to the focussing
>(or defocussing) of geodesics, and it's a key ingredient in Hawking and
>Penrose's proof that black holes (or other singularities) must form in
>certain circumstances.  Anyone who wants to beat me to it can check out
>p. 218 in Wald.
>

I'm also not 100% sure about this myself. If you've read Penrose's 
`The Emperor's New Mind' then you've probably seen that he describes
Ricci as that part of Riemann associated with uniform compression and
Weyl as the part associated with tidal forces. However, as he also points
out himself in a small note at the end of the chapter, this is not the whole 
story and is actually one of those bits of oversimplification which you
have to do when making GR 'understandable' to everyone (Weyl is the bit of
Riemann which vanishes when you contract it down to Ricci).
Of course you also have tidal forces in a 2,1 spacetime, and Weyl doesnt
play any role there. My understanding (after having discussed this with 
my supervisor briefly a few weeks ago) is that Weyl is only associated with 
some tidal forces, while Ricci is associated both with compression and 
(other) tidal forces. To see which kinds of tidal forces are due to Weyl 
and which ones to Ricci, you have to compare 2,1 and 3,1 space-times and
see which kinds of tidal effects cannot happen in 2,1 - they are due to
Weyl only. For example when you have a 'beam' of null geodesics with circular
cross section which gets deformed into an ellipse - this is due to Weyl, 
because it can't happen in a 2D space.

>>I don't know.  But going back to my crystal physics book to check on the
>>stiffness tensor,  I see that symmetry reduces its 81 components to no more
>>than 36 independent entries.  So I wonder how much information we do
>>discard here in contracting the Riemann tensor? 
>
>I don't know if you saw it yet, but in one of my posts I noted that the
>Riemann tensor has 20 independent entries (in 4d spacetime), while the
>Ricci satisfies
>
>R_{ab} = R_{ba},
>
>so it has 10.  So we lose 10 juicy facts about the curvature of spacetime
>when we contract the Riemann to get the Ricci.

Article 97856 (5 more) in sci.physics:
From: Emory F. Bunn
Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 3 Feb 1996 23:16:22 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 39
Distribution: world
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Marco de Innocentis  <[email-masked]> wrote:
>Of course you also have tidal forces in a 2,1 spacetime, and Weyl doesnt
>play any role there. 

I'm not sure I understand you here.  Are you saying that you can have
tidal forces in an *empty* region of 2+1-dimensional spacetime?  If
so, I'm enormously surprised.  Doesn't the Einstein equation say that
an empty patch of 2+1-dimensional spacetime is completely flat?

Maybe you mean that you can have tidal forces in a region of
2+1-dimensional spacetime that's filled with matter.  In that case,
I'm still not sure what you're getting at.  Can you give a precise
definition of a "tidal force"?  It seems like it needn't mean anything
more than "geodesic deviation", which pretty much just means
"curvature."  So then you'd be saying that wherever there's matter,
there's curvature.  That's true, but I'm missing the part where
insight into the nature of the Ricci and Weyl tensors takes place.

>My understanding (after having discussed this with 
>my supervisor briefly a few weeks ago) is that Weyl is only associated with 
>some tidal forces, while Ricci is associated both with compression and 
>(other) tidal forces. 

I think it would help me a lot if you could explain precisely why 
compression doesn't count as a tidal force.

>To see which kinds of tidal forces are due to Weyl 
>and which ones to Ricci, you have to compare 2,1 and 3,1 space-times and
>see which kinds of tidal effects cannot happen in 2,1 - they are due to
>Weyl only. For example when you have a 'beam' of null geodesics with circular
>cross section which gets deformed into an ellipse - this is due to Weyl, 
>because it can't happen in a 2D space.

This is definitely true, and it's why gravitational radiation doesn't
exist in fewer than 3+1 dimensions.  Circular beams get deformed into
ellipses when a gravitational wave passes by.

-Ted

Article 97865 (4 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 16:40:27 -0800
Organization: University of California, Riverside
Lines: 26
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Keit
h Ramsay) writes:

>John has avoided defining geodesics independently of parallel
>transport, but I think there may be some merit in stating how
>one can do so. 

Very much merit indeed.  Thanks for doing it.

To summarize what Keith said: in spacetime, a timelike geodesic is the
path an object moves along in free fall.  It's "timelike" because it
covers more ground in time than it does in space... since it's going
slower than light.  A geodesic of this sort is locally the *longest*
curve from one point to another, i.e., the one on which your watch makes
the most ticks.  Why?  Wiggling hither and thither creates time
dilation, so your watch would tick less than if it followed a geodesic.
Remember the "twin paradox" and all that. 

In space, a geodesic is the path a string takes when you pull it taut.
It's spacelike, of course, and it's locally the *shortest* curve from
one point to another.

In both cases there is an "extremal principle" involved.  Particles in
free fall travel the path that takes the most proper time because they are
trying to minimize the "action".  A stretched string follows the
shortest path because it's trying to minimize the "energy".

Article 98079 (46 more) in sci.physics:
From: Keith Ramsay
Subject: Re: General relativity tutorial
Date: 5 Feb 1996 00:33:02 GMT
Organization: Boston College
Lines: 101
Distribution: world
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, 
[email-masked] wrote:
[Baez:]
|>So there is trouble seeing how to reconstruct the concept of
|>geodesics from the concept of parallel transport.  And indeed, now that
|>I think about it, this is obvious; two different connections can have
|>the same geodesics, and to pick out the "good" one, the one to use in
|>Einstein's equation, we need to pick the one with no "torsion".  

He meant, of course, "trouble seeing how to reconstruct the concept
of parallel transport from the concept of geodesics". Viz, you cannot.
Geodesics have some relationship with parallel transport, but it does
not give enough information to pin down parallel transport.

|Woah, woah. Slooooow down.
|Clarification please y'er lordship, sir.
|
|1) Parallel transport seems more 'primitive' than 'geodesics'. Correct?

It depends upon how you think about it. Parallel transport is a more
"comprehensive" set of data. It's enough to tell you how to make
geodesics. Not vice-versa (unfortunately?).

|Well, I mean in the sense that you seem to imply that there is only one
|parallelly transported path (no direction changes) from a to b, but
|there may be more than one geodesic from a to b. 

No, it's not that. Parallel tranport is the general process of
what happens when you carry a vector around on some path, without
"turning" it as you go. Any sort of path, carrying any vector you
feel like starting out with.

A geodesic is a particular kind of path-- the kind with "no direction 
changes". 

The two ideas are related: You can tell the yellow brick road is a 
geodesic by carrying a vector pointing  "forward" along it, and seeing 
that as you ease it down the road, "parallelly", it keeps pointing along 
the yellow brick road.

It is true, by the way, that there is sometimes more than one geodesic
between a and b. For instance, returning to our surface-of-the-Earth
example, if the Earth were simply a sphere, then antipodal points
would be the ones which are joined by more than one geodesic. Any
geodesic from the north pole goes to the south pole. If two distinct
points on a sphere are not antipodal, then there is just one "great
circle" (i.e., geodesic) route between them.

|Hmmmm, yes, I could go
|with that even in curved 3D space. However I am probably wrong (and
|getting used to it).

Good. :-)

|2) Baez:
|>two different connections can have
|>the same geodesics, and to pick out the "good" one, the one to use in
|>Einstein's equation, we need to pick the one with no "torsion".
|
|Uh-oh. Can we forget about this and use parallelly transported, or do we
|have to figure out the concept 'geodesic' in all it's ramifications?

You can forget about the concept of "connection" and just think about
"parallel transport" which is one specific kind of connection.

If you are happy with the idea of parallel transport, then you need
not think about geodesics. (Geodesics are nice to know, though; they 
tell you what space-time paths correspond to the motions of objects in 
free-fall.)

| Lots of new words like 'torsion', 

It's purely to distinguish parallel transport (which you want to think 
about) from certain other sorts of things one could do with vectors as 
you carry them around (a.k.a. other connections).

In our highschool marching band, on parades we would have people 
twirling mock "rifles". They would be held perpendicular to the
way they were facing, but twirled around. This is *not* what we
have in mind for parallel transport. Such a way of carrying a
vector about involves torsion ("twisting"), even though they may
be keeping the angle between their rifle and the path they are
taking constant. In the plane, there isn't this problem, because
there is no way to turn a vector in place while keeping it at a
fixed angle to another one.

|'right one to use in Einstein's equation'. 

I hope none of these words is actually new to you. :-)

|Nobody's even mentioned (I think: he says quickly) where
|geodesics even come into Einstein's equation yet. I expect it's
|'obvious'. Ah, well.

Geodesics don't. It's the Ricci tensor R_ij, which can be computed
from the curvature tensor, which can be worked out by parallel
transport of vectors, which appears in Einstein's equation. Look 
at the "course outline" again.

Keith Ramsay

Article 97992 (44 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 4 Feb 1996 20:17:03 -0800
Organization: University of California, Riverside
Lines: 174
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:

>briiiiiing... 

>There's the bell...  gotta go! 

Okay.  I will now try to give a nice geometrical/physical interpretation
of the Ricci tensor.  It's possible that we'll need a little math here
and there, but I will try to make it as simple as possible.

First recall the key notions of parallel translation and Riemann
curvature.  Given a spacetime with a metric on it, we know how to
"parallel translate" a tangent vector along a curve.  Intuitively, this
means to drag it along while at every step of the way rotating and
stretching it as little as possible.  We say a spacetime is "curved" if,
after parallel translating a vector around a loop, it can come back as a
rotated version of its original self!   

Also remember what happens when we parallel translate a tangent vector
around a little parallelogram.  It will rotate by only a little bit...
approximately proportional to the area of the parallelogram, but 
also depending on the original direction the vector was pointing in, and
on the parallelogram we use.   If epsilon is a small number,
epsilon u  and  epsilon v  are tangent vectors corresponding to the
sides of a little parallelogram based at the point P,  w  is a tangent
vector at P, and  w'  is the result of parallel translating  w  around
the parallelogram, we have 

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

where R(u,v,w) is some tangent vector at P.  We call this gadget R the
"Riemann curvature".  It eats 3 vectors and spits out 1.

Now let's talk about geodesics.  A geodesic is the next best thing to a
straight line in a curved spacetime!  More precisely, any curve in
spacetime has a tangent vector at each point, so we can ask: is that
tangent vector being parallel translated as we move along the curve?  
If the answer is yes, then the curve is a geodesic.  In other words, the
geodesic is doing its best not to speed up, slow down, or take turns. 

Now physically, the importance of a geodesic is that any test particle
follows a geodesic in spacetime when there are no forces acting on it
besides gravity.  What's a "test particle"?  It's any particle small
enough that we feel okay ignoring its effect on the curvature of
spacetime.  For most practical purposes, an apple is a fine test
particle.  So what we are saying is that an apple in "free fall" follows
a geodesic.   It is doing its best to keep going the same way.  If it
seems to trac out a rather curious path in the process (a parabola,
say), well, that's just because spacetime is curved.  

Rather than trying to calculate out how this works in an example, I want
to use these ideas to explain the Ricci tensor.  

So, suppose an astronaut taking a space walk accidentally spills a can
of ground coffee.

Consider one coffee ground.  Say that a given moment it's at the point P
of spacetime, and its velocity vector is the tangent vector v.  Note:
since we are doing relativity, its velocity is defined to be the tangent
vector to its path in *spacetime*, so if we used coordinates v would
have 4 components, not 3.  

The path the coffee ground traces out in spacetime is called its
"worldline".  Let's draw a little bit of its worldline near P:

   |
  v^
   |
   P
   |
   |
   |
    
The vector v is an arrow with tail P, pointing straight up.   I've tried
to draw it in, using crappy ASCII graphics.

Now imagine a bunch of comoving coffee grounds right near our original
one.  What does this mean?  Well, it means that for any tangent vector w
at P which is orthogonal to v, if we follow a geodesic along w for a
certain while, we find ourselves at a point Q where there's another coffee
ground.  Let me draw the worldline of this other coffee ground.

   |       |
  v^       ^v'
   | w     |
   P->-----Q
   |       |
   |       |
   |       |

I've drawn w so you can see how it is orthogonal to the worldline of our
first coffee ground.  The horizontal path is a geodesic from P to Q,
which has tangent vector w at Q.  I have also drawn the worldline of the
coffee ground which goes through the point Q of spacetime, and I've also
drawn the velocity vector v' of this other coffee ground.

What does it mean to say the coffee grounds are comoving?  It means
simply that if we take v and parallel translate it over to Q along the
horizontal path, we get v'.  

This may seem like a lot of work to say that two coffee grounds are
moving in the same direction at the same speed, but when spacetime is
curved we gotta be very careful.  Note that everything I've done is
based on parallel translation!  (I defined geodesics using parallel
translation.)  

Now consider, not just two coffee grounds, but a whole swarm of comoving
coffee grounds near P.  If spacetime were flat, these coffee grounds
would *stay* comoving as time passed.  But if there is a gravitational
field around (and there is, even in space), spacetime is not flat.  So
what happens?

Well, basically the coffee grounds will tend to be deflected, relative
to one another.  It's not hard to figure out exactly how much they will
be deflected!  We just use the definition of the Riemann curvature!  We
get an equation called the "geodesic deviation equation".

But let me not do that just yet.  Instead, let me say what the Ricci
tensor has to do with all this.  Then, when we use the "geodesic
deviation equation" to work out the deflection of the coffee grounds
using the Riemann curvature, we will see what this has to do with the
usual definition of the Ricci tensor in terms of the Riemann curvature.

Imagine a bunch of coffee grounds near the coffee ground that went
through the point P.  Consider, for example, all the coffee grounds that
were within a given distance at time zero (in the local rest frame of the
coffee ground that went through P).  A little round ball of coffee
grounds in free fall through outer space!  As time passes this ball will
change shape and size depending on how the paths of the coffee gournds
are deflected by the spacetime curvature.  Since everything in the
universe is linear to first order, we can imagine shrinking or expanding,
and also getting deformed to an ellipsoid.  There is a lot of
information about spacetime curvature encoded in the rate at which this
ball changes shape and size.  But let's only keep track of the rate of
change of its volume!  This rate is basically the Ricci tensor.  

More precisely, the rate of change of volume of this little ball is
equal to 

r(v,v)

where r is a rank (0,2) tensor called the Ricci tensor, and v is as
above.  (In general r eats two vectors and spits out a number.  Here it
is eating two copies of v.)

This is actually a complete description of the Ricci tensor.  It might
not seem like it.  After all, if we know the Ricci tensor we know
r(u,v) for any tangent vectors u and v, not just things like r(v,v).
However, it turns out that

r(u,v) = r(v,u).  

Thus we can use a trick called "polarization" to figure out
r(u,v) if we just know r(v,v) for *all* vectors v.  The point is that r
is linear in each argument so:

r(u+v,u+v) = r(u,u+v) + r(v,u+v)
           = r(u,u) + r(u,v) + r(v,u) + r(v,v)
           = r(u,u) + r(v,v) + 2r(u,v)

so:

r(u,v) = (1/2)[r(u+v,u+v) - r(u,u) - r(v,v)]

The right hand side just has things like r(v,v) in it.

Well, I should explain the geodesic deviation equation a bit, so that
we'll see where the Riemann tensor fits into all this.  But I'm getting
tired out.

Article 98027 (38 more) in sci.physics:
From: Emory F. Bunn
Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 4 Feb 1996 20:09:54 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 56
Distribution: world
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>2) We throw out the Weyl tensor, but (shock horror, this must be deep)
>even the tyros don't seem too sure what the Weyl tensor represents.

I've never been too clear on exactly what information is contained
in the Weyl tensor; I'm hoping I'll find out when Baez et al. get
up to that point.

I can say one thing that may help, though.  It is true that the
Einstein equation only contains ten pieces of information, although
you need 20 to specify the curvature tensor.  So the Einstein equation
doesn't let you reconstruct the complete curvature tensor.  That
sounds disturbing at first, but it's really OK.  

As with many things in general relativity, it can help to state the
corresponding fact about electricity and magnetism.  If you know the
charge and current distributions everywhere in space, you might think
that that would let you figure out the electric and magnetic fields
everywhere.  But it doesn't.  There's extra information in the fields
beyond just what the sources of the fields can tell you.  After all,
you could have an electromagnetic wave passing by.  It needn't have
any source, but it still alters the fields.

So in electromagnetism, knowing all about the sources isn't enough to
specify the fields.  In general relativity, knowing all about the
sources (the stress-energy tensor T) isn't enough to tell you all
about the curvature.  In both cases, you can supplement the source
information with some extra initial conditions to get a unique
solution.  (For example, in electromagnetism you can specify that no
electromagnetic waves are zooming in from infinity.  That's enough to
give you a unique solution to the fields given the sources.  For
general relativity, you can perform similar feats, although it's
technically trickier.)

Anyway, I hope that makes it a little bit clearer why people say that
the Weyl part of the curvature has to do with gravitational radiation:
the Weyl tensor carries information about the kind of curvature that's
independent of the source distribution, sort of like electromagnetic
waves are fields that propagate independently of whatever sources are
around.

>3) Misreading what messrs Ted & Marco discuss, suggests that at least
>some of the Weyl stuff may describe non-physical thingies (2+1D
>spacetime is completely flat so the Weyl bit describing how it curves
>would be redundant perhaps). But exactly what the Weyl tensor *does*
>represent is not trivial.

Let me correct one thing: *empty* 2+1-dimensional spacetime is flat.
There certainly are curved 2+1-dimensional spacetimes, but they
all have matter in them.  Specifically, in 2+1 dimensions, the
Einstein equation implies that the curvature is zero wherever T is
zero.  (That's because in 2+1 dimensions the Ricci and Riemann
tensors contain the same amount of information.)

-Ted

Article 98021 (38 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Sun, 4 Feb 1996 09:37:51 +0000
Organization: Oz
Lines: 56
Distribution: world
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
MIME-Version: 1.0
X-Newsreader: Turnpike Version 1.11 <V67Vak$Q$lyzk9NR0hujdE7wCg>

Gosh, this is getting involved. Can't we just stay in 4D spacetime for a
while? No? Ok, you're the boss.

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>(Keith Ramsay) writes:
>
>>John has avoided defining geodesics independently of parallel
>>transport, but I think there may be some merit in stating how
>>one can do so. 
>
>Very much merit indeed.  Thanks for doing it.
>
>To summarize what Keith said: in spacetime, a timelike geodesic is the
>path an object moves along in free fall.  It's "timelike" because it
>covers more ground in time than it does in space... since it's going
>slower than light.  A geodesic of this sort is locally the *longest*
>curve from one point to another, i.e., the one on which your watch makes
>the most ticks.  Why?  Wiggling hither and thither creates time
>dilation, so your watch would tick less than if it followed a geodesic.
>Remember the "twin paradox" and all that. 

OK, spacetime (4D). A free-fall geodesic is (locally: obviously not
flatily, so what does it mean - parallelly transportedly?) the longest
path in spacetime. At last, something concrete to hang on to. Yup, I
follow that. One also wonders what a thingy that wiggled so much that it
got dilated to zero might look like, hmmm one is tempted to say 'light'
as it does a LOT of wiggling. Presumably this thingy would cover more
time in space than it does in time and so would be described by analogy
as 'spacelike'. Hmmm. I have a bad feeling about this.

>In space, a geodesic is the path a string takes when you pull it taut.
>It's spacelike, of course, and it's locally the *shortest* curve from
>one point to another.

OK, have I got it right. In space (3D) a taut string is the shortest
path in space. OK, not unexpected perhaps. In spacetime though, one
might dare to wonder because the string will be curved and the path
would depend on how fast you went down the string. Baez has thrashed it
in to me to treat these sorts of things with great caution.

>In both cases there is an "extremal principle" involved.  Particles in
>free fall travel the path that takes the most proper time because they are
>trying to minimize the "action".  A stretched string follows the
>shortest path because it's trying to minimize the "energy".

Hang on. Free-fall expressed above in 4D, taut string in 3D. I follow
the idea of course, but we keep being told to stay in 4D.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98014 (37 more) in sci.physics:
From: Matthew P Wiener
Subject: Re: Riemann tensor (was: Re: General relativity tutorial)
Date: 4 Feb 1996 14:24:49 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 47
NNTP-Posting-Host: sagi.wistar.upenn.edu
In-reply-to: [email-masked] (Toby Bartels)

In article <[email-masked]>, toby@ugcs (Toby Bartels) writes:
>Matthew P Wiener ([email-masked]) wrote:

>>I'd have to look this one up to get it right--ask me if interested and
>>I will give you a reference.  To give an elementary example, if all you
>>know about an inner product <x,y> is the set of <w,w> values, then as
>><x+c.y,x+c.y> = <x,x>+2c<x,y>+c^2<y,y>, you can solve for <x,y> (and
>>prove the Cauchy-Schwartz inequality to boot).  Something similar, but
>>algebraically more messy and less ASCIIble, applies here.

>Notice that this only works because we know <y,x> = <x,y>, a symmetry.
>This symmetry of the inner product is reflected geometrically
>in that there's a definition that only involves the lengths <w,w>.
>That's just what Matt's talking about below, for the Riemann tensor.

>>Remember, you don't need a full 4x4x4x4 set of values.  The above gives
>>you 8x6 numbers, so it's not a priori surprising.

(Actually, that should be 10x6, not 8x6.  So one only needs a further
factor of three redundancy to explain.)

>>                                                   In fact, one hopes
>>for some sort of geometric way to understand the Riemannian redundancies,
>>that is, a definition that generates far fewer numbers to begin with.

Well, I looked it up anyway.  I knew I had seen it within the past year,
and was wondering where.

It's in Gallot, Hulin and LaFontaine RIEMANNIAN GEOMETRY.

It turns out that (up to a factor, and maybe some notational permutation)
that R(u,v,x,y) is the mixed partial derivative with respect to S and T
(once each), evaluated at S=T=0, of

                R(u+S.x,v+T.y,","") - R(u+T.y,v+S.x,","")

(where R(a,b,","") stands for R(a,b,a,b), which can be defined directly
via the sectional curvature of the surface tangent to directions a,b.)

Similar to Toby's comment about how the symmetry of <,> allows you to
polarize inner products, explicitly halving the apparent number of
parameters in the high level description of the object, so too does the
symmetry in Riemann (in this case, including the Bianchi identities)
allow one to concoct a more efficient high level definition, and then
polarize.
-- 
-Matthew P Wiener ([email-masked])

Article 98092 (35 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 5 Feb 1996 10:19:59 -0800
Organization: University of California, Riverside
Lines: 48
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:
>In article <[email-masked]>, john baez
><[email-masked]> writes

>>The "metric" is no more than the generalization from space to spacetime
>>of the dot product you know and love.  It "measures" lengths and angles.
>>So in the following little blurb, let me not use the scary symbol

>So do I have it right. The metric is merely (!) a
>prescription or mechanism to extract 'lengths' and 'angles'
>from a pair of tangent vectors. We could chose all sorts of
>mechanisms, most of which would be 'unphysical', but for GR
>we use the g one as in g(v,w).

Yes.  For each pair of tangent vectors v,w based at the same point, 
g(v,w) is a kind of "dot product" of them.  If we were in boring flat
old Minkowski space... the land of *special* relativity... the 
vectors v and w would look like this:

v = (t,x,y,z)
w = (t',x',y',z')

and the metric would be given by:

g(v,w) = - tt' + xx' + yy' + zz'

Just the usual dot product you learn in college, with a minus sign
thrown in front of the time part to make time be different from space.
(Or else we could throw minus signs in front of all the space parts.
General relativists like to do it the way I do above, so that the
geometry of space stays as much like it used to be as possible.
Particle physicists like to do it the other way.)

>Presumably there are other metrics (not g(v,w)) that apply to
>other strange and wonderful constructs in mathematics that we
>need not consider here. Probably luckily.

Well, in the world of general relativity, "metric" means no more and no
less than a symmetric nondegenerate tensor of rank (0,2), or if you
prefer, a dot product thingie.  

In other realms, "metric" means other things, and then to be specific we
say that the metric in GR is a "Lorentzian metric" if it has that minus
sign in front of the time part, or a "Riemannian metric" if it's just
for space.  But let's not worry about those other realms just now, eh?

Article 98096 (34 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 5 Feb 1996 11:00:17 -0800
Organization: University of California, Riverside
Lines: 52
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Ke
ith Ramsay) writes:
>In article <[email-masked]>,
> [email-masked] wrote:
>|John Baez wrote:
>|>In space, a geodesic is the path a string takes when you pull it taut.
>|>It's spacelike, of course, and it's locally the *shortest* curve from
>|>one point to another.

>|OK, have I got it right. In space (3D) a taut string is the shortest
>|path in space. OK, not unexpected perhaps. In spacetime though, one
>|might dare to wonder because the string will be curved and the path
>|would depend on how fast you went down the string. Baez has thrashed it
>|in to me to treat these sorts of things with great caution.

>The string analogy starts getting less workable when the string need be
>stretched across space and time.

Yes, note that I only discussed the stretched string in SPACE, not in
SPACETIME.  By space I mean a Riemannian manifold: no minus signs around
in the metric.  We can use this without any great ambiguity to study
spacetimes where there is a chosen time coordinate, and the shape of
space does not change with time, and everything is static --- if we know
what we are doing.  (There are some technical details I'm sweeping under
the rug here, so watch out and don't trip over the bumps.)  But as soon
as we get into general curved SPACETIMES, the notion of a "string
stretched taut" becomes hopelessly ambiguous.

It's best to treat the "string stretched taut" as a okay way of thinking
about geodesics in 3d curved SPACE, but as a mere warmup for thinking
about geodesics in 4d curved SPACETIME.  There is no easy sense in which
a "string stretched taut is a geodesic in 4d spacetime"... so don't let
that thought take root.

>|>In both cases there is an "extremal principle" involved.  Particles in
>|>free fall travel the path that takes the most proper time because they are
>|>trying to minimize the "action".  A stretched string follows the
>|>shortest path because it's trying to minimize the "energy".

>|Hang on. Free-fall expressed above in 4D, taut string in 3D. I follow
>|the idea of course, but we keep being told to stay in 4D.

Come come, unless you are already perfectly comfortable with 4d
spacetime geometry, the ANALOGY of 3d space geometry is very handy.
They are not the same thing; merely nice mathematical analogs.  The serious
physics happens in 4d spacetime, but there is a lot of nice fun toying
around to be had in 3d space.

Article 98105 (33 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 5 Feb 1996 12:10:15 -0800
Organization: University of California, Riverside
Lines: 72
NNTP-Posting-Host: guitar.ucr.edu

A couple of corrections:

In article <[email-masked]> [email-masked] (john baez) writes:

>Rather than trying to calculate out how this works in an example, I want
>to use these ideas to explain the Ricci tensor.  

>So, suppose an astronaut taking a space walk accidentally spills a can
>of ground coffee.

[etc.]

>   |       |
>  v^       ^v'
>   | w     |
>   P->-----Q
>   |       |
>   |       |
>   |       |

[etc.]

>This may seem like a lot of work to say that two coffee grounds are
>moving in the same direction at the same speed, but when spacetime is
>curved we gotta be very careful.  Note that everything I've done is
>based on parallel translation!  (I defined geodesics using parallel
>translation.)  

Actually, I also used the metric in a direct way at one place: when I
said that w was orthogonal to v.

But this is not really all that big a deal, since in GR we get parallel
translation from the metric in the first place.

>A little round ball of coffee
>grounds in free fall through outer space!  As time passes this ball will
>change shape and size depending on how the paths of the coffee gournds
>are deflected by the spacetime curvature.  Since everything in the
>universe is linear to first order, we can imagine shrinking or expanding,
>and also getting deformed to an ellipsoid.  There is a lot of
>information about spacetime curvature encoded in the rate at which this
>ball changes shape and size.  But let's only keep track of the rate of
>change of its volume!  This rate is basically the Ricci tensor.  

>More precisely, the rate of change of volume of this little ball is
>equal to 

>r(v,v)

Sorry, the instantaneous rate of change of volume is its original volume
*TIMES* r(v,v)!

Article 98174 (22 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 5 Feb 1996 17:45:09 -0800
Organization: University of California, Riverside
Lines: 52
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>I've never been too clear on exactly what information is contained
>in the Weyl tensor; I'm hoping I'll find out when Baez et al. get
>up to that point.

Yes, I'm hoping I'll find out too when I get to that point.  I can
already begin to glimpse the significance of the "vile tensor" now
that I sort of understand the Ricci tensor...

Recall the definition of the Ricci tensor in terms of coffee grounds
floating through outer space.  

We consider a bunch of initially comoving coffee grounds near a point P
in spacetime, with the coffee ground that actually goes through P having
velocity v at that instant.  (Hence the term "instant coffee".)  Working
in the local rest frame of the coffee ground that goes through P, we
consider a small round ball of comoving coffee grounds centered at P,
and see what happens as time passes.  Each coffee ground moves along a
geodesic, but since spacetime is curved, the ball may shrink, expand,
rotate, and/or be deformed into an ellipsoid.

The Ricci tensor --- let's call it r --- only keeps track of the change
of volume of this ball.  Before, I said that the instantaneous
rate of change of volume of the ball is r(v,v) times the ball's original
volume.  Now I think I may have screwed up: maybe it's the SECOND time
derivative of the volume of the ball which equals r(v,v) times the
ball's volume.  I'll have to do some more calculations.  But something
like this is right: the Ricci curvature tells us how the ball changes
volume.  

So: I think the Weyl tensor tells the REST of the story about
what happens to the ball.  I.e., how much it rotates or gets deformed
into an ellipsoid.  This makes sense, if you think about what you said:

>Anyway, I hope that makes it a little bit clearer why people say that
>the Weyl part of the curvature has to do with gravitational radiation:
>the Weyl tensor carries information about the kind of curvature that's
>independent of the source distribution, sort of like electromagnetic
>waves are fields that propagate independently of whatever sources are
>around.

In short, when we are in truly empty space, there's no Ricci curvature,
so actually our ball of coffee grounds doesn't change volume (to
first order, or second order, or whatever).  But there can be Weyl
curvature due to gravitational waves, tidal forces, and the like.  
Gravitational waves and tidal forces tend to stretch things out in one
direction while squashing them in the other.  So these would correspond
to our ball changing into an ellipsoid!  Just as we hoped.

anization: Oz
Lines: 101
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In article <[email-masked]>, "Emory F. Bunn"
<[email-masked]> writes
>In article <[email-masked]>,
>Oz  <[email-masked]> wrote:
>>2) We throw out the Weyl tensor, but (shock horror, this must be deep)
>>even the tyros don't seem too sure what the Weyl tensor represents.
>
>I've never been too clear on exactly what information is contained
>in the Weyl tensor; I'm hoping I'll find out when Baez et al. get
>up to that point.
>
>I can say one thing that may help, though.  It is true that the
>Einstein equation only contains ten pieces of information, although
>you need 20 to specify the curvature tensor.  So the Einstein equation
>doesn't let you reconstruct the complete curvature tensor.  That
>sounds disturbing at first, but it's really OK.  
>
<snip>
>
>So in electromagnetism, knowing all about the sources isn't enough to
>specify the fields.  In general relativity, knowing all about the
>sources (the stress-energy tensor T) isn't enough to tell you all
>about the curvature.  In both cases, you can supplement the source
>information with some extra initial conditions to get a unique
>solution.  (For example, in electromagnetism you can specify that no
>electromagnetic waves are zooming in from infinity.  That's enough to
>give you a unique solution to the fields given the sources.  For
>general relativity, you can perform similar feats, although it's
>technically trickier.)
>
Well, since everybody knows I don't know anything, and this bit
definitely, I feel I am allowed to speculate pending authoritative
pronouncements.

If the Ricci bit allos us to calculate the curvature effect of
everything we *do* know about (or care to consider), then the Weyl
bit must (sort of by definition) be the curvature of everything
else.

This makes some sort of sense really, when you^H^H^H I think about
it.

So, as you indicated before, Weyl might be the boundary condition
bit. I am not quite sure if this is exactly the best way to think
of it since one would tend to imagine (in 4D) that it would leave
curvature within spacetime if you sort of removed the Ricci bit.
Anyway it's what's left.

End of mindless speculation.
More mindless speculation (a little later):

I have just read Baez's explanation of the Ricci tensor. 
Fancy using coffee grounds, it should be tealeaves. Tealeaves can
travel in the time direction as everyone in Britain knows. That's
why you can read the future in tea leaves.

Anyway the Ricci tensor seems to be something that operates in a
small volume. One might even suspect another infinitesimal one. In
fact infinitesimal enough that curvature due to external thingies
can be neglected. On top of that it could be infinitesimal enough
that it contains, or one can neglect, anisotropy within the
volume. In other words the curvature due to a speck of momentum
flow within the volume. Inevitably, then, we must lose curvature
due to external thingies; indeed we want to.

Right ball park? Not even wrong?

This mindless stuff is sure to bring on a Baez Bashing:

Teacher:
(Striding down the classroom, his face a contorted red mask,
lightning bolts coming from his eyes)
"You infinitesimal moron. Haven't you understood a word I said".

(Picks Oz up by the hair until the poor little mite's feet are
three feet off the ground. His eyes are an 'O' of terror.)

"The dummest clut in the class, and he tries guessing what I
meant. I am the teacher here, you learn what I say, when I say it,
and don't start thinking about it. Moronic twit".

(With a casual flick of his hand he flings Oz across the room by
the hair. Oz strikes the wall near the ceiling and he slides down
the wall to a dejected heap on the floor. Stars flick in and out
of existance round his head.)

Teacher: Now, Ed, perhaps you can explain to the class about the
Ricci tensor, since you have been doing so brilliantly of late.

Ed:
(Ed rises confidently. He gives a scornful look to the bedraggled
Oz crawling painfully along the floor and back to his seat).

"Now the Ricci tensor is a simple construct ........" says Ed.

[Apologies to Ed, at least I think so].    :-)

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98219 (44 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 5 Feb 1996 20:32:54 -0800
Organization: University of California, Riverside
Lines: 114
NNTP-Posting-Host: guitar.ucr.edu

Let me just comment on a few of many things.  I deliberately avoid 
commenting on those remarks that give me the urge to hurl a thunderbolt,
because it's getting late and I will be able to hurl bigger, badder
thunderbolts tomorrow when I am all nice and rested.

So:

In article <[email-masked]> [email-masked] (Edward 
Green) writes:

>...I have no idea what the second term in the Einstein tensor is doing
>yet.  

If you think of the Einstein tensor as concocted like this:

G_{ab} = R_{ab} - (1/2) R g_{ab}

it is indeed somewhat mysterious!  Why, as you ask, do we "subtract half
the trace" of R_{ab}???

The physical answer is this: energy and momentum are locally conserved,
so the stress-energy tensor is divergence-free.  Using some lingo I
haven't explained yet, one writes this as follows:

D^a T_{ab} = 0.

So if we are going to write down Einstein's equation 

G_{ab} = T_{ab} 

we had damn well better have

D^a G_{ab} = 0!

Now there is a wonderful identity, the Bianchi identity, which says
that D^a G_{ab} = 0, no matter what the metric is!  It's not true that
D^a R_{ab} = 0.  So we really do much better to use G_{ab}, since then:

           IN GENERAL RELATIVITY, LOCAL CONSERVATION OF ENERGY AND
              MOMENTUM IS AN AUTOMATIC CONSEQUENCE OF TAUTOLOGOUS
                    FACTS ABOUT SPACETIME CURVATURE!!!!!!

By "tautologous" I mean, "relying on mathematical identities which hold
no matter what the metric is".  

This is a truly wondrous thing.  I leave you to ponder it, and to
remember what is the analogous wonderful thing about Maxwell's
equations.  

>1)  It is not really the Ricci tensor that has the dilational
>interpretation,  but the Einstein tensor itself.  The second term is a
>"correction" occasioned by the use of a local metric that in general
>distorts the underlying *physical* properties of space time  {I sense a
>thunderbolt brewing...} 

ZAP!  Whoops, and here I said I wasn't going to hurl any thunderbolts.
I must have forgotten to disable my automatic bullshit detection system.

>We notice that this term is the only one involving our metric,  whose
>physical significance we have been rather quiet about so far... 

No, the metric is what underlies everything: we used the metric to
get parallel transport!!!!  Then we used that to get the Riemann tensor,
etc..  Everything on the left hand side of Einstein's equation is built
out of the metric.  We could write it out all explicitly in terms of g_{ab}
if we wanted.  (Take my word for it: we don't want to.)

>2)  It really *physically* modifies the Ricci tensor,  in the sense that we
>are discarding yet more information.  In this connection,  note what
>happens in 
> 
>         R^a_d = g^{ab} R_{bd}, 
> 
>         R = R^a_a 
> 
>         G_{ab} = R_{ab} - (1/2)R g_{ab}. 

We are not discarding information in passing from Ricci tensor to
Einstein tensor: we are rearranging it.  We can go backwards.  Let's do
some index gymnastics.... those who aren't in shape, please go to the
back of the room and watch.  Okay, class.  Raise an index!  

G^a_b = R^a_b - (1/2)R g^a_b

Contract!

G^a_a = R^a_a - (1/2)R g^a_a

Note that R^a_a = R and that g^a_a = 4 in 4 dimensions!

G^a_a = R - 2R = -R

Substitute back in the definition of G_{ab}!

G_{ab} = R_{ab} + (1/2) G^c_c g_{ab}    
 
                               (Whispers from the back of the room:
                                "How'd the G^a_a turn into G^c_c?"
                                "It's a dummy index, you dummy!")
                 
Solve for the Ricci tensor!

R_{ab} = G_{ab} - (1/2) G^c_c g_{ab}

So you see that the formula for Einstein in terms of Ricci tensor
is just like the formula for Ricci in terms of Einstein... at least in
dimension 4.   

Article 98394 (43 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 6 Feb 1996 13:57:41 -0800
Organization: University of California, Riverside
Lines: 91
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:

>If the Ricci bit allos us to calculate the curvature effect of
>everything we *do* know about (or care to consider), then the Weyl
>bit must (sort of by definition) be the curvature of everything
>else.

The Weyl tensor is not uninteresting.  It's not that we don't care to
consider it.  But you are right that we can understand it by
understanding the Ricci tensor, and using the fact that the Weyl tensor
says everything about the curvature at a given point that the Ricci
tensor does NOT say.  The Ricci tensor says how initially comoving
coffee grounds "focus" or "defocus", in the sense that a little
(infinitesimal) ball of them grows or shrinks in volume.  So I think the
Weyl tensor says how the ball gets squashed in some directions while
getting stretched in the others.  This can happen in a volume-preserving
way.  

Alternatively, the Ricci tensor is the part of the curvature that's
determined by the presence of energy and momentum flowing through that
very point.  The Weyl tensor is the part that's not.  So in the vacuum,
there's no Ricci, just Weyl.  If you've ever thought about "tidal
forces", this should make sense.  A bunch of coffee grounds floating
through the vacuum will experience "tidal forces", meaning that an
initially round ball will get stretched in some directions and squashed
in others.  (Or sheared, or rotated.  I need to ponder more deeply the
relation here between stretch/squashings, shears, and rotations.)

>So, as you indicated before, Weyl might be the boundary condition
>bit. I am not quite sure if this is exactly the best way to think
>of it since one would tend to imagine (in 4D) that it would leave
>curvature within spacetime if you sort of removed the Ricci bit.

Yes, you can have the universe be a complete vacuum, but still have
gravitational waves rippling through it.  These waves have no Ricci
curvature, only Weyl.  They should be determined by the "boundary
conditions at infinity".  (The math of boundary conditions for nonlinear
equations like Einstein's equations is more tricky than that for linear
equations like Maxwell's.  Still, folks know a lot about it.)

>Anyway it's what's left.

Yes.

>Anyway the Ricci tensor seems to be something that operates in a
>small volume. One might even suspect another infinitesimal one. 

Yes, all my "small", "itsy-bitsy", and "tiny" things are really meant to
be infinitesimal.   You know, take that epsilon and send it scurrying on
home to zero.

>In
>fact infinitesimal enough that curvature due to external thingies
>can be neglected. On top of that it could be infinitesimal enough
>that it contains, or one can neglect, anisotropy within the
>volume. In other words the curvature due to a speck of momentum
>flow within the volume. Inevitably, then, we must lose curvature
>due to external thingies; indeed we want to.

Hmm, well, the Ricci curvature is a tensor which takes a value at each
point of spacetime, and its value at any point is determined solely by
the flow of energy and momentum through that very point.  That's what
Einstein's equation says.  So if I understand you correctly, it sounds
right.

>Right ball park? Not even wrong?

Definitely right ballpark.

>This mindless stuff is sure to bring on a Baez Bashing:
>
>Teacher:
>(Striding down the classroom, his face a contorted red mask,
>lightning bolts coming from his eyes)
>"You infinitesimal moron. Haven't you understood a word I said".
>
>(Picks Oz up by the hair until the poor little mite's feet are
>three feet off the ground. His eyes are an 'O' of terror.)
>
>"The dummest clut in the class, and he tries guessing what I
>meant. I am the teacher here, you learn what I say, when I say it,
>and don't start thinking about it. Moronic twit".
>
>(With a casual flick of his hand he fings Oz across the room by
>the hair. Oz strikes the wall near the ceiling and he slides down
>the wall to a dejected heap on the floor. Stars flick in and out
>of existance round his head.)

What I really like about teaching is the affection and respect my
students feel for me.

Article 98395 (42 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Ricci Tensor (was: Re: General relativity tutorial)
Date: 6 Feb 1996 13:59:52 -0800
Organization: University of California, Riverside
Lines: 27
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (john baez) writes:

>Also, I screwed up in one or two ways.  First, the rate of change of
>volume is not just equal to r(v,v), it's (obviously) proportional to the
>initial volume as well.  Second, I think it's really the *second*
>derivative of the volume with respect to time which matters here.  So
>if you call V(t) the volume of the little ball at time t, we have

>d^2V/dt^2 = r(v,v) V

>(I still need to check this first derivative/second derivative stuff.) 

Yes, it's a second derivative.  I think I got it right, finally.  Or at
least close enough.

Article 98212 (40 more) in sci.physics:
From: Bruce Bowen
Subject: Re: Tensors for twits please.
Organization: Megatest Corporation
Date: Mon, 5 Feb 1996 19:40:39 GMT
Lines: 80

From article <[email-masked]>, by Oz <[email-masked].
co.uk>:
> Starting a new thread at 9.30 on a Sunday evening after several glasses
> of good wine is probably not wise, but there you go. Slurp, hmm, very
> good wine actually.
> 

Many linear physical and mathematical objects require more than a one
dimensional (vector) discription.  A tensor is a geometric object that,
like a vector, exists independent of any coordinate parameterization,
although the actual value of its components depend on the
parameterization, just like a vector.  A tensor is a generalization of
the vector concept, and a vector is a specific example of a tensor of
rank 1.  The rank of a tensor refers to the number of indices it has. 
The dimension refers to the range of the indices and is determined by
the underlying vector space (2 dimensions, 3 dimensions, etc.).  If we
let n = the dimension and r = the rank, then the number of components in
a tensor is equal to n^r. 

I will give as an example a simple minded physical model of a symmetric
sailboat.  By symmetric I mean its bow and stern are identical.  This
model assumes that, due to the viscous medium the boat travels through,
its velocity is proportional (read "linear") to the applied force.  Due
to the design of the hull a sailboat goes forward and backward much more
readily than it goes sideways, so the frictional forces are anisotropic. 
We can express the motion of the sailboat as V where Vi are the
components of velocity, and Fj are the components of the applied force. 
In general the velocity and the applied force are NOT parallel.  The
relationship between V & F is a two dimensional second rank tensor.  It
embodies the relationship between the boat's resistance to forward and
backward motion in relation to its resistance to sideways motion.  It is
purely a function of the hull design.  The motion of the boat couldn't
care less what coordinate system we use to calculate or describe its
motion. 

Another simple example of a second rank tensor is the polarization
distribution of incident light.  For unpolarized light the intensity
will be the same in all directions or circular.  As the light gets more
and more (linearly) polarized the distribution takes on the shape of an
ellipse, and approaches a line as the polarization approaches 100%. 

Note: do not confuse the above with circularly or elliptically polarized
light, which is a different thing altogether. 

A tensor can be thought of as a multi-linear machine.  You input
vectors, or even other tensors, in one side, turn the crank, and out
pops a new vector, tensor, or even a scalar on the other side. 

If your underlying coordinate basis is NOT orthornormal, then the
concepts of contravariant and covariant come into play. If your
underlying vectorspace IS orthonormal, then the concepts of covariant
and contravariant coordinates are not relevant.  The components of a
tensor are usually specified with respect to the dual basis of the
vectors it operates on, unless the metric tensor is included as part
of the equation. A tensor can have both contravariant and covariant
components. The position of the indices (vertically) denote whether a
particular set of components is contravariant or covariant.

A defining characteristic of a tensor is the way its components
transform with a change of basis.  In general, when transforming to a
new coordinate system (where T is the transformation matrix), one must
multiply by T once for each covariant index, and by (T)^-1 once for
each contravariant index.

The difference between a rank 2 tensor and a matrix is the difference
between a formula and an OBJECT!  A tensor is a description of a real
physical phenomena, object, etc.  such as the strain in a rigid solid.
Such objects exist independently of any coordinate system.

For an excellent set of examples of real world tensors in the physical
sciences (restricted to  an orthonormal basis), check out The Feynman
Lectures on Physics, vol II, chapter 31.  The whole chapter is virtually
a list of various tensors, including the polarization tensor of  a
crystal, the conductivity tensor of a crystal, the Moment of Inertia
tensor of an  asymmetric solid, the stress tensor of a solid, the strain
tensor of a solid, as well as others.

-Bruce          [email-masked]
-- 
Bruce Bowen                     Take only meat and hide, leave only guts
[email-masked]             "Just say F.O."

Article 98357 (39 more) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: Tensors for twits please.
Date: Tue, 6 Feb 1996 08:42:25 +0000
Organization: Oz
Lines: 118
Distribution: world
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:

1) In modern notation vectors are/tend-to-be 'column' vectors.
OK, that's just fine.

Er, one little teeny thought. Matrix operations don't commute. So a
'column' vector transforms differently to a 'row' vector. 
eg [a1,...,a3][b1]=/=[b1][a1,..,a3], one scalar, other [3x3]
              [c1]   [c1]
              [d1]   [d1]

Reading between the lines 
<Baez: SMACK! How many times have I *told* you!>
in your notation vectors: column, covectors: row??
Of course it would help if I had the faintest idea what a covector was.
If as above, however, I can see a difference.

>Well, let's see.  You are using a 4x4 matrix to turn row vectors into row
>vectors.  Mind if I use one to turn column vectors to column vectors?

No, go right ahead.

>So let me say:
>
>[b1,b2,b3,b4] [a1] = [b.a]
>[c1.......c4] [a2]   [c.a]
>[d1.......d4] [a3]   [d.a]
>[e1.......e4] [a4]   [e.a]
>
>Then our matrix is serving as a machine that turns tangent vectors to
>tangent vectors in a linear way, so it's a (1,1) tensor.
>
>Now, you protest, but doesn't it ALSO serve to turn cotangent vectors
>into cotangent vectors?  (I.e., row vectors into row vectors, like you
>had done.)  The answer is yes.  (1,1) tensors are multipurpose gadgets.

A very simple and obvious question. Do we have to be very caerful WHAT
our tensor is used for. Ie just row>row not col>col, or does the *same*
tensor do 'double duty' in some way. I am confident this is a
misinterpretation, but something along the lines of:

Tensor T transforms a vector V to say a rotated vector V', and the
*same* tensor T transforms (if I knew what a covector *was* it would
help) a covector C (lets say it's some angle to some other vector) to a
covector C' which is the angle V' makes with the 'other' vector. 

If you have understood what I am trying to say, you are a genius.

>Now we are getting into some of the subtleties I had wanted to avoid.

Hey, I don't want to be able to manipulate these things. Just get an
idea of tha salient points, and some idea of the mechanism. Otherwise I
can't properly follow what's going on.

>Oh well, education occurs even when one least wants it.

If I didn't want it, I would have given up weeks ago. Have you any idea
how many concepts I have tried to take in over the last few weeks?
Basically all the information I have is all here in the thread. No books
(ahem) no barroom chats, no exercises, no supervision. This doesn't make
it easier, but harder. IMHO you are all doing a surprisingly effective
job. I am surprised it's possible to do it at all.

>>What would a tensor of rank (0,2) be?
>
>>Well it turns TWO vectors into a scalar. Oh, dear.
>>Ah, well Baez drew g(u,v) out. It was a square 4x4 matrix as above.
>
>There's something like that which works, but I don't want to explain it
>in PreCambrian.  How about Jurassic?  Say I have a tangent vector v and
>another one w.  Then a (0,2) tensor like the metric g can be thought of
>as a matrix of numbers g_{ab}, and the way we get the scalar g(v,w) is
>by doing this:
>
>g(v,w) = g_{ab} v^a v^b
>
>Here I am summing over the indices a and b.

This is the problem. I have done too much (mis)reading between the lines
in the past (rubs ear). There are a whole lot of ways of defining
'summing over the indices a and b'. YOU mean in a particular way. *I*
don't know which way you mean. Since we are getting close to the
notation you are using in GR, it would help if I knew, I think.

>Actually I am afraid I am opening an enormous can of worms here, by
>trying to translate PreCambrian into more recent dialects.  PreCambrian
>is so vague that there are limitless possibilities for argument.

Well, I am perfectly happy to be brought up to pre-modern. I am (would
prefer) to use your notation, if only I knew what it was!
<Sickly grin>

>  Why don't you burn that book?

1) It's too damp to burn.
2) It would be inhumane. There are moulds on that book that have been
digesting physics for generations. God only knows what they will evolve
into. Maybe their descendents will evolve into intelligent life!
3) From what you say if I keep it for another 30 years, it will be a
collectors item.

Well, I suppose if you are going to fill in the bits, I could bury it
somewhere nutritious. <Sob, sob sob>.

NB The keyboard is b*ggered. I have washed it in copious warm water and
left it to dry in the boiler room. Pray for it. I have nicked my son's
keyboard. As long as it's back before he gets home from school, I am
safe!

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98366 (38 more) in sci.physics:
From: Matthew P Wiener
(SAME) Subject: Re: Tensors for twits please.
Date: 6 Feb 1996 17:11:09 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 68
Distribution: world
NNTP-Posting-Host: sagi.wistar.upenn.edu
In-reply-to: Oz <[email-masked]>

In article <[email-masked]>, Oz <Oz@upthorpe writes:
>Of course it would help if I had the faintest idea what a covector was.

In ordinary xyz coordinates, you have two ways of describing points.  The
ordinary way is just to measure off the x,y, and z axes.

The dual way is to count off x=?? and y=?? and z=?? planes.

These lead to the same answer, so most people ignore the difference.

But when you change to or between curvilinear coordinates, the two ways
are distinct.  This shows up in the transformations and derivatives.
Check out the last chapter of the Schaum's Outline Series on VECTOR
ANALYSIS (I think it was that one) for more details of this approach,
in slow motion.  Off-hand, I have not seen it spelled out anywhere else
as this, but I assume it was once standard knowledge.

One form of it is the MTW "bongs of a bell" description of "covectors"
aka 1-forms.  A vector is just an arrow pointing somewhere.  A covector
(in R^3) is a dissection of R^3, consisting of parallel planes, such
that they are labelled "linearly", with 0 for the plane passing through
the origin, one of the other planes labelled 1, the one twice as far
labelled 2, etc.  These may be described with 3 coordinates (a,b,c)
by letting ax+by+cz=1 be the plane labelled 1.  (0,0,0) is the plane
at infinity, so to speak.

So covectors form a 3-dimensional space.  There is a natural duality
between covectors and vectors, namely < (a,b,c) | (x,y,z) > = ax+by+cz.
In terms of the starting R^3 dissection, this just identifies the label
of the plane that vector (x,y,z) reaches.

The unit covectors are commonly denoted dx,dy,dz.  Although you were
taught dx is "differential x", here its "dual x".  But the notation
was chosen this way deliberately.  1-forms Adx+Bdy+Cdz are naturally
understood as covectors, not vectors.

For example, given a scalar function f on R^3 (ie, f(x,y,z)=some number),
we can take its gradient.  You probably learned grad f = (@f/@x).i + ...

What piffle.

"Really" grad f is df = (@f/@x).dx + ....  It forms a covector field, not
a vector field.  So what does this really look like?  It's ultimately just
a contour map of f, custom linearized at each point.

This is easier to imagine in 2D.  Take a contour map of a scalar function
"height", assumed differentiable.  Pick a point.  Kind of curvy, right?
Blow up the neighborhood 1000 times, notice how the nearby contours are
a lot straighter.  So just _redraw_ them as straight, extend to a covector
worth of lines dissecting R^2, and scale the line labelled 1 down by 1000.
This is the dual version of (F(x+.001)-F(x))/.001 we all know and love.

That this is so direct, the moral equivalent of drawing tangent lines, is
all the proof anyone really needs that gradients are covectors.

By pushing the MTW pictures to the limit, it is actually possible to work
out a pretty good visualization of k-forms.  For 2-forms, one has two sets
of parallel dissections, together forming pipes.  For 3-forms, one has
three sets of them dissection.  In n-dimensions, one works with hyperplanar
dissections instead of planar dissections.

The differential that maps k-forms to k+1-forms becomes obvious: just
more contouring.  Given a k-form field, U(p) at each point p, we want to
compare the k hyperplanar dissections of U(p) with U(p+h) for small h's.
Just draw a new hyperplane perpendicular to the direction of change, and
scale by 1/h.  As h->0, this converges to a k+1-form.
-- 
Matthew P Wiener ([email-masked])

Article 98449 (36 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Tensors for twits please.
Date: 6 Feb 1996 17:48:13 -0800
Organization: University of California, Riverside
Lines: 196
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:
>In article <[email-masked]>, john baez <[email-masked]>
>writes
>>In article <[email-masked]> [email-masked] 
>>writes:

>in your notation vectors: column, covectors: row??

Yeah.  This is sort of arbitrary; I could have said the other way
around, but let's do it this way.  

>Of course it would help if I had the faintest idea what a covector was.

I said it was just slang for a (0,1) tensor, a guy who eats a vector and
spits out a number, in a linear way.

This "co" stuff is an fundamental notion in mathematics and physics.
It's sometimes called "duality", or the "covariant/contravariant"
distinction.  

Say we have a vector v and a covector f.  Then 

f(v) = x

is a number.  We usually think of f eating v and spitting out the number
x, but we can equally well say that v is eating f and spitting out the
number x!  

So in some sense covectors are just as basic as vectors.  

Ponder this, but don't worry about it; it'll take a long time to get
this point.  

But it's lurking in the "row vector" vs "column vector" stuff you always
see in old-fashioned linear algebra books.  You can think of a row
vector as a guy that's dying to eat column vectors and spit out numbers:

[a b][c] = ac + bd
     [d]

but if you change your point of view you can think of it the other way
around: the column vector eats the row vector.

>A very simple and obvious question. Do we have to be very caerful WHAT
>our tensor is used for. Ie just row>row not col>col, or does the *same*
>tensor do 'double duty' in some way.

There are two levels to this question, because you are reading an old
book that talks about "row vectors" and "column vectors", and I am
struggling to translate it into the modern lingo of "vectors" and
"covectors", but the translation is a tricky business, in the way that
translating foreign languages always is.  In fact, I urge you to not pay
much attention to that book, and just let me explain everything.  I will
wind up doing LESS work that way.

Anyway, I can confidently say: yes, a (1,1) tensor does double duty: by
definition it serves to turn a vector into a vector, but if we are
clever we can use it to go turn a covector into a covector.

On the other hand, something like a matrix is even more
multipurpose: we can use a matrix to stand for either a (1,1) tensor, or
a (0,2) tensor (like the metric!) or a (2,0) tensor.  This is because
the PreCambrian dialect of "row vectors", "column vectors" and
"matrices" fails to make some of the finer-grained distinctions that the
modern lingo does.  Why?  Because it's deceptively easy to turn a column
vector into a row vector: just flop it over!  E.g.

[2]
[3]

becomes   

[2 3]

The modern approach is set up to make this impossible to do UNLESS you
use the metric!!!!!!!!!   You can't just "flop over" a vector and get a
covector.  You do this using the metric (in a way I'll eventually
explain.)  

Again, file this stuff away in your brain, or somewhere, but don't let
it get you down now.

>Hey, I don't want to be able to manipulate these things. Just get an
>idea of the salient points, and some idea of the mechanism. Otherwise I
>can't properly follow what's going on.

Yes, but well, I am now being forced to explain not just tensors, which
is bad enough, but the difference between the way your textbook explains
tensors and the way modern mathematical physicists think about tensors.
I don't mind doing it, but you are likely to become very confused for a
while before it all clears up.  It's as if, rather than just learning
German, you wanted to learn German out of a book written in Old Norse.

>>Oh well, education occurs even when one least wants it.

>If I didn't want it, I would have given up weeks ago. Have you any idea
>how many concepts I have tried to take in over the last few weeks?

Come come, I didn't say you didn't want some education.  I was making a
cryptic joke about how you are now busily learning about the history of
different mathematical notations for tensors!  This means you are going
to learn a whole bunch of extra concepts which you don't even need to
know to understand general relativity.  They are good to know, but will
perhaps be the straw that breaks the camel's back.

>>Say I have a tangent vector v and
>>another one w.  Then a (0,2) tensor like the metric g can be thought of
>>as a matrix of numbers g_{ab}, and the way we get the scalar g(v,w) is
>>by doing this:
>>>

>>g(v,w) = g_{ab} v^a v^b
>>
>>Here I am summing over the indices a and b.

>This is the problem. I have done too much (mis)reading between the lines
>in the past (rubs ear). There are a whole lot of ways of defining
>'summing over the indices a and b'. YOU mean in a particular way. *I*
>don't know which way you mean. Since we are getting close to the
>notation you are using in GR, it would help if I knew, I think.

Okay, let me spell it out.  I was really hoping someone else would do
it, since this involves typing long dull rows of symbols.  But okay.
Let's say we're in 4d spacetime.  Then indices like a,b etc.  stand for
0,1,2, or 3.  When we sum over them, we sum from 0 to 3.  So for
example, when I speak of "the metric g_{ab}", what I mean is a
particular batch of numbers, namely:

g_{00}   g_{01}   g_{02}   g_{03}
g_{10}   g_{11}   g_{12}   g_{13}
g_{20}   g_{21}   g_{22}   g_{23}
g_{30}   g_{31}   g_{32}   g_{33}

Note how both a and b go from 0 to 3.  Each thing like g_{21} above is
just a number.  So for example g_{ab} might be the matrix

  -1       0        0        0
   0       1        0        0                       (*)
   0       0        1        0
   0       0        0        1

in which g_{22} = 1 and so on.  Similarly, when I refer to "a vector
v^a", I mean a column vector like

  v^0
  v^1
  v^2
  v^3

A random example would be the vector

   1
   2                                                 (**)
   0
   9

In this example we'd have v^3 = 9.   Similarly, when I refer to "the
vector w^a" I mean another such thing, like

  w^0
  w^1
  w^2
  w^3

An example would be, say

   2
   2                                             (***)
   0
   0

Now when I write something like g_{ab} v^a w^b, I mean that I multiply
the number g_{ab} by the number v^a and by the number w^b, and then
SUM UP letting a and b range from 0 to 3.  So this is short for

g_{00}v^0w^0 + g_{01}v^0w^1 + g_{02}v^0w^2 + g_{03}v^0w^3 + 
.... 4 more of these bloody things .... +
.... 4 more of them ....                +
g_{30}v^3w^0 + g_{31}v^3w^1 + g_{32}v^3w^2 + g_{33}v^3w^3

Okay.  Here's a test!!!  Take the example of the tensor g_{ab} that I
gave, back at (*), and the example of v^a I gave, back at (**), and the
example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is.
The answer is some specific number; what is it?

This is how you actually work out the inner product of two vectors v and
w:

g(v,w) = g_{ab} v^a w^b.

Article 98464 (58 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Ricci Tensor (was: Re: General relativity tutorial)
Date: 6 Feb 1996 18:50:42 -0800
Organization: University of California, Riverside
Lines: 113
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> Geoffrey A. Landis <geoffrey.lan
[email-masked]> writes:

>Let's work from physics back to mathematics, instead of the other way. 
>We have a spherical ball of coffee particles.  This ball is distorted by
>the curvature of space (equivalently, we can say it is distorted by tidal
>accelerations); the distortion changes the sphere into an ellipsoid.  An
>ellipsoid in space will instantly make any physicist say "tensor"...

Yes.  The question, is, which tensor?  

>...pick a set of coordinates defined by the directions of the maximum,
>minimum, and intermediate semimajor axes, the radius tensor is diagonal. 
>[For the 0,0 component of the tensor in 4-space, we will assume that it
>is equally straightforward to define the t axis as forward in time along
>the geodesic].  Looking at the infinitesimal change in shape from sphere
>to ellipse over an infinitesimal d time, we can define a
>second-derivative-of-the-radius tensor, which is by definition the
>tidal-acceleration tensor, which is by definition a curvature-of-space
>tensor.  In our nice diagonal coordinate system, the diagonal elements
>are simply d^2r_x/dt^2, d^2r_y/dt^2, and d^2r_z/dt^2, plus a 0,0 term
>relating to differential time dilation.  For simplicity, we can normalize
>by dividing by the initial radius of the coffee sphere.

>Let me simply assert, without proof, that this
>second-derivative-of-the-radius-of-the-coffee-sphere tensor is the same
>as the Ricci tensor R_mn.

Sorry.  The "tidal-acceleration tensor" you describe is not the Ricci
tensor, but rather

R^m_{anb} v^a v^b 

where v is the 4-velocity of the coffee ground at the center of the
ball, and R^m_{anb} is the Riemann tensor.  I could explain WHY it's
this: it follows from the "geodesic deviation equation" which relates
the deviation of nearby geodesics to the curvature of spacetime.
Instead, let's see how this result jibes with my earlier claims.
>The volume of this ellipsoid is, to within factors of pi, the product V =
>r_x r_y r_z.  The (normalized) second derivative of the volume will be
>the sum of the second derivative diagonal elements (keeping only first
>order terms each time we do a derivative, and ignoring factors of pi
>again); that is, the *trace*, or, in the Einstein summing notation, the
>tensor contracted to a scalar by summing indices. 

Right.  So to get the second derivative of volume we take the *trace* of
the tensor 

R^m_{anb} v^a v^b

on the two free indices m and n, getting

R^m_{amb} v^a v^b.

Using the definition of the Ricci tensor, R_{ab} = R^m_{amb}, this is
the same as

R_{ab} v^a v^b

In coordinate-free notation this is just r(v,v), where now I write
the Ricci tensor with a lowercase r to keep it straight from the Riemann
tensor, because the indices don't keep things straight for us.

This agrees with what I was saying earlier:

"A little round ball of coffee grounds in free fall through outer space!
As time passes this ball will change shape and size depending on how the
paths of the coffee gournds are deflected by the spacetime curvature.
Since everything in the universe is linear to first order, we can
imagine it shrinking or expanding, and also getting deformed to an
ellipsoid.  There is a lot of information about spacetime curvature
encoded in the rate at which this ball changes shape and size.  But
let's only keep track of the rate of change of its volume!  This rate is
basically the Ricci tensor.

More precisely, the second time derivative of the volume of this little
ball is equal to 

r(v,v)

times its original volume, where r is a rank (0,2) tensor called the
Ricci tensor, and v is as above.  (In general r eats two vectors and
spits out a number.  Here it is eating two copies of v.)"

>Then, if we want a tensor which represents the change in *shape* of the
>initial sphere, without change in *volume*, we can subtract the change in
>volume from the tensor representing the entire change in the sphere.  To
>subtract a scalar from a tensor, we have to multiply it by the metric.  A
>factor of 2 comes in from some differentiation I lost track of.  Thus, we
>get the divergenceless Ricci tensor

[typically called the Einstein tensor]

>G^ab = R^ab - (1/2) g^ab R
>
>I think that this is correct.  Time to check the homework with the
>teacher--

Sorry... nice guess, but the gadget that represents the change of shape
of the initial sphere discounting change in volume is built from the Weyl
tensor, not the Einstein tensor.  As explained in other posts, the
Einstein tensor and the Ricci tensor are two ways of repackaging the
same information about spacetime curvature.  The Weyl tensor expresses
the information about curvature that's not in the Ricci tensor (or
Einstein tensor).  

I think you'll find all this clearer if I explain the geodesic deviation
equation.  

Article 98491 (34 more) in sci.physics:
From: Edward Green
Subject: Re: general relativity tutorial
Date: 6 Feb 1996 21:16:17 -0500
Organization: The Pipeline
Lines: 73
NNTP-Posting-Host: pipe9.nyc.pipeline.com
X-PipeUser: egreen
X-PipeHub: nyc.pipeline.com
X-PipeGCOS: (Edward Green)
X-Newsreader: The Pipeline v3.4.0

I don't want to deviate too much on this question,  but since you have
moved it here,  one more try.  Ok? 
 
[email-masked] (john baez)' wrote: 
 
>Say you have two clocks next to each other in 
>your house, at spacetime point A.  Then you move one up to a mountain 
>for a while, and then move it back down and compare the two clocks at 
>spacetime point B.  You see the one you took up the mountain is ahead of 
>the other.  
 
>Note that since I am only comparing clocks when they are right next to 
>each other, I don't need to worry about how the information was passed 
>from one to the other, and whether extra time lags were introduced in 
>the process.   
> 
>Note also that I don't need any coordinate system!  I just need two
clocks. 
> 
>In terms of the mathematics of GR, what's going on?  Well, we have two 
>paths from spacetime point A to spacetime point B, and the "length" (or 
>more precisely, "proper time") along one path is more than the other. 
>That's all.  
 
Ok.  Operationally,  I still think I can capture the idea of time running
"slower or faster" at another location,  even if you prefer not to express
it that way. 
 
This time start with three clocks,  A,  B, and  C, originally sychronized
in our house.  At t = 0,  hoist B and C slowly to the top of a mile high
tower you have constructed in your backyard.  One hundred years later by
the clock in the house (this is a multi-generational experiment),  lower
*one* of the two clocks on the tower,  B, and note the time discrepancy. 
In yet another century,  lower the second clock, C,  in the same way as B. 
Now what as happened?   Let the time gains relative to the ground be t_B
and t_C.  Without knowing what GR predicts,  I surmise either: 
 
1)   t_C = t_B,   or 
 
2)   t_C = 2 t_B 
 
In other words,  we have left the damn things at the top of tower so long, 
that any effects attributable to the trip there have become negligible, 
unless that's the only effect in town!    These results show either that 1)
the discrepancy is attributable only to the relative motion in the
gravitational field,  a one shot deal,  or 2) there is an accumulating
discrepancy attributable to their separation.  In the second case I would
say  _operationally_  that "time is moving faster at the top of the tower".
  If that's not what we mean by such a statement,  I don't know what is! 
 
Now if you tell me 2) is the predicted outcome,  then  in GR,  in cases
where the gravitational potential predicts the time discrepancy,  this
potential must correspond to some local property of space,  unlike in
newtonian gravity,  where the potential is just a summary of the global
field.   Space really is more sluggish down there in the graviational well.

>The gravitational potential can be a useful tool in *certain* problems, 
>but I prefer this other way of thinking of it, since it tells you 
>something relevant to quite general GR problems.  Also, it fits with the 
>idea that the *metric on spacetime* is what really matters: that's what 
>you'd use to compute the proper time along a curve.  Also, it seems 
>pretty simple and unmysterious. 
 
Granted.  So where is the metric doing its thing?  Just going up and down
the towers,  or while sitting on the top? 
 
-- 
 
Ed Green / [email-masked] 
 
"All coordinate systems are equal, 
 but some are more equal than others".

ject: Re: General relativity tutorial
Date: 7 Feb 1996 11:04:48 -0800
Organization: University of California, Riverside
Lines: 129
NNTP-Posting-Host: guitar.ucr.edu

While some of the stuff Ed said was wrong, it has helped me understand
the simple geometrical essence of Einstein's equation, and that's good.

In article <[email-masked]> [email-masked] (Edward 
Green) writes:

>The Ricci tensor is the strain rate derivative of spacetime.  In other
>words,  translating in any direction in spacetime,  the Ricci tensor gives
>us the resulting dilations and shears.  The Ricci tensor is symmetric, and
>the diagonal components represent stretching,  the off-diagonals,  shear. 
>However,  we could always diagonalize it by a coordinate transformation to
>principle axes.  Then we would find all information in it summarized in *4*
>numbers,  giving expansion or contraction along the 4 principle axes, 
>transforming our little hyperspherical test region into a hyper-ellipsoid. 
>Up to scale factors these axes give the locally *best* coordinate system
>for curved spacetime,  one that captures the local behavior most
>succinctly. 

Maybe that's true in some sense but I don't know what sense.  All I know
is that if one lets a little 3d ball of initially comoving test particles
freely fall along in the direction v, the ball starts to expand at a
rate proportional to r(v,v), where r is the Ricci tensor:

r(v,v) = R_{ab} v^a v^b

When I say "starts to expand, I'm talking about SECOND derivatives along the
geodesic whose tangent vector was initially v, which is what one would
expect from the fact that v appears in a QUADRATIC way in that formula.

So we have a quadratic form in 4 variables, r(v,v), and thus we can find
4 principal axis for it.  Typically, one of these is the axis along
which the ball of test particles expands the MOST, one is the axis along
which it expands, the LEAST, and then there are two "middle" ones. 

>The remaining information in the Riemann tensor,  which we have thrown out,
> captures the *rotation* of our little test sphere under translation.  In
>the same rank as our Ricci tensor,  which is symmetric,  we could add
>another antisymmetric tensor,  which would contain the pure rotations. 
>Apparently,  these rotations are *not* determined by the local energy
>density,  but may be globally determined by boundry conditions.  Dilation
>is an intrinsic property,  determined locally by the stress tensor, while
>rotation is an extrinsic property,  determined by the behavior of
>nieghboring bits of space. 

Now I guess you're no longer talking about your "little 4d
hyperspherical test region" but my "little 3d spherical ball of comoving
coffee grounds," right?  I hope so.  The Ricci tensor describes how, as
the ball freefalls, it expands and contracts.  I am struggling to figure
out whether it can start to rotate... I have some vague reasons to think
it can't.  This leaves volume-preserving transformations that stretch
out some axes and squash the others.  That sounds very much like what
gravitational radiation and tidal forces do to stuff... so I think
that's the Weyl tensor!  

>It seems reasonable from the definition R_{bd} = R^c_{bcd} that each
>component of R_{bd} is measuring how much space "oriented in the d
>direction" is spreading out as we "move in the "b direction", since we sum
>over all the side excursions the test vector might make,  and in particular
>just sum the "spreading" terms (the discrepancy in the same direction as
>the side excursion.   Just waving my hands.... 

Something sorta vaguely like this is true... you can figure it all out 
from the definition of the Riemann tensor I gave.  But I'll do it for
you, eventually: it's the "geodesic deviation equation" you're after.

>But I have no idea what the second term in the Einstein tensor is doing
>yet.  

>         G_{ab} = R_{ab} - (1/2)R g_{ab}. 

Well, here's a good way to think about.  Let's do the same index
gymnastics we did in the last class, only starting with Einstein's
equation.  Okay, everyone!  Time for index gymnastics.  Stand with your
feet slightly apart and hands loosely at your sides.  Now, assume the
Einstein equation!

G_{ab} = T_{ab}

Substitute the definition of Einstein tensor!

R_{ab} - (1/2)R g_{ab} = T_{ab}

Raise an index!

R^a_b - (1/2)R g^a_b = T^a_b

Contract!  

R^a_a - (1/2)R g^a_a = T^a_a

Remember the definition of Ricci scalar, and g^a_a = 4 in 4d!

R - 2R = T^a_a

Solve!

R = - T^a_a

Okay.  That's already a bit interesting.  It says that when Einstein's
equation is true, the Ricci scalar R is the sum of the diagonal terms of
T^a_a.  What are those terms, anyway?  Well, they involve energy density and
pressure.  But let's wait a bit on that... let's put this formula for
R back into Einstein's equation:

R_{ab} + (1/2) T^c_c g_{ab} = T_{ab}

or

R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}

In other words: you might naively have hoped that since the Ricci tensor
has such a nice interpretation in terms of convergence of geodesics, it
must be equal to the stress-energy tensor.  In fact, I think Einstein
made exactly this guess, before realizing that it'd ruin local
conservation of energy and momentum.  To get the right Einstein
equation, you need this "extra term" around, coming from the trace of
the stress-energy tensor, T^c_c.  

If we work this out a bit more, we'll see that Einstein's equation has a
simple meaning in terms of a little ball of initially comoving particles
in free fall: it says that in the rest frame of the ball, the second
time derivative of the volume of the ball is given by a simple formula
in terms of the energy density and pressure at that point! 

What's the simple formula?  Hmm, let me think...

Article 98676 (15 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 7 Feb 1996 18:42:50 -0800
Organization: University of California, Riverside
Lines: 219
NNTP-Posting-Host: guitar.ucr.edu

Okay, I claimed I was going to describe the "simple geometrical essence
of Einstein's equation, so let me do that.  I left off last time after
rerwriting Einstein's equation that it looks like this:

R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}

This is nice because we have a simple geometrical way of understanding
the Ricci tensor R_{ab}.  Namely, if v is the velocity vector of the
particle in the middle of a little ball of initially comoving test
particles in free fall, and the ball starts out having volume V, the
second time derivative of the volume of the ball is 

R_{ab} v^a v^b

times V.  Here "time" means proper time.   If we know the above quantity
for all velocities v (even all timelike velocities, which are the physically
achievable ones), we can reconstruct the Ricci tensor R_{ab}.  But we
might as well work in the local rest frame of the particle in the middle
of the little ball, and use coordinates that make things look just like
Minkowski spacetime right near that point.  Then 

g_{ab} = -1  0  0  0 
          0  1  0  0
          0  0  1  0
          0  0  0  1

and v^a = 1
          0
          0 
          0

So then --- here's a good little computation for you budding tensor
jocks --- we get 

R_{ab} v^a v^b = -R_{00}

So in this coordinate system we can say the 2nd time derivative of the
volume of the little ball of test particles is just -R_{00}.

On the other hand, check out the right side of the equation:

R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}

Take a = b = 0 and get

R_{00} = T_{00} + (1/2) T^c_c 

Note: demanding this to be true at every point of spacetime, in every
local rest frame, is the same as demanding that the whole Einstein
equation be true!   So we just need to figure out what it MEANS!

What's T_{00}?  It's just the energy density at the center of our little
ball.  How about T^c_c?   Well, remember this is just g^{ca} T_{ac},
where we sum over a and c.  So --- have a go at it, tensor jocks and
jockettes! --- it equals -T_{00} + T_{11} + T_{22} + T_{33}.  So if I
haven't made a mistake (and I darn well could've) we get

R_{00} = (1/2) T_{00} + T_{11} + T_{22} + T_{33}.

What about T_{11}, T_{22}, and T_{33}?  In general these are the flow of
x-momentum in the x direction, and so on.  

So the "simple geometrical essence of Einstein's equation" is this:

Take any small ball of initially comoving test particles in free fall.
Work in the local rest frame of this ball.  As time passes the ball
changes volume; calculate its second derivative at time zero and divide
by the original volume.  The negative of this equals 1/2 the energy
density at the center of the ball, plus the flow of x-momentum in the x
direction there, plus the flow of y-momentum in the y direction, plus
the flow of z-momentum in the z direction.

Note: all of general relativity can in principle be recovered from the
above paragraph!  I'm glad I'm "wasting time" talking about general
relativity on sci.physics, because I'd never known this formulation of
general relativity until you folks forced me to explain the Ricci
tensor.

Note that the minus sign in that paragraph is good, since it says if you
have POSITIVE energy density, the ball of test particles SHRINKS.  I.e.,
gravity is attractive.

Now you might wonder about what all that "x-momentum in the x direction"
stuff really means.   I don't have anything thrillingly insightful to
say about this except that, for a perfect fluid, each of these terms is
just the pressure!  I will quit here for now, leaving Michael Weiss to
explain why that's true.  

Next time (when I'm not too busy swatting down questions) I will explain
the "geodesic deviation equation" which relates the change of shape of
this little ball of freely falling particles to the Riemann curvature.

Okay, here's Michael and Bruce:

[etc.]

Article 98586 (52 more) in sci.physics:
From: john baez
Subject: Re: general relativity tutorial
Date: 7 Feb 1996 12:47:42 -0800
Organization: University of California, Riverside
Lines: 102
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward G
reen) writes:
>I don't want to deviate too much on this question,  but since you have
>moved it here,  one more try.  Ok? 

Okay.

>Ok.  Operationally,  I still think I can capture the idea of time running
>"slower or faster" at another location,  even if you prefer not to express
>it that way. 

Okay; if you give me the description of an experiment, I can try to say
what happens.  Then I can leave it to you to decide whether or not it
means "time is running faster" at one location than another.  Whatever
happens after that is your own fault.  :-)  If you let language
lead you astray, there's no telling where you'll wind up.

>This time start with three clocks,  A,  B, and  C, originally sychronized
>in our house.  At t = 0,  hoist B and C slowly to the top of a mile high
>tower you have constructed in your backyard.  One hundred years later by
>the clock in the house (this is a multi-generational experiment),  lower
>*one* of the two clocks on the tower,  B, and note the time discrepancy. 
>In yet another century,  lower the second clock, C,  in the same way as B. 
>Now what as happened?   Let the time gains relative to the ground be t_B
>and t_C.  Without knowing what GR predicts,  I surmise either: 

>1)   t_C = t_B,   or 
> 
>2)   t_C = 2 t_B 

And the answer is.... 2)!

>In other words,  we have left the damn things at the top of tower so long, 
>that any effects attributable to the trip there have become negligible, 
>unless that's the only effect in town!    These results show either that 1)
>the discrepancy is attributable only to the relative motion in the
>gravitational field,  a one shot deal,  or 2) there is an accumulating
>discrepancy attributable to their separation.  In the second case I would
>say  _operationally_  that "time is moving faster at the top of the tower".
>  If that's not what we mean by such a statement,  I don't know what is! 

I sympathize....   But as you begin to get more rhapsodic, I get more
nervous:

>Now if you tell me 2) is the predicted outcome,  then  in GR,  in cases
>where the gravitational potential predicts the time discrepancy,  this
>potential must correspond to some local property of space,  unlike in
>newtonian gravity,  where the potential is just a summary of the global
>field.   Space really is more sluggish down there in the graviational well.

"Space is more sluggish"... hmm, now we are really waxing rhapsodic,
aren't we!  What makes me really nervous is your talk of a "local
property of space" being different.  This can be fatally ambiguous:

1.  The equivalence principle says that to an object in free fall, every
tiny patch of spacetime looks, to first order, just like every other.
What do I mean by that?  I mean that if you work in the local rest frame
of a freely falling object at a given moment and do an experiment in its
immediate locale, the result will always be the same, as long as we
ignore higher-order effects --- namely CURVATURE --- which show up when
the experiment is not infinitesimal.  This is one sense in which
spacetime has no local properties.

2.  But of course your house and the tower are NOT in free fall, and we
are NOT doing an infinitesimal experiment in the immediate locale of one
spacetime event.  So to some extent 1 is not relevant.  I just thought
I'd mention it.

3.  If we do take into account second-order effects, we may say that
spacetime at any given point DOES have local properties, namely
curvature.  This curvature is higher in your house than up on the tower.
But the way the "time runs faster at the top of the tower" is NOT simply
proportional to the curvature or anything like that; the relation is
subtler.  So curvature is not a "local property of spacetime which makes
time run slower or faster".  A better way to think of the role of
curvature is this.  Think about a 2d surface rather than 4d spacetime.
If we know the surface is flat we know how to compute the length of any
path on it.  If we THINK it's flat but it's NOT, our computations will
be wrong.  Some paths will be mysteriously shorter or longer than we
expect.  That's what's happening in the tale of two clocks: two
worldlines in spacetime which commonsense suggests should have the same
length, do not, because spacetime is curved.

4.  In a sense, one might say the *metric* on spacetime is the "local
property" that makes paths long or short --- perhaps longer or shorter
than naive commonsense suggested --- hence makes a clock run "faster" or
"slower" than expected.  This has a certain truth to it, despite 1.
As I said:

>>... the *metric on spacetime* is what really matters: that's what 
>>you'd use to compute the proper time along a curve.  Also, it seems 
>>pretty simple and unmysterious. 

>Granted.  So where is the metric doing its thing?  Just going up and down
>the towers,  or while sitting on the top? 

Mainly while sitting on top, if the clock sits on top for a long time.  

I almost feel like showing you the formulas, but I don't like typing
formulas too much.

Article 98666 (54 more) in sci.physics:
From: Emory F. Bunn
(SAME) Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 8 Feb 1996 00:40:02 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 31
Distribution: world
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Keith Ramsay <[email-masked]> wrote:

>In article <[email-masked]>, [email-masked] (ale2) wrote:
>|For any spacetime point around, say the sun, is there an infinite
>|number of geodesics through that point?
>
>Yes, one for each tangent direction through it.

Lest anyone get confused, I just want to point out that when Keith
says "each direction" he means each direction through *spacetime*, not
just space.  As long as you've gotten into the proper relativity
spirit, that should be obvious.

So if you're sitting at a particular point in spacetime, you can't just
choose a particular direction through space (say, North) and expect
to find a unique geodesic in that direction through that point.
That's because North is just a direction in space, not a direction
through spacetime.

(Of course, before a word like "North" even makes sense, you have to
have laid down coordinates on your patch of spacetime.)

If you want to specify a unique geodesic through a particular
spacetime point, one way to do it is to specify the *velocity* that a
particle would have if it passed through that point following that
geodesic.  That's because the velocity is a 4-dimensional vector (a
tangent vector) in spacetime, so specifying the velocity is a good way
to specify a direction through spacetime.

-Ted

Article 98602 (39 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 7 Feb 1996 13:27:02 -0800
Organization: University of California, Riverside
Lines: 117
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (
Steven Hall) writes:
>In article <[email-masked]>, [email-masked] (john baez) wrote:

>John, don't despair.  There are those of us lurking out here who
>understand all about row vectors, column vectors, etc., but not about
>tensors. (In engineering, the "old-fashioned" matrix/vector notation
>dominates.)  The connections you are making between the two notations are
>certainly helping me.

That's great!  It's really a relief to know that typing in those
enormous matrices is doing *somebody* some good.  In Oz's case, I was
afraid I might simply be confusing the heck out of him unnecessarily.

>OK, let's see if I have it:
>
>> Okay, let me spell it out.  I was really hoping someone else would do
>> it, since this involves typing long dull rows of symbols.  But okay.
>> Let's say we're in 4d spacetime.  Then indices like a,b etc.  stand for
>> 0,1,2, or 3.  When we sum over them, we sum from 0 to 3.  So for
>> example, when I speak of "the metric g_{ab}", what I mean is a
>> particular batch of numbers, namely:
>> 
>> g_{00}   g_{01}   g_{02}   g_{03}
>> g_{10}   g_{11}   g_{12}   g_{13}
>> g_{20}   g_{21}   g_{22}   g_{23}
>> g_{30}   g_{31}   g_{32}   g_{33}
>> 
>> Note how both a and b go from 0 to 3.  Each thing like g_{21} above is
>> just a number.  So for example g_{ab} might be the matrix
>> 
>>   -1       0        0        0
>>    0       1        0        0                       (*)
>>    0       0        1        0
>>    0       0        0        1
>
>[more deletia]

>So to an engineer, we would just have the matrix G, with which we could do
>any number of things.  We could multiply by a column vector x, yielding a
>column vector y = G x, or we could multiply by a row vector u, to get v =
>u G, are take an inner product, p G q.

Exactly.

>In tensor notation, we can do only the last, because the tensor is g_{ab},
>and the inner product would be g_{ab} u^a v^b.  The first would be written
>y^a = g^a_b x^b, and the second would be v_a = g_a^b u_b.

Exactly!!!!!   

>If I have that right, then the interpretation is that with a matrix, you
>have to know the allowable operations from the context of the problem,
>whereas with tensor notation, the tensor tells you what to do.

Right.  Now one extra comment.  Why do engineers feel so free to treat
matrices in this multipurpose way?   Or conversely, why are general
relativists so fussy, that they DON'T want to do so?  The key thing is
that to an engineer, it's perfectly acceptable to turn a column vector into
a row vector.  It's no big deal.  But to a relativist, one should be
careful turning vectors (aka tensors of rank (1,0)) into covectors (aka
tensors of rank (0,1)).  It's not against the law, but one does it using
the METRIC.  This is a tensor g_{ab}.  We define the tensor g^{ab} in
such a way that

g_{ab}g^{bc} = delta_a^c

where delta_a^c is 0 if a is not c, and 1 if a = c.  In other words, in
matrix lingo g^{bc} is just the inverse of the matrix g_{bc}.

We can then turn a vector v^a into the corresponding covector ---
written simply v_a --- using the metric:

v_a = g_{ab}v^b

Or conversely, we can turn a covector w_a into the corresponding vector
--- written simply w^a --- as follows:

w^a = g^{ab}w_b

These are inverse processes since g_{ab} and g^{ab} are inverse matrices
(thinking of them as mere matrices).  

Now: most engineers act like they live in ordinary Euclidean
3-dimensional space.  The metric for that is just

1 0 0 
0 1 0
0 0 1

so this process of "raising and lowering indices using the metric" ---
i.e. turning vectors into covectors and vice versa --- is completely
invisible!  It just amounts to multiplying a column vector by the
identity matrix and then flopping it over to being a row vector.  

But in special relativity, the metric is

-1 0 0 0
 0 1 0 0 
 0 0 1 0
 0 0 0 1

so one has to be more careful... and in general relativity, the metric
g_{ab} can be ANY invertible symmetric matrix, and it depends on where
you are in spacetime, too!  So you gotta be *really* careful not to
blithely mix up vectors and covectors.

See some stuff Matthew Wiener wrote recently, for a more geometrical
explanation of the difference between vectors and covectors.

Article 98610 (34 more) in sci.physics:
From: john baez
Subject: Re: Gravitational Red Shifts - Real or Apparent ?
Date: 7 Feb 1996 14:04:07 -0800
Organization: University of California, Riverside
Lines: 60
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward
 Green) writes:
>[email-masked] (Jonathan Scott)' wrote: 

>>I suspect that GR in the strong field case implies that it IS possible 
>>to measure an absolute potential, in that it seems to me (without any 
>>formal mathematical basis) that a given amount of mass will apparently 
>>form a black hole more readily in the vicinity of other large masses 
>>which "lower" the potential even if the other masses are distributed so 
>>as to give rise to no overall field.  I find this unsatisfactory, and 
>>it is one reason why I think that GR is only approximately correct. 

>Then you share my original distaste.   

Ugh!  If you think YOU'RE feeling distaste, it ain't nothing, let me
tell you.  Don't take the gravitational potential too seriously.  There
is no bloody way to stand at a point in spacetime and measure the
"gravitational potential" there; this is just a mathematical tool folks
use to solve certain VERY PARTICULARRR general relativity problems in
certain SPECIAL coordinate systems... the only time I've ever seen it
used is in dealing with the spherically symmetric static case!  I have
never seen it used to analyze a case where there are a bunch of massive
objects around and I really really doubt one could do that, because that
will never be spherically symmetric and rarely static.  

Remember: there ain't no "gravitational potential" in Einstein's
equation: there's the metric and things derived from that, like parallel
transport, Riemann curvature, Ricci and Einstein tensors, etc..  You
can't understand GR by thinking about this "gravitational potential"
nonsense, because GR is not about that.

>This was precisely my objection: 
>how can matter know to act differently when local conditions are apparently
>the same?  

It doesn't.  It doesn't need to.

> How can black holes know to form more readily, or clocks run
>more slowly  (in a well defined operational sense, of course) in a lower
>gravitational potential,  independent of the field conditions?  

Oh-oh.  Look.  You talked about one very specific problem involving two
clocks at different elevations in a spherically symmetric static
gravitational field (the earth's), and came up with an experiment, and
you said if the answer was answer number "2", you would say time ran
slower on mountains.  And I warned you that everything you said after
that would be your own fault.  And now you are acting as if it was some
generally true sort of fact, not just an artifact of this one very
special situation, that there is something called the "gravitational
potential" and that clocks "run slower" when it's lower.  And that's
completely wrong, or at least it would be wrong if it were even meaningful.

>Now that I
>understand the framework a little better,  I would say:  According to GR, 
>the local conditions are *not* the same.  The potential must have a local
>physical significance. 

I'd prefer to put it this way: the potential has no bloody physical
significance.   A slight exaggeration, but quite a useful principle to
keep in mind for starters.

Article 98590 (23 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: GR Tutorial Water Cooler
Date: Wed, 7 Feb 1996 19:43:42 +0000
Organization: Oz
Lines: 49
Distribution: world
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
MIME-Version: 1.0
X-Newsreader: Turnpike Version 1.11 <V67Vak$Q$lyzk9NR0hujdE7wCg>

In article <[email-masked]>, Edward Green
<[email-masked]> writes
>[email-masked] (john baez)' wrote: 
> 
>Oh,  hi there,  professor. 
> 
>>I promise to restrain myself in the future, though I gave Ed quite a 
>>tongue-lashing for some things I overheard from that last post.  I agree 
>>that the students need to be able to talk things over freely.  There are 
>>all sorts of things that the "experts" are too familiar with to be able 
>>to explain very well. 
> 
>Tongue-lashing?  Oh, oh.  I don't think I am going to feel good when I see
>the grade for that assignment. 

Oh wow. I just caught up with the postings, at least Demon has. I must
look very slow at replying. This has just arrived. Must catch up with
the reams of stuff that's just come in so I'll be back for a chat later.

<later, panting hard from running down the corridor.>

You are too serious, Ed. I thought you would at least comment on hot
mexican tacos you have eaten that would incinerate any Sri Lankan curry,
or possibly comment on the low price of Cabanas there. Anyway, you don't
want to worry about Baez. Baez's bark is wose than his bite. 

< Oz nervously scans up and down the corridor for signs of the irascible
lecturer and, relieved to see no sign, rubs his hair cautiously.>

Well, I'm definitely less in the dark than I was a week ago. In fact,
it's starting to make some sense. I think a bit more detail on the
derivation etc of the Riemann tensor is in order. Hey Ed <with
enthusiasm> why don't YOU ask him about it. You always get good grades.

Brrrrriiiiiing

Oh dear, back we go.

<Oz turns into the lecture room to find the doughty Baez standing
thoughtfully begind the door. He doesn't like the look in his eye. No,
not at all.>

Ed, hey Ed.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98679 (13 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 7 Feb 1996 19:01:09 -0800
Organization: University of California, Riverside
Lines: 59
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>2.  The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
>>v,w and spits out a number g(v,w), which we think of as the "dot
>>product" or "inner product" of the vectors v and w.  This lets us
>>compute the length of any tangent vector, or the angle between two
>>tangent vectors. 

>>        g(v,w) = g_{ab} v^a v^b

>OK. Doing this on another thread. 
>Urk. g(v,w) is a scalar <ducks>, so one might think this has something
>to do with 'length'. <Dives under desk>.

Look up at what I wrote.  g(v,w) is a scalar, yes, and it's just what we
call the "dot product" of the vectors v and w.  You know about dot
products, right?  [Eyebrows raised expectantly, with a slightly worried
smile playing around the lips.]

>>where as usual we sum over the repeated index a.  Then we "raise an
>>index" and define
>>
>>         R^a_d = g^{ab} R_{bd},
>
>Urgh. 'Raise and index' sounds technical. At least I hope so.
>Not impossible to fathom I think.

Oh, it's mildly technical, but I just showed you how to do it, right
there!  You are always suspecting that I'm talking about something
mysterious when I'm actually laying everything right out on the table.
(Except of course when I'm not.)  

So: you have a 4x4 matrix of numbers R_{bd} (remember, b and d, and all
such indices, run over 0,1,2,3, this being 4d spacetime).  You have
another 4x4 matrix of numbers g^{ab}.  I said a while back that this
matrix is just the inverse of the matrix g_{ab}.  You do know about
inverse matrices, right?  [Nervous smile and slight twitch.]  And I also
said what the matrix g_{ab} was.  Anyway, even if you forget all that
stuff, let me point out how we just "raised an index".  Namely: we cook
up a new matrix R^a_d by multiplying the numbers g^{ab} and R_{bd}, and
then (as usual) adding them up for b = 0, 1, 2, 3, since b is a repeated
index.  We say simply:

  R^a_d = g^{ab} R_{bd}

Note that the left hand side has one index UP while the original Ricci
tensor R_{ab} had both DOWN.  Thus we have "raised an index".  

>>That's it!

>Oh, really? *I* think that's the start, for others at least.

Yeah, but that's it for me.  After you guys understand Einstein's
equation you're on your own.

Article 98680 (12 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 7 Feb 1996 19:05:35 -0800
Organization: University of California, Riverside
Lines: 25
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (ale2) writes:

>Student interupts (many classmates give him a look of disgust)...

>For any spacetime point around, say the sun, is there an infinite
>number of geodesics through that point? 

Yes indeed, basically one for each unit vector through that point.
Remember a spacetime point is a moment and a place.  At any given
moment, at any given place, there are many ways a freely falling rock
could zip through that place at that moment.  The velocity of the rock
(a 4d vector!)  could be any future-pointing timelike vector.  As it
continues to whiz along its worldline traces out a geodesic.  Then
there are the lightlike and spacelike geodesics....

>Is the geodesic of a particle
>at some point in the spacetime around the sun determined once we know
>the velocity of the particle at that spacetime point?

Yes, if we are told the geometry of spacetime, we can completely work
out a geodesic if we know a point it goes through and its tangent vector
(velocity) at that point.  It satisfies a second-order differential
equation, which I have avoided writing down, and you just solve them.

Article 98699 (48 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 7 Feb 1996 20:12:40 -0800
Organization: University of California, Riverside
Lines: 75
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:
>In article <[email-masked]>, john baez <[email-masked]>
>writes
>>In GR we always assume the connection is torsion free, so this whole
>>issue is pretty irrelevant to GR.  Don't blame me for bringing it up...
>>it's all Keith's fault.  :-)

>Why do we assume [the connection] in GR is torsion free?

>A short couple of non-sarcastic information-rich understandable
>sentences would suffice. Enough to place it in a footnote of the brain.

Non-sarcastic, eh?  You must think I'm really mean and nasty, when
actually I am the sweetest, nicest guy you could ever meet.  

Relatively few people understand why in GR we assume the connection ---
the gadget we use to do parallel translation --- is torsion-free.  It's
often presented as a technical assumption not worth trying to understand.
But here's a nice way understanding what torsion-free-ness means (in
conjunction with our other assumptions: that parallel translation be
linear and metric-compatible).  

Take a tangent vector v at P.  Parallel translate it along a very short
curve from P to Q, a curve of length epsilon.  We get a new tangent
vector w at Q.  Now let two particles free-fall with velocities v and w
starting at the points P and Q.  They trace out two geodesics.  Let me
try to draw this:

    |      |
    |      |
    |      |
    ^v     ^w
    |      | 
    P------Q

Remember, this is a picture in spaceTIME.  Here I've drawn what it might
look like in flat Minkowski spacetime, where the geodesics are boring
old straight lines, and I've drawn everything very rectilinearly, since
ASCII is so bad at drawing curves.  Okay.  Now, let's call our two
geodesics C(t) and D(t), respectively.  Here we use as the parameter t
the proper time: the time ticked out by stopwatches falling along the
geodesics.  (We set the stopwatches to zero at the points P and Q,
respectively.)  

Now we ask: what's the time derivative of the distance between C(t) and
D(t)?  Note this "distance" makes sense because C(t) and D(t) are really
close, so we can define the distance between them to be the arclength
along the shortest geodesic between them.  

If, no matter how we choose P and Q and v, the time derivative of the
distance between C(t) and D(t) at t = 0 is ZERO, up to terms
proportional to epsilon^2, then the torsion is zero!  And conversely!

If v got "rotated" a bit when we dragged it over to Q, and things looked
like this:

    |           /
    |          /
    |         /
    ^v       ^w
    |       / 
    P------Q

then the time derivative of the distance would not be zero (it'd be
proportional to epsilon).  In this case the torsion would not be zero.

In GR we assume this kind of "rotation" effect doesn't happen.  In some
other theories of gravity there is torsion.  But there's no experimental
evidence for torsion, so most people stick with GR.

Article 98736 (58 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Thu, 8 Feb 1996 06:57:37 +0000
Organization: Oz
Lines: 56
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In article <[email-masked]>, Doug Merritt <[email-masked]>
writes
>In article <[email-masked]> [email-masked] (john baez) writes:
>>Non-sarcastic, eh?  You must think I'm really mean and nasty, when
>>actually I am the sweetest, nicest guy you could ever meet.  
>
>Having actually met John in the flesh, I figure I should now take
>bids from opposing sides as to his nature. High bid wins.
>

<Oz extracts his foot from his mouth. Makes a mental note to change his
socks more often. His lips pucker in in disbelief at the taste.>

Ah .......

Well I would guess he has a smile on his face quite a bit of the time. I
also guess he likes a fast thought out response to a carelessly framed
comment, that may be cutting at times if you were stupid enough to take
it that way. Obviously he is considerate, you can see that from his
apologies when he inadvertntly steps on some delicate ego. He also does
it on occasion when he has stepped on a non-delicate ego so I guess that
must be right. Among his peers I would guess he delights in utterly
confusing them at every opportunity, probably with a real big grin. If
he has something wrong the smile vanishes and his brow puckers as he
tries to work out how an earth he could have made such a booboo.

He is worried he is getting a reputation as a grumpy and irascible
lecturer. I cannot imagine why that should be (not guilty y'r 'onnor).

He is doing this GR course primarily to make sure he has it absolutely
right for his students. He also hoped to chide them on with a comment
like "Good grief, even Oz has this figured out", but has been woefully
let down. He also hopes to avoid getting caught out by any smart-assed
students who try to do what he did in their position and catch out the
prof.

I suppose in short probably best described as a mischevous sprite.
At least from the subset of Baez as indicated by his sci.physics
persona. I bet this will cause hernias amongst his friends.

<End of Baez ego boost. He doesn't need it anyway.>

>John is cool about mathematical physics, but he's none of
>those poetical things, so I say again, make me a dollar offer. ;-)
>       Doug
>P.S. Just kidding...I like your tutorial series, keep going, by
>all means.

You owe me $5 right or wrong. Sheesh, the risk of posting this on the
group. I must find myself a shrink.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98747 (57 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Thu, 8 Feb 1996 12:56:43 +0000
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
>
>>OK. Doing this on another thread. 
>>Urk. g(v,w) is a scalar <ducks>, so one might think this has something
>>to do with 'length'. <Dives under desk>.
>
>Look up at what I wrote.  g(v,w) is a scalar, yes, and it's just what we
>call the "dot product" of the vectors v and w.  You know about dot
>products, right?  [Eyebrows raised expectantly, with a slightly worried
>smile playing around the lips.]

Sure, doc. Don't worry about it. <Confident persuasive grin>

Nomenclature seems to to be rather well defined and unambiguous. Now I
would expect tensor algebra not to be commutative so I would expect a
clear formulation so as to keep track of this. So I look around
(confusedly as usual) and note:

p(v,w)  : a scalar product of v & w.
q_{ab}  : a 2D (axb) tensor (are the curly brackets compulsory?).
v^a     : a vector. 1D column representation in 'a' dimensions.
r_{abc} : a 3D (axbxc) tensor. URK. How to manipulate this!
r_{abcd}: I don't even want to think about it. Or r_{abcdefgh}!!

Then there are some other possibilities which are also presumably
defined:
v_a     :an upsidedown vector? Covector? Row thingy??
         presumably means something physical?
q^{ab}  :Well, I've never heard of a cotensor. It's something like that.

Then we have some operations:
They all seem to be x^y m_n in structure. In other words one is a
subscript and the other a superscript. There must be rules for this,
they can't commute so m_n x^y must be different. Presumably either p^q
or p_q as the result. On the other hand some will be scalar, some
matrices, unless the rules prevent this. Hmmm, caution prevents me from
guessing a bunch of operations, or commenting on them. Neck on block
enough as it is.

>
>>>where as usual we sum over the repeated index a.  Then we "raise an
>>>index" and define
>>>
>>>         R^a_d = g^{ab} R_{bd},
>>
>>Urgh. 'Raise and index' sounds technical. At least I hope so.
>>Not impossible to fathom I think.
>
>Oh, it's mildly technical, but I just showed you how to do it, right
>there!  You are always suspecting that I'm talking about something
>mysterious when I'm actually laying everything right out on the table.
>(Except of course when I'm not.)  

My thought *exactly*.   :-)

>So: you have a 4x4 matrix of numbers R_{bd} (remember, b and d, and all
>such indices, run over 0,1,2,3, this being 4d spacetime).  You have
>another 4x4 matrix of numbers g^{ab}.  I said a while back that this
>matrix is just the inverse of the matrix g_{ab}.  You do know about
>inverse matrices, right?  [Nervous smile and slight twitch.]

I seem to remember they were easier to say and use, than to calculate.
<He says sunnily, avoiding the question in detail.>

>  And I also
>said what the matrix g_{ab} was.  Anyway, even if you forget all that
>stuff, let me point out how we just "raised an index".  Namely: we cook
>up a new matrix R^a_d by multiplying the numbers g^{ab} and R_{bd}, and
>then (as usual) adding them up for b = 0, 1, 2, 3, since b is a repeated
>index.  

Er, like R^a_d[0_1]=g[0,0]R[0,1]+g[0,1]R[1,1]+g[0,2]R[2,1]+g[0,3]R[3,1] 
and so on for the rest of R^a_d ?  
<Nervous grin, now Oz is twitching too.>

Note. You seem to have kept tabs of your sub and superscipts. 
Ie (ab)/(bd)=a/d > R^a_d, thus beating matrix algebra nicely.
<Another more nervous grin, Oz looks a little pale.>

>We say simply:
>
>  R^a_d = g^{ab} R_{bd}
>Note that the left hand side has one index UP while the original Ricci
>tensor R_{ab} had both DOWN.  Thus we have "raised an index".

<Satisfied smirk appears on Oz's face.>
>  

Yup. The simplest things are the hardest to explain.
Of course one cannot but ask the irritationg question:
"What does this mean"? 
"What is the physical description of this operation."?
Please don't say 'well we put a subscipt up, and the other down".
I mean what is the physical difference between the descriptors:
R_{ab}
R^a_b

Hmmm, the metric g^{ab} is obviously a description in some way of how
things in the g^a direction affect something in the g^b direction sort
of thingy whatsit. 
<Unconvincing smile hoping this cop-out meets with approval.>

>Yeah, but that's it for me.  After you guys understand Einstein's
>equation you're on your own.

<It has clearly been a hard session for the prof. He decides to teach
freshman physics next year. He now knows why the other profs were so
keen to offer him this important postgrad subject. He walks down the
hall shaking his head at the dumness of some of his students, sighing
deeply. It would be something if they appreciated his hard work. Sigh.>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98776 (56 more) in sci.physics:
From: john baez
(SAME) Subject: Re: general relativity tutorial
Date: 8 Feb 1996 09:47:07 -0800
Organization: University of California, Riverside
Lines: 80
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:

>I suppose that's what we are getting here on this thread. A primer.
>
>Hahahahahahahah  ROFL. 
>
>I hope a teeny weeny bit more than that at least, all in all.

In a sense it's a pre-primer, because we are avoiding all the nasty
equations that you'd see in a typical introduction to general
relativity.  As a result, you may be in for a shock when you finally
see, for example, the standard formula for the Riemann curvature in a
textbook.

On the other hand, this pre-primer is more thorough than many textbooks
in certain ways, in that we are getting into the "inner meaning" of a
lot of concepts that books typically gloss over.  Sure, you can extract
this inner meaning from the nasty equations with enough effort, but I
don't think everyone gets around to it.

For example, my attempt to gloss over the concept of torsion failed
miserably: Ed and Bronis kept asking what the hell it meant to parallel
transport a vector without rotating it, then Oz came up with a nice way
of describing, then Darryl described the "Schild's ladder" construction
for torsion-free parallel translation, and finally Oz forced an operational
definition of "torsion" out of me.  This operational definition makes it
clear (I hope) exactly what physical assumption is built into the
"torsion-free" condition on parallel transport in general relativity.
On the contrary, if you look in Wald's General Relativity, which is in
many respects an excellent book, the only thing you will see about this
is:

"5.  Torsion free: for all f, 
          
                         D_a D_b f = D_b D_a f

This condition is sometimes dropped, and indeed there are theories of
gravity where it is not imposed.  However, in general relativity the
derivative operator is assumed to satisfy condition 5, and, unless
otherwise stated, all derivative operators in this book will be assumed
to be torsion free."

This leaves reader completely in the dark as to the meaning of torsion,
and why we assume it vanishes.  

So while we won't get very far in some respects, we will be way ahead of
the crowd in terms of really understanding what we've covered.

"Petrified knowledge" is a constant danger in science: someone figures
something out that's very important and interesting, and then it becomes
reduced in the textbooks to a brief remark without any clue to the
reader about what's really going on, or turned into an equation whose
inner meaning can only be unpacked with a lot of effort.  There's no way
to get the "inner meaning" out of textbooks without making that effort!  
For example, it was only in this "tutorial" that I finally understood
the geometrical essence of the Ricci tensor; I always knew I should do
it someday, but I never saw it spelled out anywhere, and I was always
too lazy to do it.  That's pretty terrible.  The inner meaning --- or
*one* inner meaning, anyway, because there are usually many layers of
inner meaning --- lay lurking in Raychaudhuri's equation, which only a
fairly dedicated reader of Wald would ever get to.  But once it's
unearthed it's not all that complicated.  

By the way, thanks go out to Dan McGuirk for catching a factor-of-two
error.  Here's the new improved "inner meaning of Einstein's equation":

Take any small ball of initially comoving test particles in free fall.
Work in the local rest frame of this ball.  As time passes the ball
changes volume; calculate its second derivative at time zero and divide
by the original volume.  The negative of this equals 1/2 of: the energy
density at the center of the ball, plus the flow of x-momentum in the x
direction there, plus the flow of y-momentum in the y direction, plus
the flow of z-momentum in the z direction.

Or:

R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}]

Article 98782 (55 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 8 Feb 1996 10:38:55 -0800
Organization: University of California, Riverside
Lines: 106
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:

>It would be really nice to have some simple examples of a Riemann tensor
>for a suitable space. I think 2D would do. Say of a sphere or other
>straightforward object so one can get an idea of what a real one would
>look like. At some point a few simple concrete numbers is helpful for
>clarity, if they are appropriately chosen.

[The man in a sorcerer's cap hems and haws for a minute and then speaks:]

2d is good for some purposes, boring for others.  In 2d it only takes
one number to describe the Riemann curvature at each point, so there is
the same amount of information in the Riemann curvature tensor, the
Ricci tensor, and the Ricci scalar.  So we can't understand the
differences between these concepts very well in 2d.

Let me describe the Ricci scalar, R, in 2d.  This is positive at a given
point if the surface looks locally like a sphere or ellipsoid there, and
negative if it looks like a hyperboloid --- or "saddle".  If the R is
positive at a point, the angles of a small triangle there made out of
geodesics add up to a bit more than 180 degrees.  If R is negative, they
add up to a bit less.

For example, a round sphere of radius r has Ricci scalar curvature R =
2/r^2 at every point.  [With a click of his fingers, a sphere of radius
r appears on it, with a small triangle drawn on it, edges bulging
slightly.]  

But how about the Riemann *tensor* on a round sphere?  Well, for this we
need some coordinates.  Let's use the usual spherical coordinates
theta, phi.  Since there is actually disagreement at times on which is
which, I remind you that for me theta is the longitude, running round
from 0 to 2pi, while phi is the angle from the north pole, going from 0
to pi.  [Coordinate lines appear on the sphere, which floats back and
forth playfully.]

When we write something like g_{ab} or R^a_{bcd}, the indices will go
from 1 to 2, with "1" corresponding to theta and "2" corresponding to
phi.  The components of the metric g are then:

g_{11} g_{12}  =  r^2 sin^2(phi)   0
g_{21} g_{22}           0         r^2

What does this mean?  Remember this matrix lets us calculate dot
products of vectors via

g(v,w) = g_{ab} v^a v^b

So if I take the vector v to be, say, 

1
0

so that its theta component is 1 and its phi component is zero
(v^1 = 1, v^2 = 0), we get --- hurry to it, tensor jocks! ---

g(v,v) = r^2 sin^2(phi)

This means that its length squared is r^2 sin^2(phi), or in other words,
its length is r sin(phi).  That's supposed to make sense.  You do
remember your spherical coordinates, right?  [Pained, worried look.
Helpful review of trig formulas flashes by on the sphere.]

But you wanted to know the Riemann tensor of the sphere, not the metric
or the Ricci scalar!  Well, okay, so we take the metric and feed it into
this machine...  [scurries behind a curtain; loud banging noises ensue,
followed by a deafening explosion and a puff of smoke; returns somewhat
blackened but smiling]... and it computes the Riemann tensor for us.
Don't worry about that machine in the other room just yet, someday you
too may learn to use it... or maybe not.

So, the Riemann tensor has lots of components, namely 2 x 2 x 2 x 2 of
them, but it also has lots of symmetries, so let me tell just tell you
one:

R^2_{121} = sin^2(phi)

Remember what this means!  {Shuffles through a vast stack of old papers,
plucks one out, and reads:]

"Say we take a tangent vector pointing in the d direction and carry it
around a little square in the b-c plane.  We go in the b direction until
the b coordinate has changed by epsilon, then we go in the c direction
until the c coordinate has changed by epsilon, then we go back in the b
direction until the b coordinate is what it started out as, and then we
go back in the c direction until the c coordinate is what it started out
as.  Our tangent vector may have rotated a little bit since space is
curved.  Its component in the a direction has changed a bit, say

                           -epsilon^2 R^a_{bcd}  "

So R^2_{121} tells us how much a vector pointing in the theta direction
swings over towards the phi direction when we parallel transport it
around a little square.  If you visualize it... [little vector pointing
east appears near the equator on the hovering globe; moves east, then
south, then west, and north back to where it was, but has rotated a wee
bit southwards in the process]... you'll see this makes sense.  Unless I
got a sign wrong....  [North pole on globe flips over to south pole, east
flips over to west, globe lets out a snicker and disappears in a puff of
smoke.]  

Article 98817 (53 more) in sci.physics:
From: Daryl McCullough
(SAME) Subject: Re: General relativity tutorial
Date: 8 Feb 1996 10:44:51 -0500
Organization: Odyssey Research Associates, Inc., Ithaca NY
Lines: 23
Distribution: world
NNTP-Posting-Host: www.oracorp.com

I have a question or comment about the Ricci tensor and the
missing information needed to go from it to the full Riemannian
tensor. If you are in vaccuum nearby a large mass, then the
stress-energy tensor, and therefore the Ricci tensor, will be
zero. However, it seems to me that the presence of mass nearby
is indicated by geodesic deviation (falling particles tend
to get closer together as they get closer to the center of the
mass). This geodesic deviation is determined by the non-Ricci
part of the Riemannian curvature tensor (the Weyl tensor). So
information about mass-energy is really present in the Weyl
tensor as well as the Ricci tensor. Maybe it is like Gauss's
theorem for electrostatics. Although the divergence of E is
zero everywhere outside of a charged particle, knowledge of
E on a surface can tell us that there is charge inside a surface
(and therefore nonzero divergence inside). Similarly, maybe
knowledge of the Weyl tensor on a surface can tell us that
there is mass-energy inside the surface (and therefore a nonzero
Ricci tensor) even though the stress-energy tensor is zero
everywhere on the surface.

Daryl McCullough
ORA Corp.
Ithaca, NY

Article 98772 (52 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: Tensors for twits please.
Date: Wed, 7 Feb 1996 18:08:01 +0000
Organization: Oz
Lines: 236
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In article <[email-masked]>, john baez <[email-masked]>
writes

>>Oz
>>Of course it would help if I had the faintest idea what a covector was.
>
>I said it was just slang for a (0,1) tensor, a guy who eats a vector and
>spits out a number, in a linear way.
>
>This "co" stuff is an fundamental notion in mathematics and physics.
>It's sometimes called "duality", or the "covariant/contravariant"
>distinction.  
>
>Say we have a vector v and a covector f.  Then 
>
>f(v) = x
>
>is a number.  We usually think of f eating v and spitting out the number
>x, but we can equally well say that v is eating f and spitting out the
>number x!  

Yup, noticed that when I looked at matrices, which is all a bit vague.
So this is why you don't use 'rows and columns' of course, the objects
stay as objects throughout. Excellent. One minor gripe I've always had,
why does the first index go down the column, in the 'y' direction??

>So in some sense covectors are just as basic as vectors.  
>
>Ponder this, but don't worry about it; it'll take a long time to get
>this point.  

(I re-read my reply. Just a note to point out that I follow that in
f(v)=x one can look at it that *either* v eats f, or f eats v)

Yes, but I feel that covectors should do slightly different things to
vectors. For one thing if you take your example above then f(v) is a
quite different beastie to v(f) and indeed one or the other may not even
exist. Er, I suspect. (Prays hard)

>But it's lurking in the "row vector" vs "column vector" stuff you always
>see in old-fashioned linear algebra books.  You can think of a row
>vector as a guy that's dying to eat column vectors and spit out numbers:
>
>[a b][c] = ac + bd
>     [d]
>
>but if you change your point of view you can think of it the other way
>around: the column vector eats the row vector.

Agreed. How about [c][a b]=[2x2] it's different.
                  [d]

>>A very simple and obvious question. Do we have to be very caerful WHAT
>>our tensor is used for. Ie just row>row not col>col, or does the *same*
>>tensor do 'double duty' in some way.
>
>There are two levels to this question, because you are reading an old
>book that talks about "row vectors" and "column vectors", and I am
>struggling to translate it into the modern lingo of "vectors" and
>"covectors", but the translation is a tricky business, in the way that
>translating foreign languages always is.  In fact, I urge you to not pay
>much attention to that book, and just let me explain everything.  I will
>wind up doing LESS work that way.

OK, boss. Consider it banned. Well just a *little* peek. <smack: OW!>

>Anyway, I can confidently say: yes, a (1,1) tensor does double duty: by
>definition it serves to turn a vector into a vector, but if we are
>clever we can use it to go turn a covector into a covector.
>
>On the other hand, something like a matrix is even more
>multipurpose: we can use a matrix to stand for either a (1,1) tensor, or
>a (0,2) tensor (like the metric!) or a (2,0) tensor.  

Ugh. But these are *different*! You'll get mixed up terribly if you let
this go on. Need a new notation - and fast, Batman!

>This is because
>the PreCambrian dialect of "row vectors", "column vectors" and
>"matrices" fails to make some of the finer-grained distinctions that the
>modern lingo does.  Why?  Because it's deceptively easy to turn a column
>vector into a row vector: just flop it over!  E.g.
>
>[2]
>[3]
>
>becomes   
>
>[2 3]
>
>The modern approach is set up to make this impossible to do UNLESS you
>use the metric!!!!!!!!!   You can't just "flop over" a vector and get a
>covector.  You do this using the metric (in a way I'll eventually
>explain.)  
>
>Again, file this stuff away in your brain, or somewhere, but don't let
>it get you down now.

No, I like this. I didn't bother with matrices years ago because they
didn't 'feel' right, and anyway I could the things I needed to do with
vanilla flavoured vectors. This looks better, at least vectors stay as
vectors.

>>>Oh well, education occurs even when one least wants it.
>
>>If I didn't want it, I would have given up weeks ago. Have you any idea
>>how many concepts I have tried to take in over the last few weeks?
>
>Come come, I didn't say you didn't want some education.  I was making a
>cryptic joke about how you are now busily learning about the history of
>different mathematical notations for tensors!  This means you are going
>to learn a whole bunch of extra concepts which you don't even need to
>know to understand general relativity.  They are good to know, but will
>perhaps be the straw that breaks the camel's back.

Hey, it's a nice break from GR for a while, let some of those conceots
sink in a bit. Anyway it's rather fun, I am just a little disappointed
that in a few months I won't remember it well enough and get told off!
Anyway, at least there's no exam at the end!

>Okay, let me spell it out.  I was really hoping someone else would do
>it, since this involves typing long dull rows of symbols.  But okay.
>Let's say we're in 4d spacetime.  Then indices like a,b etc.  stand for
>0,1,2, or 3.  When we sum over them, we sum from 0 to 3.  So for
>example, when I speak of "the metric g_{ab}", what I mean is a
>particular batch of numbers, namely:
>
>g_{00}   g_{01}   g_{02}   g_{03}
>g_{10}   g_{11}   g_{12}   g_{13}
>g_{20}   g_{21}   g_{22}   g_{23}
>g_{30}   g_{31}   g_{32}   g_{33}
>
>Note how both a and b go from 0 to 3.  Each thing like g_{21} above is
>just a number.  So for example g_{ab} might be the matrix
>
>  -1       0        0        0
>   0       1        0        0                       (*)
>   0       0        1        0
>   0       0        0        1
>
>in which g_{22} = 1 and so on.  Similarly, when I refer to "a vector
>v^a", I mean a column vector like
>
>  v^0
>  v^1
>  v^2
>  v^3
>
>A random example would be the vector
>
>   1
>   2                                                 (**)
>   0
>   9
>
>In this example we'd have v^3 = 9.   Similarly, when I refer to "the
>vector w^a" I mean another such thing, like
>
>  w^0
>  w^1
>  w^2
>  w^3
>
>An example would be, say
>
>   2
>   2                                             (***)
>   0
>   0
>
>Now when I write something like g_{ab} v^a w^b, I mean that I multiply
>the number g_{ab} by the number v^a and by the number w^b, and then
>SUM UP letting a and b range from 0 to 3.  So this is short for
>
>g_{00}v^0w^0 + g_{01}v^0w^1 + g_{02}v^0w^2 + g_{03}v^0w^3 + 
>.... 4 more of these bloody things .... +
>.... 4 more of them ....                +
>g_{30}v^3w^0 + g_{31}v^3w^1 + g_{32}v^3w^2 + g_{33}v^3w^3
>
>Okay.  Here's a test!!!  Take the example of the tensor g_{ab} that I
>gave, back at (*), and the example of v^a I gave, back at (**), and the
>example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is.
>The answer is some specific number; what is it?

Oooh. I didn't expect this. No exam indeed, humpf. OK, this looks to be
fun. Now I see it out longhand I can see that its:

--        a b
\  g    v  w            oh dear. You have said this before, 
/   ab                  but now I follow the notation.
--

so g_{ab} above is mostly 0's. Only the diagonal has non-zero indices so
we can simplify it as:

g_{00}v^0w^0 + g_{11}v^1w^1 + g_{22}v^2w^2 + g_{33}v^3w^3

Hmmmm, only the diagonal is non-zero. So no nasty cross terms. Hmmm this
wouldn't be flat space (I suppose in spacetime this would be Minkowski
space) would it? Must check later. Anyway

g_{ab}v^aw^b = -1x1x2 + 1x2x2 + 1x0x0 + 1x9x0 = -2 + 4 = 2

Hope to god this is right. Hmmm also get some idea why you are so
pernickity about your indices and whatsits. Like

g(uv) is different to g_{ab} and to a vector g^a, I presume.

Anyway lets just check using (for the last time, promise) the old
fashioned matrix formulation.
[1,2,0,9][2]= whaddayaknow=1x2+2x2+0x0+9x0=2+4=6 BUT no g_{00}=-1
         [2]               ^time thingy should be negative 
         [0]               so works out as +2 as before.
         [0]               so it looks like flat spacetimey thingy.

Phew, at least he stared with an easy one <wipes sweat from brow>.

>This is how you actually work out the inner product of two vectors v and
>w:
>
>g(v,w) = g_{ab} v^a w^b.

Er, ahem, cough, cough, er, "Prof, prof, er, prof sir?"
don't like to ask, but how do I handle a tensor that turns two vectors
into another vector, and what is the notation, please sir? 

<smack! smack!: ooooowch! Oz hurriedly leaves the lecture hall.>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 98786 (51 more) in sci.physics:
From: john baez
(SAME) Subject: Re: Tensors for twits please.
Date: 8 Feb 1996 11:20:03 -0800
Organization: University of California, Riverside
Lines: 195
NNTP-Posting-Host: guitar.ucr.edu

I answer only a random subset of the vast list of questions posed by Oz:

In article <[email-masked]> [email-masked] writ
es:
>One minor gripe I've always had,
>why does the first index go down the column, in the 'y' direction??

To keep people who worry about this sort of thing out of mathematics.

Seriously, there's no deep reason to this or any other "row/column",
"up/down", "right/left" convention.  My student James Dolan uses
completely different conventions in almost every respect, and in
particular I bet he has the first index go across the rows.  But he
still seems to get along perfectly fine.  I greatly value the ability to
communicate with the brainwashed masses, so I just learn how do whatever
everyone else does, as far as notation is concerned.  This may involve
learning several incompatible different notations, but so what?  When in
Rome, do as the Romans do.

There is probably an interesting historical answer to your question,
but I don't know it.

>>So in some sense covectors are just as basic as vectors.  
>>
>>Ponder this, but don't worry about it; it'll take a long time to get
>>this point.  

>(I re-read my reply. Just a note to point out that I follow that in
>f(v)=x one can look at it that *either* v eats f, or f eats v)

>Yes, but I feel that covectors should do slightly different things to
>vectors. For one thing if you take your example above then f(v) is a
>quite different beastie to v(f) and indeed one or the other may not even
>exist. Er, I suspect. (Prays hard)

Oh yes, vectors and covectors are very very different in character, this
is a big theme in modern mathematics.  When you have a metric around you
can freely convert one to the other, by lowering or raising indices.
But they have a different geometrical significance... and the process of
raising and lowering indices has a very geometrical description as well.
If you think the "index gymnastics" I'm describing seems somewhat
mechanical and removed from the physics and geometry, that's just
because I was focussing on getting you going on the mechanics.  There is
also a lot to say about the inner meaning of it all.  Matthew Wiener
said a bunch so I quote him... this may or may not make sense to you
just yet:

"A vector is just an arrow pointing somewhere.  A covector
(in R^3) is a dissection of R^3, consisting of parallel planes, such
that they are labelled "linearly", with 0 for the plane passing through
the origin, one of the other planes labelled 1, the one twice as far
labelled 2, etc.  These may be described with 3 coordinates (a,b,c)
by letting ax+by+cz=1 be the plane labelled 1.  (0,0,0) is the plane
at infinity, so to speak.

So covectors form a 3-dimensional space.  There is a natural duality
between covectors and vectors, namely < (a,b,c) | (x,y,z) > = ax+by+cz.
In terms of the starting R^3 dissection, this just identifies the label
of the plane that vector (x,y,z) reaches.

The unit covectors are commonly denoted dx,dy,dz.  Although you were
taught dx is "differential x", here its "dual x".  But the notation
was chosen this way deliberately.  1-forms Adx+Bdy+Cdz are naturally
understood as covectors, not vectors.

For example, given a scalar function f on R^3 (ie, f(x,y,z)=some number),
we can take its gradient.  You probably learned grad f = (@f/@x).i + ...

What piffle.

"Really" grad f is df = (@f/@x).dx + ....  It forms a covector field, not
a vector field.  So what does this really look like?  It's ultimately just
a contour map of f, custom linearized at each point.

This is easier to imagine in 2D.  Take a contour map of a scalar function
"height", assumed differentiable.  Pick a point.  Kind of curvy, right?
Blow up the neighborhood 1000 times, notice how the nearby contours are
a lot straighter.  So just _redraw_ them as straight, extend to a covector
worth of lines dissecting R^2, and scale the line labelled 1 down by 1000.
This is the dual version of (F(x+.001)-F(x))/.001 we all know and love.

That this is so direct, the moral equivalent of drawing tangent lines, is
all the proof anyone really needs that gradients are covectors."

To continue.....

>>On the other hand, something like a matrix is even more
>>multipurpose: we can use a matrix to stand for either a (1,1) tensor, or
>>a (0,2) tensor (like the metric!) or a (2,0) tensor.  
>
>Ugh. But these are *different*! You'll get mixed up terribly if you let
>this go on. Need a new notation - and fast, Batman!

Yes, well, that's why we write (1,1) tensors as things like g^a_b, 
(0,2) tensors as things like g_{ab}, and (2,0) tensors as things like
g^{ab}.   

>Hey, it's a nice break from GR for a while, let some of those conceots
>sink in a bit. Anyway it's rather fun, I am just a little disappointed
>that in a few months I won't remember it well enough and get told off!

What, you mean you're planning on forgetting all this stuff?

>Anyway, at least there's no exam at the end!

That's what YOU think.  In fact...

>>So for example g_{ab} might be the matrix
>>
>>  -1       0        0        0
>>   0       1        0        0                       (*)
>>   0       0        1        0
>>   0       0        0        1

>>[v^a might be the vector]
>>
>>   1
>>   2                                                 (**)
>>   0
>>   9

>>[and w^b might be the vector]
>>
>>   2
>>   2                                             (***)
>>   0
>>   0

>>Okay.  Here's a test!!!  Take the example of the tensor g_{ab} that I
>>gave, back at (*), and the example of v^a I gave, back at (**), and the
>>example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is.
>>The answer is some specific number; what is it?

>Oooh. I didn't expect this. No exam indeed, humpf. OK, this looks to be
>fun. Now I see it out longhand I can see that its:
>
>--        a b
>\  g    v  w            oh dear. You have said this before, 
>/   ab                  but now I follow the notation.
>--
>
>so g_{ab} above is mostly 0's. Only the diagonal has non-zero indices so
>we can simplify it as:
>
>g_{00}v^0w^0 + g_{11}v^1w^1 + g_{22}v^2w^2 + g_{33}v^3w^3
>
>Hmmmm, only the diagonal is non-zero. So no nasty cross terms. Hmmm this
>wouldn't be flat space (I suppose in spacetime this would be Minkowski
>space) would it? Must check later.

Yes, that metric g^{ab} I wrote down is the metric for Minkowski
spacetime.  I was subtly trying to get this metric to seem familiar.

>Anyway
>
>g_{ab}v^aw^b = -1x1x2 + 1x2x2 + 1x0x0 + 1x9x0 = -2 + 4 = 2

Yes indeed.  Correct!!  You are picking up the art of index juggling.  

>Hope to god this is right. Hmmm also get some idea why you are so
>pernickity about your indices and whatsits. Like
>
>g(uv) is different to g_{ab} and to a vector g^a, I presume.

Aheh, yes, by the way we write g(u,v) with a comma in there when we
think of the metric g as a guy waiting to swallow two hapless vectors v
and w and spit out their inner product g(u,v).  

>Er, ahem, cough, cough, er, "Prof, prof, er, prof sir?"
>
>I don't like to ask, but how do I handle a tensor that turns two vectors
>into another vector, and what is the notation, please sir? 

><smack! smack!: ooooowch! Oz hurriedly leaves the lecture hall.>

You have a strong masochistic streak, Oz.   This may actually come in
handy when learning about tensors.  The notation (using indices) for a
tensor that turns two vectors into one is --- well, say our tensor is
called H --- H^c_{ab}.  One index downstairs for each input vector, one
upstairs for each output vector!  If we were in 4d spacetime this
thing could be viewed in a lowbrow way as a 4 x 4 x 4 array of numbers.
How does it act on two vectors u and v to give the vector H(u,v)?  
Note: now H(u,v), the output, is a vector!  So it has components
H(u,v)^c where c goes from 0 to 3.  And what are they?  Here:

H(u,v)^c = H^c_{ab} u^a v_b

where as usual we sum on repeated indices (and, to keep us from screwing
up, repeated indices only come in pairs, one up and one down).

Article 98845 (3 more) in sci.physics:
From: Emory F. Bunn
Subject: Re: GR Tutorial Water Cooler
Followup-To: sci.physics
Date: 8 Feb 1996 20:47:26 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 43
Distribution: world
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Edward Green <[email-masked]> wrote:
>We've been bandying about the "Weyl" tensor recently,  but I seem to have
>missed its definition.  Did anybody write it down? 

I don't think so.  That's because the actual definition is kind of
nasty to write down.  You take the full Riemann tensor R^a_{bcd} and
subtract off a few things from it (I forget exactly what), and you get
the Weyl tensor C^a_{bcd}.  There must be a nice geometric definition
of it in terms of javelins and coffee grounds, but I don't know what
it is; I've only seen the ugly, index-based definition.

>I picked up on the claim that it captures the *other* ten independent
>components of the Riemann,  the ones we didn't use in the Einstein. 
> 
>But what's its rank?  Same as the Einstein?  Is it antisymmetric? 

If you only knew about four-dimensional geometry, you would be tempted
to guess that it was rank 2 like the Ricci tensor, since in four
dimensions Ricci and Weyl have the same number of pieces of
information.  But things aren't that simple, as you can see if you
look at how many pieces of information the various tensors contain in
different numbers of dimensions:

Dimensions   Riemann  Ricci  Weyl

    1           0       0     0
    2           1       1     0
    3           6       6     0
    4          20      10    10

and so on.  (Do I have all these numbers right?  Something doesn't
look right to me.  Anybody want to confirm them?)

It turns out that Weyl is a rank-4 tensor like Riemann, but it has
some extra conditions it has to satisfy, which is why it has fewer
bits of information hidden in it.  Basically, Weyl is "trace-free,"
which means that if you try to contract Weyl to a rank-2 tensor in the
same way you contract Riemann to get Ricci, you always get zero.
(Ricci is R^a_{bad}.  Try to perform the same trick on Weyl, and you
get C^a_{bad}, which turns out to be identically zero.)

-Ted

Article 98867 (7 more) in sci.physics:
From: john baez
Subject: Re: GR Tutorial Water Cooler
Date: 8 Feb 1996 18:00:55 -0800
Organization: University of California, Riverside
Lines: 78
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>,
>Edward Green <[email-masked]> wrote:
>>We've been bandying about the "Weyl" tensor recently,  but I seem to have
>>missed its definition.  Did anybody write it down? 

No. 

>I don't think so.  That's because the actual definition is kind of
>nasty to write down.  You take the full Riemann tensor R^a_{bcd} and
>subtract off a few things from it (I forget exactly what), and you get
>the Weyl tensor C^a_{bcd}.  There must be a nice geometric definition
>of it in terms of javelins and coffee grounds, but I don't know what
>it is; I've only seen the ugly, index-based definition.

I don't yet know a beautiful geometrical definition, though I'm working
on it (in 4d).  So far I think of it as just "the stuff about curvature
that's not in the Ricci tensor" --- as Ed recalls:

>>it captures the *other* ten independent
>>components of the Riemann,  the ones we didn't use in the Einstein. 

But another way to think of it is that we take R^a_{bcd} and subtract
off stuff to make it trace-free on all its indices, while still keeping
all the symmetries the Riemann tensor has.  We get a gadget they call
C^a_{bcd}.

>>But what's its rank?  Same as the Einstein?  Is it antisymmetric? 

Same rank as the Einstein, same symmetries, only now it's "trace-free on
all the indices", meaning that C^a_{bad} = 0, and C^a_{acd} = 0, and
C^a_{bca} = 0.  The Einstein tensor satisfied the last of these
properties already, by virtue of its symmetries, but R^a_{bad} = R_{bd}
was the Ricci tensor, and R^a_{bca} was just minus the Ricci tensor.  So
what we have in effect done in forming the Weyl tensor is to take the
Riemann tensor and subtract out the stuff that gives the Ricci.

>If you only knew about four-dimensional geometry, you would be tempted
>to guess that it was rank 2 like the Ricci tensor, since in four
>dimensions Ricci and Weyl have the same number of pieces of
>information.  But things aren't that simple, as you can see if you
>look at how many pieces of information the various tensors contain in
>different numbers of dimensions:

>Dimensions   Riemann  Ricci  Weyl

>    1           0       0     0
>    2           1       1     0
>    3           6       6     0
>    4          20      10    10
>
>and so on.  (Do I have all these numbers right?  Something doesn't
>look right to me.  Anybody want to confirm them?)

It looks right to me.  For dimension n, the Riemann has n^2(n^2 - 1)/12
independent components.  (I wrote my book on general relativity in part
so I could easily look up this kind of thing!)  The Ricci, being an n x
n symmetric guy, has n(n+1)/2 --- except in dimensions 1 and 2,
amusingly!!  Perhaps those exceptions in dimension 1 and 2 were what was
bugging Ted.  In dimension 1 there is no curvature at all.  In dimension
2 it turns out that R_{ab} = (1/2)R g_{ab}, so there is just one degree
of freedom in the Ricci tensor.  (By the way, this identity wreaks havoc
for GR in 2 dimensions --- guess why!)

I think there is something mystically important about Weyl and Ricci
having the same number of independent entries in dimension 4, maybe
something to do with the "duality" symmetry present in 4d, but I dunno.

>It turns out that Weyl is a rank-4 tensor like Riemann, but it has
>some extra conditions it has to satisfy, which is why it has fewer
>bits of information hidden in it.  Basically, Weyl is "trace-free,"
>which means that if you try to contract Weyl to a rank-2 tensor in the
>same way you contract Riemann to get Ricci, you always get zero.
>(Ricci is R^a_{bad}.  Try to perform the same trick on Weyl, and you
>get C^a_{bad}, which turns out to be identically zero.)

Oh, whoops, Ted already said this stuff.  Well --- it's like he said!

Article 98861 (82 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 8 Feb 1996 17:32:19 -0800
Organization: University of California, Riverside
Lines: 62
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:
>
>Yup. The simplest things are the hardest to explain.
>Of course one cannot but ask the irritating question:
>"What does this mean"? 
>"What is the physical description of this operation."?
>Please don't say 'well we put a subscipt up, and the other down".
>I mean what is the physical difference between the descriptors:
>R_{ab}
>R^a_b

Well, consider the simplest case first: what's the difference between
the vector v^a and the covector

v_a = g_{ab}v^a ?                                      *

The vector v^a is a tangent vector.  It's a little arrow pointing
somewhere.  Easy to visualize THAT.

The covector v_a is really a (0,1) tensor, a tensor that eats one vector
for breakfast and spits out a number.  Say it eats the vector w^b.  What
number does it spit out?

It spits out the number v_a w^a.  That's by definition... but this definition
is really supposed to remind Oz of how he took a row vector and a column
vector and got a number from them.

What is this number?  Well, by equation * up there, it's

g_{ab} v^a w^b.

But remember, this is just the inner product of the tangent vectors v
and w!  So:

Given a vector v, the corresponding covector is the machine that
eats a vector w and spits out the inner product g(v,w) of v and w.

So how do we visualize that?  Well, one nice way is to visualize the
covector v_a is as a bunch of planes stacked up, all orthogonal to v^a.
I can't draw this but in my book (and in MTW) there are lots of pictures
of this sort of thing, which gets you to the real geometrical
significance of covectors.  Matt Wiener has already explained it on
this thread, so I won't say more now... even though it's really important.

>Hmmm, the metric g^{ab} is obviously a description in some way of how
>things in the g^a direction affect something in the g^b direction sort
>of thingy whatsit. 

What the hell is the "g^a direction"?  The metric is g_{ab}; you appear
to have raised an index (fine) and ripped the other one off and thrown
it out (not fine). 

Maybe you mean just "the a direction" or the "b direction" --- a random
couple of coordinate directions?

><Unconvincing smile hoping this cop-out meets with approval.>

Not quite.

Article 98908 (52 more) in sci.physics:
From: Keith Ramsay
Subject: Re: General relativity tutorial
Date: 8 Feb 1996 19:49:25 GMT
Organization: Boston College
Lines: 28
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, [email-masked] (john baez) wrote:

|Now you might wonder about what all that "x-momentum in the x direction"
|stuff really means.   I don't have anything thrillingly insightful to
|say about this except that, for a perfect fluid, each of these terms is
|just the pressure!  I will quit here for now, leaving Michael Weiss to
|explain why that's true.  

I have a prosaic example to add to the explanation. You know those little
toys they have, with a row of metal balls hung on threads? You pull the first
little ball to the side, and let it go. The mechanics of the toy are such
that the ball you let go comes to a stop in its resting position, and the
other balls remain where they were (more or less), excepting the last one, 
which swings up on the other side.

This is an example of momentum in a given direction being "transported"
in that direction. It is of course by means of pressure of the balls
in the series upon each other that the momentum is "transported" from
one end to the other.

When you have pressure, you can think of it as the collective effect
of a lot of little particles being bounced back and forth. When you
switch to other coordinate systems, the energy of the system is
different, in a way which depends upon the distribution of velocities 
of the particles, even if the average is fixed.

Keith Ramsay

Article 99051 (39 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 9 Feb 1996 13:40:06 -0800
Organization: University of California, Riverside
Lines: 181
NNTP-Posting-Host: guitar.ucr.edu

[Oz knocks on the wizard's door, and then opens it.  He sees the wizard
juggling dozens of glowing Greek and Roman letters in complicated,
everchanging patterns.  The wizard catches sight of Oz, frowns, and the
letters explode in a blinding white flash.  He motions for Oz to come
in.]

So, how did you do on your homework?  You were learning about
tensors... right?  What did you learn?

In article <[email-masked]>
[email-masked] writes:  

>So the first index of the rank is the
>output form(s) (with 0=scalar), and the second is the input forms(s).

>In the notation the subscript is the number of input vectors and the
>superscript the number of output vectors with a scalar unscripted.

Right.

>There also seems to be rules about taking products.

Yes, you just multiply numbers the usual way, and then sum over
repeated indices when one is up and one is down.  

Some extra advice, young fellow: never repeat an index more than
twice, and when you do repeat one, always have one up and one down.
E.g., never have two b's that are both superscripts, or two c's that are
both subscripts, or three d's.  This rule has some mathematical
significance, but also it allows one to catch 90% of the infinite number
of errors one makes in complicated tensor calculations, since most such
errors involve accidentally writing the wrong letter as a superscript or
subscript somewhere.
  
>>What, you mean you're planning on forgetting all this stuff?

>*Certainly not* planning. No sir. Realistically if they aren't used
>regularly (particularly without doing a few thousand examples) the
>knowledge fades. Dammit!

Okay, every month or so I'll ask you a few questions, then.

>I read Wieners piece when he posted it. I decided to look into it more
>carefully when my mind was completely clear. Whilst not impossible it is
>not entirely trivial. 

Yes, it deals with another approach to the subject than the one I have
been taking, so it'll take some work to see how it all relates.  It's
not particularly urgent to understand why covectors look like little
stacks of hyperplanes, but it's very important in the long run.  Being
able to visualize stuff is very handy.

>...no tensors in economics in her day, not even vectors...

Hmm, there sure are these days!  Back in school I took an intro to
mathematical economics and there was plenty of such stuff.

>>H(u,v)^c = H^c_{ab} u^a v_b

>Stunned! I suppose you expect me to work out a few terms to see if I
>have it right? 

Only if you want to.

>B*astard!!!!  

Ah, there's nothing as inspiring as the eagerness of the human mind to
learn. 

>Now, lets see. Hmmmm. OhMyGodIwillHaveToUseSomePen&Paper
>Sheef. This must be getting technical.

Yes, you'll find that paper and pencil do come in handy occaisionally when
studying general relativity.  

>Hmmm, odd. I would have expected H(u,v)^c=H^c_{ab}u^a v^b so the indices
>sort of cancel, a & b that is leaving a component H(u,v)^c. I presume a
>notation as neat as this seems to be would do things like that, but I'm
>obviously wrong. Again. 

No, I slipped!  [Clutches his forehead and grimaces in digust.]   Sorry,
I meant exactly what you expected: 

H(u,v)^c= H^c_{ab} u^a v^b 

It was a typo!

><sigh> 

>Of course rotten profs that have the
>tensor called H^c_{ab} and the result called H(u,v)^c with components
>H(u,v), well. Not only that, is there a teeny typo? Surely the
>components must be scalars [H(u,v)], but the vector must be a vector
>H^c?? Anyway for sanity I shall call the tensor K^c_{ab}

Okay, that's fine.  Let me clarify.  In slick, abstract notation, u is a
vector, v is a vector, H is a (1,2) tensor, and when H eats u and v we
get a vector H(u,v).  Nice and clean, no?

Okay, but to compute these things, we might want to use coordinates.
So then u has components u^a and v has components v^a.  If we are being
conceptually sloppy the way physicists like to be, we might say u^a "is
a vector".  That can be misleading!  u is the vector, and it has
components u^a where a = 0,1,2,3.  But once one is an expert one can get
away with being sloppy, and get away with saying "the vector u^a".  

Anyway, so u and v are vectors with components u^a and v^b, say (notice
I just changed the superscript on v -- it doesn't matter since we aren't
doing anything with it yet!), and we want to compute H(u,v), so we want
to know ITS components too.  It has components H(u,v)^c.  And the
formula for these is

H(u,v)^c = H^c_{ab} v^a w^b

where H^c_{ab} is some big 4 x 4 x 4 array of numbers.  And how do we
interpret this formula?  Well, it just means multiply the number
H^c_{ab} by the number v^a and the number v^b, and then sum over all
repeated indices (namely a and b).  

>Anyway, with extraordinary lack of confidence, I assume a typo. 

Yup.

>Please note that this is at risk of being decapitated. 

Well, you lucked out this time.  [Takes out large double-bladed axe and
polishes it fondly.]

>Hmm, lets just do H(0). This seems to imply c=0 to me.

True, but folks usually call it H(u,v)^0 or H^0, since that c was a
superscript.  Anyway:

>H(0)=
>K^0(0,0)u^0v^0 + K^0(0,1)u^0v^1 + K^0(0,2)u^0v^2 + K^0(0,3)u^0v^3 +
>K^0(1,0)u^1v^0 + K^0(1,1)u^1v^1 + K^0(1,2)u^1v^2 + K^0(1,3)u^1v^3 +
>K^0(2,0)u^2v^0 + etc
>K^0(3,0)u^3v^0 + ............................... + K^0(3,3)u^3v^3

Very good!   You got it, despite my typo, which was a deliberate error
thrown in to confuse you (just kidding).  I would have said:

H(u,v)^0 = 

H^0_{00}u^0v^0 + H^0_{01}u^0v^1 + H^0_{02} u^0v^2 + H^0_{03}u^0v^3 +
H^0_{10}u^1v^0 + H^0_{11}u^1v^1 + H^0_{12} u^1v^2 + H^0_{13}u^1v^3 +
.......
H^0_{30}u^3v^0 + H^0_{31}u^3v^1 + H^0_{32} u^3v^2 + H^0_{33}u^3v^3 +

but the difference is purely notational, no big deal.

>Well, I have to say I vastly prefer this notation to a 'do what you like
>even if it's wrong' matrix method. I would probably like it even more if
>I knew what I was doing, too. Although I have no wish to do it, it would
>appear that this notion would allow things that produced tensors of any
>rank from other tensors of any rank. I think a computer program would,
>ahem, do it with less effort.

Indeed the terse notation lets you immediately know what sums you would
have to do to compute, say

R^a_{bcd} u^a v^b w^d

or 

R^a_{bcd}R_a^{bcd}

or any such thing that might come up in general relativity.  The beauty
of the notation is that it tells you what sum you would have to do, in a
very terse way, without actually making you do the sum.  Actually DOING
the sums is often best left to a computer, as you note.

So now when I write stuff using tensor notation you will, at least in
some sense, know what it means.  At least you'll know what sum it
describes.  That leaves the more interesting aspect, namely what the
hell we're doing it for.

Article 98956 (32 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Fri, 9 Feb 1996 08:56:26 +0000
Organization: Oz
Lines: 82
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Yes, if we are told the geometry of spacetime, we can completely work
>out a geodesic if we know a point it goes through and its tangent vector
>(velocity) at that point.  It satisfies a second-order differential
>equation, which I have avoided writing down, and you just solve that.

Honestly, a red flag to a bull, "which I have avoided writing down". 

Any chance of seeing the form of this? After all one of the 'things'
about GR seems to be the ability to calculate geodesics. We don't really
have any idea as yet. Probably, on reflection never will, but still.

[NB Cattle are colourblind so it doesn't matter what colour flag you
wave. Bull is probably an appropriate word. Most bulls are more
interested in 'other things' than chasing people.]

A sort of moronic thought occured to me. (I never have problems with
having these). If 4D space is curved in a GR-like fashion what do we
really mean by this? Rather basic, I know, and we have lots of
mathematical thingies to describe it but what are they actually
describing? Maybe it would be a good idea to be clear about it?

How about some thoughts like:

1) An unaccelerated body follows it's tangent vector. Since we have been
told this is a geodesic, this would seem right.

2) The Riemann tensor 'derivation' we were given involves rotation of a
tangent vector when parallely transported to another point in spacetime
on a path that is not entirely on its geodesic. It would seem to me that
we cannot say that anything is parallel to something else unless they
start from the same point in spacetime because any other description
will be path dependent (although I expect geodesics have some 'nice'
properties I have excluded them above). The only way to get to another
part of spacetime (as defined above) is to accelerate, no? This strikes
me as meaning something, but I'm not sure exactly what.

3) It is valid to describe distances measured along an (arbitarily long)
path. Of course you won't get the same distance if you choose a
different path, but a distance as measured along a path is a valid
distance. I think this is valid even if the path has accelerations along
it. <Down, Baez, down> Indeed I suspect that this is the ONLY valid
distance you can define. Hmmm, or is it simply the definition of a
'distance'. Note that this path need not be a geodesic. Is this right???

3) In a torsion-free GR then an infinitesimal sphere of massless test
points surrounding an (even more) infinitesimal speck of momentum-flow
will behave with spherical symmetry under 'control' of the Ricci tensor.
We note that this essentially says that the *local* spacetime in this
infinitesimal volume can be considered flat. We also note that this is
also the case for a torsion-free parallel transportation: it is in
locally flat spacetime. This sort of implies that GR models a world
where a suitably small piece of spacetime can be considered locally
flat. On a philosophoical note one wonders if a black hole horizon is
such a place for example, and if any extension of GR will remove this
simplification resulting in a truly complex piece of mathematics that
will require new tools to manage it. <End of daydream>

4) We don't seem to have any relativistic effects modelled into our
version of GR yet. I expect they are hidden in the notation somewhere or
maybe we need to discuss geodesics to find them. I have a feeling that
the math may become complex at this point. <Oh dear>.

5) It's kind of interesting that considering energy as momentum flow in
the time direction, we can dispense mentally with both mass and energy.
We only need to consider momentum flow to describe space curvature, and
everything else. Indeed it would be 'nice' to remove that nasty -1 in
our metric and make it +1 which I suspect would make us view the
momentum flow in the time direction as something slightly different. Has
anybody done this, and how would you view momentum flow in the time
direction if it had a metric of (+,+,+,+)? (Ie still modeling the real
world).

Hmmm, 'nuff for now I think.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 99055 (36 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 9 Feb 1996 14:00:46 -0800
Organization: University of California, Riverside
Lines: 32
NNTP-Posting-Host: guitar.ucr.edu

[The wizard and Oz are talking....  the wizard says:]

>>Yes, if we are told the geometry of spacetime, we can completely work
>>out a geodesic if we know a point it goes through and its tangent vector
>>(velocity) at that point.  It satisfies a second-order differential
>>equation, which I have avoided writing down, and you just solve that.

[Excited, Oz replies:]

>Honestly, a red flag to a bull, "which I have avoided writing down". 
>Any chance of seeing the form of this? 

WHAT?!?!  

[The wizard's eyebrows bristle and daggers of light shoot out
from them.]  

You want to see THAT??  NO!!!  I keep that equation in the room back
there, behind that curtain, where I do my work.  That's where I keep all my
heavy-duty equipment.  I can't let you go back there, because you don't
know how to use that machinery and it's VERY DANGEROUS.  There's really
no telling what trouble you might get into if I let you back there!
So, just remember this: never, NEVER go back there, especially when I'm
not around.

Now get out of here!  The impudence!!

[Sheepishly, Oz skulks off, not without a peek to see what's back behind
the curtain.]

Article 99005 (137 more) in sci.physics:
From: Matthew P Wiener
Subject: Applied Covectors and Symmetric Tensors
Date: 9 Feb 1996 16:49:32 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 97
NNTP-Posting-Host: sagi.wistar.upenn.edu

The story so far....

A nonzero covector (a,b,c) can be visualized as the plane ax+by+cz=1.
This plane can be thought of as part of a dissection of R^3 into all
the parallel planes, each labelled by the value ax+by+cz.

Covectors and vectors are dual, with < (a,b,c) | (x,y,z) > = ax+by+cz,
the label of the plane in the (a,b,c) dissection that the vector (x,y,z)
reaches out to.

One thing I didn't spell out before is that scalar multiples of (a,b,c)
are found simply by multiplying the labels by the scalar in question.
If you scale by 10, the 1-labeled plane is now 10-labeled, with the
new 1-labeled plane .1 the way down.

And I mentioned that gradients are really covectors, with grad f properly
thought of as df = @f/@x dx + ..., where dx = "dual x" = (1,0,0) etc are
the unit covectors.

So let's start using them.  The most common use of gradient is for a
potential: force = - grad U.  So this says that force is a covector
all along.  And if force is a covector, and force = d/dt (momentum),
momentum too is a covector.

You thought position/momentum duality was a quantum thing?  No, it was
part of classical mechanics, all along.  You can see this in relativity
too, where the 4-momentum has the inverse Lorentz transformation from
the 4-position.

What's amazing is that the covector dissection, in the right units, is
just a de Broglie wave.  Momentum is the inverse of wavelength, so a
covector for very high momentum corresponds, as per the scaling above,
to a dissection with the integral labelled planes packed very close
together.

The de Broglie p=h/l relationship just means that the momentum push is
packed in the positive covector direction, and its magnitude is a count
%%%of the covector planes with integral labels from here to here + h%%%.
Simple, right?  For p in the x direction, what happens to a wave function?
If the position operator is multiplication by x, how about the momentum
operator just being multiplication--of a covector 1-off push--by p:

                                W(x+l)-W(x)     @W
              p.W(x+l)-W(x) = h ----------- = h --   ?
                                     l          @x

This is one standard heuristic derivation in disguise, by the way, but in
covector language it goes over much better.

Now that you're convinced that quantum mechanics could have been figured
out much faster had people back then known that momentum was a covector,
let's see more about those classical covectors.

I started out with a conservative force being the gradient of a potential:
F=-dU.  What if there's a constraint, some surface H(r)=0?  The equilibrium
can be found as follows.

The constraint force, which you thought was the normal vector sticking out,
is actually a tangent covector.  Right away, you know this *has* to be, a
priori, the only proper invariant description: perpendicularity is _not_
invariant in relativity, but parallelism _is_.

The actual constraint force is thus l.dH, some scalar l.  At equilibrium,
the net force is zero: so F+l.dH=0 or equivalently dU=l.dH.  Look familiar?
Yes!  It's a Lagrange multiplier!

So with covectors firmly established, it is natural to expect physics to
occur relating vectors with covectors.  A simple example is Hooke's law:
F=kx, some k.  This is _not_ a vector equation.  It only works out that way
in the simplest of cases.  In full generality, it is F=K.x, where F is the
force covector, x is the displacement vector, and K is the _strain tensor_.

So what's a tensor?

In this case, just a linear map vectors to covectors.  Mathematically,
covectors are duals to vectors: they work by taking vectors to numbers.
(See above: (a,b,c)--our R^3 planar dissection--sits and waits for
(x,y,z) so as to get ax+by+cz.)  So what we have here is something
that takes a vector, and then sits and waits for another vector to
show up, so as to get a number.  In other words, we have a bilinear
map K, sending pairs of vectors to numbers.  That just takes a matrix
to describe.

In the Hooke case, K is symmetric and positive definite, and thus can
be described as an ellipsoid.  Geometrically, if you are given the
displacement, starting at the ellipsoid's center and pointing out
somewhere, find the cone with vertex the vector's far end and which
is tangent to the ellipsoid.  The tangency occurs in a single plane,
which is the 1-labelled plane for the covector F=K.v.

If the vector is short, scale it up first, and then compensate.  Note
that one can reverse the process.  The usual Euclidean metric (where
the bilinear map is the dot product) is just the unit sphere.

And in the Minkowski case, one can do this using a hyperboloid.
-- 
-Matthew P Wiener ([email-masked])

Article 99043 (136 more) in sci.physics:
From: Edward Green
Subject: Re: General relativity tutorial
Date: 8 Feb 1996 09:01:56 -0500
Organization: The Pipeline
Lines: 96
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[email-masked] (john baez)' wrote: 
 
>IN GENERAL RELATIVITY, LOCAL CONSERVATION OF ENERGY AND 
>MOMENTUM IS AN AUTOMATIC CONSEQUENCE OF TAUTOLOGOUS 
>FACTS ABOUT SPACETIME CURVATURE!!!!!! 
 
yeah.  yeah.   so physics becomes geometry and topology.  there is no need
to shout about it.  I knew that was the game a long time ago,  (before I
ever met you,  sir).  But I'm glad you feel the same way!   :-)    
 
>By "tautologous" I mean, "relying on mathematical identities which hold 
>no matter what the metric is".   
 
Well,  like I always said,  when you really understand some deep fact,  it
usually becomes a tautology,  ie, so blindingly obvious,  that you can't
imagine how it cannot be true.  Since I am only getting second hand
accounts of the math here (my fault,  not yours)  of course I can't quite
share in the joy of this particular tautology with you.... yet. 
 
>This is a truly wondrous thing.  I leave you to ponder it, and to 
>remember what is the analogous wonderful thing about Maxwell's 
>equations.   
 
That the Poynting vector doesn't transform properly?  No, that can't be it.
 Hmmm....     :-)  
 
<shows how to go back and forth from Einstein to Ricci> 
 
Thanks for plowing through that far.  Too bad you didn't get a little
farther,  because: 
 
On Feb 05, 1996 17:45:09 in article <Re: General relativity tutorial>,
[email-masked] (john baez)' wrote: 
> 
> ...  something 
>like this is right: the Ricci curvature tells us how the ball changes 
>volume.   
> 
>So: I think the Weyl tensor tells the REST of the story about 
>what happens to the ball.  I.e., how much it rotates or gets deformed 
>into an ellipsoid.  This makes sense, if you think about what you said: 
 
keyword:  "rotates" 
 
While On Feb 05, 1996 12:15:41 in article <Re: General relativity
tutorial>, [email-masked] (Edward Green)' wrote: 
 
>The remaining information in the Riemann tensor ... 
>captures the *rotation* of our little test sphere under translation.  In 
>the same rank as our Ricci tensor,  which is symmetric,  we could add 
>another antisymmetric tensor,  which would contain the pure rotations.  
>Apparently,  these rotations are *not* determined by the local energy 
>density,  but may be globally determined by boundry conditions.  Dilation 
>is an intrinsic property,  determined locally by the stress tensor, while 
>rotation is an extrinsic property,  determined by the behavior of 
>nieghboring bits of space.  
> 
 
So you see,  out of the mouths of babes...   :-)  
 
[We did publish the same day... must have been in the air   ;-) ] 
 
I still have a hunch the distortions of the test sphere into the ellipsoid
are captured by the Einstein part,  not the leftovers,  a la some kind of
principle axis construction.  In other words,  I think you can't know the
volume dilation without knowing this also.   Please take one more look at
the first paragraph of this B.S. (appended) and tell me if it rings a bell?

How many numbers do you need to capture a general rotation of a body in
four space? 
 
-- 
 
Ed Green / [email-masked] 

"The Ricci tensor is the strain rate derivative of spacetime.  In other 
words,  translating in any direction in spacetime,  the Ricci tensor gives 
us the resulting dilations and shears.  The Ricci tensor is symmetric, and 
the diagonal components represent stretching,  the off-diagonals,  shear.  
However,  we could always diagonalize it by a coordinate transformation to 
principle axes.  Then we would find all information in it summarized in *4*

numbers,  giving expansion or contraction along the 4 principle axes,  
transforming our little hyperspherical test region into a hyper-ellipsoid. 

Up to scale factors these axes give the locally *best* coordinate system 
for curved spacetime,  one that captures the local behavior most 
succinctly."  

 
"All coordinate systems are equal, 
 but some are more equal than others".

Article 99174 (134 more) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 00:17:35 +0000
Organization: Oz
Lines: 32
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>
>>4) We don't seem to have any relativistic effects modelled into our
>>version of GR yet. 
>
>HUH?
>

Oh, dear. I must have made a *serious* booboo.
Anyway it's past midnight and my apology for a brain isn't functioning
very well. I will go away and commit hari-kari.

Have a nice weekend. I will speak again when I'm sober. Hoc. I mean Hic.

<Oz realises that somewhere he has seriously blown it>
<He hopes it will be clear in the morning>
<he has a horrible feeling it won't be>

Oh, dear.

Have a good weekend,

from Oz.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 99133 (133 more) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 00:23:58 +0000
Organization: Oz
Lines: 40
Distribution: world
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In article <[email-masked]>, Geoffrey A <?@?.landis>
writes
>In article <[email-masked]>
>[email-masked] writes:
>>>4) We don't seem to have any relativistic effects modelled into our
>>>version of GR yet. 
>
>In article <[email-masked]> john baez, [email-masked]
>replies:
>>HUH?
>
>In this case, I think that the professor's answer, while correct, is a
>little to concise.
>
>What I think he means to say here is, by using a metric with a diagonal
>[-1,1,1,1], known as the "Lorentz metric", *all* of the special
>relativistic effects are already included in the geometry.

(Message timed at 12.15 at night)

What!!!!!!!

That does it all!!!

Wow. Oh dear, a whole lotta questions. <Sigh>
Whoever thought this up was, indisputably, a genius!
Why didn't they tell us before?
Why do they even *bother* with SR!!!

This is *brilliant*!!! Frabjabjous!

I will post again when sober. Many thanks for your post. Many thanks. I
think Baez, atypically, has missed something ever so slightly important
out of his description. Obvious to some, maybe. But important.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

(SAME) Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 08:07:15 +0000
Organization: Oz
Lines: 46
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In article <[email-masked]>, Geoffrey A <?@?.landis>
writes
>In article <[email-masked]>
>[email-masked] writes:
>>>4) We don't seem to have any relativistic effects modelled into our
>>>version of GR yet. 
>
>In article <[email-masked]> john baez, [email-masked]
>replies:
>>HUH?
>
>In this case, I think that the professor's answer, while correct, is a
>little to concise.

Hey, I have managed to get Baez to do a post ENTIRELY in capitals. It's
also completely devoid of content. It must carry a record Baez crank
index at least. Er, unless of course it's a strange notation he has that
encompasses the entire universe. Well I guess that "HUH?" is as good an
explanation for it as any. It is at least in keeping with 99.99% of the
population's explanation of the entire universe, so it's got that
'common touch' to it.

Makes for a mighty short publication though.   :-(

>
>What I think he means to say here is, by using a metric with a diagonal
>[-1,1,1,1], known as the "Lorentz metric", *all* of the special
>relativistic effects are already included in the geometry.

Yeah. Braindead, that's me. I even bet that there could be a little 'c'
somewhere that has been downgraded to a '1', I think that was mentioned
waaaaaay up the thread near when it started, some seconds after the big
bang.

I *knew* the metric had been glossed over. What I don't quite see is how
this tensor produces relativistic effects. I am sure it's quite simple
and obvious to you tensor tyros, you have done a full and proper course.
Can anyone spell it out in words of one syllable, oh well Ok then three
syllables, how this happens?

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 99202 (131 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 08:46:16 +0000
Organization: Oz
Lines: 81
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In article <[email-masked]>, john baez <[email-masked]>
writes
>[The wizard and Oz are talking....  the wizard says:]
>
>>>Yes, if we are told the geometry of spacetime, we can completely work
>>>out a geodesic if we know a point it goes through and its tangent vector
>>>(velocity) at that point.  It satisfies a second-order differential
>>>equation, which I have avoided writing down, and you just solve that.
>
>[Excited, Oz replies:]
>
>>Honestly, a red flag to a bull, "which I have avoided writing down". 
>>Any chance of seeing the form of this? 
>
>WHAT?!?!  
>
>[The wizard's eyebrows bristle and daggers of light shoot out
>from them.]  
>
>You want to see THAT??  NO!!!  I keep that equation in the room back
>there, behind that curtain, where I do my work.  That's where I keep all my
>heavy-duty equipment.  I can't let you go back there, because you don't
>know how to use that machinery and it's VERY DANGEROUS.  There's really
>no telling what trouble you might get into if I let you back there!
>So, just remember this: never, NEVER go back there, especially when I'm
>not around.
>
>Now get out of here!  The impudence!!
>
>[Sheepishly, Oz skulks off, not without a peek to see what's back behind
>the curtain.]
>

In another dimension, far, far away ........

[Quiet, doleful music plays in the background.]
[From Ducas "The Sorcerer's Apprentice of course: what else?]

The aged retainer shuffles back to his hole in the wall, dragging his
broom behind him over the solid milk quartz floor. From time to time he
peers wistfully over his shoulder at the Great Wizard who ignores him
and is busy curving space with a wave of his arms and producing black
holes with a flick of his fingers. He arrives at his cramped, cold and
damp cave in the corner, pulls the tattered curtain across the doorway
and sits on his straw filled bunk. It's ichy. He is not so sure that
getting sorcery lessons and food (food: ha!) for keeping the Great
Wizards cave tidy is such a good deal as it seemed originally, but
determines to soldier on.

The Great Wizard watches his aged retainer close the curtain. "At last,"
he thinks, "he has gone, time for some proper work. Time to cut this
simple child's stuff. Got to get up the wizard rankings.".

[Music changes to Borodin: Night on a Bare Mountain: fortissimo.]

The yelps from the Great Wizard and occasional fireballs wake the aged
retainer from his reverie. He peers cautiously through one of the many
holes in his curtain. The Great Wizard seems to be working on something
terribly small that keeps slipping from his fingers. He just can't hold
it in one place, every time he does it zooms away painfully taking a
chunk of wizard with it. Fireballs erupt from the clear shielding the
Great Wizard has thrown up around himself. "Oh well." thinks the aged
retainer "It's not always that much fun being a Great Wizard, either.".

[Music changes to Brahms lullaby]

The aged retainer falls asleep on his itchy bunk. He dreams of creating
black holes with a flick of his finger, but always ends up being dragged
in and painfully turned into sorcerer's spaghetti. He dreams that he
goes behind the curtain and is immediately filled with knowledge without
any effort at all. If only he could remember it all in the morning.

[Music fades]

Dream on.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"
Article 99235 (129 more) in sci.physics:
From: Jerry Freedman
(SAME) Subject: Re: General relativity tutorial
Date: 9 Feb 1996 17:06:03 GMT
Organization: GTE Government Systems Corporation
Lines: 8
NNTP-Posting-Host: 155.95.68.41

  I am just a poor software type whose mathematical education was cut
short and I don't really read this group much but I LIKE this thread,
particularly Mr. Baez' contributions. If this, or something like it were
captured, organized and printed Id buy it. Its the first time I have had
any chance of comprehending this stuff so at the least, continue gentle
people while I go back to polite lurking

                          Jerry Freedman,Jr

Article 99318 (127 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 10 Feb 1996 23:01:25 GMT
Organization: University of California, Riverside
Lines: 23
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (john baez) writes:
>In article <[email-masked]> [email-masked] wri
tes:
>
>>4) We don't seem to have any relativistic effects modelled into our
>>version of GR yet. 
>
>HUH?

To clarify: we never did anything that WASN'T perfectly "relativistic".
Not only did we never introduce any yucky "absolute rest", or any yucky
split of spacetime into space and time, we didn't even introduce any
yucky "global inertial frames"!  So not only we have avoided the errors
of Aristotelian and Newtonian mechanics, we have avoided the errors of
special relativity.  So we are doing general relativity.

Perhaps you mean that I haven't said any cool stuff about black holes or
the big bang?  Actually we're almost ready for some of that fun
stuff.

Article 99319 (126 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 10 Feb 1996 23:03:12 GMT
Organization: University of California, Riverside
Lines: 92
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:
>[email-masked] (john baez)' wrote: 

>>IN GENERAL RELATIVITY, LOCAL CONSERVATION OF ENERGY AND 
>>MOMENTUM IS AN AUTOMATIC CONSEQUENCE OF TAUTOLOGOUS 
>>FACTS ABOUT SPACETIME CURVATURE!!!!!! 

>>This is a truly wondrous thing.  I leave you to ponder it, and to 
>>remember what is the analogous wonderful thing about Maxwell's 
>>equations.   

>That the Poynting vector doesn't transform properly?  No, that can't be it.
> Hmmm....     :-)  

A vector that doesn't transform properly, isn't.  Don't worry about
that fiddle-faddle.  No, I mean how local conservation of charge is an 
automatic consequence of Maxwell's equations!  Start with

div E = rho
div B = 0
curl B = j + dE/dt
curl E = -dB/dt

and derive the equation for local conservation of charge, 

d rho/dt = - div j.

That's a prototype of how you derive local conservation of energy-
momentum from Einstein's equation.  

Einstein was not as dumb as most people think.  He knew that one of the
coolest features of Maxwell's equations is how conservation of charge is
an automatic spinoff, and made damn sure something like this worked in
general relativity.  In fact, in late 1915 he first wrote down a wrong
version of his equation which included only a piece of the Ricci tensor,
and then ANOTHER wrong version that went (in modern notation)

R_{ab} = T_{ab}.

But by November 25th he corrected this to (in modern notation)

R_{ab} - (1/2) R g_{ab} = T_{ab}, 

saying "With this step, general relativity is finally completed as a
logical structure."

>I still have a hunch the distortions of the test sphere into the ellipsoid
>are captured by the Einstein part,  not the leftovers,  a la some kind of
>principle axis construction.
 ^principal

>In other words,  I think you can't know the
>volume dilation without knowing this also.  Please take one more look at
>the first paragraph of this B.S. (appended) and tell me if it rings a bell?

It rings a bell; I know what you're thinking and I read what you wrote
with interest the first time.  I have been trying to better understand
the separation of transformations of a sphere into dilations,
stretching-squishings, and rotations, so as to see how it corresponds to
the separation of the curvature into Ricci and Weyl.  E.g., I thought,
perhaps we don't need to worry about one of the 3 kinds of
transformations of the sphere too much for some reason?  But the
stretching-squishings seem possible even in the absence of Ricci
curvature --- that's what gravitational waves do when they go through
empty space!  So I don't hope to calculate THOSE using the Ricci tensor
as you suggest.  Instead, I hope to see why, even though it seems there
are more sorts of stretching-squishings and rotations than dilations,
the Weyl tensor has just as many degrees of freedom as the Ricci.
And stuff like that.

>How many numbers do you need to capture a general rotation of a body in
>four space? 

Hmm, I didn't know we were talking about that, but the answer is 
(4 x 3)/2 = 6.  

To capture a general linear transformation of a body in 3-space it takes
3 x 3 = 9 numbers.  1 of these keeps track of dilation.  3 keep track of
rotation.  The remaining 5 keep track of stretching-squishings.  (That's
what I was worrying about.)  For the mathematically macho, it's easiest
to understand this using the Lie algebra gl(3).  This splits up into:
scalar multiples of the identiy (1 dim), skew-symmetric matrices (3
dim), and symmetric traceless matrices (5 dim).

Article 99320 (125 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 10 Feb 1996 23:03:49 GMT
Organization: University of California, Riverside
Lines: 42
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward
 Green) writes:
>[email-masked] (john baez)' wrote: 
>> 
>>>Comprehension of the geometry is important, but not as important as the 
>>>physics. 

>>You're just saying that to get in good with these guys.  :-)  I'm 
>>playing bad cop, and trying to force them to get used to the geometry of 
>>curved spacetime in all its abstract splendor.  I'd say both aspects are 
>>very important.   

>Huh?   Oh,  I am so disappointed.   I thought the geometry *was* the
>physics.  Remember,  sir?  Remember all our brave talk of tautologous
>relations?   Stand up to them!  Don't let them sway you! 

Don't worry.  There, there.

>"The physics" in this sense is just code for any aspect of physics we have
>not *yet* succeeded in reducing to tautologous geometric relations.  You
>know it's true,   don't you?   

Of course.  

>You frighten me when you talk like this.  I know you are just saying it to
>placate the unenlightened.  Yes,  that must be it.   You couldn't mean it. 

Yes, of course.  Indeed, I considered taking the hard line and saying
"what do you mean, geometry not as important as the physics, what else
is there besides geometry??"  But there are limits even to my proclivity
for taking principled stances, and Wiener meant something quite
reasonable: if you don't know how to derive Newton's laws of gravitation
as a limiting case of Einstein's equation, all the deep truths about
physics reducing to geometry will sail over you like a 747 over an
ostrich with its head in the sand.  As a Buddhist might say, the
illusory distinction between physics and geometry can only be overcome
by learning lots of physics and geometry!

Article 99321 (124 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 10 Feb 1996 23:04:47 GMT
Organization: University of California, Riverside
Lines: 150
Distribution: world
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writ
es:

>If 4D space is curved in a GR-like fashion what do we
>really mean by this?

I told you this once a while ago, but you'll probably understand it a
lot better now.  Here's an operational definition of the sentence
"spacetime is curved": we say spacetime is curved at the point p if, in
a very small neighborhood of p, initially comoving geodesics accelerate
relative to one another.

I could make that more precise but actually I already did.  Remember those
coffee grounds?  We set them up very carefully to follow "initially
comoving geodesics", and then a little ball of them started rotating and
changing shape.  If, regardless of the velocity v of the central coffee
ground at the point P, the ball does NOT start rotating or change shape,
then space is flat at P.  Moreover, every bit of information about the
Riemann curvature tensor at P can be recovered from knowing how these
balls change shape (for all possible velocities v).  So we can quite
rightly say: "Operationally, curvature of spacetime is just a way of talking
about the relative acceleration of initially comoving nearby geodesics."

And that's good, right, because curvature of spacetime is all about
GRAVITY, and gravity is what you need to understand about WHAT HAPPENS 
IN FREE FALL.

>1) An unaccelerated body follows it's tangent vector. Since we have been
>told this is a geodesic, this would seem right.

Yeah.  More precisely: it follows a geodesic, which is a curve whose own
tangent vector is parallel translated along itself.

>2) The Riemann tensor 'derivation' we were given involves rotation of a
>tangent vector when parallely transported to another point in spacetime
>on a path that is not entirely on its geodesic. It would seem to me that
>we cannot say that anything is parallel to something else unless they
>start from the same point in spacetime because any other description
>will be path dependent (although I expect geodesics have some 'nice'
>properties I have excluded them above). The only way to get to another
>part of spacetime (as defined above) is to accelerate, no? This strikes
>me as meaning something, but I'm not sure exactly what.

I really don't follow this.

>3) It is valid to describe distances measured along an (arbitarily long)
>path. Of course you won't get the same distance if you choose a
>different path, but a distance as measured along a path is a valid
>distance. I think this is valid even if the path has accelerations along
>it. <Down, Baez, down> Indeed I suspect that this is the ONLY valid
>distance you can define. Hmmm, or is it simply the definition of a
>'distance'. Note that this path need not be a geodesic. Is this right???

WOW!  YES!  You seem to be seeing exactly why I would always scold you for
talking about the distance between P and Q without specifying a path
between P and Q!  YES!  The "arclength" or "proper time" along a
path is well-defined both operationally and in our mathematical
formalism for GR.   Operationally: if the path is spacelike you can use
rulers; if it is timelike you can use watches.  Mathematically: the
metric lets us compute the length of a tangent vector, and we integrate
this length along a given path from P to Q to compute the length of the
path.

And YES!  Distance between points is ONLY defined given a path between
them; this is what distance means in GR.  

>3) In a torsion-free GR then an infinitesimal sphere of massless test
>points surrounding an (even more) infinitesimal speck of momentum-flow
>will behave with spherical symmetry under 'control' of the Ricci tensor.
>We note that this essentially says that the *local* spacetime in this
>infinitesimal volume can be considered flat. We also note that this is
>also the case for a torsion-free parallel transportation: it is in
>locally flat spacetime. This sort of implies that GR models a world
>where a suitably small piece of spacetime can be considered locally
>flat. 

If I understand you aright, yes.  This is what they call the
"equivalence principle".

>On a philosophical note one wonders if a black hole horizon is
>such a place for example, and if any extension of GR will remove this
>simplification resulting in a truly complex piece of mathematics that
>will require new tools to manage it. <End of daydream>

Well, in GR the black hole horizon is very much such a place.  We could
be on a black horizon right now and not know it if the black hole was
big enough!!!   I would hate to think this version of the equivalence
principle was a "simplification" in a "bad" sense.  I would prefer to
think it's an important principle which is trying to tell us something.

>4) We don't seem to have any relativistic effects modelled into our
>version of GR yet. I expect they are hidden in the notation somewhere or
>maybe we need to discuss geodesics to find them. I have a feeling that
>the math may become complex at this point. <Oh dear>.

Actually, having scolded you for this sentence a couple times, let me
say something nice.  Namely, the conceptually hard part is to learn all
the geometry you need to understand Einstein's equation, and you have
almost done all that!  Then you are in the position to look at some
solutions and know what they mean!  For example, I can go into my back
room, solve Einstein's equation, work out some geodesics, and show them
to you.  Then you'll see what the big bang is really like, or black
holes, or whatever.  This will be fun and easy in comparison with what
we've been doing.  But just remember: never ever try to go into that
back room.  It's VERY DANGEROUS in there... lots of nasty, scary
MATHEMATICS back there.

>5) It's kind of interesting that considering energy as momentum flow in
>the time direction, we can dispense mentally with both mass and energy.
>We only need to consider momentum flow to describe space curvature, and
>everything else. Indeed it would be 'nice' to remove that nasty -1 in
>our metric and make it +1 which I suspect would make us view the
>momentum flow in the time direction as something slightly different. Has
>anybody done this, and how would you view momentum flow in the time
>direction if it had a metric of (+,+,+,+)? (Ie still modeling the real
>world).

A chap named Hawking did that once.  He called this trick "imaginary time"
because (it)^2 = -t^2, so you can get rid of the minus sign in the
metric by making the substitution t -> it.  In the world of imaginary
time, time is no different from space.  

Why did he do this?  Well, he was wondering about the question: "what
happened right at the moment of the big bang, or before?"  Of course,
classically this question doesn't make sense at all.  But what about
when you take quantum gravity into account?  That's what Hawking was
wondering about.

Unfortunately, to answer this question, Hawking had to go way back into
that back room where we keep the mathematical machinery.  [The wizard
gestures with his staff to the curtain, which looks blacker than ever,
shadows seeping from it and filling the room.  Oz suddenly notices it
has grown very late and is dark outside.]  Way, way back where they keep
the REALLY scary mathematics, stuff you wouldn't believe.  And
unfortunately to understand his answer, you'd have to go in there too.
Because, you see, he never came out!  

And if you went in too --- not that you'd ever even think of it, of
course --- but if, *if* you went in, and went THAT far back, it's
very likely YOU TOO MIGHT NEVER COME OUT AGAIN.  

[Deep sinister laughter emanates from behind curtain.  Oz bids a hasty
goodbye and runs all the way home.]

Article 99310 (120 more) in sci.physics:
From: Edward Green
(SAME) Subject: Re: GR Tutorial Water Cooler
Date: 10 Feb 1996 16:41:37 -0500
Organization: The Pipeline
Lines: 40
NNTP-Posting-Host: pipe10.nyc.pipeline.com
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X-PipeGCOS: (Edward Green)
X-Newsreader: The Pipeline v3.4.0

<[email-masked]>' wrote: 
 
>You are too serious, Ed. I thought you would at least comment on hot 
>mexican tacos you have eaten that would incinerate any Sri Lankan curry, 
>or possibly comment on the low price of Cabanas there. Anyway, you don't 
>want to worry about Baez. Baez's bark is wose than his bite.  
 
Ok,  ok.    I know a little Indian restaurant in New Jersey that serves
chili pakoras.  You know,  like an eggplant pakora would be a little
batter-dipped deep-fried ball of eggplant?  That's right,  little
deep-fried balls of green chilies.   Then,  I recently stopped in a little
place here while I was working called the "Punjabi Deli",  a hole in the
wall whose chief product lines are Indian audio cassettes,  Indian
Newspapers,  Indian food,  and motor oil  (I think most of their clientele
is Punjabi cabbies).  Their food is only mildly spicy (possibly cooked in
the motor oil),  but they had a little bowl of whole green chilies out on
the counter.  Appetizers?   Candy?  I didn't ask what they were for.  Maybe
toothpicks. 
 
You are right,  Baez is a pretty nice guy.  I seem to keep testing it.   
 
I am getting overwhelmed keeping up with the volume here... I'm  behind to
about the middle of last week right now,  if you want to know the truth.  I
guess I'll just have to give up sifting through sci-physics to find threads
to make snide comments in,  like "haven't you ever heard of a little thing
called a correlation, buddy?",  and getting flamed by the irate;  enjoyable
as that activity is.   Next I'll have to give up Monday night wrestling!   
 
About that cabana...  The prices remind me of a famous tourist destination
in the north of India,  where beautiful houseboats on the lake can be
rented for a song,  most especially after the guerillas kidnapped and
killed that American tourist...  Still,  he was walking in the mountains. 
The guerillas *never* come down to the houseboats.  So they say. 
  
-- 
 
Ed Green / [email-masked] 
 
"All coordinate systems are equal, 
 but some are more equal than others".

Article 99322 (17 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 10 Feb 1996 23:05:04 GMT
Organization: University of California, Riverside
Lines: 25
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> Grai <[email-masked]> w
rites:
>[email-masked] (john baez) wrote:
>>
>>See some stuff Matthew Wiener wrote recently, for a more geometrical
>>explanation of the difference between vectors and covectors.

>Just a question on nomenclature. I always used the terms vector and
>1-form. Is covector a new, more general entity applicable to other
>vector spaces, or just exactly the same thing as a 1-form.

If you hand me any vector space I could call its elements vectors and
call the elements of the dual vector space "covectors".  It's certainly
interesting to think about this.  But I wasn't really trying to be so
general here... I was mainly talking about the tangent space, whose
elements I call "tangent vectors" or (for short) "vectors", and its dual
space, the cotangent space, whose elements I call "cotangent vectors" or
(for short) "covectors".  If I have a tangent vector at each point of
spacetime I say I have a "vector field", while if I have a cotangent
vector at each point of spacetime I might say I have a "1-form".

Article 99130 (17 more + 1 Marked to return) in sci.physics:
From: Steve Carlip
(SAME) Subject: Re: General relativity tutorial
Date: 9 Feb 1996 22:28:01 GMT
Organization: University of California, Davis
Lines: 69
NNTP-Posting-Host: dirac.ucdavis.edu
X-Newsreader: TIN [version 1.2 PL2]

john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] writes:

:[...] : I think the Weyl tensor tells the REST of the story about
: what happens to the ball.  I.e., how much it rotates or gets deformed
: into an ellipo.   [...]

: >Anyway, I hope that makes it a little bit clearer why people say that
: >the Weyl part of the curvature has to do with gravitational radiation:
: >the Weyl tensor carries information about the kind of curvature that's
: >independent of the source distribution, sort of like electromagnetic
: >waves are fields that propagate independently of whatever sources are
: >around.

: In short, when we are in truly empty space, there's no Ricci curvature,
: so actually our ball of coffee grounds doesn't change volume (to
: first order, or second order, or whatever).  But there can be Weyl
: curvature due to gravitational waves, tidal forces, and the like.  
: Gravitational waves and tidal forces tend to stretch things out in one
: direction while squashing them in the other.  So these would correspond
: to our ball changing into an ellipsoid!  Just as we hoped.

Yes, this is at least mostly right.  (The "mostly" comes because
there's a part I don't understand---I'll explain below.)  Start
with your ball of coffee grounds, with an initial four-velocity
u^a.  (In the rest frame, u^0=1 and the rest of the components
are zero.)  If you look at the configuration slightly later, it
will typically have changed in three ways.  First, it may have
expanded or contracted (its volume may have changed).  Second,
it may have twisted.  Third, it may have sheared---that is, become
distorted from a sphere to an ellipsoid without changing volume.
The Weyl tensor (in empty space) gives the rate of change of the
shear.

Specifically, you can describe shear by a three by three symmetric
matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX
users).  The three eigenvectors of \sigma give three spatial axes,
and the eigenvalues give the rate of expansion along each axis.
The fact that \sigma is traceless means that the sum of the rates
of expansion is zero, i.e., the total volume is remaining constant.
In empty space, if your ball of coffee grounds has no initial shear,
rotation, or expansion, the rate of change of \sigma is given by the
Weyl tensor contracted twice with u.  (If you want that in symbols,
I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

This means that the Weyl tensor is the thing gravitational radiation
detectors are designed to measure.  A laser interferometer detector
like LIGO or VIRGO (now under construction) is just a very large, very
accurate interferometer with two perpendicular arms.  If the Weyl tensor
changes, one arm will expand while the other one contracts, changing the
relative lengths and thus the interference pattern.

Now, the part I don't understand.  The Weyl tensor really splits into
two parts, and "electric" part (with five components) and a "magnetic"
part (also with five components).  The contraction C(*,u,*,u) that I
described above is the electric part, in the rest frame of the coffee
grounds.  Like ordinary electric and magnetic fields, the electric and
magnetic components of the Weyl tensor mix under coordinate changes.
But it would be nice to have a direct geometrical picture of the magnetic
part.  Does anybody know one?  (Presumably it's hidden in what's called
the Newman-Penrose formalism, but I've never learned that very well.)

(The obvious guess is that it has to do with rotation, but this is
wrong---rotation isn't produced directly by geometry, but only
indirectly by expansion and shear.)

Steve Carlip
[email-masked]

Article 99465 (9 more + 1 Marked to return) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 11 Feb 1996 19:23:22 GMT
Organization: University of California, Riverside
Lines: 43
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Steve Carlip)
 writes:

>In empty space, if your ball of coffee grounds has no initial shear,
>rotation, or expansion, the rate of change of \sigma is given by the
>Weyl tensor contracted twice with u.  (If you want that in symbols,
>I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

I see, so it's really that simple!  My problem was that I didn't
see why this gives us a 3x3 *symmetric* traceless matrix, describing
only the *shears* and not the rotations of our little ball of coffee
grounds.  Now I see.  Let me just say it rather quickly so the experts
like Steve can see it's finally penetrated my thick skull; later I may
attempt more of a "pop" exposition of this stuff.

Working in the rest frame of the center of the little ball of coffee, so
that u = (1,0,0,0) and the metric is the Minkowski one (at that point),
the 2nd derivative of the shape of the ball is given by R^i_{0j0}, and
when we subtract out the part of this that describes the 2nd derivative
of the volume we get C^i_{0j0}.  (Here i,j = 1,2,3.)  We may harmlessly
lower an index and get C_{0j0i}, using the convention that the top index
goes to the back.  The point is: this is *symmetric* since the Weyl
tensor, like the Riemann, has C_{abcd} = C_{cdab}.

>Now, the part I don't understand.  The Weyl tensor really splits into
>two parts, and "electric" part (with five components) and a "magnetic"
>part (also with five components).  The contraction C(*,u,*,u) that I
>described above is the electric part, in the rest frame of the coffee
>grounds.  Like ordinary electric and magnetic fields, the electric and
>magnetic components of the Weyl tensor mix under coordinate changes.
>But it would be nice to have a direct geometrical picture of the magnetic
>part.  Does anybody know one?  (Presumably it's hidden in what's called
>the Newman-Penrose formalism, but I've never learned that very well.)

Perhaps the magnetic part could be understood in terms of the shear
of a ball of tachyonic coffee grounds, but I bet that's not quite
the answer you're looking for.

>(The obvious guess is that it has to do with rotation, but this is
>wrong---rotation isn't produced directly by geometry, but only
>indirectly by expansion and shear.)

That's what I hadn't been able to see, though I suspected it.

Article 99406 (16 more + 1 Marked to return) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: General relativity tutorial
Date: Sun, 11 Feb 1996 09:13:33 +0000
Organization: Oz
Lines: 42
Distribution: world
NNTP-Posting-Host: upthorpe.demon.co.uk
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] (john baez) writes:
>>In article <[email-masked]> [email-masked] 
>writes:
>>
>>>4) We don't seem to have any relativistic effects modelled into our
>>>version of GR yet. 
>>
>>HUH?
>
>To clarify: we never did anything that WASN'T perfectly "relativistic".
>Not only did we never introduce any yucky "absolute rest", or any yucky
>split of spacetime into space and time, we didn't even introduce any
>yucky "global inertial frames"!  So not only we have avoided the errors
>of Aristotelian and Newtonian mechanics, we have avoided the errors of
>special relativity.  So we are doing general relativity.
>
>Perhaps you mean that I haven't said any cool stuff about black holes or
>the big bang?  Actually we're almost ready for some of that fun
>stuff.

Nah, nah, nah. Big bang black holestuff is for practicals, not theory.

I don't quite see how the metric produces relativistic-like effects.
Other posters have pointed out it is due to the metric. (-,+,+,+). Here
we arrive at the situation where I should have already done 10,000
examples of tensors so I would immediately see this. I bet there is a
lecture just on this bit. It's all to do with it defining a hyperbolic
geometry, posters have said. I just *knew* that a "metric" meant more
than a bundle of signs. 

But it doesn't say this to me. It's just a tensor. Could someone walk me
through (real easy) how I get to see this metric as representing a
hyperboloid space. I think this is VERY VERY important to follow this
clearly. It musn't be glossed over.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 99414 (15 more + 1 Marked to return) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: General relativity tutorial
Date: Sun, 11 Feb 1996 13:12:20 +0000
Organization: Oz
Lines: 78
Distribution: world
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X-Newsreader: Turnpike Version 1.11 <V67Vak$Q$lyzk9NR0hujdE7wCg>

In article <[email-masked]>, Doug Merritt <[email-masked]>
writes
>
>Like Geoffrey said, once you've got the Lorentz metric, you've
>got relativistic effects. That metric specifies a roughly
>hyperbolic geometry, and that's all it takes. The interesting
>stuff comes from all the modulations of the basic hyperboloid.
>
>The metric means you've got s ~= sqrt(dx^2 + dy^2 + dz^2 - dt^2)
>to first order. Mass/energy/stress/cosmological constants modulate
>those terms.

OK. Demon will be clogged up for days so no replies can be expected.
This should give me enough clues to have a look see.

<Oz takes a bit of burnt twig and starts scrawling on the polished white
quartz floor. The Great Wizard is busy behind the curtain cursing and
calling up demons like Hamilton and Riemann. Sometime the sound of a
500A electric arc is heard, shortly followed by a yelp of irritation.
The G. Wiz (not to be confused with Gee Whiz, and never, but NEVER, to
be called GW) will be busy for hours. Oz hopes his eyebrows don't get
singed, 'cos he gets ratty for days afterwards.>

Hmmm, lets look at this metric thingy. Lets do it in 2D as well just to
make it simple. Lets take a vector a and see how a^0 and a^1 vary if it
stays at length 1.

So the metric will be

[-1  0] and a will be [a0]
[0  +1]               [a1]

Soooo the length = 1 = g_{00}a^0a^0 +g_{11}a^1a^1 = -a^0^2 + a^1^2

Oz decides that these superscripts are boring, and since the G Wiz is
never going to see this, he decides to call a^0, a0 for simplicity.

OK so a1^2 = 1 + a0^2

Hmmm, perhaps we ought to graph this because it looks like a1 and a0 can
have a whole bundle of values. Ok so let's see.

when a0 = 0 then a1^2 = 1, so a1 = +-1. That's two points on the graph.

when a1 = 0 then  a0=sqrt(-1) so it's imaginary. Hmmm, so a1 can never
have a real value between +1 and -1. Hokay.

As a0 goes to infinity, a1 goes to infinity so a couple of infinite
crossed line at 45 degrees are azimtotes. So it looks like this:

<Bungled attempt at an ascii picture. Oz scuffles it out with his foot.>

Basically it starts at a1=oo, a0=-oo zooms down keeping above the
azimtote to cross over the a0=0 axis at a1=+1 and then zooms off to it's
azimtote at a0=oo, a1=oo. Then there's the curve mirrored in the a1=0
axis.

Anyway, it certainly has a hyperbolic feel to it, even in Minkowski
spacetime. So what does it mean? Well, firstly that it ain't Euclidian
space at all. So we bin all those nice triangles that give us a nice
vector additions. There's 10 years knowledge down the shute straight
away. Good job I *have* forgotten it all. I mean, you can have a vector
with two near infinite co-ordinates and a length of 1, or .000001 for
that matter. Or looking at it the other way round, if one co-ordinate is
eversobig and the length is 1, the other has to be eversobig positive or
negative.

Needs a bit more thought, dunnit?

<Oz is so engrossed he fails to hear G. Wiz coming up behind him. Boy is
HE in for a shock.>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From: john baez
Subject: Re: General relativity tutorial
Date: 11 Feb 1996 19:41:57 GMT
Organization: University of California, Riverside
Lines: 69
Distribution: world
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] write
s:
>In article <[email-masked]>, Geoffrey A <?@?.landis>
>writes
>>In article <[email-masked]>
>>[email-masked] writes:
>>>>4) We don't seem to have any relativistic effects modelled into our
>>>>version of GR yet. 

>>In article <[email-masked]> john baez, [email-masked]
>>replies:

>>>HUH?

>>In this case, I think that the professor's answer, while correct, is a
>>little too concise.

>>What I think he means to say here is, by using a metric with a diagonal
>>[-1,1,1,1], known as the "Lorentz metric", *all* of the special
>>relativistic effects are already included in the geometry.

>Why didn't they tell us before?

Hmm, I thought I remember Ted Bunn and I saying stuff about how
special relativity was all about "flat Minkowski spacetime" with its
"usual Lorentz metric".  I guess I should've waited 'til you learned all
about the metric, and then emphasized that special relativity is all
about a very special metric, the Lorentz metric:

-1  0  0  0
 0  1  0  0
 0  0  1  0
 0  0  0  1

This metric has zero Riemann curvature... since parallel translating a
vector around a loop doesn't ever affect it.  Thus the Einstein tensor
is zero, so special relativity is a very special solution of the
equations of general relativity with the gravitational constant set to
zero --- i.e., neglecting the fact that energy-momentum curves
spacetime.

Remember, if we stop hiding the gravitational constant k, Einstein's
equation is

G_{ab} = 8 pi k T_{ab}

(Of course c is still 1 here.)

And, the cool part, while the geometry of Minkowski spacetime is 
boringly flat, it still gives you all the usual stuff about special 
relativity.  That's how special relativity should be taught: not
as a melange of "relativistic effects", but as the natural outgrowth
of Lorentzian geometry.

>This is *brilliant*!!! Frabjabjous!

Yes.  

>I will post again when sober. Many thanks for your post. Many thanks. I
>think Baez, atypically, has missed something ever so slightly important
>out of his description. Obvious to some, maybe. But important.

Yes, very important; for some reason I assumed you knew it.  

Article 99483 (10 more) in sci.physics:
From: john baez
Subject: Re: Minkowski Metrics for Tensor Twits
Date: 11 Feb 1996 12:40:19 -0800
Organization: University of California, Riverside
Lines: 37
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] write
s:

>EUREKA!!!!!!!
>
>Oh, I just gotta say it again. Louder.
>
>EEEEE  U   U   RRR    EEEE   K   K      A       !!
>E      U   U   R  R   E      K  K      A A      !!
>EEEEE  U   U   RRR    EEEE   KK       A   A     !!
>E      U   U   R R    E      K K      AAAAA     !!
>E      U   U   R  R   E      K   K    A   A
>EEEEE   UUU    R   R  EEEE   K    K   A   A     !!

[The wizard pokes his head into Oz's room, glaring with annoyance.]

Could you keep it down?  I'm trying to get some serious thinking done,
and here you are making lots of noise because you finally got around
to learning the prerequisites for the general relativity course you're
taking!  

[He smiles ironically.]

I'm glad you're catching on, though.  Why don't you work out the Lorentz
contraction with your newfound knowledge?  When you figure it out, just
yell your head off.

[He disappears and Oz frantically starts trying to remember exactly what
the Lorentz contraction is and how one could possibly derive it from the
metric tensor.]

From: Oz <[email-masked]>
Subject: Re: General relativity tutorial
Date: Mon, 12 Feb 1996 14:06:24 +0000
Organization: Oz
Lines: 179
Distribution: world
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
MIME-Version: 1.0

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>
>>It would be really nice to have some simple examples of a Riemann tensor
>>for a suitable space. I think 2D would do. Say of a sphere or other
>>straightforward object so one can get an idea of what a real one would
>>look like. At some point a few simple concrete numbers is helpful for
>>clarity, if they are appropriately chosen.
>
>[The man in sorcerer's cap hems and haws for a minute and then speaks:]
>
>2d is good for some purposes, boring for others.  In 2d it only takes
>one number to describe the Riemann curvature at each point, so there is
>the same amount of information in the Riemann curvature tensor, the
>Ricci tensor, and the Ricci scalar.  So we can't understand the
>differences between these concepts very well in 2d.

Fair enough. On the other hand before understanding the differences, how
about trying to understand the things themselves a bit? Good idea, no?

>But how about the Riemann *tensor* on a round sphere?  Well, for this we
>need some coordinates.  Let's use the usual spherical coordinates
>theta, phi.  Since there is actually disagreement at times on which is
>which, I remind you that for me theta is the longitude, running round
>from 0 to 2pi, while phi is the angle from the north pole, going from 0
>to pi.  [Coordinate lines appear on the sphere, which floats back and
>forth playfully.]

OK, now lets get this quite clear. Without a metric, you can't measure
distances, so you can't have an epsilon for your Riemann tensor and so
you gotta have a metric first. Now G.Wiz insists at this stage that a
co-ordinate system is not required, so we have to distinguish between a
co-ordinate system and a metric even though they do look very similar.
So what's the difference between a metric(theta,phi) and a co-ordinate
system(theta,phi)? Well the only thing I can see is that
metric(theta,phi) is a relative thingy. A co-ordinate(theta,phi) is wrt
a set of fixed axes. Never properly discussed, but there you go.

>This means that its length squared is r^2 sin^2(phi), or in other words,
>its length is r sin(phi).  That's supposed to make sense.  You do
>remember your spherical coordinates, right?  [Pained, worried look.
>Helpful review of trig formulas flashes by on the sphere.]

Hey, spherical geometry at this level is kid's stuff. Even I remember
it. I have only regressed to age 17ish, this stuff you learn at 14!! I
have to say that it's a much easier formalism than the old fashioned
way. If there are any schoolteachers listening in (particularly UK ones)
could they comment if this stuff (tensors) is done at school nowadays?
Even if they aren't called tensors.

Presumably we could pick some other strange metric that isn't
orthogonal, and all that would happen is that we would get some cross
terms coming into the metric and probably they would all be nasty
expressions. But hey, who cares, plug it into the computer and out comes
the answer.

>But you wanted to know the Riemann tensor of the sphere, not the metric
>or the Ricci scalar!  Well, okay, so we take the metric and feed it into
>this machine...  [scurries behind a curtain; loud banging noises ensue,
>followed by a deafening explosion and a puff of smoke; returns somewhat
>blackened but smiling]... and it computes the Riemann tensor for us.
>Don't worry about that machine in the other room just yet, someday you
>too may learn to use it... or maybe not.
>
>So, the Riemann tensor has lots of components, namely 2 x 2 x 2 x 2 of
>them, but it also has lots of symmetries, so let me tell just tell you
>one:
>
>R^2_{121} = sin^2(phi)

<snip>

>Its component in the a direction has changed a bit, say
>
>                           -epsilon^2 R^a_{bcd}  "
>

        2    rotation in this direction (only)
-epsilon    R
             {starting direction,second direction,tangent direction}

Ok, so now we see why it has 2^4 components.

so R^2_{121}, 1=theta, 2=phi

        2   phi
-epsilon   R                    = sin^2(phi)
            {theta, phi, theta}                               ahhhhh

No, no, no, no,no. This really will not do. How do you expect us to get
a proper grasp, nay even a basic vague concept, if we can't even see how
to work out one element of what is likely the simplest non-trivial
Riemann tensor. I know, I don't like it either, but it's no good fudging
it. It's just gotta be done. You just have to put guards on the machine,
hand out hard hats, dark goggles, and make sure we all stay clear, keep
our fingers out of the way, and just watch.

I have to say, in my heart of hearts, that I don't really like this
definition very much. Not really. For one thing each leg is not epsilon
long. I would rather prefer to see it as going in a little square and
finding I was *not* back where I started. Then I would have a little
path back to where I *had* started from which would be a vector I could
'easily calculate' and would (I think) give me a measure of curvature. I
also rather suspect that it would give me the same measure of curvature.
Not the actual vector itself, of course, you would have to fiddle with
it a bit to get the curvature.

>Let me describe the Ricci scalar, R, in 2d.  This is positive at a given
>point if the surface looks locally like a sphere or ellipsoid there, and
>negative if it looks like a hyperboloid --- or "saddle".  If the R is
>positive at a point, the angles of a small triangle there made out of
>geodesics add up to a bit more than 180 degrees.  If R is negative, they
>add up to a bit less.
>
>For example, a round sphere of radius r has Ricci scalar curvature R =
>2/r^2 at every point.  [With a click of his fingers, a sphere of radius
>r appears on it, with a small triangle drawn on it, edges bulging
>slightly.]  

Well, I don't like to mention this, but nobody has mentioned a Ricci
SCALAR before. I take it we are still talking about a spherical shell
too. Anyway, not wanting to quibble too much, lets see what we can find
out from this. Well, the first thing that strikes you is that if r is
very small then R becomes very large. As r tends to infinity, R rends to
zero so one expects R of an infinite plane to be zero, which is not
unexpected. One has the irresistable urge to throw down a tiny circle on
the spherical shell and measure it's area, and feel that the Ricci
scalar reflects the ratio of the (difference between the area of the cap
and the area of the circle) and the circle. Hmmm. 

<Goes upstairs and nicks son's math book>

Area of a circular cap is 2pi.r.h     (height of cap=h) ...........(1)

If half the chord of a segment of a circle = d (radius of our circle)
the the radius of the circle (our sphere) has radius r then

2r = d^2/h + h Hmmm so h^2 - 2r.h + d^2 = 0 and so

h= + r -sqrt(r^2 - d^2) ..........................................(2)

Substituting (2) into (1) gives us
Area of cap= 2pi.r^2[1 - (1 - d^2/r^2)^0.5]         oh dear, anyway

d is small << r so expanding we get 

2pi.r^2(1 - 1 + d^2/2r^2 - d^4/{8r^4} + ...) shouldn't have started

Dropping off terms in d^6 onwards we get 

2pi(d^2/2 - d^4/{8r^2})                  it isn't going to come off
pi.d^2 - 2pi.d^4/{8r^2}

Take off the circle area pi.d^2               dunno, looking better

to get -2pi.d^4/{8r^2}        difference between a cap and a circle

And the ratio to the circle area

-2pi.d^4/{8r^2} / pi.d^2                I don't like the constants

gives a ratio of -d^2/4r^2                 which is wrong, dammit!

well, if done properly I expect the constants would come out right.

so for a small circle of constant diam the curvature varies as 1/r^2??

I have a nasty feeling that most 2D Ricci scalars come out at 1/r^2 and
that the little constant is ever so important. Why *do* I bother? <sigh>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Summary: 
Expires: 
References: <[email-masked]> <[email-masked]> <[email-masked]>
Sender: 
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Keywords: 

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>But how about the Riemann *tensor* on a round sphere?  Well, for this we
>>need some coordinates.  Let's use the usual spherical coordinates
>>theta, phi.  Since there is actually disagreement at times on which is
>>which, I remind you that for me theta is the longitude, running round
>>from 0 to 2pi, while phi is the angle from the north pole, going from 0
>>to pi.  [Coordinate lines appear on the sphere, which floats back and
>>forth playfully.]

>OK, now lets get this quite clear. Without a metric, you can't measure
>distances, so you can't have an epsilon for your Riemann tensor and so
>you gotta have a metric first. 

It's true that we've got to have a metric before we get our Riemann
tensor.  But metric is NOT required to define the "epsilon" in our
definition of the Riemann tensor.  Rather, it's required to define
parallel transport!   

Recall the course outline.  We said:

RIEMANN CURVATURE TENSOR: Take the vector w, and parallel translate it
around a wee parallelogram whose two edges point in the directions
epsilon u and epsilon v , where epsilon is a small number.  The vector w
comes back a bit changed by its journey; it is now a new vector w'.  We
then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Thus the Riemann tensor keeps track of how much parallel translation
around a wee parallelogram changes the vector w."

Nowhere is distance or angle mentioned, so we're not using the metric in
any explicit way!  To march off along a teeny tangent vector epsilon v
we don't need to know about *distance*.  After all, the tangent vector
is already an "infinitesimal arrow", so there's no ambiguity in how to
take a step in that direction.

The only place the metric is *implicit* here is in the definition of parallel
translation!  Recall: 

PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  

>Now G.Wiz insists at this stage that a
>co-ordinate system is not required, so we have to distinguish between a
>co-ordinate system and a metric even though they do look very similar.

WHAT?????  A metric is just a gadget g that eats two tangent vectors v
and w and spits out a number g(v,w)!  How does that look like a
coordinate system, which is a gadget that assigns coordinates to every
point?? 

Look: if I hand you a perfectly round sphere of a certain size, you know
the metric: you know how to compute dots products of tangent vectors,
you know how to compute lengths of tangent vectors and angles between
them, etc..  This is very different than giving you an abstract sphere with a
coordinate system, e.g. a longitude-latitude grid or some other weird
coordinate system!  Knowing a metric does NOT tell you a coordinate
system.  Knowing the coordinate system does NOT tell you a metric.

The metric is a very physical aspect of spacetime: in GR, it completely
describes the gravitational field.   A coordinate system is a very
unphysical thing that we lay down on a patch of spacetime to help us do
calculations.  

Given a metric, we can work out its components in ANY coordinate system.
Here I have taken the round sphere of radius r and worked out the
components of the metric in ONE PARTICULAR coordinate system, namely,
spherical coordinates.  

Remember, FIRST my sphere with its metric appeared:

"With a click of his fingers, a sphere of radius r appears on it, with a
small triangle drawn on it, edges bulging slightly."

And THEN I decided to start computing things...

"Well, for this we need some coordinates.  Let's use the usual spherical
coordinates theta, phi....  [Coordinate lines appear on the sphere,
which floats back and forth playfully.]

When we write something like g_{ab} or R^a_{bcd}, the indices will go
from 1 to 2, with "1" corresponding to theta and "2" corresponding to
phi.  For example, the components g_{ab} of the metric in this
coordinate system are:

g_{11} g_{12}  =  r^2 sin^2(phi)   0
g_{21} g_{22}           0         r^2    "

So hopefully you see how confused you were when you wrote:

>So what's the difference between a metric(theta,phi) and a co-ordinate
>system(theta,phi)? Well the only thing I can see is that
>metric(theta,phi) is a relative thingy. A co-ordinate(theta,phi) is wrt
>a set of fixed axes. Never properly discussed, but there you go.

>Presumably we could pick some other strange metric that isn't
>orthogonal, and all that would happen is that we would get some cross
>terms coming into the metric and probably they would all be nasty
>expressions. But hey, who cares, plug it into the computer and out comes
>the answer.

Note, it's not the metric itself which is or is not "orthogonal" --- which
I guess is your new terminology for "diagonal" --- instead, its the
COMPONENTS of the metric IN A PARTICULAR COORDINATE SYSTEM which are, or
aren't, diagonal.   We could take the usual round metric on the sphere
and work in some screwy wiggly coordinate system, and then its
components in that coordinate system would not be diagonal.

>>But you wanted to know the Riemann tensor of the sphere, not the metric
>>or the Ricci scalar!  Well, okay, so we take the metric and feed it into
>>this machine...  [scurries behind a curtain; loud banging noises ensue,
>>followed by a deafening explosion and a puff of smoke; returns somewhat
>>blackened but smiling]... and it computes the Riemann tensor for us.
>>Don't worry about that machine in the other room just yet, someday you
>>too may learn to use it... or maybe not.

>>So, the Riemann tensor has lots of components, namely 2 x 2 x 2 x 2 of
>>them, but it also has lots of symmetries, so let me tell just tell you
>>one:
>>
>>R^2_{121} = sin^2(phi)

By the way, the above is only good for a unit sphere.  For a sphere of
radius r, which is what I really had, we have

R^2_{121} = sin^2(phi)/r^2

A small sphere is more curved!

>No, no, no, no,no. This really will not do. How do you expect us to get
>a proper grasp, nay even a basic vague concept, if we can't even see how
>to work out one element of what is likely the simplest non-trivial
>Riemann tensor. I know, I don't like it either, but it's no good fudging
>it. It's just gotta be done. You just have to put guards on the machine,
>hand out hard hats, dark goggles, and make sure we all stay clear, keep
>our fingers out of the way, and just watch.

NO!!!!   You must NEVER, NEVER go back into that room where I actually
compute things.  The machines are VERY dangerous.  If they could only
cut off your fingers, frankly I wouldn't mind; I'd let you go in.  But
they can destroy your very soul!  I was AFRAID you'd start wanting to go
back there.  

[Wizard pauses, frowns, fingers his beard fretfully as he ponders what
to do.]

Let me show you what I mean.  Hold on a second.  

[He walks over to the black curtain and slips behind it, motioning for
Oz to keep his distance.  A click is heard and then a long, high
creaking noise as of a rusty door opening.  Then Oz some thumping around
and a loud clang as of a door slamming shut.  After some more noise, 
the wizard lifts the curtain and rolls out a stretcher.  Oz is shocked
to find on the stretcher a human figure completely covered with white
crystals of ice.]

This, my friend, is what I meant.  See this poor fellow?  He is frozen
stiff.  The worst thing is, he's still alive under all that ice!  Do you
want to know how this fellow got that way?

[Oz nods, getting over his shock and gradually moving closer towards the
stretcher.]

This was once a student of mine, like you, eager to learn general
relativity.  He was an excellent student, much better than SOME, and he
had progressed to the point where he was writing code to do numerical
simulations of Einstein's equation.  He was trying to figure out what
happened when two black holes collide --- a problem, by the way, that is
still not fully understood.  

Anyway, he noticed a lot of problems.  When he reduced the mesh size ---
never mind, this is just some jargon --- sometimes his answers seemed to
converge, other times not.  He asked me about it so I suggested that he
read a bit on numerical analysis.  Oh, had I only known!  [The wizard
pauses sadly a moment.]

To understand the numerical analysis he realized he needed to learn a
bit of analysis.  After all, how could you compute the answer to
something if you weren't sure the solution existed in the first place??
Pretty soon he was quite an expert on existence and uniqueness for
nonlinear hyperbolic PDE.  He studied Sobolev spaces, and energy bounds,
and the work of Choquet-Bruhat....

But as he did he noticed something funny happening.  Occaisionally he
would feel a slight chill.  He disregarded it and kept on working,
delving ever more deep into nonlinear analysis.  He lost interest in his
original goal of simulating black hole collisions.  Proving existence of
solutions to equations seemed much more interesting than actually
solving them.  After a while he noticed frost forming on his glasses.
He just wiped it off and kept on proving theorems.  Unfortunately he
failed to notice the icicles growing on his desk....  By the time we
found him, it was too late.  He was frozen solid, but still thinking
about existence and uniqueness of solutions of nonlinear PDE....

And here he is, still that way, in a condition of... rigor mortis.

[Oz leaned forwards and touched the frozen figure with his finger.
He felt a strange longing, but also a chill, which seemed to seep up
his finger and into his heart.]

DON'T DO THAT!!!!!   [The wizard raised his staff and aimed it at Oz,
sending a fireball at him, and wheeled the stretcher away from Oz.]
BACK!!!!  Little do you know the dangers!!!!  I am protected from the
infectious chill by many magic spells, but you are not.  Oh, you fool!
Let me put this back into the vault.  Stay there.  [He wheels the
stretcher into the back room again, and Oz again hears the clanging of a
great metal door being slammed shut.  The wizard then reappeared....]

So, you may think it a little thing, a trifling thing, to learn how I
calculated the Riemann tensor of a sphere, but I assure you it is not.
Very few know that secret.  And if you learn that, you may be
irresistably drawn to mathematics, and thence towards RIGOR, and you too
may, like my poor student, perish in the icy splendor thereof.

[Oz, abashed, decides not to press the point.  But after a long pause, he
starts thinking about the Riemann tensor of the sphere again... and
says:]

>I have to say, in my heart of hearts, that I don't really like this
>definition very much. Not really. For one thing each leg is not epsilon
>long. I would rather prefer to see it as going in a little square and
>finding I was *not* back where I started. Then I would have a little
>path back to where I *had* started from which would be a vector I could
>'easily calculate' and would (I think) give me a measure of curvature. I
>also rather suspect that it would give me the same measure of curvature.
>Not the actual vector itself, of course, you would have to fiddle with
>it a bit to get the curvature.

Hmm.  I'm not sure what you are saying.   First of all, as I said, the
LENGTH of the edges of the "little square" plays no role in the
definition of the curvature.   To compute

R^2_{121}                            1 = theta, 2 = phi

we simply take the vector in the theta direction at a point P, parallel
translate it from P over to the point whose theta coordinate is epsilon
more, then over to the point whose phi coordinate is epsilon more, then
over to the point whose theta coordinate is epsilon less, and then back
where we started, and we see how much it now points in the phi
direction.  Then we divide by epsilon^2, throw in a minus sign for good
luck, and take the limit as epsilon goes to zero.

There are various things about the above paragraph which are a wee bit
subtle and may confuse you... but anyway, I'm just parrotting the recipe
for curvature, and if this is confusing, that means perhaps you didn't
quite understand the original definition of curvature.  Which is nothing
to be ashamed of, because often things look simple in the abstract and
then mysteriously become confusing in any concrete special case.

[Oz picks up some old notes and reads them:]

>>Let me describe the Ricci scalar, R, in 2d.  This is positive at a given
>>point if the surface looks locally like a sphere or ellipsoid there, and
>>negative if it looks like a hyperboloid --- or "saddle".  If the R is
>>positive at a point, the angles of a small triangle there made out of
>>geodesics add up to a bit more than 180 degrees.  If R is negative, they
>>add up to a bit less.
>>
>>For example, a round sphere of radius r has Ricci scalar curvature R =
>>2/r^2 at every point.  [With a click of his fingers, a sphere of radius
>>r appears on it, with a small triangle drawn on it, edges bulging
>>slightly.]  

>Well, I don't like to mention this, but nobody has mentioned a Ricci
>SCALAR before. 

WHAT??????????  [The wizard, obviously still stressed from the previous
incident, goes ballistic.  He waves his staff about and shoots fireballs
in all four directions of the compass, cursing with anger.]   Listen
here, Oz!  I keep careful notes on EVERYTHING I EVER TELL YOU, so don't
say I never mentiond the Ricci scalar.  Here's what I said, and I
quote....  [He ruffles around on his desk through enormous stacks of
yellowing papers, pauses, scratches his head, and then yanks out one
from the middle of the very tallest pile.]  Ahem:

"7.  The EINSTEIN TENSOR.  The matrix g_{ab} is invertible
and we write its inverse as g^{ab}.  We use this to cook up some tensors
starting from the Riemann curvature tensor and leading to the Einstein
tensor, which appears on the left side of Einstein's marvelous equation
for general relativity.  We will do this using coordinates to save
time... though later we should do this over again without coordinates.
This part is the only profound and mysterious part, at least to me.

Okay, starting from the Riemann tensor, which has components R^a_{bcd},
we now define the RICCI TENSOR to have components

         R_{bd} = R^c_{bcd}

where as usual we sum over the repeated index c.  Then we "raise an
index" and define

         R^a_d = g^{ab} R_{bd},

and then we define the RICCI SCALAR by

         R = R^a_a

The Riemann tensor knows everything about spacetime curvature, but these
gadgets distill certain aspects of that information which turn out to be
important in physics.  Finally, we define the Einstein tensor by

         G_{ab} = R_{ab} - (1/2)R g_{ab}."

Now GET OUT!  How do you expect to become a sorcerer at this rate?

[Oz skulks out, thinking the old fellow must have drank too much coffee
today... "What an old fart," he mutters.]  

Article 99503 (40 more) in sci.physics:
From: Edward Green
Subject: Re: Tensors for twits please.
Date: 11 Feb 1996Article 99503 (40 more) in sci.physics:
From: Edward Green
Subject: Re: Tensors for twits please.
Date: 11 Feb 1996 18:07:05 -0500
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[email-masked] (john baez)' wrote: 
 
Pursuing my modest and realistic programme of learning math and physics by
osmosis... 
 
>If you hand me any vector space I could call its elements vectors and 
>call the elements of the dual vector space "covectors".  It's certainly 
>interesting to think about this.  But I wasn't really trying to be so 
>general here... 
 
I assume you refer to the fact that the space of all linear functions on a
vector space is itself a vector space;  the dual space...  (and the dual of
the dual is the original space,  up to isomorphism and a cavil or two...). 
 Yes, that must be it...  
 
>I was mainly talking about the tangent space, whose 
>elements I call "tangent vectors" or (for short) "vectors", and its dual 
>space, the cotangent space, whose elements I call "cotangent vectors" or 
>(for short) "covectors".   
 
So the covectors associated with a point are just the linear functions on
the vectors,  which in 4 dimensions look a hell of a lot like vectors, 
being represented as ordered quadruples (quartets?). 
 
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[email-masked] (john baez)' wrote: 
 
Pursuing my modest and realistic programme of learning math and physics by
osmosis... 
 
>If you hand me any vector space I could call its elements vectors and 
>call the elements of the dual vector space "covectors".  It's certainly 
>interesting to think about this.  But I wasn't really trying to be so 
>general here... 
 
I assume you refer to the fact that the space of all linear functions on a
vector space is itself a vector space;  the dual space...  (and the dual of
the dual is the original space,  up to isomorphism and a cavil or two...). 
 Yes, that must be it...  
 
>I was mainly talking about the tangent space, whose 
>elements I call "tangent vectors" or (for short) "vectors", and its dual 
>space, the cotangent space, whose elements I call "cotangent vectors" or 
>(for short) "covectors".   
 
So the covectors associated with a point are just the linear functions on
the vectors,  which in 4 dimensions look a hell of a lot like vectors, 
being represented as ordered quadruples (quartets?). 
 
>If I have a tangent vector at each point of 
>spacetime I say I have a "vector field", while if I have a cotangent 
>vector at each point of spacetime I might say I have a "1-form". 
 
Oh... so that's what a 1-form is...   so much nomenclature,  more than
concepts even!  I've got a little list: 
 
     tangent/cotangent,   direct/dual,   covariant/contravariant... 
 
And don't forget the 1-forms...  
 
[Q]:  Am I right that through an historical quirk we say tangent vectors
are "covariant" under coordinate transformations while cotangent vector are
"contravariant"?   Are we that fortunate?   :-)   Because,  if Y is a
(coordinate representation of a) tangent vector and Y' = TY  our
transformed representation,  and if X a (coordinate representation of a)
covector,  then we must have:   X' = XT^(-1) .  That way; 
 
    X'.Y' = X'T^(-1)TY = X.Y        i.e.,  scalars are invariant 
 
(I am thinking in the bad old evil matrix notation... I will get this index
stuff down presently)  So it seems reasonable that since the inverse
transformation appears next to the cotangent vector,  it "contravaries"...
Or have I stated this backwards and created confusion?  (As usual). 
 
-- 
 
Ed Green / [email-masked] 
 
"All coordinate systems are equal, 
 but some are more equal than others".

Article 99219 (38 more) in sci.physics:
From: Edward Green
Subject: Re: Tensors for twits please.
Date: 9 Feb 1996 12:24:43 -0500
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[email-masked] (Matthew P Wiener)' wrote: 
 
>In ordinary xyz coordinates, you have two ways of describing points.  The 
>ordinary way is just to measure off the x,y, and z axes. 
> 
>The dual way is to count off x=?? and y=?? and z=?? planes. 
> 
>These lead to the same answer, so most people ignore the difference. 
 
Why God bless you sir!   I am glad somebody else noticed this.  I had to
figure this out for myself,  years ago.  I always assumed it was standard
knowledge, somewhere, once,  too.    
 
>But when you change to or between curvilinear coordinates, the two ways 
>are distinct. 
 
Well just for the sake of accuracy,  or to be a twit,  whichever,  I'd
point out that you don't have to go so far as curvilinear coordinates to
make this distinction,  just general linear coodinates.   
 
For me it falls out of the two ways we can think of a linear transformation
represented by a matrix. 
 
In the first case we can think of AX as taking the inner product of the
vector X with each of the row vectors of A.  This amounts to the dual
approach,  as you say,  measuring off X against sets of parallel planes to
get the new coordinates.  But we can also think of AX as telling us to take
components of X to tell us how much to take!  This is the direct approach. 
 
So AX is either: 
 
1)  A list of coordinates formed from X by taking the inner product with
covectors. 
 
2)  The coordinates of a vector formed by using the old coordinates to form
a linear combination of vectors. 
 
We can actually think of A as representing two *different* linear
transformations this way...  one operates in the direct space,  and returns
a list of numbers to be used in the dual space,  the other uses the
preexisting list of numbers to build an object in the dual space. 
 
When we picture a matrix as "rotating and stretching" a vector,  we are
implicitly mapping the dual space back to the direct space... or in other
words,  just failing to make the distinction. 
 
Ok,  maybe I am being just a *tad* mystical... 
 
And we haven't even started to ponder A^-1 ! 
 
>Off-hand, I have not seen it spelled out anywhere else 
>as this, but I assume it was once standard knowledge. 
 
Some relatively useful ways of looking at things seem to have fallen off
the standard technical curriculum at some point.  There used to be such a
subject as "spherical trigonometry" once upon a time,  did there not?  Well
I never met it,  and though I do not think it is the ne-plus-ultra of
sophistication,  it probably could be useful to know about. 
 
For example,  the kind of Stokes theorem we noticed for parallel
transporting a vector around the surface of a sphere;  rotation is
proportional to the enclosed area (I assume the ambiguity in the enclosed
area gives us the compliment w.r.t. pi);  must be a factoid from this area.

-- 
 
Ed Green / [email-masked] 
 
"All coordinate systems are equal, 
 but some are more equal than others".

Article 99607 (39 more) in sci.physics:
From: Matthew P Wiener
(SAME) Subject: Re: Tensors for twits please.
Date: 12 Feb 1996 16:13:28 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 80
NNTP-Posting-Host: sagi.wistar.upenn.edu
In-reply-to: [email-masked] (Edward Green)

In article <[email-masked]>, egreen@nyc (Edward Green) write
s:

>>If I have a tangent vector at each point of spacetime I say I have a
>>"vector field", while if I have a cotangent vector at each point of
>>spacetime I might say I have a "1-form".

>Oh... so that's what a 1-form is...   so much nomenclature,  more than
>concepts even!  I've got a little list: 

>     tangent/cotangent,   direct/dual,   covariant/contravariant... 

>And don't forget the 1-forms...  

The standard examples of a vector field on a manifold (for example,
spacetime) are sums of the form A.@/@x + B.@/@y + ....  The standard
examples of a covector field are sums P.dx + Q.dy + ....

Are these different?

Most assuredly!

If you change coordinates, you get different results.

Consider X=X(x,y,...), Y=Y(x,y,...), ....  Then for any function f on the
manifold (meaning f(x,y,...)=f(X,Y,...) and so on in coordinates):

    @f     @f            @X @f   @Y @f            @X @f   @Y @f
  A.-- + B.-- + ... = A.(--.-- + --.-- +...) + B.(--.-- + --.-- +...) + ...
    @x     @y            @x @X   @x @Y            @y @X   @y @Y

                         @X     @X        @f      @Y     @Y        @f
                    = (A.-- + B.-- + ...).-- + (A.-- + B.-- + ...).-- + ...
                         @x     @y        @X      @x     @y        @Y

In contrast,
                         @x      @x                 @y      @y
 P.dx + Q.dy + ... = P.( -- dX + -- dY + ...) + Q.( -- dX + -- dY + ...) + ...
                         @X      @Y                 @X      @Y

                        @x     @y                @x     @y
                   = (P.-- + Q.-- + ...).dX + (P.-- + Q.-- + ...).dY + ....
                        @X     @X                @Y     @Y

The assignment of n-tuples, one for each coordinate system, such that they
relate under a change of coordinates as in the first expression above, is
called a "contravariant" vector field.  If they relate under a change of
coordinates as in the second expression above, they are called a "covariant"
vector field.

The @/@x, @/@y, ... form a basis for the contravariant vectors, while dx,
dy, ... form a basis for the covariant vectors.

This is old-style talk.  New style talk is to call the former vectors, and
the latter covectors.  Covector fields are also called 1-forms.

>[Q]:  Am I right that through an historical quirk we say tangent vectors
>are "covariant" under coordinate transformations while cotangent vector are
>"contravariant"?

Exactly backwards.  (Although, as you could have guessed, there are people
who use the backwards convention.)

@/@x strikes out a tangent direction, along the direction of increasing x
coordinate.  Note that I am not assuming that x is *the* x coordinate.

Rather, I am assuming that we have coordinates on our manifold, meaning
a map from some portion of our manifold to a vector of R^n values, and
the first coordinate is x=x(p), the second coordinate is y=y(p), etc.
If I fix a point p0 on the manifold, with coordates x0,y0,..., and then
I freeze the y0,z0,... coordinates but let the x vary, I get a curve that
passes through p0.  @?/@x, to be evaluated at x0, perfectly captures the
notion of tangent in the x direction at p0.

In contrast dx pushes, not in the x direction, but perpendicular to the
other tangent directions @/@y, @/@z, ....  The picture of covectors as
a hyperplane dissection still applies here: the 0-labelled hyperplace
is the one containing @/@y, @/@z, ...., and the 1-labelled hyperplane
is parallel to this but an infinitesimal distance dx over.
-- 
-Matthew P Wiener ([email-masked])

Article 99356 (37 more) in sci.physics:
From: Edward Green
Subject: Re: GR Tutorial Water Cooler
Date: 10 Feb 1996 19:51:30 -0500
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<[email-masked]>' wrote: 
 
> 
>Well, it could be one, er, or the other. I dunt reely hunderstand der 
>question. Trouble as I see it is dat dere's bits of the universe as like 
>wot haven't got 'ere yet. So I guess they mayunt 'av got to your scrap 
>of spacetime. Like the hobservehable huniverse is getting bigger hand 
>more bits of hunivese is affecting our bit of spacetime hinnit? But they 
>didn't before. But hour bit hov spacetime is like my room for der last 
>ten minoots hand some distant galaxy sumwere will haffect my spacetime 
>in ten munoots time, but it didnt before. Oh, it might be like a taut 
>string as wot sumone av plucked sumwere far away. Its dead flat until 
>the triangular wave comes flying past. I bet der speed of lite av summit 
>to do wiv it. 
> 
>[Apologies. I didn't have the *faintest* idea of how to start answering 
>Ed's question, so to keep the ball rolling I thought I would be one of 
>the local very dim ball players. However I seem to have stumbled on a 
>plausible answer as I wrote. Unfortunately I omitted to change dialect. 
>This rather b*gg**d the joke.] 
 
I'm afraid it did.  Would like to translate that?  I've been to St. Mary le
Bow's,  but it didn't take.  Too long out in the world,  I guess. 
 
No,  wait,  I've got it now.  You just transcribed this from a random
dockworkers' conversation you overheard in a pub!  It was hard to hear over
the soccer results,  but I think you got the gist of it.   Then the other
beefy guy said  "No,  ere,  you take that back what you said about my
universe!", with a hurt look on his dull but honest face,  and a fistfight
ensued.  Fortunately,  you were able to duck out the back door without
spilling your pint. 
   
Serious mode:  Let me try another analogy.  Spacetime is like a drumhead. 
A source is like a pencil poking up under the drumhead.  Boundry conditions
are like the rim of the drum.  The pencil is like a region where the
stress-energy tensor is non-zero.  G = T tells us how the drumhead has to
respond there (well partially... ignore this qualification for a moment). 
But... what tells us the shape of the rest of the drumhead? 
 
G = 0.  Well, that's part of the story.  I suppose any value for the
Riemann tensor consistent with G = 0 is ok then?  But how much does this
limit us?  Someone said there were "gravitational wave solutions".  I
assume these are analogous to electromagnetic waves... source-free,  we can
take any superposition of them, and have a source-free solution. 
 
So are these the only ambiguity?  How far can we push the drum head
analogy?  We are also tempted to notice that the drumhead stretched by the
pencil can vibrate,  and equate these with the gravitational waves (except
that in space time,  nothing ever really happens...  it's timeless,
changeless, etern...  whoops... I hear a rhapsody...  anyway space
vibrates,  spacetime just is,  but ignore this. 
 
So on the drumhead,  when we specify "there are no waves" we then have a
unique solution for the shape of the head poked from underneath by the
pencil. Is the same sort of thing true in GR?  Is there a "base" solution
corresponding to the undistrubed drumhead,  shaped only by nearby mass
distributions?  What *is* the equation determing the "propagation" of the
Riemann in empty space time (T = 0)...  is "G = 0" a sufficient
description?  Does this tie the Riemann down enough so that "the answer is
specified up to a linear superposition of gravity waves",  whatever that
might mean? 
 
The Weyl is supposed to have 10 independent components.  So another way of
asking this is,  are all ten of these tied down by picking an element out
of the gravity wave Hibert space... (zap!... sorry,  I have no idea if they
form a Hilbert space...)  Actually,  Oz  (are you still there?),  maybe
this is what you were getting at... excuse me,  what "Bert" was gettin at
by "Oh, it might be like a taut string as wot sumone av plucked sumwere far
away. Its dead flat until the triangular wave comes flying past." 
 
But what is this dern "tidal force" thing some of these guys are talking
about...   there seems to seems to be some feeling this should be
associated with the Weyl too...  seems to mess up this picture a bit...  
this is due to local mass distribution... 
 
Hey!   Here comes a passing expert!  Let's ask HIM!!!  (Volunteers?) 
 
-- 
 
Ed Green / [email-masked] 
 
"All coordinate systems are equal, 
 but some are more equal than others". 

Article 99481 (36 more) in sci.physics:
From: Oz <[email-masked]>
(SAME) Subject: Re: GR Tutorial Water Cooler
Date: Sun, 11 Feb 1996 19:34:51 +0000
Organization: Oz
Lines: 126
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In article <[email-masked]>, Edward Green
<[email-masked]> writes
><[email-masked]>' wrote: 
> 
>> 
>>Well, it could be one, er, or the other. I dunt reely hunderstand der 
>>question. Trouble as I see it is dat dere's bits of the universe as like 
>>wot haven't got 'ere yet. So I guess they mayunt 'av got to your scrap 
>>of spacetime. Like the hobservehable huniverse is getting bigger hand 
>>more bits of hunivese is affecting our bit of spacetime hinnit? But they 
>>didn't before. But hour bit hov spacetime is like my room for der last 
>>ten minoots hand some distant galaxy sumwere will haffect my spacetime 
>>in ten munoots time, but it didnt before. Oh, it might be like a taut 
>>string as wot sumone av plucked sumwere far away. Its dead flat until 
>>the triangular wave comes flying past. I bet der speed of lite av summit 
>>to do wiv it. 
>> 
>>[Apologies. I didn't have the *faintest* idea of how to start answering 
>>Ed's question, so to keep the ball rolling I thought I would be one of 
>>the local very dim ball players. However I seem to have stumbled on a 
>>plausible answer as I wrote. Unfortunately I omitted to change dialect. 
>>This rather b*gg**d the joke.] 
> 
>I'm afraid it did.  Would like to translate that?  I've been to St. Mary le
>Bow's,  but it didn't take.  Too long out in the world,  I guess.

Oh, dear, Ed. I rather hoped you would treat this with the derision it
deserved, and simply distain to answer. Unfortunately you have hoist me
with my own petard and er, well, dropped me in the er um thingy whatsit
stuff. I don't suppose that American Football players talk like this, do
they? Quite an unlikely accent. 
<Oz goes out and shoots himself, in embarrasement, which was painful.>

>No,  wait,  I've got it now.  You just transcribed this from a random
>dockworkers' conversation you overheard in a pub!  It was hard to hear over
>the soccer results,  but I think you got the gist of it.   Then the other
>beefy guy said  "No,  ere,  you take that back what you said about my
>universe!", with a hurt look on his dull but honest face,  and a fistfight
>ensued.  Fortunately,  you were able to duck out the back door without
>spilling your pint. 
>   
>Serious mode:  Let me try another analogy.  Spacetime is like a drumhead. 
>A source is like a pencil poking up under the drumhead.  Boundry conditions
>are like the rim of the drum.  The pencil is like a region where the
>stress-energy tensor is non-zero.  G = T tells us how the drumhead has to
>respond there (well partially... ignore this qualification for a moment). 
>But... what tells us the shape of the rest of the drumhead? 

I think we just gotta measure it. Keeps the experimentalists in work.

>G = 0.  Well, that's part of the story.  I suppose any value for the
>Riemann tensor consistent with G = 0 is ok then?  But how much does this
>limit us?  Someone said there were "gravitational wave solutions".  I
>assume these are analogous to electromagnetic waves... source-free,  we can
>take any superposition of them, and have a source-free solution. 

G=0 would surely correspond to a situation where space bending was due
to another cause, and anyway outside of our (observable?) volume. Unless
it's just a local G=0.

>So are these the only ambiguity?  How far can we push the drum head
>analogy?  We are also tempted to notice that the drumhead stretched by the
>pencil can vibrate,  and equate these with the gravitational waves (except
>that in space time,  nothing ever really happens...  it's timeless,
>changeless, etern...  whoops... I hear a rhapsody...  anyway space
>vibrates,  spacetime just is,  but ignore this. 

Yes, but we can still describe dx/dt, dx/dy and so on. So it doesn't
stop us working up some messy maths to keep the mathematicians in work.
And that's before we start projecting things onto planes, onto curves
even, and oh dear, do we really want to go through with this?

>So on the drumhead,  when we specify "there are no waves" we then have a
>unique solution for the shape of the head poked from underneath by the
>pencil. Is the same sort of thing true in GR?  Is there a "base" solution
>corresponding to the undistrubed drumhead,  shaped only by nearby mass
>distributions?  What *is* the equation determing the "propagation" of the
>Riemann in empty space time (T = 0)...  is "G = 0" a sufficient
>description?  Does this tie the Riemann down enough so that "the answer is
>specified up to a linear superposition of gravity waves",  whatever that
>might mean? 

Wow Ed, you have lost me here. I dunno, you are way past me.

I just hope all this stuff is linear so we can just add curvatures or
whatever in a simple way. Hahahaha, ...... ha?

>The Weyl is supposed to have 10 independent components.  So another way of
>asking this is,  are all ten of these tied down by picking an element out
>of the gravity wave Hibert space... (zap!... sorry,  I have no idea if they
>form a Hilbert space...)  Actually,  Oz  (are you still there?),  maybe
>this is what you were getting at... excuse me,  what "Bert" was gettin at
>by "Oh, it might be like a taut string as wot sumone av plucked sumwere far
>away. Its dead flat until the triangular wave comes flying past." 

Me I just thought these 10 independent components were the 'external'
forms of the Ricci. One describes whatsits due to the momentum flow
inside, the other the curvature coming from the outside.

>But what is this dern "tidal force" thing some of these guys are talking
>about...   there seems to seems to be some feeling this should be
>associated with the Weyl too...  seems to mess up this picture a bit...  
>this is due to local mass distribution... 

Hey, I am not too sure that the Ricci IS due to AN infinitesimal speck
of momentum flow. It feels like it should explain how to calculate
groups of bits of momentum flow due to gravitational curvature. In this
case tidal and wave expressions all have an effect. We ought to ask prof
in the lecture and see if he can explain. Maybe, if we get to working
stuff out, it will become clear?

>Hey!   Here comes a passing expert!  Let's ask HIM!!!  (Volunteers?) 

Dunno about that Ed. There's experts, and then there's experts.
Particularly in this college. Some of them float 6" above the floor, and
they are STRANGE.

<An orange glow appears round Oz, and with a horrified look on his face
he is swept down the corridor into a room marked AA+VV. Screams emanate
from it. Ed shrugs his shoulders and goes into class.>

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 99422 (34 more) in sci.physics:
From: Matthew P Wiener
Subject: Minkowski Metrics for Tensor Twits
Date: 11 Feb 1996 15:24:59 GMT
Organization: The Wistar Institute of Anatomy and Biology
Lines: 96
Distribution: world
NNTP-Posting-Host: sagi.wistar.upenn.edu
In-reply-to: Oz <[email-masked]>

In article <[email-masked]>, Oz <Oz@upthorpe writes:
>In article <[email-masked]>, Geoffrey A <?@?.landis>

>>What I think he means to say here is, by using a metric with a diagonal
>>[-1,1,1,1], known as the "Lorentz metric", *all* of the special
>>relativistic effects are already included in the geometry.

>Yeah. Braindead, that's me. I even bet that there could be a little 'c'
>somewhere that has been downgraded to a '1', I think that was mentioned
>waaaaaay up the thread near when it started, some seconds after the big
>bang.

In long form, let's work with the metric [1,-1/c^2,-1/c^2,-1/c^2], ie,

                2        2    1    2    1    2    1    2
              ds   =   dt  - --- dx  - --- dy  - --- dz  .
                             c^2       c^2       c^2

>I *knew* the metric had been glossed over. What I don't quite see is how
>this tensor produces relativistic effects.

You're in luck: it's very trivial.  ds is infinitesimal proper time.

That is really all you need to know to do special relativity.

The Lorentz transformations can be derived by noticing that ds^2=0 for a
photon, and that this is invariant between inertial frames.

Total proper time is easily computed as the integral of itsy bitsy proper
times totalled together.

For example, consider a trip from (0,0,0,0) to (20,0,0,0) along the stay
at home path x=y=z=0.  Then dx=dy=dz=0 and elapsed proper time is just
the integral of dt from 0 to 20, which is just 20.

Now consider the path x(t)=ct/2, y=z=0 for t=0 to 10, and x(t)=10c-ct/2,
y=z=0 for t=10 to 20.  That is, travel at half the speed of light in
the x direction for 10 years, and then back home for the next 10 years.
(Note that "10c" in the formula above is really "10 light-(time-unit)s".)

dx/dt=c/2 going out, and -c/2 coming back.  (dx/dt)^2=c^2/4 either way.
So ds^2=dt^2-dx^2/c^2=(1-1/4)dt^2, and ds=sqrt(1-1/4)dt.  Integrating
from t=0 to 10 gives sqrt(3/4).10=8.66 years.  This is sometimes called
the "twin paradox".  If you want to replace x=x(t) with a more realistic
path, be my guest.  The integral is much messier, of course, but the idea
is the exact same.

The sqrt(1-1/4) factor is just the world famous sqrt(1-v^2/c^2) factor, as
you've no doubt noticed already.

>                                           I am sure it's quite simple
>and obvious to you tensor tyros, you have done a full and proper course.
>Can anyone spell it out in words of one syllable, oh well Ok then three
>syllables, how this happens?

This was Minkowski's great contribution.  The mathematics of relativity
existed before Einstein.  Einstein rederived it, but more importantly,
he realized that it was secondary to the relative principle, and not, as
others were thinking, a conspiracy of mathematics to force relativistic
invariance.  (So backwards was the thinking pre-Einstein, that no one
really noticed that they _had_ invariance.)  Nevertheless, Einstein's
version of the mathematics was piecemeal and not particularly conceptual
on its own.  It fit the physics, and that was that.

Minkowski realized relativity was a certain skewed but still very clear
form of geometric thinking.  That is what the metric encapsulates.

For example, to compute arc length of a curve, one merely has to write
down ds^2=du^2+dv^2 and identify the appropriate u and v and integrate.
Advanced calculus stuff, very Pythagorean.  Any beginning book on the
differential geometry of curves and surfaces should contain this in
gory detail somewhere.

What you _generally_ get, though, is ds^2 = A.du^2 + 2B.du.dv + C.dv^2.
Think of ds^2=du^2+dv^2 as special geometry and the above as general
geometry.  The kinematic passage from SR to GR is pretty much the same.

As Einstein realized from the equivalence principle, relativistic gravity
is simply not compatible with Euclidean geometry.  In his writings, he
liked to give the example of the relativistic merry-go-round.  But the
mere business of light--which follows a geodesic--bending because of
gravity is already non-Euclidean.

So the first step on the path to GR is doing physics on a metric.  So
long as the metric is [1,-1,-1,-1], you are only doing SR.  But you are
doing SR in a form ready to jump to GR.
-- 
        I don't know how often he told me, "You're stupid" and
        suchlike. That helped me a lot.         --Heisenberg on Pauli

-Matthew P Wiener ([email-masked])

Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Summary: 
Expires: 
References: <[email-masked]> <[email-masked]>
Sender: 
Followup-To: 
Distribution: 
Organization: University of California, Riverside
Keywords: 

In article <[email-masked]> [email-masked] (Edward Green) writes:

>Here goes:  For a stationary distribution of masses,  it is possible to
>define a function of space in the prefered coordinate frame, called the
>"gravitational potential",  which has the property of predicting elapsed
>proper time for test points transported quasi-stationarily about our
>system, provided we make the additional physical assumption that space is
>flat at infinity. 

>Okey doke?   Have I dotted the t's and crossed the i's so far?   :-)  

Pretty good, though instead of "for a stationary distribution of masses"
you should say "for a static spacetime geometry", since that's what
counts.  After all, it's really the fact that the geometry of spacetime
is not rollicking around, and behaves like flat Minkowski spacetime at
infinity, which lets you do your clock experiment and define a
"gravitational potential" with the properties you desire.  The
distribution of masses, or even the whole stress-energy tensor, could in
principle be static without the spacetime geometry being static.
Remember, the stress-energy tensor does NOT completely determine the
geometry of spacetime, since gravitational waves with no Ricci curvature
but plenty of Weyl curvature could be zipping past --- even in an
asymptotically flat spacetime.

Of course, properly defining a "static, asymptotically flat spacetime
geometry" is a wee bit subtle, but let's not worry about that.  

So: I'm back, Ed seems to have got cleared up on the "gravitational
potential" business, and Oz has relearnt his special relativity from
the modern viewpoint, in which everything is based on the metric.
Moreover, everybody has spent the weekend boning up on their tensors and
is ready to go.  

That means it's time for a new improved course outline.  It looks like
the old one, but mysteriously it has gotten a bit longer in various
places....

1.  A TANGENT VECTOR or simply VECTOR at the point p of spacetime may be
visualized as an infinitesimal arrow with tail at the point p.  The
tangent vectors at p form a vector space called the TANGENT SPACE; in
other words, we can add them and multiply them by real numbers.

Suppose we work in a local coordinate system with coordinates
(x^0,x^1,x^2,x^3).  (Since we are working in 4d spacetime there are 4
coordinates; we may think of x^0 as the time coordinate t and the other
3 as x, y, and z, but we don't need to think of them this way, since
we're using an utterly arbitrary coordinate system.)  Then we can
describe a tangent vector v by listing its components (v^0,v^1,v^2,v^3)
in this coordinate system.  For short we write these components as v^a,
where the superscript a, like all of our superscripts and subscripts,
goes from 0 to 3. 

2.  A COTANGENT VECTOR or simply COVECTOR at the point p is a function f
that eats a tangent vector v and spits out a real number f(v) in a linear way.
Cotangent vectors can be viewed as ordered stacks of parallel planes in the
tangent space at p.  They don't "point" like tangent vectors do;
instead, they "copoint".  

Working in local coordinates, we define the components of a covector f
to be the numbers (f_0,f_1,f_2,f_3) you get you get when you evaluate f
on the basis vectors:

f_0 = f(1,0,0,0)
f_1 = f(0,1,0,0)
f_2 = f(0,0,1,0)
f_3 = f(0,0,0,1)

3.  A TENSOR of "rank (0,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a number.  The output must depend linearly on each input.   

A TENSOR of "rank (1,k)" at a point p of spacetime is a function that
takes as input a list of k tangent vectors at the point p and returns as
output a tangent vector at the point p.  The output must depend linearly
on each input.    

More generally, a TENSOR of "rank (j,k) at a point p of spacetime is a
function that takes as input a list of j cotangent vectors and k tangent
vectors and returns as output a number.  The output must depend linearly
on each input.   Note that this definition is compatible with the
previous ones!  This is obvious for the rank (0,k) tensors, but for the
rank (1,k) ones we need to check that a function that eats k vectors and
spits out a vector v can be reinterpreted as a function that eats k
vectors and one covector f and spits out a number.  We just let the
covector f eat the vector v and spit out f(v)!

Similarly, note that a vector can be reinterpreted as a tensor of rank
(1,0), and a covector can be reinterpreted as a tensor of rank (0,1).

In local coordinates we write the components of a tensor T of rank (j,k)
as a monstrous array T^{ab....c}_{de....f} with j superscripts and k
subscripts.   Again, all superscripts and subscripts range from 0 to 3; 
each number T^{ab....c}_{de....f} is simply the number the tensor spits
out when fed a suitable wad of basis vectors and covectors.  I will
describe this in more detail in the following example:

4.  The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
v,w and spits out a number g(v,w), which we think of as the "dot
product" or "inner product" of the vectors v and w.  This lets us
compute the length of any tangent vector, or the angle between two
tangent vectors.  Since we are talking about spacetime, the metric need
not satisfy g(v,v) > 0 for all nonzero v.  A vector v is SPACELIKE if 
g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

The inner product g(v,w) of two tangent vectors is given by

        g(v,w) = g_{ab} v^a w^b

for some matrix of numbers g_{ab}, where we sum over the repeated
indices a,b (this being the so-called EINSTEIN SUMMATION CONVENTION).
Another way to think of it is that our coordinates give us a basis of
tangent vectors at p, and g_{ab} is the inner product of the basis
vector pointing in the x^a direction and the basis vector pointing in
the x^b direction.

The metric is the star of general relativity.  It describes everything
about the geometry of spacetime, since it lets us measure angles and
distances.  Einstein's equation describes how the flow of energy and
momentum through spacetime affects the metric.  What it actually affects
is something about the metric called the "curvature".  The biggest job
in learning general relativity is learning to understand curvature.

5.  PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  We can define the "torsion tensor", with
           components t_{ab}, as follows.  Take a little vector of size
           epsilon pointing in the a direction, and a little vector of
           size epsilon pointing in the b direction.  Parallel translate
           the vector pointing in the a direction by an amount epsilon
           in the b direction.  Similarly, parallel translate the vector
           pointing in the b direction by an amount epsilon in the a
           direction.  (Draw the resulting two vectors.)  If the tips
           touch, up to terms of epsilon^3, there's no torsion!
           Otherwise take the difference of the tips and divide by epsilon^2.
           Taking the limit as epsilon -> 0 we get the torsion t_{ab}.
           We say that parallel translation is "torsion-free" if t_{ab}
           = 0.

A GEODESIC is a curve whose tangent vector is parallel transported along
itself.  I.e., to follow a geodesic is to follow ones nose while never
turning ones nose.  A particle in free fall follows a geodesic in
spacetime.

6.  The CONNECTION is a mathematical gadget that describes "parallel
translation along an infinitesimal curve in a given direction".  In
local coordinates the connection may be described using the components
of the CHRISTOFFEL SYMBOL Gamma_{ab}^c.  There is an explicit formula
for these components in terms of components g_{ab} of the metric, which
may be derived from the assumptions a-c above.  However, this formula is
very frightening, so I will only describe it to those who have passed
certain tests of courage and valor.

7.  The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point
of spacetime.  Thus it takes three tangent vectors, say u, v, and w as
inputs, and outputs one tangent vector, say R(u,v,w).  The Riemann tensor
is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number.  The vector w comes back a bit changed by its
journey; it is now a new vector w'.  We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Thus the Riemann tensor keeps track of how much parallel translation
around a wee parallelogram changes the vector w.             

In addition to this simple coordinate-free definition of the Riemann
tensor, we may describe its components R^a_{bcd} using coordinates.  Namely, 
the vector R(u,v,w) has components  

        R(u,v,w)^a = R^a_{bcd} u^b v^c w^d

where we sum over the indices b,c,d.  Another way to think of this is
that if we feed the Riemann tensor 3 basis vectors in the x^b, x^c, x^d
directions, respectively, it spits out a vector whose component in the
x^a direction is R^a_{bcd}. 

There is an explicit formula for the components R^a_{bcd} in terms of
the Christoffel symbols.  Together with the aforementioned formula for
the Christoffel symbols in terms of the metric, this lets us compute the
Riemann tensor of any metric!  Thus to do computations in general
relativity, these formulas are quite important.  However, they are not
for the faint of heart, so I will not describe them here!  

7.  The RICCI TENSOR.  The matrix g_{ab} is invertible
and we write its inverse as g^{ab}.  We use this to cook up some tensors
starting from the Riemann curvature tensor and leading to the Einstein
tensor, which appears on the left side of Einstein's marvelous equation
for general relativity.  We will do this using coordinates to save
time... though later we should do this over again without coordinates.

Okay, starting from the Riemann tensor, which has components R^a_{bcd},
we now define the RICCI TENSOR to have components

         R_{bd} = R^c_{bcd}

where as usual we sum over the repeated index c.  

The physical significance of the Ricci tensor is best explained by an
example.  So, suppose an astronaut taking a space walk accidentally
spills a can of ground coffee.

Consider one coffee ground.  Say that a given moment it's at the point P
of spacetime, and its velocity vector is the tangent vector v.  Note:
since we are doing relativity, its velocity is defined to be the tangent
vector to its path in *spacetime*, so if we used coordinates v would
have 4 components, not 3.  

The path the coffee ground traces out in spacetime is called its
"worldline".  Let's draw a little bit of its worldline near P:

   |
  v^
   |
   P
   |
   |
   |
    
The vector v is an arrow with tail P, pointing straight up.   I've tried
to draw it in, using crappy ASCII graphics.

Now imagine a bunch of comoving coffee grounds right near our original
one.  What does this mean?  Well, it means that for any tangent vector w
at P which is orthogonal to v, if we follow a geodesic along w for a
certain while, we find ourselves at a point Q where there's another coffee
ground.  Let me draw the worldline of this other coffee ground.

   |       |
  v^       ^v'
   | w     |
   P->-----Q
   |       |
   |       |
   |       |

I've drawn w so you can see how it is orthogonal to the worldline of our
first coffee ground.  The horizontal path is a geodesic from P to Q,
which has tangent vector w at Q.  I have also drawn the worldline of the
coffee ground which goes through the point Q of spacetime, and I've also
drawn the velocity vector v' of this other coffee ground.

What does it mean to say the coffee grounds are comoving?  It means
simply that if we take v and parallel translate it over to Q along the
horizontal path, we get v'.  

This may seem like a lot of work to say that two coffee grounds are
moving in the same direction at the same speed, but when spacetime is
curved we gotta be very careful.  Note that everything I've done is
based on parallel translation!  (I defined geodesics using parallel
translation.)  

Now consider, not just two coffee grounds, but a whole swarm of comoving
coffee grounds near P.  If spacetime were flat, these coffee grounds
would *stay* comoving as time passed.  But if there is a gravitational
field around (and there is, even in space), spacetime is not flat.  So
what happens?

Well, basically the coffee grounds will tend to be deflected, relative
to one another.  It's not hard to figure out exactly how much they will
be deflected!  We just use the definition of the Riemann curvature!  We
get an equation called the "geodesic deviation equation".

But let me not do that just yet.  Instead, let me say what the Ricci
tensor has to do with all this.  Then, when we use the "geodesic
deviation equation" to work out the deflection of the coffee grounds
using the Riemann curvature, we will see what this has to do with the
usual definition of the Ricci tensor in terms of the Riemann curvature.

Imagine a bunch of coffee grounds near the coffee ground that went
through the point P.  Consider, for example, all the coffee grounds that
were within a given distance at time zero (in the local rest frame of the
coffee ground that went through P).  A little round ball of coffee
grounds in free fall through outer space!  As time passes this ball will
change shape and size depending on how the paths of the coffee gournds
are deflected by the spacetime curvature.  Since everything in the
universe is linear to first order, we can imagine shrinking or expanding,
and also getting deformed to an ellipsoid.  There is a lot of
information about spacetime curvature encoded in the rate at which this
ball changes shape and size.  But let's only keep track of the rate of
change of its volume!  This rate is basically the Ricci tensor.  

More precisely, the second time derivative of the volume of this little
ball is approximately

R_{ab} v^a v^b

times the original volume of the ball.  This approximation becomes
better and better in the limit as the ball gets smaller and smaller.  

8.  The RICCI SCALAR.  Starting from the Ricci tensor, we define

         R^a_d = g^{ab} R_{bd}.

As always, we follow the Einstein summation convention and sum over
repeated indices when one is up and the other is down.  This process,
which turned one subscript on the Ricci tensor into a superscript, is
called RAISING AN INDEX.  Similarly we can LOWER AN INDEX, turning
any superscript into a subscript, using g_{ab}.

Then we define the RICCI SCALAR by

         R = R^a_a.

This process, whereby we get rid of a superscript and a subscript in a
tensor by summing over them a la Einstein, is called CONTRACTING.

9.  The EINSTEIN TENSOR.  Finally, we define the Einstein tensor by

         G_{ab} = R_{ab} - (1/2)R g_{ab}.

You should not feel you understand why I am defining it this way!!
Don't worry!  That will take quite a bit longer to explain; the point is
that with this definition, local conservation of energy and momentum
will be an automatic consequence of Einstein's equation.  To understand
this, we need to know Einstein's equation, so we need to know about:

10.  The STRESS-ENERGY TENSOR.  The stress-energy is what appears on the
right side of Einstein's equation.  It is a tensor of rank (0,2), and it
defined as follows: given any two tangent vectors u and v at a point p,
the number T(u,v) says how much momentum-in-the-u-direction is flowing
through the point p in the v direction.  Writing it out in terms of
components in any coordinates, we have

         T(u,v) = T_{ab} u^a v^b

In coordinates where x^0 is the time direction t while x^1, x^2, x^3 are
the space directions (x,y,z), we have the following physical
interpretation of the components T_{ab}:

The top row of this 4x4 matrix, keeps track of the density of energy ---
that's T_{00} --- and the density of momentum in the x,y, and z
directions --- those are T_{01}, T_{02}, and T_{03} respectively.  This
should make sense if you remember that "density" is the same as "flow in
the time direction" and "energy" is the same as "momentum in the time
direction".  The other components of the stress-energy tensor keep track
of the flow of energy and momentum in various spatial directions.

11.  EINSTEIN'S EQUATION: This is what general relativity is all about.
It says that

          G = T

or if you like coordinates and more standard units,

          G_{ab} = 8 pi k/c^2 T_{ab}

where k is Newton's gravitational constant and c is the speed of light.
So it says how the flow of energy and momentum through a given point of
spacetime affect the curvature of spacetime there.

But what does it mean?  To see this, let's do some "index gymnastics".
Stand with your feet slightly apart and hands loosely at your sides.
Now, assume the Einstein equation!

G_{ab} = T_{ab}

Substitute the definition of Einstein tensor!

R_{ab} - (1/2)R g_{ab} = T_{ab}

Raise an index!   

R^a_b - (1/2)R g^a_b = T^a_b

Contract!  

R^a_a - (1/2)R g^a_a = T^a_a

Remember the definition of Ricci scalar, and note that g^a_a = 4 in 4d!

R - 2R = T^a_a

Solve!

R = - T^a_a

Okay.  That's already a bit interesting.  It says that when Einstein's
equation is true, the Ricci scalar R is the sum of the diagonal terms of
T^a_a.  What are those terms, anyway?  Well, they involve energy density and
pressure.  But let's wait a bit on that... let's put this formula for
R back into Einstein's equation:

R_{ab} + (1/2) T^c_c g_{ab} = T_{ab}

or

R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}.

This equation is equivalent to Einstein's equation.  What does it mean?
Well, first of all, it's nice because we have a simple geometrical way
of understanding the Ricci tensor R_{ab} in terms of convergence of
geodesics.  Remember, if v is the velocity vector of the particle in the
middle of a little ball of initially comoving test particles in free
fall, and the ball starts out having volume V, the second time
derivative of the volume of the ball is

-R_{ab} v^a v^b

times V.  If we know the above quantity for all velocities v (even all
timelike velocities, which are the physically achievable ones), we can
reconstruct the Ricci tensor R_{ab}.  But we might as well work in the
local rest frame of the particle in the middle of the little ball, and
use coordinates that make things look just like Minkowski spacetime
right near that point.  Then

g_{ab} = -1  0  0  0 
          0  1  0  0
          0  0  1  0
          0  0  0  1

and v^a = 1
          0
          0 
          0

So then --- here's a good little computation for you budding tensor
jocks --- we get 

R_{ab} v^a v^b = R_{00}

So in this coordinate system we can say the 2nd time derivative of the
volume of the little ball of test particles is just -R_{00}.

On the other hand, check out the right side of the equation:

R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}

Take a = b = 0 and get

R_{00} = T_{00} + (1/2) T^c_c 

Note: demanding this to be true at every point of spacetime, in every
local rest frame, is the same as demanding that the whole Einstein
equation be true!   So we just need to figure out what it MEANS!

What's T_{00}?  It's just the energy density at the center of our little
ball.  How about T^c_c?   Well, remember this is just g^{ca} T_{ac},
where we sum over a and c.  So --- have a go at it, tensor jocks and
jockettes! --- it equals -T_{00} + T_{11} + T_{22} + T_{33}.  So we get

R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] 

What about T_{11}, T_{22}, and T_{33}?  In general these are the flow of
x-momentum in the x direction, and so on.  In a typical fluid at rest, these
are all equal to the pressure.  

So the "simple geometrical essence of Einstein's equation" is this:

--------------------------------------------------------------------------
Take any small ball of initially comoving test particles in free fall.
Work in the local rest frame of this ball.  As time passes the ball
changes volume; calculate its second derivative at time zero and divide
by the original volume.  The negative of this equals 1/2 the energy
density at the center of the ball, plus the flow of x-momentum in the x
direction there, plus the flow of y-momentum in the y direction, plus
the flow of z-momentum in the z direction.
--------------------------------------------------------------------------

Or, if you want a less precise but more catchy version:

--------------------------------------------------------------------------
Take any small ball of initially comoving test particles in free fall.
Work in the local rest frame of this ball.  As time passes, the rate at
which the ball begins to shrink in volume is proportional to 
the energy density at the center of the ball plus the flow of
x-momentum in the x direction there plus the flow of y-momentum in the
y direction plus the flow of z-momentum in the z direction.
--------------------------------------------------------------------------

Note: all of general relativity can in principle be recovered from the
above paragraph!  Also note that the minus sign in that paragraph is
good, since it says if you have POSITIVE energy density, the ball of
test particles SHRINKS.  I.e., gravity is attractive.

Article 101363 (14 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 21 Feb 1996 17:23:18 -0800
Organization: University of California, Riverside
Lines: 58
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:
>[email-masked] (Matthew P Wiener)' wrote: 
>>>>du^2+u^2.dv^2 is polar coordinates, and while it doesn't look like 
>>>>Mr Pythagoras would approve, it's clear that that's just because he 
>>>>hadn't mastered elementary trigonometry first. 

>>>>But dp^2+sin(p)^2.dq^2, not only doesn't look pythagorean, it can never 
>>>>be made to look so, since it's actually spherical.  

>>>I'm scratching my head here...  

>>You've never taken a metric in polar or spherical coordinates?  Look it 
>>up in an advanced calculus text.  What I've written above are, up to a 
>>change of alphabet, polar coordinates and spherical coordinates after 
>>setting the radius to 1. 

>Well... Now that you put it that way.  Of course I screwed about in polar
>and spherical coordinates when I was a lad...  and wrote down little
>"volume elements" and "line elements",  but the word "metric" never came
>up.  I suppose it would more or less correspond to the latter. 

The length of a tangent vector v is g_{ab} v^a v^b, but in lots of
situations one runs into in college, the matrix g_{ab} is diagonal, so
you might never have thought much about other cases.  On a round
2-sphere of radius r, for example, where we use coordinates theta and
phi, the metric is

g_{11} g_{12}  =  r^2 sin^2(phi)   0
g_{21} g_{22}           0         r^2

with "1" corresponding to theta and "2" corresponding to
phi.  

Now --- here's the sneaky part --- g is a (0,2) tensor so we can
alternatively write just

g = r^2 sin^2(phi) (d theta)^2 + r^2 (d phi)^2.

Here we are using the mysterious fact that  d theta  and  d phi  are
cotangent vectors, and the mysterious fact that some sort of product
of two cotangent vectors is a (0,2) tensor, to express g in terms of
d theta and d phi.  

Alternatively, we can call the metric "ds^2" --- this is the "line
element" thingie you know and love, hopefully --- and then we get

ds^2 = r^2 sin^2(phi) (d theta)^2 + r^2 (d phi)^2.

However, ds^2 is really just an older-fashioned way of talking about the
metric g, so this is no big deal.

>Now I am getting the distinct flavor that to really follow your
>description,  I would actually have to seek a text,  and work some
>problems,  and all would become clear...

Oh, hopefully the above makes enough sense so that you won't have to
resort to such drastic measures.

Article 101295 (1 more) in sci.physics:
From: Oz <[email-masked]>
Subject: Re: Gravitational Red Shifts - Real or Apparent ?
Date: Wed, 21 Feb 1996 18:19:25 +0000
Organization: Oz
Lines: 63
Distribution: world
NNTP-Posting-Host: upthorpe.demon.co.uk
X-NNTP-Posting-Host: upthorpe.demon.co.uk
MIME-Version: 1.0
X-Newsreader: Turnpike Version 1.11 <V67Vak$Q$lyzk9NR0hujdE7wCg>

Unfortunately due to a 'feature' in my newsreader this thread got
killed. I missed the beginning, so I will probably be justly jumped
upon. However this is as far back as I can go, so:

In article <[email-masked]>, john baez <[email-masked]>
writes
>
>
>Yes.   Of course, when you do quantum gravity like I do, results
>that apply to weak fields in the asymptotically flat case aren't worth
>beans. 

I suppose you did the math suitably simplified for us poor souls that
only vaguely understand it? I wonder if you could resend this?

>For the beginner, drawing grand over-generalized conclusions from a very
>special case is always dangerous.  Suppose we just learned how to solve
>the problem of the falling body in classical mechanics.  The height h is
>the following function of time:
>
>h = h_0 - mgt^2/2
>
>We start to philosophize: 
>
>"Wow, so height always decreases with time!"
>
>"In fact, time is a just a way of measuring height, because
>t = sqrt(2(h_0 - h)/mg)!"
>
>"Cool, so time is really just basically the same thing as height!"
>
>Etc. etc..  

Ooo. Straight to the heart. I bet I am the worst culprit for this.
However your point is very well made. If viciously well observed :-(

The whole point as far as I was concerned in starting out following this
GR course for idiots was to have some rather better understanding of
what GR was, and what it predicted. Do we have nearly enough math yet to
follow it at an ever-so-slightly deeper level. The simplest form would
seem to be an isolated mass in otherwise empty space with minkowski
space at infinity. Presumably there are some expressions that we can
dissect. Say for an observer falling from infinity, and one hovering at
a fixed r from the mass at the origin? Hopefully the expressions will
not be excessively complex such that they can reasonably be written down
in a posting (somehow I have a bad feeling about this).

I have to admit that I did try to make a start on this, but my pathetic
efforts simply weren't replied to. I suspect that they were either so
pathetic or in the 'not even wrong' category that kind people decided
not to reply. It is nice to know that people here are actually a
considerate bunch.

It is clear from Baez's posting that it is easier to get the wrong end
of things than the right end, so hopefully he, Wiener et al can pull the
rabbits from the hat and not ask us to pull them out wrongly. At least
not too often, that is.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Article 101406 (4 more) in sci.physics:
From: john baez
Subject: Re: Gravitational Red Shifts - Real or Apparent ?
Date: 21 Feb 1996 22:34:24 -0800
Organization: University of California, Riverside
Lines: 105
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] write
s:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Yes.   Of course, when you do quantum gravity like I do, results
>>that apply to weak fields in the asymptotically flat case aren't worth
>>beans. 

>I suppose you did the math suitably simplified for us poor souls that
>only vaguely understand it? I wonder if you could resend this?

I didn't actually post a proof that a 'gravitational potential'
determining the 'rate of flow of time' made sense in the static
asymptotically flat case, or a proof that it didn't make sense in
general.  Proving the former isn't hard.  Proving the latter requires
making the rules of the game very precise.  I.e., you gotta state very
precisely what it is you're trying to do, before you can prove you can't
do it.  So I just tried to pin Ed down on what exactly his
'gravitational potential' was supposed to do and how it was supposed to
do it, until it seems he saw there wasn't any way.  (I missed a bunch of
posts, so I'm not sure just how he became enlightened.)

>>For the beginner, drawing grand over-generalized conclusions from a very
>>special case is always dangerous.  Suppose we just learned how to solve
>>the problem of the falling body in classical mechanics.  The height h is
>>the following function of time:
>>
>>h = h_0 - mgt^2/2
>>
>>We start to philosophize: 
>>
>>"Wow, so height always decreases with time!"
>>
>>"In fact, time is a just a way of measuring height, because
>>t = sqrt(2(h_0 - h)/mg)!"
>>
>>"Cool, so time is really just basically the same thing as height!"
>>
>>Etc. etc..  

>Ooo. Straight to the heart. I bet I am the worst culprit for this.
>However your point is very well made. If viciously well observed :-(

I just thought that a little viciousness would make my point clearer.
Of course, science is all about making generalizations, most of which
are false.  It's all a matter of *levels* of naivete.  When studying
general relativity, however, it's good to start with the presumption
that nothing need work like it did in Newtonian mechanics or special
relativity unless there's a damn good reason.  That's because it's
really a revolutionary theory.
 
>The whole point as far as I was concerned in starting out following this
>GR course for idiots was to have some rather better understanding of
>what GR was, and what it predicted. Do we have nearly enough math yet to
>follow it at an ever-so-slightly deeper level.

We're getting there.  Check out the new course outline.

>The simplest form would
>seem to be an isolated mass in otherwise empty space with minkowski
>space at infinity. Presumably there are some expressions that we can
>dissect. Say for an observer falling from infinity, and one hovering at
>a fixed r from the mass at the origin? Hopefully the expressions will
>not be excessively complex such that they can reasonably be written down
>in a posting (somehow I have a bad feeling about this).

Well, the simplest interesting solution of Einstein's equation is the
"static point mass" solution you describe, better known as the
Schwarzschild solution.  Actually, though, there's not really a point
mass!  The solution describes a black hole with the stress-energy tensor
equal to zero everywhere.

I could write down the metric for this solution, and then I could write
down what the geodesics look like in this solution, and so on.  Of
course, to know that it *is* a solution, you need to be able to compute
the Riemann tensor from the metric!   And this, of course, means we need
to be able to compute parallel transport (or more precisely, the
"connection") starting from the metric.  

Similarly, to know why the geodesics of a given metric *are* geodesics,
we need to know how to compute parallel transport (or more precisely,
the connection) starting from the metric.  

I have been avoiding telling you how do do these things, since I didn't
want to scare you silly until I had already lulled you into the false
impression that learning GR was just a bowl of cherries.  In short: 
yes, your bad feeling was a premonition: it all has to do with the
mathematics going on in the back room, that the wizard suspects you don't
really want to see.

More on that wizard business later....

Anyway, you have two alternatives.  Either take a peek at the Schwarzschild
metric and start to see how some of the weird time-warping,
space-bending properties of black holes work, taking my word for it that
all the math actually works out like I say.  Or get a bit deeper
understanding of what we skimmed over in the earlier version of the
course outline, so you'll at least know what sort of calculations one
must do to figure these things out.... even if we don't actually DO the
calculations.  (It's too yucky to do these calculations in ASCII.)

Article 101630 (32 more) in sci.physics:
From: john baez
Subject: Re: GR Tutorial Water Cooler
Date: 22 Feb 1996 21:50:07 -0800
Organization: University of California, Riverside
Lines: 144
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward 
Green) writes:

>Now,  suppose we have a PDE:    {A}X = S 

>Here "{A}" is just some general differential operator, written as a matrix,
> X is a vector of unknown functions,  and S is a vector of given functions,
> which we may think of as the "source".    

>Generally,  X is not completely specified by this equation.  We must add
>"boundry conditions"...  but THIS  however is not what I was originally
>asking,  although some people supplied this (good) piece of information.  I
>was supposing we had fixed on a *particular* solution,  X.  Then I asked
>the following:   What would allow us to continue X to infinity
>unambiguously?  Just this little piece of X and _no_other_information_...? 
>Not very damn likely, is it!? 

>But given a little piece of X,  AND  RETAINING the information contained in
>{A}X = S,  THEN can we unambiguously extend X to the complete solution??   

What do you mean, "retaining the information contained in {A}X = S"?
That's a bit vague.  Do you mean, 

"Given a little piece of X, and all of S, and knowing that {A}X = S, can
we uniquely determine X?"

I'll assume that's what you mean.  Okay?  What I have to say should at
least be somewhat interesting, even if that's not what you meant.

>I am taking a very strong GUESS that for all the classical PDE's of
>mathematical physics the answer is YES,  and in fact this is just what we
>mean by saying "the problem is well posed". 

No.  (Not if I understand you.)

Let's break down and consider examples.  First let's consider a typical
elliptic PDE, in fact the archetypal one, Poisson's equation:

           Laplacian f = g

where f and g are functions.  How this works depends on where f and g live,
but let's suppose f and g are functions on good old Euclidean space,
R^n.  One fact, typical of elliptic equations, is that if g is slightly
"nice" in the sense of being "not too jagged", then any solution f is at
least equally nice.  For example, if g is continuous, then f is.  If 
g is smooth, then f is.  ("Smooth" means infinitely differentiable with
all derivatives being continuous.)  If g is analytic, then f is.
("Analytic" means smooth plus the fact that Taylor series converge
within some nonzero radius of convergence.)  Those are just a few of the
many gradations of niceness, but you get the idea.  Actually f will
typically be nicer than g.  This is known as "elliptic regularity".  

Now in particular if g is zero, g is very very nice!  So if 

           Laplacian f = 0

then f is analytic.  

Now as you note, g does not determine f.  Say we have

           Laplacian f_1 = Laplacian f_2 = g

for two different functions f_1, f_2.  Then subtracting,

           Laplacian (f_1 - f_2) = 0

so f_1 - f_2 is analytic.  But you can recover an analytic function if
you know it on any nonempty open set.  So if we have

          Laplacian f_1 = Laplacian f_2 = g

and f_1 and f_2 agree on some nonempty open set, they agree everywhere.

So for *this* PDE the answer to your question is "yes", if I understand
it.  This is typical of elliptic PDE.  

But now let's consider the archetypal hyperbolic PDE: the wave equation

          D'Alembertian f = g

where f and g are functions on good old Minkowski spacetime, R^{n+1}.
Remember, the D'Alembertian is the spacetime analog of the Laplacian;
it's

          d^2/dt^2 - d^2/dx_1^2 - ... - d^2/dx_n^2

The only difference is some minus signs.  But what a difference!  Why?
Well, if 

         D'Alembertian f = g

and g is nice in the sense of "not too jagged", f is under NO obligation
to be nice at all!  It can be arbitrarily nasty!  For example, suppose g
is utterly nice: suppose it's zero.  Let's consider the case n = 1 to
keep life simple.  Then writing f as a function of t and one variable x,
we could have 

             f(t,x) = delta(t - x)

where delta is the Dirac delta "function" --- so jagged it's not even a
function!  (It's a distribution.)   You could be sitting at x = 0 and
never expect this singular wave would crash in upon you until t = 0.

So: elliptic regularity fails miserably for hyperbolic PDE!  It's one of those
sad facts of life everyone must learn eventually!

In particular, if 

          D'Alembertian f_1 = D'Alembertian f_2 = g

and we know f_1 = f_2 on some open set, it is NOT in general true that
f_1 = f_2 everywhere.

If you've followed me thus far you are in for the treat: this fact about
hyperbolic vs elliptic PDE is really one of the main differences between
SPACE and SPACETIME.  Elliptic PDE are the PDE that show up in *static*
situations, like electrostatics, magnetostatics, etc..  This is because
SPACE is a RIEMANNIAN manifold and the PDE that naturally arise on
Riemannian manifolds are elliptic.  Hyperbolic PDE are the PDE that show
up in *relativistic, dynamic* situations, like electrodynamics.  This is
because SPACETIME is a LORENTZIAN manifold and the PDE that naturally
arise on Lorentzian manifolds are hyperbolic.  Note those changes of
sign when we went from the Laplacian to the D'Alembertian.  Those came
from the changes of sign in the metric when we went from a Riemannian
manifold to a Lorentzian one.  

And what is this saying?  It's saying that an essential aspect of
spacetime is that unexpected things can happen: you may know what's
going on in a little region of spacetime, but at any moment some wave
may crash in upon you, Dirac-delta-like in its jaggedness and its
suddenness.  This is not the case with space, where elliptic regularity
reigns.  But that's always what time was supposed to do: to allow
surprises to occur, to make the future something one would have to wait
for, rather than merely predict from local observations!  One triumph of
special relativity is thus the fact that when spacetime is a Lorentzian
manifold, the laws of physics are hyperbolic PDE, elliptic regularity
fails, and the future constantly brings new information crashing in upon
us.  

>Maybe that's enough for one post.  Now somebody ask me how the
>Cauchy-Riemann equations fit into this.  Come on,  ask me.  I dare you. 

The Cauchy-Riemann equations are another primordial elliptic PDE.

Article 101754 (15 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 23 Feb 1996 13:45:37 -0800
Organization: University of California, Riverside
Lines: 182
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Doug Merr
itt) writes:

>Poor John. No matter how he tries to make that back room of mysterious
>calculations tantalizing, no one takes the bait and rushes off to
>do actual problems.

I wasn't expecting that.  I just want to make sure people are properly
warned before they try to get me to explain how to compute the Riemann
tensor, etc..

>Although it did generate some interesting vignettes about sorcerer's
>apprentices sneaking around. :-)

Ah yes...  I recall it well....

[An arpeggio in a natural scale simulates "going into a dream" as your
screen begins to blur.]

One day Oz was sweeping up the spent debris from one of G. Wiz's more
extravagant attempts at quantising time. The floor was littered with
shrivelled up discarded electrons, photons and failed universes. Bundles
of fibrous ectoplasm still writhed through multiple internal dimensions
as they died a ghastly and grisly death. Microscopic black holes popped
randomly amongst the detrius as they gave up the last of their energy in
a whimper of unbeingness.

"What a bl**dy mess." Thought Oz, 
"All that power and he just scatters everything around. You'd think he
would at least organise it to collect itself into a heap at the end".

G. Wiz staggered through the curtain of his inner sanctum. (You know,
THAT curtain.) He looked grey and worn. His usually unlined and jovial
face was covered with more wrinkles than a 'father of the college' and
he was haggard and grim.

"I'm out of magic herbs," said G. Wiz
"I shall have to go off to the high mountainside and collect some more,
or I'll be a goner. Tidy this up and have the rest of the day off. And
DON'T do anything stupid whilst I am away."

"Right-oh, your magicness sir." said Oz, pretty pi**sed off. 
"No sooner said than done y'r wizardness."

"OK, see you at sunset." said G. Wiz as he tottered out of the door.

"Hours of work here," thought Oz
"No sooner said than done, indeed! Humph, he's never even held a broom!"

Oz swept up, put the debris into the black hole recycler and cleaned the
larger scorch marks from the white quartzite floor. Then he ate the dry
and stale bread as supplied by the wizard. "Good for the brain" the wiz
always said, although Oz had never actually seen the Wiz ever eat any
himself.

The G. Wiz wasn't due back for hours. 

"I wonder HOW he works out the Riemann curvature of spacetime?"
thought Oz.
"Maybe that'll be powerful and general enough magic to put a spell on
that wench in the village pub? I bet it's really simple. Perhaps if I
just had a peek inside the sanctum I could spot a tome, read it, and who
knows what I'll be able to do then?"

Oz cautiously walked over to the curtain and lifted a corner. A cold and
hostile wind blew out and the hair on the back of Oz's neck stood on
end. A blue corona flickered round the edges of the curtain but Oz still
looked in.

Books and papers were scattered all over. Some flickered in and out of
existence and others were supported by massive steel frames although
they looked quite small and light.

"Ahah," though Oz "some of these are weighty tomes indeed."

Oz walked cautiously in. Unbeknownst to him the sky outside the cavern
went black, and small thunderbolts played space invaders with little
black clouds in the erie and supernatural gloom.

At the back he saw a book. 

"Riemann curvature of transdimensional spacial entities in 26
dimensional quarternion metaspace", was embossed in gold on its aged
and worn dragonskin spine. It seemed to have a lot more pages inside
than you could get between the covers.

"Ahah," thought Oz, "just the thing for me."

He opened the book and froze. There was a heavy and hostile breathing
right behind his left ear. "OhmygodthewizisbackandIamforthehighjump"
thought Oz. Last time he had been turned into a frog for a week. He
hadn't minded the flies (they were crunchy), or even the moths (juicy
but a bit dusty) but the dry air had played hell with his skin.

Oz slowly, very slowly, turned his head to the left and found himself
eye to eye, and nose to nose, with a very fit, healthy and young looking
G. Wiz. He didn't look pleased at all. Restrained thunderbolts could be
seen limbering up through his glowing red pupils. Oz decided to say
nothing, and hoped it wouldn't hurt too much. His hair slowly and
deliberately lifted off his head in an effort to get as far away from Oz
as possible.

"Well, well, well," growled the G. Wiz in technicolour surround sound
with accentuated near lethal subsonics.  "What have we here?" 
Oz was silent. Very silent. Mostly because he didn't dare to breath.

Then the Wiz said, "A budding mathematical physicist, eh?  Reading up on
STRING THEORY, eh?"  As he spoke, he lifted his staff, and myriads of
small glowing loops, vibrating and glowing, appeared near its tip.  
Oz took a step back without thinking, mesmerized yet frightened by the 
sight.  

"In a rush to leave?" asked the wizard, his voice dripping sarcasm.  "I
thought you would ENJOY seeing some strings, being such an expert on
them."  The strings grew and shone ever more blindingly, expanding and
rippling, humming and singing at incredibly high frequencies.  As Oz
stared at them he was shocked to see that they did not live in
ordinary 3-dimensional space: they vibrated in directions that he had
never seen or conceived of before.  He tried to see clearly what was
going on but the more they expanded, the more dimensions they seemed to
inhabit.  

"Beautiful, no?  So... you did not heed my warnings about going into the
back room.  You want to compute Riemann curvature, no doubt.  But do you
have what it takes?  That's the question."  The shimmering vibrations now
filled the room, which shook with their energy.  The very fabric of
space seemed to start dissolving as they encircled Oz... the very laws
of classical logic seemed to dissolve (or was it just that Oz was scared
witless) leaving behind some deeper, more powerful substratum of reality
following its own profound logic, crystal clear yet unspeakable.  "If
you just wanted to grind away at calculations there would be no problem,
but I'm afraid you want INSIGHT.  You do want INSIGHT into physics,
don't you?"   Speechless, Oz was barely able to nod, barely able to hear
the wizard or follow what he said.  

The wizard laughed.  "Yes, you want INSIGHT.  But are you prepared to pay
the price?  You want to understand the laws of the universe --- but would
you know yourself any longer if you did?"  The wizard exploded into
light, Oz staggered over, and everything went black.

Oz awoke to find himself in his cave, lying on the straw-filled bunk.
He didn't remember going there... he had a splitting headache... the
last thing he remembered was opening a book on something or other in 26
dimensions, and the wizard catching him.  Ugh!  He propped himself up,
and it felt as if his head was about to burst.  He looked around and saw
a piece of paper lying on the ground near his bed, with glowing writing
on it, which faded and dissolved as he read:

"Dear Oz --

If you wish to learn more general relativity I am afraid you will need
to pass a test of your valor.  So: answer me the following questions:

1.  Explain why, when the energy density within a region of space is
sufficiently large, a black hole must form, no matter how much the
pressure of whatever substance lying within that region attempts to
resist the collapse.

2.  Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may is
determined at each point.  

3.  In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation?

I have taught you enough to answer these questions on your own.  Slip
the answers under my door --- I'm busy.

Best wishes,
G. Wiz

(Hint: read the course outline, you fool.)"

Oz fell back into bed with a groan.  

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!gatech!newsfeed.internetmci.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Fri Feb 23 18:51:14 PST 1996
Article: 101572 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!gatech!newsfeed.internetmci.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz
From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Gravitational Red Shifts - Real or Apparent ?
Date: Thu, 22 Feb 1996 18:29:53 +0000
Organization: Oz
Lines: 34
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]>
 <[email-masked]> <[email-masked]>
 <[email-masked]>
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NNTP-Posting-Host: upthorpe.demon.co.uk
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X-Newsreader: Turnpike Version 1.11 <V67Vak$Q$lyzk9NR0hujdE7wCg>

In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Anyway, you have two alternatives.  Either take a peek at the Schwarzschild
>metric and start to see how some of the weird time-warping,
>space-bending properties of black holes work, taking my word for it that
>all the math actually works out like I say. 

This would (caution to the winds) sound good. Make a nice thread, I
think. Basically, in the end, I gotta take your word for it. It would be
easy to fool me as my knowledge of the subject will never be good enough
to properly check the maths. There are bound to be some interesting
questions and suitably mind expanding conceptual things to digest.

Anyway, Wiener et al wouldn't let you get away with fibbing. 
Ha - so there!

> Or get a bit deeper
>understanding of what we skimmed over in the earlier version of the
>course outline, so you'll at least know what sort of calculations one
>must do to figure these things out.... even if we don't actually DO the
>calculations.  (It's too yucky to do these calculations in ASCII.)

Yup, this sounds good too. Sort of theory and practice. I expect to
drown regularly, but doubtless Ed will drag me out to safety. And then
throw me in again. Well, one can only do one's best. Perhaps just a
teeny bit of calculation, just so we can get some idea of the flavour?
You know liqwufiqwufg=;khdf;oqh, which we solve to find a^2=42.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Fri Feb 23 19:26:47 PST 1996
Article: 101774 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Gravitational Red Shifts - Real or Apparent ?
Date: 23 Feb 1996 18:49:42 -0800
Organization: University of California, Riverside
Lines: 37
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Anyway, you have two alternatives.  Either take a peek at the Schwarzschild
>>metric and start to see how some of the weird time-warping,
>>space-bending properties of black holes work, taking my word for it that
>>all the math actually works out like I say. 

>This would (caution to the winds) sound good. Make a nice thread, I
>think. Basically, in the end, I gotta take your word for it. 

Okay, in a little bit I'll start up on that sort of thing.  Actually I
may wait until I need to get ready to teach the kids in my course here
at UCR about black holes.  That'll be a couple of weeks I guess.  Then
I'll be really motivated to remember all the formulas and such.

>> Or get a bit deeper
>>understanding of what we skimmed over in the earlier version of the
>>course outline, so you'll at least know what sort of calculations one
>>must do to figure these things out.... even if we don't actually DO the
>>calculations.  (It's too yucky to do these calculations in ASCII.)

>Yup, this sounds good too. Sort of theory and practice. I expect to
>drown regularly, but doubtless Ed will drag me out to safety. And then
>throw me in again. 

Okay.  Well, while I get ready to teach you about black holes, you can
see the wizard about learning more about Riemann curvature.
Unfortunately, that nasty fellow caught you trying to find the answer in
a book in his back room, so now you're going to have to meet some challenges
before he'll explain that stuff.  Bad move!  Good luck, though.

From galaxy.ucr.edu!not-for-mail Sat Feb 24 21:15:23 PST 1996
Article: 101513 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Gravitational Energy (was: Re: General relativity tutorial)
Date: 22 Feb 1996 11:17:55 -0800
Organization: University of California, Riverside
Lines: 353
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>

>>Yes.  There is only that niggling point about defining what
>>"gravitational energy" actually IS!  As one tries to do so, one learns
>>that it's not so easy.  There are various possible definitions, but they
>>are all either 1) strange in their implications

>Like for example?

I explained this a few times so far.  It will probably make more sense
this time now that you know about the stress-energy tensor, Einstein
tensor and the like.  If you use the usual recipes for calculating
stress-energy tensors you find that that stress-energy tensor of the
gravitational field basically works out to -G_{ab}.  Here G_{ab} is the
Einstein tensor and I am ignoring annoying factors of pi, Newton's
constant and the like.  I explain what "basically" means below, but
let's ignore it for the moment.

Einstein's equation says G_{ab} = T_{ab}, where T_{ab} is the
stress-energy tensor of all the matter around.  Thus the energy density
of matter, namely T_{00}, PLUS the "energy density of gravity", namely
G_{00}, is ZERO.  Energy density of matter exactly cancelled by that of
gravity!  That seems strange to some people, nice to others.  

Now for a technical remark on the "basically" business.  Actually the
stress-energy tensor of gravity works out to be G_{ab} plus a "total
divergence".  Sometimes folks don't bother taking this extra term into
account, but sometimes they do, and sometimes they really should.  If
you do, it means that the energy density of matter plus that of gravity
doesn't really equal zero, but still if you integrate it over a *closed*
spacelike slice you get zero, so if the universe is closed its total
energy works out to be zero by this definition.

This is extremely important in quantum gravity.  

I don't really want to talk about this more because its rather slippery
and one must be a bit technical to keep from conveying misimpressions.

There is more in the FAQ, which I know you, Oz, have read, but Landis
wanted it, and I think I forgot to repost it last time, so here it
comes.   

Item 7.

IS ENERGY CONSERVED IN GENERAL RELATIVITY?            original by Michael Weiss
------------------------------------------                    and John Baez
	
	In special cases, yes.  In general--- it depends on what you mean
by "energy", and what you mean by "conserved". 

	In flat spacetime (the backdrop for special relativity) you can
phrase energy conservation in two ways: as a differential equation, or as
an equation involving integrals (gory details below).  The two formulations
are mathematically equivalent.  But when you try to generalize this to
curved spacetimes (the arena for general relativity) this equivalence
breaks down.  The differential form extends with nary a hiccup; not so the
integral form. 
	
	The differential form says, loosely speaking, that no energy is
created in any infinitesimal piece of spacetime.  The integral form says
the same for a finite-sized piece.  (This may remind you of the
"divergence" and "flux" forms of Gauss's law in electrostatics, or the
equation of continuity in fluid dynamics.  Hold on to that thought!) 

	An infinitesimal piece of spacetime "looks flat", while the effects
of curvature become evident in a finite piece.  (The same holds for curved
surfaces in space, of course).  GR relates curvature to gravity.  Now, even
in Newtonian physics, you must include gravitational potential energy to
get energy conservation.  And GR introduces the new phenomenon of
gravitational waves; perhaps these carry energy as well?  Perhaps we need
to include gravitational energy in some fashion, to arrive at a law of
energy conservation for finite pieces of spacetime? 

	Casting about for a mathematical expression of these ideas,
physicists came up with something called an energy pseudo-tensor. (In fact,
several of 'em!)  Now, GR takes pride in treating all coordinate systems
equally.  Mathematicians invented tensors precisely to meet this sort of
demand--- if a tensor equation holds in one coordinate system, it holds in
all.  Pseudo-tensors are not tensors (surprise!), and this alone raises
eyebrows in some circles. In GR, one must always guard against mistaking
artifacts of a particular coordinate system for real physical effects. 
(See the FAQ entry on black holes for some examples.) 

	These pseudo-tensors have some rather strange properties.  If you
choose the "wrong" coordinates, they are non-zero even in flat empty
spacetime.  By another choice of coordinates, they can be made zero at any
chosen point, even in a spacetime full of gravitational radiation.  For
these reasons, most physicists who work in general relativity do not
believe the pseudo-tensors give a good *local* definition of energy
density, although their integrals are sometimes useful as a measure of
total energy. 

	One other complaint about the pseudo-tensors deserves mention. 
Einstein argued that all energy has mass, and all mass acts
gravitationally.  Does "gravitational energy" itself act as a source of
gravity?  Now, the Einstein field equations are 

            G_{mu,nu} = 8pi T_{mu,nu}

	Here G_{mu,nu} is the Einstein curvature tensor, which encodes
information about the curvature of spacetime, and T_{mu,nu} is the
so-called stress-energy tensor, which we will meet again below.  T_{mu,nu}
represents the energy due to matter and electromagnetic fields, but
includes NO contribution from "gravitational energy".  So one can argue
that "gravitational energy" does NOT act as a source of gravity.  On the
other hand, the Einstein field equations are non-linear; this implies that
gravitational waves interact with each other (unlike light waves in
Maxwell's (linear) theory).  So one can argue that "gravitational energy"
IS a source of gravity. 

	In certain special cases, energy conservation works out with fewer
caveats. The two main examples are static spacetimes and asymptotically
flat spacetimes. 

	Let's look at four examples before plunging deeper into the math.
Three examples involve redshift, the other, gravitational radiation. 

(1) Very fast objects emitting light.

	According to *special* relativity, you will see light coming from a
receding object as redshifted.  So if you, and someone moving with the
source, both measure the light's energy, you'll get different answers. 
Note that this has nothing to do with energy conservation per se.  Even in
Newtonian physics, kinetic energy (mv^2/2) depends on the choice of
reference frame.  However, relativity serves up a new twist.  In Newtonian
physics, energy conservation and momentum conservation are two separate
laws.  Special relativity welds them into one law, the conservation of the
*energy-momentum 4-vector*.  To learn the whole scoop on 4-vectors, read a
text on SR, for example Taylor and Wheeler (see refs.)  For our purposes,
it's enough to remark that 4-vectors are vectors in spacetime, which most
people privately picture just like ordinary vectors (unless they have
*very* active imaginations). 

(2) Very massive objects emitting light.

	Light from the Sun appears redshifted to an Earthbound astronomer. 
In quasi-Newtonian terms, we might say that light loses kinetic energy as
it climbs out of the gravitational well of the Sun, but gains potential
energy.  General relativity looks at it differently.  In GR, gravity is
described not by a "potential" but by the "metric" of spacetime.  But "no
problem", as the saying goes.  The Schwarzschild metric describes spacetime
around a massive object, if the object is spherically symmetrical,
uncharged, and "alone in the universe".  The Schwarzschild metric is both
static and asymptotically flat, and energy conservation holds without major
pitfalls.  For further details, consult MTW, chapter 25. 

(3) Gravitational waves.

	A binary pulsar emits gravitational waves, according to GR, and one
expects (innocent word!) that these waves will carry away energy.  So its
orbital period should change.  Einstein derived a formula for the rate of
change (known as the quadrapole formula), and in the centenary of
Einstein's birth, Russell Hulse and Joseph Taylor reported that the binary
pulsar PSR1913+16 bore out Einstein's predictions within a few percent. 
Hulse and Taylor were awarded the Nobel prize in 1993. 

	Despite this success, Einstein's formula remained controversial for
many years, partly because of the subtleties surrounding energy
conservation in GR.  The need to understand this situation better has kept
GR theoreticians busy over the last few years.  Einstein's formula now
seems well-established, both theoretically and observationally. 

(4) Expansion of the universe leading to cosmological redshift.

	The Cosmic Background Radiation (CBR) has red-shifted over billions
of years.  Each photon gets redder and redder.  What happens to this
energy? Cosmologists model the expanding universe with
Friedmann-Robertson-Walker (FRW) spacetimes.  (The familiar "expanding
balloon speckled with galaxies" belongs to this class of models.)  The FRW
spacetimes are neither static nor asymptotically flat.  Those who harbor no
qualms about pseudo-tensors will say that radiant energy becomes
gravitational energy.  Others will say that the energy is simply lost. 

	It's time to look at mathematical fine points.  There are many to
choose from!  The definition of asymptotically flat, for example, calls for
some care (see Stewart); one worries about "boundary conditions at
infinity".  (In fact, both spatial infinity and "null infinity" clamor for
attention--- leading to different kinds of total energy.)  The static case
has close connections with Noether's theorem (see Goldstein or Arnold).  If
the catch-phrase "time translation symmetry implies conservation of energy"
rings a bell (perhaps from quantum mechanics), then you're on the right
track. (Check out "Killing vector" in the index of MTW, Wald, or Sachs and
Wu.) 

	But two issues call for more discussion.  Why does the equivalence
between the two forms of energy conservation break down?  How do the
pseudo-tensors slide around this difficulty? 

	We've seen already that we should be talking about the
energy-momentum 4-vector, not just its time-like component (the energy). 
Let's consider first the case of flat Minkowski spacetime.  Recall that the
notion of "inertial frame" corresponds to a special kind of coordinate
system (Minkowskian coordinates). 

	Pick an inertial reference frame.  Pick a volume V in this frame,
and pick two times t=t_0 and t=t_1.  One formulation of energy-momentum
conservation says that the energy-momentum inside V changes only because of
energy-momentum flowing across the boundary surface (call it S).  It is
"conceptually difficult, mathematically easy" to define a quantity T so
that the captions on the Equation 1 (below) are correct.  (The quoted
phrase comes from Sachs and Wu.) 

  Equation 1:  (valid in flat Minkowski spacetime, when Minkowskian
                coordinates are used) 

                                               t=t_1
       /                  /                    /
       |                  |                    |
       | T dV     -       | T dV       =       | T dt dS
       /                  /                    /
      V,t=t_0           V,t=t_1               t=t_0

   p contained       p contained            p flowing out through
   in volume V    -  in volume V       =    boundary S of V
   at time t_0       at time t_1            during t=t_0 to t=t_1

   (Note: p = energy-momentum 4-vector)

T is called the stress-energy tensor.  You don't need to know what
that means! ---just that you can integrate T, as shown, to get
4-vectors.  Equation 1 may remind you of Gauss's theorem, which deals
with flux across a boundary.  If you look at Equation 1 in the right
4-dimensional frame of mind, you'll discover it really says that the
flux across the boundary of a certain 4-dimensional hypervolume is
zero.  (The hypervolume is swept out by V during the interval t=t_0
to t=t_1.)  MTW, chapter 7, explains this with pictures galore.  (See
also Wheeler.)

	A 4-dimensional analogue to Gauss's theorem shows that Equation 1
is equivalent to: 

  Equation 2:  (valid in flat Minkowski spacetime, with Minkowskian
                coordinates)

       coord_div(T) = sum_mu (partial T/partial x_mu) = 0

We write "coord_div" for the divergence, for we will meet another
divergence in a moment.  Proof?  Quite similar to Gauss's theorem: if
the divergence is zero throughout the hypervolume, then the flux
across the boundary must also be zero.  On the other hand, the flux
out of an infinitesimally small hypervolume turns out to be the
divergence times the measure of the hypervolume.

	Pass now to the general case of any spacetime satisfying Einstein's
field equation.  It is easy to generalize the differential form of
energy-momentum conservation, Equation 2: 

  Equation 3:  (valid in any GR spacetime)

        covariant_div(T) = sum_mu nabla_mu(T) = 0    

                    (where nabla_mu = covariant derivative)

(Side comment: Equation 3 is the correct generalization of Equation 1 for
SR when non-Minkowskian coordinates are used.)

	GR relies heavily on the covariant derivative, because the
covariant derivative of a tensor is a tensor, and as we've seen, GR loves
tensors.  Equation 3 follows from Einstein's field equation (because
something called Bianchi's identity says that covariant_div(G)=0). But
Equation 3 is no longer equivalent to Equation 1! 

	Why not?  Well, the familiar form of Gauss's theorem (from
electrostatics) holds for any spacetime, because essentially you are
summing fluxes over a partition of the volume into infinitesimally small
pieces.  The sum over the faces of one infinitesimal piece is a divergence.
 But the total contribution from an interior face is zero, since what flows
out of one piece flows into its neighbor.  So the integral of the
divergence over the volume equals the flux through the boundary.  "QED". 

	But for the equivalence of Equations 1 and 3, we would need an
extension of Gauss's theorem.  Now the flux through a face is not a scalar,
but a vector (the flux of energy-momentum through the face). The argument
just sketched involves adding these vectors, which are defined at different
points in spacetime.  Such "remote vector comparison" runs into trouble
precisely for curved spacetimes. 

	The mathematician Levi-Civita invented the standard solution to
this problem, and dubbed it "parallel transport".  It's easy to picture
parallel transport: just move the vector along a path, keeping its
direction "as constant as possible".  (Naturally, some non-trivial
mathematics lurks behind the phrase in quotation marks.  But even
pop-science expositions of GR do a good job explaining parallel transport.)
 The parallel transport of a vector depends on the transportation path; for
the canonical example, imagine parallel transporting a vector on a sphere. 
But parallel transportation over an "infinitesimal distance" suffers no
such ambiguity. (It's not hard to see the connection with curvature.) 

	To compute a divergence, we need to compare quantities (here
vectors) on opposite faces.  Using parallel transport for this leads to the
covariant divergence.  This is well-defined, because we're dealing with an
infinitesimal hypervolume.  But to add up fluxes all over a finite-sized
hypervolume (as in the contemplated extension of Gauss's theorem) runs
smack into the dependence on transportation path. So the flux integral is
not well-defined, and we have no analogue for Gauss's theorem. 

	One way to get round this is to pick one coordinate system, and
transport vectors so their *components* stay constant.  Partial derivatives
replace covariant derivatives, and Gauss's theorem is restored.  The energy
pseudo-tensors take this approach (at least some of them do).  If you can
mangle Equation 3 (covariant_div(T) = 0) into the form: 

       coord_div(Theta) = 0

then you can get an "energy conservation law" in integral form.
Einstein was the first to do this; Dirac, Landau and Lifshitz, and
Weinberg all came up with variations on this theme.  We've said
enough already on the pros and cons of this approach.

	We will not delve into definitions of energy in general relativity
such as the Hamiltonian (amusingly, the energy of a closed universe always
works out to zero according to this definition), various kinds of energy
one hopes to obtain by "deparametrizing" Einstein's equations, or
"quasilocal energy".  There's quite a bit to say about this sort of thing! 
Indeed, the issue of energy in general relativity has a lot to do with the
notorious "problem of time" in quantum gravity.... but that's another can
of worms. 

References (vaguely in order of difficulty):

  Clifford Will, "The renaissance of general relativity", in "The New
    Physics" (ed. Paul Davies) gives a semi-technical discussion of the
    controversy over gravitational radiation.
  Wheeler, "A Journey into Gravity and Spacetime".  Wheeler's try at
    a "pop-science" treatment of GR.  Chapters 6 and 7 are a
    tour-de-force: Wheeler tries for a non-technical explanation of
    Cartan's formulation of Einstein's field equation.  It might be
    easier just to read MTW!)
  Taylor and Wheeler, "Spacetime Physics".
  Goldstein, "Classical Mechanics".
  Arnold, "Mathematical Methods in Classical Mechanics".
  Misner, Thorne, and Wheeler (MTW), "Gravitation", chapters 7, 20,
    and 25 
  Wald, "General Relativity", Appendix E.  This has the Hamiltonian
     formalism and a bit about deparametrizing, and chapter 11
     discusses energy in asymptotically flat spacetimes.
  H. A. Buchdahl, "Seventeen Simple Lectures on General Relativity Theory"
    Lecture 15 derives the energy-loss formula for the binary star, and
    criticizes the derivation.
  Sachs and Wu, "General Relativity for Mathematicians", chapter 3
  John Stewart, "Advanced General Relativity".  Chapter 3 ("Asymptopia")
    shows just how careful one has to be in asymptotically flat spacetimes
    to recover energy conservation.  Stewart also discusses the Bondi-Sachs
    mass, another contender for "energy".
  Damour, in "300 Years of Gravitation" (ed. Hawking and Israel). Damour
    heads the "Paris group", which has been active in the theory of
    gravitational radiation.
  Penrose and Rindler, "Spinors and Spacetime", vol II, chapter 9.  The
    Bondi-Sachs mass generalized.
  J. David Brown and James York Jr., "Quasilocal energy in general
    relativity", in "Mathematical Aspects of Classical Field Theory".

Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Summary: 
Expires: 
References: <[email-masked]> <[email-masked]>
Sender: 
Followup-To: 
Distribution: world
Organization: University of California, Riverside
Keywords: 

Oz knocked hesitantly at the wizard's door.  "Ahem?"

From within he heard a mumbled "I'm busy.  Go away."  

He stood there for a second, and then said "Umm, but I had some problems
with the course notes." 

The door popped open and a very grumpy-looking face leaned out.
"Problems?" asked the wizard.  "You have PROBLEMS with them???"

"Ah, err, yes sir.  First of all, you say that 

R_{00} = T_{00} + (1/2) T^c_c 

and later: 

T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}

and conclude that

R_{00} = (1/2) T_{00} + T_{11} + T_{22} + T_{33}.  

Shouldn't that be 

R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] ?"

The wizard didn't reply, and only stared at Oz.  St. Elmo's fire
began flickering around his head.  Oz continued, "Also, umm, sir, you said
that R_{ab} v^a v^b when v = (1,0,0,0), but 
I don't see this.  v(0)=1, so the only non zero term should be

R_{00}.v(0).v(0) = R_{00} not -R_{00}

It would appear that g_{ab} ought to make an appearance here, but I
don't see where.  Help!"

The last word, "Help!", was uttered by Oz when the wizard, in a rage,
began to hurl fireballs hither and thither, cursing up a storm.
"Factors of two and sign errors!  Factors of two and sign errors!  Here
I am trying to quantize gravity using n-categories and you bother me
with FACTORS OF TWO AND SIGN ERRORS???  When I get through with you,
you'll wish you were a planarium, you ninny!  You'll even wish you were
a dung beetle!  There are things worse than that, you know!"  He
advanced towards Oz, swinging his staff, and Oz retreated, desperately
pleading for mercy.  At the last minute, just as Oz felt the whishing of
the staff right before his nose, a brightly lit sphere appeared in the
air.  The staff bounced off it and fell to the ground.

"Hi," said the sphere.  Oz recalled it now... it was the sphere of radius
r!  "Say, G. Wiz, aren't you being a trifle petulant?  Don't you
remember when you were an apprentice like Oz is now, and your master
hided you for making sign errors?  Are you saying it's okay to make sign
errors now?"  

Oz glared at the sphere.  "Ah yes, my professor.  He always was a real 
nitpicker."  He looked down thoughtfully for a minute.  "Still, he did 
teach me well."  He sighed.  "All right, I won't turn Oz into an 
intestinal nematode just yet.  Yes, Oz, the factor of 1/2 goes in front of 
the whole thing:

R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] 

As for that other thing, yes,

R_{00} v^0 v^0 = R_{00},

not -R_{00}.  I just got it wrong: the second time derivative of the
volume of the ball of coffee grounds is MINUS R_{ab} v^a v^b times the
volume of the ball, where v is the velocity 4-vector of the ball."  A
swarm of minus signs appeared around the wizard's head, buzzing like
midges.  He glared at them and swatted them with his staff.  "Minus
signs!  I *hate* them!"  The sphere, hoping to keep the wizard from
becoming angry again, chased after the minus signs, and they all flew away
down the tunnels of the wizard's keep.

"By the way," the wizard said, and here he picked up his staff and waved
it threateningly at Oz once more, "don't write v(0) when you mean v^0,
since then it's ambiguous whether you mean v^0 or v_0, and these are
different... in the example I gave, we'd have v_0 = -v^0."

Oz asked, "Umm, by the way, how do you derive that bit about the second time
derivative of the volume of the ball being, umm, MINUS R_{ab} v^a v^b?
It seems like it should follow from something or other about curvature,
but..."  

"I didn't explain that part yet!" interrupted the wizard.  "Just take my
word for it for now!  I'll explain it sometime when I'm less busy, if
you pass that test.  Now get out of here... and hand in the answers to 
those questions SOON."  He went back into his room and slammed the door.  

Oz thought, "Hmm, so he is human after all.  In fact, he makes just as
many mistakes as I do, I bet!  He sure gets upset about it, though."

The wizard's door creaked open, though the wizard remained hunched over
his desk scribbling.  "What was that you were thinking?" asked the
wizard, not turning around.

"Oh, nothing," said Oz, and scurried away.

From galaxy.ucr.edu!not-for-mail Thu Feb 29 17:05:26 PST 1996
Article: 102534 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 29 Feb 1996 13:35:57 -0800
Organization: University of California, Riverside
Lines: 142
Message-ID: <[email-masked]>
References: <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

Recall one of the previous episodes...

Oz awoke to find himself in his cave, lying on the straw-filled bunk.
He didn't remember going there... he had a splitting headache... the
last thing he remembered was opening a book on something or other in 26
dimensions, and the wizard catching him.  What a fool!  He propped
himself up, and it felt as if his head was about to burst.  He looked
around and saw a piece of paper lying on the ground near his bed, with
glowing writing on it, which faded and dissolved as he read:

"Dear Oz --

If you wish to learn more general relativity I am afraid you will need
to pass a test of your valor.  So: answer me the following questions:

1.  Explain why, when the energy density within a region of space is
sufficiently large, a black hole must form, no matter how much the
pressure of whatever substance lying within that region attempts to
resist the collapse.

2.  Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may is
determined at each point.  

3.  In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation.

I have taught you enough to answer these questions on your own.  Slip
the answers under my door --- I'm busy.

Oz fell back into bed with a groan.  

"Absolutely typical." though Oz
"The deal was that I sweep the floors, and the Wiz answers the
questions. Now I sweep the floors AND answer the questions."

Oz muttered dark thoughts under his breath and kicked angrily at his
bunk, which promptly collapsed in a cloud of dust. Oz looked at it
gloomily. 

"Trouble is that all the Wiz's notes fade away after a week or two so he
can reuse the paper. Hmmmm. OK, so I can go and look at my Hard Drawn
Descriptions (or HDD's) that I carved on the rocks outside, but it's
blowing a blizzard. In any case some have been taken by the trolls who
used the rocks for Gremlin Pancake Fodder (or GPF's) so they aren't what
you could call complete." muttered Oz despairingly to himself.

"Oh, well, I suppose I had better try" he thought, and walked over to
the cavern's mouth. The snow was blowing a force 12 and the temperature
was 250K. No way was Oz going out there! Oz returned glumly to his hole
in the rock and started to think. This unexpected mental activity was
not nice. In fact the pain was pretty bad, but he tried hard. Synapses
that had been asleep for years complained bitterly about being woken up
by a sharp electric shock, and out of condition neurons puffed around
frantically trying to make sense of the unfamiliar impulses.

"OK, so why must a black hole always form when the energy density within
a region of space becomes large enough. Hey, this is practical stuff. We
haven't done any practical stuff! The bast**d! Now if we had seen some
worked out expressions for the Schwarzchild metric, then it would have
been much easier. It would also have been nice if someone had given some
relationship between force and curvature, no matter, doubtless it will
come up in due time."

At this moment, the wizard was in the back room reading a parchment
that his slaves had fetched him... a recently written disquisition on
n-categories, by a wizard far across the seas.  He frowned.  It seemed
his competitor was catching up... he had been spending too much time on
that apprentice of his, and neglecting his serious work.  He decided to
put in a good several hours trying to see just what the other fellow had
done.  Just as he was reaching for his quill, to take some notes, a
large bell off in the corner began ringing.  "Damn!  What is it now?"
the wizard cried.  "Someone is making a conceptual error somewhere...
Bell!  Tell me, who is screwing up?"

"It's Oz, sir," said the bell, in a melodious high-pitched voice.  "He's
trying to solve a general relativity by treating gravity as a force!"

"Gravity as a FORCE???  What?  Are you sure it's Oz?  I told him
millions of times that in GR, freely falling objects simply follow
geodesics!  The whole point of GR is that gravity is not described as a
force!  I'm sure he knows that.  It can't possibly be Oz!  Tell me it's
not!"

"Yes, it's Oz sir, I can sense his aura very clearly.  He's in his cave."

"FORCE?  What the hell is he trying to do, anyway?"

"He's trying to solve that problem you gave him, on why sufficiently
dense matter collapses to form a black hole."  

"God's wounds, why doesn't he just use the course notes!  I told him the
exact formula for how the stress energy tensor curves spacetime and
makes initially parallel geodesics converge!  Doesn't he see what that
implies?"  

"Apparently not, sir."

"Why, I'll go out and hide that fellow, teach him a thing or two about
force..."  The wizard stormed towards the door, cape swirling behind him
and static electricity building for an enormous discharge, when all of a
sudden he halted... he had a thought.  Why not get that other fellow, Ed
Green, to help Oz out?  Why do all the work himself?  True, Ed never
seemed to enter into the wizard fantasy realm, but if he could be
reached in his own realm, he could probably be coaxed into helping Oz on
these test problems.  It would probably be good for Ed, too.

"Courier!"  The wizard clapped his hands, and a phoenix appeared in a
puff of smoke.  "Tell Ed Green to help Oz with those test questions.
You will have to leave this realm and take on a form suitable for the
realm of Late-Twentieth Century America.  I think he lives in
Manhattan.  Perhaps you should be bicycle courier."

"Is he wired?"

"Yes, certainly.  I've communicated with him that way before."  

"Well then," said the Courier, "With your permission, I will simply take
a little snippet of the spacetime continuum, say the last five minutes
in Oz's cave and this room here.  Don't worry, I won't actually cut it
out; I'll just copy it, edit and compress it a bit, translate it into
ASCII form, and email it to him.  That's more efficient than sending me
off to Manhattan, no?"

The wizard smiled.  "You're so clever.  I can never get used to these
labor-saving conveniences.  Go right ahead.  Post it to sci.physics
while you're at it.  Ask him to help Oz solve this GR problem without
referring to that archaic and misleading notion of the "force" of gravity." 

From galaxy.ucr.edu!ihnp4.ucsd.edu!ames!purdue!lerc.nasa.gov!magnus.acs.ohio-state.edu!math.ohio-state.edu!uwm.edu!newsfeed.internetmci.com!in2.uu.net!psinntp!psinntp!psinntp!pipeline!not-for-mail Thu Feb 29 18:58:42 PST 1996
Article: 102571 of sci.physics
Path: galaxy.ucr.edu!ihnp4.ucsd.edu!ames!purdue!lerc.nasa.gov!magnus.acs.ohio-state.edu!math.ohio-state.edu!uwm.edu!newsfeed.internetmci.com!in2.uu.net!psinntp!psinntp!psinntp!pipeline!not-for-mail
From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Is the equivalence principle true in homo. grav. fields?
Date: 27 Feb 1996 08:31:10 -0500
Organization: The Pipeline
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X-Newsreader: The Pipeline v3.4.0

I would like to restate the question I believe started this thread, 
because after a long detour arguing about gravitational potential,  I at
least finally understand the question that was asked! 
 
I think we are mostly agreed that in the case of static gravitational
fields,  GR predicts  in effect,  whatever we may think of this
terminology,  that a "relative rate of time" may be established spatially, 
for the reference frame in which the fields are stationary.  The clock at
the top of the tower runs...  either slower or faster... heck,  I've
forgotten which!  No matter. 
 
This certainly should apply to a gravitatational field arbitrarily close to
"homogeneous",  since we can take an arbitarilty large spherical
gravitating body,  and look at the field on on arbitrarily fine scale.  So,
 even though we may object that a completely "flat" gravitational field is
unphysical,  we do not believe this has any practical significance for this
question. 
 
Now we consider a rocket far from local sources of gravitation,  undergoing
a uniform acceleration,  for a long period of time.  On this rocket,  we
can perform the same sorts of experiments we might perform on the surface
of the earth with towers.  If the rocket is a mile long,  and we have two
initially sychronized clocks at the midpoint,  we may then move them apart
to the ends of the ship for a year or so,  then reunite them at the center,
 and compare readings. 
 
So what do we predict?   On what physical basis?  
 
-- 
 
E. R. (Ed) Green /  [email-masked] 
  "All coordinate systems are equal, 
   but some are more equal than others".

From galaxy.ucr.edu!not-for-mail Thu Feb 29 18:58:59 PST 1996
Article: 102375 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Is the equivalence principle true in homo. grav. fields?
Date: 28 Feb 1996 17:56:22 -0800
Organization: University of California, Riverside
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References: <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] (Edward Green) writes:

>Now we consider a rocket far from local sources of gravitation,  undergoing
>a uniform acceleration,  for a long period of time.  On this rocket,  we
>can perform the same sorts of experiments we might perform on the surface
>of the earth with towers.  If the rocket is a mile long,  and we have two
>initially sychronized clocks at the midpoint,  we may then move them apart
>to the ends of the ship for a year or so,  then reunite them at the center,
> and compare readings. 

>So what do we predict?   On what physical basis?  

Second question first: this is a special relativity question since
there's no gravity, so we might as well solve it using special relativity.

Now: which part of the rocket do you want to be uniformly accelerating?
The tip?  The tail?  Every piece?  Just remember, "uniformly
accelerating extended objects" are a bit subtle in special relativity.
(Remember how a string stretched between two uniformly acccelerating
rockets will snap!)  

I'll assume that the tail and tip accelerate so as to each feel a
constant acceleration A in their own instantaneous rest frame.  Then a
little calculation using special relativity shows that after a year, the
clock at the tip will be ahead of the clock at the tail.  Actually I
didn't do a calculation, I just drew a spacetime diagram of what was
going on... the vertical axis being time, the horizontal axis being the
direction in which the rocket was accelerating... and then the answer is
evident.

This is really just a version of the twin nonparadox, by the way.
It's the same basic sort of calculation.

If your dangerous dallyings with "gravitational potential" haven't
drained all the sense out of you, you'll remember that in the clock
tower experiment, the clock on the top of the tower was ahead of the
clock at the base, when we compared them after a year.  In both cases
it's the same basic idea --- though in one case it's GR, in the other
SR.  

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Article: 102567 of sci.physics
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From: [email-masked] (Michael Weiss)
Newsgroups: sci.physics
Subject: Re: Is the equivalence principle true in homo. grav. fields?
Date: 27 Feb 1996 20:40:04 GMT
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cc:  [email-masked]

Ed Green asks:

    Now we consider a rocket far from local sources of gravitation,
    undergoing a uniform acceleration, for a long period of time.  On
    this rocket, we can perform the same sorts of experiments we might
    perform on the surface of the earth with towers.  If the rocket is
    a mile long, and we have two initially sychronized clocks at the
    midpoint, we may then move them apart to the ends of the ship for
    a year or so, then reunite them at the center, and compare
    readings.

    So what do we predict?   On what physical basis?  

The clock that spent the trip up with Captain Picard on the bridge
will read a later time than the one stuck in the back with Geordi in
engineering. 

Why?  Well, must be those tachyon fields--- or maybe the dilithium
crystals.... 

OK, seriously, the forward clock reads a later time than the aft clock
at the joyous reunion.  There are two ways to see this.

Spacetime diagrams:  I can't draw these here, so get out a low-tech
pencil and paper and follow my directions.  First draw your good old
horizontal x-axis and your vertical t-axis.  (No, I *don't* know why
every relativity book in the world uses this convention, while every
elementary calculus book in the world uses the *opposite* convention
for problems about falling balls.)

Draw a curve for the worldline of "10-forward" (the midpoint of the
spaceship).  This starts out headed straight up the page, if the
spaceship is initially at rest in our (t,x) coordinate system.  Then
it bends smoothly over towards the right.  In fact, the right curve to
use is a hyperbola, but let's not get into that right now.

Now draw the worldlines of the two clocks.  Really, with real pencil
and paper!  (Or on a chalk-board, or on one of those hideous modern
whiteboards, if (like me) you're stuck with that.)

Do they *look* like they're the same length?

Uh-oh, you caught me: the worldline that looks *longer* --- the one to
the convex side of the hyperbola --- is for the clock with the
*shorter* elapsed time.

OK, Minkowski diagrams can be tricky.  The pitfall here is that
sneaky sign change:

     Minkowski:     ds^2 = dt^2 - dx^2

     Pythagoras:    ds^2 = dx^2 + dy^2

We have Pythagorean eyes, and paper, and chalkboards and whiteboards,
so visual estimates aren't reliable when it comes to judging elapsed
proper time along a worldline.  It *is* possible to develop an
intuition for Minkowskian "distances", but it takes some work.  I
won't attempt it in ASCII, that's a sure route to carpal tunnel
syndrome.

But at least we've made a prima facie case that the two clocks won't
read the *same* when they get back together to share a swig of Tholian
ale.

Second approach: light signals.  This is (nearly) Einstein's original
argument. The forward clock sends out regular flashes; the aft clock
picks them up.  The rest of the tale sounds just like the usual
elevator story: if the forward clock thinks it's pulsing once per
second, then the aft clock will record them as arriving more
frequently, say once per 0.88 seconds.

This is easiest to see near the beginning of the trip--- say just
after the clocks have reached the fore and aft of the ship.  Look at
it from the perspective of a nearby inertial observer, say
space-station DS9 "at rest".  The reference frame of DS9 is just the
(t,x) coordinate system in "flesh and blood" (or I suppose, silicon
and duranium).  The DS9-measured velocities of the two clocks are
still so small that the (special) relativistic time dilation factors
can be ignored.  I.e., according to the DS9 observers, both clocks
aboard the Enterprise are keeping nearly correct time (near the
beginning of the trip).

So it looks a little like this to the folk on DS9: the forward clock
is heaving light-pulses at the aft clock, which *speeds up to catch
them*.  OK, let's switch channels from Star Trek for a moment.  This
station has an old silent movie, the Keystone cops chasing some
robbers.  The robbers hurl a couple of bags at the cops, one after the
other.  The cops speed up to catch the bags.  The time interval
between throws is longer than the time interval between catches.

This post is long enough already --- but it really needs to make four
more points to finish.  I'll just leave them as "exercises":

1.  Convince yourself that the total number of pulses *sent* equals
    the total number of pulses *received*, if we start the
    light-pulses when both clocks are at the ship's midpoint and stop
    them after the reunion.

2.  Convince yourself that so few pulses are sent at the very
    beginning and end of the trip, compared to those sent during the
    main part of the journey (one year, I think you said) that we
    don't need to worry about emission and reception times for those.

3.  After a while, the ship is travelling so fast that the SR time
    dilation factors are no longer negligible.  Take care of this by
    imagining *another* inertial observer, say a cloaked Romulan
    Warbird.  The accelerating Enterprise is, for an instant, at rest
    in the reference frame of the Warbird.  So the conclusion about
    emission intervals vs. reception intervals holds for the whole
    journey, not just the beginning.

4.  Finally, add in the light-signals to the spacetime diagram you
    drew above.  They should look like parallel straight lines at a 45
    degree angle.  Figure out what the elevator argument means in
    terms of this diagram.

Conclusion:  the forward clock sent 1 jillion pulses, each 1 second
apart; the aft clock received 1 jillion pulses, each 0.88 seconds apart.
The forward clock records a greater time lapse than the aft clock.

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Article: 102191 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Mon, 26 Feb 1996 12:35:54 +0000
Organization: Oz
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Oz has returned from his weekend labours. Now he has to answer the three
vital questions. The second was:

2.  Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may be(?)
determined at each point.  

Well, obviously the Riemann tensor is required. So we have to go in our
little loop and find out the rotation of our iddy-biddy vector. Now this
sounds absolutely fine except that it's not very easy to go on a little
rectangular loop whilst simultaneously zooming along the t-axis at c.
Let's try and figure out how it might be done. Well, the first thing to
realise is that we can choose our basis vectors and luckily they need
not be orthogonal, although one could imagine that 'various corrections'
could be made to make them so. Hmmmm ....

Ok, how about this way. You have two identical stable 'clock and laser'
flashers at your origin.  You take one a suitable distance away from
your origin and set it down in the same frame as your origin. You know
you have done this because a laser beaming from the origin has exactly
the same frequency as when you had the equipment side by side at the
origin. Ok, now you have a point in spacetime. You know how far you went
because you measured it locally with your 1mm ruler. You do this very
slowly. Heck, it's a 1mm ruler and you are going light seconds at least,
in any case you should be able to count the flashes from your laser at
the origin, both out and back, to 'synchronise' time somehow or other.
Anyway, whatever, you send off a laser pulse to say all is 'at rest'.
Now you have the equivalent of a distance in, well, let's call it the
'out' direction. Now you trundle back to your origin and see what
transpired. Well, you (in curved spacetime) will see two things. Firstly
the distant laser started off with the same frequency as the origin
laser (once you set it up), but slowly it will drift as spacetime
curvature moves it away (or towards). You will also see a change in the
time pulses (much the same thing really) as the distance gets further
away. So at the very least you have a little parallelogram in the time-
out plane.

        *
        |\      The '*' are my little parallelogram
        | \     for my Riemann loop.
        |  \    The light pulses are at 45 deg as req'd.
        *   *
        .\  |
        . \ |
        .  \|
        .   *   All set up point.
        .  .
        .  .
        . . Trundle out leg.
        ..
        ..
        .
        .

Now exactly how you interpret this as a rotation of your tangent vector,
I am not rightly sure. My guess is that you deduce that the distant leg
is 'really' at a small angle determined by it's relevant velocity
accumulated over the time it travelled (vertically) , which being a
dx/dt thingy, looks like a slope to me.

Oh well, maybe worth 25%?? .....   20%...... 

Right ball park? .. Not even wrong <gloom>.

Notes: 
a) The universe has to be uniform and isotropic to avoid you interacting
with a bendy, lumpy bit of spacetime that throws everying off.

b) This give you a bit of the Riemann tensor, obviously it should be the
same for vectors in the other space/time directions by symmetry. How you
do it in the space-space planes I am less certain. Well, I don't have a
clue, actually. Probably so simple you can't see it.

c) I suppose somewhere I am assuming the underlying metric is a 
ds^2 = -a dt^2 + dx^2 + dy^2 + dz^2 
sort of thingy with appropriate choice of co-ordinates etc. 
Mostly 'cos that's what Ted sed.

Waiting in trepidation,

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 102544 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 28 Feb 1996 09:02:58 -0500
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[email-masked] (Steve Carlip)' wrote: 
 
>: Suppose we followed a time-like geodesic "forever"...  
>: ... the "local rate of volume expansion" must 
>: always have a negative second derivative!  If it starts positive,  it
must 
>: get less positive,  then possibly go through zero,  then go negative. 
Is 
>: this a cosmological principle? 
> 
>Well, in one sense there's nothing too deep here---it's just the 
>statement that gravity is always attractive, that is, that it 
>always makes objects tend to converge.   
 
Wow.  But there must be a bit more to this.  This "convergence of the
coffee grounds",  a purely local effect,  purely attributable to the local
value of T,  seems to globalize to "the convergence of a stone to the
center of the earth",  an effect *not* attributable to the local value of
T,  but to the continuation of the local effects of all that "T" wrapped up
in the earth.  In general T can be zero in the vicinity of a test particle,
 and it will attempt to converge quite happily with other objects in its
vicinity. 
 
That is why I was groping so long with the idea of "continuation". 
*Somehow* the physical world seems to be able to extrapolate a pretty
definite "gravitational field" in the vicinity of the earth, 
notwithstanding the possibility of "gravitational waves abruptly crashing
in from infinity",  (which no one has observed yet,  anyway  :-)  
Soooo....  all this business of elliptic vs. hyperbolic PDE's doesn't seem
to bother mother nature too much in setting up a perfectly predictable
scheme of things in our locality,  nicely attributable to local,  (but not
too local),  influences,  and I have been trying to understand how GR
manages this.  Can it be that the definite gravitational effects we all
know and love are contained in the 4 elliptic PDE's of GR, while the
possibility of unforetold waves suddenly crashing on our gravitational
shores is contained in the other 6?  
 
>If the expansion goes 
>to negative infinity, in particular, this is an indication of 
>a "caustic," a point in spacetime at which two geodesics cross. 
>For example, consider the longitude lines on a sphere; the 
>expansion goes to negative infinity at the north pole (assuming 
>a choice of sign convention to distinguish the north and south 
>poles). 
 
Sounds reasonable,  though I am not sure exactly what you mean by "the
expansion going to negative infinity".   Something simple like "E =
ln(V/V_0)",  where E is the expansion,  V_0 the volume now...?  (Just a
guess... don't shoot) . 
 
>But you're also right in connecting this behavior to cosmology 
>(among other things)---the fact that an initially negative 
>expansion goes to negative infinity, combined with a fairly 
>complicated argument about global topology, is the basis for 
>most "singularity theorems" in cosmology: for example, for the 
>theorem that if the Universe is now expanding everywhere, and 
>if energy density is always positive, then there must have been  
>an initial singularity a finite time in the past.  A similar argument 
>implies that the current expansion of the Universe must be decelerating, 
>though the question of whether it eventually turns around (i.e., whether 
>the expansion becomes negative) depends on the mass density. 
 
Way cool!  So,  a mere 8 weeks or so into the tutorial,  and we have
deduced the necessity of the big bang...  er, ah, sorta.    :-)  
 
-- 
 
E. R. (Ed) Green /  [email-masked] 
  "All coordinate systems are equal, 
   but some are more equal than others".

From galaxy.ucr.edu!not-for-mail Thu Feb 29 19:32:16 PST 1996
Article: 102576 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 29 Feb 1996 17:14:04 -0800
Organization: University of California, Riverside
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References: <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] (Edward Green) writes:

>"Still,  is GR nothing but a wave equation?  No,  that's not quite right. 
>Or maybe it is.  What if we put a source term with the D'Alembertian in
>this toy equation?  Well,   I guess it *still* basically tells us how
>information propagates in time and space,  but now time and space are
>somewhat lumpy!  Hmm...  Begins to sound something like GR after all."

GR is really 10 equations, since G_{ab} is a 4x4 symmetric matrix.

Split spacetime into time and space, arbitrarily. 

Then of these 10 equations, 4 can be viewed as constraints on the metric
on space and its first time derivative at time zero.  These are of a generally
"elliptic" flavor, though they are sufficiently nonlinear and otherwise
tricky that they are not elliptic in the most technical sense.  The
remaining 6 can be viewed as a wave equation describing how the metric
on space evolves in time.  (Note, the metric on space has 6 components.)
These are of a generally hyperbolic flavor, although again they are
nasty enough as to fail to be hyperbolic in the most technical sense of
the term.  

GR is the partial differential equation tyro's dream, or nightmare.

From galaxy.ucr.edu!not-for-mail Fri Mar  1 17:09:55 PST 1996
Article: 102753 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 1 Mar 1996 17:00:58 -0800
Organization: University of California, Riverside
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References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

Oz returned from his weekend labours. Now he has to answer the three
vital questions. The second was:

2.  Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may be(?)
determined at each point.  

Oz ponders this for an hour or two, and walks over to the wizard's door
and knocks on it.  "Come on in!" cries the wizard.  As Oz enters, G. Wiz
slips out from behind a curtain, shaking some dust off his hands. 
He then wipes his brow and asks, "So, how are you doing on those
questions?  I presume you're starting with question 2, right?"

That mind-reading ability always takes Oz aback.  He nods and says,
"Well, obviously the Riemann tensor is required. So we have to go in our
little loop and find out the rotation of our iddy-biddy vector. Now this
sounds absolutely fine except that it's not very easy to go on a little
rectangular loop whilst simultaneously zooming along the t-axis at c.
Let's try and figure out how it might be done. Well, the first thing to
realise is that we can choose our basis vectors and luckily they need
not be orthogonal, although one could imagine that 'various corrections'
could be made to make them so. Hmmmm ...."

The wizard also says "Hmmm ...."

Oz continues.  "Ok, how about this way. You have two identical stable
'clock and laser' flashers at your origin.  You take one a suitable
distance away from your origin and set it down in the same frame as your
origin. You know you have done this because a laser beaming from the
origin has exactly the same frequency as when you had the equipment side
by side at the origin. Ok, now you have a point in spacetime. You know
how far you went because you measured it locally with your 1mm ruler.
You do this very slowly. Heck, it's a 1mm ruler and you are going light
seconds at least, in any case you should be able to count the flashes
from your laser at the origin, both out and back, to 'synchronise' time
somehow or other.  Anyway, whatever, you send off a laser pulse to say
all is 'at rest'.  Now you have the equivalent of a distance in, well,
let's call it the 'out' direction. Now you trundle back to your origin
and see what transpired. Well, you (in curved spacetime) will see two
things. Firstly the distant laser started off with the same frequency as
the origin laser (once you set it up), but slowly it will drift as
spacetime curvature moves it away (or towards). You will also see a
change in the time pulses (much the same thing really) as the distance
gets further away. So at the very least you have a little parallelogram
in the time- out plane."

He scratches the following picture in the dust on the floor:

        *
        |\      The '*' are my little parallelogram
        | \     for my Riemann loop.
        |  \    The light pulses are at 45 deg as req'd.
        *   *
        .\  |
        . \ |
        .  \|
        .   *   All set up point.
        .  .
        .  .
        . . Trundle out leg.
        ..
        ..
        .
        .

"Now exactly how you interpret this as a rotation of your tangent vector,
I am not rightly sure. My guess is that you deduce that the distant leg
is 'really' at a small angle determined by it's relevant velocity
accumulated over the time it travelled (vertically) , which being a
dx/dt thingy, looks like a slope to me."

The wizard again says "Hmm ...." and stares off abstractedly in a rather
unnerving way. 

Oz asks: "Oh well, maybe worth 25%?? .....   20%...... Right ball
park?... Not even wrong?"   

The wizard turns back to Oz, smiles and says "Well, it has a certain
charm to it."  Oz breathes a sigh of relief, but then notices a curious
gleam in the wizard's eye... and begins to fear he will not be let off
quite so easy.

"Maybe you should think of it this way," says the wizard.  "Start with
two clocks next to each other, out in the wilderness of empty space!"  
As he speaks, the room dims and Oz seeks two clocks next to each other,
floating in starry emptiness.  "Now drag one a foot away from the other
and do your best to leave it at rest relative to the first."  One clock
moves over a foot, starting at rest, but then the clocks begin to drift
away from each other at an accelerating rate.  "You you let them float
out there in the wilderness of space.  They are in free fall, so they
trace out geodesics.  These geodesics may converge or diverge, and the
rate of this "geodesic deviation" may be measured as you suggest, by
shining a laser from one to the other and measuring the redshift."

"By the way: note that since the clocks are relatively close to each
other the notion of the distance between them, as measured between two
clocks that start out at rest with respect to each other --- and indeed
the notion of starting out at rest with respect to each other! --- are
unproblematic.  Really we should do this with clocks that are
infinitesimally close to each other, but a distance of say, a
lightsecond is darn close to infinitesimal on cosmic scales."

"Now, what does geodesic deviation have to do with curvature, exactly?
Well, let's look at a spacetime diagram of the situation.  Hang on while
I equip you with 4-dimensional vision."  He passes his hands over Oz's
head and mutters an almost inaudible syllable, and all of sudden Oz is
shocked to find that he can see the clocks, not just in space, but in
spacetime.  They trace out curves in spacetime, and he sees these curves
in their entirety as a static whole!  Yet he is simultaneously able to
see how at any given "moment" --- i.e., any given slice of spacetime ---
the clocks occupy positions in "space" just as they did in the previous
scene!  Furthermore, as before, he can see, as in a movie, how as time
passes the clocks begin to accelerate away from each other.  Somehow he
is seeing not just space but spacetime!

"Hey, how are you doing that?" he asks.  "I bet if I could do this
myself these problems would be a whole lot easier!"  

"I'm just using my limited telepathic ability to show you how I think of
these things," the wizard says.  "For you to do it too, all you need is
practice."  

Oz frowns, unsure as to how he'd practice this.  He decides to keep
quiet and pay careful attention; maybe he'll get the habit of 4d vision.

"Wise move," says the wizard, smiling.  

"Now, let's use "v" to denote the velocity vector of the first clock at
time zero, and let "w" denote the vector from the first clock to the
second."  Oz sees something like the following, but of course everything
he sees is in 4 dimensions:

   |       |
   |       |
  v^       ^
   | w     |
   P->-----Q

"I've labelled the initial positions of the two clocks at rest by P and
Q.  Note that the velocity vector of the clock at Q is just the result
of PARALLEL TRANSPORTING v in the direction w."

Oz asked, "Wait?  Is w the path from P to Q, or just a tangent vector at
P?"

The wizard smiled.  "Good question!  Roughly speaking, we can pretend
the path from P to Q is a vector because the path is so short... it
should really be infinitesimally short, of course."

"Now, suppose we let each clock wait a second.  They now have new
positions (in spacetime) P' and Q'."   Oz now sees something like this:

   P'->----Q'
   |       |
   |       |
  v^       ^
   | w     |
   P->-----Q

Oz asks, "Say, shouldn't v be infinitesimal too?"  

"Right, if we want to think of the path from P to P' as a vector we
should really be waiting just an *infinitesimal* amount of time to get
from P to P', not a second; but a second is close enough for practical
purposes."

They stare at the infinitesimal rectangle a while and then the wizard
continues: "Now, what's the velocity vector of the clock at Q'?  Well,
think how we got it: first we parallel transported v over to Q along w.
Then we parallel tranported the result over to Q', since the curve from
Q to Q' is a geodesic, which means its velocity vector is parallel
translated along itself."  Oz sees a helpful green dot go across from P
to Q and then forwards in time from Q to Q', carrying a green copy of
the tangent vector v with it.  

"Now for the big question!  We want to know if the clock at Q' is moving
away from the clock at P'.  To answer this, we compare its velocity
vector to the following vector: what we get by first parallel
translating v along itself over to P', and then over to Q'.  That's the
velocity the clock at Q' would have it it were at rest relative to the
clock at P'."  Oz now sees a red dot go forwards in time from P to P'
and then across from P' to Q', carrying a red copy of the tangent vector
v with it.

   P'->----Q'
   |       |
   |       |
  v^       ^
   | w     |
   P->-----Q

The resulting red tangent vector at Q' is a bit different from
the green one representing the actual velocity of the clock at Q'.

"Curvature!" cries Oz in a moment of revelation.

"Right!  We are taking the vector v and parallel translating it two
different ways from P to Q' and getting two slightly different answers.
If the answers were the same, the second clock would remain at rest
relative to the first.  But in fact they are not, and the difference
tells us how the second one begins accelerating away from the first."

"Now remember how curvature works: the result of

               dragging v from P to Q to Q'

minus the result of 

               dragging v from P to P' to Q'

is going to be

                   -R(w,v,v)

where R is the Riemann tensor.  Right?"

Oz nods unconvincingly, and promises himself that he'll reread the
course notes for the definition of the Riemann tensor.

"Good," says G. Wiz.  "But this is just the same as 

                    R(v,w,v)

since the Riemann tensor is defined so that it's skew-symmetric in the
first two slots."

"Cool," says Oz.  

"This is called the GEODESIC DEVIATION EQUATION, by the way.  I've
cheated in 2 or 3 places in deriving it here --- for example, I hid some
epsilons under the table --- but the result is correct.  Let me
summarize."  The wizard had a tendency towards pedagogical pedantry
which Oz forgave him only because of his tendency to hurl thunderbolts
when interrupted.  "Two initially comoving particles in free fall will
accelerate relative to one another in a manner determined by the
curvature of space.  Suppose the velocity of one particle is v, and the
initial displacement from it to the second is small, so that it may be
represented as a vector w.  Then the acceleration A of the second
relative to the first is given by R(v,w)v.  Or if you like indices,

                    A^a = R^a_{bcd} v^b w^c v^d

So..." the wizard paused.  "So, we can really determine the Riemann
curvature using experiments as you proposed, and using this formula."

"Now," asked the wizard, "remember how there are 20 components to the
Riemann tensor, of which 10 are determined by the Ricci curvature and 10
by the Weyl?  If we are doing the case of a homogeneous isotropic big
bang model, most of those darn components should be redundant, thanks to
the symmetry.  Can you figure out how many components we really need to
worry about in this big bang model?"

Oz gulped.  

"By the way, you may enjoy remembering the definition of the Ricci and
Weyl tensor in terms of how a bunch of initially comoving test particles
begins to change volume and shape.  We have a nice formula for the Ricci
tensor along these lines.  A while back you asked we derived it.  I said
"wait and see".  Well, now you can derive it from the geodesic deviation
equation, at least if you are better at index juggling than I suspect
you are.  After all the geodesic deviation equation says exactly what
those coffee grounds are going to do."

Oz suddenly got the feeling G. Wiz was going to assign him more
homework, so he casually looked at the grandfather clock in the corner
and said "Gee whiz!  It's late!  I have to start polishing the
doorknobs!  You told me to finish them all by tonight, remember?"
G. Wiz smiled and nodded as Oz beat a hasty retreat.  "Anyway," he
called after Oz, "try to see what you can say about the Ricci and Weyl
curvature in the big bang cosmology, using the coffee grounds business
and all the symmetry.  That would really improve your grade on this
question!"  

From galaxy.ucr.edu!not-for-mail Sat Mar  2 12:04:38 PST 1996
Article: 102965 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 2 Mar 1996 11:06:15 -0800
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] writes:

>Could you explain a little more?  I kind of see why that's true..  all
>these index gymnastics are just raising and lowering with the metric,
>and Lorentz geometry is determined by the metric (although I'm not sure
>how you get the Lorentz transformations solely from the signature of the
>metric, as I think someone else alluded to.)  At any rate, is there some
>slick mathematical way to prove that all the index-gymnastic
>transformations are Lorentz invariant?

Say we have a vector space V with a metric g of signature (+++-).
Actually any signature will do, but let's pick that one just to be
specific.  The Lorentz transformations are defined as those linear
transformations of V which preserve the metric, i.e. all L: V -> V with
g(Lv,Lw) = g(v,w) for all v,w in V.  

(If you don't like taking this as a definition, work out what this
definition gives you, and you'll see that precisely all products of
rotations and boosts meet this definition.  But clearly this is the best
definition; it explains exactly what is special about products of rotations and
boosts.)

Now let V* be the dual vector space, i.e. the space of linear functions
from V to R (the real numbers).  Guys in V* are called covectors.  A
Lorentz transformation L acts on a covector f as follows:

(Lf)(v) = f(L^{-1}v).

So now we know how to Lorentz transform vectors in V and covectors in
V*.  All more complicated tensors are formed by tensoring vectors and
covectors.  E.g. a fancy tensor like the Riemann tensor, written using
indices as R^a_{bcd}, lives in the tensor product

V tensor V* tensor V* tensor V*

Thus we can Lorentz transform all tensors if we know how to transform
vectors and covectors.  (Details left to reader.)

Now we need only check that the basic operations of "index gymnastics"
are Lorentz invariant: i.e., if we first do the operation and then do a
Lorentz transform, we get the same thing as if we first do the Lorentz
transform and then do the operation.  We may take four operations as
basic: index raising, index lowering, contraction, and tensoring with
the metric.  Let's check that these are Lorentz invariant.

Index raising is the map

#: V -> V*

taking the vector v to the covector v# given by

v#(w) = g(v,w).  

Note: only mathematicians who really like avoiding indices use this
sharp and flat notation for index raising and lowering.  It's a bit
effete, but it's nice to see how it can be done.  You wanted slick, I'm
giving you slick!  To check that this is invariant note that

(Lv)# (w) = g(Lv,w) = g(v,L^{-1}w) = L(v#)(w)

where in the middle step we use the definition of Lorentz
transformations.  

Index lowering is the inverse map

b: V* -> V; 

it's obviously invariant since index raising is.  

Contraction is the map 

c: V* tensor V -> R

given by 

c(f tensor v) = f(v).

This is Lorentz invariant since the Lorentz transformations don't do
anything to scalars (R), and 

c(Lf tensor Lv) = Lf(Lv) = f(L^{-1}Lv) = f(v) = c(f tensor v).

Tensoring with the metric is Lorentz invariant since the metric.  

If the above seems overly abstract I urge you to rewrite all the
equations above using indices, which may help if you are familiar with
those.  

From galaxy.ucr.edu!ihnp4.ucsd.edu!swrinde!howland.reston.ans.net!math.ohio-state.edu!pacific.mps.ohio-state.edu!freenet.columbus.oh.us!magnus.acs.ohio-state.edu!csn!news-1.csn.net!csus.edu!news.ucdavis.edu!landau.ucdavis.edu!rmiller Sat Mar  2 15:48:48 PST 1996
Article: 102989 of sci.physics
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From: [email-masked] (Roger Miller)
Newsgroups: sci.astro.amateur,sci.astro,alt.journalism,sci.physics
Subject: Re: Today is the last Feb 29th of the millenium
Followup-To: sci.astro.amateur,sci.astro,alt.journalism,sci.physics
Date: 1 Mar 1996 23:11:33 GMT
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Lori Eichelberger ([email-masked]) wrote:
: [email-masked] (Jedidiah Whitten) wrote:
: >Emory F. Bunn ([email-masked]) wrote:
: >
: >: put the turn of the millennium at the end of 1999.  If a usage becomes
: >: common enough, then it becomes silly to fight it.  I think the pedants
: >: have already lost this particular battle.
: >
: >I don't think so.  I don't think I know anyone personally who is under 
: >the impression that the 21st century starts in 2000, and most people I've 
: >talked to are aware that it doesn't start until 2001.
: >
: Gotta disagree here--many, many people think the new millenium starts in 
: 2000 (regardless of mathematics, logic, etc.).

: Also, it seems to be popular with the "popular" media''"Strange Days" 
: (film), "Outside" (music), etc.  It sounds good--Year 2000=new Millenium.

: To be sure, most educated people know that "you start counting with Year 
: One, not Year Zero", but the "popular usage" may have overrun the 
: educated usage.  

: I'll wager there'll be one Hell of a party in 1999, whether (The Artist 
: Formerly Known as Prince) is there or not!  Many, many, many people now 
: think that 1999 is the last of the 20th Century, whether true or not.

: Does belief=truth?

I wish I remember the source, but some time in the last 6 years I either 
read an article or heard a report about this topic relating it to the
turn of the century back in 1900.  Apparantly the popular press was filled
with debates and editorials about whether 1900 was the last year of the
19th century or the first year of the 20th.  I imagine we'll get through
it.  On the other hand, the turn of the 20th century didn't have a million?
computer programs that were about to go wonky on January 1, 1900 as
we do for January 1, 2000; independent of whether it's the last year of this
millenia or the first year of the next.

Roger 

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!gatech!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun Mar  3 15:05:20 PST 1996
Article: 103183 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
Organization: Oz
Lines: 163
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>"Now for the big question!  We want to know if the clock at Q' is moving
>away from the clock at P'.  To answer this, we compare its velocity
>vector to the following vector: what we get by first parallel
>translating v along itself over to P', and then over to Q'.  That's the
>velocity the clock at Q' would have it it were at rest relative to the
>clock at P'."  Oz now sees a red dot go forwards in time from P to P'
>and then across from P' to Q', carrying a red copy of the tangent vector
>v with it.
>
>   P'->----Q'
>   |       |
>   |       |
>  v^       ^
>   | w     |
>   P->-----Q
>
>The resulting red tangent vector at Q' is a bit different from
>the green one representing the actual velocity of the clock at Q'.

Well, of course in his heart of hearts Oz rather felt that the Wiz was
having it both ways. He rather felt that if he had drawn little square
rightangled thingies then G. Wiz would have said, 

"Aha, using tachions again I see. Yes, well it *does* make it all much
easier doesn't it? Now go away and do it properly. zzzZZZAP!!!"

On the other hand, Wiz being very wise, always knows which bit of
approximation he can get away with. Hmmmm something about "experience"
Oz thought. Oz rather hoped he could catch a dose of experience, but it
did sound unpleasant.

>"Curvature!" cries Oz in a moment of revelation.

[Well, it paid to humor the old boy occasionally]

>"Right!  We are taking the vector v and parallel translating it two
>different ways from P to Q' and getting two slightly different answers.
>If the answers were the same, the second clock would remain at rest
>relative to the first.  But in fact they are not, and the difference
>tells us how the second one begins accelerating away from the first."
>
>"Now remember how curvature works: the result of
>
>               dragging v from P to Q to Q'
>
>minus the result of 
>
>               dragging v from P to P' to Q'
>
>is going to be
>
>                   -R(w,v,v)
>
>where R is the Riemann tensor.  Right?"
>
>Oz nods unconvincingly, and promises himself that he'll reread the
>course notes for the definition of the Riemann tensor.

Oz went red. Very, very red. He looked like he was going to explode.
Steam started coming out of his ears and his eyes started slowly and
deliberately popping out of his head. A small, very tiny, blue corona
started flickering faintly around his head. It looked like he was going
to explode. With a monumental effort of will he managed to slam down the
dampers before he went critical. It would probably only have resulted on
a pzzzittt instead of a zap, or probably only a ..... phit. 

"Ahem, Wiz, er Wi-hiz, woo-hoo, Wizziness. That's exactly what I have
been going on about when I said that going in a little RECTANGLE will
leave you not back where you started. The vector joining where you end
up to where you started from must surely relate to the Riemann tensor.
In this case the little vector is directly related to the velocity
picked up by Q as it went to Q'. I will leave you (Wiz) to work this
out. Oz knew his questions were mostly rubbish, but sometimes he did
wish people would listen. HUMPF!!!"

>
>"Good," says G. Wiz.  "But this is just the same as 
>
>                    R(v,w,v)
>
>since the Riemann tensor is defined so that it's skew-symmetric in the
>first two slots."
>
>"Cool," says Oz. 

Wondering what *exactly* "skew-symmetric in the first two slots" really
meant!

>"This is called the GEODESIC DEVIATION EQUATION, by the way.  I've
>cheated in 2 or 3 places in deriving it here --- for example, I hid some
>epsilons under the table --- but the result is correct.  Let me
>summarize."  The wizard had a tendency towards pedagogical pedantry
>which Oz forgave him only because of his tendency to hurl thunderbolts
>when interrupted.  "Two initially comoving particles in free fall will
>accelerate relative to one another in a manner determined by the
>curvature of space.  Suppose the velocity of one particle is v, and the
>initial displacement from it to the second is small, so that it may be
>represented as a vector w.  Then the acceleration A of the second
>relative to the first is given by R(v,w)v.  Or if you like indices,
>
>                    A^a = R^a_{bcd} v^b w^c v^d
>
>So..." the wizard paused.  "So, we can really determine the Riemann
>curvature using experiments as you proposed, and using this formula."

Was Oz pleased about this. You betcha. As soon as he had done the
washing up, he would have to have a look at R^a_{bcd} v^b w^c v^d. Well,
maybe after a beer or two with Ed, first. Well, maybe several. Oz noted
that the Wiz dropped into acceleration, rather assuming that curvature
only really existed in the t-direction. He also noted that all these
little deltas wandering about had differentiated themselves into
acceleration rather nicely. He kinda felt that these tensor jobbies
tended to do this without you really noticing. He would have to watch
out for that. It was the sort of thing that Wiz tended to assume was
obvious. He hoped they hid a lot of integration too. Preferably as much
as possible.

>"Now," asked the wizard, "remember how there are 20 components to the
>Riemann tensor, of which 10 are determined by the Ricci curvature and 10
>by the Weyl?  If we are doing the case of a homogeneous isotropic big
>bang model, most of those darn components should be redundant, thanks to
>the symmetry.  Can you figure out how many components we really need to
>worry about in this big bang model?"
>
>Oz gulped.  

Well the BB couldn't almost by definition have any Weyl bits since it
must surely be described entirely by itself, so to speak. Of course it
would help if we knew which components in the Weyl were zero. Since the
Ricci tensor can be derived from the Stress-energy tensor we only (ha)
need to sort that out. Well, obviously T_{00} is important but you can
see that T_{11} = T_{22} = T_{33} and that these won't be zero, in fact
these should be rather large and possibly even comparable to T_{00}. 

Then we have the spacial cross terms. Now I don't really know what these
represent. These are momentum in one direction being transported in
another. Now the only thing I can think of that might do that would be a
spinning particle. I suppose if we had a lot of spinning particles then
they might well cancel each other out if they were nice and random, but
I'm not totally convinced. I wonder what Wiz thinks.

Then we have the space-time cross terms T_{0X} and T_{X0}. I expect it's
obvious once you know, and probably also to do with spinning thingies
(actually I might have all this mixed up with the spacial ones now I
think about it) with luck they will all cancel out too.

I hope Wiz can discuss these cross thingies without getting, ahem,
cross.

Oz remembered all the chores he had to do. Sometimes they looked
positively attractive. He picked up his broom and ......

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!ihnp4.ucsd.edu!ames!uhog.mit.edu!news.mathworks.com!newsfeed.internetmci.com!in2.uu.net!pipeline!not-for-mail Sun Mar  3 14:09:44 PST 1996
Article: 103139 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 1 Mar 1996 12:26:22 -0500
Organization: The Pipeline
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References: <[email-masked]>
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_In which a new character is introduced_ 
 
Rudely seized from his mundane life in a late twentieth century metropolis
by the wizard's messenger,  Ed finds himself hurled through some
mind-bending special effects thingies involving swirling colors and the end
of the popular song by the dung-beetles,  'I am the Large Marine Mammal'. 
["Hmm..." thinks Ed,  "that doesn't sound quite right...  must be some sort
of reality dislocation"]   His trip ends in a thunk as he falls into a snow
drift,  and rolls over to see the messenger receeding,  laughing to
himself... "ASCII...  Applied Strategem to Corral Innocent Inebriates....  
ha ha ha ha... " 
 
[The wizard employed many creatures of dubious antecendents bound by debts
of service,  but he was not always able to fully control their impish
aspects] 
 
"Ow!"  said Ed,  "What have I been drinking?" 
 
He finds himself wearing the scratched and dented armor of an apprentice, 
one who has served many masters but never completed his apprenticeship.  A
bit graying around the temples for an apprentice...  On the breast plate is
the curious device "erg",  held on by a few rivets. 
 
Before erg has time to realize just how cold this feebly insulated suit of
armor is in a snow drift,  he is roused from his post translational reverie
by an approaching troll,   one who has been eating stone tablets,  and has
a touch of indigestion. 
 
"ARGGHHHESTH" comments the troll,  drolly.   Ed leaps up and attempts to
dislodge his sword from the scabbard,  but apparently his fantastic
alter-ego has not kept his equipment in shape,  and it is rusted solidly
into place.  "Ohmygod"  he thinks,  giving one last desperate pull at the
weapon.  The impulse is sufficient to overcome the coefficient of static
friction between steel boots and snow,  an he falls in a heap before the
charging troll,  who stubs his toes on the armor,  leaving fresh toe-dents,
 trips over the mighty warrior and rolls into a ball, gaining momentum on
the icy slope and finally flying out into space like a boulder,  smashing
into smithereens on the rocks below.  Seeing that the fearsome warrior
knows JUDO,  the other trolls beat a tasty retreat (munching the remaining
tablets). 
 
"Whoa!"  thinks the hero,  nursing his bruised arm and slowly pulling
himself up.  "This place is dangerous!"   Just then he spies a feeble bit
of firelight in the mouth of a nearby cavern,  and painfully heads in that
direction.   In the cavern,  of course,  is his old friend,  Oz. 
 
"What,  you too?"  says Oz,  once a wizard in his own right,  in a distant
story.  "Has your geodesic entered this place of no return?" 
 
"Yes" says the erg,  pulling a stone up to the fire,  and drawing his sword
to dry it on some straw,  finding that the fall in the heroic trollomachy
was sufficient to dislodge it from the scabbard,  "and the odd thing is,  I
am not even sure I remember crossing the horizon.      
 
And Oz said to him.... 
 
[Cheap b*stard... ain't I...  :-) ]

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Article: 103060 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Fri, 1 Mar 1996 21:59:55 +0000
Organization: Oz
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In article <[email-masked]>, Edward Green
<[email-masked]> writes
>"Whoa!"  thinks the hero,  nursing his bruised arm and slowly pulling
>himself up.  "This place is dangerous!"   Just then he spies a feeble bit
>of firelight in the mouth of a nearby cavern,  and painfully heads in that
>direction.   In the cavern,  of course,  is his old friend,  Oz. 
> 
>"What,  you too?"  says Oz,  once a wizard in his own right,  in a distant
>story.  "Has your geodesic entered this place of no return?" 
> 
>"Yes" says the erg,  pulling a stone up to the fire,  and drawing his sword
>to dry it on some straw,  finding that the fall in the heroic trollomachy
>was sufficient to dislodge it from the scabbard,  "and the odd thing is,  I
>am not even sure I remember crossing the horizon.      
> 

"Well," said Oz 
"it's quite nice here really. In the summer you can walk over to the
village and have a drink at the village pub. Of course at the moment
it's a mite chilly."

"You gotta Wizard here?" said Ed

"Yup," said Oz "not a bad one really. Usual bad temper on occasion and
he has this tendency to drift off when you are talking to him. He starts
thinking about some higher thing and you gotta hold him down from fair
floating away".

"How powerful is he, then." asked Ed "Hmm, a fair sized keep, but not
one of them with a whole faculty to terrorise."

"Oh, he's pretty good." said Oz "Not that old either. He has to use a
wizening spell to look the part. Really enjoys his subject too, which
makes a change. You should see the pleasure he got from turning me into
a turd^H^H^H^H toad. He is trying to teach me about General Relativity,
but I'm making heavy weather of it. He *says* it's really powerful, I
hope it's powerful enough to hex the wench at the pub, must be ..
surely. Anyway he does it in a funny way, we have been at it for weeks
and we still haven't got to any sargents, let alone any generals.
There's all this stuff he gives me that I have to carve onto these stone
tablets." 

Oz gestures casually at a mighty pile of inscribed stone tablets
balanced precariously over the chasm. 

"It's a full time job keeping the Trolls away from them. These local
trolls claim to have a sophisticated palate. Humpf" sighed Oz.

"Anyway, Wiz is having a thaumaturlogical battle with some smart-ass
upstart from over the water at the moment. No work for me as he is glued
to his desk producing magic spells to win the battle. You want to have a
look at some of this stuff? There's a few questions that he doesn't like
my answers to, no sir, not at all." Groaned Oz

Ed walked over to the heap of tablets and pulled one out. On it, in
flickering glowing letters, were the Wizard's questions.

"Hey, no sweat." said Ed "Lets go through it slowly."

And then Ed said .......

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Sun Mar  3 15:44:35 PST 1996
Article: 102999 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 2 Mar 1996 15:48:37 -0800
Organization: University of California, Riverside
Lines: 158
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References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>OK, start from the beginning.

>1) If v is the velocity vector of a particle in the middle of a little
>ball of blah blah .... of volume V.
>
>so d^2V/dt^2 = -R_{ab} v^a v^b. Hokay.

It's actually 

d^2V/dt^2 = -R_{ab} v^a v^b V, 

since the rate of change of volume is proportional to the initial volume
as well as everything else.  But this is no big deal here.

>2) Now to find R_{ab}
>
>R_{ab} = T_{ab} - (1/2)T^c_c g_{ab}
>
>and T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
>
>Now since it's uniform and isotrophic we should be able to say that 

>Tc = T_{11} = T_{22} = T_{33} which also head towards infinity as V->0

As we discussed a few times, the diagonal entries T_{ii} (i = 1,2,3) are
just the pressure in the x, y, and z directions.  Certainly isotropy
implies that they are equal, so we can just think of them all as
"pressure".  Ted pointed out that your intuition is right; the pressure
approaches infinity as we go back in time towards the big bang,
particularly in the "radiation-dominated era" when there was a lot more
energy in the form of light.  For light, pressure is comparable to
energy density (in sensible units where G = c = hbar = 1).  In the
current "matter-dominated epoch", where the stars don't bump into each
other all that much, the pressure is negligible compared to the energy
density.  

>and probably the cross terms are equal, at least some of them.
>Well, surely
>T_{12} = T_{21} = T_{13} = T_{31} = T_{23} = T_{32} = Td
>must be equal too in this situation. They presumably give the density of
>momentum flow in one direction travelling in another. Maybe something to
>do with things that spin or suchlike. 

Can you use the isotropy to say more about these off-diagonal terms
T_{ij} (i,j = 1,2,3)?  Hint: yes, you can!  Certainly what you have said
so far is true, but it doesn't nearly exhaust the consequences of
rotational symmetry.  You have only used rotations that switch the x, y,
and z axes.  (Please, folks, don't post the answer to this question; let
my victims Oz and Ed tackle this one first.)

>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms.  What they
>represent physically is not too clear. 

Remember, T_{ab} is the flow in the a direction of b-momentum.  So the
component T_{0i} represents the density of i-momentum (i.e., x-momentum,
y-momentum, or z-momentum), while T_{i0} represents the flow of energy
in the i direction.  Can you use isotropy to say something about these?
Hint: yes!

>So assuming we can take a minkowski metric in a small area, which seems
>unlikely...

You can't do it in a small finite region unless spacetime is flat in
that region --- since the Minkowski metric is flat.  But you *can* do it
at a *single point*.  Indeed, we may as well take the definition of a
Lorentzian metric --- the sort of metric we're interested in --- as one
which takes the form

-1  0  0  0 
 0  1  0  0
 0  0  1  0
 0  0  0  1

in some coordinates at any given point.  This is the essential
mathematical content of the "equivalence principle": at every point,
there are coordinate systems in which the metric looks just like that of
good old Minkowski space.  

So, what you do below is fine, since you're doing it at a single
(arbitrary) point:

>but still, we get

R_{ab} = 

[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}),   T_{12}, T_{13}]
[T_{20} ,  T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}),  T_{23}]
[T_{30} , T_{31}, T_{32} ,  (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]

=

[(1/2)(T_{00} + 3Tc) , T_{01} , T_{02} , T_{03}]
[T_{10} ,  (1/2)(T_{00} - Tc),   T_{12}, T_{13}]
[T_{20} ,  T_{21}, (1/2)(T_{00} - Tc),  T_{23}]
[T_{30} , T_{31}, T_{32} ,  (1/2)(T_{00} - Tc)]

>Well, a bit untidy...

Yes, it's a bit untidy, so use isotropy to the hilt and boil it down to
something a lot simpler!

>however we can see that
>
>R_{11} = (1/2)(T_{00} - Tc) [Nb Tc=T_{11} etc]

I don't like calling that "Tc"; it reminds me too much of the critical
temperature of a superconductor or something.  Since you know that
T_{11} = T_{22} = T_{33} is equal to the pressure in this isotropic
situation, why don't you just call it 3P?

>Now, as the universe gets hotter and smaller, one could imagine that Tc
>gets bigger. In fact one could imagine that Tc could even get to be as
>big as T_{00}. Lots of interactions producing massless particles and
>all. In this case R_{11} *might* become zero. With T_{00} going
>infinite, this is attractive. <groan>

Yes indeed, the pressure can dominate the energy.  

>3) Now let's see. 
>
>d^2V/dt^2 = R_{ab} v^a v^b.
>
>Now I can pick out each term by my choice of a & b. However it's not
>very clear, without explanation, what for example R_{12} means. So I
>will skip this for now. One optimistically hopes they turn out to be
>zero. 

As for a simple conceptual understanding, it's easiest to understand
R_{00}, as we've done.  As for your optimism, you will have to follow my
hints above to see if it's well-founded.

By the way, you can't "pick" a and b, exactly.  Remember, we sum over
all a and b in the formula above!  You can pick v, if you like.  

>One is tempted to imagine a cone filled with tangential
>spheres from an itsy bitsy one at t=epsilon to a big one centered at
>t=10^10 years. However this can't be quite right because the curvature
>of a sphere is constant. In the 4D thingy the curvature is forever
>decreasing from t=0 to t=10^10y.

It's certainly true that we can picture a closed big-bang universe as
a sphere (a 3-sphere) whose radius increases with time.  The radius is
not exactly a linear function of time so it's not exactly a cone.  But
you are right that as the radius increases, the curvature decreases.
Note how that fits with the fact that pressure and energy density are
decreasing.  

>I need more 'obvious' information to proceed further.

I hope to have provided some.  But what you really need to proceed
further is to ponder isotropy a bit more.

You're doing fine, by the way.

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Article: 103142 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 17:10:31 +0000
Organization: Oz
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Message-ID: <[email-masked]>
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In article <[email-masked]>, Edward Green
<[email-masked]> writes
>
>Suddenly it becomes clear our apprentice hasn't a clue how to answer these
>questions quantitatively,   or even sharply qualitatively.  Perhaps one of
>the spectators will throw a rock with a clue wrapped around it into the
>fantasy world of G. Wiz. 
> 
>But finally the time shift seems to be clearing up, so let's hear what Oz
>has to say...    
> 

Oz re-appears in a kaleidoscope of naked women and strange goings on.

"Hey!" said Oz "There's a serious transdimensional time warping going on
in the link to Wiz-time over in the United Plates. A whole lotta demons
are using the lode-lines over there for their own nefarious purposes,
and the lode-masters have gone on short time working. "

"Gosh" said Ed "you had better be careful. You definitely wouldn't want
to cross-link with some of those. You couldn't IMAGINE what you might
drop into."

"It's really nice to talk to someone who doesn't think a quick ZAP! will
make you think better." sighed Oz. "I guess we can chat over this in
peace and quiet. What we gotta show Wiz is that d^2V/dt^2 heads negative
as V-> something small no matter what the momentum flow is."

Now d^2V/dt^2 = -(1/2)[T_{00} + T_{11} + T_{22} + T_{33}]

lets say T_{11} + T_{22} + T_{33} = T_{ss} for simplicity.

and as you say there must be a relationship between T_{00} and T_{ss}
and for expansion T_{ss} has gotta be negative. Now it seems to me that
T_{00} is a volume term so if we keep the same amount of energy in the
volume V the energy density will vary as 1/V or perhaps better 1/r^3 if
r is the characteristic radius of the volume V. If T_{ss} is really a
pressure term then you would expect it to vary as the surface area or as
1/r^2. The trouble is that I don't know of this is right. Dimensionally
it doesn't look right, but there are so many c's that have dissapeared
off as 1, that it's hard to be sure.

Anyway if this is about right then as r gets very small the T_{00} term
will always win out in the end over the T_{ss} term. Then d^2V/dt^2 will
become negative and the volume will start to decrease (as long as
dV/dt=0 at some point) and as it decreases so it will decrease even
faster, and eventually become (horror) zero.

Then there's this E^2 = p^2 + m^2 thingy I came across years ago. I'm
sure we ought to be able to work this in somewhere. What do you think?

Ed wiped his rusty sword, jumped up and swiped it around impressively.
Oz dived at the floor as it swung round over his thinning hair.

"Well," interjected Ed "what I think is this ......"

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!gatech!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun Mar  3 15:48:45 PST 1996
Article: 103183 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
Organization: Oz
Lines: 163
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]>
 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>"Now for the big question!  We want to know if the clock at Q' is moving
>away from the clock at P'.  To answer this, we compare its velocity
>vector to the following vector: what we get by first parallel
>translating v along itself over to P', and then over to Q'.  That's the
>velocity the clock at Q' would have it it were at rest relative to the
>clock at P'."  Oz now sees a red dot go forwards in time from P to P'
>and then across from P' to Q', carrying a red copy of the tangent vector
>v with it.
>
>   P'->----Q'
>   |       |
>   |       |
>  v^       ^
>   | w     |
>   P->-----Q
>
>The resulting red tangent vector at Q' is a bit different from
>the green one representing the actual velocity of the clock at Q'.

Well, of course in his heart of hearts Oz rather felt that the Wiz was
having it both ways. He rather felt that if he had drawn little square
rightangled thingies then G. Wiz would have said, 

"Aha, using tachions again I see. Yes, well it *does* make it all much
easier doesn't it? Now go away and do it properly. zzzZZZAP!!!"

On the other hand, Wiz being very wise, always knows which bit of
approximation he can get away with. Hmmmm something about "experience"
Oz thought. Oz rather hoped he could catch a dose of experience, but it
did sound unpleasant.

>"Curvature!" cries Oz in a moment of revelation.

[Well, it paid to humor the old boy occasionally]

>"Right!  We are taking the vector v and parallel translating it two
>different ways from P to Q' and getting two slightly different answers.
>If the answers were the same, the second clock would remain at rest
>relative to the first.  But in fact they are not, and the difference
>tells us how the second one begins accelerating away from the first."
>
>"Now remember how curvature works: the result of
>
>               dragging v from P to Q to Q'
>
>minus the result of 
>
>               dragging v from P to P' to Q'
>
>is going to be
>
>                   -R(w,v,v)
>
>where R is the Riemann tensor.  Right?"
>
>Oz nods unconvincingly, and promises himself that he'll reread the
>course notes for the definition of the Riemann tensor.

Oz went red. Very, very red. He looked like he was going to explode.
Steam started coming out of his ears and his eyes started slowly and
deliberately popping out of his head. A small, very tiny, blue corona
started flickering faintly around his head. It looked like he was going
to explode. With a monumental effort of will he managed to slam down the
dampers before he went critical. It would probably only have resulted on
a pzzzittt instead of a zap, or probably only a ..... phit. 

"Ahem, Wiz, er Wi-hiz, woo-hoo, Wizziness. That's exactly what I have
been going on about when I said that going in a little RECTANGLE will
leave you not back where you started. The vector joining where you end
up to where you started from must surely relate to the Riemann tensor.
In this case the little vector is directly related to the velocity
picked up by Q as it went to Q'. I will leave you (Wiz) to work this
out. Oz knew his questions were mostly rubbish, but sometimes he did
wish people would listen. HUMPF!!!"

>
>"Good," says G. Wiz.  "But this is just the same as 
>
>                    R(v,w,v)
>
>since the Riemann tensor is defined so that it's skew-symmetric in the
>first two slots."
>
>"Cool," says Oz. 

Wondering what *exactly* "skew-symmetric in the first two slots" really
meant!

>"This is called the GEODESIC DEVIATION EQUATION, by the way.  I've
>cheated in 2 or 3 places in deriving it here --- for example, I hid some
>epsilons under the table --- but the result is correct.  Let me
>summarize."  The wizard had a tendency towards pedagogical pedantry
>which Oz forgave him only because of his tendency to hurl thunderbolts
>when interrupted.  "Two initially comoving particles in free fall will
>accelerate relative to one another in a manner determined by the
>curvature of space.  Suppose the velocity of one particle is v, and the
>initial displacement from it to the second is small, so that it may be
>represented as a vector w.  Then the acceleration A of the second
>relative to the first is given by R(v,w)v.  Or if you like indices,
>
>                    A^a = R^a_{bcd} v^b w^c v^d
>
>So..." the wizard paused.  "So, we can really determine the Riemann
>curvature using experiments as you proposed, and using this formula."

Was Oz pleased about this. You betcha. As soon as he had done the
washing up, he would have to have a look at R^a_{bcd} v^b w^c v^d. Well,
maybe after a beer or two with Ed, first. Well, maybe several. Oz noted
that the Wiz dropped into acceleration, rather assuming that curvature
only really existed in the t-direction. He also noted that all these
little deltas wandering about had differentiated themselves into
acceleration rather nicely. He kinda felt that these tensor jobbies
tended to do this without you really noticing. He would have to watch
out for that. It was the sort of thing that Wiz tended to assume was
obvious. He hoped they hid a lot of integration too. Preferably as much
as possible.

>"Now," asked the wizard, "remember how there are 20 components to the
>Riemann tensor, of which 10 are determined by the Ricci curvature and 10
>by the Weyl?  If we are doing the case of a homogeneous isotropic big
>bang model, most of those darn components should be redundant, thanks to
>the symmetry.  Can you figure out how many components we really need to
>worry about in this big bang model?"
>
>Oz gulped.  

Well the BB couldn't almost by definition have any Weyl bits since it
must surely be described entirely by itself, so to speak. Of course it
would help if we knew which components in the Weyl were zero. Since the
Ricci tensor can be derived from the Stress-energy tensor we only (ha)
need to sort that out. Well, obviously T_{00} is important but you can
see that T_{11} = T_{22} = T_{33} and that these won't be zero, in fact
these should be rather large and possibly even comparable to T_{00}. 

Then we have the spacial cross terms. Now I don't really know what these
represent. These are momentum in one direction being transported in
another. Now the only thing I can think of that might do that would be a
spinning particle. I suppose if we had a lot of spinning particles then
they might well cancel each other out if they were nice and random, but
I'm not totally convinced. I wonder what Wiz thinks.

Then we have the space-time cross terms T_{0X} and T_{X0}. I expect it's
obvious once you know, and probably also to do with spinning thingies
(actually I might have all this mixed up with the spacial ones now I
think about it) with luck they will all cancel out too.

I hope Wiz can discuss these cross thingies without getting, ahem,
cross.

Oz remembered all the chores he had to do. Sometimes they looked
positively attractive. He picked up his broom and ......

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!gatech!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun Mar  3 18:42:09 PST 1996
Article: 103183 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
Organization: Oz
Lines: 163
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]>
 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>"Now for the big question!  We want to know if the clock at Q' is moving
>away from the clock at P'.  To answer this, we compare its velocity
>vector to the following vector: what we get by first parallel
>translating v along itself over to P', and then over to Q'.  That's the
>velocity the clock at Q' would have it it were at rest relative to the
>clock at P'."  Oz now sees a red dot go forwards in time from P to P'
>and then across from P' to Q', carrying a red copy of the tangent vector
>v with it.
>
>   P'->----Q'
>   |       |
>   |       |
>  v^       ^
>   | w     |
>   P->-----Q
>
>The resulting red tangent vector at Q' is a bit different from
>the green one representing the actual velocity of the clock at Q'.

Well, of course in his heart of hearts Oz rather felt that the Wiz was
having it both ways. He rather felt that if he had drawn little square
rightangled thingies then G. Wiz would have said, 

"Aha, using tachions again I see. Yes, well it *does* make it all much
easier doesn't it? Now go away and do it properly. zzzZZZAP!!!"

On the other hand, Wiz being very wise, always knows which bit of
approximation he can get away with. Hmmmm something about "experience"
Oz thought. Oz rather hoped he could catch a dose of experience, but it
did sound unpleasant.

>"Curvature!" cries Oz in a moment of revelation.

[Well, it paid to humor the old boy occasionally]

>"Right!  We are taking the vector v and parallel translating it two
>different ways from P to Q' and getting two slightly different answers.
>If the answers were the same, the second clock would remain at rest
>relative to the first.  But in fact they are not, and the difference
>tells us how the second one begins accelerating away from the first."
>
>"Now remember how curvature works: the result of
>
>               dragging v from P to Q to Q'
>
>minus the result of 
>
>               dragging v from P to P' to Q'
>
>is going to be
>
>                   -R(w,v,v)
>
>where R is the Riemann tensor.  Right?"
>
>Oz nods unconvincingly, and promises himself that he'll reread the
>course notes for the definition of the Riemann tensor.

Oz went red. Very, very red. He looked like he was going to explode.
Steam started coming out of his ears and his eyes started slowly and
deliberately popping out of his head. A small, very tiny, blue corona
started flickering faintly around his head. It looked like he was going
to explode. With a monumental effort of will he managed to slam down the
dampers before he went critical. It would probably only have resulted on
a pzzzittt instead of a zap, or probably only a ..... phit. 

"Ahem, Wiz, er Wi-hiz, woo-hoo, Wizziness. That's exactly what I have
been going on about when I said that going in a little RECTANGLE will
leave you not back where you started. The vector joining where you end
up to where you started from must surely relate to the Riemann tensor.
In this case the little vector is directly related to the velocity
picked up by Q as it went to Q'. I will leave you (Wiz) to work this
out. Oz knew his questions were mostly rubbish, but sometimes he did
wish people would listen. HUMPF!!!"

>
>"Good," says G. Wiz.  "But this is just the same as 
>
>                    R(v,w,v)
>
>since the Riemann tensor is defined so that it's skew-symmetric in the
>first two slots."
>
>"Cool," says Oz. 

Wondering what *exactly* "skew-symmetric in the first two slots" really
meant!

>"This is called the GEODESIC DEVIATION EQUATION, by the way.  I've
>cheated in 2 or 3 places in deriving it here --- for example, I hid some
>epsilons under the table --- but the result is correct.  Let me
>summarize."  The wizard had a tendency towards pedagogical pedantry
>which Oz forgave him only because of his tendency to hurl thunderbolts
>when interrupted.  "Two initially comoving particles in free fall will
>accelerate relative to one another in a manner determined by the
>curvature of space.  Suppose the velocity of one particle is v, and the
>initial displacement from it to the second is small, so that it may be
>represented as a vector w.  Then the acceleration A of the second
>relative to the first is given by R(v,w)v.  Or if you like indices,
>
>                    A^a = R^a_{bcd} v^b w^c v^d
>
>So..." the wizard paused.  "So, we can really determine the Riemann
>curvature using experiments as you proposed, and using this formula."

Was Oz pleased about this. You betcha. As soon as he had done the
washing up, he would have to have a look at R^a_{bcd} v^b w^c v^d. Well,
maybe after a beer or two with Ed, first. Well, maybe several. Oz noted
that the Wiz dropped into acceleration, rather assuming that curvature
only really existed in the t-direction. He also noted that all these
little deltas wandering about had differentiated themselves into
acceleration rather nicely. He kinda felt that these tensor jobbies
tended to do this without you really noticing. He would have to watch
out for that. It was the sort of thing that Wiz tended to assume was
obvious. He hoped they hid a lot of integration too. Preferably as much
as possible.

>"Now," asked the wizard, "remember how there are 20 components to the
>Riemann tensor, of which 10 are determined by the Ricci curvature and 10
>by the Weyl?  If we are doing the case of a homogeneous isotropic big
>bang model, most of those darn components should be redundant, thanks to
>the symmetry.  Can you figure out how many components we really need to
>worry about in this big bang model?"
>
>Oz gulped.  

Well the BB couldn't almost by definition have any Weyl bits since it
must surely be described entirely by itself, so to speak. Of course it
would help if we knew which components in the Weyl were zero. Since the
Ricci tensor can be derived from the Stress-energy tensor we only (ha)
need to sort that out. Well, obviously T_{00} is important but you can
see that T_{11} = T_{22} = T_{33} and that these won't be zero, in fact
these should be rather large and possibly even comparable to T_{00}. 

Then we have the spacial cross terms. Now I don't really know what these
represent. These are momentum in one direction being transported in
another. Now the only thing I can think of that might do that would be a
spinning particle. I suppose if we had a lot of spinning particles then
they might well cancel each other out if they were nice and random, but
I'm not totally convinced. I wonder what Wiz thinks.

Then we have the space-time cross terms T_{0X} and T_{X0}. I expect it's
obvious once you know, and probably also to do with spinning thingies
(actually I might have all this mixed up with the spacial ones now I
think about it) with luck they will all cancel out too.

I hope Wiz can discuss these cross thingies without getting, ahem,
cross.

Oz remembered all the chores he had to do. Sometimes they looked
positively attractive. He picked up his broom and ......

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!psgrain!iafrica.com!pipex-sa.net!plug.news.pipex.net!pipex!tube.news.pipex.net!pipex!lade.news.pipex.net!pipex!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun Mar  3 18:42:19 PST 1996
Article: 103257 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 3 Mar 1996 10:52:07 +0000
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>>and probably the cross terms are equal, at least some of them.
>>Well, surely
>>T_{12} = T_{21} = T_{13} = T_{31} = T_{23} = T_{32} = Td
>>must be equal too in this situation. They presumably give the density of
>>momentum flow in one direction travelling in another. Maybe something to
>>do with things that spin or suchlike. 
>
>Can you use the isotropy to say more about these off-diagonal terms
>T_{ij} (i,j = 1,2,3)?  Hint: yes, you can!  Certainly what you have said
>so far is true, but it doesn't nearly exhaust the consequences of
>rotational symmetry.  You have only used rotations that switch the x, y,
>and z axes.  (Please, folks, don't post the answer to this question; let
>my victims Oz and Ed tackle this one first.)

Well, I took 'isotrophic' to mean isotrophic in space. It's evolving in
time so I don't feel that I ought to start making it isotrophic there
without a good argument.

>>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms.  What they
>>represent physically is not too clear. 
>
>Remember, T_{ab} is the flow in the a direction of b-momentum.  So the
>component T_{0i} represents the density of i-momentum (i.e., x-momentum,
>y-momentum, or z-momentum), while T_{i0} represents the flow of energy
>in the i direction.  Can you use isotropy to say something about these?

Oops. I thought I had set T_{0i} (i=1..3) equal and T_{i0} (i=1..3)
equal, but I hadn't. Clearly they are equal. It would be worth
investigating if we could plausibly make these zero as this would
simplify things quite a bit. Well T_{i0} representing the flow of energy
in the i direction looks a good candidate. There should be as much going
in the -i direction as the +i direction so I vote for T_{i0} = 0.

Now this makes R_{ab} look terribly asymmetric with those T_{0i} terms
representing the density of i-momentum. Not good for an isotrophic
uniform universe. Oh, well the argument is the same, momentum is a
vector. So in any small volume if we add all the little vectors up, that
are travelling in the i-direction, +ve and -ve, we will get zero net
density. Er, I think.

How did I miss this before? Is it right? Sounds plausible anyway. 

Oh dear. Now all those T_{ij} (i,j = 1 to 3) look terribly out of place.
Wouldn't it be nice to have them zero as well? Let's see. T_{ij} must be
the flow in the i direction of momentum in the j direction. Now we gotta
be careful here or we will convince ourselves that T_{ii} equals zero
too. I have the distinct feeling that this will result in a period of
time spent as a colonic parasite, which is even worse than an intestinal
one. Well, T_{ii} is a pressure. It's a little hard to see how T_{ij}
(j=/=i) could be a pressure as long as we keep our hands waving at all
times since the flow is perpendicular to the action, so to speak. Of
course if we knew the derivation of T_{ii} being a pressure then it would
almost certainly be perfectly clear. Anyway, throwing caution to the
winds, and without a really proper explanation, lets set T_{ij} (i,j=1 to
3; j=/=i) = 0 too. In for a penny in for a pounding.

so R_{ab} = 

[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}),   T_{12}, T_{13}]
[T_{20} ,  T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}),  T_{23}]
[T_{30} , T_{31}, T_{32} ,  (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]

or 2R_{ab} =

[(T_{00} + 3P),0            ,0            ,0            ]
[0            ,(T_{00} - 3P),0            ,0            ]
[0            ,0            ,(T_{00} - 3P),0            ]
[0            ,0            ,0            ,(T_{00} - 3p)]

Which is noticeably simpler. Also possibly wrong, anyway 

d^2V/dt^2 = -2V R_{ab} v^a v^b 

so now we want to know what this ball is doing.

No, I'm going to need some hints as to where to go next. Obviously I can
now pick various vectors v=(1,0,0,0) and v=(0,1,0,0) or even if one felt
masochistic v=(1,1,1,1). Well, let's do two to show willing.

1) v=(1,0,0,0) Or how the volume varies in the t-direction?

d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} + 3P)

>>Now, as the universe gets hotter and smaller, one could imagine that 3P
>>gets bigger. In fact one could imagine that 3P could even get to be as
>>big as T_{00}. Lots of interactions producing massless particles and
>>all. 

I assume that 3P is in fact negative. In other words it resists the
contraction of the ball. Now of course we haven't considered dV/dt. One
has the intuitive feel that 3P and dV/dt ought to be related rather
strongly, more so if much of the energy is composed of mass. If all the
energy were photons then one might not be astonished if somebody who had
considered this carefully were to suggest that d^2V/dV^2 = 0 or even
became positive. 

Then there is the problem of blackholiness. Now this (one above)
derivation (ho-ho-ho) is odd in that we are enforcing a minkowski metric
at every small point. We are also (at least I am at the moment) setting
the Weyl tensor at zero. So we are 'sort of' saying that curvature due to
distant thingies cancels out due to the uniform isotrophy of everything.
This doesn't seem to be the same case I would expect from a blackhole
derivation where 'distant' space is minkowskian and the Weyl tensor
presumably dominates due to the 'point' source. In other words the
blackhole situation is very far from uniform and isotrophic.

2) v=(0,1,0,0) Or how the volume varies in the x-direction???????

d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} - 3P)

Well, I have a conceptual problem here. I have a gut feeling that the
volume here ought to be a (0,1,1,1) type volume and we should be
considering d^2V/dt^2 or possibly dV/dt.dv/dx. Anyway because so many
derivations are skipped (probably thankfully) it is not very clear what
this represents. So someone will have to tell me. One can guess all over
the place, but it's better to be told. Or perhaps hinted at heavily, very
heavily.

PS I hope I am not toooooooo far out.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Sun Mar  3 20:00:34 PST 1996
Article: 103267 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 3 Mar 1996 18:40:16 -0800
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Oz writes: 

>Well, I took 'isotrophic' to mean isotrophic in space. It's evolving in
>time so I don't feel that I ought to start making it isotrophic there
>without a good argument.

By the way, it's ISOTROPIC, not "isotrophic"... my gentle hints (like
spelling the word correctly) seem not to be getting through...

Yes, the big bang universe is only isotropic in space, and using this
you should be able to conclude a lot about the space components
T_{ij} (i,j = 1,2,3) of the stress-energy tensor... and similarly for
the Ricci tensor!

>>>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms.  What they
>>>represent physically is not too clear. 

>>Remember, T_{ab} is the flow in the a direction of b-momentum.  So the
>>component T_{0i} represents the density of i-momentum (i.e., x-momentum,
>>y-momentum, or z-momentum), while T_{i0} represents the flow of energy
>>in the i direction.  Can you use isotropy to say something about these?

>Oops. I thought I had set T_{0i} (i=1..3) equal and T_{i0} (i=1..3)
>equal, but I hadn't. Clearly they are equal. It would be worth
>investigating if we could plausibly make these zero as this would
>simplify things quite a bit. Well T_{i0} representing the flow of energy
>in the i direction looks a good candidate. There should be as much going
>in the -i direction as the +i direction so I vote for T_{i0} = 0.

Yes, indeed, isotropy makes these zero!  The net flow of energy in any
given direction through a point is zero, or there'd be a preferred
direction there.  So T_{i0} = 0.  

>Now this makes R_{ab} look terribly asymmetric with those T_{0i} terms
>representing the density of i-momentum. Not good for an isotrophic
>uniform universe. Oh, well the argument is the same, momentum is a
>vector. So in any small volume if we add all the little vectors up, that
>are travelling in the i-direction, +ve and -ve, we will get zero net
>density. Er, I think.

>How did I miss this before? Is it right? Sounds plausible anyway. 

It's right.  The density of momentum in the i-direction must be zero, or
there'd be a preferred direction. 

Okay, so now you've got rid of those irksome T_{i0} and T_{0i} guys.

>Oh dear. Now all those T_{ij} (i,j = 1 to 3) look terribly out of
>place.

Indeed.

>Wouldn't it be nice to have them zero as well?

Indeed.

>Let's see. T_{ij} must be
>the flow in the i direction of momentum in the j direction. Now we gotta
>be careful here or we will convince ourselves that T_{ii} equals zero
>too. I have the distinct feeling that this will result in a period of
>time spent as a colonic parasite, which is even worse than an intestinal
>one. Well, T_{ii} is a pressure. It's a little hard to see how T_{ij}
>(j=/=i) could be a pressure as long as we keep our hands waving at all
>times since the flow is perpendicular to the action, so to speak. Of
>course if we knew the derivation of T_{ii} being a pressure then it would
>almost certainly be perfectly clear.

The derivation is quite simple and I think I said it before.  For
example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
for visual vividness that this flow is carried by little ball-shaped
atoms.  Then if you put a wall in their way they push on the wall in
the x direction, with a certain pressure.  This pressure is a force per
unit area, which is just the same as "momentum per time per unit area".
This is given by the flow of x-momentum in the x-direction!

>Anyway, throwing caution to the
>winds, and without a really proper explanation, lets set T_{ij} (i,j=1 to
>3; j=/=i) = 0 too. In for a penny in for a pounding.

Good!  Your intuitions are leading you right here... those off-diagonal
terms T_{ij} (i,j=1,2,3) are also zero.

By the way, a tensor like T_{ij} --- a symmetric (0,2) tensor on space
--- can only be invariant under rotations if it is diagonal and all the
diagonal entries are equal.  That is how the math experts out there
would have figured this out.  But you triumphed using just the physics
of this example!  And I knew you could... believe it or not, I am not
setting you off on an impossible quest... just testing your valor,
that's all.

>so R_{ab} = 

>[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
>[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}),   T_{12}, T_{13}]
>[T_{20} ,  T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}),  T_{23}]
>[T_{30} , T_{31}, T_{32} ,  (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]

>or 2R_{ab} =

>[(T_{00} + 3P),0            ,0            ,0            ]
>[0            ,(T_{00} - 3P),0            ,0            ]
>[0            ,0            ,(T_{00} - 3P),0            ]
>[0            ,0            ,0            ,(T_{00} - 3P)]

>Which is noticeably simpler. Also possibly wrong...

It looks right to me.  Why don't we call T_{00} something like E, the
energy density.  So, you've seen what the Ricci tensor is in terms of
E and P.  Now you are pretty close to answering the question.  What was
the question, anway?  Oh yes...   

3. In the big bang model, what happens to the Ricci tensor as you go back in
past all the way to the moment of creation? 

So, all you need is to tell me what happens to E and P.  

>d^2V/dt^2 = -2V R_{ab} v^a v^b 

Let's see, I don't think there should be a 2 there.  

>so now we want to know what this ball is doing.

>No, I'm going to need some hints as to where to go next. Obviously I can
>now pick various vectors v=(1,0,0,0) and v=(0,1,0,0) or even if one felt
>masochistic v=(1,1,1,1). Well, let's do two to show willing.

>1) v=(1,0,0,0) Or how the volume varies in the t-direction?

>d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} + 3P)

Yes, this is the most interesting case.  If you keep with it, by the
way, you can do a lot more than I hoped for... you can figure out pretty
much everything about the big bang solution!  But let's see, I'm not
sure this stuff about the ball of particles will help you figure out
what happens to R_{ab} as you approach the initial singularity.  (It
will help you understand what it *means* though.)  

>I assume that 3P is in fact negative. 

Why should the pressure be negative?  Pressure is usually positive.

>In other words it resists the contraction of the ball.

Hmm.  Sounds fishy to me.

>Now of course we haven't considered dV/dt. 

>Then there is the problem of blackholiness. 

Aren't we talking about the big bang in this problem?  Have you shifted
over to considering a black hole? I'm confused.

>We are also (at least I am at the moment) setting the Weyl tensor at
>zero. 

Oh yeah?  Why are you doing that?  I'm not saying it's bad, but explain
why you're doing that?  (Hint: isotropy, isotropy, isotropy.)

From galaxy.ucr.edu!noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!zombie.ncsc.mil!news.duke.edu!hookup!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun Mar  3 20:11:44 PST 1996
Article: 103241 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Friedmann Models
Date: Thu, 29 Feb 1996 06:36:06 +0000
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In article <[email-masked]>, "Emory F. Bunn"
<[email-masked]> writes

<snip>

>dr^2+r^2 dphi^2.
>
>A uniform curved two-dimensional surface is either a sphere or
>two-dimensional hyperbolic space.  If we label the sphere with
>polar coordinates (theta,phi), the line element is
>
>dtheta^2+(sin theta)^2 dphi^2.
>
<snip>
>
>The line element for two-dimensional hyperbolic space is
>
>dr^2+(sinh r)^2 dphi^2.
>
>So (up to random renamings of coordinates) hyperbolic space is what
>you get when you replace trig functions by hyperbolic ones.

I was wondering why I couldn't see exactly how you derived this. Then I
remembered that Baez hasn't actually expressly indicated how you
calculate curvature from a real geometry. Now my integration has rusted
right away. About the only thing I remember about sinh(x) is that
d[sinh(x)]/dx = cosh(x) and d[cosh(x)]/dx = sinh(x). As against sin(x)
and cos(x) where they flip sign. However, surely an abbreviated
description relating a constant curvature and the metric is possible for
2D space?

>So much for two dimensions.  On to three.  Flat three-dimensional
>space has line element dx^2+dy^2+dz^2, or in spherical coordinates,
>
>dr^2 + r^2 (dtheta^2 + (sin theta)^2 dphi^2).
>
>The three-sphere has line element
>
>dr^2 + (sin r)^2 (dtheta^2 + (sin theta)^2 dphi^2),
>
>and you get the answer for negatively curved three-space by replacing
>sin r by sinh r.

This looks like integrating ds is going to be a real pig, mess
everywhere. I suppose it is too much to hope that it ends up neat and
tidy? No, I thought not. This sort of tree-killer usually ends up with a
named set of functions so the poor old mathematicians don't actually
have to write anything much down. For mere mortals, however, this
doesn't speak to us.

>>>2. You need to know a(t) for all time.  This tells you the size of the
>>>Universe as a function of time.  Its derivative tells you about the
>>>expansion rate of the Universe.  Specifically, (da/dt)/a is nothing
>>>other than the Hubble constant.
>>
>>Er, hang on a tick. If (da/dt)=ka, where k=hubble factor= 'a constant'
>>then a(t) = b exp(kt),  b & k constants. This is a very specific
>>function. 
>
>Aha.  You've fallen into our clever trap.

No, no, no. OK, *I* called it the Hubble *factor*, instead of parameter,
but you should have guessed what I was driving at.

>>So in cosmological terms, can I assume (as might be expected) that the
>>T_{11} to T_{33) terms are negligeable? Would this have been the case in
>>the very early universe when one might have expected high kinetic
>>energies?
>
>Exactly.  Nowadays T_{00} is the dominant component.  At early times,
>when things were hot, all of the diagonal components of T were about
>the same magnitude.  Cosmologists refer to that epoch as the time of
>"radiation domination," and say that nowadays the Universe is "matter
>dominated."

I am just trying to get a handle on this. In an energetic and relatively
low density object, such as a sun, one would expect T_{00} still to
dominate. However one might expect the T_{11} terms might still have
important effects if they equate to pressure. Could you comment?

>Remember how you measure curvature?  You drag your tangent vector
>around a little loop and you see how much it deviates.  If you do that
>in a flat Friedmann model, and all of the points on your loop lie at a
>single value of t, you won't get any deviation.  That's because, even
>though the full four-dimensional spacetime is curved, the
>three-dimensional slice through spacetime that your loop lies in
>is flat.  In order to demonstrate that space was curved, you'd
>have to use a loop that had excursions in the t direction, not
>just in the spatial directions.
>
>(I hope it's obvious that in this last paragraph, I'm thinking of
>"you" as a mathematician studying this particular spacetime, not as a
>hapless denizen of it.  Obviously you yourself can't carry a tangent
>vector around a loop at a fixed instant of time, since you're
>constantly whizzing forward through time!)

That's what I thought you meant. However, I have learned to be very
careful!. I take it that one might expect the *real* universe not to be
flat Friedmann, but to have curvature appropriate to it's momentum
density at the very least. This brings us into the entertaining mind
expanding bit about universes without end or centre that curve in the
space directions too. ALL of the space directions at once. Oh dear, and
the time direction too is not impossible. Yerk! And us poor peasants
still trying to cope with minkowski space's idiosyncracies. Oh, well,
onwards and upwards. (NB any chance of a comment or two on this
paragraph?).

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!ihnp4.ucsd.edu!swrinde!cs.utexas.edu!howland.reston.ans.net!agate!physics12.Berkeley.EDU!ted Mon Mar  4 10:24:18 PST 1996
Article: 103303 of sci.physics
Path: galaxy.ucr.edu!ihnp4.ucsd.edu!swrinde!cs.utexas.edu!howland.reston.ans.net!agate!physics12.Berkeley.EDU!ted
From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Friedmann Models
Followup-To: sci.physics
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>OK, I wonder if you could answer a very simple question that I suspect I
>am supposed to know already. Excluding any cosmological constants, but
>including energy-momentum stress, should we see our universe as a
>minkowski space but with curvature impressed by it's contents. In other
>words is the metric minkowski. 

I'm afraid I don't understand the question.  Minkowski space is flat.
If there's energy and momentum around, space isn't flat, so it's not
Minkowski.  That much I'm sure you know.  You seem to mean something
by the phrase "Minkowski with curvature," but I'm hard pressed to
guess what it might be, since flatness is the defining characteristic
of Minkowski space.

If you confine yourself to an infinitesimal neighborhood of spacetime,
that neighborhood always looks like Minkowski space.  To be a little
less vague, if your neighborhood is of size epsilon, deviations
between the real metric and the flat Minkowski metric go to zero as
epsilon goes to zero.

>To take another case of a mass in
>otherwise flat spacetime 

I haven't even gotten to the end of your sentence yet and already my
soul is troubled.  The spacetime surrounding a mass is not flat.
Maybe it's "otherwise flat," but before I can tell you whether it is
or not you have to tell me what that phrase means.

>would the situation be the same. I rather
>suspect that you could also equivalently express the space as one having
>effectively no curvature due to T_{ab}, but having a metric that varied
>from place to place in spacetime.
>
>I can see a technical problem with dragging the little vector round a
>loop to measure the curvature. It would typically be a little tricky to
>come back to where you started. For one thing your path lengths would
>not be the same on each leg of the loop. I would have thought that if
>your loop was small, and you did in fact do equal length legs in two
>locally orthogonal directions then you would end up a very small
>distance from where you started. Then the vector joining your starting
>and ending points should relate to the curvature in a reasonably direct
>way. Why is this not so?

Here we have to play games where we count powers of epsilon.  If you
walk a distance epsilon North, then epsilon East, then epsilon South,
then epsilon West, you won't get back to exactly where you started.
Of course, as epsilon goes to zero, the distance by which the loop
fails to close also goes to zero.  Furthermore, this distance goes to
zero fairly fast.  (That is, it goes like a high power of epsilon.)
So for small loops, you get to neglect this discrepancy.  In
particular, for small loops it's a smaller effect than the fact that
tangent vectors rotate as you parallel transport them.

In fact, if you want a homework problem, it might be nice to work this
out.  Start at the North pole of a sphere.  Walk a distance epsilon
South.  (What other direction could you go, after all?)  Turn left,
and walk a distance epsilon.  Turn left again, and walk a distance
epsilon.  Turn left one last time, and walk a distance epsilon.  If
you were on a flat surface, you'd be back home now.  On the sphere,
you'll be a little ways away.  You can work out how far away you'll
be, and you'll find it goes like epsilon cubed or something.  That
means it drops off faster than the amount of rotation suffered by a
parallel-transported tangent vector.  So when you're measuring
curvature, you can choose a loop that's small enough that the former
effect can be neglected, and then the latter effect tells you the
curvature.

-Ted

From galaxy.ucr.edu!not-for-mail Mon Mar  4 17:01:50 PST 1996
Article: 103494 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 4 Mar 1996 15:43:38 -0800
Organization: University of California, Riverside
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One day G. Wiz was in his study, busily juggling indices.  Greek and
Roman superscripts and subscripts were flying through the air in neat,
orderly paths... a miracle of precision.  Then Oz burst into the study
all of a sudden.  Shocked, the wizard dropped all his indices, and they
scurried away across the floor and under the curtain that led to that
mysterious back room.  

"Damn!" cried the wizard.  "Now look what you've made me done.  I have
to start all over!  Who said you could come in here, anyway?" 

"Hi!" said Oz cheerfully.  "I want to see how the Ricci tensor to
the change in volume of a small ball of test particles, just like you
said, using the geodesic deviation equation!"

The wizard scowled and asked, "Didn't I say something like 
you can derive it from the geodesic deviation equation, AT LEAST IF YOU
ARE BETTER AT INDEX JUGGLING THAN I SUSPECT YOU ARE?"  He pulled out a
yellowing sheet of parchment from a towering stack on his desk and
glanced at it.  "Yes, I believe those were my exact words!"  He stuck it
back in the exact same spot.  

Undaunted, Oz said, "Well, okay, but anyway... you said that

          Riemann = Ricci + Weyl,

right?"

The wizard nodded.  "In some vague sense, yeah."

Oz added "I thought R^a_{bcd} = R^c_{bcd} + W^a_{bcd} was given
somewhere or other." 

The wizard hurled a fireball in Oz's general direction.  "Wait a
minute!  You are violating the basic law of index juggling: all the
indices appearing on the left hand side of the equation must appear on
the right.  For the purposes of this rule, repeated indices which are
summed over --- like the c in R^c_{bcd} --- do not count!"

Oz said "You never told me that!"

The wizard hurled another, bigger fireball, and Oz stepped back.  The
wizard said "I know I never told you this rule, but I *warned* you not
to stick your fingers into the machinery, so don't blame me!  
Think about it: this rule is obvious!  Something like R^a_{bcd} is a
tensor of rank (3,1), right?  While something like R^c_{bcd}..."

"Well, R^c_{bcd} = R_{bd} is the Ricci tensor, right?" asked Oz.

"Yes, damn it, that's the point, it's a tensor of rank (0,2)!  You
can't go around adding tensors of different ranks; that's like adding
vectors and numbers... which I bet you used to do as a kid, right?"  The
wizard scowled and hurled a still bigger fireball, this time singing
Oz's left ear.  "You're just the sort who would.  I know your type.... 
So," the wizard continued, "the equation 

R^a_{bcd} = R^c_{bcd} + W^a_{bcd} 

makes no sense!  By the way," he said, his voice dripping sarcasm as he
hurled yet another fireball, singing Oz's right ear this time, "THIS IS
EXACTLY WHY NOBODY EVER TOLD YOU THIS EQUATION."  

"Gee whiz, G. Wiz!  Give me a break!" cried Oz.  "Students learn by making
mistakes!"  

"Is that what they think where you come from?" asked the wizard,
shooting a few lightening bolts from the fingers of both hands in a
casual gesture of impatience.  "Around here, students learn by GETTING
THINGS RIGHT!  And if they don't learn fast, they..."  

"Okay, okay!  So what IS the right equation?"

"Hmm," said the wizard, thinking a minute, and seeming to lose interest
in scolding Oz.  "Hmm.  I think it's something like 

R^a_{bcd} = R_{bd} g^a_c + W^a_{bcd}.

Notice that the metric tensor g^a_c provides the indices which the 
Ricci tensor is missing.  By the way, for any metric g we have g^a_c =
delta^a_c, that is, it's 1 if a = c and 0 otherwise --- we call this
delta gadget the "Kronecker delta".  So we would usually write
delta^a_c instead of g^a_c.  But I don't think this formula for
the Riemann tensor is exactly right.  I'm pretty sure that

R^a_{bcd} = K R_{bd} delta^a_c + W^a_{bcd} 

for *some* constant K, but we need to work out what the constant is. 
Just set a = c and sum over the resulting repeated index.  Note that
delta^c_c = 4 in 4 dimensions, so we get

R^c_{bcd} = 4K R_{bd} + W^c_{bcd}

Now W is defined to be the "trace free" part of the Riemann tensor,
meaning that *by definition* W^c_{bcd} is zero, so we get

R^c_{bcd} = 4K R_{bd}.

But the left side is R_{bd} by definition so we must have K = 1/4.
So I think

R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}."

When G. Wiz looked up, Oz was mysteriously nowhere to be found, so
the wizard never got to finish explaining his teaching philosophy.

From galaxy.ucr.edu!library.ucla.edu!agate!howland.reston.ans.net!nntp.coast.net!news.kei.com!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Mon Mar  4 17:01:57 PST 1996
Article: 103335 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 3 Mar 1996 14:59:31 +0000
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Oz gulped.  
>
>"By the way, you may enjoy remembering the definition of the Ricci and
>Weyl tensor in terms of how a bunch of initially comoving test particles
>begins to change volume and shape.  We have a nice formula for the Ricci
>tensor along these lines.  A while back you asked we derived it.  I said
>"wait and see".  Well, now you can derive it from the geodesic deviation
>equation, at least if you are better at index juggling than I suspect
>you are.  After all the geodesic deviation equation says exactly what
>those coffee grounds are going to do."

Well, Oz was well known for not having enough common sense to keep his
fingers out of the fire. So he came back with a few questions.

1) Riemann = Ricci + Weyl

R^a_{bcd} = R^c_{bcd} + W^a_{bcd} was given somewhere or other.

and R^c_{bcd} = R_{bd)

So lets look at the Ricci tensor. This says that if we have three
vectors (lets make them basis for simplicity) then the Ricci tensor will
output a vector giving the curvature along one of the *input* vectors.
Anyway something related to curvature. Now this is a subset of the
Riemann tensor. It's some sort of subset of diagonals of the Riemann
tensor but being 4D it's a little hard to pick it out. What we can say
is that it has 16 elements. Now as you can imagine I am slowly sinking,
but in for a penny.

Now you all say there are only ten independent items in the 16 elements.
This makes one want to set the cross terms equal (ie R_ij = R_ji). Well,
one would anyway since this is nothing more than switching your input
vectors about so it looks reasonable.

So one imagines that something *analogous* to the following is going on:

[00,01,02,03] = [00,01,02,03] + [- ,- ,- , -] in a wierd but clear 
[10,11,12,13]   [10,- ,- ,- ]   [- ,11,12,13] notation.
[20,21,22,23]   [20,- ,- ,- ]   [- ,21,22,23]
[30,31,32,33]   [30,- ,- ,- ]   [- ,31,32,33]

              = [00,01,02,03] +  
                [10,- ,- ,- ]   [11,12,13] 
                [20,- ,- ,- ]   [21,22,23]
                [30,- ,- ,- ]   [31,32,33]

Now lets look a what Wiz says about the Riemann tensor R^a_{bcd}
If x^a,x^b,x^c are basis vectors then it spits out a 

vector a *whose component in the x^a direction* is R^a_{bcd}

so presumably the component in the x^c direction is R^c_{bcd} which is
the Ricci tensor. Whew! Took me a bl**dy long time to work that out!!!

Now what does this mean. Well, it means that if I go in a little loop
along v^b and w^d carrying a vector u^c then u^c, ooops. Thats what
skew-symmetric in the first two slots means. It means R(wvv)=-R(vwv).

Anyway, we can begin to see why it's a rate of change of volume term. If
we define a small volume as being of radius (or side of a cube if you
like) epsilon. Then the volume is epsilon^3. Basis vectors x^a,x^b,x^c
spacially, and x^d temporally.

Now let's evolve this little system dt along the time axis. Compared to
the center of our little sphere, in a distance dt x^a has evolved into a
new vector whose incremental change in the x^a direction is epsilon dt
R(x^a,x^d,x^d) and similarly for x^b and x^c. So ignoring a few higher
order terms the new volume is (e=epsilon)

[e+ e dt R(x^a,x^d,x^d)][e+ e dt R(x^b,x^d,x^d)][e+ e dt R(x^c,x^d,x^d)]

or dV=[e dt R(x^a,x^d,x^d)][e dt R(x^b,x^d,x^d)][e dt R(x^c,x^d,x^d)]

or dV=e^3 dt^3 R(x^a,x^d,x^d) R(x^b,x^d,x^d) R(x^c,x^d,x^d)

Oh shit! I'm nearly there but I have this darned dt^3 that should be dt
and it's not a rate of change. Somewhere I should have had a dx/dt type
term. I am not quite grasping this yet. It's a velocity that I get out
of the Riemann tensor really and not a displacement. Anyway, I haven't
any more time I can spend on this now. 

I humbly request some wizardly help.

And how you do it using index juggling I haven't a clue. I am pretty
certain I know why. This index juggling is differentiating I think, but
without a course or help, I don't quite see just yet. 

Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp!

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Mon Mar  4 19:28:11 PST 1996
Article: 103506 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 4 Mar 1996 16:51:52 -0800
Organization: University of California, Riverside
Lines: 178
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes
>>By the way, it's ISOTROPIC, not "isotrophic"... my gentle hints (like
>>spelling the word correctly) seem not to be getting through...

>Ah, one of those deeply ingrained spelling errors. 

Well, just think what it means.  "Iso" means "equal" and "tropos" means
something like "to change, or turn" --- remember how plants that turn to
meet the light are "heliotropic", and the layer of atmosphere where
the weather keeps turning and churning is called the "troposphere".  So
"isostropic" means that no matter how you turn, things are equal.  

On the other hand, "trophos" is all about food, growth, and the like:
muscles can "atrophy" or "hypertrophy", for example.  So "isotrophic"
might refer to something like "equally well developed" muscles.  

You may argue that the word "catastrophic" means, etymologically, a
"downturn" --- but here the suffix is not "trophic" but really
"strophic", from "strophein", which also means to turn.  

So if you'd said "isostrophic", I would understand and forgive.  But
"isotrophic"??  Never!! 

>it incorrectly have failed to make an impression, your gentle chiding is
>unlikely to have much effect. I will make an effort to spot it to avoid
>you in too much agitation. 

Hopefully the annoying pedantry above will stick in your mind and make
it utterly inconceivable that you'd ever repeat the error.

---------------------------------------------------------------------------

Okay, on to the role of pressure in the big bang cosmology:

>>>I assume that 3P is in fact negative. 

>>Why should the pressure be negative?  Pressure is usually positive.

>>>In other words it resists the contraction of the ball.

>>Hmm.  Sounds fishy to me.

>Eh? Here you are looking at a ball of hot gas. It's a mini-universe. If
>3P is positive then R_{00} is always positive and the ball will always
>contract.

Don't mix up first and second time derivatives of V, my friend.  It's 
d^2V/dt^2 that's proportional to R_{00}, not dV/dt.

>Oh-ho well, perhaps ...., after all our universe is expanding
>but at a decreasing rate. 

Exactamente.  Though the universe may be expanding, pressure slows the
rate of expansion of the universe!  Ed Green, flown in specially to help
you, has already noted this ironic truth: though they always analogize
the big bang universe to a big balloon, it ain't true that pressure
inflates this "balloon".  Au contraire.

By the way, don't forget this fact --- it'll come in handy for one of
the other questions on your test, too!

>Well, I'm still not convinced. Lets take an
>almost massless ball of hot gas stuck all alone in minkowski space. Hey
>looky here, it expands, presto! Oz gets a large cannon and blows his
>head clean off. OK, you are right. The CURVATURE is positive. That's
>what I had in mind way back when I first partially answered way up the
>thread. I said that R_{00} was always positive so would go to infinity.
>Forget the coffee grains. Let's stick to curvature.

>R_{00} = (1/2)(E + 3P) Both E & P must -> infinity as size->0

>So the curvature sure tends to infinity in the time direction.

Okay, that shows that R_{00} goes to infinity.   Now a while back you
showed that all the off-diagonal terms in R_{ab} are zero in the
big-bang cosmology (at least in the coordinates we are implicitly
using here!).  So that leaves the guys like R_{11}, R_{22}, R_{33}.  What
happens to these?

You're almost done computing the Ricci curvature!

But now we digress into computing the Weyl curvature...

>>>We are also (at least I am at the moment) setting the Weyl tensor at
>>>zero. 

>>Oh yeah?  Why are you doing that?  I'm not saying it's bad, but explain
>>why you're doing that?  (Hint: isotropy, isotropy, isotropy.)

>Eh? I thought I gave a quasi-plausible reason way way back at the
>beginning of the thread. My argument went as follows:

>a) There ain't no boundary conditions since we include the entire
>universe. From what has been said about the Weyl tensor, some of it's
>terms relate to boundary conditions. Maybe tidal effects 'n stuff. These

This is quasi-plausible.  Indeed, Penrose makes a big deal out of his
quasi-plausible "Weyl curvature hypothesis", which says that as we
approach the big bang, the Weyl curvature must go to zero, for reasons
similar to what you say, and so the Weyl curvature serves as a kind of
"arrow of time", increasing as time passes.  

But this is merely speculation, and your argument is merely
quasi-plausible.  Why don't you concoct a 100% rigorous argument based
on what we know about the Weyl curvature?  Remember: the Ricci
curvature tells how your little ball of freely falling coffee grounds
changes volume, while the Weyl curvature tells how it changes shape into
an ellipsoid.

>b) The universe is uniformly isotroph^Hic so there ain't no waves or
>other transient stuff wandering about. This should take out another
>whole load of terms from the Weyl.

Again, this is true as far as it goes, but why not see if you can kill
the Weyl curvature *completely* using isotropy?

>c) So far we have only considered a teeny point-like bit of this
>universe. I have a vague memory that we have, at some point, made use of
>the approximation that in a teeny volume, spacetime is minkowskian.
>However I can't recall exactly where. On the other hand the R_{ab} we
>have derived is far from flat so hopefully this is a valid approximation
>in the limit. 

No, we've never made that approximation, we've only at times used
coordinates in which the metric is Minkowskian AT A POINT.  "At a point"
is a lot smaller than "a teeny volume"!  Anyway, I don't see how this 
helps us with the Weyl curvature.  It seems pretty irrelevant.

>I am a mite concerned that the Weyl does include static curvature from
>distant bits. Hopefully in this situation these end up zero too.
><nervous grin>

Ah, yes.  Indeed, in the earth's gravitational field, or that of a black
hole, in any region where there is no matter, all the curvature is Weyl
curvature.  Why?  Well, there's certainly curvature, since otherwise
there'd be no "gravitational field", but there's no Ricci curvature at points
where there is no matter --- by Einstein's equation.  So black holes
work very differently from the big bang.  Let's just talk about the big
bang, eh, in this thread?  You'll be dealing with black holes in the
other problem.

>2) In the spacial directions.

>R_{11} = (1/2)(E - 3P) Now hang on, what IS *R*_{11}?

Okay, yes, now what do you think happens to R_{11} as we approach time
zero?

>R_{ab} = R^c_{acb} aaaah, this is the key. 

>We have calculated a subset of the Riemann tensor, and if all the other
>elements (if the Weyl elements =0) are zero then we have calculated the
>Riemann tensor itself. I think I'll go and lie down.

Yes indeed!  You deserve it!  Once you've fully understood the Ricci
curvature in the big bang cosmology, if you can figure out the Weyl
curvature, you can nail down the entire Riemann tensor.  In a memorably
unpleasant meeting with the wizard, you will by now have learned the
exact formula for the Riemann tensor in terms of the Ricci and Weyl...
but just as important, for now, is the simple point that if we understand
Ricci and Weyl, we in principle understand Riemann.

>The problem now is to get a proper handle on what curvature 'feels'
>like. I mean actual results. Now a curvature of 1 ought to mean that we
>end up with ordinary flat pythagorean space stuff, I guess. No that's
>too general a statement, all those cross terms. Let's call on Riemann
>himself, perhaps.

WHAT THE HECK???  FLAT MEANS CURVATURE IS ZERO!!!!  R^a_{bcd} = 0!!!!
WHEN YOU CARRY A VECTOR AROUND A LOOP, IT DOESN'T CHANGE AT ALL!!!!!!

Excuse me, I just had a heart attack.  Hopefully I'll post some
more from the intensive care unit tomorrow.  

From galaxy.ucr.edu!ihnp4.ucsd.edu!munnari.OZ.AU!news.ecn.uoknor.edu!paladin.american.edu!hookup!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Tue Mar  5 15:59:06 PST 1996
Article: 103642 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Tue, 5 Mar 1996 07:20:18 +0000
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>But this is merely speculation, and your argument is merely
>quasi-plausible.  Why don't you concoct a 100% rigorous argument based
>on what we know about the Weyl curvature?  Remember: the Ricci
>curvature tells how your little ball of freely falling coffee grounds
>changes volume, while the Weyl curvature tells how it changes shape into
>an ellipsoid.

Aw, you gave it away. Ok, ok it's a better way of saying it than
'boundary conditions' and stuff. The universe couldn't be or stay
uniformly isotroph^hic if Weyl was anything but zero. Almost by
definition, well probably actually by definition.

<Minor lightbulb from Oz snipped>
<The shock caused him to carry on, mindlessly>
<A BIG mistake>

>
>WHAT THE HECK???  FLAT MEANS CURVATURE IS ZERO!!!!  R^a_{bcd} = 0!!!!
>WHEN YOU CARRY A VECTOR AROUND A LOOP, IT DOESN'T CHANGE AT ALL!!!!!!

Um, er, ahem, toodle-oodle-tee-tee-tum, Oh Hi Wiz, wotcha doing
thrashing on the ground tearing your hair out? Uh-ho, it's not often Wiz
is speechless with rage. Er, I think I'll just take a walk so the
mountain is between me and Wiz, OK. Teedle-tidal-toodle-tum .....

<Oz's happy go lucky little song fades slowly into the distance.>
<Once out of sight he legs it round the mountainside at full speed.>

Unfortunately my tiny scrap of brain was drifting into how the metric
and the Riemann curvature relate. I was thinking that a flat *metric*
diagonalises as all ones (not the R-curvature), a sort of mental typo.
Somehow, and it's almost seeable, we must be able to merge the
underlying minkowski metric with the curvature to end up with another
metric that encodes the curvature. On the other hand I suppose Riemann
itself encodes the (-+++) essence of minkowski already so maybe it's
much simpler.

I will send a 'get well' tablet, as soon as possible.
Maybe Ed will take it for me?

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Tue Mar  5 16:43:51 PST 1996
Article: 103694 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 5 Mar 1996 14:05:24 -0800
Organization: University of California, Riverside
Lines: 136
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, [email-masked] (john baez) wrote:

>> Now let V* be the dual vector space, i.e. the space of linear functions
>> from V to R (the real numbers).  Guys in V* are called covectors.  A
>> Lorentz transformation L acts on a covector f as follows:

>> (Lf)(v) = f(L^{-1}v).

>This is the main thing that would have kept me from working this out
>abstractly like you did below.

>We originally had L:V->V, and now you're using L:V*->V*.  There must be
>pullbacks or pushforwards involved here.  
>When you write 'L' in (Lf)(v), do you really mean the dual of the
>pushforward of L? 

Hmm, errr, well, let me just say that if I had wanted to use standard
jargon I would have written that map from V* to V* as (L^{-1})* rather
than merely L.  I was trying to keep the notation down to a dull roar.  
As for what you call pushforwards and pullbacks and so on...

>And L^{-1} is the pushforward of L?

...hmmm, well, let me give a long answer, not a "yes or no" answer,
since the latter would be misleading.

In the stuff above, we're just doing linear algebra.  Let me explain the
general picture.  If we have any linear map L: V -> W between vector
spaces, we get a linear map, the ADJOINT, L*: W* -> V*, given by

L*(f)(v) = f(L*v).

Note that here L doesn't need to be invertible.  Now there is a
ever-so-slightly annoying thing about adjoints, namely that they "go the
other way".  This implies that

(LM)* = M*L*

whenever either side makes sense (namely, whenever M: U -> V is some
other linear map).  So things get backwardized.  It's no big deal,
just a minor nuisance at times.  

If we are dealing with invertible linear maps we can debackwardize
things using their inverses.  If L: V -> W is invertible, then we get
(L^{-1})*: V* -> W*.  So we get

(L^{-1})* (M^{-1})* = ((LM)^{-1})*

I.e., things no longer get twisted around.  This is handy when you are
talking about group representations like I was above.  In
the situation I was talking about above, I had a representation of the
Lorentz group on vectors.  In other words, for any vector v and any
Lorentz group element L, I get a vector Lv, and

(LM)(v) = L(M(v)).

That's what representations satisfy.  Then I wanted to get a
representation on covectors.  If I just used L*, I would have gotten 

(LM)*(v) = M*(L*(v)).

So this wouldn't have been representation... it would be some other
thing called a "right representation"!  No big deal, but a minor
nuisance.  So I corrected it, the way people typically do, by using
(L^{-1})*.  

-------------------------------------------------------------------------

Now say you're studying tangent vectors and cotangent vectors on
manifolds.  Say you have a smooth map f: M -> N between manifolds.
Then for any x in M you get a linear map

df: T_x M -> T_{f(x)} N

called the pushforward.  This linear map might not be invertible; if
it's not, we can't use that trick above to "debackwardize" things, so we
just use the pullback

(df)*: T*_{f(x)} N -> T*_x M

and accept the fact that

(d (fg))* = (dg)* (df)*.

-----------------------------------------------------------------------
>Even after all this GR tutorial stuff, I still don't feel totally
>comfortable with vectors, covectors, pullbacks, and the like.  I see
>mathematically what these things are, but I don't really have a feel
>for it. 

Well, as far as I can tell, it helps if you solve lots of physics and
math problems using these ideas; then you get a feel for it.  Work out
some problem using some coordinates, and then use this stuff to change
coordinates, and it'll start making more sense.  

>It's easy for me to see that L is a linear transformation on
>V, but not as easy for me to see why what you wrote is the most
>natural, canonical way of turning L into a linear transformation on V*.

Well, as you see, it's not!  What I wrote is only applicable to the case
where L is invertible, and there is another very important thing to use
when L is not necessarily invertible.

>> Tensoring with the metric is Lorentz invariant since the metric.  

>Umm, did something happen to this sentence between there and here?  :)

Whoops: 

Tensoring with the metric is Lorentz invariant since the metric is
Lorentz invariant.

>What exactly is "tensoring with the metric," anyway?

Break out the indices!  If you have a tensor like R_{ab}, you can get a
new tensor with more indices like X_{abcd} = R_{ab} g_{cd}.  This is
"tensoring with the metric".   You will note that I did this when
explaining to Oz the formula for the Riemann tensor in terms of the
Ricci and Weyl tensors.

Abstractly, we have a tensor R in the space V* tensor V* (where V is
some tangent space), and a tensor g in the space V* tensor V*, so there
is a kind of product of them, R tensor g, in V* tensor V* tensor V*
tensor V*.  This is called "taking the tensor product" or simply
"tensoring".  I describe this on p. 368 of my book, by the way.

>> If the above seems overly abstract I urge you to rewrite all the
>> equations above using indices, which may help if you are familiar with
>> those.  

>Indices?  I don' NEED no stinkin' indices!

Good!  (Though it's also important to know how to do everything using
indices.)  

From galaxy.ucr.edu!library.ucla.edu!agate!howland.reston.ans.net!swrinde!newsfeed.internetmci.com!in1.uu.net!pipeline!not-for-mail Tue Mar  5 17:04:45 PST 1996
Article: 103436 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 4 Mar 1996 09:35:26 -0500
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Scene:  Cave of the Apprentices 
 
Erg is discovered cleaning his armor.   Oz is in conference with the wizard
above,  and the stray spacetime eddies that continually drift from his back
rooms carry little snippets of the conversation...  
 
Wiz>"Now for the big question!  We want to know if the clock at Q' is 
>moving away from the clock at P'.  To answer this, we compare its 
>velocity vector to the following vector: what we get by first parallel 
>translating v along itself over to P'... 
 
Yow!   This Oz must really know his stuff...  I don't even know what the
Wiz is talking about anymore...   I don't know what *Oz* is talking
about...  though I suspect he is getting assistance from his brothers Ox
and Oy (they shared a common origin).  I liked it better when *I* got to
ask the questions... "Help Oz"...  ha ha ha ha,   right.  Oh,  well,  I
suppose I'd better try to bring myself up to speed here... 
 
Erg sets out to retrieve some of the extant tablets from outside the cave; 
as an afterthought,  he decides to put on his armor;  there may be trolls
lurking about.  Sure enough,  no sooner has he put his head out the opening
then a cry is raised:  They had been waiting for a return engagement.   
 
The meeker ones cringe away,  but a band of young troll-toughs urge each
other on,  eager to settle the score.  Brandishing his newly cleaned sword
seems to interest them only mildly,  and then slowly,  as their brains
react with their characteristic geological wit,  to amuse.  Slowly, and
horribly,  individual trolls let out squawks and burps,  which gradually
begin to resolve themselves into that most frightening of troll utterances:
 Laughter!  A noise like a rock sifter working on aggregate became
deafening,  and some of the trolls actually began to crack smiles,  the
fissures extending from the corners of their mouths and throwing off rock
splinters. 
 
Dumbfounded,  our hero let the sword hang limp and began backing up, while
the cacaphonous trolls scented victory and closed in for the kill.  And
then,  he did the only thing that could possibly have saved him,  whether
by luck or design we cannot know,  though with each retelling the history
drifted torwards the latter.  Erg tripped over backwards and again
collapsed in a heap! 
 
Instantly,  silence.  The expressions of amusement turned to horror as they
all remembered where *that* had led to.  Stumbling over themselves in their
eagerness to be away,  lest they too be hurled over the cliff,  the young
toughs soon showed their superior mettle by outdistancing all other trolls
in the rout.   Indeed,  some of their fearsome but oddly alluring mates,
the trollops,  who had been hanging back during the confrontation,  now
seemed in little hurry to leave the field,  but shambled off giving the
hero suggestive moues over their brawny shoulders, which made him blush
under his visor.  But soon they too were gone, leaving the field
unconstested,  for which the reader may be grateful,  for he is beginning
to suspect I have overextended the franchise of sci.physics...  Our hero
shrugs, and gathering the remaining tablets under his arm  (some with troll
bite marks),  returns to the cave.   
 
Here he begins to take notes. 
 
[But now,  having exhausted the time available to pursue this in frivolity,
 yet eager to offer up this humble leavening of amusement while it is still
fresh,   I must beg the gentle reader's indulgence if I post an item
totally free of physics and resume note-taking on the morrow...  No more
trolls,  I promise.]

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Article: 103449 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Mon, 4 Mar 1996 08:02:30 +0000
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>By the way, it's ISOTROPIC, not "isotrophic"... my gentle hints (like
>spelling the word correctly) seem not to be getting through...

Ah, one of those deeply ingrained spelling errors. If years of spelling
it incorrectly have failed to make an impression, your gentle chiding is
unlikely to have much effect. I will make an effort to spot it to avoid
you in too much agitation. I can provide plenty of that simply by
struggling to answer your tests!

R_{ab} =

[(E + 3P),0            ,0            ,0            ] . (1/2)
[0            ,(E - 3P),0            ,0            ]
[0            ,0            ,(E - 3P),0            ]
[0            ,0            ,0            ,(E - 3P ]

>
>3. In the big bang model, what happens to the Ricci tensor as you go back in
>past all the way to the moment of creation? 
>
>So, all you need is to tell me what happens to E and P.  
>
>>d^2V/dt^2 = -V R_{ab} v^a v^b 
>
>>1) v=(1,0,0,0) Or how the volume varies in the t-direction?
>
>>d^2V/dt^2 = -V R_{ab} v^a v^b = -(1/2)V(E + 3P)
>
>Yes, this is the most interesting case.  If you keep with it, by the
>way, you can do a lot more than I hoped for... you can figure out pretty
>much everything about the big bang solution!  But let's see, I'm not
>sure this stuff about the ball of particles will help you figure out
>what happens to R_{ab} as you approach the initial singularity.  (It
>will help you understand what it *means* though.)  
>
>>I assume that 3P is in fact negative. 
>
>Why should the pressure be negative?  Pressure is usually positive.
>
>>In other words it resists the contraction of the ball.
>
>Hmm.  Sounds fishy to me.

Eh? Here you are looking at a ball of hot gas. It's a mini-universe. If
3P is positive then R_{00} is always positive and the ball will always
contract. Oh-ho well, perhaps ...., after all our universe is expanding
but at a decreasing rate. Well, I'm still not convinced. Lets take an
almost massless ball of hot gas stuck all alone in minkowski space. Hey
looky here, it expands, presto! Oz gets a large cannon and blows his
head clean off. OK, you are right. The CURVATURE is positive. That's
what I had in mind way back when I first partially answered way up the
thread. I said that R_{00} was always positive so would go to infinity.
Forget the coffee grains. Let's stick to curvature.

R_{00} = (1/2)(E + 3P) Both E & P must -> infinity as size->0

So the curvature sure tends to infinity in the time direction.

>
>>Now of course we haven't considered dV/dt. 

Yeah. The coffee grounds are not an appropriate analogy.

>
>>Then there is the problem of blackholiness. 
>
>Aren't we talking about the big bang in this problem?  Have you shifted
>over to considering a black hole? I'm confused.

I was trying to pick out how the BB differs from the BH.
This way I avoid making too many stupidities.

>>We are also (at least I am at the moment) setting the Weyl tensor at
>>zero. 
>
>Oh yeah?  Why are you doing that?  I'm not saying it's bad, but explain
>why you're doing that?  (Hint: isotropy, isotropy, isotropy.)

Eh? I thought I gave a quasi-plausible reason way way back at the
beginning of the thread. My argument went as follows:

a) There ain't no boundary conditions since we include the entire
universe. From what has been said about the Weyl tensor, some of it's
terms relate to boundary conditions. Maybe tidal effects 'n stuff. These
gotta be zero.

b) The universe is uniformly isotroph^Hic so there ain't no waves or
other transient stuff wandering about. This should take out another
whole load of terms from the Weyl.

c) So far we have only considered a teeny point-like bit of this
universe. I have a vague memory that we have, at some point, made use of
the approximation that in a teeny volume, spacetime is minkowskian.
However I can't recall exactly where. On the other hand the R_{ab} we
have derived is far from flat so hopefully this is a valid approximation
in the limit. 

I am a mite concerned that the Weyl does include static curvature from
distant bits. Hopefully in this situation these end up zero too.
<nervous grin>

2) In the spacial directions.

R_{11} = (1/2)(E - 3P) Now hang on, what IS *R*_{11}?

R_{ab} = R^c_{acb} aaaah, this is the key. 

We have calculated a subset of the Riemann tensor, and if all the other
elements (if the Weyl elements =0) are zero then we have calculated the
Riemann tensor itself. I think I'll go and lie down.

The problem now is to get a proper handle on what curvature 'feels'
like. I mean actual results. Now a curvature of 1 ought to mean that we
end up with ordinary flat pythagorean space stuff, I guess. No that's
too general a statement, all those cross terms. Let's call on Riemann
himself, perhaps.

So if we input a trio of vectors into the Riemann we have derived.

Well, let's try v = (1,0,0,0) = a = b = c.

This should output a vector in the c-direction giving the curvature in
that direction. Oooooooo dear.

Perhaps I'll post later on this.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Tue Mar  5 17:05:03 PST 1996
Article: 103495 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Friedmann Models
Date: 4 Mar 1996 15:56:34 -0800
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, "Emory F. Bunn"
><[email-masked]> writes

>>>To take another case of a mass in
>>>otherwise flat spacetime 

>>I haven't even gotten to the end of your sentence yet and already my
>>soul is troubled.  The spacetime surrounding a mass is not flat.
>>Maybe it's "otherwise flat," but before I can tell you whether it is
>>or not you have to tell me what that phrase means.

>Yes, I see the problem. What I am perhaps trying to grasp is how one
>might go about making use of what (little) I have picked up about GR.
>Now (probably to keep it simple) John keeps dropping back to a minkowski
>metric for the trivial example tests (that I struggle to do, but
>still).

No.  I keep on saying you can always find coordinates in which the
metric to have the Minkowski form *at a single point*, but this only
useful for certain special calculations.  For understanding how a
spactime looks, it's useless... because the shape of the spacetime
depends not just at the metric at a single point, but all over.  

To repeat: no matter how wiggly and bumpy the metric, if you pick a single
point, we can arrange the coordinates so that the metric looks like

-1 0 0 0 
 0 1 0 0 
 0 0 1 0 
 0 0 0 1

*at that particular point*.  It won't look like it right near that
point!  If it did, the spacetime would not be curved there: it would be
the same as flat old Minkowski space.

Pass those tests of valor; then you will learn how to compute the
Riemann curvature, and then your knowledge of GR will become practical.
Right now you don't know enough to do problems by thoughtless
calculation; you need to be very careful at every step, never
overstepping what's allowed.  Once you know how to calculate everything,
you will no longer need to think.  Well, not nearly as much.

From galaxy.ucr.edu!library.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!gatech!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Tue Mar  5 17:28:42 PST 1996
Article: 103603 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Mon, 4 Mar 1996 21:34:38 +0000
Organization: Oz
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In article <[email-masked]>, Oz
<[email-masked]> writes
>
>R_{ab} =
>
>[(E + 3P),0            ,0            ,0            ] . (1/2)
>[0            ,(E - 3P),0            ,0            ]
>[0            ,0            ,(E - 3P),0            ]
>[0            ,0            ,0            ,(E - 3P ]
>
>R_{ab} = R^c_{acb} aaaah, this is the key. 
>
>We have calculated a subset of the Riemann tensor, and if all the other
>elements (if the Weyl elements =0) are zero then we have calculated the
>Riemann tensor itself. I think I'll go and lie down.

Poor old Oz is depressed. He fears that the Wiz is going to be terribly,
terribly cross with him. Her needs that most exquisitly boring of
subjects, an elementary Tensor revision course. <abject embarrasement>

The Wiz, rightly, tends to gloss over details. Oz also has a poor memory
at times and it takes a little while for things to sink in.

Basically Oz has been looking at ways to find which elements of
R^a_{bcd} correspond to his 16 elements of R_{ab}. He realises that he
is not as sure as he should be about some basic tensor stuff. He might
have come a fair way, but his knowledge is still appallingly elementary.
Basically he does it mechanically, and not very well at that.

He hopes that Wiz will find the time to answer a few basic questions, or
at least some other wandering Wiz can deign to.

Q1) Example T^c_c = g^(ca)T_{ac} = scalar

Oz feels that there should also be a product that is another tensor. He
also wonders if the reversal of the sub and superscript order is
important. He also wonders if g_{ca}T^(ac) is different, and why? He
suspects that g_{ab}T_{cd} would be a tensor of form U_{abcd}.

Q2) R^a_{bcd} This takes in three vectors and outputs one.

He thinks this could be evaluated as something that takes in three
vectors u^b,v^c,w^d to output a vector q^a like this:

q^n=sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3)<R(n,i,j,k).u^i.v^j.w^k>]}

Is this right (if you can work out what I actually said!)?

In other words R^a_{bcd} is a 4x4x4x4 matrix of 256 elements.

Q3) Now when we raise an index how does that alter the terms in the
array? I suspect it's not trivial.

Q4) Now I suspect that to get from R_{ab} to a form like R^a_{bcd) I
would need to blend a g_{cd} with R_{ab} to get a R_{abcd} then raise an
index to get R^a_{bcd}. I think this will be very, very messy. Also
g_{ab} is which metric. Can it be simply the minkowskian one, despite
the evidence that the metric is likely to be non-minkowskian?

Q5) I also suspect that this brute force way of doing it is silly. All
we need to know is in R_{ab}, and a diagonalised form at that. How
should I proceed trying to see what this means?

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Tue Mar  5 20:27:08 PST 1996
Article: 103708 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 15:58:57 -0800
Organization: University of California, Riverside
Lines: 60
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>But this is merely speculation, and your argument is merely
>>quasi-plausible.  Why don't you concoct a 100% rigorous argument [for
>>why the Weyl curvature vanishes in the big bang cosmology] based
>>on what we know about the Weyl curvature?  Remember: the Ricci
>>curvature tells how your little ball of freely falling coffee grounds
>>changes volume, while the Weyl curvature tells how it changes shape into
>>an ellipsoid.

>Aw, you gave it away. Ok, ok it's a better way of saying it than
>'boundary conditions' and stuff. The universe couldn't be or stay
>uniformly isotroph^hic if Weyl was anything but zero. Almost by
>definition, well probably actually by definition.

Sorry, I should have given you a sneakier hint and let you have more of
the fun of figuring it out.  Yes, if a little sphere of initially
comoving particles in free fall turned into an ellipsoid, there would
necessarily be some preferred directions (the axes of the ellipsoid),
hence anisotropy.  

So you have worked out the Weyl curvature of the big bang universe and
also the Ricci curvature --- given the energy density E and the pressure
P.  No heavy-duty calculations, just symmetry!  Cool, huh?

Now, if you read that nasty lecture from G. Wiz on the formula for the
Riemann tensor in terms of the Ricci and the Weyl, you will know how to
calculate the Riemann tensor for the big bang universe as a
function of E and P.  These in turn depend on time, but not on position
in space, since things are homogeneous.  Figuring out exactly *what*
functions of time they are requires one to make some assumptions as to
what the universe is made of... freely falling dust (no pressure),
radiation, or whatever.  

I would say you're almost done with questions 2 and 3:

2.  Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may is
determined at each point.  

3.  In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation?

I still would like to hear a bit more on what happens to the components
R_{ii} of the Ricci as we approach the moment of creation.  (Note: it's
standard to use letters like i,j,k to range from 1 to 3.  You told me
R_{00} went to infinity, so I'm asking about the other diagonal terms.)

But that's mainly it.  You have come so close to completely
working out the big bang model, that when you answer question number 1,
I will reward you by finishing it up.  By the way, I'm tremendously
enjoying this.  I never really understood the big bang model as well as
I do now... I'll have to tell you what I've realized, after you answer
question number 1.

From galaxy.ucr.edu!not-for-mail Tue Mar  5 20:28:02 PST 1996
Article: 103713 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 16:37:50 -0800
Organization: University of California, Riverside
Lines: 85
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[Starting with some stuff Oz wrote, and seguing seamlessly to stuff of
mine...] 

Wiz came over and sat down. "So, how's it all going my noble men?" he
asked, to Ed and Oz's astonishment. "I feel like some idle chatter, my
work is going well, and that Wizard over the water hasn't got it quite
right yet."

"Wellll," said Oz, "it would be quite nice just to talk over the subject
a bit. You know, philosophise a bit about life, the universe and general
relativity."

Ed dived for cover.  He had experience of the wizard, mostly electrifying.

"Calm down there, Ed!" laughed the wizard.  "I'm really not as bad as
you guys think.  True, I do resort to some seemingly severe measures at
times to further your education, but think: both of you are still alive
and well, safe and sound, and growing more knowledgeable every day...
and by the way, you have most delightful cave here!  What more could you
ask?"   

"Well," said Oz (ignoring Ed, who was peeking around a boulder, putting
his finger to his lips in a desperate attempt to keep Oz from saying
something that would send the wizard into a rage), "a little heat now
and then on winter nights would be nice!" 

"Heat?" chuckled the wizard.  "What do you need heat for?  It's just
disorganized energy... only keeps one from thinking clearly.  There's
really nothing like a nice chilly winter night to focus ones attention
on important matters, like theoretical physics!  Why, when I was a lad,
we never had any heat in our caves, it would have been unthinkable.
People are getting soft these days.  Never fear, you'll have all the
heat you want, come summer."

"True," said Ed, cautiously emerging from behind the boulder.  
G. Wiz seemed indeed to be in a good mood.  Perhaps if he humored him
everything would be okay.  

"So, you wanted to talk things over, eh?" the wizard asked Oz.  "You've
been making good progress on the last two questions... I really hadn't
expected you to go into nearly so much detail... of course you still
need to figure out that black hole question... but on the whole I'm
quite pleased.  You're beginning to absorb the basic principles: that
gravity in general relativity is not a force, that all you need to know
is how energy-momentum affects the geometry of spacetime, and then to
see what a particle does in free fall, you just calculate the geodesic
it follows..."

"Umm, which we still haven't learned to do," added Oz.  Ed looked
distressed, this being a touchy subject.

"NOT UNTIL YOU PASS THE TEST!" thundered the wizard.  Ed jumped off the
log he was sitting on and fell over onto his back, panting with terror,
and the wizard exploded in laughter.  "Yes, indeed, the main thing you
are missing is how to compute parallel transport, or the so-called
"connection", given the metric.  Once I tell you this --- assuming you
pass the test --- you will know how to work out geodesics given the
metric, since a geodesic is just a curve whose velocity is parallel
transported along itself.  And you will be able to compute the Riemann
tensor, starting from its definition in terms of parallel translation
around a wee parallelogram.  And so you won't need to go skulking around
and peeking at my books!"  He winked and nudged Oz.  

"But even before you know how to calculate all these things, I hope the
principles are becoming clear.  They're very beautiful, are they not?"
asked the wizard, staring abstractedly at the dripping icicles.  "It's
all very simple, in the end: the geometry of spacetime was always known
to affect the motion of matter... without spacetime geometry, how would matter
know which way to go?  But here we are seeing a wonderful instance of
the principle that "if A affects B, B must affect A."  The motion of
matter in turn affects the geometry of spacetime!   And it's all wound
up into one equation..."  He scratched it on the dirt of the cave floor,

                            G_{ab} = T_{ab}.

"Spacetime geometry on the left, the energy and momentum of matter on
the right... they turn out to be exactly the same, or two different ways
of talking about the same thing.  Well, I'm getting a bit carried away
here," he added.  "But you should remind me to explain sometime how
local conservation of energy and momentum follows from this equation!
After you've passed that test and learned a bit more math."

From galaxy.ucr.edu!not-for-mail Tue Mar  5 20:30:52 PST 1996
Article: 103714 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 17:04:33 -0800
Organization: University of California, Riverside
Lines: 91
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] writes:

>Anyway, we can begin to see why it's a rate of change of volume term. If
>we define a small volume as being of radius (or side of a cube if you
>like) epsilon. Then the volume is epsilon^3. Basis vectors u,v,w
>spacially, and t temporally.

>Now let's evolve this little system dt along the time axis. Compared to
>the center of our little sphere, in a distance dt then v has evolved
>into a new vector whose incremental change in the v direction is epsilon
>dt R(v,t,t) and similarly for u and w. So ignoring a few higher order
>terms the new volume after time dt is (e=epsilon) is V' =

>[e+ e dt R(w,t,t)][e+ e dt R(v,t,t)][e+ e dt R(w,t,t)]

>so dV = V - V'= e^3 - (that lot above) in time dt

>or dV/dt = e^3 dt[ R(u,t,t)+ R(v,t,t) + R(w,t,t)]

>so d^2V/dt^2 = V [ R(u,t,t)+ R(v,t,t) + R(w,t,t)]

>Well, I think I've fiddled this a bit but it kinda looks plausible.

>Er, doesn't it??  Ahem, cough cough.

You've got the basic idea, there are just a few mistakes here and there.
You want to start with an orthonormal basis of spacelike vectors u, v, w.
Consider one of them, say u.  Then you want to look at this picture

   P'->----Q'
   |       |
   |       |
  t^       ^
   | u     |
   P->-----Q

where following your notation I'm using t for a unit timelike vector,
the velocity of the particle at P.  NOTA BENE: this is what I had been
calling *v*, since it's a *velocity*!  This terminological confusion can
get us confused if we let it.  I'll switch to your notation, but below
you actually get a bit tangled up.

So P and Q are two initially comoving grounds of coffee separated by u;
we want to work out their relative acceleration, so we use the geodesic
equation I told you last time and get

            R(t,u,t)

I think you got the order wrong when you wrote down R(u,t,t).  

Okay, but now we want to see how the ball of coffee grounds expands or
contracts; that is, we want to know d^2V/dt^2.  So first we want to see how
much the acceleration R(t,u,t) is *pointing in the u direction*, to see
how Q starts accelerating *away from P*.  
That's

         g(u,R(t,u,t)),

the "dot product" of R(t,u,t) with u.  Then we want to do this for the
other directions, v and w.  Then we add them up, and get

g(u,R(t,u,t))+ g(v,R(t,v,t)) + g(v,R(t,v,t)).

This should be proportional to d^2V/dt^2.

>So now all I have to do is show that 
>R(u,t,t)+ R(v,t,t) + R(w,t,t) is equivalent to R^a_{bcd} v^a w^c v^d
>At which point I roll over and play dead. Oh, I don't know, if I squint
>my eyes enough .........
>This isn't kosher. I think it sort of signposts the way to go though.
>Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp!!

Okay, let me say what it is you REALLY want to show, after making the 
corrections above.  You want to show that

g(u,R(t,u,t))+ g(v,R(t,v,t)) + g(v,R(t,v,t)) 

is the same as

R_{ab} t^a t^b, 

which by definition is just

R^c_{acb} t^a t^b

Remember, now following your notation we are writing t for the velocity
vector of the the coffee ground at P.  

This looks like it'll work.  Why don't you try it... after making sure
you understand all the corrections I made.

From galaxy.ucr.edu!not-for-mail Tue Mar  5 20:31:00 PST 1996
Article: 103715 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 17:28:32 -0800
Organization: University of California, Riverside
Lines: 120
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, Oz
><[email-masked]> writes

>Poor old Oz is depressed. He fears that the Wiz is going to be terribly,
>terribly cross with him. Her needs that most exquisitly boring of
>subjects, an elementary Tensor revision course. <abject embarrasement>

Boring, boring, boring.  Next we'll be reviewing partial differential
equations and Fourier transforms.  

>Q1) Example T^c_c = g^(ca)T_{ac} = scalar
>
>Oz feels that there should also be a product that is another tensor. 

A product of g and T that's a tensor?  Sure, like g_{ab}T_{cd}.
That's rank (0,4): eats four vectors for breakfast and spits out a
scalar at noon.

By the way, why are you using a mix of {}'s and ()'s?  There's no point
in doing that here, and it could even be confusing if I didn't know that
you couldn't possibly know what () meant in this context.  We just use
^{}'s and _{}'s as a substitute for writing batches of superscripts or
subscripts, respectively.

>He
>also wonders if the reversal of the sub and superscript order is
>important. He also wonders if g_{ca}T^(ac) is different, and why? 

Well, the order is very important in general, but the metric and the
stress energy just HAPPEN to be symmetric: g_{ab} = g_{ba}, and T_{ab} =
T_{ba}.  Figure out why.

Show that g_{ac} T^{ac} = g^{ac} T_{ac}, using the fact that g^{ab} is
the inverse matrix of g_{ab}, and maybe some other facts.

>He
>suspects that g_{ab}T_{cd} would be a tensor of form U_{abcd}.

Yup.

>Q2) R^a_{bcd} This takes in three vectors and outputs one.

>He thinks this could be evaluated as something that takes in three
>vectors u^b,v^c,w^d to output a vector q^a like this:

>q^n=sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3)<R(n,i,j,k).u^i.v^j.w^k>]}

>Is this right (if you can work out what I actually said!)?

Almost; you mean R^n_{ijk} where you wrote R(n,i,j,k).   Also, the Einstein
summation convention is designed to prevent such an obscene proliferation
of summation signs.  The quick way to write what you wrote is:

q^n = R^n_{ijk} u^i v^j w^k

>In other words R^a_{bcd} is a 4x4x4x4 matrix of 256 elements.

Yes, but thankfully, it possesses enough symmetries to reduce down to
only 20 independent elements.  Exercise: show from the the definition
that R^a_{bcd} = -R^a_{cbd}.  You know, that "skew-symmetric in the
first two slots" businesss.

>Q3) Now when we raise an index how does that alter the terms in the
>array? I suspect it's not trivial.

Well, index raising is done using the metric and the Einstein summation
convention, so e.g. if we want to raise the first index on X_{abcd} we
do this

X^a_{bcd} = g^{an} X_{nbcd}

where we sum over n.  So yes, this can alter the terms in a severe way.

>Q4) Now I suspect that to get from R_{ab} to a form like R^a_{bcd) I
>would need to blend a g_{cd} with R_{ab} to get a R_{abcd} then raise an
>index to get R^a_{bcd}. 

That's right.  

>I think this will be very, very messy.

Nah, it's easy.  Maybe my message didn't get across the Atlantic yet,
but the formula for the Riemann in terms of the Weyl and the Ricci is

R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}

You can also think of this the way you suggest.  Starting with

R_{abcd} = (1/4) R_{bd} g_{ac} + W_{abcd}

we can raise the index a and get

R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}

using the fact that g^a_c = delta^a_c.

>Also g_{ab} is which metric. 

It's whatever the hell metric is the actual metric on spacetime that you
happen to be studying!!!!!!!!

Crucial, crucial point.  Remember, the metric describes the geometry of
spacetime.  All sorts of things depend on the metric.  You gotta use the
metric you actually are studying, you can't just pull one out of the hat.

>Can it be simply the minkowskian one, despite
>the evidence that the metric is likely to be non-minkowskian?

No way.  That'd be like randomly picking an electric field and
stuffing it into the problem you're solving, instead of using the
physically correct one.

>Q5) I also suspect that this brute force way of doing it is silly. All
>we need to know is in R_{ab}, and a diagonalised form at that. How
>should I proceed trying to see what this means?

It depends on what you're doing.  

From galaxy.ucr.edu!not-for-mail Wed Mar  6 09:38:34 PST 1996
Article: 103715 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 17:28:32 -0800
Organization: University of California, Riverside
Lines: 120
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, Oz
><[email-masked]> writes

>Poor old Oz is depressed. He fears that the Wiz is going to be terribly,
>terribly cross with him. Her needs that most exquisitly boring of
>subjects, an elementary Tensor revision course. <abject embarrasement>

Boring, boring, boring.  Next we'll be reviewing partial differential
equations and Fourier transforms.  

>Q1) Example T^c_c = g^(ca)T_{ac} = scalar
>
>Oz feels that there should also be a product that is another tensor. 

A product of g and T that's a tensor?  Sure, like g_{ab}T_{cd}.
That's rank (0,4): eats four vectors for breakfast and spits out a
scalar at noon.

By the way, why are you using a mix of {}'s and ()'s?  There's no point
in doing that here, and it could even be confusing if I didn't know that
you couldn't possibly know what () meant in this context.  We just use
^{}'s and _{}'s as a substitute for writing batches of superscripts or
subscripts, respectively.

>He
>also wonders if the reversal of the sub and superscript order is
>important. He also wonders if g_{ca}T^(ac) is different, and why? 

Well, the order is very important in general, but the metric and the
stress energy just HAPPEN to be symmetric: g_{ab} = g_{ba}, and T_{ab} =
T_{ba}.  Figure out why.

Show that g_{ac} T^{ac} = g^{ac} T_{ac}, using the fact that g^{ab} is
the inverse matrix of g_{ab}, and maybe some other facts.

>He
>suspects that g_{ab}T_{cd} would be a tensor of form U_{abcd}.

Yup.

>Q2) R^a_{bcd} This takes in three vectors and outputs one.

>He thinks this could be evaluated as something that takes in three
>vectors u^b,v^c,w^d to output a vector q^a like this:

>q^n=sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3)<R(n,i,j,k).u^i.v^j.w^k>]}

>Is this right (if you can work out what I actually said!)?

Almost; you mean R^n_{ijk} where you wrote R(n,i,j,k).   Also, the Einstein
summation convention is designed to prevent such an obscene proliferation
of summation signs.  The quick way to write what you wrote is:

q^n = R^n_{ijk} u^i v^j w^k

>In other words R^a_{bcd} is a 4x4x4x4 matrix of 256 elements.

Yes, but thankfully, it possesses enough symmetries to reduce down to
only 20 independent elements.  Exercise: show from the the definition
that R^a_{bcd} = -R^a_{cbd}.  You know, that "skew-symmetric in the
first two slots" businesss.

>Q3) Now when we raise an index how does that alter the terms in the
>array? I suspect it's not trivial.

Well, index raising is done using the metric and the Einstein summation
convention, so e.g. if we want to raise the first index on X_{abcd} we
do this

X^a_{bcd} = g^{an} X_{nbcd}

where we sum over n.  So yes, this can alter the terms in a severe way.

>Q4) Now I suspect that to get from R_{ab} to a form like R^a_{bcd) I
>would need to blend a g_{cd} with R_{ab} to get a R_{abcd} then raise an
>index to get R^a_{bcd}. 

That's right.  

>I think this will be very, very messy.

Nah, it's easy.  Maybe my message didn't get across the Atlantic yet,
but the formula for the Riemann in terms of the Weyl and the Ricci is

R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}

You can also think of this the way you suggest.  Starting with

R_{abcd} = (1/4) R_{bd} g_{ac} + W_{abcd}

we can raise the index a and get

R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}

using the fact that g^a_c = delta^a_c.

>Also g_{ab} is which metric. 

It's whatever the hell metric is the actual metric on spacetime that you
happen to be studying!!!!!!!!

Crucial, crucial point.  Remember, the metric describes the geometry of
spacetime.  All sorts of things depend on the metric.  You gotta use the
metric you actually are studying, you can't just pull one out of the hat.

>Can it be simply the minkowskian one, despite
>the evidence that the metric is likely to be non-minkowskian?

No way.  That'd be like randomly picking an electric field and
stuffing it into the problem you're solving, instead of using the
physically correct one.

>Q5) I also suspect that this brute force way of doing it is silly. All
>we need to know is in R_{ab}, and a diagonalised form at that. How
>should I proceed trying to see what this means?

It depends on what you're doing.  

From galaxy.ucr.edu!ihnp4.ucsd.edu!swrinde!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Wed Mar  6 15:58:40 PST 1996
Article: 103826 of sci.physics
Path: galaxy.ucr.edu!ihnp4.ucsd.edu!swrinde!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz
From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Wed, 6 Mar 1996 15:44:11 +0000
Organization: Oz
Lines: 52
Distribution: world
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 <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>You've got the basic idea, there are just a few mistakes here and there.
>You want to start with an orthonormal basis of spacelike vectors u, v, w.
>Consider one of them, say u.  Then you want to look at this picture
>
>   P'->----Q'
>   |       |
>   |       |
>  t^       ^
>   | u     |
>   P->-----Q
>
>where following your notation I'm using t for a unit timelike vector,
>the velocity of the particle at P.  NOTA BENE: this is what I had been
>calling *v*, since it's a *velocity*! 
              ^^^^^^^^^^^^^^^^^^^^^^

> This terminological confusion can
>get us confused if we let it.

Too true. However it has confused me mightily, and this has offered up a
philosophical question that bears discussion.

I am seeing what is effect a static volume of spacetime by dropping one
spacial dimension and fitting in a time dimension in it's place.

So what I have marked vector t in the time dimension is a *distance*.
How do I know how long it is? Well, it's ct long in P's restframe, or I
suppose P's geodesic. Of course if you choose units where c=1, then it's
t long. I don't really have velocities as such in this universe, I have
dx/dt's. Really there isn't a flow of time, just a path. I am seeing it
exactly the same way as I see an x-y graph on paper. I can read off the
x-position and get back a y-position. I can get the slope, it's dy/dx,
and move about at will. So whilst velocity has a meaning, it's not got
the uniquness that time has in calculus. It's just an ordinary axis
(well fairly ordinary) and I can read off distances with my ruler.

Of course it lives in a rather odd space where transformation of co-
ordinates are not linear in the sense that if I transform to a new
origin that is moving wrt another the transformation is non-linear. On
top of that the co-ordinates may not be equally spaced and may curve,
but no bother, I just get out my bent ruler and measure stuff off.

Now what could be easier, so where am I going wrong?

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!europa.chnt.gtegsc.com!news.kreonet.re.kr!news.nuri.net!imci2!imci3!imci4!newsfeed.internetmci.com!newsxfer2.itd.umich.edu!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Wed Mar  6 19:19:44 PST 1996
Article: 103842 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!europa.chnt.gtegsc.com!news.kreonet.re.kr!news.nuri.net!imci2!imci3!imci4!newsfeed.internetmci.com!newsxfer2.itd.umich.edu!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz
From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Wed, 6 Mar 1996 08:08:30 +0000
Organization: Oz
Lines: 39
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]>
 <[email-masked]> <[email-masked]>
 <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>"So, you wanted to talk things over, eh?" the wizard asked Oz.  "You've
>been making good progress on the last two questions... I really hadn't
>expected you to go into nearly so much detail...

Well. We could have waffled usefully round the edges waving our hands in
an encouraging manner. This is good for the early stages. Ted did a
brilliant job on cosmology and Milne, without this paradigm change GR
would all have been very difficult. As I said at the beginning an
explanation with the *minimum* maths is required. I think I am somewhat
below the bare minimum required, but there is this time limitation. By
the end I hope to be vaguely able to follow an argument. To express it
with no maths would be a great shame and hey if Newton needed calculus
and Einstein needs tensors, then we gotta know a teeny weeny bit of
tensor stuff. Sort of 'you do the sums - we note the answers' stuff as
we have been doing. Personally I find tensors a real neat set of tricks,
even if I suspect that actually getting real answers out might be a real
serious pig of a job. Luckily we leave that for the experts to while
their time away doing, it's what they are paid for, after all. :-)

> of course you still
>need to figure out that black hole question... but on the whole I'm
>quite pleased.  You're beginning to absorb the basic principles: that
>gravity in general relativity is not a force, that all you need to know
>is how energy-momentum affects the geometry of spacetime, and then to
>see what a particle does in free fall, you just calculate the geodesic
>it follows..."

Well, Wiener indicated how it could be done, just so long as you knew
the metric and had a small Cray to work the answers out.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Wed Mar  6 19:28:26 PST 1996
Article: 103845 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 6 Mar 1996 15:58:31 -0800
Organization: University of California, Riverside
Lines: 110
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>it's a *velocity*! 
  ^^^^^^^^^^^^^^^^^^

>So what I have marked vector t in the time dimension is a *distance*.
>How do I know how long it is? Well, it's ct long in P's restframe, or I
>suppose P's geodesic. Of course if you choose units where c=1, then it's
>t long. I don't really have velocities as such in this universe, I have
>dx/dt's. Really there isn't a flow of time, just a path. I am seeing it
>exactly the same way as I see an x-y graph on paper. I can read off the
>x-position and get back a y-position. I can get the slope, it's dy/dx,
>and move about at will. So whilst velocity has a meaning, it's not got
>the uniquness that time has in calculus. It's just an ordinary axis
>(well fairly ordinary) and I can read off distances with my ruler.

You are right that I have not said much about how to compute the
velocity of a particle... or mathematically speaking, the "tangent
vector of a curve".  It *does* seem quite mysterious at first.  But of
course, you learned how to do it in special relativity class; it works
the same way here.

Heh.  I'm just trying to annoy Oz, since I *know* he snuck into this
special relativity class without the prerequisites (a good modern course
in special relativity).

I'll explain it below.

>Of course it lives in a rather odd space where transformation of co-
>ordinates are not linear in the sense that if I transform to a new
>origin that is moving wrt another the transformation is non-linear. On
>top of that the co-ordinates may not be equally spaced and may curve,
>but no bother, I just get out my bent ruler and measure stuff off.

Don't freak yourself out by worrying about change of coordinates here.
That's sort of like learning German while simultaneously worrying how to
translate every phrase into Thai, Navajo, Urdu and all other languages.  The
recipe I will give for computing the velocity will work in *any*
coordinate system.  Remember, all truly physical concepts possess a
lofty indifference to the coordinate system we mortals happen to use to
study them.  

Okay.  You are getting confused because in the good old days of Newton
we computed velocities like this:

(dx/dt,dy/dt,dz/dt)

thinking of x, y, and z as a function of time, t.

But now we realize we live in spacetime!  As Minkowski said, "Hence
space of itself, and time of itself, will sink into mere shadows, and
only a union of the two shall survive."  So t shouldn't play a dramatically
different role from x, y, and z.  

Indeed, when we compute the velocity of a curve, we should be able to do
it in ANY coordinates... so whatever recipe we come up with, if some guy
decides to use coordinates p, q, r, and s, it should work just as well.
In fact, let's use those coordinates, just to avoid any tendency to favor "t"
over "x, y and z".  Just to emphasize that it doesn't make a damn bit of
difference what coordinates we use.

Okay.  Suppose we have a particle moving through spacetime, and we want
to compute its velocity.  It traces out a curve in spacetime.  Note:
this curve has a preferred parameter, the PROPER TIME.  The proper time
is the time a watch moving along that clock would tick off.  Let's call
it T.  Folks often use the Greek letter tau.  Sometimes they even use
"t", but proper time this has nothing, nada, zilch, zip, to do with that
yucky "t" coordinate we were discussing earlier.  Well, I'm
exaggerating... there is *some* relation in *certain* circumstances.
But don't get them mixed up.  This proper time T is only defined ON THE
CURVE, and it's a very physically significant thing, since it describes
how a clock will actually tick.

Now say the clock is at the point C(T) at proper time T.  Roughly
speaking, to compute the velocity, we simply let T change a wee bit, say
epsilon, draw a little tangent vector from where C(T) started out to
where it moved, and divide this vector by epsilon.  Then take the limit
as epsilon goes to zero.

That's a coordinate-free description of the velocity of a curve.  To do
calculations it can be nice to use coordinates.  So say we are working
with local coordinates (p,q,r,s).  Then we can describe the point C(T)
by listing its coordinates, say (p(T),q(T),r(T),s(T)).  Then the
velocity vector is

                (dp(T)/dT,dq(T)/dT,dr(T)/dT,ds(T)/dT).

That's it.

But now you ask, how the heck do I actually compute things?  To compute
things I need to know how the proper time T changes along the curve!
I can't parametrize the curve using proper time unless I know how this
works!  

True.  I suppose (sigh) I will have to explain that too.  Or maybe some
kind soul could step in and explain these more practical aspects
of how one computes velocities in general relativity?  I just wanted to
start off by getting the concepts straight....

By the way, note that we have done very well so far without knowing how
to do this stuff.  It is only Oz's insane zeal to know how to actually
COMPUTE things that is making me discuss this subject.  

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Article: 103859 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 5 Mar 1996 21:17:16 -0500
Organization: The Pipeline
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References: <[email-masked]>
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Well,  Ed is back from his daily labors,  and rematerialized in the
wizard's study,  this time wearing more comfortable clothing,  and having
bypassed the trolls.   As usually happens in these cases,   only about five
minutes has passed in wizard land (which is moving very fast relative to
mundania).  The wizard has left,  and Oz is musing... 
 
<[email-masked]>' wrote: 
 
>Basically Oz has been looking at ways to find which elements of 
>R^a_{bcd} correspond to his 16 elements of R_{ab}. 
 
Well,  it's not so much a what, as a whom...  Huh?   Delete this line. 
 
They all do... and we have lost information in "contracting" the Riemann to
make the Ricci.  The simplest kind of "contraction" we know is the
Euclidean metric.  In R^3 
 
              |a|^2 == a.a = a_1 a_1 + a_2 a_2 + a_3 a_3 
 
Now,  you might as well ask "which component of a corresponds to |a|^2". 
 
Ok,  so recall the definition of the Ricci: 
                
                        R_{bd} = R^c_{bcd} 
 
We have summed c from 0 to 4 here,  by the Einstein summation convention
for repeated indices.  By this convention,  we could have written above: 
 
                          |a|^2 = a_i a_i        (here i = 1,2,3) 
 
Now I realize I have made a slight mistake,  but I will let it stand above,
 because we are just chatting,  and I'm not trying to write a text book!  
:-)  
 
When I said the dot product was the simplest example of a contraction,  I
may have misspoke.  It doesn't look exactly analogous to the formation of
the Ricci,  does it?   There is no multiplication of elements in forming
the Ricci,  just summation.  So it actually looks more like the *trace* of
a matrix. 
Now let A = {A_ij} be a three by three matrix.  Then we define 
 
                       Tr A == A_ii  = a_1 + a_2 + a_3 
 
Okay?  (I am using "==" to mean "is defined as") 
 
So this looks a lot more like the type of thing that we do in making the
Ricci.  We take a "trace" on *two* of the four indices of the Riemann. 
It's as if we sliced up the Riemann into 16 matrices labeled by the indices
b and d,   each matrix then having 16 components labeled by a and c,  and
then took the trace of each of these matrices:  R_{bd} = R^c_{bcd} 
 
(Psst -- the wizard would not like this,  too *coordinate-bound*,  nudge, 
nudge,  wink). 
 
Hey...  I misspoke again!  How about that.   I said the Ricci involved
*all* of the components of the Riemann,  like the dot product eats all the
components of a vector.  But it doesn't;   only 4 of the 16 components of
each of these submatrices get to play in the trace...   But... I don't
think we are throwing out quite as much information as we might think, 
because of the symmetries of the Riemann.  This bears closer study.   Not
now!   
 
Let me think about your very excellent questions in another place (I know, 
I know,  you wanted a real *wizard* to answer them... still,  maybe I can
make the odd connection...) 
 
-erg

From galaxy.ucr.edu!not-for-mail Wed Mar  6 21:20:04 PST 1996
Article: 103878 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 6 Mar 1996 20:04:46 -0800
Organization: University of California, Riverside
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References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward Green) writes:

>Now,  I am perfectly sure T^c_c = T_c^c ,  but right now I can only offer
>my "proof by it would just be too weird otherwise".     

See, erg has good mathematical intuition; he knows when something would be
too weird to be true.  But he's wrong about it not being able to prove
it's true.  Earlier in the post I'm quoting he wrote both:

>                          T_c^c = g^{ca}T_{ac}

and 

>                          T^c_c = g^{ca}T_{ac} 

Since both of these are true, we definitely have T^c_c = T_c^c!

Of course T_{ab} is symmetric, i.e. T_{ab} = T_{ba}.  This allows us to
get away with certain sins that would otherwise be mortal.  But let's
show that for any old tensor X_{ab} we have X^c_c = X_c^c.  Here we will
be very careful about orders of things.  We will only use the symmetry
of the metric: g_{ab} = g_{ba}.

Okay.  Time for index gymnastics!  Stand upright with arms loosely at
your sides... do a few neck rolls, and then, note:

X_c^c = g^{cd} X_{cd} 

while on the other hand

X^c_c = g^{cd} X_{dc} 

But on the right side of the second equation we can switch the "dummy
indices" c and d --- dummies because they are summed over and are purely
arbitrary names --- and get

X^c_c = g^{dc} X_{cd}

So then we get

X^c_c = g^{cd} X_{cd}

using the symmetry of the metric.

>I think this is a rule of thumb:   Almost all tensors in GR are symmetric
>or antisymmetric,  so we can always sloppily sway indices,  and at worst
>introduce a stray minus sign,  which can be fudged out later.  Well, 
>maybe. 

That rule has a certain truth to it, though of course if you don't
happen to already know the right answer, it can be important to be able
to get the right sign.  Someday, if you folks get really interested in
the inner workings of curvature, I'll go over the symmetries of the
Riemann tensor.  If we define

R_{abcd} = g_{ax} R^x_{bcd}

(note, this is a controversial convention and perhaps a bad one, but
it's the one I used in my book), we have

R_{abcd} = -R_{dbca}

and

R_{abcd} = -R_{acbd},

and

R_{abcd} = R_{cdab}.

I don't think these are all the symmetries (I know others which I don't
think are consequences of these), but they suffice to prove "Green's
Theorem", namely that if you switch any two indices of the Riemann
tensor when all indices are down, at most you get a sign error.

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Article: 103842 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Wed, 6 Mar 1996 08:08:30 +0000
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>"So, you wanted to talk things over, eh?" the wizard asked Oz.  "You've
>been making good progress on the last two questions... I really hadn't
>expected you to go into nearly so much detail...

Well. We could have waffled usefully round the edges waving our hands in
an encouraging manner. This is good for the early stages. Ted did a
brilliant job on cosmology and Milne, without this paradigm change GR
would all have been very difficult. As I said at the beginning an
explanation with the *minimum* maths is required. I think I am somewhat
below the bare minimum required, but there is this time limitation. By
the end I hope to be vaguely able to follow an argument. To express it
with no maths would be a great shame and hey if Newton needed calculus
and Einstein needs tensors, then we gotta know a teeny weeny bit of
tensor stuff. Sort of 'you do the sums - we note the answers' stuff as
we have been doing. Personally I find tensors a real neat set of tricks,
even if I suspect that actually getting real answers out might be a real
serious pig of a job. Luckily we leave that for the experts to while
their time away doing, it's what they are paid for, after all. :-)

> of course you still
>need to figure out that black hole question... but on the whole I'm
>quite pleased.  You're beginning to absorb the basic principles: that
>gravity in general relativity is not a force, that all you need to know
>is how energy-momentum affects the geometry of spacetime, and then to
>see what a particle does in free fall, you just calculate the geodesic
>it follows..."

Well, Wiener indicated how it could be done, just so long as you knew
the metric and had a small Cray to work the answers out.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Thu Mar  7 11:29:43 PST 1996
Article: 104027 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 7 Mar 1996 09:53:08 -0800
Organization: University of California, Riverside
Lines: 35
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] (Jan Bielawski) writes:

>Ouch, this notation is the other way around:  #: V* -> V  and
>b: V -> V*  (as the names "raising" and "lowering" (of the coordinate
>indices) suggests).

Yes, you're right, the standard notation goes as you say.  If we have
a vector field we typically write its components as v^a, with the
indices up, and if we have a 1-form we typically write its components as
w_a, with the indices down.  Thus we can think of the map from V
(vectors) to V* (covectors) as "lowering indices".

I think I get everything upside down and backwards because I have truly
accepted the rotational invariance of the laws of physics.  Seriously, a
rich source of confusion is the fact that physicists sometimes call
vectors "contravariant" and 1-forms "covariant", while mathematicians
now call vectors "covariant" and 1-forms "contravariant".  The reason is
that a typical vector field is something like

v = v_a partial^a 

where partial^a are the basis vector fields, while a typical 1-form is
something like 

w = w^a dx_a

where dx_a are the basis 1-forms.  Whether you think the indices are up
or down depends on whether you focus on the components (like v_a) or the
basis vector fields (partial^a).  

But you're right that in the present case it really does make more sense
to use the sharp and flat notation as you suggest.  (I leave the
diligent reader to ferret out exactly why, apart from convention.)

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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 7 Mar 1996 09:11:07 -0500
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And yet more...   
 
[email-masked] (john baez)' wrote: 
 
>>[I am a little unclear on whether this "trace" (sum of the diagonal
terms) 
>>of T is supposed to be invariant or not...  every atom of my being cries 
>>out "Yes! It's invariant!  Lorentz transformations are trace
preserving!", 
>>but  I don't know if that's true or not:  The atoms of my being are long
on 
>>slogans and short on calculation.  In a specific case,  I just don't see 
>>it.]  
 
>Trust the atoms of your being!  Any scalar you can cook up by index 
>gymnastics is automatically Lorentz-invariant, which is the beauty of 
>this index notation. 
 
Cool.  Bears repeating: 
 
"Any scalar you can cook up by index gymnastics is automatically 
Lorentz-invariant, which is the beauty of this index notation." 
 
>So the trace T^a_a is invariant under Lorentz 
>transformations.  I could explain why but only mathematicians care about 
>that crap. 
 
>>Now this trace is "positive definite";  positive where we got stuff,  and

>>zero otherwise,  since all the terms are non-negative,  unless we got
some 
>>negative mass roaming about.  
> 
>Well, ahem, the pressure in the x direction is really the pressure in 
>the x direction, not the energy density.  <under consideration>  And ditto

>for y and z.  And the pressure can be negative!  That's what they call 
>"tension". 
 
Whoops.   And after all that discussion with G. Hurst about water under
tension in trees.   So tension is kind of a negative energy density < :-)
>?  I think tension is always resolvable into shear stresses.   Which may
or may not mean anything here:  Hey,  I know!  GR seems to be formulated
for matter as a mildly anisotropic fluid.  Let's go for a new,  generally
worthless, formulation,  where matter is treated as an arbitary anisotropic
solid!  Think of the indices!  Oh,  the ink!   The trees!   
 
>Secondly, ahem ahem, you seem to be ignoring some minus signs in the 
>trace T^a_a, that result from raising indices.  Not that I'm so great at 
>keeping track of minus signs either, mind you.   But recall from the 
>course outline: 
> 
>"... But we might as well work in the 
>local rest frame of the particle in the middle of the little ball, and 
>use coordinates that make things look just like Minkowski spacetime 
>right near that point.  Then 
> 
>g_{ab} = -1  0  0  0  
>0  1  0  0 
>0  0  1  0 
>0  0  0  1 
> 
>and v^a = 1 
>0 
>0  
>0 
> 
>So R_{ab} v^a v^b = R_{00}, so so in this coordinate system we can say 
>the 2nd time derivative of the volume of the little ball of test 
>particles is just -R_{00}. 
> 
>On the other hand, check out the right side of the equation: 
> 
>R_{ab} = T_{ab} - (1/2) T^c_c g_{ab} 
> 
>Take a = b = 0 and get 
> 
>R_{00} = T_{00} + (1/2) T^c_c  
> 
>Note: demanding this to be true at every point of spacetime, in every 
>local rest frame, is the same as demanding that the whole Einstein 
>equation be true!   So we just need to figure out what it MEANS! 
> 
>What's T_{00}?  It's just the energy density at the center of our little 
>ball.  How about T^c_c?   Well, remember this is just g^{ca} T_{ac}, 
>where we sum over a and c.  So --- have a go at it, tensor jocks and 
>jockettes! --- it equals -T_{00} + T_{11} + T_{22} + T_{33}.  So we get 
> 
>R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}]  
 
Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 
 
I was right.   :-)   After we get through juggling minus signs,  the
physical effect does only depend on the trace,  well,  not exactly  T^c_c, 
but something invariant that looks like the plain vanilla un-minkowskized
trace. 
Excuse me,  but I have to shout again... 
 
Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!! 

Physics can be cool sometimes.   I think I'll stop here and go play a
fugue,  and try to remember this feeling the next time I hit a wall. 

-- 
 
E. R. (Ed) Green /  [email-masked] 
  "All coordinate systems are equal, 
   but some are more equal than others".

From galaxy.ucr.edu!ihnp4.ucsd.edu!ames!uhog.mit.edu!news.kei.com!wang!uunet!in2.uu.net!pipeline!not-for-mail Sat Mar  9 15:40:57 PST 1996
Article: 104165 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 6 Mar 1996 09:15:37 -0500
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Refreshed,  and moving on to the next question... 
 
<[email-masked]>' wrote: 
 
>Q2)  R^a_{bcd} This takes in three vectors and outputs one. 
> 
>He thinks this could be evaluated as something that takes in three 
>vectors u^b,v^c,w^d to output a vector q^a like this: 
> 
>q^n=sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3)<R(n,i,j,k).u^i.v^j.w^k>]} 
> 
>Is this right (if you can work out what I actually said!)? 
> 
>In other words R^a_{bcd} is a 4x4x4x4 matrix of 256 elements. 
 
Yes.  In a particular set of coordinates.   Except Mr. Einstein,  in a
brilliant bit of work,  which is absolutely the sole reason we have ever
heard of him, and that he did not die an obscure Swiss patent clerk with a
sheaf of unpublished crackpot theories about curved space and such, 
thought  "You know,  in my crackpot theories,  we are forever taking sums
over indices going from 0 to 3.  So I think that whenever I write one of my
crackpotty equations and an index is *repeated*,  either inside one thingie
or in two adjacent thingies,  then we are supposed to *sum* over that
index!   
Therefore we can write what you wrote above (I think)  as 
 
         q^n = R^n_{ijk} u^i v^j w^k 
 
except that we seem to prefer the indices a, b, c, d,  no doubt to
distinguish this from the bad old matrix days,  when we used those *other*
conventional indicies,   and I also must repeat the wizard's caution  
(again passing by the door,  I hope,  and nodding approvingly)   against
writing things like 
"R(n,i,j,k)",  since this obscures whether these index postitions are
"covariant" or "contravariant",  which to avoid having to remember which is
which,  we simply refer to as upper and lower (or vis. versa). 
 
Anyway,  hearing of this brilliant piece of work,  Einstein's peers at the
patent office carried him out into the street on their shoulders with loud
huzzahs,   and soon the scientific establishment got wind of it,  and he
was installed as a grey eminence at Princeton.   I think that's how the
story goes,  anyway. 
 
Maybe I should have stopped at "Yes",  but it seemed kind of niggardly. 
 
>Q3) Now when we raise an index how does that alter the terms in the 
>array? I suspect it's not trivial. 
 
We have to use the metric,   though in particularly nice situations,  the
metric may be  diag ( -1, 1, 1, 1)  ...  ie,  a diagonal array (matrix)
with these elements.   So in that case the difference between a
"vectorlike"  (raised) index and a "covectorlike" (lowered) index is a
factor of -1 in (local) time. 
 
Suppose we had some arbitrary tensor   U_{a,b,c,d}  and we wanted to raise
index b,  for example  (Let's raise an index to the Queen!).   So,  our
host gets out a copy of the metric,  g^{,} -- where I have intentionally
left the names of the indices blank for now,  except to note that we
correctly have chosen the "upper/upper" version of the array --  and he
rolls it into position,   alligns the slots,  and: 
 
                      U^{e}_{a,c,d} = g^{e,b} U_{a,b,c,d} 
 
Huzzah!   To the Queen!   <much clinking of index glasses>    Now you may
wonder that I have pulled the raised index out of order,  ie,  shouldn't I
have written  "U_{a}^{e}_{c,d}   (youch!)....  well,  I think the
explanation is,  while these indices have separate meanings,  and in any
real calculation we had better keep them straight,  we just take "keeping
them straight" as read,  and never want to write something so kludgy as the
above... 
 
Another long-winded answer bites the dust. 
 
>Q4) Now I suspect that to get from R_{ab} to a form like R^a_{bcd) I 
>would need to blend a g_{cd} with R_{ab} to get a R_{abcd} then raise an 
>index to get R^a_{bcd}. I think this will be very, very messy. 
 
Oz,   it's going to be even messier if the wizard hears you talking like
this!   Shhh!   You're just saying this to make me feel good,  right? 
 
You are pretending to forget that R_{ab} is the *Ricci* tensor and
R^a_{bcd}  is the *Riemann*.  I don't think the wizard is going to find
this species of humor very funny.   It was very historically awkward of
these characters to have names beginning with the same letter,  but they
did!  Now it happens that the Ricci is directly derived from the Riemann, 
so that when we write   
        
                     R_{bd} = R^c_{bcd} 
 
it may look like we are writing a different "form" of the Riemann.  But
we're not.  We are writing down a brand new tensor.  And in particular, 
while the Riemann determines the Ricci,  you can't go backwards;  you've
lost information,   right?   Forgetting about symmetries for the moment, 
the Ricci only has 16 components,  while the Riemann has 256!  (Actually, 
I seem to recall that these 256 components boil down to a grand total of
*20* independent quantities,  but we still see it can not be recovered from
the Ricci...  which I strongly suspect has less than 16 independent
components anyway...   10 ?    Yes,  that's it.  Both the Ricci and it's
friend the Weyl have 10 independent components...   The Ricci is symmetric
and the Weyl anti-symmetric?   Better check this one out...    It *sounds*
like knowing the Ricci and the Weyl together may be sufficient to recover
the Riemann,  doesn't it?) 
 
>Also g_{ab} is which metric. Can it be simply the minkowskian one, despite

>the evidence that the metric is likely to be non-minkowskian? 
 
Well,  locally the coordinates can always be chosen to make the metric
"minkowskian" at one point.  In these coordinates I think the deviation of
the metric from Minkowskian as we move away from that point has some rather
interesting interpretation  in terms of gravitational acceleration,  or so
I remember one of the other wizards telling me.  I think this coordinate
system is a rest frame for a body locally in free-fall...  thought come to
think of it,  there could be lots of bodies in free-fall through a point of
spacetime with different relative velocities...   but that's fine,   this
means we have a bunch of local coordinate systems all related by lorentz
trasformations...  which is what we usually have in "flat" Minkowski space.

So the answer to this question is "yes";   in doing some tensorial
manipulation *at a single point of spacetime*,  we can always assume we
have "diagonalized the metric"  to make it look like a minkowski metric, 
just to keep life simple. 
 
Almost through these excellent questions,   which you pretended you didn't
know,   but  really just wanted to help me... you sly gentleman.  But this
is dry work,  where does the wizard keep his tea set...  (whiskey?  a bit
early in the day,  don't you think?)...

From galaxy.ucr.edu!library.ucla.edu!agate!remarque.berkeley.edu!doug Sat Mar  9 22:25:29 PST 1996
Article: 104479 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!agate!remarque.berkeley.edu!doug
From: [email-masked] (Doug Merritt)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 9 Mar 1996 19:26:06 GMT
Organization: University of California, Berkeley
Lines: 49
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: remarque.berkeley.edu

In article <[email-masked]>,
Edward Green <[email-masked]> wrote:
>[email-masked] (john baez)' wrote: 
> 
>>>I am a mite concerned that the Weyl does include static curvature from 
>>>distant bits. Hopefully in this situation these end up zero too. 
>> 
>>Ah, yes.  Indeed, in the earth's gravitational field, or that of a black 
>>hole, in any region where there is no matter, all the curvature is Weyl 
>>curvature.  Why?  Well, there's certainly curvature, since otherwise 
>>there'd be no "gravitational field", but there's no Ricci curvature at
>>points where there is no matter --- by Einstein's equation.
>
>*How* does the Weyl know to be non-zero near the earth,  in a way that
>falls off with distance?  What equations are we solving to tell us this? 
>Einstein's equation says precisely nothing about this...  so is Einstein's
>equation all of GR or what?   What assumptions are you hiding in the back
>room,  oh Wizard??    

He already said in the course notes: The Riemann curvature tensor
is derived via the connection (Christoffel symbol gamma^c_ab) from
the metric g_ab, but he isn't revealing how that works until we pass
the valor test and prove that we're not faint of heart. :-)

But that's enough to go on for the first level answer to your question:
the metric specifies Riemann, which is composed of Ricci and Weyl.
Since Ricci is zero where there is no matter, yet the metric is
going to be forcing curvature where there is no matter (just by
common sense, a good metric must do so), therefore Weyl must be what forces
non-zero curvature in the absence of matter.

So it'll end up being the choice of metric that tells Weyl how to
do this. (A really thorough metric would of course specify all matter
in the universe, but this margin is too small... ;-)

A deeper explanation than that must wait upon the introduction of
computations with the connection.

Reading ahead, the motivation for gamma^c_ab is to generalize
partial differentiation to give a suitably linear tensor differential
operator. But G. Wiz has been cleverly hiding issues of differentiation
so far, which Oz thanked him for a time or two. It's that valor test
again.
	Doug
-- 
Doug Merritt                            [email-masked]
Professional Wild-eyed Visionary        Member, Crusaders for a Better Tomorrow
Unicode Novis Cypherpunks Gutenberg Wavelets Conlang Logli Alife Anthro
Computational linguistics Fundamental physics Cogsci Egyptology GA TLAs

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Article: 104491 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 9 Mar 1996 19:38:05 +0000
Organization: Oz
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In article <[email-masked]>, Edward Green
<[email-masked]> writes
>[email-masked] (john baez)' wrote: 
> 
>>>I am a mite concerned that the Weyl does include static curvature from 
>>>distant bits. Hopefully in this situation these end up zero too. 
>>><nervous grin> 
>> 
>>Ah, yes.  Indeed, in the earth's gravitational field, or that of a black 
>>hole, in any region where there is no matter, all the curvature is Weyl 
>>curvature.  Why?  Well, there's certainly curvature, since otherwise 
>>there'd be no "gravitational field", but there's no Ricci curvature at
>points 
>>where there is no matter --- by Einstein's equation.  So black holes 
>>work very differently from the big bang.  Let's just talk about the big 
>>bang, eh, in this thread?  You'll be dealing with black holes in the 
>>other problem. 
> 
>Harumph!   I was trying to understand this a while back,  and instead of
>somebody taking me by the hand in a friendly manner,  and saying "Well,  as
>you correctly note,  Einstein's equation simply says the Ricci is 0 in
>regions with no matter,  and yet there is a 'quasi-local' effect near
>massive bodies,  and that effect is mediated by the Weyl",  instead of
>that,  I was left to try all sorts of doors in the castle of mathematical
>physics,  until I found one marked "analytic continuation",  which had been
>left unlocked,  so I wandered down that corridor a bit until I fell over an
>interesting truth about the wave equation,  so the trip was not totally
>wasted.  But now I have found my way out into the relatively well lighted
>main corridor again,  and I still don't know how to answer this question!! 
> 
>*How* does the Weyl know to be non-zero near the earth,  in a way that
>falls off with distance?  What equations are we solving to tell us this? 
>Einstein's equation says precisely nothing about this...  so is Einstein's
>equation all of GR or what?   What assumptions are you hiding in the back
>room,  oh Wizard??    
> 
>--erg 
> 
>Q:  How much can Oz and the Wizard take of me in one sitting? 
> 
>A:  6.22 x 10^(-27) erg-sec 

Oh Ed. It's truly a hard life for a poor apprentice. However one must
learn to be patient and suitably grovelling. When the Great Wizard
deigns to answer our pitifully pathetic questions in a suitably simple
language is up to the gods. When one trecks through the outer reaches of
space without a spacesuit one can only expect to be considered somewhat
odd, if not downright loony. Well, we (or at least I) am doing exactly
that in attempting to begin to fathom this subject out with absolutely
no mathematics worth a candle (nuked or not). You must have faith that,
after all the hard work sweeping the keep, the Great Wizard will come
out and flourish a bundle of (correct) equations for the BB scenario and
a brief but enlightening philosophical description of what the equation
is really telling us. One can only keep sweeping and hope he comes home
feeling in a good mood, and isn't kept awake all night by the Trolls
muching on hard disks.

I slunk into his cave one night. Yup, his home cave, and absconded with
various texts hidden therein. What the FAQ did I find? Well, they seemed
somehow a bit elementary (not a tensor in sight), whilst a year ago they
seemed complex. Have we learned something?

I am really looking forward to the Black Hole. Wow this stuff is fun.
Even if G. Wiz makes you *think* godammit! 

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 104508 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 9 Mar 1996 21:43:18 +0000
Organization: Oz
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 <[email-masked]>
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In article <[email-masked]>, Doug Merritt
<[email-masked]> writes
>
>He already said in the course notes: The Riemann curvature tensor
>is derived via the connection (Christoffel symbol gamma^c_ab) from
>the metric g_ab, but he isn't revealing how that works until we pass
>the valor test and prove that we're not faint of heart. :-)

Ahh. But will 'we' pass? What we need is a second level (or maybe third,
or even some grade of wizard) to post a neat and understandable answer.
(Hint, hint). Or would there be no merit in this?

>But that's enough to go on for the first level answer to your question:
>the metric specifies Riemann, which is composed of Ricci and Weyl.
>Since Ricci is zero where there is no matter, yet the metric is
>going to be forcing curvature where there is no matter (just by
>common sense, a good metric must do so), therefore Weyl must be what forces
>non-zero curvature in the absence of matter.

Well, even I guessed this. However we have been given few clues as to
how this is done. One might presume that it may work something like
this. The stress-energy of the Ricci causes curvature where there is
energy. However this would result in a discontinuity immediately outside
the volume with energy unless the curvature could 'gently decay' as you
went away. One might imagine that the metric might determine how you
stitch a curved bit of space with, say, a minkowski at infinity.
Doubtless all these Christoffel thingys might connect with this. One
waits and hopes to learn. Perhaps we could use a black hole as a simple
(ha!) example. Are you listening Mr Great Wizard, sir, y'r 'onnor m'lud?

>So it'll end up being the choice of metric that tells Weyl how to
>do this. (A really thorough metric would of course specify all matter
>in the universe, but this margin is too small... ;-)
>
>A deeper explanation than that must wait upon the introduction of
>computations with the connection.

You know, Sir magical Douglas, I think you know more than you let on.
Perhaps you could brave the wrath of the irascible but sometimes
friendly wizard and, as they say, help out a bit?

>Reading ahead, the motivation for gamma^c_ab is to generalize
>partial differentiation to give a suitably linear tensor differential
>operator. But G. Wiz has been cleverly hiding issues of differentiation
>so far, which Oz thanked him for a time or two. It's that valor test
>again.

Oh Oh Oh dear. "a suitably linear tensor differential operator" sounds
really technical. Still (feigning a happy go lucky brightness) at least
it sounds as if it *might* be followable. After all, it could be " a
suitably linear tensor integral operator". There aren't *too* many
integrations that nasty old wizard might ask us. (Says Oz with absolute
confidence that there will be. GROAN!>

Anyone got a book of integrals? I bet Ed has.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 104469 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 9 Mar 1996 11:34:43 -0500
Organization: The Pipeline
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[email-masked] (john baez)' wrote: 
 
>>I am a mite concerned that the Weyl does include static curvature from 
>>distant bits. Hopefully in this situation these end up zero too. 
>><nervous grin> 
> 
>Ah, yes.  Indeed, in the earth's gravitational field, or that of a black 
>hole, in any region where there is no matter, all the curvature is Weyl 
>curvature.  Why?  Well, there's certainly curvature, since otherwise 
>there'd be no "gravitational field", but there's no Ricci curvature at
points 
>where there is no matter --- by Einstein's equation.  So black holes 
>work very differently from the big bang.  Let's just talk about the big 
>bang, eh, in this thread?  You'll be dealing with black holes in the 
>other problem. 
 
Harumph!   I was trying to understand this a while back,  and instead of
somebody taking me by the hand in a friendly manner,  and saying "Well,  as
you correctly note,  Einstein's equation simply says the Ricci is 0 in
regions with no matter,  and yet there is a 'quasi-local' effect near
massive bodies,  and that effect is mediated by the Weyl",  instead of
that,  I was left to try all sorts of doors in the castle of mathematical
physics,  until I found one marked "analytic continuation",  which had been
left unlocked,  so I wandered down that corridor a bit until I fell over an
interesting truth about the wave equation,  so the trip was not totally
wasted.  But now I have found my way out into the relatively well lighted
main corridor again,  and I still don't know how to answer this question!! 
 
*How* does the Weyl know to be non-zero near the earth,  in a way that
falls off with distance?  What equations are we solving to tell us this? 
Einstein's equation says precisely nothing about this...  so is Einstein's
equation all of GR or what?   What assumptions are you hiding in the back
room,  oh Wizard??    
 
--erg 
 
Q:  How much can Oz and the Wizard take of me in one sitting? 
 
A:  6.22 x 10^(-27) erg-sec 

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Article: 104512 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Followup-To: sci.physics
Date: 9 Mar 1996 22:56:22 GMT
Organization: Physics Department, U.C. Berkeley
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In article <[email-masked]>,
Edward Green <[email-masked]> wrote:
>*How* does the Weyl know to be non-zero near the earth,  in a way that
>falls off with distance?  What equations are we solving to tell us this? 
>Einstein's equation says precisely nothing about this...  so is Einstein's
>equation all of GR or what?   What assumptions are you hiding in the back
>room,  oh Wizard??    

Let me see if I can confuse the situation a little bit more.  The
thing that's bothering you, I take it, is this: there are only ten
independent Einstein equations, but there are 20 independent
components of the Riemann tensor.  (Specifically, there are 10 Ricci
components and 10 Weyl components, and the Einstein equation ells you
about only the Ricci part, not the Weyl part.)  So do the ten Einstein
equations manage to completely determine the 20 Riemann components,
and if so, how?

First, let me point out that the problem goes away if you choose to
think of the unknown quantities in the Einstein equation as the metric
components g_{ab} instead of the Riemann components R^a_{bcd}.  After
all, there are 10 g_{ab}'s to go along with the 10 Einstein
equations.  So you might hope that you could take the Einstein
equation (together with a sufficient amount of information about
initial and boundary conditions) and use it to solve for g_{ab}
everywhere (and everywhen).  Once you've got g_{ab}, there's an as yet
unspecified prescription for getting R^a_{bcd}.

There's an analogy to be made with electromagnetism, which might be
helpful here.  It goes something like this:

Gravity                       Electromagnetism
-------                       ----------------

Stress-energy                 Charge and current density
Riemann tensor                E and B fields
Metric                        Potentials (A and Phi, or better yet 
                                          the four-vector A)
Einstein equation             Maxwell equations with source terms
                                     (Gauss's law and Ampere's law)

So why have I written down only half of the Maxwell equations?  Where
are the source-free ones (div B=0 and Faraday's law)?  I don't need to
write them down, because they're implicit in the relation between the
fields and the potentials.  The fact that I can write E and B in terms
of the potentials A and Phi implies that the source-free Maxwell
equations are satisfied.  So the gravitational analogue of the
source-free Maxwell equations has something to do with the mysterious
procedure by which you compute Riemann from the metric.

I'll finish by giving you a little puzzle.  I said that you could
think of the Einstein equation as a differential equation to be solved
for the 10 quantities g_{ab}.  But that actually seems a little bit
strange when you think about it: you shouldn't ever be able to come up
with a unique solution for g_{ab}, since there should be infinitely
many g_{ab}'s that describe any particular spacetime.  (The g_{ab}'s
are the components of the metric in a particular coordinate system; if
you used a different coordinate system, you'd get different g_{ab}'s.)

In fact, this is true.  Before you can narrow down to a unique
solution for the g_{ab}'s, you have to impose some additional
constraints.  The puzzle is simply this: tell me what the analogous
phenomenon in electromagnetism is called.

-Ted

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Article: 104607 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 10 Mar 1996 09:24:16 +0000
Organization: Oz
Lines: 44
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Anyway, Oz can certainly satisfy me as to why black holes must form
>under certain conditions without getting into this stuff... and then,
>having passed his test, I can show him (and the rest of you) some of the
>scary formulas I have been hiding thus far, and we can see a bit more
>about how one actually uses them to do physics.

The Wiz had come out of his mathematicave for a cup of coffee. Whilst
Wiz and Oz waited for the alembic to boil they fell into a short
discussion.

"Um." said Oz hesitantly in his usual lucid way "You know this black
hole question you asked?"

"Yeeess." said Wiz suspiciously "the one you are making very heavy
weather over. Yes."

"Well," continued Oz, "You remember at the beginning of the course you
told me to forget all I thought I knew about Gr, space, time and
anything else even vaguely connected with it."

"No I don't, but it's the sort of thing I might have said. There is so
much misinformation going about, you would be better off forgetting it."
Said Wiz sternly and even more suspiciously.

"Well I did, as per instructions. So I figure that you only want answers
more or less derivable from your course notes." 

mumbled Oz as he blew hard on the embers to get a bit of flame going. 

"Ed seems to think the same way and it's not really very
straightforward, see. I even tried handwaving but that didn't meet with
your approval. It didn't even get a quick blast of ire back."

"So?" snapped Wiz curtly as he picked up the alembic, poured himself
some coffee, and swept into his cave. Oz thought it was nice to see Wiz
sweeping occasionally, if only regally, and went to get his broom.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Sun Mar 10 14:13:11 PST 1996
Article: 104663 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 10 Mar 1996 13:08:52 -0800
Organization: University of California, Riverside
Lines: 63
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

Said G. Wiz, "Anyway, Oz can certainly satisfy me as to why black holes
must form under certain conditions without getting into this stuff...
and then, having passed his test, I can show him (and the rest of you)
some of the scary formulas I have been hiding thus far, and we can see a
bit more about how one actually uses them to do physics."

The Wiz had come out of his mathematicave for a cup of coffee. Whilst
Wiz and Oz waited for the alembic to boil they fell into a short
discussion.

"Um." said Oz hesitantly in his usual lucid way "You know this black
hole question you asked?"

"Yeeess." said Wiz suspiciously "the one you are making very heavy
weather over. Yes."

"Well," continued Oz, "You remember at the beginning of the course you
told me to forget all I thought I knew about Gr, space, time and
anything else even vaguely connected with it."

"No I don't, but it's the sort of thing I might have said. There is so
much misinformation going about, you would be better off forgetting it."
Said Wiz sternly and even more suspiciously.

"Well I did, as per instructions. So I figure that you only want answers
more or less derivable from your course notes." 

mumbled Oz as he blew hard on the embers to get a bit of flame going. 

"Ed seems to think the same way and it's not really very
straightforward, see. I even tried handwaving but that didn't meet with
your approval. It didn't even get a quick blast of ire back."

"So?  You gotta wave your hands the right way for the magic to happen,"
snapped Wiz curtly as he picked up the alembic, poured himself some
coffee, and swept into his cave. Oz thought it was nice to see Wiz
sweeping occasionally, if only regally, and went to get his broom.

Just as he picked up, the Wiz poked his head out again.  "Ahem. 
While you're sweeping, why don't you meditate a bit on the concept of 
dust."

"Ashes to ashes, dust to dust, that sort of thing?" asked Oz.  It
surprised him that G. Wiz would be giving him spiritual exercises
instead of the usual mathematical ones.  

"Not exactly.  Imagine a sphere of dust particles in outer space
interacting only through gravity.  Each one in free fall.  What will
happen?"

"Well, they'll fall in..." blurted out Oz.

"Right.  Can you show that a singularity will form?  A place where the
curvature becomes infinite?  You know.... a BLACK HOLE?"

Oz thought and took a sip of coffee.  Yech!  The coffee was full of
coffee grounds, since there weren't any filters around.  Coffee
grounds... what did that remind him of....

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Article: 104607 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 10 Mar 1996 09:24:16 +0000
Organization: Oz
Lines: 44
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Anyway, Oz can certainly satisfy me as to why black holes must form
>under certain conditions without getting into this stuff... and then,
>having passed his test, I can show him (and the rest of you) some of the
>scary formulas I have been hiding thus far, and we can see a bit more
>about how one actually uses them to do physics.

The Wiz had come out of his mathematicave for a cup of coffee. Whilst
Wiz and Oz waited for the alembic to boil they fell into a short
discussion.

"Um." said Oz hesitantly in his usual lucid way "You know this black
hole question you asked?"

"Yeeess." said Wiz suspiciously "the one you are making very heavy
weather over. Yes."

"Well," continued Oz, "You remember at the beginning of the course you
told me to forget all I thought I knew about Gr, space, time and
anything else even vaguely connected with it."

"No I don't, but it's the sort of thing I might have said. There is so
much misinformation going about, you would be better off forgetting it."
Said Wiz sternly and even more suspiciously.

"Well I did, as per instructions. So I figure that you only want answers
more or less derivable from your course notes." 

mumbled Oz as he blew hard on the embers to get a bit of flame going. 

"Ed seems to think the same way and it's not really very
straightforward, see. I even tried handwaving but that didn't meet with
your approval. It didn't even get a quick blast of ire back."

"So?" snapped Wiz curtly as he picked up the alembic, poured himself
some coffee, and swept into his cave. Oz thought it was nice to see Wiz
sweeping occasionally, if only regally, and went to get his broom.

-------------------------------
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Sun Mar 10 20:20:19 PST 1996
Article: 104751 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 10 Mar 1996 20:07:05 -0800
Organization: University of California, Riverside
Lines: 42
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>Oz has been looking back to the beginning of this thread.

>Suddenly he is puzzled. A post seems to be missing. In fact his second
>post. 

>Now if he remembers rightly his original answer was not (to him) very
>satisfactory because it didn't offer a finite radius where 'interesting
>things' happened. It went something like this:

>We have a little ball of coffee grains. Naturally, being massive, they
>come together. However, being pointlike, they can get as close together
>as they like. Now
>
>d^2V/dt^2 = - (1/2)(E+3P) V
>
>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>positive, so the volume should tend to zero. The density will thus tend
>to infinity. I guess this might (just) be called a singularity.
>Certainly the curvature gets to be infinite which is good enough for me.

>Personally I find this too glib and too simplistic. 

However, it's exactly along the lines I wanted!!!!  I wanted something
glib, simplistic, yet correct.

I never saw this one!  

So, keeping in mind this stuff, think a bit about the wizard's comment
about a ball of dust... and tell me what you can about *curvature
singularities*... places where the Riemann curvature blows up.  Also,
how are your coffee grounds moving... are they perchance following
geodesics?  (I guess so, since you're using the formula that applies
then.)  What can you say about the more realistic case where, after they
get close enough together, they "bump into each other".  Doesn't this keep
the black hole from forming?

From galaxy.ucr.edu!library.ucla.edu!nnrp.info.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!scramble.lm.com!news.math.psu.edu!psuvax1!uwm.edu!vixen.cso.uiuc.edu!newsfeed.internetmci.com!iol!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Mon Mar 11 12:03:27 PST 1996
Article: 104805 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!nnrp.info.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!scramble.lm.com!news.math.psu.edu!psuvax1!uwm.edu!vixen.cso.uiuc.edu!newsfeed.internetmci.com!iol!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz
From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Mon, 11 Mar 1996 09:31:38 +0000
Organization: Oz
Lines: 142
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]>
 <[email-masked]> <[email-masked]>
Reply-To: [email-masked]
NNTP-Posting-Host: upthorpe.demon.co.uk
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X-Newsreader: Turnpike Version 1.12 <V67Vak$Q$lyzk9NR0hujdE7wCg>

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>>We have a little ball of coffee grains. Naturally, being massive, they
>>come together. However, being pointlike, they can get as close together
>>as they like. Now
>>
>>d^2V/dt^2 = - (1/2)(E+3P) V
>>
>>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>>positive, so the volume should tend to zero. The density will thus tend
>>to infinity. I guess this might (just) be called a singularity.
>>Certainly the curvature gets to be infinite which is good enough for me.
>
>>Personally I find this too glib and too simplistic. 
>
>However, it's exactly along the lines I wanted!!!!  I wanted something
>glib, simplistic, yet correct.
>

Well, it was not on my newsreader and I'm pretty sure I have set all
this thread to 'keep', yet I have a record of posting a similar one. I
think it never got past the Demons. The problem I have now is
remembering to do it from scratch, ignoring any posts I may have posted.
So this time we will start from the beginning.

>So, keeping in mind this stuff, think a bit about the wizard's comment
>about a ball of dust... and tell me what you can about *curvature
>singularities*... places where the Riemann curvature blows up.  Also,
>how are your coffee grounds moving... are they perchance following
>geodesics?  (I guess so, since you're using the formula that applies
>then.)  What can you say about the more realistic case where, after they
>get close enough together, they "bump into each other".  Doesn't this keep
>the black hole from forming?

See! I was right! It *was* too glib and simplistic. At least I got
something right, even if it's nothing to do with the question.

OOooo dear. If we aren't careful we is going to get into trouble
defining energy again. These poor coffee grains that started at rest
with each other are, after a period, flying towards each other at
significant speed. (Oz hears the dratted minus signs revving their wings
in their nest high in the wizards cave). Yes, yes, I know, they are
still following geodesics but this is what is bringing them together
ever faster.

Hmmmm. I think a good wheeze is to do it in little steps and see if we
are heading in the right direction each time. This is called fireball
delaying tactics.

Well we have:

R_{00} = (1/2)(E + 3P) 

Well, the momentum flow in our little ball of particles would seem to be
increasing but I would guess that the ball is isotropic and uniform in
the spacial dimensions so the above looks a reasonable guide to the
curvature. We note that the density (E & P) are getting bigger and
bigger as the ball contracts, so the curvature is getting bigger and
bigger too.

Well, after a while our little ball of coffee grounds (iron filings
would have been better) is so dense that the grounds start bumping into
each other. Some of them at the surface get ejected and start to head
away from the ball. Unfortunately for them they are now in very curved
space. Since the ejected grounds now follow geodesics away from the ball
we can use our little formula:

d^2V/dt^2 = -(1/2)(E + 3P) 

except this time as the particle speeds outwards the volume of space
enclosing the mass is increasing so the effective density is decreasing
and so d^2V/dt^2 is decreasing as it heads outwards. (Well, nobody has
actually said this, indeed they implied that it was a naughtly thing to
say, however this way we should get a definitive fireball reply one way
or the other).

Now some lucky speedy little grounds will escape all the way to
infinity, but other slower doomed grounds will find that their velocity
wrt the main ball eventually becomes negative and the curvature bends
them round to head inexorably back to the still-contracting main ball.
In this way the velocity between grounds will be reduced. Essentially by
evaporation of high velocity coffee grounds, to below the ever
increasing escape velocity of the ball. We note in passing that these
coffee grounds feel no acceleration as they travel along their curved
path.

So the relative velocities of the coffee grounds (note that they are
infinitely small coffee grounds) within the ball can increase as the
curvature increases. As the ball gets smaller it takes faster and faster
coffee grounds to escape from the increasingly curved spacetime that
surrounds the ball. Now lets give these relative velocities a more
useful name, say 'temperature'. 

Now in passing it's clear that any given curvature will have some
characteristic temperature where it will be in equilibrium. In effect
all the coffee grounds are orbiting the centre. It will inevitably lose
some energy continually by evaporation. If you had an energy source that
put heat into the ball at the rate it is lost, it could sit at that
curvature until the power source ran out. It would then continue to
contract. These are called 'suns'. Lo and behold we note that low
density suns have a low temperature, and high density suns have a high
temperature. This is why a red giant is red, it has very low density.
Massive stars have to have a high temperature because of their high
density. We ignore neutron stars and dwarves because they have a
different mechanism to prevent collapse.

However our coffee grounds have no such source. They continue to
contract and the temperature continues to rise as the coffee grounds
have to get faster and faster to escape to infinity due to the ever
increasing curvature of the ever increasing density of the ever
decreasing ball. I feel that there should be a mechanism that allows the
ball to completely evaporate, but starting where we started I am not too
sure there is one.

One might wonder if there is a maximum velocity. If there were then at
some point the curvature gets so curved that nothing can escape to
infinity. This would be the event horizon, and I *still* think that is
where the curvature is one (or something like that). The worldline of a
photon then goes from 45 degrees to being parallel to the time axis.

Anyway, our hot little ball of incinerated coffee grounds would still
continue to contract, and contract, and contract, until it was
infinitely small and the space around it infinitely curved. I guess this
counts as a Baez curvature 'blow up'.

This is nothing I couldn't have said before this course started (except
for swapping curvature for 'gravitational field'). It's a lot easier, of
course, if I don't have to pull this stuff out of equations I only
partly understand (well, you gotta be optimistic). So where have I used
my (ho, ho, ho, ROTFL) newfound knowledge? 
Nowhere. <sob, sob, booo-hoo.>

I hope this will do for starters. It's not, however, an adequate answer
for this course. IMHO.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!not-for-mail Mon Mar 11 14:01:36 PST 1996
Article: 104897 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Is the equivalence principle true in homo. grav. fields?
Date: 11 Mar 1996 12:53:59 -0800
Organization: University of California, Riverside
Lines: 40
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward Green) writes:

>But what most interests me here though is the conclusion that a
>'gravitational time dilation' can be derived,  in the acceleration
>equivalent case,  purely by SR.  It's not that gravity or acceleration
>affects the way clocks run,  apparently....   Actually,   I find this a
>surprising assumption.   It would seem to me the most natural thing in the
>world to expect an atom undergoing acceleration would emit light at
>different frequencies than an unaccelerated atom.     

Well, sure, accelerated atoms act differently than unaccelerated ones.
Accelerated clocks act differently than unaccelerated ones, too: when
you accelerate them too much, they break!

I.e., in practice any "ticker" gets busted when you try to accelerate it
too much.  This is *not* what special relativity and general relativity
is about... in special relativity and general relativity thought
experiments we typically assume (for simplicity) that the mechanisms of
our clocks are not affected by acceleration.  

This may seem overly abstract and unphysical, but it bears repeating:

"Gravitational time dilation" is simply due to the fact that the amount
of time it takes to get from A to B along a certain path is given by an
integral along that path involving the metric.  It's simply the "length"
of the path, as it were... though people use the term "proper time"
here.  To compute this length, we compute the tangent vector of the
path, say v(t), then work out the length of that vector, which is
sqrt(-g(v(t),v(t)), and then integrate this along the path.  

The fact that some paths from A to B are longer than others should not
be surprising, and the math behind how it works is not too surprising
either.

From galaxy.ucr.edu!not-for-mail Mon Mar 11 16:42:16 PST 1996
Article: 104910 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 11 Mar 1996 15:16:18 -0800
Organization: University of California, Riverside
Lines: 112
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

"Ack!" says the wizard, swatting madly.  "It's these damn signs.  It's
impossible to think when they're around."

"Let's just wait around," said Oz, "until the smoke returns.  After
all, if it really wants to know the answer, it'll come back.  Why waste
our time fighting these damn signs?"

"Gee whiz!" said G. Wiz.  "A great idea."

So they hung out by the campfire and talked about other subjects until
the signs, seeing there would be no more food, buzzed off.  Eventually,
hours later, the smoke returned!

"I hate those signs too", said the smoke.  "I've found that raising
and lowering indices serves as a breeding ground for them, which is
why I suggest this process be avoided whenever possible!!
That's why I recommend using T^a_b, G^a_b and R^a_b"

"Well," said the wizard, "that's fine as long as you don't ever need to
use R_{ab}, but typically you do.  One needs to battle signs occaisionally in
this business... perhaps your method would help mimimize it, but not
eliminate it."

"I am now getting:

T^a_b =

-E 0 0 0
 0 P 0 0
 0 0 P 0
 0 0 0 P

i.e. T^0_0 = -E, T^i_i = P (pressure) for i=1,2,3."

The wizard said "Good, that's what I get."

"Did you really work it out," asked the smoke, "or are you just taking
my word for it?  I have a feeling you're not all that eager to do this
sort of calculation."

The wizard smiled.  "You're right, but if you look in the course notes
you'll see I have that in there somewhere.  Check out 

http://math.ucr.edu/home/baez/outline2.html "

The smoked approximated a nod.  "Furthermore, I find that R = R^c_c = -Tc_c

i.e. R= -(T^0_0 + T^1_1 + T^2_2 + T^3_3) = E-3P"

"Yes, that's what I got too," said G. Wiz.

Since R^a_b = T^a_b + .5 R g^a_b  . (g^a_b = 1 iff a=b, 0 otherwise)

R^0_0 = .5^(T^0_0 - T^1_1 - T^2_2 - T^3_3) = .5(E+3P)

"This indicates that P makes things contract! (Or at least go in the
same direction as E.)   This seems weird, but" the cloud turns slightly
pinkish, "we are talking about d^2 V/dt^2, not dV/dt.  It's obvous that
dV/dt is positive when a high pressure area is in empty space, but that
says nothing about d^2V/dt^2"

"Wait a minute!" said G. Wiz.  "Your formula for R^0_0 agrees with the
stuff in the course outline, and you are right that it means the
gravitational effect of pressure is to cause things to contract.  In
layman's lingo: like energy, pressure causes an attractive gravitational
force."

Oz added "I thought you said gravity wasn't considered a `force' in
general relativity."  

"It's not," admitted the wizard, "that's why I said, `in layman's
lingo'."

"BUT," added G. Wiz, "the formula relating R^0_0 to d^2V/dt^2 only applies
to a ball of particle following geodesics!  That is, particles in free
fall, feeling no force other than gravity (which, ahem, is not really a
force either).  In reality, when we have a gas with different pressures
at different regions, the atoms of the gas *do* feel forces other than
gravity.  So pressure does have the usual effect of making stuff want to
expand, in addition to the special GR effect of making it want to
contract.  Just don't mix them up; they are very different."

"Something else that is interesting.  If we take a simple particle with
no pressure term, the stress energy tensor is"

T^0_0 = -E

all other terms are zero."

"If we take this and boost half of it in the -x direction at some velocity
v, (energy E/2) and boost half of it in the +x direction at the same
velocity v so that the average momentum/velocity is zero and add, we get"

T^0_0 = -gamma^2E
T^1_1 = v^2gamma^2E

where gamma = 1/sqrt(1-v^2) [c assumed to be 1]  

"The other terms all cancel out."  The cloud swirls as it attempts to cross
it's nonexistent fingers as it makes this remark.

"This gives us some insight into the pressure term, the above anisotropic
case has a positive pressure in the x direction."

"It seems odd at first that T^0_0 scales as gamma^2, but it's 
energy / volume, not energy, and volume is defintely not Lorentz
invariant!"

"Hmm," said the wizard.  "Interesting.  I'll have to ponder that."

From galaxy.ucr.edu!not-for-mail Mon Mar 11 16:44:56 PST 1996
Article: 104663 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 10 Mar 1996 13:08:52 -0800
Organization: University of California, Riverside
Lines: 63
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

Said G. Wiz, "Anyway, Oz can certainly satisfy me as to why black holes
must form under certain conditions without getting into this stuff...
and then, having passed his test, I can show him (and the rest of you)
some of the scary formulas I have been hiding thus far, and we can see a
bit more about how one actually uses them to do physics."

The Wiz had come out of his mathematicave for a cup of coffee. Whilst
Wiz and Oz waited for the alembic to boil they fell into a short
discussion.

"Um." said Oz hesitantly in his usual lucid way "You know this black
hole question you asked?"

"Yeeess." said Wiz suspiciously "the one you are making very heavy
weather over. Yes."

"Well," continued Oz, "You remember at the beginning of the course you
told me to forget all I thought I knew about Gr, space, time and
anything else even vaguely connected with it."

"No I don't, but it's the sort of thing I might have said. There is so
much misinformation going about, you would be better off forgetting it."
Said Wiz sternly and even more suspiciously.

"Well I did, as per instructions. So I figure that you only want answers
more or less derivable from your course notes." 

mumbled Oz as he blew hard on the embers to get a bit of flame going. 

"Ed seems to think the same way and it's not really very
straightforward, see. I even tried handwaving but that didn't meet with
your approval. It didn't even get a quick blast of ire back."

"So?  You gotta wave your hands the right way for the magic to happen,"
snapped Wiz curtly as he picked up the alembic, poured himself some
coffee, and swept into his cave. Oz thought it was nice to see Wiz
sweeping occasionally, if only regally, and went to get his broom.

Just as he picked up, the Wiz poked his head out again.  "Ahem. 
While you're sweeping, why don't you meditate a bit on the concept of 
dust."

"Ashes to ashes, dust to dust, that sort of thing?" asked Oz.  It
surprised him that G. Wiz would be giving him spiritual exercises
instead of the usual mathematical ones.  

"Not exactly.  Imagine a sphere of dust particles in outer space
interacting only through gravity.  Each one in free fall.  What will
happen?"

"Well, they'll fall in..." blurted out Oz.

"Right.  Can you show that a singularity will form?  A place where the
curvature becomes infinite?  You know.... a BLACK HOLE?"

Oz thought and took a sip of coffee.  Yech!  The coffee was full of
coffee grounds, since there weren't any filters around.  Coffee
grounds... what did that remind him of....

From [email-masked] Mon Mar 11 17:36 PST 1996
Date: Mon, 11 Mar 1996 16:28:14 +0000
To: john baez <[email-masked]>
From: Oz <[email-masked]>
Reply-To: [email-masked]
Subject: Re: Riemann Awakes
Content-Length: 7535
Status: OR

This has been posted, slightly differently.

This is more Wizardworlded for your WWW page.
A bit more drama etc etc.

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>>We have a little ball of coffee grains. Naturally, being massive, they
>>come together. However, being pointlike, they can get as close together
>>as they like. Now
>>
>>d^2V/dt^2 = - (1/2)(E+3P) V
>>
>>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>>positive, so the volume should tend to zero. The density will thus tend
>>to infinity. I guess this might (just) be called a singularity.
>>Certainly the curvature gets to be infinite which is good enough for me.
>
>>Personally I find this too glib and too simplistic. 
>
>However, it's exactly along the lines I wanted!!!!  I wanted something
>glib, simplistic, yet correct.
>
>So, keeping in mind this stuff, think a bit about the wizard's comment
>about a ball of dust... and tell me what you can about *curvature
>singularities*... places where the Riemann curvature blows up.  Also,
>how are your coffee grounds moving... are they perchance following
>geodesics?  (I guess so, since you're using the formula that applies
>then.)  What can you say about the more realistic case where, after they
>get close enough together, they "bump into each other".  Doesn't this keep
>the black hole from forming?

Oz stared at the Wizard. He was speechless. Slowly he placed both hands
on his hips and, careless of bodily danger, faced the Wizard up. He was
cool and controlled, but you could feel the exasperation boiling up
inside him. He had just worked through a whole lotta junk, and he
suddenly realised the wizard just wanted the old fashioned stuff.

Oz opened his mouth and spoke:

"Hey Wiz, now you just listen here and cop a dose of this!

We have:

R_{00} = (1/2)(E + 3P) 

Well, the momentum flow in our little ball of particles would seem to be
increasing but I would guess that the ball is isotropic and uniform in
the spacial dimensions so the above looks a reasonable guide to the
curvature. We note that the density (E & P) are getting bigger and
bigger as the ball contracts, so the curvature is getting bigger and
bigger too.

After a while our little ball of coffee grounds (iron filings
would have been better) is so dense that the grounds start bumping into
each other. Some of them at the surface get ejected and start to head
away from the ball. Unfortunately for them they are now in very curved
space. Since the ejected grounds now follow geodesics away from the ball
we can use our little formula:

d^2V/dt^2 = -(1/2)(E + 3P) 

except this time as the particle speeds outwards the volume of space
enclosing the mass is increasing so the effective density is decreasing
and so d^2V/dt^2 is decreasing as it heads outwards. (Well, nobody has
actually said this, indeed they implied that it was a naughtly thing to
say, however this way we should get a definitive fireball answer one way
or the other).

Now some lucky speedy little grounds will escape all the way to
infinity, but other slower doomed grounds will find that their velocity
wrt the main ball eventually becomes negative and the curvature bends
them round to head inexorably back to the still-contracting main ball.
In this way the velocity between grounds will be reduced. Essentially by
evaporation of high velocity coffee grounds, to below the ever
increasing escape velocity of the ball. We note in passing that these
coffee grounds feel no acceleration as they travel along their curved
path.

So the relative velocities of the coffee grounds (note that they are
infinitely small coffee grounds) within the ball can increase as the
curvature increases. As the ball gets smaller it takes faster and faster
coffee grounds to escape from the increasingly curved spacetime that
surrounds the ball. Now lets give these relative velocities a more
useful name, say 'temperature'. 

In passing it's clear that any given curvature will have some
characteristic temperature where it will be in equilibrium. In effect
all the coffee grounds are orbiting the centre. It will inevitably lose
some energy continually by evaporation. If you had an energy source that
put heat into the ball at the rate it is lost, it could sit at that
curvature until the power source ran out. It would then continue to
contract. These are called 'suns'. Lo and behold we note that low
density suns have a low temperature, and high density suns have a high
temperature. This is why a red giant is red, it has very low density.
Massive stars have to have a high temperature because of their high
density. We ignore neutron stars and dwarves because they have a
different mechanism to prevent collapse.

However our coffee grounds have no such source. They continue to
contract and the temperature continues to rise as the coffee grounds
have to get faster and faster to escape to infinity due to the ever
increasing curvature of the ever increasing density of the ever
decreasing ball. I feel that there should be a mechanism that allows the
ball to completely evaporate, but starting where we started I am not too
sure there is one.

One might wonder if there is a maximum velocity. If there were then at
some point the curvature gets so curved that nothing can escape to
infinity. This would be the event horizon, and I *still* think that is
where the curvature is one (or something like that). The worldline of a
photon then goes from 45 degrees to being parallel to the time axis.

Anyway, our hot little ball of incinerated coffee grounds would still
continue to contract, and contract, and contract, until it was
infinitely small and the space around it infinitely curved. I guess this
counts as a 'blow up' even for ornery old Wizards. HUMPF!"

Oz paused for breath. He had let it all come out. Blow the infernal
Ricci's and Riemann's and Einsteinian giff-gaff. He just said it how he
saw it. You could see the defiance all over his face.

"AND while we're at it, " bellowed Oz in intense irritation,
"this is nothing I couldn't have said before this course started (except
for swapping curvature for 'gravitational field'). It's a lot easier, of
course, if I don't have to pull this stuff out of equations I only
partly understand (well, you gotta be optimistic). So where have I used
my newfound knowledge?"

"Nowhere!" he cried.

The strain all became too much for Oz and he fell on the quartzite floor
and sobbed as he waited for the fireball to end all fireballs.

**************
Notes: 

1) This has been pepped up a bit, for dramatic effect, you understand.
Well, Oz always was a bit of a ham actor. Always overacting.
2) Whilst it was originally written in one swoop, I didn't feel any
antipathy to you whatsoever. I just figured out you wanted something
along these lines. Actually it was a useful recap, I think I had been
losing sight of the physics a bit. Well, quite a bit actually.
3) I hope to god I haven't made some stupid error and it's the sort of
thing you were after. (Fingers crossed!)
4) I think this has more pazzaz than the original I posted, don't you?

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From [email-masked] Tue Mar 12 08:56 PST 1996
From: [email-masked]
Subject: Re: smoke
To: [email-masked] (john baez)
Date: Tue, 12 Mar 96 10:55:26 CST
Mailer: Elm [revision: 70.85]
Content-Length: 7638
Status: OR

> 
> Do you have the original "smoke" post of yours and my first reply?
> I can't find them!  I want to add this to the "Oz and the Wizard"
> collection on my website.
> 
> jb
> 
> 

------------------
Some smoke from the campfire drifts in - and assumes a vaguely
human form.

"Sorry I couldn't make it in person", says bhv, "My sending will
have to do.  Besides, in this form, the worst thing that can
happen to me is being dispelled - which is much less stressful
than experiencing in person lightning bolts and fireballs and
whatnot that seem to fly around here!"

"Anyway, I have some ideas.  It seems to me that pressure should
cause things to expand - if you have a ball of a perfect gas,
and it has a positive pressure, it should want to expand, right?"

"And gravity should want to make things contract!"

Now if we finally decided that

R^0_0 = T^0_0 - .5 T^c_c
"Drat - I need a corrected version of the classnotes.  Why G. Wiz
likes those dratted R_ab and T_ab I'll never understand, he should
stick to R^a_b and T^a_b if you ask me.  Not that he will, of
course.  Use the dratted g^a_b for the dratted kronecker delta
function if need be!  Oh well, if I was a powerful wizard like G. wiz,
I might not listen to a ball of smoke, either."

"Anyway this means that R^0_0 = .5(T^0_0 - T^1_1 - T^2_2 - T^3_3)
so that R^0_0 = .5(density - 3*pressure) if we have a perfect fluid."

"And d^2V/dt^2 was proportional to -R^0_0.  Which means we've just
proved that positive density causes things to contract and positive
pressure causes them to expand!"

"This is a nice start, but I was hoping that pressure would
contribute to the energy somehow.  I sure can't see why a sufficient 
pressure can't make R^0_0 negative (which indicates
expansion) with this formulation!"

"Oh well, back to the drawing board.  Maintaining this sending isn't
free, I'll stop back in a bit to see what happens".

The puff of smoke dissipates.

"Hmmm - I don't see why R^0_0 can't be negative either", said Oz.
Do you think that R^0_0 was negative for that puff of smoke, Ed?"

----------------
Some smoke from the campfire drifts in - and assumes a vaguely
human form.

"Damn," says Oz!  "It's vaguely human!"  

"Vaguely human!" agrees G. Wiz. "It must be a sci.physics reader.  I
wonder what it wants?" 

"Sorry I couldn't make it in person", says bhv, "My sending will
have to do.  Besides, in this form, the worst thing that can
happen to me is being dispelled - which is much less stressful
than experiencing in person lightning bolts and fireballs and
whatnot that seem to fly around here!"

"The worst thing, eh?" asks the wizard, who knows a lot about flaming.
"Have no fear, ask away."

"Anyway, I have some ideas.  It seems to me that pressure should
cause things to expand - if you have a ball of a perfect gas,
and it has a positive pressure, it should want to expand, right?"

"And gravity should want to make things contract!"
The wizard nods, "Ah, well, looking ahead into the future I sense a
forthcoming proliferation of possibilities for sign errors, but I feel
pretty sure that pressure creates gravity which makes things want to
CONTRACT.  Of course, pressure makes gases such as yourself expand, for
the usual reasons; that has nothing to do with general relativity.  But
the reason why black holes must form under certain circumstances is
that, no matter how much a star tries to resist collapse with pressure,
that pressure merely generates more gravity, which makes it collapse all
the more.  Ironic, ain't it?"

"Now if we finally decided that

R^0_0 = T^0_0 - .5 T^c_c...

Drat - I need a corrected version of the classnotes.  Why G. Wiz
likes those dratted R_ab and T_ab I'll never understand, he should
stick to R^a_b and T^a_b if you ask me.  Not that he will, of
course.  Use the dratted g^a_b for the dratted kronecker delta
function if need be!  Oh well, if I was a powerful wizard like G. Wiz,
I might not listen to a ball of smoke, either."

"I listen to all sorts of hot air, here on sci.physics" notes the Wiz.
In your present form, you fit right in."  

The smokes swirls about in amusement and says, "Anyway this means that
R^0_0 = .5(T^0_0 - T^1_1 - T^2_2 - T^3_3) so that R^0_0 = .5(density -
3*pressure) if we have a perfect fluid."  

"And d^2V/dt^2 was proportional to -R^0_0.  Which means we've just
proved that positive density causes things to contract and positive
pressure causes them to expand!"  Oz notes a buzzing swarm of minus
signs in the distance, heading for the campfire.  

"This is a nice start, but I was hoping that pressure would
contribute to the energy somehow.  I sure can't see why a sufficient 
ppressure can't make R^0_0 negative (which indicates
expansion) with this formulation!"  With a buzzing roar, the swarm of
signs reaches the campfire.  Oz and the Wiz can barely hear over the
noise of the signs.  They swat this way and that, to no avail.

"Oh well, back to the drawing board.  Maintaining this sending isn't
free, I'll stop back in a bit to see what happens".

The puff of smoke dissipates.  But alas, the minus signs he appears to
have brought with him remain.  

"Hmmm - I don't see why R^0_0 can't be negative either", said Oz.
Do you think that R^0_0 was negative for that puff of smoke, Ed?"

"Ack!" says the wizard, swatting madly.  "It's these damn signs.  It's
impossible to think when they're around."

[To be continued...Don't forget, to read the full adventures of Oz and the Wizard, together
with their whacky sidekick erg, tune in to

http://math.ucr.edu/home/baez/gr/gr.html

This also has the course outline and other handy documents to help you
learn general relativity.]

------------
and the latest (dunno if it made it over there yet?)

-------------------
)"Ack!" says the wizard, swatting madly.  "It's these damn signs.  It's
)impossible to think when they're around."
)
)[To be continued...

"I hate those signs too", said the smoke.  "I've found that raising
and lowering indices serves as a breeding ground for them, which is
why I suggest this process be avoided whenever possible!!.
That's why I recommend using T^a_b, G^a_b and R^a_b"

I am now getting:

T^a_b =

-E 0 0 0
 0 P 0 0
 0 0 P 0
 0 0 0 P

i.e. T^0_0 = -E, T^i_i = P (pressure) for i=1,2,3

Furthermore, I find that R = R^c_c = -Tc_c

i.e. R= -(T^0_0 + T^1_1 + T^2_2 + T^3_3 = E-3P

Since R^a_b = T^a_b + .5 R g^a_b  . (g^a_b = 1 iff a=b, 0 otherwise)

R^0_0 = .5^(T^0_0 - T^1_1 - T^2_2 - T^3_3) = .5(E+3P)

"This indicates that P makes things contract! (Or at least go in the
same direction as E.)   This seems weird, but" the cloud turns slightly
pinkish, "we are talking about d^2 V/dt^2, not dV/dt.  It's obvous that
dV/dt is positive when a high pressure area is in empty space, but that
says nothing about d^2V/dt^2"

"Something else that is interesting.  If we take a simple particle with
no pressure term, the stress energy tensor is"

T^0_0 = -E

all other terms are zero.

"If we take this and boost half of it in the -x direction at some velocity
v, (energy E/2) and boost half of it in the +x direction at the same
velocity v so that the average momentum/velocity is zero and add, we get"

T^0_0 = -gamma^2E
T^1_1 = v^2gamma^2E

where gamma = 1/sqrt(1-v^2) [c assumed to be 1]  

"The other terms all cancel out."  The cloud swirls as it attempts to cross
it's nonexistent fingers as it makes this remark.

"This gives us some insight into the pressure term, the above anisotropic
case has a positive pressure in the x direction."

"It seems odd at first that T^0_0 scales as gamma^2, but it's 
energy / volume, not energy, and volume is defintely not Lorentz
invariant!"

From galaxy.ucr.edu!library.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!rochester!udel!news.mathworks.com!newsfeed.internetmci.com!solaris.cc.vt.edu!csugrad.cs.vt.edu!csugrad.cs.vt.edu!not-for-mail Tue Mar 12 20:06:56 PST 1996
Article: 105054 of sci.physics
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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 12 Mar 1996 02:28:31 -0500
Organization: Virginia Polytechnic Institute and State University
Lines: 82
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
Reply-To: [email-masked]

In article <[email-masked]>, [email-masked] (john baez) wrote:

> Hmm, errr, well, ...

followed by:

> ...hmmm, well, ...
 
Aww, just come out and say I don't know what I'm talking about.  :)

Basically, I knew that pushforwards and pullbacks could be used to turn
things around.. and thought that if you took their definitions and
replaced every occurence of a smooth map with a linear transformation,
and every manifold with a vector space, then you might end up with
something (conveniently also called "pushforwards" and "pullbacks", of
course) that might do what you were actually doing with adjoints.

Guess I was wrong.  :)

[...]

> spaces, we get a linear map, the ADJOINT, L*: W* -> V*, given by
 
> L*(f)(v) = f(L*v).
 
I'm going to assume you meant:

  L*(f)(v) = f(Lv).

since that's the only thing that makes sense to me in this context.

[some adjoint stuff]

Okay, I follow it so far.

> I.e., things no longer get twisted around.  This is handy when you are
> talking about group representations like I was above.  In
> the situation I was talking about above, I had a representation of the
> Lorentz group on vectors.  In other words, for any vector v and any
> Lorentz group element L, I get a vector Lv, and
 
> (LM)(v) = L(M(v)).
 
Hmm, okay, wait a minute.  Why, above, were you talking about a
representation of the Lorentz group on vectors?  Why are we dragging
representations into this?  A representation is just a homomorphism
from the group to linear transformations on the vector space, but
elements of the Lorentz group already _are_ linear transformations on
the vector space. (Aren't they?)  It seems to me that the only
representation you'd want to use here is the identity map (or at least,
an isomorphism).

> That's what representations satisfy.  Then I wanted to get a
> representation on covectors.  If I just used L*, I would have gotten 
 
> (LM)*(v) = M*(L*(v)).
 
> So this wouldn't have been representation... it would be some other
> thing called a "right representation"!  No big deal, but a minor
> nuisance.  So I corrected it, the way people typically do, by using
> (L^{-1})*.  

What are the consequences of trying to use a right representation where
you should be using a representation?  If I understood why we're using
representations in the first place, I guess I'd be able to figure that
out..
 
> >It's easy for me to see that L is a linear transformation on
> >V, but not as easy for me to see why what you wrote is the most
> >natural, canonical way of turning L into a linear transformation on V*.
 
> Well, as you see, it's not!  What I wrote is only applicable to the case
> where L is invertible, and there is another very important thing to use
> when L is not necessarily invertible.
 
Ah, now you're just taunting me!  Okay, what is this very important
thing called, so I can at least look it up?
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
--------------------------------------------------------------------------

From galaxy.ucr.edu!library.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!rochester!udel!news.mathworks.com!newsfeed.internetmci.com!solaris.cc.vt.edu!csugrad.cs.vt.edu!csugrad.cs.vt.edu!not-for-mail Tue Mar 12 20:19:24 PST 1996
Article: 105054 of sci.physics
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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 12 Mar 1996 02:28:31 -0500
Organization: Virginia Polytechnic Institute and State University
Lines: 82
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
Reply-To: [email-masked]

In article <[email-masked]>, [email-masked] (john baez) wrote:

> Hmm, errr, well, ...

followed by:

> ...hmmm, well, ...
 
Aww, just come out and say I don't know what I'm talking about.  :)

Basically, I knew that pushforwards and pullbacks could be used to turn
things around.. and thought that if you took their definitions and
replaced every occurence of a smooth map with a linear transformation,
and every manifold with a vector space, then you might end up with
something (conveniently also called "pushforwards" and "pullbacks", of
course) that might do what you were actually doing with adjoints.

Guess I was wrong.  :)

[...]

> spaces, we get a linear map, the ADJOINT, L*: W* -> V*, given by
 
> L*(f)(v) = f(L*v).
 
I'm going to assume you meant:

  L*(f)(v) = f(Lv).

since that's the only thing that makes sense to me in this context.

[some adjoint stuff]

Okay, I follow it so far.

> I.e., things no longer get twisted around.  This is handy when you are
> talking about group representations like I was above.  In
> the situation I was talking about above, I had a representation of the
> Lorentz group on vectors.  In other words, for any vector v and any
> Lorentz group element L, I get a vector Lv, and
 
> (LM)(v) = L(M(v)).
 
Hmm, okay, wait a minute.  Why, above, were you talking about a
representation of the Lorentz group on vectors?  Why are we dragging
representations into this?  A representation is just a homomorphism
from the group to linear transformations on the vector space, but
elements of the Lorentz group already _are_ linear transformations on
the vector space. (Aren't they?)  It seems to me that the only
representation you'd want to use here is the identity map (or at least,
an isomorphism).

> That's what representations satisfy.  Then I wanted to get a
> representation on covectors.  If I just used L*, I would have gotten 
 
> (LM)*(v) = M*(L*(v)).
 
> So this wouldn't have been representation... it would be some other
> thing called a "right representation"!  No big deal, but a minor
> nuisance.  So I corrected it, the way people typically do, by using
> (L^{-1})*.  

What are the consequences of trying to use a right representation where
you should be using a representation?  If I understood why we're using
representations in the first place, I guess I'd be able to figure that
out..
 
> >It's easy for me to see that L is a linear transformation on
> >V, but not as easy for me to see why what you wrote is the most
> >natural, canonical way of turning L into a linear transformation on V*.
 
> Well, as you see, it's not!  What I wrote is only applicable to the case
> where L is invertible, and there is another very important thing to use
> when L is not necessarily invertible.
 
Ah, now you're just taunting me!  Okay, what is this very important
thing called, so I can at least look it up?
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
--------------------------------------------------------------------------

From galaxy.ucr.edu!not-for-mail Tue Mar 12 21:00:54 PST 1996
Article: 105136 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 12 Mar 1996 20:06:50 -0800
Organization: University of California, Riverside
Lines: 131
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, [email-masked] (john baez) wrote:

>>... we get a linear map, the ADJOINT, L*: W* -> V*, given by

>> L*(f)(v) = f(L*v).

>I'm going to assume you meant:

>  L*(f)(v) = f(Lv).

>since that's the only thing that makes sense to me in this context.

Yes, sorry, an egregious typo!

>Hmm, okay, wait a minute.  Why, above, were you talking about a
>representation of the Lorentz group on vectors?  Why are we dragging
>representations into this?  A representation is just a homomorphism
>from the group to linear transformations on the vector space, but
>elements of the Lorentz group already _are_ linear transformations on
>the vector space. (Aren't they?)  It seems to me that the only
>representation you'd want to use here is the identity map (or at least,
>an isomorphism).

Well, remember what were trying to show: that all index-juggling
operations on tensors (contraction, index-raising, index-lowering, etc.)
were Lorentz invariant.  But we don't know what "Lorentz-invariant"
means until we know how the Lorentz group acts on the tensors in
question!  That's why we are talking about all sorts of representations
of the Lorentz group: its representations on tensors.

More precisely:

You are right that the Lorentz group has a very obvious and important
"identity representation", namely the representation on vectors.  But it
also has lots of other important representations, namely its
representations on tensors of various sorts.  Indeed, for any j and k,
the Lorentz group has a representation on the space of tensors of rank
(j,k).  Let's use [j,k] to denote the space of tensors of rank (j,k).  

Index-juggling operations are linear maps between these various spaces.
For example: contraction is a linear operator from [j,k] to [j-1,k-1].
Actually there are *lots* of contraction operators like this, depending
on which pair of indices we contract.  But just pick one and call it

                     F: [j,k] -> [j-1,k-1].

Okay, so what does it mean for this operation to be Lorentz-invariant?
It means that

                     F(Lx) = L(Fx)

for all x in [j,k] and all L in the Lorentz group.  In this equation,
over on the left we are using the representation of the Lorentz group on
[j,k], and over on the right we are using its representation on
[j-1,k-1].

So we need to talk about all these representations!

And that's what I was doing, when I proved that all index-juggling
maneuvers were Lorentz-invariant.

>What are the consequences of trying to use a right representation where
>you should be using a representation?  

Oh, endless minor nuisances like getting g^{-1} where you wanted to get
g, or getting gh where you wanted to get hg.   

>> >It's easy for me to see that L is a linear transformation on
>> >V, but not as easy for me to see why what you wrote is the most
>> >natural, canonical way of turning L into a linear transformation on V*.

>> Well, as you see, it's not!  What I wrote is only applicable to the case
>> where L is invertible, and there is another very important thing to use
>> when L is not necessarily invertible.

>Ah, now you're just taunting me!  Okay, what is this very important
>thing called, so I can at least look it up?

It's the "pullback" or "adjoint" that you know and love.  But I wasn't
taunting you: I explained it in the very post you are replying to.  Let
me repost a portion, and you'll see that I first said what to do when L
is not invertible, and then what you can do when it is:

-------------------------------------------------------------------------

Let me explain the general picture.  If we have any linear map L: V -> W
between vector spaces, we get a linear map, the ADJOINT, L*: W* -> V*,
given by

L*(f)(v) = f(L*v).

Note that here L doesn't need to be invertible.  Now there is a
ever-so-slightly annoying thing about adjoints, namely that they "go the
other way".  This implies that

(LM)* = M*L*

whenever either side makes sense (namely, whenever M: U -> V is some
other linear map).  So things get backwardized.  It's no big deal,
just a minor nuisance at times.  

If we are dealing with invertible linear maps we can debackwardize
things using their inverses.  If L: V -> W is invertible, then we get
(L^{-1})*: V* -> W*.  So we get

(L^{-1})* (M^{-1})* = ((LM)^{-1})*

I.e., things no longer get twisted around.  This is handy when you are
talking about group representations like I was above.

[...]

Now say you're studying tangent vectors and cotangent vectors on
manifolds.  Say you have a smooth map f: M -> N between manifolds.
Then for any x in M you get a linear map

df: T_x M -> T_{f(x)} N

called the pushforward.  This linear map might not be invertible; if
it's not, we can't use that trick above to "debackwardize" things, so we
just use the pullback [or "adjoint"]

(df)*: T*_{f(x)} N -> T*_x M

and accept the fact that

(d (fg))* = (dg)* (df)*.

From galaxy.ucr.edu!library.ucla.edu!newsfeed.internetmci.com!news.mathworks.com!zombie.ncsc.mil!admaix.sunydutchess.edu!ub!dsinc!netnews.upenn.edu!msunews!agate!physics12.Berkeley.EDU!ted Tue Mar 12 21:56:56 PST 1996
Article: 104961 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Followup-To: sci.physics
Date: 12 Mar 1996 05:17:29 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 17
Sender: [email-masked]
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
Reply-To: [email-masked]
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>I am utterly confident that Ted knows!

Sure do.  But I think you're supposed to figure it out for yourself.
I'm not about to risk Baez's wrath by telling you the answer.  For
what it's worth, you're getting very close.

You've already established that in the absence of pressure a ball of
dust will collapse in on itself.  All you have to do now is explain
why, once you've gotten above some threshold of density, even pressure
can't stop further collapse.

Uh oh.  I hope I haven't said too much.  I'd just as soon avoid a
thunderbolt.

-Ted

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!newsserver.jvnc.net!raffles.technet.sg!nuscc.nus.sg!matmcinn Tue Mar 12 22:29:41 PST 1996
Article: 105170 of sci.physics
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From: [email-masked] (Brett McInnes)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 6 Mar 1996 05:19:45 GMT
Organization: National University of Singapore
Lines: 31
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: [email-masked]
X-Newsreader: TIN [version 1.2 PL2]

john baez ([email-masked]) wrote:
: Sorry, I should have given you a sneakier hint and let you have more of
: the fun of figuring it out.  Yes, if a little sphere of initially
: comoving particles in free fall turned into an ellipsoid, there would
: necessarily be some preferred directions (the axes of the ellipsoid),
: hence anisotropy.  

True, but we are only talking about spacelike directions here. 
What about the timelike components? No doubt this is an easy 
consequence of the symmetries of the Weyl tensor.

This is an interesting way of looking at the FRW metric! If indeed
one can get conformal flatness directly from local isotropy, then
the foliation of the spacetime [ie the splitting into spacelike
slices] follows immediately and doesn't have to be assumed [or 
derived in a tricky way].

: So you have worked out the Weyl curvature of the big bang universe and
: also the Ricci curvature --- given the energy density E and the pressure
: P.  No heavy-duty calculations, just symmetry!  Cool, huh?

: 2.  Explain how, in the standard big bang model, where the universe is
: homogeneous and isotropic 

I can never understand why people say "homogeneous AND isotropic", 
especially when there is so little evidence for homogeneity. Better
to say [locally] isotropic, since that is what we see. ["locally",
meaning that given any point and any pair of spacelike unit vectors
there, there exists an open set and an isometry of that set mapping
either vector to the other. Of course, there is no justification 
for assuming global isotropy.]

From galaxy.ucr.edu!not-for-mail Wed Mar 13 00:09:13 PST 1996
Article: 105177 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 12 Mar 1996 22:29:34 -0800
Organization: University of California, Riverside
Lines: 36
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Brett McInnes) writes:
>john baez ([email-masked]) wrote:
>: Sorry, I should have given you a sneakier hint and let you have more of
>: the fun of figuring it out.  Yes, if a little sphere of initially
>: comoving particles in free fall turned into an ellipsoid, there would
>: necessarily be some preferred directions (the axes of the ellipsoid),
>: hence anisotropy.  

>True, but we are only talking about spacelike directions here. 
>What about the timelike components? 

Yes... I overlooked this at first but then realized this fly in the
ointment.  In one of my G. Wiz posts I hinted darkly at this problem...
but I never yet got around to figuring out its resolution.

>No doubt this is an easy 
>consequence of the symmetries of the Weyl tensor.

I wish it were.  I'm a bit nervous about that, though.  We can recover
the Weyl tensor if we know how balls of dust moving along at arbitrary
velocity change shape into ellipsoids (in their own rest frame).  But we
can only apply the isotropy argument to balls of dust which are at rest
in the "cosmic rest frame" of the big bang.  I think we only get 5
components of the Weyl tensor to vanish by this argument.  (There are 
5 linearly independent 3x3 symmetric traceless matrices.)  5 out of 10
ain't bad... but it doesn't look like we'll be able to get all 10
components of the Weyl to vanish without some extra reasoning.

Can anyone think of a way to fix my argument?

>This is an interesting way of looking at the FRW metric! If indeed
>one can get conformal flatness directly from local isotropy [....]

If!

From [email-masked] Wed Mar 13 01:00 PST 1996
Date: Wed, 13 Mar 1996 08:23:13 +0000
To: john baez <[email-masked]>
From: Oz <[email-masked]>
Reply-To: [email-masked]
Subject: Re: Defeat
Content-Length: 4025
Status: OR

In message <[email-masked]>, john baez
<[email-masked]> writes
>> I think Merritt would like to continue on. He told me he had kept off
>> because he wasn't invited. I said I thought you would prefer more people
>> on the thread anyway. Once I bow out, maybe more people will come in? I
>> will try very hard to keep quiet, but ....
>
>I told him to join in.  He posted something which gives away 99% of the
>answer.  You are really trying to make much more out of this black hole
>problem than it was meant to be.  

Yeah. I get this feeling too. But what???

>> I sometimes wonder if we should have 'done' a simple example first, like
>> a sattelite orbitting a massive body for example. Probably a mite
>> trivial from your lofty position but .......
>
>No, that problem is much harder than what I'm asking you to do.

I believe you.

>> I am beginning to get the really horrible feeling that the answer Wiz
>> wants is so straightforward that one almost wouldn't deign to mention
>> it. 
>
>Maybe you wouldn't deign to mention it... you haven't so far!  :-)

Well, I tried all the 'so obvious it's not worth the mention' one way
and another. Quite a few in my final post.

>
>> Either that or we are looking in the wrong direction. How about the
>> other terms. John has studiously avoided even making yes-no comments to
>> a fair number of the posts on the thread, so it's a mite hard to see
>> where to go without positive or negative reinforcement. We derived
>> R_{11} for the BB, at least I think we did, I don't know that anyone
>> said if it was right or not. Did it tend to zero, or -infinity? Nobody
>> said either way.
>
>Sure, you derived it, I said it was okay, and then I asked you what it
>did as we went to the BB....

Yes, but the big delays that occur (your posting came way after your
email) maybe confuse me too. Perhaps I ought to collate some then you
can go down the list saying "yes", "no", "stupid", "sort of", "not even
wrong".

On the other hand maybe my memory is failing.

On the other hand maybe my memory is failing.

On the other hand ..........

>> So what do we know about singularities, or more important (I suppose)
>> things heading that way? Well, I suppose this E+3P should have something
>> to do with it. The only thing that springs to mind is that the more E &
>> P you keep shoving into a volume of space to get it to prevent collapse,
>> the more curvature you produce that needs more P to resist it but this
>> produces more curvature. This is trivial and I've said it before. 
>
>That's the answer!  I don't remember you saying it --- are you
>sure this is not another of the "unposted works of Oz"?

AAaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrggggggggggggghhhhhhhhhhhhhh!!!!!!!
NNNNNNNNNnnnnnnnnnnnoooooooooooooooooooooooo, no, no, no, no, no!

No, it can't be. Tell me it's not true. <Oz reaches for shotgun>

No, no. Surely you are just letting me down gently? 
You are just being humane. Aren't you?

Of course I have said it before. It's basic and obvious. I'll give you a
reference........

Well, I just looked back and I can find several places where I thought
of it, and I certainly implied it, but I don't think I ever actually
*said* it. I talked about almost all the energy being in P for the early
BB and how that would make E-3P negative and E+3P very positive and the
curvature finally infinite ....   ..     .  

Er, ho-hum, ooops.

Well, come on, it is pretty obvious isn't it? Isn't it? 

**Please, please, please can I come back on the course again?**

Actually I have had half a dozen emails from people who I have never
even seen on S.physics who have asked me to hurry up and answer the
question because they are enjoying the course. They mostly seem to be
undergraduates. I have the horrible feeling that I have been playing the
fool in front of an audience of 200 or so.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!ihnp4.ucsd.edu!agate!newsxfer2.itd.umich.edu!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Wed Mar 13 13:40:11 PST 1996
Article: 105204 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Wed, 13 Mar 1996 07:49:31 +0000
Organization: Oz
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It was night. Oz had pushed his final bunch of answers under the 
Wizards door. He had put down any old stuff from the 'blatantly 
obvious' to the 'not even wrong', but his confidence had cracked. 
There was only one thing left to be done and he went back to his 
cave to pack. He was beaten. He couldn't answer the Wizard's 
final question. He would have to be gone in the morning before 
the Wizard awoke.

Oz, his shoulders hunched despondently, slowly pushed his few 
meagre belongings into his old patched bag. Finally all that was 
left was the few tattered and worn scraps of paper he had made 
his notes on. He looked at them one by one. One by on he 
carefully and reverently placed them on the fire. Ah, well, what 
did it matter now. Slowly the pile grew smaller and the fire 
brighter till only one paper was left. It was the Curse Notes 
that had materialised out of the air long ago. On it he had 
written "Number Two". Ah, those hopeful enthusiastic days when 
there were no difficult questions. Oz sighed deeply. He went to 
throw them on the burning pyre, but couldn't bring himself to. 
With a nervous glance at his door he hurriedly thrust them to the 
bottom of his old bag and rose to go.

He walked slowly over the white quartzite flags and glanced for 
the last time with longing at the Wizards curtain. Now he would 
never have the faintest idea what wonders lay in the books behind 
it. He turned sadly round as he took his final look at the 
Wizards cave. Then he walked over to the entrance and looked out 
into the night, at the heavens covered in bright stars. The 
snowdrops had come out over the last day or so and made a cheery 
and springlike display by the cave, but Oz was too sorrowful to 
notice. With his shoulders hunched and with a doleful pace he set 
off for the village. He slowly faded into the darkness as he 
stepped out into the night, alone, away from the Wizards Keep to 
a future, who could tell.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!guitar!baez Wed Mar 13 18:12:15 PST 1996
Article: 105330 of sci.physics
Path: galaxy.ucr.edu!guitar!baez
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 14 Mar 1996 00:13:52 GMT
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>>No doubt this is an easy 
>>>consequence of the symmetries of the Weyl tensor.

>>I wish it were.  I'm a bit nervous about that, though.  We can recover
>>the Weyl tensor if we know how balls of dust moving along at arbitrary
>>velocity change shape into ellipsoids (in their own rest frame).  But we
>>can only apply the isotropy argument to balls of dust which are at rest
>>in the "cosmic rest frame" of the big bang.  I think we only get 5
>>components of the Weyl tensor to vanish by this argument.  (There are 
>>5 linearly independent 3x3 symmetric traceless matrices.)  5 out of 10
>>ain't bad... but it doesn't look like we'll be able to get all 10
>>components of the Weyl to vanish without some extra reasoning.

>Could anyone say what the physical significance of these five are?

It takes 5 numbers to describe the different ways a round ball can
change shape into an ellipsoid while keeping the same volume.  The slick
way to see this is to note the above parenthetical remark about
symmetric traceless matrices.  But we can be grungy: two numbers tell us
which direction the longest axis of the ellipsoid points, one more tells
how much that direction gets stretched, one more tells us where the
shortest axis points (just one needed since we know it's perpendicular
to the longest axis), and one more tells us how much that axis gets
scrunched.  (Since the volume of the ellipsoid equals the volume of the
ball, we can determine from all this how much the 3rd axis must be
stretched or scrunched.)

>Gravitational waves?

As a gravitational wave passes through empty space, an initially round
ball of dust in free fall will turn into an ellipsoid of the same
volume, and these 5 numbers describe exactly what the ellipsoid looks
like.

Gravitational waves are somewhat similar to electromagnetic waves.  
These 5 components are the called the "electric" part of the Weyl
curvature, but there are also 5 "magnetic" components.  Recall what
Steve Carlip said about this a while back.  I had said:

: In short, when we are in truly empty space, there's no Ricci curvature,
: so actually our ball of coffee grounds doesn't change volume (to
: first order, or second order, or whatever).  But there can be Weyl
: curvature due to gravitational waves, tidal forces, and the like.  
: Gravitational waves and tidal forces tend to stretch things out in one
: direction while squashing them in the other.  So these would correspond
: to our ball changing into an ellipsoid!  Just as we hoped.

And he replied:

Yes, this is at least mostly right.  (The "mostly" comes because
there's a part I don't understand---I'll explain below.)  Start
with your ball of coffee grounds, with an initial four-velocity
u^a.  (In the rest frame, u^0=1 and the rest of the components
are zero.)  If you look at the configuration slightly later, it
will typically have changed in three ways.  First, it may have
expanded or contracted (its volume may have changed).  Second,
it may have twisted.  Third, it may have sheared---that is, become
distorted from a sphere to an ellipsoid without changing volume.
The Weyl tensor (in empty space) gives the rate of change of the
shear.

Specifically, you can describe shear by a three by three symmetric
matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX
users).  The three eigenvectors of \sigma give three spatial axes,
and the eigenvalues give the rate of expansion along each axis.
The fact that \sigma is traceless means that the sum of the rates
of expansion is zero, i.e., the total volume is remaining constant.
In empty space, if your ball of coffee grounds has no initial shear,
rotation, or expansion, the rate of change of \sigma is given by the
Weyl tensor contracted twice with u.  (If you want that in symbols,
I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

This means that the Weyl tensor is the thing gravitational radiation
detectors are designed to measure.  A laser interferometer detector
like LIGO or VIRGO (now under construction) is just a very large, very
accurate interferometer with two perpendicular arms.  If the Weyl tensor
changes, one arm will expand while the other one contracts, changing the
relative lengths and thus the interference pattern.

Now, the part I don't understand.  The Weyl tensor really splits into
two parts, and "electric" part (with five components) and a "magnetic"
part (also with five components).  The contraction C(*,u,*,u) that I
described above is the electric part, in the rest frame of the coffee
grounds.  Like ordinary electric and magnetic fields, the electric and
magnetic components of the Weyl tensor mix under coordinate changes.
But it would be nice to have a direct geometrical picture of the magnetic
part.  Does anybody know one?  (Presumably it's hidden in what's called
the Newman-Penrose formalism, but I've never learned that very well.)

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!nntp.club.cc.cmu.edu!cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!scramble.lm.com!hookup!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Thu Mar 14 19:43:32 PST 1996
Article: 105454 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Thu, 14 Mar 1996 06:52:10 +0000
Organization: Oz
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In article <[email-masked]>, Doug Merritt
<[email-masked]> writes
>
>You are much closer than you seem to think, though. The first 2/3rds
>of John's questions above are simply probing you to state slightly
>more explicitly what you've already said. The final question is
>merely pointing out that you haven't said explicitly why pressure 
>doesn't prevent the black hole from forming...but you did say why
>*implicitly* (very implicitly :-), 

Yes. Well before I wrote my "Oz trudges off into the night" piece, I DID
shove an email 'under his door'. It had a selection of answers including
the trivial answer John required. So now it's up to the Wizard to catch
Oz before he reaches the village to tell him he's passed the test.

>I think you can answer all of John's remaining questions in a
>single paragraph that is just as glib as the other one. :-)
>
>Go for the obvious.

It wasn't so much obvious as trivial. Indeed I was absolutely certain
that I had already mentioned it several times. In a considerable huff I
went back over the postings to prove same. You are right. I never
explicitly stated it anywhere, I had simply assumed it as obvious. I
mean, when you say the curvature/volumetric acceleration goes as (E+3P)
several times, and say that most of the energy will go into P but that
doesn't matter ....  Oh, well. The Wizard likes his little tricks :-)

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!ihnp4.ucsd.edu!munnari.OZ.AU!news.hawaii.edu!ames!uhog.mit.edu!news.mathworks.com!newsfeed.internetmci.com!info.ucla.edu!csulb.edu!csus.edu!news.ucdavis.edu!dirac.ucdavis.edu!carlip Thu Mar 14 19:44:12 PST 1996
Article: 105572 of sci.physics
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From: [email-masked] (Steve Carlip)
Newsgroups: sci.physics
Subject: Re: Combined fields: E&M + gravi
Date: 14 Mar 1996 19:12:21 GMT
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Emory F. Bunn ([email-masked]) wrote:

: Those two differences between gravity and electromagnetism are
: actually related to each other.  If you try to form a theory of
: gravity in which the field equations are linear, you find that you
: can't do it in a self-consistent way.  For some reason, which I don't
: really understand, a theory that has as its source a rank-2 tensor and
: that reduces to a nice inverse-square law in the appropriate limit
: must be nonlinear.

There are a few ways of seeing why this is true.  Here's one:

In special relativity, the divergence of the stress-energy
tensor is zero---this is simply the statement of energy
conservation.  If you use the stress-energy tensor as the
source for linear field equations for a symmetric rank-two
tensor, this conservation law is no longer an extra assumption,
but follows from the field equations, in the same way that
current conservation follows from Maxwell's equations.  (It's
technically a consequence of gauge invariance and Noether's 
theorem.) 

But if the equations are linear, then by definition the stress-
energy tensor does not contain a term for gravitational energy.  
This means the theory predicts conservation of *nongravitational* 
energy.  If you now work out equations of motion for particles,
you find that they simply don't respond to a gravitational field.  
This is what you would expect---after all, the theory tells you 
that particles can't exchange energy with the gravitational field.  

Obviously this is not a very good model of nature.  The reason
is also obvious---the source term needs to include a piece that
describes gravitational energy.  This makes the equations nonlinear
---the gravitational field interacts with itself---and for internal
consistency you must also include the energy of this self-interaction
as a source of gravity.  The surprising result of Deser, Weinberg, 
and others is that once you begin this "bootstrap" process, you
are led almost uniquely to the field equations of general relativity.

Note, by the way, that if you run this argument backwards, the purely
linear approximation of general relativity is inconsistent, or at
least inconsistent with the equations of motion of gravitating bodies.
There's actually a published (!) paper that claims this as a disproof
of GR.  (At least it's in an obscure journal.)

It's also worth pointing out that there is direct observational
evidence that gravitational energy interacts gravitationally.  The
Earth and the Moon have different proportions of gravitational
binding energy, and if this energy didn't interact in the same
way that other forms of energy did, the Earth and Moon would fall
towards the Sun with slightly different accelerations.  Such an
effect (the Nordtvedt effect) has been looked for, and ruled out 
to a high degree of accuracy.

: >Also how do we know (experimentally)
: > that the source of gravitation is the stress-energy tensor rather than
: > the mass-momentum vector? TIA. Jim Graber  [email-masked]

: Well, mass density and momentum density don't form a vector in the
: same way that charge and current density do.  (Neither do energy
: density and momentum density).  They form part of the stress-energy
: tensor, which is rank 2.  That's just how those quantities transform
: under coordinate transformations.  If you want to make a manifestly
: coordinate-independent theory that has energy as its source, your
: theory is going to have to involve the stress-energy tensor, since
: that's the "covariant object" (so to speak) that energy is a part of.

You can try to define a vector-like theory of gravity, in which you use,
say, the rest mass current as a source.  But the result does not agree
with experiment.  In particular, it violates the equivalence principle
(since only rest mass, and not energy, acts as a source of gravity); it
does not predict the gravitational red shift; and it typically gives
the wrong perihelion precessions and light-bending results.  Furthermore,
to make gravity attractive in such a theory, you must require that the
gravitational field has negative energy, which (apart from the obvious
instabilities) would drastically disagree with binary pulsar observations.

You'll find a nice, though dated, description of some of the problems in
Robertson and Noonan, _Relativity and Cosmology_.

Steve Carlip
[email-masked]

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Article: 105333 of sci.physics
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From: [email-masked] (Bronis Vidugiris)
Subject: Re: Riemann Awakes
Organization: What - me organized?
Date: Wed, 13 Mar 1996 06:47:14 GMT
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In article <[email-masked]>, john baez <[email-masked]> wrote:

)"Wait a minute!" said G. Wiz.  "Your formula for R^0_0 agrees with the
)stuff in the course outline, and you are right that it means the
)gravitational effect of pressure is to cause things to contract.  In
)layman's lingo: like energy, pressure causes an attractive gravitational
)force."

...

)"BUT," added G. Wiz, "the formula relating R^0_0 to d^2V/dt^2 only applies
)to a ball of particle following geodesics!  That is, particles in free
)fall, feeling no force other than gravity (which, ahem, is not really a
)force either).  In reality, when we have a gas with different pressures
)at different regions, the atoms of the gas *do* feel forces other than
)gravity.  So pressure does have the usual effect of making stuff want to
)expand, in addition to the special GR effect of making it want to
)contract.  Just don't mix them up; they are very different."

"Hmmm - well, we need a basic eq. to relate pressure back to d^2V/dt^2
then, I guess.  The only one I can think of is dE = Pressure dV"
mused the cloud.  "But what about the dt^2?"

"I wish I had my old blue spellbooks handy, written by Haliday and Resnick.
I left them unattended too long, and the magic in them seems to have caused
them to vanish".

"It certainly seems that as the volume of the ideal gas cloud contracts,
both the energy and the pressure will go up, sort of a 'double whammy'.
But I can't quite seem to see how to handle the inconvenient dt^2 to
get a numerical relationship, it seems to me that the relation between
energy and pressure is dE/dt = pressure dV/dt, there's a dt factor missing
somewhere."

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Article: 105813 of sci.physics
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From: [email-masked] (Chris Hecker)
Subject: Re: Tensors for Twits (Redux)
Message-ID: <[email-masked]>
Organization: NETCOM On-line Communication Services (408 261-4700 guest)
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[email-masked] (john baez) writes:
>In lower-level math and physics courses the distinction between V and V*
>is often obscured because the vector spaces one plays with come equipped
>with a "dot product".

I think I need to realize that a vector space doesn't automatically
have a dot product.  My understanding is that a vector space has addition
and multiplication by a scalar, but since I learned linear algebra
through computer graphics, I think I added a dot product in the mix as
well.

I'm still slightly confused though.  First, a dot produt and an inner
product are the same thing, right?  So, when I've been learning tensors,
I've thought of this:

r = v_i u^i

as an inner product, but I think that's the wrong way of thinking about
it.  Would it be more correct to think of it as a plain old tensor
contraction (is that the right word?), and in Euclidean R^3 it happens
to be a dot product (because there's no difference between vectors and
covectors?

Basically, I think I need to get it through my head that r = v_i u_i is
simply not defined for most vector (tensor?) spaces, right?  It happens
to be definied for Euclidean R^3 as the dot/inner product.

>When one is careful to keep track of the difference between vector
>spaces and their duals, various things become clearer.  If you think of
>position as living in V then momentum lives in V*.

I saw people talking about this before, but I don't understand how
momentum and position are duals given my newfound understanding of
duality.  Was this covered in another thread (I luckily happened upon
this tensor thread but sci.physics gets hundreds of articles a day, so
I'm a bit wary of exploring)?  Since momentum is mv (in my phyics
books, at least), wouldn't that put v and mv in the same vector space
(since m is a scalar) and so v would be in the dual space of position
as well?  I'm all confused.

>It really depends what you want to do.  Certainly you need the
>full-fledged tensor calculus to really understand special relativity and
>(even more so) general relativity.

My next big thing after I've nailed rigid body mechanics is fields and
deformable body mechanics and maybe some fluid dynamics...basically
non-rigid stuff.  Are the tensors still Cartesian here?

Chris

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!csulb.edu!csus.edu!news.ucdavis.edu!dirac.ucdavis.edu!carlip Fri Mar 15 19:51:56 PST 1996
Article: 105820 of sci.physics
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From: [email-masked] (Steve Carlip)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 16 Mar 1996 02:40:09 GMT
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I have a partial answer to the question I raised about
the Weyl tensor a while ago.  I had written

: Now, the part I don't understand.  The Weyl tensor really splits into
: two parts, and "electric" part (with five components) and a "magnetic"
: part (also with five components).  The contraction C(*,u,*,u) that I
: described above is the electric part, in the rest frame of the coffee
: grounds.  Like ordinary electric and magnetic fields, the electric and
: magnetic components of the Weyl tensor mix under coordinate changes.
: But it would be nice to have a direct geometrical picture of the magnetic
:part.

The term "magnetic" has two hints in it.  Remember that in
ordinary electromagnetism,

1) To detect a magnetic field, you need a moving particle
---uniform motion will do---or a current.  The force exerted
by a magnetic field is proportional to the velocity, F=qvxB.

2) According to special relativity, on the other hand, there's
no absolute meaning to the phrase "uniform motion."  A Lorentz
transformation can take a uniformly moving particle that is 
experiencing a magnetic field, and transform it to a particle
"at rest," which can't feel a magnetic field.  Electromagnetism
cleverly escapes this problem by allowing electric and magnetic 
fields to mix under Lorentz transformations; the force that
looks "magnetic" in one frame will be "electric" in another.

Now let's turn back to gravity, and look at John's "sphere of
coffee grounds" a little more carefully.  John wants his sphere 
to start off at rest, and use five components of the Weyl tensor 
to describe its subsequent distortion.  But this is cheating---
there is no absolute meaning to "at rest," even locally.  In a 
small enough region, physics in a freely falling reference frame 
in general relativity looks like the physics of special relativity.
So just as in special relativity, there are an infinite number of
local "inertial frames" moving at constant velocities relative
to one another, any one of which is as good as any other.  The
five "electric" components of the Weyl tensor in one frame will
look like mixtures of the "electric" and "magnetic" components
in another.  So by considering a collection of spheres of coffee
grounds, moving relative to one another at constant initial
velocities, you should be able to pick up the whole tensor.

A colleague recommended a paper by Szekeres, "The Gravitational
Compass" (J. Math. Phys. 6 (1965) 1387), which helped me think
about this.  It's a little terse, but it's worth a look.

Steve Carlip
[email-masked]

Newsgroups: sci.physics
Subject: General relativity tutorial - Weyl tensor
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In article <[email-masked]>, Edward Green
<[email-masked]> writes

>>I bet you wish you had thought of having some people nailed to the Tree of
>>Woe in the General Relativity Tutorial sometimes... 

Actually, they usually nail themselves to the tree of woe and leave me
to pry them off.

Anyway, I'm briefly back on the net before another week away, and I have
a bit to say about the Weyl tensor.  I spent last week at the CGPG
(Center for Gravitational Physics and Geometry) at Penn State.  Luckily,
Roger Penrose was visiting; he has a part-time position there.  He's an
expert on Weyl curvature and the "Weyl spinor", which is something folks
here have been advising me to learn about.

I told him I was trying to explain general relativity in very simple
terms and this had forced me to attempt to actually understand the
subject.  I mentioned the "Ricci tensor causes change in total volume of
a ball of comoving particles in freefall; Weyl tensor causes change in
its shape" idea.  He noted that while the first part was true, the
change in shape is due NOT ONLY to Weyl curvature, but also to Ricci.
As he put it: if a fluid is flowing by, the Ricci curvature it causes
will make a ball of comoving particles in freefall distort into an
ellipsoid.

He also said that this annoying nuance goes away if we consider
particles moving along *lightlike* geodesics... for example, light.
Lightlike geodesics introduce some other nuances, though: we should
consider not a 3d ball of light at time zero, but a 2d disc that is
orthogonal to the direction the light is moving.  The mathematical
details here are a wee bit subtle, but imagine turning on a flashlight
for a really really brief moment; such a disc of light would be emitted.
Then, as time passes, the disc of light will begin to distort into an
ellipse.  The rate of distortion (more precisely, the 2nd derivative) is
determined by the Weyl tensor alone.  No Ricci stuff.

I told him about my attempt to prove that the Weyl curvature of the big
bang cosmology was zero.  Remember, this "proof" used only the spatial
isotropy of the big bang cosmology; the idea was that by symmetry, no
sphere could turn into an ellipsoid.  The original "proof" considered a
ball of freely falling particles which were initially at rest.  But this
only showed that 5 of the 10 components of the Weyl tensor were zero:
the so-called "electric" components.  I asked him if this argument could
be saved.  He said: if we shine a disc of light in any direction, we
know by isotropy that it cannot get squashed into an ellipsoid, so we
know certain components of the Weyl tensor are zero.  Since we can shine
the light in *any* direction, we get a lot of components this way ---
and in fact, we can conclude that the Weyl tensor is zero! 

So it's true that in any spatially isotropic cosmology the Weyl tensor
is zero, just by a symmetry argument.

He also told me a bunch of stuff about the "Weyl spinor", which is an
mathematically elegant way of encoding this information about what the
Weyl curvature does to discs of light... involving the fact that we can
think of the disc of light as a piece of the complex plane.  This is
part of a big framework he developed for studying general relativity
using spinors and complex analysis.  But I won't go into this now...
folks can read Penrose and Rindler's book Spinors and Space-Time for
this stuff.

From galaxy.ucr.edu!not-for-mail Thu Apr  4 20:10:15 PST 1996
Article: 109496 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: GR tutorial ?
Date: 4 Apr 1996 11:40:39 -0800
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Oz complains: "The poor bloody apprentice is getting real fed up with
sweeping the floors. He wants some action! Time to make a Dutchman
vanesch from one world and bring him to a parallel dimension of wizards,
trolls and things."

"Else he is going to go behind the poxy curtain and torch all the Wizards
mighty tomes and kick the black holes out of their holders and see if
the really DO eat up the world."

Someone knocks at the door.  Oz blanches and says .... "come in?"

The wizard leans in.  "Watch it, pal.  You should be busily reviewing
the stuff you know, so that when I start explaining how you actually
solve Einstein's equation you'll know what I'm talking about.  Right now
I have to prepare for a SERIOUS general relativity class I'm teaching...
curiously, we are just about to do black holes and the big bang.  Later
today I might get around to satisfying your thirst for knowledge.  But
I've been a bit distracted lately, trying to finish off a paper on
2-Hilbert spaces.  Don't get the impression that you're my ONLY duty."

He slams the door.  

Oz says "Humpf!"

"Now where was that broom?"

Newsgroups: sci.physics
Subject: General relativity tutorial
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Okay, where were we?

Perhaps a few scenes from previous episodes will help remind you, while
we run the credits: 

.... He walked slowly over the white quartzite flags and glanced for the last
time with longing at the Wizard's curtain. Now he would never have the
faintest idea what wonders lay in the books behind it. He turned sadly
round as he took his final look at the Wizards cave. Then he walked over
to the entrance and looked out into the night, at the heavens covered in
bright stars. The snowdrops had come out over the last day or so and
made a cheery and springlike display by the cave, but Oz was too
sorrowful to notice. With his shoulders hunched and with a doleful pace
he set off for the village. He slowly faded into the darknes s as he
stepped out into the night, alone, away from the Wizard's Keep to a
future, who could tell....

.... The Wizard read: "Anyway. I admit defeat. I have stopped learning.
The fun has gone. I preferred tensors. Great disappointment. I am
leaving this night."

"WHAT!" shouted the wizard. "Leaving? The fool! Leaving!? How can he
have given up just when he figured out the right answer?" He cursed
loudly and strode back to the bedroom, hurriedly tossing on his cape and
putting on his hat. He would have to make good time to catch the fellow.
Had he been too hard on Oz?  Not enough encouragement?  Or was Oz simply
too hard on himself? He'd made so much progress in such a short time,
and here he was giving up in defeat at his very moment of triumph!  "The
absolute ninny! Just when I was getting used to having him around!"

Picking up his staff, he rushed out....

...."Ooo my, but you're ever so cold," she said.  "Perhaps I should warm
you up a bit. I don't suppose you have anywhere to stay, fancy staying
in my rooms for the night?"  

Oz very nearly choked on his pastie. He wondered if perhaps he had been
a bit remiss in leaving the Wizard so precipitously. Obviously GR was
powerful magic indeed. On the other hand the Wizard's wrath was a
fearsome thing too, and he felt very comfortable where he was. Well,
very comfortable indeed, actually.

"Why, that's uncommonly kind of you," said Oz "I would be delighted to." 

"We'll soon have you warmed up then," said Rosie "I think you will enjoy it." 

At that minute the Inn doors burst open. In flew the Wizard (literally,
his feet were a foot off the floor). He whisked off his goggles and
turned off his navigation lights. He turned to Oz and said "You did it.
You answered the questions. You passed, now you can come straight back
and I'll show you how I calculate the Riemann tensor, isn't that good?" 

At that particular moment Oz didn't think it was good at all. In fact it
was the last thing on his mind. He wished the Wizard had better timing.

"Ooo" said Rosie "Are you going to be a Wizard?" 

Oz's brow furrowed. What an impossible decision. To go, or to stay? A
night with Rosie or Riemann tensors.  Well, what would you do? (NB
Please don't answer this). With a heavy heart Oz to ok the only possible
choice.  It's best to humour Wizards.

"Why thank you, Great Wizard." he said more brightly than he felt "I
wonder if you could give me a lift?" (NB Derisory boos from off left)

"Ooo" said Rosie "I can't wait for you to come back again." 

But Oz hadn't heard, the Wizard had flicked up his goggles, turned on
his navigation lights, grabbed Oz round the waist and shot out through
the door towards his Keep. It would have been more impressive if he had
opened the door first.

On the way back, Oz decided that learning was a difficult thing, and
required many sacrifices. Somehow he didn't think the Wizard really
appreciated this.

Oh, and it was damned cold at 10,000 ft and 800KPH....

-------------------------------------------------------------------------

When they reached the Wizard's keep, the Wiz invited Oz to his room for
a cup of coffee.  The sun was beginning to rise, and Oz felt rather
exhausted and chilled.  Luckily the fire was already lit, and the water
soon began to boil.  "I already had a cup," said the Wiz.  "But I could
use another.  That ridiculous episode you just wrote, where you had us
flying through the air at an ungodly speed, was rather wearing."

Oz smiled and said nothing.  They both knew, of course, that they were
merely fictional characters of their own devising, but it wasn't wise to
dwell on it too much.   

Grinding some coffee, the Wiz smiled at Oz.  "You seemed to be hitting
it off pretty well with Rosie."  

Oz blushed.  "Umm, yes, and the strange thing is, I'm not quite sure
why.  Some sort of... magic?"  

The Wizard raised his eyebrows.  "Don't you know?  It's the general
relativity."  

Oz had heard this, but it was hard to believe.  "Really??  How does it
work?"  

"Well, it's simple really.  Rosie was one of my students, once upon a
time.  But you know how hard it is to get jobs in physics.  So she took
up bartending.  She knows a lot more than she lets on - that manner of
hers, she just does that to get good tips.  I'm sure she'd love nothing
more than a good intimate chat about the symmetries of the Riemann
tensor.  But most of the men down there are uneducated oafs.  Well, what
do you expect?  It's just a typical village.  Anyway, she *knows* you've
been learning some general relativity... rumors travel fast here... so
I'm sure she's been dying to meet you."

Oz didn't know what to say.

"Here, have some coffee.  Anyway," said the Wiz, "To please her you're
going to need to learn a lot more about tensors than you have so far!"

Oz frowned.  So Rosie was a physicist?  It seemed hard to believe for
some reason.  Could the Wiz be saying this just to motivate him to
study?  Still, Rosie *had* seemed to fall for him awfully easily... he
couldn't think of any other explanation.

Unconcerned with Oz's musings, G. Wiz continued, "So: you wanted to look
behind the curtain and see how to compute the Riemann curvature of a
metric.  Let's do it!  It's not going to be easy.  Gird yourself.  It
will take a while of trying, but let's get started."  He strolled over
towards the curtain, and then stopped dead in his tracks.

"Hmm," said the Wiz.  "Wait a minute.  This calls for some ominous
music."  He clapped his hands and a waft of low, dissonant cello music 
drifted over from the back room.  A cloud drifted over the sun, and it
began to rain. 

"Now, before I let back there," said the Wiz, "You must promise not to
reveal what you learn to the uninitiated.  This knowledge is DANGEROUS."
A flash of lightening and a thunderclap served to accentuate his
warning.

"Okay, sure, I promise," said Oz, eager to finally see what was back
there.

"Hmm," said the Wizard.  "You don't seem to be taking this sufficiently
seriously.  But it's been a long night.  I'll just show you some
equations now, and then we'll dig into them more deeply later."  He
pulled back the curtain, and the cello music worked its way to a
crescendo.  "Come on in!"

Oz stepped in and saw to his surprise that the back room was not too
different from the room they had just been in!  True, there was a large
tome resting on a podium, with six-foot-tall black candles on either
side.  But besides that, nothing much... a couple of chairs, a desk...
but nothing like what he had expected, or what he had seen when he had
snuck in there that fateful night.  Then he noticed that at the rear
of *this* room there was *another* black curtain, just like the one he
had stepped through, apparently leading to still another room!

"Oh yes," the Wiz said, following Oz's glance with his eyes.  "I moved
most of the REALLY dangerous stuff back there so you wouldn't get hurt.
Whatever you do, DON'T GO BACK THERE, okay?"

The Wiz motioned to the large tome.  "Now, what you'll do in the weeks to
come is to work with the following formulas.  First, you will
familiarize yourself with the metric for a big bang universe... a
special case, actually, the "spatially flat" or "critical" case.  It
looks like this..."  He turned a page and Oz saw, in large print, the
formula

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

Oz felt slightly scared, but not too scared... it seemed vaguely
familiar.  He was about to ask a question when the Wiz said "No, not
now, we'll talk all about it soon enough.  I just want to show you a few
things first."

"Then," said the Wizard, "I am going to teach you how to compute the
Christoffel symbols, or connection, given any metric."

"Christoffel symbols?  Connection?  Huh?" asked Oz.

"That is the way we talk about parallel translation, here in the back
room," said the Wizard.  "Don't worry, I'll explain it all in due time.
For now I just want to show you the explicit formula for the Christoffel
symbols in terms of the metric.  It is..." and he turned the page:

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

"... that!"

Oz blanched with fear.  What were those commas doing down there with the
subscripts?  What did this all MEAN?  

Unperturbed, the Wizard continued, "Usually people write the Christoffel
symbols as Gamma, but we're in an ASCII environment here so we'll call
them C... for Christoffel, or connection.  For now, all you need to know is
there exists this explicit formula, but your goal will be to understand
it, so that you can apply it to the metric on the previous page."

"And then," the Wiz continued, "we will compute the Riemann tensor from
the Christoffel symbols.  This was what you wanted, remember?  You
wanted to compute the Riemann curvature of a metric... well, we do it
using the Christoffel symbols as an intermediate step."

Oz wondered why he had ever wanted to know these things.  The formula
for the connection seemed to expand and grow ever more fearsome as he
stared at it.

"So," continued the Wizard, "we will need to learn, and understand, the
following formula for the Riemann tensor in terms of the Christoffel
symbols."  And he turned the page, revealing the following formula,
written, it seemed, in blood:

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

Oz's hair stood on end.  He backed away, slowly.  "No," he said.  "No.
There is no way I am EVER going to understand THAT."

The Wizard smiled.  "That's what most people say when they first see it.
I felt that way myself."   But somehow this did not reassure Oz.
Another stroke of lightening cracked through the air, and he jumped.

"NO!" cried Oz.  "I will NOT understand it!  NEVER!"  

The Wizard laughed.  "I won't force you to, if you decide you don't
want to.  But remember, I *warned* you not to come back here, but
you insisted!"  

"It was a mistake!  I take it back," said Oz.  "Can I go now?"

The Wizard let out a guffaw.  "Hey!  I'll EXPLAIN all this stuff.  When
you approach it correctly, it's not so hard.  Right now I'm just SHOWING
it to you..."

"... just to scare the wits out of me," said Oz.

"Right," agreed the Wizard.  "Hey, look, you were expecting thrills and
chills!  Who am I to let you down?  And it's actually a tradition, in
general relativity, to scare people with these formulas.  Actually,
you're having an easy time compared to me, back when I was a kid."  He
shut the book.  "Come on, you've had a long night, go to bed.  We'll
start when you get rested up.  When we're finally done, you'll be able
to compute the Riemann curvature of that big bang metric and actually
solve Einstein's equation!  Think of that!  But first, we'll get used to
that big bang metric, and also get used to this Christoffel symbol
business, and, yes, even get used to that formula for the Riemann tensor.
So don't get too scared just yet."

But just at this moment, some sinister laughter poured forth from behind
the curtain leading to the back room of the back room.  The Wizard
frowned and cast a thunderbolt in the general direction of the curtain.
"Not yet," he repeated... and Oz slowly walked out, and then ran
down the hallway and out the door to his cave, and fell exhausted on
his straw mat, and fell asleep immediately.

Newsgroups: sci.physics,sci.math
Subject: Re: Lorentz transform of a rank-2 tensor
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In article <[email-masked]> [email-masked] writes:

>So:  let L be the matrix for a Lorentz transformation that takes vectors
>from V to V'.  Let T be the matrix for a rank-2 tensor T_ab in frame V.
>Let g be the matrix for the standard Minkowski metric of signature
>(-,+,+,+).  What is the matrix equation that will yield the matrix T'
>for T_ab in frame V', in terms of these matrices?  (And how do you
>derive it?)

What are V and V'?  You call them "frames", but that's a bit vague.
Let's say they are vector spaces!  So then you are thinking of a
Lorentz transform as a linear transformation L: V -> V' from one vector
space to another.  Alternatively you could think of it as going from a
vector space to itself, like L: V -> V.  Either viewpoint is acceptable.
But let me use the latter, okay?  It's just a wee bit easier that way,
since then we can use one coordinate system on V and think of L as an
active coordinate transformation.  Fear not, I am really not changing
the intent of your question one whit.  Just a half-whit.

Okay, so you derive the answer to your question by taking the linear
transformation L: V -> V, and first of all, working out its
corresponding contravariant version

(L^{-1})*: V* -> V*

(I stick in the inverse here for the reason mentioned when we talked
about this stuff last time.   Whether you put it in or not depends on
whether you set your clock forwards or backwards last night... i.e.,
it's sort of a matter of convention, but there are reasons for doing it
the way other folks do.)

Then you tensor this with itself and get a linear transformation
from V* tensor V* to V* tensor V*.  Then you apply this to T, which
lives in V* tensor V*.  

Okay, but what does it all boil down to in index-ridden notation?
Let me do it in painfully painstaking detail.  It's going to take
forever, but hopefully you will learn lots of handy things en route.

Physicists like to write a vector v in V by listing its components in
their own personal coordinate system.  (Every physicist always carries a
coordinate system in their pocket, which they automatically use whenever
they do a calculation.  Mathematics is less well-funded, so
mathematicians can only use coordinates if you *give them* a coordinate
system.)

So a physicist will call the vector v^a, and call its Lorentz transform

(v')^b = L^b_a v^a

Now how do we compute that (L^{-1})* thingie?  Well, if we have ANY
linear transformation L: V -> V, written L^b_a in coordinates, its
adjoint L*: V* -> V* is defined by 

(L*f)(v) = f(Lv) 

for all vectors v and covectors f.  I.e.

(L*)_a^b f_b v^a = f_b L^b_a v^a 

so

(L*)_a^b = L^b_a.

Note that no indices are raised or lowered here, since this is
something we can do without any metric around.

But L ain't just any transformation, it's a Lorentz transform: it
preseves the metric.  This means that 

g(u,v) = g(Lu,Lv) 

for all vectors u,v in V.  If we express this in terms of coordinates 
we get:

g_{ab} u^a v^b = g_{ab} (Lu)^a (Lv)^b
               = g_{ab} L^a_c u^c L^b_d v^d
               = g_{cd} L^c_a u^a L^d_b v^b

where in the last step we reshuffle dummy indices to get u^a and v^b
to appear, so we can conclude

g_{ab} = g_{cd} L^c_a L^d_b

That's what a Lorentz transform preserving the metric says, in
components.  

Now by raising or lowering an index or two we get

delta^a_b = g_{cd} L^{ca} L^d_b
          = delta^c_d L_c^a L^d_b
          = L_d^a L^d_b

What does this mean?  It means that the two matrices over on the right
at the end are inverses.  But what does THAT mean?  Well, you might
remember that a 3x3 matrix R represents a rotation in Cartesian
coordinates iff its inverse is its transpose.  This is just the same
sort of characterization of Lorentz transformations.

Here's another way to think of the above equation.  The inverse of L has

delta^a_b = (L^{-1})^a_d L^d_b

so it's saying that

(L^{-1})^a_d = L_d^a

This makes it easy to compute the (L^{-1})* thingie.  Go back and look
at the formula for computing the adjoint of a transformation... using
it, we get

((L^{-1})*)_a^b = (L^{-1})^b_a = L_a^b

So, the (L^{-1})* thingie is actually no big deal, it's just L
itself with the up index pushed down and the down index pushed up!

In short, if we have a covector f in V*, its components are f_a, and if
we transform it we get a covector f' with components

f'_b = L_b^a f_b

So, finally, if we have a tensor T in V* tensor V*, it transforms using
(L^{-1})* tensor (L^{-1})*, which means that we use two copies of the
above recipe for transforming guys with indices downstairs:

T'_{ab} = L_a^c L_b^d T_{cd}

The above was more long-winded than I intended, and it's possible I made
some typos and even some thinkos, but I tried not to.  As a consistency
check, you might want to take a vector, Lorentz-transform it, then turn
it into a covector by lowering its index, VERSUS first turning it into a
covector and then Lorentz-transforming that.  We're supposed to get the
same thing, and I think we do.

From [email-masked] Sun Mar  3 17:52 PST 1996
Date: Sun, 3 Mar 1996 20:51:47 -0500
To: [email-masked]
Subject: (Fwd) Re: Riemann Awakes
From: [email-masked] (Edward Green)
Content-Length: 6620
Status: OR

>There is one of yours I'm missing, or hasn't shown up yet... after 
 
Well,  this is the only other one I've posted so far...    
 
>----- Forwarded message ([email-masked](Edward Green)) -----< 

Newsgroups: sci.physics 
From: [email-masked](Edward Green) 
Subject: Re: Riemann Awakes 
Organization: The Pipeline 
X-PipeUser: egreen 
X-PipeHub: nyc.pipeline.com 
X-PipeGCOS: (Edward Green) 
X-Newsreader: The Pipeline v3.4.0 
 
Just as Oz was about to reply to the dented apprentice's last remark,  it  
befell that a stray piece of spacetime dilation drifted in from one of the 

wizard's experiments in his fell castle,  high up on the troll infested  
mountain on a bleak outcropping.   As this disturbance dropped into the
cave  
the part where Oz stood kind of slowed down,  then seemed to black out  
altogether! Only a few faint sparks jumping across the seams in erg's rusty
 
armor told him that the odd microwave and radio signal were still coming
from  
that part of the room. 
 
"Oh great",   erg thought,  "Just what we need;  a usenet timelag" 
(Uniform  
Spacelike Entrapment;  a wizardly technology using the coarse woven fabric
of  
space as a net to catch the unwary.  Stray blobs of redshift were either an
 
industrial waste products or failed prototypes,  it's not clear which.) 
 
"Well", mused the knight erring,  "normally it's bad form to hold a  
conversation with one's self,  but there is no telling how long this could 

last, or in whose proper time."  So,  looking around the visible portion of
the  
cave's floor,  and careful not to put his hand in the menacing looking red
aura  
surrounding the missing region,  he came upon the fell paper: 
 
>"Dear Oz -- 
 
>If you wish to learn more general relativity I am afraid you will need 
>to pass a test of your valor.  So: answer me the following questions: 
 
"Hmm",  thinks the sorcerer's lesser apprentice,  "I wonder what these
wedge  
shaped runes in the margin mean?  Maybe if I held it sideways..." 
 
A image of the wizard suddenly appeared on the sheet,  laughing in a rather
 
unpleasant manner.  "And stop saying 'fell',  nitwit!" 
 
"Yow!"  he remarked,  dropping the paper.  "Maybe I should just try to
answer  
the questions.  Let's see...   
 
"1.  Explain why, when the energy density within a region of space is 
sufficiently large, a black hole must form, no matter how much the 
pressure of whatever substance lying within that region attempts to 
resist the collapse." 
 
I bet the "no matter how much the pressure" is kind of a red-herring.  In  
General Relativity pressure is just a form of energy,  and a very high
pressure  
is just the same as a very high energy density;  so "pressure" sort of gets
 
hoisted on its own petard:  An extremely high pressure *provides* its own  
containment.  Or something.    
 
Somehow I feel the answer should involve geodesics...   It's a most
peculiar  
thing.  A blackhole is a region in space where geodesics can get *into* but
 
cannot get *out*.  But soft...  isn't a geodesic a geodesic in either  
direction? Of course it is.  But one direction is "physical"  and the other
is  
not.  In particular,  one direction of a timelike geodesic describes a
particle  
moving "forward in time" and the other,  well,  you know...   So GR
evidently  
has an arrow of time build into it.  Fair statement?   And what about  
"spacelike" geodesics.  I suppose they would represent a superluminal  
trajectory in all reference frames...  or should I say, all "proper"
reference  
frames,  the way timelike geodesics represent sub-luminal trajectories in
all  
proper reference frames...?   
 
Is there an anti-black hole?  A region of space(time) where stuff can only 

*leave* from,  but never get back in?  Sure!  The complement of a blackhole
is  
such a region...  but is it strange that all blackholes are roughly
spherica in  
space, and all "anti-blackholes"  are the complement of such regions,  or
in  
fact the *one* anti-blackhole is the complement of all such regions... 
 
What happens to geodesics after they get *into* a blackhole?  I suppose
they  
all wind up at a singularity.  Is a black hole a mini big bang in reverse?
What  
if we embedded another big bang into *this* universe... would that look
like a  
spherical anti-blackhole...?    
 
But all this makes blackholes sound like complicated global results about  
 
tracing geodesics...  I don't think it was supposed to be this
complicated... 

"2.  Explain how, in the standard big bang model, where the universe is 
homogeneous and isotropic --- let us assume it is filled with some fluid 
(e.g. a gas) --- the curvature of spacetime at any point is determined at
each  
point." 
 
Oh damn.  This is *just* what got me started on all that analyticity stuff.
  
*Part* of the curvature of spacetime at each point is dertermined by local 

conditions.  I think it was 10 of the 20 independent components of the
Riemann  
tensor,  the 10 in the Ricci,  the Weyl being free.  So...  maybe in the  
standard big bang model,  we just set the Weyl equal to zero?  Why?  It
seems  
too simple.  Maybe it turns out that the partial differential equations  
governing these components are the non-elliptic ones?  That they therefore 

represent something propagating,  rather than something that is locally  
evolving?  After all,  the "standard big bang model" must just be a name
for a  
simplified pet solution to the general equations...  but how do we enforce
this  
simplification.  Boundry conditions?   Arbitrary assignment? 

"3.  In the big bang model, what happens to the Ricci tensor as you go 
back in past all the way to the moment of creation." 
 
Um... becomes singular?   Somehow I think a little more detail is called
for.   
No doubt the density of stuff becomes infinite,  so the Ricci becomes
infinite,  
and so forth.  I know!   It's a trick question.  "Happens" implies
evolution  
forward in time...  Nothing "happens" going back in the past.   :-)  

Suddenly it becomes clear our apprentice hasn't a clue how to answer these 

questions quantitatively,   or even sharply qualitatively.  Perhaps one of
the  
spectators will throw a rock with a clue wrapped around it into the fantasy
 
world of G. Wiz. 
 
But finally the time shift seems to be clearing up, so let's hear what Oz
has  
to say...    
 
-- 
 
E. R. (Ed) Green /  [email-masked] 
"All coordinate systems are equal, 
but some are more equal than others". 
-- 
 
E. R. (Ed) Green /  [email-masked] 
  "All coordinate systems are equal, 
   but some are more equal than others".

From galaxy.ucr.edu!not-for-mail Tue Apr  9 12:22:27 PDT 1996
Article: 110357 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 9 Apr 1996 11:54:41 -0700
Organization: University of California, Riverside
Lines: 136
Message-ID: <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu
Summary: Here we go!

Hmm.  No post from Oz yet in response to my attempt to start him up on the
next phase of learning general relativity, although he did email me
saying he found the following equations suitably intimidating:

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

Don't tell me he has lost his knack for asking questions?

Well, let's start with the easiest formula, the first one.  This is just a
formula for a particular metric, a special case of the "Robertson-Walker
metric".  Remember, that's the technical name for the solution of
general relativity describing the simplest sort of big bang cosmology.  
I'll probably save keystrokes and call it the "big bang metric" most of
the time.  But... 

Robertson and Walker looked at homogeneous, isotropic cosmologies.  In
other words, those in which there was some "cosmic time coordinate" t,
such that any 3-dimensional slice of spacetime of the form
     
                         t = constant

is a Riemannian manifold that looks the same at every point, and the
same in every direction.  We call one of these t = constant
hypersurfaces a "spacelike slice" or simply "space" for short.  There
turn out to be 3 cases to consider: the case where space is flat, the
case where space is positively curved, and the case where space is
negatively curved.  (Watch out: even in the case where space is flat at
each moment, it's not necessarily true that spaceTIME is flat.  We'll
see that very vividly as we work our way through this series of
lessons.)  

When space is flat (as well as homogeneous and isotropic) it is
basically just R^3 with its usual Eucliean metric.  I hope all you folks
out there know what the heck "R^3" is.  Mathematicians use "R" to mean
the set of real numbers, and R^n to mean the set of ordered n-tuples of
real numbers, so R^3 is the set of ordered tripes of real numbers, like
(x,y,z).  This is a nice 3-dimensional manifold, and we often like to
give it the metric

                    dx^2 + dy^2 + dz^2

or if you prefer, 

                     1 0 0 
            g_{ij} = 0 1 0
                     0 0 1

where as usual i and j run over 1,2,3, which correspond to x, y, and z.
With this metric, R^3 becomes good old "Euclidean space".

When space is positively curved (as well as homogeneous and isotropic)
it is just the 3-sphere, S^3, with its usual metric.  S^3 can be
visualized as a sphere of radius 1 in R^4, given by the equation 
w^2 + x^2 + y^2 + z^2 = 1.  It then gets a metric coming from the
Euclidean metric on R^4.  

When space is negatively curved (as well as homogeneous and isotropic)
it is just 3-dimensional hyperbolic space, H^3.  You can think of this
as R^3 with a screwy nonEuclidean metric on it, or you can think of it
as a hyperboloid in R^4 given by the equation t^2 - x^2 - y^2 - z^2 = 1.
It then gets a metric coming from the Minkowski metric on R^4.

Anyway, these are the 3 big possibilities!  Whether the universe is open
or closed depends on whether space is flat or negatively curved --- then
it's open --- or positively curved --- then it's closed.  This has a lot
to do with whether the universe recollapse in a "big crunch" or not.
There is a lot of fascinating stuff to say about the quest of
astronomers to determine WHICH of these 3 big possibilities actually
matches our universe, if any.  

However, Ted Bunn spends vast amounts of time explaining this stuff on
sci.physics, so I'm sure you're all experts on it by now, and I don't
need to go into it again here.  (I suppose could be persuaded, if you
happen to have forgotten.)  So instead, I want to get into some of the
nitty-gritty details, and look at the simplest case --- the spatially 
flat case --- and actually solve Einstein's equation in that case.

Okay.  So I said space at any given time is flat.  But it's "size" can
vary with time!  So the metric on space is not just

                      dx^2 + dy^2 + dz^2

for all time; instead, there's a time-dependent scale factor

                   R(t)^2 (dx^2 + dy^2 + dz^2)

Note what this means!  It means that two galaxies which stay at the same
position in space --- i.e., their (x,y,z) coordinates are constant as a
functioon of t --- will actually get closer or farther apart, depending
on how R(t) changes.  This should make you think of the perennial
puzzlement, the perennial yapping and squalling, the perennial
confused arguments concerning "whether the galaxies are really moving
away from us or not" and "whether space is expanding or not".  Here is
where the answer to that puzzlement lies.  Think about it.  

But there is more to the story, I hasten to add.  More than a
time-dependent Riemannian metric on space!  There is an honest
Lorentzian metric on spacetime!  (Remember, "Lorentzian" means that
there's a minus sign lurking in there somewhere.)  It is:

                 -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

or
                     -1   0       0   0
            g_{ab} =  0   R(t)^2  0   0
                      0   0    R(t)^2 0
                      0   0       0  R(t)^2

where as usual a and b range over 0,1,2,3, corresponding to t,x,y,z.

So: if R(t) = 1 for all t, this is just the usual Minkowski metric, but
if R(t) changes then --- roughly speaking, remember this is just a
bunch of words! --- space "expands" or "contracts" as time passes.  

Okay, does this make sense?  Can you see how we'd use it to model a big
bang?  R(t) would start out at zero and get bigger as time passes.  Can
you see why it would make redshifts and all that jazz happen?  Can you
see why it simply FINESSES the question of whether the galaxies are
"moving away from each other" or not?  I hope so. 

Anyway, once we comprehend this metric in our very bones, we can start
trying to compute its Riemann tensor.  That will take a while, since we
need to learn how to compute the Riemann tensor!  But then we can try to
solve Einstein's equation to see how R(t) actually changes with time.
That'll depend, of course, on what sort of *matter* is around.... dust,
radiation, or whatever.  

From galaxy.ucr.edu!not-for-mail Wed Apr 10 12:20:47 PDT 1996
Article: 110475 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 9 Apr 1996 20:43:01 -0700
Organization: University of California, Riverside
Lines: 114
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Patrick van Esch) writes:
>: In the spatially flat big bang, the metric on space is not just
>
>:                       dx^2 + dy^2 + dz^2
>
>: for all time; instead, there's a time-dependent scale factor
>
>:                    R(t)^2 (dx^2 + dy^2 + dz^2)

>: Note what this means!  It means that two galaxies which stay at the same
>: position in space --- i.e., their (x,y,z) coordinates are constant as a
>: functioon of t --- will actually get closer or farther apart, depending
>: on how R(t) changes.  This should make you think of the perennial
>: puzzlement, the perennial yapping and squalling, the perennial
>: confused arguments concerning "whether the galaxies are really moving
>: away from us or not" and "whether space is expanding or not".  Here is
>: where the answer to that puzzlement lies.  Think about it.  

>I have a question here.  After all, the way we lay our (x,y,z,t) grid
>onto the spacetime manifold (eg, defining a coordinate system) is 
>arbitrary.  So I have a bit of a difficulty to try to imagine what you
>mean by: "two galaxies which stay at the same position in space ---
>i.e., their (x,y,z) coordinates are constant as a function of t".  

Good!

All I mean is what I said in the "i.e." part: their (x,y,z) coordinates
are constant as a function of t.  The yapping and squalling begins as
soon as we try to attribute any more meaning to my phrase "stay at the
same position in space" than that!

To repeat: "the galaxies stay at the same position in space" means nothing
until we have ARBITRARILY chosen coordinates (x,y,z) on space for all
times t.  And then, it means simply that the worldline of a galaxy
traces out a curve in spacetime for which these coordinates (x,y,z) are
independent of t.  

Let me repeat this once more, not for Van Esch --- who obviously gets
the point --- but for all the unfortunates out there who never really
understood this business about whether the galaxies are moving away from
each other, or not.  It's a crucial point:

"The galaxies stay at the same position in space" is a COORDINATE-
DEPENDENT statement, and must not be taken to have any significance
apart from our choice of coordinates.  Above I happened to choose coordinates
in which this statement is true.  Someone else, choosing different
arbitrary coordinates, may equally rightly say that "the galaxies do NOT
stay at the same position in space".  That's fine!  

>(well, I understand, technically, what you mean)

Okay, good.  All I mean is that.  

>But one could easily claim: yes, you just happened to choose a very
>unfortunate grid, one with narrower spacings for the x,y,z coordinates
>if t increases (meaning: your ruler becomes shorter as t increases).
>So according to your ruler, yes, the universe expands (or contracts, I
>should think about this one :).  But it was just an unfortunate choice
>of coordinate grid.

Ah yes.

It happens that this coordinate system I have chosen will make the
computations I have in mind easy for me to do.  But one might suspect
that some other coordinate system would shed more light on this problem.
Remember the Milne metric, as discussed by Bunn et al?  It looked fancy
in one coordinate system, and then it boiled down the Minkowski metric
on part of Minkowski space in other coordinates.  So let's see!

>So comes the question: is it possible to define an (T,X,Y,Z) so that
>the function R(t) gets absorbed ?
>First guess was: X = R(t).x etc, but that is a bit too cheap, as 
>taking dX will involve terms in dt too.

You can certainly define new coordinates 

X = R(t)x, 
Y = R(t)y, 
Z = R(t)z, 

and then *part* of the metric simplifies... but the metric doesn't become
the Minkowski metric, since as you note we will have

dX = R'(t)xdt + R(t)dx

and so on.  I urge people learning the subject to figure out what the
metric looks like in Esch's coordinates (t,X,Y,Z)!

>But in general, could it be
>done?  I suppose not, because then the whole thing would fall apart
>(as we've just flat spacetime), but somewhere I just have this feeling
>there is some arbitrariness in R(t)...

When we compute the Riemann curvature of this metric, we will not get
zero.  Since the Riemann curvature is a tensor, having it come out
nonzero in one coordinate system means it can't come out zero in any
other coordinate system.  In short, our spacetime is not flat, and no
coordinate system will flatten it out into Minkowski spacetime.

>Well, it would be clear if somehow I could convince myself these x,y,z
>are "space as we know it".
>But I still have this weird feeling of arbitrariness in R(t)...

There's a lot of arbitrariness in my coordinate system, but it'll make
computing the curvature nice and easy, and it has lots of other
properties that make it a favorite among cosmologists.  That's all I can
say.  A coordinate system can make the math more or less convenient, but
the actual physical predictions are coordinate-independent, so there is
a limit to how much one should argue about coordinates.  

From galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!vixen.cso.uiuc.edu!newsfeed.internetmci.com!in1.uu.net!fu-berlin.de!news.dfn.de!news.dkrz.de!dscomsa.desy.de!jamaica!vanesch Wed Apr 10 12:49:14 PDT 1996
Article: 110535 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!vixen.cso.uiuc.edu!newsfeed.internetmci.com!in1.uu.net!fu-berlin.de!news.dfn.de!news.dkrz.de!dscomsa.desy.de!jamaica!vanesch
From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 10 Apr 1996 09:29:18 GMT
Organization: DESY
Lines: 96
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: jamaica.desy.de
X-Newsreader: TIN [version 1.2 PL2]

john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] (Patrick van Esch) writes:
: >: In the spatially flat big bang, the metric on space is not just
: >
: >:                       dx^2 + dy^2 + dz^2
: >
: >: for all time; instead, there's a time-dependent scale factor
: >
: >:                    R(t)^2 (dx^2 + dy^2 + dz^2)

: >: Note what this means!  It means that two galaxies which stay at the same
: >: position in space --- i.e., their (x,y,z) coordinates are constant as a
: >: functioon of t --- will actually get closer or farther apart, depending
: >: on how R(t) changes.  This should make you think of the perennial
: >: puzzlement, the perennial yapping and squalling, the perennial
: >: confused arguments concerning "whether the galaxies are really moving
: >: away from us or not" and "whether space is expanding or not".  Here is
: >: where the answer to that puzzlement lies.  Think about it.  

: >I have a question here.  After all, the way we lay our (x,y,z,t) grid
: >onto the spacetime manifold (eg, defining a coordinate system) is 
: >arbitrary.  So I have a bit of a difficulty to try to imagine what you
: >mean by: "two galaxies which stay at the same position in space ---
: >i.e., their (x,y,z) coordinates are constant as a function of t".  

: Good!

: All I mean is what I said in the "i.e." part: their (x,y,z) coordinates
: are constant as a function of t.  The yapping and squalling begins as
: soon as we try to attribute any more meaning to my phrase "stay at the
: same position in space" than that!

: To repeat: "the galaxies stay at the same position in space" means nothing
: until we have ARBITRARILY chosen coordinates (x,y,z) on space for all
: times t.  And then, it means simply that the worldline of a galaxy
: traces out a curve in spacetime for which these coordinates (x,y,z) are
: independent of t.  

Ah yes, thanks.  I was indeed getting the idea I missed some "deeper meaning"
but I didn't (I hope :): "staying at the same point in space" is of the same
level as: staying at the same x in an Euclidean plane with an xy coordinate
system.

: >But one could easily claim: yes, you just happened to choose a very
: >unfortunate grid, one with narrower spacings for the x,y,z coordinates
: >if t increases (meaning: your ruler becomes shorter as t increases).
: >So according to your ruler, yes, the universe expands (or contracts, I
: >should think about this one :).  But it was just an unfortunate choice
: >of coordinate grid.

: Ah yes.

: It happens that this coordinate system I have chosen will make the
: computations I have in mind easy for me to do.  But one might suspect

Ok.  Again, I thought I missed a clue about the meaning of this metric.
So for the moment I'll just assume we're doing some abstract gymnastics
and suppose you'll point out when these things get a physical meaning
(in the sense of being related to observable quantities).

: >But in general, could it be
: >done?  I suppose not, because then the whole thing would fall apart
: >(as we've just flat spacetime), but somewhere I just have this feeling
: >there is some arbitrariness in R(t)...

: When we compute the Riemann curvature of this metric, we will not get
: zero.  Since the Riemann curvature is a tensor, having it come out
: nonzero in one coordinate system means it can't come out zero in any
: other coordinate system.  In short, our spacetime is not flat, and no
: coordinate system will flatten it out into Minkowski spacetime.

Yes, that is convincing.  Nevertheless, I still think PART of R(t) is
arbitrary.  For example, I could multiply R(t) with an arbitrary constant
c (well, eh :-) no, C, and redefine X = x/C ; Y = y/C, Z = z/C, T = t
and everything would be the same.  Now this is trivial, but maybe other
freedoms are still possible.  Ok, maybe this isn't important.  It is just
that I'd like to know what parts of R(t) are somehow essential, and what parts
are just a consequence of choice of coordinates...

: There's a lot of arbitrariness in my coordinate system, but it'll make
: computing the curvature nice and easy, and it has lots of other
: properties that make it a favorite among cosmologists.  That's all I can
: say.  A coordinate system can make the math more or less convenient, but
: the actual physical predictions are coordinate-independent, so there is
: a limit to how much one should argue about coordinates.  

Well, I'd better wait until we're there....

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

From galaxy.ucr.edu!not-for-mail Wed Apr 10 19:07:48 PDT 1996
Article: 110602 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Black holes within black holes
Date: 10 Apr 1996 10:43:53 -0700
Organization: University of California, Riverside
Lines: 59
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>NB Please note that this is not "GR tutorial" so heavy handed firebolts, 
>electrical discharges, and turning into colonic parasites isn't fair.

Okay, I'll be nice.

>Oh, weird and wonderful! Some nice mental contortions to exercise the
>brain. One might (almost certainly wrongly) wonder what this means
>because at first sight one would imagine that an infinite radius would
>seem to imply to the ignorant (me) that lines on this *surface* are
>straight and uncurved when seen locally. This sort of has some sort of
>flatness to it. 

Hmm.   

(See how subdued I'm being?)

>The next question is what happens to the time co-
>ordinate? It is presumably in the process of becoming spacelike but is
>it doing so through zero, or by nipping round the back of infinity?

The radial space coordinate r becomes timelike and time coordinate t
becomes spacelike as we pass through the event horizon.  The
Schwarzschild metric is

g = -(1 - 2M/r) dt^2 + (1 - 2M/r)^{-1} dr^2 + 
    familiar-looking d theta and d phi stuff

So the length of the vector d/dr (that's the vector in the r direction;
forgive my mathematical style --- we identify a vector with the partial
derivatives in the direction *of* the vector) blows up as we come into
the event horizon, and then it "nips back around infinity" and becomes
negative.  On the other hand, the length of d/dt goes to zero and then
becomes positive.  (Remember, folks, positive length = spacelike,
negative length = timelike in my conventions.)  

>It would seem to me that once 'through' the event horizon it is no 
>longer appropriate to define a volume (whatever that means) using the 
>Schwarzchild 'r' since it is now timelike. One 'ought' perhaps to use 
>only spacelike ones. Now my immediate reaction is that the Schwarzchild 
>co-ordinates are no longer very appropriate, and some better ones would 
>be more use.

Indeed!  Schwarzshild coordinates are at their best far outside the
event horizon, where the Schwarzschild metric approaches the Minkowski
metric.  They go berserk at the event horizon and are sort of
inside-out-and-backwards within.  Kruskal coordinates are much nicer for
understanding the inside of a black hole.  I was just trying to
illustrate the problems with naively implementing Landis' suggestion for
computing the "density of a Schwarzschild black hole"... as a rhetorical
lead-in to my suggestion that it's best not to bother!

From galaxy.ucr.edu!not-for-mail Wed Apr 10 19:09:52 PDT 1996
Article: 110622 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 10 Apr 1996 12:48:57 -0700
Organization: University of California, Riverside
Lines: 38
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

>Ok.  Again, I thought I missed a clue about the meaning of this metric.
>So for the moment I'll just assume we're doing some abstract gymnastics
>and suppose you'll point out when these things get a physical meaning
>(in the sense of being related to observable quantities).

That's one acceptable attitude.  You could put it this way: what's the
the simplest sort of Lorentzian metric on R^4 that we can write down
which makes space be flat at any given time t, but where the metric on
space can change with time?  Well, 

-dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

springs to mind.  So we might as well see what solutions of Einstein's
equation we get that are of this form.  

But: it'll work out that we get solutions where the stress-energy tensor
is diagonal in these coordinates, so the matter is static with respect
to these coordinates, in the sense that there is no net flow of energy
in the x, y, or z directions.  We'll eventually consider the example of
"dust", meaning pressure-free matter.  This is a good example because
one can approximate our current universe by a homogeneous sprinkling of
such "dust" in the form of galaxies.  (I leave it to the cosmologists in
the crowd to say just how accurate this approximation is.  I'm just
concerned with working out a simple example, making no claims that it's
the real universe.)  Then when we solve Einstein's equation we'll see
that the dust must be static with respect to the above coordinates.

Pretty soon I should start telling folks how to compute the curvature
of this metric.  But before I do, are there any other questions about
it?  Do y'all see how you could compute the redshift of one galaxy as
seen by another, if both are at rest with respect to these coordinates?
Do you see how if R(t) is increasing with t, there'll be a redshift?

From galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Fri Apr 12 09:41:39 PDT 1996
Article: 110757 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted
From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Black holes within black holes
Followup-To: sci.physics
Date: 11 Apr 1996 06:32:50 GMT
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In article <[email-masked]>, john baez <[email-masked]> wrote:
>Personally, I see no need to talk about the density at the singularity
>of a black hole; it's a vague and pretty useless concept.

I agree with this completely.  I regret ever trying to say anything
about it at all, but since I did, maybe I should state explicitly the
angle from which I was approaching the subject.

As John says, quantities like density and curvature are simply
undefined at singularities, because singularities aren't even points
in the spacetime manifold at all.  Nonetheless, people seem to like to
say things like "the big-bang singularity had infinite density" or
"the curvature is infinite at the singularity of a black hole."  I was
trying to suggest a way to make such statements meaningful, at least
under certain circumstances.

The idea goes like this.  One of the defining characteristics of a
singularity is that there are curves that can't be extended
arbitrarily far.  They get so far, and then they "hit the singularity"
and you have to stop.  In fact, I believe that this is the way the
word "singularity" is defined for purposes of proving the famous
singularity theorems: roughly speaking, you say that a spacetime has a
singularity if there are curves that can't be extended arbitrarily
far.

Well, since a singularity is identified by this feature of having
paths that lead right up to it and then stop, it seems to me that one
sensible way to assign a value to something like the density at the
singularity would be to take the limit of that quantity as you go
along one of those paths that stops at the singularity.  If you got
the same answer for that limit along all possible paths, then that
would be a reasonable definition.

With that definition, the Schwarzschild black-hole singularity has
zero density but infinite curvature.  (Specifically, the scalar
quantity R_{abcd}R^{abcd} goes to infinity.)  Also, the big-bang
singularity really does have infinite density with this definition.

I guess that ultimately what I'm saying is this: you should probably
avoid talking about the values of physical quantities at singularities
at all, but if you absolutely can't resist, then a limit-along-paths
definition like this seems to make the most sense.

-Ted

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Article: 110824 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: Thu, 11 Apr 1996 15:55:28 +0100
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Thursday 11th April 1996. It's arrived at last.
How many times has it gone round the world??

In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Okay.  So I said space at any given time is flat.  But it's "size" can
>vary with time!  So the metric on space is not just
>
>                      dx^2 + dy^2 + dz^2
>
>for all time; instead, there's a time-dependent scale factor
>
>                   R(t)^2 (dx^2 + dy^2 + dz^2)
>
>Note what this means!  It means that two galaxies which stay at the same
>position in space --- i.e., their (x,y,z) coordinates are constant as a
>functioon of t --- will actually get closer or farther apart, depending
>on how R(t) changes.  This should make you think of the perennial
>puzzlement, the perennial yapping and squalling, the perennial
>confused arguments concerning "whether the galaxies are really moving
>away from us or not" and "whether space is expanding or not".  Here is
>where the answer to that puzzlement lies.  Think about it.  

God. Think about it .. already. Well, I suppose it is at least page two.
Is it quite as simple as this? Well, we have the problem of what exactly 
you mean by distance, what paths you tread to decide distance, but I 
think (hope) that in this rather empty isomorphic and reasonably 
homogenous universe one could have a sensible measure of distance on at 
least slices at constant t. On this measure then things initially  at 
rest wrt each other will indeed get further apart as time evolves 
although neither will feel any acceleration. It's going to play hell 
with working out any celestial mechanics. Oh, I suppose that's why you 
define your set of co-ordinates to be co-moving as being 'convenient'. 
In this co-ordinate system 'relative accelerations' due to R(t) are 
zero, so at least to first order it's a bit simpler when you come to 
dealing with other forces and accelerations. Hmmm, if so that's quite a 
good idea. So, I conclude that galaxies are indeed moving away from us.

Is space expanding or not? Well, in time honoured GR fashion one is 
tempted to say "it depends on what you mean by expanding", but this is 
being devious. If I go and measure a volume of space marked by eight 
free-falling bodies then after a time come back and do it again (with 
appropriate corrections of course) I would find a bigger volume is 
enclosed if I was careful to site them appropriately. So, I would 
conclude that space is expanding . 

But on the other hand this sort of thing is easier said than done. What 
if one of my test bodies was near a sun, orbiting it even? Then I am in 
real trouble. Do I start to say that bits of the universe are not 
expanding, or anyway doing weird things in the expansion line? 
Heellllllp.

>But there is more to the story, I hasten to add.  More than a
>time-dependent Riemannian metric on space!  There is an honest
>Lorentzian metric on spacetime!  

Oh my goodness, no, no, no!

>(Remember, "Lorentzian" means that
>there's a minus sign lurking in there somewhere.)  It is:
>
>                 -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)
>
>or
>                     -1   0       0   0
>            g_{ab} =  0   R(t)^2  0   0
>                      0   0    R(t)^2 0
>                      0   0       0  R(t)^2
>
>where as usual a and b range over 0,1,2,3, corresponding to t,x,y,z.
>
>So: if R(t) = 1 for all t, this is just the usual Minkowski metric, but
>if R(t) changes then --- roughly speaking, remember this is just a
>bunch of words! --- space "expands" or "contracts" as time passes.  
>
>Okay, does this make sense?  Can you see how we'd use it to model a big
>bang?  R(t) would start out at zero and get bigger as time passes.  Can
>you see why it would make redshifts and all that jazz happen?  Can you
>see why it simply FINESSES the question of whether the galaxies are
>"moving away from each other" or not?  I hope so. 

Er. Um. Ooooh.
What does it mean to have a metric with R(t)=0??; <1; >1 ???

Equally what would be the effect of having a metric 
-R'(t)dt^2 + dx^2 + dy^2 + dz^2  where R'(t) decreases from 00 through 1 
to -OO??? I would have thought this might ease the maths a little at the 
very least.

Er, please sir, why can't we simply say that space IS expanding? I mean, 
you get the same answer and it's rather easier to visualise. OK, it's 
not expanding in the sense that each little part feels no acceleration, 
but that's not so hard to mentally tack in.

>Anyway, once we comprehend this metric in our very bones, we can start
>trying to compute its Riemann tensor.  

Colonic parasites don't have any bones, sir. <ouch>

>That will take a while, since we
>need to learn how to compute the Riemann tensor!  But then we can try to
>solve Einstein's equation to see how R(t) actually changes with time.
>That'll depend, of course, on what sort of *matter* is around.... dust,
>radiation, or whatever.  

Lets go for it!

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Fri Apr 12 10:54:16 PDT 1996
Article: 110941 of sci.physics
Path: galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted
From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Black holes within black holes
Followup-To: sci.physics
Date: 12 Apr 1996 00:30:52 GMT
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>Well, you know what's coming of course. What are the Kruskal co-
>ordinates? Do I want to know?

They're really quite nice.  Here's what you do.  You take the ordinary
Schwarzschild coordinates, and you define two new coordinates u and v
in terms of r and t:

u = f(r,t)
v = g(r,t)

where f and g are some functions you can look up in a textbook.  (I'm
not going to look them up and type them in right now.  If you
desperately want to know, I'll tell you, but the formulae themselves
aren't especially enlightening.)  Now you can describe the spacetime
around a Schwarzschild black hole in terms of the four coordinates
(u,v,theta,phi) instead of (t,r,theta,phi).

In terms of these coordinates, a lot of things are easier to
understand.  First, u is a timelike coordinate on both sides of the
horizon, and v is spacelike on both sides.  (Either that or v is
always timelike and u is always spacelike; I don't remember if there's
a standard convention for the names of these guys.)  Also, nothing
nasty happens at the event horizon.  The coordinates are perfectly
nicely behaved there.  

Finally, Kruskal coordinates have the great property that incoming and
outgoing radial light rays have slope plus or minus 1.  That is, if
you plot the path of a light ray in Kruskal u,v coordinates, it always
comes out as a straight line at a 45 degree angle.  So the path of
anything that goes slower than light is always a curve whose slope in
the (u,v) plane is greater than 1.

Now, it turns out that the horizon of a black hole is the set
of points for which u = v or u = -v.  Note that these are lines with
slope 1, which means that the horizon is a lightlike surface.  But
you knew that already.  A photon sitting on the horizon trying
to escape just remains on the horizon.  It doesn't fall in, but it
doesn't escape either.

In these coordinates, the singularity is located on the hyperbola
u^2 - v^2 = 1.  Here's a crude ASCII diagram showing the horizon
(slashes) and the singularity (dots):

\.           ./
 \..       ../
  \ ..   .. /
   \  ...  /
    \    */
     \   /
      \ /
       X       
      / \
     /   \
    /     \  +
   /  ...  \
  / ..   .. \
 /..       ..\
/.           .\

Up is the u direction (towards the future), and right is the v
direction (radially out).  Only the part between the two sets of dots
exists.  There isn't anything "beyond" the two singularities.  The
upper quadrant of the diagram is inside the black hole; the lower
quadrant is inside the white hole, and the other two are "the
outside."  This whole diagram describes a black-hole white-hole pair.
A realistic black hole, that formed from the collapse of a star or
something, would consist of only the upper part of the diagram; there
wouldn't be a white-hole part.

If you're an observer who can't move faster than light, then your path
through spacetime has to constantly go up (toward the future) with
slope greater than 1.  (You're supposed to imagine that the slashes
making up the horizon have slope 1.)  So if you start out outside of
the black hole (say at the place marked with a "+"), you can follow a
path that will take you across the horizon (say to the "*").  But once
you've reached that point, you can't get back out without going faster
than light.  In fact, once you've crossed the horizon, every possible
path you can take winds up hitting the singularity.

Kruskal coordinates give you a much more realistic picture of the
overall geometric structure of a black hole than Schwarzschild
coordinates.  For example, people often have a naive picture that the
singularity of a black hole is there "right now" in the center of the
black hole.  But in Kruskal coordinates you can see that really the
singularity of the black hole lies in the future; it doesn't exist
yet.  (Any path you can draw from your location, say the "+", to the
singularity is a future-pointing path.)

And, using Kruskal coordinates, you can see exactly how Schwarzchild
coordinates go bad at the singularity.  You could draw a series of
curves corresponding to the surfaces r=constant and t=constant on the
above diagram.  The r=constant surfaces are hyperbolae of the form 
v^2 - u^2 = constant.  The surface r = Schwarzschild-radius
corresponds to the "degenerate hyperbola" v^2 - u^2 = 0, that is, the
horizon.  And the surfaces of constant t are straight lines through
the origin: u/v = constant.  But it turns out that as t tends to
infinity, that constant tends to 1.  So the constant-t surfaces "bunch
up" infinitely close together as you get close to the line u=v.

Outside the horizon, the r=constant surfaces are timelike (they go up
faster than they go across), and inside they're spacelike.  The
opposite is true for the t=constant surfaces.

I think you said that you had to get rid of your copy of Misner,
Thorne, and Wheeler.  But if you can borrow a copy and stare at the
pictures in it for a while, I think it'll help a lot.  They have lots
of great pictures of black holes in Kruskal coordinates.

-Ted

From galaxy.ucr.edu!not-for-mail Fri Apr 12 10:56:04 PDT 1996
Article: 111038 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 00:46:04 -0700
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In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

>I don't know yet if (x0,y0,z0,t) is a geodesic !  (well, it better be,
>but I hope John could comment on this one...)

Yes, these are geodesics in the universe with metric

-dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

When we get a little better at math we'll be able to show this.
But anyway, that's one of the pleasant features of this coordinate
system: if we "stay at the same place", in the coordinate-dependent
sense that our worldline traces out a curve in spacetime for which
(x,y,z) remains constant as t varies, then we are in free-fall. 

>If you define "acceleration" as second derivative of the distance,
>then (except if R(t) = a + b.t) there are, for sure, relative accelerations.

Certainly.  Here you are presumably measuring the distance between two
points (x0,y0,z0) and (x1,y1,z1) at the same time t.  

Quickie quiz, to make sure everyone is awake: what *is* the distance
between the points (x0,y0,z0) and (x1,y1,z1) at time t, if the spacetime
metric is 

-dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2) ?

Make sure to tell me what you MEAN by "the distance", while you're at
it.

>>Hmmm, if so that's quite a 
>: good idea. So, I conclude that galaxies are indeed moving away from us.

>they might be closing in or vibrating back and forth, depending on R(t) :-)

It would not be giving too much away to reveal that in practice, R(t)
will be positive and will increase with time.

>Well, that was my question before !  Can't we think up a coordinate grid
>in which we absorb R(t).  Then we would end up with flat Minkowski space.
>But I'm pretty sure it cannot be done (as John said, but we're not supposed
>to look behind the curtain :-) we'll end up with a non-zero curvature.
>So it cannot be equivalent to a flat Minkowski space.

Right.  Besides, use psychology.  If this metric it WERE equivalent to
flat Minkowski space, do you think I'd be using it as my main example of
a metric in this second part of the general relativity course?  

By the way, Oz was allowed to go behind the curtain, and this is what he
saw:

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

So you can use this to compute the Riemann tensor whenever you want ---
at least, after I tell you want those commas mean in things like
g_{cd,b} and C^a_{bd,c}.  They mean partial differentiation!  So for
example

g_{cd,b} = partial g_{cd} / partial x^b

where "partial" is a feeble ASCII equivalent of one of those curly
d's they use for partial derivatives.  Here x^b is just the coordinate
t, x, y, or z depending on whether the superscript b is 0, 1, 2, or 3.

Similarly,  

C^a_{bd,c} = partial C^a_{bd} / partial x^c

and so on.  

So whenever you have a little spare time you can compute the Riemann
curvature of this big bang metric and see for yourself that it's
nonzero.

Actually one enterprising fellow has already computed the Christoffel
symbols C^a_{bc} --- and, even better, explained what they mean!

More on that later.

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Article: 111067 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: Fri, 12 Apr 1996 09:31:54 +0100
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In article <[email-masked]>, Patrick van Esch
<[email-masked]> writes
>I'm playing a dangerous game here.  I'm going to try to answer 
>some of Oz's questions... of course such a mixing into business
>can easily provoke thunderbolts and fire bullets out of nervous
>Wizzards, but then at least I know I'm making a mistake somewhere
>and will learn form it :-) 

Oh-ho. So the Grand Wizard has got himself a journeyman wizard. Now
there will be twice the amount of sweeping up. Oh dear. Still, this
jourenyman seems quite keen, so Oz might be able to avoid having to do
the more difficult (read: easy) maths that he can't do anyway.

>Oz ([email-masked]) wrote: not hard to work out Oz's bits  :-)

>Well, you're probably too much impressed by all those insecurities
>that come along with curved spaces :-)  

Carefully and lovingly nurtured by Baez and Bunn, I might add.

>As the space part is FLAT
>(the metric being proportional to dx^2+dy^2+dz^2), it is just ordinary
>Euclidean 3-dimensional space, and the grid x,y,z is just an orthonormal 
>coordinate grid.

Oh, yes. So it is.

>The points (x0,y0,z0) and (x1,y1,z1) being supposed to be fixed (independent
>of t), we have that the distance S = R(t) . (some fixed number).
>So all distances just grow (or shrink) with R(t).

OK, a good comment at the start to remind us. The co-ordinates and the
distances are related, but we must remember that they are not the same.
Yes, worth saying.

>Well, I don't know if "relative accelerations" due to R(t) are zero.
>I don't know yet if (x0,y0,z0,t) is a geodesic !  (well, it better be,
>but I hope John could comment on this one...)
>If you define "acceleration" as second derivative of the distance,
>then (except if R(t) = a + b.t) there are, for sure, relative accelerations.

We really ought to have different words for acceleration and d^2S/dt^2
since they are not the same.

>Hmmm, if so that's quite a 
>: good idea. So, I conclude that galaxies are indeed moving away from us.
>
>they might be closing in or vibrating back and forth, depending on R(t) :-)

You fit in well on this thread, I can see that. Yes, the van Esch
vibrating universe, hmmmm.

>I wouldn't.  I still don't know what free-falling bodies are supposed to
>do in our coordinate system.

Oh. Oh dear. Blasted assumptions again. Humpf!

>: What does it mean to have a metric with R(t)=0??; <1; >1 ???
>
>no, short or long meter sticks, I suppose :-)

Hang on. A meter is a distance, an S. So it might properly stay the same
length (er, whatever that means). It's it's relationship to our co-
ordinate system that changes. OK you could look at it either way, but I
think that we are tending towards a co-ordinate system of fixed points,
but whose relationship as measured by the parameter 'distance' varies
with time.

>: Equally what would be the effect of having a metric 
>: -R'(t)dt^2 + dx^2 + dy^2 + dz^2  where R'(t) decreases from 00 through 1 
>: to -OO??? I would have thought this might ease the maths a little at the 
>: very least.
>
>Ah, tricky !  I wanted to say, this is just a coordinate transformation,
>but I think it isn't !  I think this form isn't equivalent to John's 
>proposal. (well, it might turn out to be so, but to me it isn't evident).
>Could the Wizard comment on this ?

I think you are right. It's not the same. For one thing 3-distances
DON'T vary with time as related to the co-ordinates. On the other hand,
since I haven't done the required 10,000 example questions on Lorenzian
metrics, I am not entirely certain that the observable result is not the
same. (See, perhaps, Milne universe).

>: Er, please sir, why can't we simply say that space IS expanding? I mean, 
>: you get the same answer and it's rather easier to visualise. OK, it's 
>: not expanding in the sense that each little part feels no acceleration, 
>: but that's not so hard to mentally tack in.
>
>Well, that was my question before !  Can't we think up a coordinate grid
>in which we absorb R(t).  Then we would end up with flat Minkowski space.

I don't think we would, you know. At the end of the day the distance
between fixed points on our co-ordinate system must still vary with t,
so it can surely never be Minkowskian. (Oz ducks behind large boulder).

>But I'm pretty sure it cannot be done (as John said, but we're not supposed
>to look behind the curtain :-) 

Certainly not when he is anywhere near. Say on the same planet, for
example. It is however, VERY tempting.

>Awaiting the thunderbolts :-)

I wish. However I rather doubt it this time. You haven't yet got the
knack of posting *really* inane questions. ... Yet...

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 111090 of sci.physics
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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 12:38:19 GMT
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john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

: >I don't know yet if (x0,y0,z0,t) is a geodesic !  (well, it better be,
: >but I hope John could comment on this one...)

: Yes, these are geodesics in the universe with metric

: -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

: When we get a little better at math we'll be able to show this.
: But anyway, that's one of the pleasant features of this coordinate
: system: if we "stay at the same place", in the coordinate-dependent
: sense that our worldline traces out a curve in spacetime for which
: (x,y,z) remains constant as t varies, then we are in free-fall. 

Right.  But I'm playing it a bit hard here, and I was not _supposed to be able
to find out_, except if I'm missing something.  Currently, your metric (and that
is all we're considering) contains an arbitrary function R(t) - which for the 
moment can be anything, and we don't officially know how to relate a metric
to parallel transport and hence geodesics.  So I wanted to make the point that
it is not because it "looks ok" to pin down galaxies to a fixed (x0,y0,z0)
that this automatically implies that this is a free fall situation.
If you would have specified a completely different metric, lines of "constant
x,y,z" are maybe no geodesics.  I'm even not sure if some strange functions
R(t) cannot make the line (x0,y0,z0,t) into a non-geodesic.
Sorry if this sounds like nitpicking, it is not intended that way.  I always
used to learn most in trying to play the devil's advocate.

: >If you define "acceleration" as second derivative of the distance,
: >then (except if R(t) = a + b.t) there are, for sure, relative accelerations.

: Certainly.  Here you are presumably measuring the distance between two
: points (x0,y0,z0) and (x1,y1,z1) at the same time t.  

Eh, yes.  I wouldn't even know now how to measure distance for a t0 and a t1,
as I don't know yet how to trace a geodesic from (x0,y0,z0,t0) to (x1,y1,z1,t1).
It was only for a constant t that it was easy as it was a flat 3-dim euclidean
space.

: It would not be giving too much away to reveal that in practice, R(t)
: will be positive and will increase with time.

Ok, but this wasn't set out yet :-)  I suppose you'll need some behavioural
equation for the stuff that fills the universe to find R(t).

: Right.  Besides, use psychology.  If this metric it WERE equivalent to
: flat Minkowski space, do you think I'd be using it as my main example of
: a metric in this second part of the general relativity course?  

Of course not !  But I thought it might have been a good exercise 
to try and see that it doesn't work out.

: By the way, Oz was allowed to go behind the curtain, and this is what he
: saw:

:               g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

:          C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

: R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

: So you can use this to compute the Riemann tensor whenever you want ---
: at least, after I tell you want those commas mean in things like
: g_{cd,b} and C^a_{bd,c}.  They mean partial differentiation!  So for
: example

: g_{cd,b} = partial g_{cd} / partial x^b

: where "partial" is a feeble ASCII equivalent of one of those curly
: d's they use for partial derivatives.  Here x^b is just the coordinate
: t, x, y, or z depending on whether the superscript b is 0, 1, 2, or 3.

: Similarly,  

: C^a_{bd,c} = partial C^a_{bd} / partial x^c

Ah, so now it is "official" :-)

Question: are we supposed to swallow these expressions, or can
we expect an explanation of how you got there ?

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

From galaxy.ucr.edu!not-for-mail Fri Apr 12 16:52:42 PDT 1996
Article: 111121 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 10:01:51 -0700
Organization: University of California, Riverside
Lines: 43
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>Arrived Thursday 11th April 1996. It's improving!

>In article <[email-masked]>, john baez <[email-masked]>
>writes
>>But before I do, are there any other questions about
>>it?  Do y'all see how you could compute the redshift of one galaxy as
>>seen by another, if both are at rest with respect to these coordinates?
>>Do you see how if R(t) is increasing with t, there'll be a redshift?

>I am not convinced that I do. At least in the terms that you seem to be 
>setting out. We have been told that this can be seen as the 'expanding 
>of a wavetrain', and this looks quite acceptable as a naive model, but I 
>am not sure that a naive model is quite good enough here. I would expect 
>to have to calculate a lightlike path through 4-space by integrating 
>with our time dependent metric from point to point and find .....

>Well, that I can't do the maths, of course.

Why can't you do the maths?  I bet you actually can.  You certainly have
the right idea: calculate how a lightlike path goes moves along, and
then suppose you have two such paths starting from a given point in
space and going in a given direction, one starting off at time 0 and one
starting at time T.  Then see when they arrive at a distant point.  The
difference between their arrival times should be more than T if the
universe is expanding.  After all, that would mean that a ticking watch
looks to be ticking slow when seen from a distance, or --- if your watch
happens to be an atom emitting light at a given frequency --- light is
redshifted!  

So, let's see, which way does light go if the metric is

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)    ?

You did want to understand this cosmology stuff, right?  You may find
this calculation intimidating, but it's not too hard and I'm sure folks
out there will be glad to help.  Look at the bright side: you are now in
the position to be able to really figure out how this redshift business
works!

From galaxy.ucr.edu!not-for-mail Fri Apr 12 16:57:54 PDT 1996
Article: 111131 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 10:53:26 -0700
Organization: University of California, Riverside
Lines: 241
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Okay.  So I said space at any given time is flat.  But it's "size" can
>>vary with time!  So the metric on space is not just

>>                      dx^2 + dy^2 + dz^2

>>for all time; instead, there's a time-dependent scale factor

>>                   R(t)^2 (dx^2 + dy^2 + dz^2)

>>Note what this means!  It means that two galaxies which stay at the same
>>position in space --- i.e., their (x,y,z) coordinates are constant as a
>>function of t --- will actually get closer or farther apart, depending
>>on how R(t) changes.  This should make you think of the perennial
>>puzzlement, the perennial yapping and squalling, the perennial
>>confused arguments concerning "whether the galaxies are really moving
>>away from us or not" and "whether space is expanding or not".  Here is
>>where the answer to that puzzlement lies.  Think about it.  

>God. Think about it .. already. Well, I suppose it is at least page
>two.

Having passed your test of valor, you will now be treated as the
fearless sort you have revealed yourself to be.

>Is it quite as simple as this? Well, we have the problem of what exactly 
>you mean by distance, what paths you tread to decide distance, but I 
>think (hope) that in this rather empty isomorphic and reasonably 
>homogenous universe one could have a sensible measure of distance on at 
>least slices at constant t. 

Please tell me that you are now spelling "isotropic" as "isomorphic" on
purpose, just to get back at me for complaining when you spelled it as
"istrophic".  Please.  

You are very right to worry a bit about "which path we tread" when
measuring distance between two points.  I see I have finally put the
fear of God, or rather Riemann in you!  (In the U.S., teachers to whip
their students "to put the fear of God in them", a noble tradition I am
upholding in this course.)

Anyway, when we are computing the distance between two points in space
at a given time t, using the metric

                   R(t)^2 (dx^2 + dy^2 + dz^2),

we need to find the *shortest* path between the two points, and work out
the length of that.  

What path do you think that is?  

Note, here we are NOT taking into account the passage of time at all: we
are simply considering space at one moment of time --- time t as defined
by our particular coordinate system (t,x,y,z).  So what we are doing now
does NOT have any *immediate* relevance to the problem of "how far away
is that ancient galaxy whose light is just now reaching me".  What we
are doing now is of a more purely mathematical nature.  Nonetheless it
is good to do, to help figure out what the hell people really mean when
they maunder on about "space expanding", "galaxies moving away from us"
and the like.

So, I claim you can easily figure out what path is the shortest between
two points in R^3 with the metric 

                   R(t)^2 (dx^2 + dy^2 + dz^2),

and what the shortest distance actually is.  I claim you can easily do
it, because this metric is closely related to another metric on R^3, one
that you have studied since grade school.

>On this measure then things initially  at 
>rest wrt each other will indeed get further apart as time evolves 
>although neither will feel any acceleration.

That's right.  Of course we need to work this out, someday.  We need to
check that if a galaxy stays at a point (x,y,z) as time t passes, it's
following a geodesic, and is thus unaccelerated.  We'll be able to do
this as soon as we understand those Christoffel symbols.  But anyway,
it's true.

>It's going to play hell with working out any celestial mechanics.

No it's not!!  It's going to play hell with the nature human desire to
chatter about things without having any precise meaning in mind.  But we
won't get in trouble as long as we recognize the following are both true
in our setup:

1)  A galaxy remaining at the point (x,y,z) in space as time passes
feels no acceleration, because it's following a geodesic.  

2)  The distance between two such galaxies at points (x0,y0,z0) and
(x1,y1,z1), computed at time t, increases as R(t) increases.  

We just need to get used to this.  

>Oh, I suppose that's why you 
>define your set of co-ordinates to be co-moving as being 'convenient'. 
>In this co-ordinate system 'relative accelerations' due to R(t) are 
>zero, so at least to first order it's a bit simpler when you come to 
>dealing with other forces and accelerations. Hmmm, if so that's quite a 
>good idea. 

It's definitely good to have coordinates where remaining "at rest" ---
i.e., at constant (x,y,z) --- means you follow a geodesic.  Also, in
this coordinate system we'll actually have our galaxies remaining at
rest in this sense.  

>So, I conclude that galaxies are indeed moving away from us.

The distance from us to them increases, but it is not clear what you
mean by saying that they are "moving".  They are not moving any more, or
less, than we are: we are all AT REST with respect to the coordinate
system (t,x,y,z).  Remember, motion and rest are defined only with
respect to a coordinate system!  

So if you go up to someone and say "the galaxies are moving away from
us", they are likely to get the wrong idea, even if you have something
precise in mind.  Say what you really mean: the distance between us and
them, computed in the spacelike slice at time t, increases as t increases.

Of course, most people who say "the galaxies are moving away from us"
have nothing precise in mind at all!  They are simply chattering away.

>Is space expanding or not? Well, in time honoured GR fashion one is 
>tempted to say "it depends on what you mean by expanding", but this is 
>being devious. If I go and measure a volume of space marked by eight 
>free-falling bodies then after a time come back and do it again (with 
>appropriate corrections of course) I would find a bigger volume is 
>enclosed if I was careful to site them appropriately. So, I would 
>conclude that space is expanding . 

Definitely your thought experiment works out as you say.  More
precisely: if you take 8 bodies at rest as defined above -- NOT just
randomly free-falling -- and you repeatedly measure, or compute, the
volume of the cube they form, the volume would increase as time passes
(assuming R(t) increases as time passes).  

How does the volume depend on R(t), exactly?

To conclude that "space is expanding" is fine as long as you make clear
what exactly you REALLY MEAN when you say that: you mean your thought
experiment will work out as you say!  But to just go up to someone and
say "space is expanding" without further explanation is begging for
trouble.  They may think you mean something else.  They may get freaked
out and think you mean something very vague and mysterious.  Again, most
people who say "space is expanding" don't necessarily have any precise
thought experiment in mind.  

>But on the other hand this sort of thing is easier said than done. What 
>if one of my test bodies was near a sun, orbiting it even? Then I am in 
>real trouble. Do I start to say that bits of the universe are not 
>expanding, or anyway doing weird things in the expansion line? 
>Heellllllp.

Well, now you see the dangers of imprecise formulations.

>>But there is more to the story, I hasten to add.  More than a
>>time-dependent Riemannian metric on space!  There is an honest
>>Lorentzian metric on spacetime!  (Remember, "Lorentzian" means that
>>there's a minus sign lurking in there somewhere.)  It is:

>>                 -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

>>or
>>                     -1   0       0   0
>>            g_{ab} =  0   R(t)^2  0   0
>>                      0   0    R(t)^2 0
>>                      0   0       0  R(t)^2

>>where as usual a and b range over 0,1,2,3, corresponding to t,x,y,z.

>>So: if R(t) = 1 for all t, this is just the usual Minkowski metric, but
>>if R(t) changes then --- roughly speaking, remember this is just a
>>bunch of words! --- space "expands" or "contracts" as time passes.  

>>Okay, does this make sense?  Can you see how we'd use it to model a big
>>bang?  R(t) would start out at zero and get bigger as time passes.  Can
>>you see why it would make redshifts and all that jazz happen?  Can you
>>see why it simply FINESSES the question of whether the galaxies are
>>"moving away from each other" or not?  I hope so. 

>Er. Um. Ooooh.
>What does it mean to have a metric with R(t)=0??; <1; >1 ???

When R(t) = 0 our metric is "degenerate": it's not really an invertible
matrix any more, so it's not technically a Lorentzian metric.  This
corresponds to the fact that when R(t) equals zero, the distance between
different points in space becomes ZERO!  This sucks, so we don't usually
consider the case R(t) = 0 as legitimate.  Strictly speaking, in the big
bang R(t) -> 0 as t -> 0, but we don't talk about what's happening AT t
= 0, despite the our natural desire to blab about the "moment when it
all started".

>Equally what would be the effect of having a metric 
>-R'(t)dt^2 + dx^2 + dy^2 + dz^2  where R'(t) decreases from 00 through 1 
>to -OO??? I would have thought this might ease the maths a little at the 
>very least.

Believe me, I am doing everything to make your ride as comfortable as
possible.   You can try various coordinate transformations on
the metric I gave, to see if you can get it to look like the metric you
suggest.   I don't see offhand how to do it, but you should try various
ways of stretching and squashing the t,x,y,z coordinates.  (I certainly
don't think you want R'(t) negative, by the way: it needs to be
positive for us to have a Lorentzian metric.)  

>Er, please sir, why can't we simply say that space IS expanding?

You can say it.  Just don't expect people to understand what you
mean.

>I mean, 
>you get the same answer and it's rather easier to visualise. OK, it's 
>not expanding in the sense that each little part feels no acceleration, 
>but that's not so hard to mentally tack in.

There are all sorts of fun ways to think about things and visualize
them, I am not opposed to that.  Just do it in the privacy of your own
brain.  Visualizing physics is sort of like sexual fantasizing.  We all
do it and we all enjoy it, but it's impolite to talk about it to
strangers.

>>That will take a while, since we
>>need to learn how to compute the Riemann tensor!  But then we can try to
>>solve Einstein's equation to see how R(t) actually changes with time.
>>That'll depend, of course, on what sort of *matter* is around.... dust,
>>radiation, or whatever.  

>Lets go for it!

Okay!  Any minute now....

From galaxy.ucr.edu!not-for-mail Fri Apr 12 17:16:12 PDT 1996
Article: 111163 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 13:25:27 -0700
Organization: University of California, Riverside
Lines: 68
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Patrick van Esch) writes:
>: Right.  Besides, use psychology.  If this metric it WERE equivalent to
>: flat Minkowski space, do you think I'd be using it as my main example of
>: a metric in this second part of the general relativity course?  

>Of course not !  But I thought it might have been a good exercise 
>to try and see that it doesn't work out.

Right.  It's definitely worthwhile to try to find a coordinate
transformation that will change the big bang metric

               g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

into the Minkowski metric, and discover that you can't (unless R is
constant).  

It's also good to try to compute the curvature now and see that we
don't get zero, using this stuff:

>:          C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

>: R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

where

>: g_{cd,b} = partial g_{cd} / partial x^b

and 

>: C^a_{bd,c} = partial C^a_{bd} / partial x^c

>Ah, so now it is "official" :-)

Yes, just to give you folks something to calculate this weekend.

>Question: are we supposed to swallow these expressions, or can
>we expect an explanation of how you got there ?

Oh no, of course you shouldn't swallow these horrible formulas!  I mean,
you should trust me that they are probably right, except for maybe a few
typos :-), and try computing some stuff with them, but certainly I owe
you a good explanation of where the hell they come from: that is going
to be the main goal of this part of the GR tutorial.

This is what I'm going to do: 1) get you folks to feel at home with the
spatially flat big bang metric 

               g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2),

and get you to understand all that nonsense about "galaxies moving
apart" and what it really means, 2) get you to understand the
Christoffel symbols, what they have to do with parallel transport and
how to compute them, 3) get you to understand how to compute the Riemann
tensor.  Then we can compute the Riemann curvature of the big bang
metric, compute the Einstein tensor, and solve Einstein's equation (assuming
the universe is full of dust or radiation or something).  Then we will
all be at home in the big bang cosmology, and we will all have a
somewhat practical understanding of a bit of general relativity, and
then we'll be DONE.  That is, I will call it quits.  

We are just playing around with 1) now, and haven't gotten to the more
substantial 2) and 3), but we can still look ahead a bit, as it were,
and compute some stuff.

Newsgroups: sci.physics
Subject: General Relativity Tutorial
Summary: 
Expires: 
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Why can't you do the maths?  I bet you actually can.  

>I am pretty sure I could understand the maths. Doing it from scratch
>with no textbook is, I am afraid, rather more of a problem.

Well, I'll help you out.  We want to see how light travels in the 
spacetime with the metric

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2),

so we can work out how it gets redshifted: i.e., how a clock "at rest"
will be seen as ticking slow by somebody else far away who is also "at rest".

Note: I will stop putting quotes "at rest" as soon as it's clear
everyone understands I really mean "following a worldline for which x,y,
and z remain constant as t varies."  The point is that "at rest" is a
coordinate-dependent notion.  There is no "absolute rest"!  When I
say "at rest" it not meant to have any grand metaphysical significance;
it's simply a handy quick way of speaking.  

>>Look at the bright side: you are now in
>>the position to be able to really figure out how this redshift business
>>works!

>Yes, that's the point. So far I am amazed how, using mostly words and
>only a tiny tad of maths, the understanding is becoming (slowly) much
>clearer. Why, I even follow much of what you and Ted are talking about!
>Sometimes.

Great.  But...

>We need two lightlike paths in this spacetime. Hmm what might that mean.
>Well locally we ought still to be Minkowskian so light travels at 45deg.
>I guess that it would mean that dx/dt=1 and we can set dy,dz=0. 
>(Shaky ground feeling)

WHAT?????????   Since when is the metric here Minkowskian?!?!?!  

(The wizard hurls thunderbolts left and right.)

Remember, our metric here is NOT the Minkowski metric

              -dt^2 + dx^2 + dy^2 + dz^2,

it's 

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)!

Now I am paying the price --- and you are, too --- for me having coddled
you earlier.  Yes, one can always CHOOSE coordinates such that at a
SINGLE POINT the metric looks like the Minkowski metric.  But here we
are not doing a calculation at a SINGLE POINT of spacetime; we are
trying to work out how light rays move through spacetime.  And
furthermore, I'm not gonna let you CHOOSE the coordinates; I have
already chosen the coordinates we are going to work in, and the
metric is decidedly not the Minkowski one, in these coordinates.  

I have repeatedly warned you of the dangers of getting carried away with this
"assume it's Minkowskian" stuff.  Do you see why, even though you can
CHOOSE coordinates so that the metric is Minkowskian at a SINGLE POINT,
here we are in a situation where we have already chosen coordinates and
are not working things out at a single point, so that this trick is of
no use?  I hope so.  If you make this mistake again I will actually get
a bit upset.   

(The wizard winks.)  

Anyway, at least you had that "shaky ground feeling".  It was trying to
tell you something.  

So let's see, why is it that in the Minkowski metric 

              -dt^2 + dx^2 + dy^2 + dz^2

light can move along with dx/dt = 1, dy/dt = dz/dt = 0?  And how will it
move in the big bang metric

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)?

Well, remember, a tangent vector is "lightlike" if its length is zero.
The worldline traced out by light is always, like, light-like.  In other
words, its tangent vector has length zero.  So for example in Minkowski
space light can move along the curve 

              (t,x,y,z) = (s,s,0,0)

where s is an arbitrary parameter.  Why?  Well, the tangent vector of
this curve is 

              d(t,x,y,z)/ds = (dt/ds, dx/ds, dy/ds, dz/ds) 
                            = (1,1,0,0)

Since t = s, this is just another way of saying that dt/dt = dx/dt = 1,
while dy/dt = dz/dt = 0.  I have just introduced the parameter s to
confuse you, or possibly enlighten you, since it lets us see that t
isn't really any more fundamental than x, y, or z.  

Anyway, the tangent vector (1,1,0,0) is lightlike: if we work out its
length^2 using the Minkowski metric we get -1^2 + 1^2 + 0^2 + 0^2 = 0.
Zero length!  Lightlike!

I hope this is clear as day.  The Minkowski metric 

              -dt^2 + dx^2 + dy^2 + dz^2

is making light take one step in the x direction for every one step in
the t direction, as long as it doesn't budge in the y and z directions.

But now

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

doesn't work quite the same way.  What'll light do in this metric?
Well, a hint: the vector (R(t),1,0,0) will be lightlike in this metric
since its length^2 is 

                  -1 R(t)^2 + R(t)^2 (1^2 + 0^2 + 0^2) = 0

So the worldline of light moving in the x direction should have its
tangent vector pointing in the direction (R(t),1,0,0).

I hope this too is clear as day.  The metric

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

is making light take one step in the x direction for every R(t) steps in
the t direction, as long as it doesn't budge in the y and z directions.

So now how does the light move?

I think this is worth working out, since it'll let you really understand
redshifts.  

From noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!tank.news.pipex.net!pipex!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Mon Apr 15 15:32:13 PDT 1996
Article: 9292 of sci.physics
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From: Oz <[email-masked]>
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Subject: Re: General relativity tutorial
Date: Sat, 13 Apr 1996 13:49:51 +0100
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>Why can't you do the maths?  I bet you actually can.  

I am pretty sure I could understand the maths. Doing it from scratch
with no textbook is, I am afraid, rather more of a problem.

>You certainly have
>the right idea: calculate how a lightlike path goes moves along, and
>then suppose you have two such paths starting from a given point in
>space and going in a given direction, one starting off at time 0 and one
>starting at time T.  Then see when they arrive at a distant point.  The
>difference between their arrival times should be more than T if the
>universe is expanding.  After all, that would mean that a ticking watch
>looks to be ticking slow when seen from a distance, or --- if your watch
>happens to be an atom emitting light at a given frequency --- light is
>redshifted!  

Exactly what I had in mind.

>
>So, let's see, which way does light go if the metric is
>
>              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)    ?
>
>You did want to understand this cosmology stuff, right?  

Spot on. You *have* noticed then?  :-)

>You may find
>this calculation intimidating, but it's not too hard and I'm sure folks
>out there will be glad to help.  

Ah, well that's another matter. That means I can struggle on and then
some nice soul can tell me what an idiot I am and then do it properly
for me to see how it should be done.

>Look at the bright side: you are now in
>the position to be able to really figure out how this redshift business
>works!

Yes, that's the point. So far I am amazed how, using mostly words and
only a tiny tad of maths, the understanding is becoming (slowly) much
clearer. Why, I even follow much of what you and Ted are talking about!
Sometimes.

>
>So, let's see, which way does light go if the metric is
>
>              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)    ?
>
>You did want to understand this cosmology stuff, right?  

Well the only guide I have is a short post of Matthew Wiener's. Now he
kindly did it the idiot's way, but I feel that there really should be a
better method. Also the straightforward technique he used give me an S
of zero for light, hmmm well I suppose that's right for SR.

We need two lightlike paths in this spacetime. Hmm what might that mean.
Well locally we ought still to be Minkowskian so light travels at 45deg.
I guess that it would mean that dx/dt=1 and we can set dy,dz=0. 
(Shaky ground feeling)

so      dS^2= -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)  
becomes dS^2= -dt^2 + R(t)^2 dx^2               and substituting dx=dt
        dS^2= -dt^2 + R(t)^2 dt^2 = (R(t)-1)dt^2        or
        dS=(R(t)-1)^0.5 dt                      which looks promising

Now all we would have to do is integrate. Now since both points are
stationary in x,y,z so long as our light beam sets off along the x-axis
(shake, shake) from x0 it will eventually arrive at the other point x1
(as long as it's not too far away, that is). So we could call them
(T1,x0) to (ta,x1) as one path, and (T2,x0) to (tb,x1). However we don't
know ta and tb and I have the feeling that if I suggested that given an
R(t) we could do it by a numerical integration the old Wiz's eyes would
pop out. Doubtless there is an appropriate mathematical method.

However it is fairly easy to see that so long as R(t) is a variable one
would typically get a different S for each path. Actually I am not at
all happy with this, I don't think I am handling a lightlike path in the
best way and I am rather uncomfortable with the idea that locally the
elapsed time for the lightbeam is zero, but when integrated it isn't,
although it doesn't feel as stupid as it ought to for some reason.

OK, done my effort. Now let's see how it *should* be done!

I just hope to goodness the journeyman wizard comes back and checks this
out so the Great Wizard never has to see this......fat chance I suppose.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

Newsgroups: sci.physics
Subject: Re: Riemann Awakes
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>>No doubt this is an easy 
>>>consequence of the symmetries of the Weyl tensor.

>>I wish it were.  I'm a bit nervous about that, though.  We can recover
>>the Weyl tensor if we know how balls of dust moving along at arbitrary
>>velocity change shape into ellipsoids (in their own rest frame).  But we
>>can only apply the isotropy argument to balls of dust which are at rest
>>in the "cosmic rest frame" of the big bang.  I think we only get 5
>>components of the Weyl tensor to vanish by this argument.  (There are 
>>5 linearly independent 3x3 symmetric traceless matrices.)  5 out of 10
>>ain't bad... but it doesn't look like we'll be able to get all 10
>>components of the Weyl to vanish without some extra reasoning.

>Could anyone say what the physical significance of these five are?

It takes 5 numbers to describe the different ways a round ball can
change shape into an ellipsoid while keeping the same volume.  The slick
way to see this is to note the above parenthetical remark about
symmetric traceless matrices.  But we can be grungy: two numbers tell us
which direction the longest axis of the ellipsoid points, one more tells
how much that direction gets stretched, one more tells us where the
shortest axis points (just one needed since we know it's perpendicular
to the longest axis), and one more tells us how much that axis gets
scrunched.  (Since the volume of the ellipsoid equals the volume of the
ball, we can determine from all this how much the 3rd axis must be
stretched or scrunched.)

>Gravitational waves?

As a gravitational wave passes through empty space, an initially round
ball of dust in free fall will turn into an ellipsoid of the same
volume, and these 5 numbers describe exactly what the ellipsoid looks
like.

Gravitational waves are somewhat similar to electromagnetic waves.  
These 5 components are the called the "electric" part of the Weyl
curvature, but there are also 5 "magnetic" components.  Recall what
Steve Carlip said about this a while back.  I had said:

: In short, when we are in truly empty space, there's no Ricci curvature,
: so actually our ball of coffee grounds doesn't change volume (to
: first order, or second order, or whatever).  But there can be Weyl
: curvature due to gravitational waves, tidal forces, and the like.  
: Gravitational waves and tidal forces tend to stretch things out in one
: direction while squashing them in the other.  So these would correspond
: to our ball changing into an ellipsoid!  Just as we hoped.

And he replied:

Yes, this is at least mostly right.  (The "mostly" comes because
there's a part I don't understand---I'll explain below.)  Start
with your ball of coffee grounds, with an initial four-velocity
u^a.  (In the rest frame, u^0=1 and the rest of the components
are zero.)  If you look at the configuration slightly later, it
will typically have changed in three ways.  First, it may have
expanded or contracted (its volume may have changed).  Second,
it may have twisted.  Third, it may have sheared---that is, become
distorted from a sphere to an ellipsoid without changing volume.
The Weyl tensor (in empty space) gives the rate of change of the
shear.

Specifically, you can describe shear by a three by three symmetric
matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX
users).  The three eigenvectors of \sigma give three spatial axes,
and the eigenvalues give the rate of expansion along each axis.
The fact that \sigma is traceless means that the sum of the rates
of expansion is zero, i.e., the total volume is remaining constant.
In empty space, if your ball of coffee grounds has no initial shear,
rotation, or expansion, the rate of change of \sigma is given by the
Weyl tensor contracted twice with u.  (If you want that in symbols,
I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

This means that the Weyl tensor is the thing gravitational radiation
detectors are designed to measure.  A laser interferometer detector
like LIGO or VIRGO (now under construction) is just a very large, very
accurate interferometer with two perpendicular arms.  If the Weyl tensor
changes, one arm will expand while the other one contracts, changing the
relative lengths and thus the interference pattern.

Now, the part I don't understand.  The Weyl tensor really splits into
two parts, and "electric" part (with five components) and a "magnetic"
part (also with five components).  The contraction C(*,u,*,u) that I
described above is the electric part, in the rest frame of the coffee
grounds.  Like ordinary electric and magnetic fields, the electric and
magnetic components of the Weyl tensor mix under coordinate changes.
But it would be nice to have a direct geometrical picture of the magnetic
part.  Does anybody know one?  (Presumably it's hidden in what's called
the Newman-Penrose formalism, but I've never learned that very well.)

Newsgroups: sci.physics
Subject: Re: Riemann Awakes
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>>No doubt this is an easy 
>>>consequence of the symmetries of the Weyl tensor.

>>I wish it were.  I'm a bit nervous about that, though.  We can recover
>>the Weyl tensor if we know how balls of dust moving along at arbitrary
>>velocity change shape into ellipsoids (in their own rest frame).  But we
>>can only apply the isotropy argument to balls of dust which are at rest
>>in the "cosmic rest frame" of the big bang.  I think we only get 5
>>components of the Weyl tensor to vanish by this argument.  (There are 
>>5 linearly independent 3x3 symmetric traceless matrices.)  5 out of 10
>>ain't bad... but it doesn't look like we'll be able to get all 10
>>components of the Weyl to vanish without some extra reasoning.

>Could anyone say what the physical significance of these five are?

It takes 5 numbers to describe the different ways a round ball can
change shape into an ellipsoid while keeping the same volume.  The slick
way to see this is to note the above parenthetical remark about
symmetric traceless matrices.  But we can be grungy: two numbers tell us
which direction the longest axis of the ellipsoid points, one more tells
how much that direction gets stretched, one more tells us where the
shortest axis points (just one needed since we know it's perpendicular
to the longest axis), and one more tells us how much that axis gets
scrunched.  (Since the volume of the ellipsoid equals the volume of the
ball, we can determine from all this how much the 3rd axis must be
stretched or scrunched.)

>Gravitational waves?

As a gravitational wave passes through empty space, an initially round
ball of dust in free fall will turn into an ellipsoid of the same
volume, and these 5 numbers describe exactly what the ellipsoid looks
like.

Gravitational waves are somewhat similar to electromagnetic waves.  
These 5 components are the called the "electric" part of the Weyl
curvature, but there are also 5 "magnetic" components.  Recall what
Steve Carlip said about this a while back.  I had said:

: In short, when we are in truly empty space, there's no Ricci curvature,
: so actually our ball of coffee grounds doesn't change volume (to
: first order, or second order, or whatever).  But there can be Weyl
: curvature due to gravitational waves, tidal forces, and the like.  
: Gravitational waves and tidal forces tend to stretch things out in one
: direction while squashing them in the other.  So these would correspond
: to our ball changing into an ellipsoid!  Just as we hoped.

And he replied:

Yes, this is at least mostly right.  (The "mostly" comes because
there's a part I don't understand---I'll explain below.)  Start
with your ball of coffee grounds, with an initial four-velocity
u^a.  (In the rest frame, u^0=1 and the rest of the components
are zero.)  If you look at the configuration slightly later, it
will typically have changed in three ways.  First, it may have
expanded or contracted (its volume may have changed).  Second,
it may have twisted.  Third, it may have sheared---that is, become
distorted from a sphere to an ellipsoid without changing volume.
The Weyl tensor (in empty space) gives the rate of change of the
shear.

Specifically, you can describe shear by a three by three symmetric
matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX
users).  The three eigenvectors of \sigma give three spatial axes,
and the eigenvalues give the rate of expansion along each axis.
The fact that \sigma is traceless means that the sum of the rates
of expansion is zero, i.e., the total volume is remaining constant.
In empty space, if your ball of coffee grounds has no initial shear,
rotation, or expansion, the rate of change of \sigma is given by the
Weyl tensor contracted twice with u.  (If you want that in symbols,
I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

This means that the Weyl tensor is the thing gravitational radiation
detectors are designed to measure.  A laser interferometer detector
like LIGO or VIRGO (now under construction) is just a very large, very
accurate interferometer with two perpendicular arms.  If the Weyl tensor
changes, one arm will expand while the other one contracts, changing the
relative lengths and thus the interference pattern.

Now, the part I don't understand.  The Weyl tensor really splits into
two parts, and "electric" part (with five components) and a "magnetic"
part (also with five components).  The contraction C(*,u,*,u) that I
described above is the electric part, in the rest frame of the coffee
grounds.  Like ordinary electric and magnetic fields, the electric and
magnetic components of the Weyl tensor mix under coordinate changes.
But it would be nice to have a direct geometrical picture of the magnetic
part.  Does anybody know one?  (Presumably it's hidden in what's called
the Newman-Penrose formalism, but I've never learned that very well.)

Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Summary: 
Expires: 
References: <[email-masked]> <[email-masked]> <[email-masked]>
Sender: 
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Keywords: 

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>>No doubt this is an easy 
>>>consequence of the symmetries of the Weyl tensor.

>>I wish it were.  I'm a bit nervous about that, though.  We can recover
>>the Weyl tensor if we know how balls of dust moving along at arbitrary
>>velocity change shape into ellipsoids (in their own rest frame).  But we
>>can only apply the isotropy argument to balls of dust which are at rest
>>in the "cosmic rest frame" of the big bang.  I think we only get 5
>>components of the Weyl tensor to vanish by this argument.  (There are 
>>5 linearly independent 3x3 symmetric traceless matrices.)  5 out of 10
>>ain't bad... but it doesn't look like we'll be able to get all 10
>>components of the Weyl to vanish without some extra reasoning.

>Could anyone say what the physical significance of these five are?

It takes 5 numbers to describe the different ways a round ball can
change shape into an ellipsoid while keeping the same volume.  The slick
way to see this is to note the above parenthetical remark about
symmetric traceless matrices.  But we can be grungy: two numbers tell us
which direction the longest axis of the ellipsoid points, one more tells
how much that direction gets stretched, one more tells us where the
shortest axis points (just one needed since we know it's perpendicular
to the longest axis), and one more tells us how much that axis gets
scrunched.  (Since the volume of the ellipsoid equals the volume of the
ball, we can determine from all this how much the 3rd axis must be
stretched or scrunched.)

>Gravitational waves?

As a gravitational wave passes through empty space, an initially round
ball of dust in free fall will turn into an ellipsoid of the same
volume, and these 5 numbers describe exactly what the ellipsoid looks
like.

Gravitational waves are somewhat similar to electromagnetic waves.  
These 5 components are the called the "electric" part of the Weyl
curvature, but there are also 5 "magnetic" components.  Recall what
Steve Carlip said about this a while back.  I had said:

: In short, when we are in truly empty space, there's no Ricci curvature,
: so actually our ball of coffee grounds doesn't change volume (to
: first order, or second order, or whatever).  But there can be Weyl
: curvature due to gravitational waves, tidal forces, and the like.  
: Gravitational waves and tidal forces tend to stretch things out in one
: direction while squashing them in the other.  So these would correspond
: to our ball changing into an ellipsoid!  Just as we hoped.

And he replied:

Yes, this is at least mostly right.  (The "mostly" comes because
there's a part I don't understand---I'll explain below.)  Start
with your ball of coffee grounds, with an initial four-velocity
u^a.  (In the rest frame, u^0=1 and the rest of the components
are zero.)  If you look at the configuration slightly later, it
will typically have changed in three ways.  First, it may have
expanded or contracted (its volume may have changed).  Second,
it may have twisted.  Third, it may have sheared---that is, become
distorted from a sphere to an ellipsoid without changing volume.
The Weyl tensor (in empty space) gives the rate of change of the
shear.

Specifically, you can describe shear by a three by three symmetric
matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX
users).  The three eigenvectors of \sigma give three spatial axes,
and the eigenvalues give the rate of expansion along each axis.
The fact that \sigma is traceless means that the sum of the rates
of expansion is zero, i.e., the total volume is remaining constant.
In empty space, if your ball of coffee grounds has no initial shear,
rotation, or expansion, the rate of change of \sigma is given by the
Weyl tensor contracted twice with u.  (If you want that in symbols,
I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).)

This means that the Weyl tensor is the thing gravitational radiation
detectors are designed to measure.  A laser interferometer detector
like LIGO or VIRGO (now under construction) is just a very large, very
accurate interferometer with two perpendicular arms.  If the Weyl tensor
changes, one arm will expand while the other one contracts, changing the
relative lengths and thus the interference pattern.

Now, the part I don't understand.  The Weyl tensor really splits into
two parts, and "electric" part (with five components) and a "magnetic"
part (also with five components).  The contraction C(*,u,*,u) that I
described above is the electric part, in the rest frame of the coffee
grounds.  Like ordinary electric and magnetic fields, the electric and
magnetic components of the Weyl tensor mix under coordinate changes.
But it would be nice to have a direct geometrical picture of the magnetic
part.  Does anybody know one?  (Presumably it's hidden in what's called
the Newman-Penrose formalism, but I've never learned that very well.)

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Mon Apr 15 16:22:41 PDT 1996
Article: 9006 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 12 Apr 1996 00:46:04 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

>I don't know yet if (x0,y0,z0,t) is a geodesic !  (well, it better be,
>but I hope John could comment on this one...)

Yes, these are geodesics in the universe with metric

-dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

When we get a little better at math we'll be able to show this.
But anyway, that's one of the pleasant features of this coordinate
system: if we "stay at the same place", in the coordinate-dependent
sense that our worldline traces out a curve in spacetime for which
(x,y,z) remains constant as t varies, then we are in free-fall. 

>If you define "acceleration" as second derivative of the distance,
>then (except if R(t) = a + b.t) there are, for sure, relative accelerations.

Certainly.  Here you are presumably measuring the distance between two
points (x0,y0,z0) and (x1,y1,z1) at the same time t.  

Quickie quiz, to make sure everyone is awake: what *is* the distance
between the points (x0,y0,z0) and (x1,y1,z1) at time t, if the spacetime
metric is 

-dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2) ?

Make sure to tell me what you MEAN by "the distance", while you're at
it.

>>Hmmm, if so that's quite a 
>: good idea. So, I conclude that galaxies are indeed moving away from us.

>they might be closing in or vibrating back and forth, depending on R(t) :-)

It would not be giving too much away to reveal that in practice, R(t)
will be positive and will increase with time.

>Well, that was my question before !  Can't we think up a coordinate grid
>in which we absorb R(t).  Then we would end up with flat Minkowski space.
>But I'm pretty sure it cannot be done (as John said, but we're not supposed
>to look behind the curtain :-) we'll end up with a non-zero curvature.
>So it cannot be equivalent to a flat Minkowski space.

Right.  Besides, use psychology.  If this metric it WERE equivalent to
flat Minkowski space, do you think I'd be using it as my main example of
a metric in this second part of the general relativity course?  

By the way, Oz was allowed to go behind the curtain, and this is what he
saw:

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

So you can use this to compute the Riemann tensor whenever you want ---
at least, after I tell you want those commas mean in things like
g_{cd,b} and C^a_{bd,c}.  They mean partial differentiation!  So for
example

g_{cd,b} = partial g_{cd} / partial x^b

where "partial" is a feeble ASCII equivalent of one of those curly
d's they use for partial derivatives.  Here x^b is just the coordinate
t, x, y, or z depending on whether the superscript b is 0, 1, 2, or 3.

Similarly,  

C^a_{bd,c} = partial C^a_{bd} / partial x^c

and so on.  

So whenever you have a little spare time you can compute the Riemann
curvature of this big bang metric and see for yourself that it's
nonzero.

Actually one enterprising fellow has already computed the Christoffel
symbols C^a_{bc} --- and, even better, explained what they mean!

More on that later.

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!newsfeed.internetmci.com!howland.reston.ans.net!math.ohio-state.edu!magnus.acs.ohio-state.edu!lerc.nasa.gov!purdue!haven.umd.edu!hecate.umd.edu!mojo.eng.umd.edu!pascal.eng.umd.edu!not-for-mail Mon Apr 15 23:49:17 PDT 1996
Article: 8790 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 11 Apr 1996 09:43:15 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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In article <[email-masked]>, john baez <[email-masked]> wrote:
>Hmm.  No post from Oz yet in response to my attempt to start him up on the
>next phase of learning general relativity, although he did email me
>saying he found the following equations suitably intimidating:
>
>              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)
>
>         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})
>
>R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}
>

Since I have too much time on my hands (well, not really, but we can
pretend) I decided to get adventuresome and calculate out the
Cristoffel symbols for this metric.  (If you want to do it all by
yourself, stop reading here and save the rest of this article to read
later)

So, here's what I got.

i can be 1,2, or 3 in the following, no Einstein summation notation

C^0_ii = R(t)R'(t)
C^i_0i = C^i_i0 = R'(t) / R(t)
all others = 0

At the moment, I'm trying to understand the physical meaning of all of
these.  For those who don't know what the Cristoffel symbols mean (I
don't think John has come out and said it), C^a_cd is the component in
the a-direction of the covarient derivative of the vector d in the c
direction.  I think that makes no sense what-so-ever to anyone who
doesn't already know what it means.  (BTW, I'm not really way ahead of
the class, I just happen to be taking an actual 'real' class at the
moment where we are talking about the same type of stuff).  So, here's
an attempt to translate that into english.  Note:  I may not get this
quite right, but I'm sure John will correct me if I screw up.

So, C^a_cd is a tensor which takes two vectors and spits out a third.

So w^a = C^a_cd u^c v^d

Specifically, we need a (smooth?) vector field V near a point of interest.

We do this by starting at a point with the member of the vector field
at the point v.  We then do a parallel transport of v for an
infinitesimal distance in the u direction.  We look at how v has
changed compared to the vector field at the new point and define w as
that change.  This sounds a little circular, because we are trying to
define parallel transport, but use the notion in defining C^a_cd.  But
really, given the above formula, we are specifying exactly what
parallel transport is on a particular set of basis vectors, and from
there we can figure it out for an arbitrary set of vectors.

I hope that was better (and not hideously wrong)

So, based on that, what is the physical meaning of the Cristoffel
symbols for this metric

C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
vector in the x direction.  Then if we move forward in time a little
bit, this vector has 'stretched' in the x direction by a factor of
R'(t)/R(t).  This, I think, just describes the Hubble expansion and
R'(t)/R(t) is, I think, the Hubble constant.

C^i_i0 makes less sense to me right away, except that I know that from
a purely mathematical analysis that C^a_cd is symmetric wrt c and d.
But the physical meaning...OK, we take a unit time vector and, holding
time constant, we move this vector a little in the x direction.  We
then find that it is slanted a little in the x direction

	t				t
	.				.
	.				.
	.				.
	.				.
 .......|.......x  becomes        ......./.....x
	.				.
	.				.
	.				.

Oh, wait is this maybe just also saying that the unit time vectors
don't look quite parallel, but instead they seem to be carrying nearby
points apart, once again by that factor R'(t)/R(t).  What do other
people think?

OK, last one

C^0_ii = R(t)R'(t)

So we take a unit x vector, push it in the x direction, and find it
has gained a t component.

	t				t
	.				.
	.				.
	.				.
	.				.   _/
 ........___....x  becomes        ........./.....x
	.				.
	.				.
	.				.

(That one's harder to draw in ASCII)

So, nearby unit x vectors seem to be diverging in time?  I don't get
this one.

Hmm...food for thought.

Kevin Scaldeferri

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue Apr 16 10:51:29 PDT 1996
Article: 9698 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 15 Apr 1996 23:15:43 -0700
Organization: University of California, Riverside
Lines: 57
Message-ID: <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:
>In <[email-masked]>, Oz <[email-masked]> writes:
> :What 
> :if one of my test bodies was near a sun, orbiting it even? Then I am in 
> :real trouble. Do I start to say that bits of the universe are not 
> :expanding, or anyway doing weird things in the expansion line? 
> :Heellllllp.

>I also always wondered about that. Here are my views on this:

>A test body orbiting the sun due to the gravitational field must
>be described by the GR, of course. This means that we loose our
>homogenity and isotropy. To describe a test body in orbit around
>a sun requires the Riemann metric, and to describe that interacting
>with the big bang metric outlined above is certainly non trivial.
>I therefore leave it to the experts to figure it out.

It's very nontrivial, and not an "exactly solvable in closed form" type
problem, but the basic rule of thumb is that gravitationally bound
systems don't expand along with the expansion of the universe.  For a
wee bit more on this, see the FAQ (appended below).

>...So let's look at a classical 1/r^2 force
>field and a test particle orbiting the origin. What do we mean when
>we say the force is proportional to 1/r squared? What is r?
>Is it sqrt(x^2 + y^2) or is it R(t)^2 that value? I'm tempted to
>say that r should be the distance to the origin, and therefore
>the latter value. 

Of course gravity isn't exactly described by a 1/r^2 force in general
relativity.  But if you are willing to settle for an approximation,
which is after all very good lots of the time, you should choose r^2 to
be the "distance in space squared" not sqrt(x^2 + y^2 + z^2), because
the latter is blatantly something that depends on our more or less
random choice of coordinates.

For more on "the distance in space", see my recent batch of posts where
I'm getting Oz and Esch to compute it in a simple case.  Note that the
"distance in space" depends on a choice of time function t, since it's
only after we choose this function that we know what "space at a given
time t" means!  So even the "distance in space squared" is a
coordinate-dependent thing, though it's LESS coordinate-dependent than 
sqrt(x^2 + y^2 + z^2).  What this means is that even it cannot always be
the right thing to use in the problem you gave.  But it's okay to use
this when studying a planet orbiting a star --- in the Newtonian
approximation --- in the big bang universe we are considering.  

You guys sure like to make things complicated!  :-)  Putting stars and
planets in my perfectly homogeneous, isotropic, simplified, bland cosmology.  

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue Apr 16 10:52:44 PDT 1996
Article: 9705 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 15 Apr 1996 23:49:07 -0700
Organization: University of California, Riverside
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NNTP-Posting-Host: guitar.ucr.edu

Sorry, it took me a while to respond to this, because it's rather
substantial.

In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:

>>              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

>>         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

>>R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

>So, here's what I got.

>i can be 1,2, or 3 in the following, no Einstein summation notation
>
>C^0_ii = R(t)R'(t)
>C^i_0i = C^i_i0 = R'(t) / R(t)
>all others = 0

That looks about right, but Oz, or Esch, or maybe Ed "algebra is the great
stupefier" Green -- had better check it to be sure.

>At the moment, I'm trying to understand the physical meaning of all of
>these.  For those who don't know what the Christoffel symbols mean (I
>don't think John has come out and said it), C^a_cd is the component in
>the a-direction of the covarient derivative of the vector d in the c
>direction. 

Yes, that sounds right.  

>I think that makes no sense what-so-ever to anyone who
>doesn't already know what it means. 

That also sounds right.  I'd put it this way: say we WORK IN LOCAL
COORDINATES and take a unit vector pointing in the d direction and
parallel transport it an amount epsilon in the c direction.  Now it
won't be pointing in the "same direction" anymore!  It'll point a bit
more in the a direction than it had been.  How much?  Well, about

                      - epsilon C^a_{cd} 

much.  

Note that annoying minus sign.  This minus sign is gonna come back and
haunt us at various points, and at some point I may get fed up and
decide to change my conventions so that it's not there.  But for now
let's have it be there.

Note also the quotes around "same direction".  Remember, it doesn't
exactly make sense to speak of two tangent vectors at different points
pointing in the "same direction".  That's the basic fact underlying
parallel transport and curvature!  So how come I am using the phrase
"same direction" like I do above?  Well, the point is that even though
it doesn't really make sense to talk about two tangent vectors at
different points being the same, we are using local coordinates, so we
can ask whether two vectors at two different points have the same
components in our coordinate system.  If they do, we might sloppily say
they are "the same".  BUT BEWARE!!  This is a coordinate-dependent thing
to do.  Someone else working in different coordinates would have
completely different, equally correct opinions.

So: it's bad form to talk about two tangent vectors at different points
being "the same".  But working in our coordinates, can certainly work
out the components of a tangent vector at the point p, and then parallel
transport it over to a point q, and see how much the components change.
This is what the Christoffel symbols record.  

I think I should pound my point in once more: the Christoffel symbols
are BLATANTLY COORDINATE-DEPENDENT.   We cannot define them in a
coordinate-free way as we could with the Riemann tensor!  This is why I
put off discussing them until Oz passed his test of valor.  They are
sneaky, fiendish, NON-TENSORIAL entities, despite the indices dangling
from them.  In other words, they transform in more messy ways when
we change coordinate systems than tensors do, owing to their "blatant
coordinate-dependence". 

>Note:  I may not get this
>quite right, but I'm sure John will correct me if I screw up.

Yes.  For example:

>So, C^a_cd is a tensor which takes two vectors and spits out a third.

WRONG!  WRONG!  WRONG!   THE CHRISTOFFEL SYMBOLS ARE ***NOT** A
TENSOR!!!!!   REPEAT: THEY ARE  ***NOT** A TENSOR!!!!! 

Ahem.  Actually, it's really great that you made this slip, because it 
gave me a good excuse to write all that stuff up there about why we
can't define the Christoffel symbols in a coordinate-free way like we
can the Riemann tensor, which is secretly because they are not a tensor.

Don't mind me. 

>So w^a = C^a_cd u^c v^d
>
>Specifically, we need a (smooth?) vector field V near a point of interest.
>
>We do this by starting at a point with the member of the vector field
>at the point v.  We then do a parallel transport of v for an
>infinitesimal distance in the u direction.  We look at how v has
>changed compared to the vector field at the new point and define w as
>that change.  This sounds a little circular, because we are trying to
>define parallel transport, but use the notion in defining C^a_cd.  But
>really, given the above formula, we are specifying exactly what
>parallel transport is on a particular set of basis vectors, and from
>there we can figure it out for an arbitrary set of vectors.

>I hope that was better (and not hideously wrong)

That's fine.  I would say it this way: if we already know what parallel
transport is, we can use what you say to define the Christoffel symbols.
This is the approach I'm following: my students learned about parallel
transport on day 1 of the course, but I put off Christoffel symbols
until after they already learned Einstein's equation.  

Conversely, if you happen to have a bunch of Christoffel symbols
lying around --- like if the person who used to work in your office left
a bunch of them written up on the blackboard --- you can use what you
say to define parallel transport.

>So, based on that, what is the physical meaning of the Christoffel
>symbols for this metric

>C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>vector in the x direction.  Then if we move forward in time a little
>bit, this vector has 'stretched' in the x direction by a factor of
>R'(t)/R(t).  This, I think, just describes the Hubble expansion and
>R'(t)/R(t) is, I think, the Hubble constant.

Yes, that sounds about right.  I had never thought of it that way!
Very cool!  It's always great to try to make all these computations make
physical sense.

>C^i_i0 makes less sense to me right away, except that I know that from
>a purely mathematical analysis that C^a_cd is symmetric wrt c and d.
>But the physical meaning...OK, we take a unit time vector and, holding
>time constant, we move this vector a little in the x direction.  We
>then find that it is slanted a little in the x direction
>
>	t				t
>	.				.
>	.				.
>	.				.
>	.				.
> .......|.......x  becomes        ......./.....x
>	.				.
>	.				.
>	.				.
>
>Oh, wait is this maybe just also saying that the unit time vectors
>don't look quite parallel, but instead they seem to be carrying nearby
>points apart, once again by that factor R'(t)/R(t).  What do other
>people think?
>
>
>OK, last one
>
>C^0_ii = R(t)R'(t)
>
>So we take a unit x vector, push it in the x direction, and find it
>has gained a t component.
>
>	t				t
>	.				.
>	.				.
>	.				.
>	.				.   _/
> ........___....x  becomes        ........./.....x
>	.				.
>	.				.
>	.				.
>
>
>(That one's harder to draw in ASCII)
>
>So, nearby unit x vectors seem to be diverging in time?  I don't get
>this one.

I don't "grok" these other ones either, not yet anyway.

This will be good to keep in mind for later, though!!

From noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!newsfeed.internetmci.com!info.ucla.edu!agate!physics12.Berkeley.EDU!ted Tue Apr 16 14:11:54 PDT 1996
Article: 9796 of sci.physics
Path: noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!newsfeed.internetmci.com!info.ucla.edu!agate!physics12.Berkeley.EDU!ted
From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 15 Apr 1996 23:08:42 GMT
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In article <[email-masked]>,
Paul G. White <[email-masked]> wrote:

>You want REALLY inane. I'll give you really inane. The thing that
>bothers me is that as R(t) is doing this expansion, here's the earth
>following its geodesic path around the sun. Is it blindly unaware of
>the expansion in the outer reaches of the universe or, during the five
>thousand million years of its existance has it been slowly creeping
>outward? If not, why not? If so, where's the evidence?

If you think that's really inane, you haven't been reading sci.physics
for very long!

The first thing to remember is that the simple Robertson-Walker
solution is a highly idealized and simplified version of the way
things really are.  As John will eventually get around to showing,
this solution describes a Universe with uniform density.  So it
doesn't have any big lumps of excess density (like, say, the Sun)
in it.  

So if we want to put the Earth-Sun system into our cosmology, we have
to do a shade better than the simple Robertson-Walker spacetime.  We
have to incorporate lumpiness.  Most of the work in applying general
relativity to cosmology comes in working out solutions that correspond
to "lumpy Robertson-Walker models."  By that I mean spacetimes that
have roughly uniform density when averaged over large scales (say, a
few hundred million light-years), but which are lumpy on small
scales.  A lot is known about such spacetimes.  Roughly speaking, they
tend to behave like this: the "average" behavior of spacetime on large
scales looks like Robertson-Walker, but in the neighborhood of an
overdense region, local gravity takes over and space isn't expanding.
The Earth-Sun system, and indeed our entire galaxy, is so overdense
that no effects of expansion occur are observable within it.

A nice homework exercise for general relativity students to work out
is the behavior of an expanding Robertson-Walker spacetime with a
single spherically symmetric overdense lump in it.  Because of the
spherical symmetry, this situation is still mathematically tractable
(although it's harder than solving the simple Robertson-Walker
equations), and it illustrates the behavior I describe above.  For
more realistic situations, in which the lumpiness is everywhere and is
not spherically symmetric, one has to resort to big computer
calculations to see what the geometry of spacetime really looks like.

This question is treated in the sci.physics FAQ.  I'll append the
relevant entry below.

-Ted

********************************************************************************
Item 12.

The Expanding Universe                          original by Michael Weiss
----------------------				updated 5-DEC-1994 by SIC 

Here are the answers to some commonly asked questions about exactly 
what it means to say that the Universe is expanding.

(1) IF THE UNIVERSE IS EXPANDING, DOES THAT MEAN ATOMS ARE GETTING BIGGER?
IS THE SOLAR SYSTEM EXPANDING?

Mrs. Felix:  Why don't you do your homework?
Allen Felix: The Universe is expanding.  Everything will fall
             apart, and we'll all die.  What's the point?
Mrs. Felix:  We live in Brooklyn.  Brooklyn is not expanding!
             Go do your homework.

                        -from "Annie Hall" by Woody Allen.

Mrs. Felix is right.  Neither Brooklyn, nor its atoms, nor the solar
system, nor even the galaxy, is expanding.  The Universe expands
(according to standard cosmological models) only when averaged over a very
large scale.

	The phrase "expansion of the Universe" refers both to experimental
observation and to theoretical cosmological models.  Lets look at them one
at a time, starting with the observations.

Observation
-----------

	The observation is Hubble's redshift law.  

	In 1929, Hubble reported that the light from distant galaxies is
redshifted.  If you interpret this redshift as a Doppler shift, then the
galaxies are receding according to the law: 

    (velocity of recession) = H * (distance from Earth)

H is called Hubble's constant; Hubble's original value for H was 550
kilometers per second per megaparsec (km/s/Mpc).  Current estimates range
>from 40 to 100 km/s/Mpc.  (Measuring redshift is easy; estimating distance
is hard.  Roughly speaking, astronomers fall into two "camps", some
favoring an H around 80 km/s/Mpc, others an H around 40-55). 

	Hubble's redshift formula does *not* imply that the Earth is in
particularly bad oder in the universe.  The familiar model of the universe
as an expanding balloon speckled with galaxies shows that Hubble's alter
ego on any other galaxy would make the same observation. 

	But astronomical objects in our neck of the woods--- everything
>from the Sun to galaxies belonging to the so-called "local group"--- show
no such Hubble redshifts.  Nearby stars and galaxies *do* show motion with
respect to the Earth (known as "peculiar velocities"), but this does not
look like the "Hubble flow" that is seen for distant galaxies.  For
example, the Andromeda galaxy shows blueshift instead of redshift.  So the
verdict of observation is: our solar system, our galaxy, our local group,
is not expanding. 

	By the way, Hubble's constant, is not, in spite of its name,
constant in time. In fact, it is decreasing.  Imagine a galaxy D
light-years from the Earth, receding at a velocity V = H*D.  D is always
increasing because of the recession.  But does V increase?  No.  In fact, V
is decreasing because gravitational attraction is slowly bringing the
expansion of the Universe to a halt.  So H is going down over time.  But it
*is* constant over space, i.e., it is the same number for all distant
objects as we observe them today. 

Theory 
------

The theoretical models are, typically, Friedmann-Robertson-Walker (FRW) 
spacetimes.  

	Cosmologists model the universe using "spacetimes", that is to say,
solutions to the field equations of Einstein's theory of general
relativity.  The Russian mathematician Alexander Friedmann discovered an
important class of global solutions in 1923.  The familiar image of the
universe as an expanding balloon speckled with galaxies is a "movie
version" of one of Friedmann's solutions.  Robertson and Walker later
extended Friedmann's work, so you'll find references to
"Friedmann-Robertson-Walker" (FRW) spacetimes in the literature. 

	FRW spacetimes come in a great variety of styles--- expanding,
contracting, flat, curved, open, closed, ....  The "expanding balloon"
picture corresponds to just a few of these. 

	A concept called the metric plays a starring role in general
relativity. The metric encodes a lot of information; the part we care about
(for this FAQ entry) is distances between objects.  In an FRW expanding
universe, the distance between any two "points on the balloon" does
increase over time. However, the FRW model is NOT meant to describe OUR
spacetime accurately on a small scale--- where "small" is interpreted
pretty liberally! 

	You can picture this in a couple of ways.  You may want to think of
the "continuum approximation" in fluid dynamics--- by averaging the motion
of individual molecules over a large enough scale, you obtain a continuous
flow.  (Droplets can condense even as a gas expands.)  Similarly, it is
generally believed that if we average the actual metric of the universe
over a large enough scale, we'll get an FRW spacetime. 

	Or you may want to alter your picture of the "expanding balloon". 
The galaxies are not just painted on, but form part of the substance of the
balloon (poetically speaking), and locally affect its "elasticity". 

	The FRW spacetimes ignore these small-scale variations.  Think of a
uniformly elastic balloon, with the galaxies modelled as mere points.
"Points on the balloon" correspond to a mathematical concept known as a
*comoving geodesic*.  Any two comoving geodesics drift apart over time, in
an expanding FRW spacetime. 

	At the scale of the Solar System, we get a pretty good
approximation to the spacetime metric by using another solution to
Einstein's equations, known as the Schwarzschild metric.  Using evocative
but dubious terminology, we can say this models the gravitational field of
the Sun.  (Dubious because what does "gravitational field" mean in GR, if
it's not just a synonym for "metric"?)  The geodesics in the Schwarzschild
metric do NOT display the "drifting apart" behavior typical of the FRW
comoving geodesics--- or in more familiar terms, the Earth is not drifting
away from the Sun. 

	The "true metric" of the universe is, of course, fantastically
complicated; you can't expect idealized simple solutions (like the FRW and
Schwarzschild metrics) to capture all the complexity.  Our knowledge of the
large-scale structure of the universe is fragmentary and imprecise. 

	In old-fashioned, Newtonian terms, one says that the Solar System
is "gravitationally bound" (ditto the galaxy, the local group).  So the
Solar System is not expanding.  The case for Brooklyn is even clearer: it
is bound by atomic forces, and its atoms do not typically follow geodesics.
So Brooklyn is not expanding.  Now go do your homework. 

References: (My thanks to Jarle Brinchmann, who helped with this list.)

Misner, Thorne, and Wheeler, "Gravitation", chapters 27 and 29.  Page 719
discusses this very question; Box 29.4 outlines the "cosmic distance
ladder" and the difficulty of measuring cosmic distances; Box 29.5 presents
Hubble's work.  MTW refer to Noerdlinger and Petrosian, Ap.J., vol. 168
(1971), pp. 1--9, for an exact mathematical treatment of gravitationally
bound systems in an expanding universe.

M.V.Berry, "Principles of Cosmology and Gravitation".  Chapter 2 discusses
the cosmic distance ladder; chapters 6 and 7 explain FRW spacetimes.

Steven Weinberg, "The First Three Minutes", chapter 2.  A non-technical
treatment.

Hubble's original paper: "A Relation Between Distance And Radial
Velocity Among Extra-Galactic Nebulae", Proc. Natl. Acad. Sci., Vol. 15,
No. 3, pp. 168-173, March 1929.

Sidney van den Bergh, "The cosmic distance scale", Astronomy & Astrophysics
Review 1989 (1) 111-139.

M. Rowan-Robinson, "The Cosmological Distance Ladder", Freeman.

A new method has been devised recently to estimate Hubble's constant, using
gravitational lensing.  The method is described in:

\O Gr\on and Sjur Refsdal, "Gravitational Lenses and the age of the
universe", Eur. J. Phys. 13, 1992 178-183.

S. Refsdal & J. Surdej, Rep. Prog. Phys. 56, 1994 (117-185)

and H is estimated with this method in:

H.Dahle, S.J. Maddox, P.B. Lilje, to appear in ApJ Letters.

Two books may be consulted for what is known (or believed) about the
large-scale structure of the universe: 

P.J.E.Peebles, "An Introduction to Physical Cosmology".
T. Padmanabhan, "Structure Formation in the Universe".

======================================================================

(2) WHAT CAUSES THE HUBBLE REDSHIFT?  ARE THE LIGHT-WAVES "STRETCHED" AS 
THE UNIVERSE EXPANDS, OR IS THE LIGHT DOPPLER-SHIFTED BECAUSE DISTANT 
GALAXIES ARE MOVING AWAY FROM US?

	In a word: yes.  In two sentences: the Doppler-shift explanation is
a linear approximation to the "stretched-light" explanation.  Switching
>from one viewpoint to the other amounts to a change of coordinate systems
in (curved) spacetime. 

	A detailed explanation requires looking at Friedmann-Robertson-Walker 
(FRW) models of spacetime.  The famous "expanding balloon speckled with
galaxies" provides a visual analogy for one of these; like any analogy, it
will mislead you if taken too literally, but handled with caution it can
furnish some insight. 

	Draw a latitude/longitude grid on the balloon.  These define
*co-moving* coordinates.  Imagine a couple of speckles ("galaxies")
imbedded in the rubber surface.  The co-moving coordinates of the speckles
don't change as the balloon expands, but the distance between the speckles
steadily increases.  In co-moving coordinates, we say that the speckles
don't move, but "space itself" stretches between them. 

	A bug starts crawling from one speckle to the other.  A second
after the first bug leaves, his brother follows him.  (Think of the bugs as
two light-pulses, or successive wave-crests in a beam of light.)  Clearly
the separation between the bugs will increase during their journey.  In
co-moving coordinates, light is "stretched" during its journey. 

	Now we switch to a different coordinate system, this one valid only
in a neighborhood (but one large enough to cover both speckles).  Imagine a
clear, flexible, non-stretching patch, attached to the balloon at one
speckle.  The patch clings to the surface of the balloon, which slides
beneath it as the balloon inflates.  (The bugs crawl along *under* the
patch.)  We draw a coordinate grid on the patch.  In the patch coordinates,
the second speckle recedes from the first speckle.  And so in patch
coordinates, we can regard the redshift as a Doppler shift. 

	Is this visually appealing?  I think so.  However, this explanation
glosses over one crucial point: the time coordinate.  FRW spacetimes come
fully-equipped with a specially distinguished time coordinate (called the
co-moving or cosmological time).  For example, a co-moving observer could
set her clock by the average density of surrounding speckles, or by the
temperature of the Cosmic Background Radiation.  (From a purely
mathematical standpoint, the co-moving time coordinate is singled out by a
certain symmetry property.) 

	We have many choices of time-coordinate to go with the
space-coordinates drawn on our patch.  Let's use cosmological time.  Notice
that this is *not* the choice usually made in Special Relativity: though
the two speckles separate rapidly, their cosmological clocks remain
synchronized. Bugs embarking on their journey from the "moving" speckle
appear to crawl "upstream" against flowing space as they head towards the
"home" speckle. The current diminishes as they approach home.  (In other
words, bug-speed is anisotropic in these coordinates.)  These differences
>from the usual SR picture are symptoms of a deeper fact: besides the
obvious "spatial" curvature of the balloon's surface, FRW spacetimes have
"temporal" curvature as well.  Indeed, not all FRW spacetimes exhibit
spatial curvature, but (with one exception) all have temporal curvature. 

	You can work out the magnitude of the redshift using patch
coordinates.  I leave this as an exercise, with a couple of hints. (1)
Since bug-speed is anisotropic far from the home speckle, consider also a
patch attached to the "moving" speckle.  Compute the initial distance
between the bugs (the "wavelength") in both patch coordinate systems, using
the standard *non-relativistic* Doppler formula for a stationary source,
moving receiver.  (2) Now think about how the bug-distance changes as the
bugs journey to the home speckle (this time sticking with home patch
coordinates).  The bug-distance does *not* propagate unchanged.  Consider
instead the analog of the period of a lightwave: the time between
bug-crossings of a grid line on the patch.  This *does* propagate almost
unchanged, *provided* the rate of balloon expansion stays pretty much the
same throughout the bugs' perilous trek.  The final result: the magnitude
of the redshift, computed using Doppler's formula, agrees to first-order
with magnitude computed using the "stretched-light" explanation.  (To the
cognoscenti: the assumptions are that Hx<<1 and (dH/dt)x<<1, where
H(t)=dR(t)/dt, R(t) is the scale factor, t is cosmological time, and x is
the average distance between the "speckles" (co-moving geodesics) during
the course of the journey.) 

	(This long-winded "proof of equivalence" between the Doppler and
"stretched-light" explanations substitutes a paragraph of imagery for a
half-page of calculus.)

	Let me close by emphasizing the word "approximation" from the first
paragraph of this entry.  The Doppler explanation fails for very large
redshifts, for then we must consider how Hubble's "constant" changes over
the course of the journey. 

References:

Misner, Thorne, and Wheeler, "Gravitation", chapter 29.

M.V.Berry, "Principles of Cosmology and Gravitation", chapter 6.

Steven Weinberg, "The First Three Minutes", chapter 2, especially pp. 13
and 30.

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Article: 9770 of sci.physics
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From: Geoffrey A. Landis <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 15 Apr 1996 21:24:38 GMT
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In article <[email-masked]> Oz,
[email-masked] writes:
>We need two lightlike paths in this spacetime. Hmm what might that mean.
>Well locally we ought still to be Minkowskian so light travels at 45deg.
>I guess that it would mean that dx/dt=1 and we can set dy,dz=0. 
>(Shaky ground feeling)
>
>so      dS^2= -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)  
>becomes dS^2= -dt^2 + R(t)^2 dx^2               and substituting dx=dt
>        dS^2= -dt^2 + R(t)^2 dt^2 = (R(t)-1)dt^2        or
>        dS=(R(t)-1)^0.5 dt                      which looks promising
>...

It "looks promising", but it's not the approach I'd take.

If I were doing the problem, I would say that, for a lightlike path dS=0.
(This is by definition.)

Given this, then dS^2 = 0 = -dt^2 + R(t)^2 dx^2
leading to
   dt^2 = R(t)^2 dx^2
and hence
   dx/dt = R(t)
for light traveling in the x direction. (with a plus or minus for the two
roots of R(t)^2).

I'm sure that there's an elegant way to solve the problem of doppler
shift, but I don't know it.  Here is an inelegant way.

Suppose a light source at x=0 emits light, with the "peak" of a wave at
some time t0.
The next peak of a wave occurs at a time t=t0+delta-t0 (where delta-t0
equals 1/f, f the frequency).

since dx/dT = R(t), this can be integrated.
x = integral [t0 to t1] R(t) dt  for the light emitted at t0
x = integral [t0+delta-t0 to t1 plus delta-t1] R(t) dt  for the light
emitted at t0+delta-t0

Where t1 is the time the light is received at a distance x from the
emitter.

Since the receiver at x is fixed in its own reference frame, the two
equations can be set equal.  The only difference between them is the
little bit of the integral between t0 and t0+delta-t0, and the bit
between t1 and t1+delta-t1.  With the usual calculus relationships
between integrals and derivitaves, assuming that R(t) has a smooth
derivative at t0 and t1,

delta-t1 R(t1) = delta-t0 R(t0)

This means that the relationship of f0, the frequency at t=t0, when the
light is emitted, to the frequency f1 at t=t1, when the light is
received, is

delta-t0/delta-t1 = f1/f0 = R(t0)/R(t1)

The doppler shift is thus directly proportional to the ratio of R when
the light was emitted, to R when the light was received.  [The doppler
shift for wavelength is just the inverse of this, of course.]

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center
physicist and part-time science fiction writer

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From: Geoffrey A. Landis <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 15 Apr 1996 21:24:38 GMT
Organization: Ohio Aerospace Institute
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In article <[email-masked]> Oz,
[email-masked] writes:
>We need two lightlike paths in this spacetime. Hmm what might that mean.
>Well locally we ought still to be Minkowskian so light travels at 45deg.
>I guess that it would mean that dx/dt=1 and we can set dy,dz=0. 
>(Shaky ground feeling)
>
>so      dS^2= -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)  
>becomes dS^2= -dt^2 + R(t)^2 dx^2               and substituting dx=dt
>        dS^2= -dt^2 + R(t)^2 dt^2 = (R(t)-1)dt^2        or
>        dS=(R(t)-1)^0.5 dt                      which looks promising
>...

It "looks promising", but it's not the approach I'd take.

If I were doing the problem, I would say that, for a lightlike path dS=0.
(This is by definition.)

Given this, then dS^2 = 0 = -dt^2 + R(t)^2 dx^2
leading to
   dt^2 = R(t)^2 dx^2
and hence
   dx/dt = R(t)
for light traveling in the x direction. (with a plus or minus for the two
roots of R(t)^2).

I'm sure that there's an elegant way to solve the problem of doppler
shift, but I don't know it.  Here is an inelegant way.

Suppose a light source at x=0 emits light, with the "peak" of a wave at
some time t0.
The next peak of a wave occurs at a time t=t0+delta-t0 (where delta-t0
equals 1/f, f the frequency).

since dx/dT = R(t), this can be integrated.
x = integral [t0 to t1] R(t) dt  for the light emitted at t0
x = integral [t0+delta-t0 to t1 plus delta-t1] R(t) dt  for the light
emitted at t0+delta-t0

Where t1 is the time the light is received at a distance x from the
emitter.

Since the receiver at x is fixed in its own reference frame, the two
equations can be set equal.  The only difference between them is the
little bit of the integral between t0 and t0+delta-t0, and the bit
between t1 and t1+delta-t1.  With the usual calculus relationships
between integrals and derivitaves, assuming that R(t) has a smooth
derivative at t0 and t1,

delta-t1 R(t1) = delta-t0 R(t0)

This means that the relationship of f0, the frequency at t=t0, when the
light is emitted, to the frequency f1 at t=t1, when the light is
received, is

delta-t0/delta-t1 = f1/f0 = R(t0)/R(t1)

The doppler shift is thus directly proportional to the ratio of R when
the light was emitted, to R when the light was received.  [The doppler
shift for wavelength is just the inverse of this, of course.]

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center
physicist and part-time science fiction writer

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Article: 9984 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 16 Apr 1996 17:35:48 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] (Patrick van Esch) writes:
>Ok, I tried to get to the Riemann tensor over lunch, didn't 
>really get there, only the Christoffel symbols....

The Riemann tensor needs a lot of napkins.

>Apart from arithmetic errors, this is what I find:

>C^0_ii = 1/2 . R'
>C^i_0i = 1/2.R'/R
>C^i_i0 = 1/2.R'/R

>i = 1, 2 and 3

>all the rest = 0.

>This is different from what Kevin had, but then I calculated it very 
>hastily, so I might very well be wrong...

It looks sort of roughly right --- but I think we're gonna have to call
in Oz to straighten this out. 

>: I think I should pound my point in once more: the Christoffel symbols
>: are BLATANTLY COORDINATE-DEPENDENT.   We cannot define them in a
>: coordinate-free way as we could with the Riemann tensor!  This is why I
>: put off discussing them until Oz passed his test of valor.  They are
>: sneaky, fiendish, NON-TENSORIAL entities, despite the indices dangling
>: from them.  In other words, they transform in more messy ways when
>: we change coordinate systems than tensors do, owing to their "blatant
>: coordinate-dependence". 

>So their messiness will have to cancel when they are combined 
>to make up the Riemann tensor....

Exactly.  One might think that's rather surprising.  We do all this
blatantly coordinate-dependent stuff to define the Christoffel symbols,
and then using a horrid formula like 

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

we concoct the Riemann tensor, and then somehow the Riemann tensor winds
up transforming under coordinate transformations in a nice, simple way.

But the above horrid formula is not the conceptual definition of the
Riemann tensor --- far from it!  I might as well remind folks of our
definition of the Riemann tensor and the Christoffel symbols:

RIEMANN: 

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number. The vector w comes back a bit changed by its
journey; it is now a new vector w'. We then have

    w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3 

CHRISTOFFEL:

FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the d
direction and parallel transport it an amount epsilon in the c
direction.  See how much its component in the a direction changes.  
(This component started out being 0 unless a = d, in which case it was
1.)  It changes by 

     - epsilon C^a_{cd} + terms of order epsilon^2

Note what's going on here.  On the one hand, the definition of the
Christoffel symbols involves parallel transporting a vector along a
little path.  It thus involves comparing vectors *different points*,
which we can only do by introducing a coordinate system and comparing
their components in these coordinates.  On the other hand, the Riemann
tensor involves parallel transporting a vector around a little loop.  It
thus involves comparing vectors at *the same point*, which we can do
without coordinates.  (The space of tangent vectors at a given point is
a vector space, so we have no trouble making sense of the difference of
two of them.)

For this reason, even though the Riemann tensor is more complicated in
some ways, it doesn't require any coordinate system to define, which
winds up meaning that it transforms very simply under coordinate
transformations.  

We could also figure out how the Christoffel symbol transforms under
coordinate changes, and use the horrid formula for the Riemman in terms
of the Christoffel to see how the Riemann transforms, and observe that, as
if by magic, all sorts of cruft cancels out!  But this is less
enlightening.  

The advantage of loops over paths, which we see above, is also the
rationale behind the "loop representation of quantum gravity", in which
we take parallel transport around loops as the basic entity to work
with, rather than the metric or Christoffel symbols.

>: >C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>: >vector in the x direction.  Then if we move forward in time a little
>: >bit, this vector has 'stretched' in the x direction by a factor of
>: >R'(t)/R(t).  This, I think, just describes the Hubble expansion and
>: >R'(t)/R(t) is, I think, the Hubble constant.
>
>: Yes, that sounds about right.  I had never thought of it that way!
>: Very cool!  It's always great to try to make all these computations make
>: physical sense.
>
>Funny you say this.  I thought it was more or less the definition
>of the Hubble constant: the rate of distance increase over the
>distance distance ?  Even without all the metric stuff...

No, I sort of knew R'(t)/R(t) was the Hubble constant.  What I meant is,
I'd never realized that the Hubble constant was just the Christoffel
symbol C^i_{0i}, expressing how "space is stretching as time passes".
(Note to the the philosophically oversensitive: the quoted phrase is a
piece of poetic whimsy not to be taken seriously.)

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Article: 10155 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 16 Apr 1996 22:23:10 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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In article <[email-masked]>,
Patrick van Esch <[email-masked]> wrote:
>
>Ok, I tried to get to the Riemann tensor over lunch, didn't 
>really get there, only the Christoffel symbols:
>
>Apart from arithmetic errors, this is what I find:
>
>C^0_ii = 1/2 . R'
>C^i_0i = 1/2.R'/R
>C^i_i0 = 1/2.R'/R
>
>i = 1, 2 and 3
>
>all the rest = 0.
>
>This is different from what Kevin had, but then I calculated it very 
>hastedly, so I might very well be wrong...
>

I think we differ in what we are using for the metric.  I am using

    ( -1    0    0    0   )
g = ( 0  R(t)^2  0    0   )
    ( 0     0  R(t)^2 0   )
    ( 0     0    0  R(t)^2)  

I think  you used

    ( -1    0    0    0   )
g = ( 0   R(t)   0    0   )
    ( 0     0  R(t)   0   )
    ( 0     0    0   R(t) )  

I'm pretty sure the first is correct, but I could be wrong.

I've actually got the Riemann tensor calculated, but I'd like to make
sure we've agreed on the Christoffel symbols first. In addition, I
tried to calculate what the geodesics have to be based on these
C. symbols, but my results are kind of disturbing.  I may be doing
something wrong, so we'll wait on that.

>
>: >C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>: >vector in the x direction.  Then if we move forward in time a little
>: >bit, this vector has 'stretched' in the x direction by a factor of
>: >R'(t)/R(t).  This, I think, just describes the Hubble expansion and
>
>Except for the nitpicking detail that the factor is probably 1 + R'/R
>since I still want to keep a piece of vector when there is no
>expansion (and hence R' = 0)...
>

Yes, of course.  The difference has that R'/R factor.  Bad use of
language on my part.

Kevin Scaldeferri

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Article: 10142 of sci.physics
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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 08:14:25 GMT
Organization: DESY
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Kevin Anthony Scaldeferri ([email-masked]) wrote:
: In article <[email-masked]>,
: Patrick van Esch <[email-masked]> wrote:
: >
: >Ok, I tried to get to the Riemann tensor over lunch, didn't 
: >really get there, only the Christoffel symbols:
: >
: >Apart from arithmetic errors, this is what I find:
: >
: >C^0_ii = 1/2 . R'
: >C^i_0i = 1/2.R'/R
: >C^i_i0 = 1/2.R'/R
: >
: >i = 1, 2 and 3
: >
: >all the rest = 0.
: >
: >This is different from what Kevin had, but then I calculated it very 
: >hastedly, so I might very well be wrong...
: >

: I think we differ in what we are using for the metric.  I am using

:     ( -1    0    0    0   )
: g = ( 0  R(t)^2  0    0   )
:     ( 0     0  R(t)^2 0   )
:     ( 0     0    0  R(t)^2)  

: I think  you used

:     ( -1    0    0    0   )
: g = ( 0   R(t)   0    0   )
:     ( 0     0  R(t)   0   )
:     ( 0     0    0   R(t) )  

: I'm pretty sure the first is correct, but I could be wrong.

YES ! Right.  So if I replace my R with your R^2 it should fit :-)
I now remember doing this with the idea that the ^2 would only 
complicate issues during the calculation, and forgot about it.

BTW, I was also a bit too much in a hurry concerning the possible
non-zero terms in Riemann.  I was cheating there because the
conclusion I had (only one non-zero index) can not follow from
the condition on the Christoffel symbols.
Somebody pointed this out to me and I thought I knew what I was
doing.  But the price of haste is error and embarrasment :)

So there are additional components of Riemann that aren't 
zero. 

These are:

R^i_ijj = - A(t).B(t)

R^i_jij = A(t).B(t)

I haven't got your post anymore with your riemann components.
We should compare... (or look 'em up in MTW :)

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 16 Apr 1996 15:22:04 GMT
Organization: DESY
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Patrick van Esch ([email-masked]) wrote:

: Ok, I tried to get to the Riemann tensor over lunch, didn't 
: really get there, only the Christoffel symbols:

: Apart from arithmetic errors, this is what I find:

: C^0_ii = 1/2 . R'
: C^i_0i = 1/2.R'/R
: C^i_i0 = 1/2.R'/R

: i = 1, 2 and 3

: all the rest = 0.

: This is different from what Kevin had, but then I calculated it very 
: hastedly, so I might very well be wrong...

Ok, in the mean time I did some messing around with the
Riemann tensor:

Since we still disagree on the Christoffel symbols, I tried
to do things in a more general way.

Let us suppose that the Christoffel symbols are such that:

    C^0_ii = A(t)
    
    C^i_0i = B(t)
    
    C^i_i0 = B(t)
    
    all the others are 0 (here, i = 1,2,3)
    
 The Riemann tensor:   
   
 R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}
   
Now, there are 2 special cases for indexes: 0 and 1,2,3   
Since no Christoffel symbol is ever different from zero when
we have 2 DIFFERENT non-zero indexes, the same will apply to
the Riemann tensor.

So we will use i and 0.

First case: a = 0.

we can try for bcd:   

      iii  --> 0
      ii0  --> C^i_i0.C^0_ii - C^i_i0.C^0_ii = 0
      i0i  --> C^0_ii,0 - C^i_0i.C^0_ii
      0ii  --> - C^0_ii,0 + C^i_0i.C^0_ii
      i00  --> 0
      0i0  --> 0
      00i  --> 0
      000  --> 0
      
      So remains: R^0_i0i = A'(t) - B(t).A(t)
       
                  R^0_0ii = - A'(t) + B(t).A(t)
                  
Second case: a = i.

      iii  --> 0
      ii0  --> 0
      i0i  --> 0
      0ii  --> 0
      i00  --> C^i_i0,0 + C^i_i0.C^i_i0 
      0i0  --> - C^i_i0,0 - C^i_i0.C^i_i0
      00i  --> C^i_0i,0 - C^i_0i,0 + C^i_0i.C^i_i0 - C^i_0i.C^i_i0 = 0
      000  --> 0
      
      So remains: R^i_i00 = B'(t) + B(t)^2
                  R^i_0i0 = - B'(t) - B(t)^2

 
If we can now come to an agreement to what is A(t) and B(t), and
if I didn't make too many errors above, we're getting closer :)

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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Article: 10054 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 16 Apr 1996 21:51:05 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
Lines: 115
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In article <[email-masked]>, john baez <[email-masked]> wrote:
>Sorry, it took me a while to respond to this, because it's rather
>substantial.
>
>In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
>
>>Note:  I may not get this
>>quite right, but I'm sure John will correct me if I screw up.
>
>Yes.  For example:
>
>>So, C^a_cd is a tensor which takes two vectors and spits out a third.
>
>WRONG!  WRONG!  WRONG!   THE CHRISTOFFEL SYMBOLS ARE ***NOT** A
>TENSOR!!!!!   REPEAT: THEY ARE  ***NOT** A TENSOR!!!!! 
>

Aaaahhh!  <dives for cover>

>Ahem.  Actually, it's really great that you made this slip, because it 
>gave me a good excuse to write all that stuff up there about why we
>can't define the Christoffel symbols in a coordinate-free way like we
>can the Riemann tensor, which is secretly because they are not a tensor.
>

OK, we're going to have to back up here.  I think I'm blurring the
Christoffel symbols and the covariant derivative, but...the C. symbols
describe how the covarient derivative acts on a basis, yes?  And the
covariant derivative is multi-linear.  So why doesn't all this act
tensorial?  I guess I was under the impression that any multilinear
map defined a tensor.  (Or is that only true for (0,n) tensors?).
You're explanation seems to go back to the first definition of a
tensor I ever heard which is "a tensor is something which transforms
like a tensor".  (I always hated that definition, ever after it made
more sense).  So, the Christoffel symbols don't transform right under
a coordinate change, you say?  Yet it's multilinear?  Am I making a
really dumb mistake or did someone teach me things wrong at some
point?  (Not you, I knew (or thought I knew) about tensors before this
tutorial started).  I'm confused.

>
>That's fine.  I would say it this way: if we already know what parallel
>transport is, we can use what you say to define the Christoffel symbols.
>This is the approach I'm following: my students learned about parallel
>transport on day 1 of the course, but I put off Christoffel symbols
>until after they already learned Einstein's equation.  
>

Aah, I think I skipped that day of class.  Sorry

>
>>C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>>vector in the x direction.  Then if we move forward in time a little
>>bit, this vector has 'stretched' in the x direction by a factor of
>>R'(t)/R(t).  This, I think, just describes the Hubble expansion and
>>R'(t)/R(t) is, I think, the Hubble constant.
>
>Yes, that sounds about right.  I had never thought of it that way!
>Very cool!  It's always great to try to make all these computations make
>physical sense.
>

I'm glad you approve.  I was getting worried because I was talking
with someone today who seemed to think I was foolish to try to extract
meaning directly from the C symbols without going through the field
equation and geodesic equation and some other stuff I don't know about
yet.

>>
>>OK, last one
>>
>>C^0_ii = R(t)R'(t)
>>
>>So we take a unit x vector, push it in the x direction, and find it
>>has gained a t component.
>>
>>	t				t
>>	.				.
>>	.				.
>>	.				.
>>	.				.  _/
>> .....___....x  becomes        ........./.....x
>>	.				.
>>	.				.
>>	.				.
>>
>>
>>(That one's harder to draw in ASCII)
>>
>>So, nearby unit x vectors seem to be diverging in time?  I don't get
>>this one.
>
>I don't "grok" these other ones either, not yet anyway.
>

Yeah, I'd really like to understand this one.  It has me a little
worried.  It suggests that a straight line in space is not
autoparallel and thus not a geodesic.  But elsewhere on the thread, we
are talking about the distance between two points at the same time
just being the familiar Euclidean straight line distance.  And
shouldn't this mean that that straight line is a geodesic?  Am I
setting myself up to make the wizard angry again, or what is going on
here?  Has anyone checked my math to make sure this is actually the
right answer for C^0_ii?

All for now.

Kevin Scaldeferri

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Wed Apr 17 21:29:50 PDT 1996
Article: 10229 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 12:29:25 -0700
Organization: University of California, Riverside
Lines: 137
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
>In article <[email-masked]>, john baez <[email-masked]> wrote:
>>In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:

>>>So, C^a_cd is a tensor which takes two vectors and spits out a third.

>>WRONG!  WRONG!  WRONG!   THE CHRISTOFFEL SYMBOLS ARE ***NOT** A
>>TENSOR!!!!!   REPEAT: THEY ARE  ***NOT** A TENSOR!!!!! 

>Aaaahhh!  <dives for cover>

>OK, we're going to have to back up here.  I think I'm blurring the
>Christoffel symbols and the covariant derivative, but...the C. symbols
>describe how the covarient derivative acts on a basis, yes?  And the
>covariant derivative is multi-linear.  So why doesn't all this act
>tensorial?  I guess I was under the impression that any multilinear
>map defined a tensor.  (Or is that only true for (0,n) tensors?).
>You're explanation seems to go back to the first definition of a
>tensor I ever heard which is "a tensor is something which transforms
>like a tensor".  (I always hated that definition, ever after it made
>more sense).  So, the Christoffel symbols don't transform right under
>a coordinate change, you say?  Yet it's multilinear?  Am I making a
>really dumb mistake or did someone teach me things wrong at some
>point?  (Not you, I knew (or thought I knew) about tensors before this
>tutorial started).  I'm confused.

Part of the problem is that I'm trying to avoid talking about covariant
derivatives, since Oz and others are undoubtedly already reeling from an
overload of mathematics.  This means I am talking about the Christoffel
symbols without the proper infrastructure to discuss why they are, or
aren't, components of a tensor.  The question is actually a bit subtle!
I may be forced to get into the covariant derivative stuff, in order to
make everything clear in the weeks to come.

But since you already know about the covariant derivative, let me
briefly explain what's up.  

I indeed did define tensors as multilinear maps, earlier in the course:
for example, a tensor at a point may eat some tangent vectors at that
point and spit out a number, or tangent vector, depending on the inputs
in a multilinear way.  The covariant derivative does not do this!  

The covariant derivative of the vector field u with respect to the
vector field v is again a vector field D_v u.  BUT: the value of D_v u
at a given point p does NOT just depend on the value of u at p!  The
covariant derivative measures how u changes as we move in the direction
of v, so it clearly depends on the value of u at points other than the
point p.

So the covariant derivative is not a tensor.

This is related to the stuff I recently said about how the Christoffel
symbols involve comparing vectors at different points.  

Now, we can compute the Christoffel symbols C^a_{bc} in any coordinate
system as described earlier.  And we can certainly cook up a tensor
which in the given coordinate system has these C^a_{bc} (or anything) as
components.  

BUT: if we do this whole process in a different coordinate system, we
get a different tensor.  So it's dangerous to think of there being a
"Christoffel tensor": there's no coordinate-independent way to define
such a thing.  

Why is this true, though?  Well, if you understand the relation between
the Christoffel symbols and the covariant derivative, it follows from
the fact that the covariant derivative isn't a tensor.  Another way to
see it is to work out how the Christoffel symbols change when we change
coordinates, and note that they don't transform like a tensor.

>I was getting worried because I was talking
>with someone today who seemed to think I was foolish to try to extract
>meaning directly from the C symbols without going through the field
>equation and geodesic equation and some other stuff I don't know about
>yet.

It's foolish UNLESS you understand the physical significance of the
coordinates you happen to be using.  All coordinate systems are equal,
but some are more equal than others.  I.e., in principle we can do any
computation in any coordinate system, always getting the same
physics.  But it's often nice to work with coordinates specially
adapted to the problem at hand.  In our case, the coordinates (t,x,y,z)
we're using turn out to be the coordinates for which the galaxies stay at
more or less constant (x,y,z) as time t passes.  So the Christoffel
symbols in these coordinates have a comprehensible meaning... at least
some of them.  

But what about Kevin's claim that C^0_{11} and the like are nonzero?

>Yeah, I'd really like to understand this one.  It has me a little
>worried.  It suggests that a straight line in space is not
>autoparallel and thus not a geodesic. 

It definitely would imply that.

>But elsewhere on the thread, we
>are talking about the distance between two points at the same time
>just being the familiar Euclidean straight line distance.  And
>shouldn't this mean that that straight line is a geodesic?  

Well, let me use "space at time T" to stand for the hypersurface t = T,
where T is some number.  This is what folks call a "spacelike slice" of
spacetime.

Okay: there is certainly no paradox in what you suggest.  It's perfectly
possible for a curve in space at time T to be a geodesic in space at
time T, but not be a geodesic in spacetime!   

A good analogy is a surface in ordinary 3-dimensional Euclidean space.
Consider a sphere, for example.  A great circle is a geodesic in the
sphere, but not in 3-dimensional space.

This is why I said: 

`Note, here we are NOT taking into account the passage of time at all: we
are simply considering space at one moment of time --- time t as defined
by our particular coordinate system (t,x,y,z).  So what we are doing now
does NOT have any *immediate* relevance to the problem of "how far away
is that ancient galaxy whose light is just now reaching me".  What we
are doing now is of a more purely mathematical nature.  Nonetheless it
is good to do, to help figure out what the hell people really mean when
they maunder on about "space expanding", "galaxies moving away from us"
and the like.'

>Has anyone checked my math to make sure this is actually the
>right answer for C^0_ii?

Hmm.  So you claim 

C^0_ii = R(t)R'(t),

but I think Esch got something else.  I'm afraid we're going to have to
call in an unbiased third party, like Oz, to figure out which of you is
right.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Wed Apr 17 21:45:25 PDT 1996
Article: 10235 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 12:39:12 -0700
Organization: University of California, Riverside
Lines: 21
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>ER, I'm rather embarrassed to ask but R(t) seems to be something more 
>definite than 'any old function of t'. I wonder what it might be?

Well, I'm afraid you're going to have to solve Einstein's equation to
figure out what it is!   You've got to work out the Christoffel symbols
from the metric, then work out the Riemann tensor from the Christoffel
symbols, then work out the Einstein tensor... and then you need to
decide whether you want your universe to be made of dust, or radiation,
or whatever, so that we can figure out the stress-energy tensor.  And
then we can use Einstein's equation and figure out R(t).

Obviously for a big bang universe R(t) will start out at zero and get
bigger, at least for a while.  But that's all I'm going to say right
now.  You'll just have to figure it out.  A bunch of people are helping
you out... they are starting to compute the Christoffel symbols and so
on.  However, they don't seem to agree on the answer!!   So you're gonna
have to step in.

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!howland.reston.ans.net!swrinde!sgigate.sgi.com!enews.sgi.com!decwrl!pa.dec.com!decuac.dec.com!haven.umd.edu!hecate.umd.edu!mojo.eng.umd.edu!forth.eng.umd.edu!not-for-mail Wed Apr 17 21:54:49 PDT 1996
Article: 10336 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 18:40:34 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
Lines: 60
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: forth.eng.umd.edu

In article <[email-masked]>,
Patrick van Esch <[email-masked]> wrote:
>
>YES ! Right.  So if I replace my R with your R^2 it should fit :-)
>I now remember doing this with the idea that the ^2 would only 
>complicate issues during the calculation, and forgot about it.
>

So we are now in agreement on the Christoffel symbols, yes?  BTW, a
friend of mine showed me another way to calculate them using a
Lagrangian with the metric as the action and the geodesic equation.
Kinda a neat approach.  At any rate, we got the same answers as I
got.  So, once again just to have them in this post, we have:

C^0_ii = R(t)R'(t)
c^i_0i = C^i_i0 = R'(t) / R(t)

>BTW, I was also a bit too much in a hurry concerning the possible
>non-zero terms in Riemann.  I was cheating there because the
>conclusion I had (only one non-zero index) can not follow from
>the condition on the Christoffel symbols.
>Somebody pointed this out to me and I thought I knew what I was
>doing.  But the price of haste is error and embarrasment :)
>
>So there are additional components of Riemann that aren't 
>zero. 
>
>These are:
>
>R^i_ijj = - A(t).B(t)
>
>R^i_jij = A(t).B(t)
>
>I haven't got your post anymore with your riemann components.
>We should compare... (or look 'em up in MTW :)
>

After checking mine, I fine that I should of had these too.  So, I
think we are in agreement on the Riemann tensor with:

R^i_i00 = - R^i_0i0 = R"/R
R^0_i0i = - R^0_0ii = RR"
R^i_jij = - R^i_ijj = (R')^2

And, once again I post my appeal to those who understand the
symmetries of the Riemann tensor.  Please explain.  I see antisymmetry
in b and c, but no others are obvious.  And speaking of MTW, I don't
know what they mean when they state the symmetries:

R^a_bcd = R^a_[bc]d		apparently denotes the b/c antisymmetry

R^a_[bcd] = 0			means what?

Kevin Scaldeferri

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Wed Apr 17 22:05:47 PDT 1996
Article: 10370 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted
From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
Date: 18 Apr 1996 03:29:04 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 44
Sender: [email-masked]
Distribution: world
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References: <[email-masked]> <[email-masked]> <[email-masked]>
Reply-To: [email-masked]
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>, Doug Merritt <[email-masked]> wrote:
>In article <[email-masked]> [email-masked] writes:
>>ER, I'm rather embarrassed to ask but R(t) seems to be something more 
>>definite than 'any old function of t'. I wonder what it might be?
>
>You pick R(t) in a way such that the resulting expansion/contraction
>of the universe matches observation and theory.

Right.  Let me expand a little on what Doug is saying.

Remember that we haven't yet applied Einstein's equation to the
Robertson-Walker spacetime.  When we do, it'll turn out to be a
differential equation that tells us what R(t) has to be.  Well,
almost.  Like most differential equations, it has multiple solutions
and you have to specify some initial conditions before you can pick
out a unique solution.  (Of course, since Einstein's equation has that
stress-energy tensor T in it, we'll also have to know something about
what kind of matter our spacetime is filled with.)

By the way, when you apply Einstein's equation to the particular case
of a Robertson-Walker spacetime, the resulting equation is called "the
Friedmann equation."

For the moment, since we haven't written down the Friedmann equation
and solved it for R(t), we're just thinking of it as an arbitrary
function.  So we're really considering a very broad class of spacetime
geometries -- a different geometry for each choice of the function R!
This would include even silly choices, like ones where R increases
exponentially, stops and remains constant for a while, wiggles up and
down like a cosine a few times, and then drops toward zero like t to
the power minus one million.

Most of those geometries won't turn out to satisfy the Einstein
equation, and so we won't be interested in them, but we haven't yet
gotten to the point of worrying about that.  When we do, we'll find
that R(t) is often a fairly simple function of t.  For instance, in
one of the most popular cosmological models, R(t) = t^(2/3).  But for
purposes of building up your intuition about what these spacetimes
really "look like," it's probably best to try to understand the case
where R is a more or less arbitrary function of t first.
 
-Ted

From noise.ucr.edu!news.service.uci.edu!ihnp4.ucsd.edu!munnari.OZ.AU!news.hawaii.edu!ames!uhog.mit.edu!news.mathworks.com!fu-berlin.de!zrz.TU-Berlin.DE!news.dfn.de!news.dkrz.de!dscomsa.desy.de!jamaica!vanesch Thu Apr 18 16:15:54 PDT 1996
Article: 10383 of sci.physics
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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 21:37:56 GMT
Organization: DESY
Lines: 41
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X-Newsreader: TIN [version 1.2 PL2]

Kevin Anthony Scaldeferri ([email-masked]) wrote:
: In article <[email-masked]>,
: Patrick van Esch <[email-masked]> wrote:
: >
: >Ok, I tried to get to the Riemann tensor over lunch, didn't 
: >really get there, only the Christoffel symbols:
: >
: >Apart from arithmetic errors, this is what I find:
: >
: >C^0_ii = 1/2 . R'
: >C^i_0i = 1/2.R'/R
: >C^i_i0 = 1/2.R'/R
: >
: >i = 1, 2 and 3
: >
: >all the rest = 0.
: >
: >This is different from what Kevin had, but then I calculated it very 
: >hastedly, so I might very well be wrong...
: >

Ok, so let us introduce R --> R^2 (the thing I had forgotten to
do at the end):

C^0_ii = 1/2.(R^2)' = 1/2.2R.R' = R.R'

C^i_0i = 1/2.(R^2)'/R^2 = R.R'/R^2 = R'/R

Guess that is what you had too...

   ?

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Thu Apr 18 16:16:06 PDT 1996
Article: 10391 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 17 Apr 1996 21:54:41 -0700
Organization: University of California, Riverside
Lines: 41
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
>>I haven't got your post anymore with your riemann components.
>>We should compare... (or look 'em up in MTW :)

No cheating!  Let me put it this way, you guys seem to be converging to
the right answer.  (But if you *do* cheat, keep in mind that different
people use different conventions for the order in which to list the
subscripts of the Riemann tensor R^a_{bcd}.)

>And, once again I post my appeal to those who understand the
>symmetries of the Riemann tensor.  Please explain.  I see antisymmetry
>in b and c, but no others are obvious.  And speaking of MTW, I don't
>know what they mean when they state the symmetries:
>
>R^a_bcd = R^a_[bc]d		apparently denotes the b/c antisymmetry
>
>R^a_[bcd] = 0			means what?

When people put square brackets around a bunch of n subscripts,
they mean to "totally antisymmetrize them": sum over all n!
permutations, multiplied by +1 or -1 depending on whether it's an even
or odd permutation, and then divide by n!.

So 

R^a_{[bc]d} = 0

(note: I use {}'s to group things, as in LaTeX) means

R^a_{bcd} - R^a_{cbd} = 0

while 

R^a_{[bcd]} = 0 

means that a certain sum of 6 things is zero.  But using the first
identity this is equivalent to a sum of 3 being zero:

R^a_{bcd} + R^a_{cdb} + R^a_{dbc} = 0

Article 10617 (27 more) in sci.physics:
From: [email-masked] (john baez)
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 16:22:57 -0700
Organization: University of California, Riverside
Lines: 31
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (john baez) writes:
>Unless we make our path from (t,x0,y0,z0) to (t,x0,y0,z0) stay on the
>slice of constant t, we can make it arbitrarily short by making it
>wiggle in the time direction (still keeping it spacelike).  This is the
>spacelike analog of the "twin paradox" effect where you can make the
>proper time along a path from (t0,x,y,z) to (t1,x,y,z) be arbitrarily
>small by making it wiggle in the space direction (still keeping it
>timelike).

>So it is not very interesting to ask about the shortest spacelike
>path between two spacelike separated points: there are always spacelike
>paths from one which are arbitrarily short!  The notion of distance only
>gets interesting when we specify a spacelike surface containing our two
>points, and ask what is the shortest path *on that surface* from one
>point to another.

I think I was getting a wee bit carried away here, because even if this
is true, there is still a "nicest" path between two sufficiently nearby
spacelike separated points, namely the shortest *geodesic* from one to
the other.  The length of this would give a notion of "distance" between
spacelike separated points that doesn't require specifying a spacelike
surface containing our two points.  

However, it's worth remembering that anything we actually see is not
spacelike separated from us, but lightlike separated (at the moment the
light was emitted).  So you have to make up some precise definition
before you can say "the galaxy we are seeing is X light-years away"....
as should be clear to those who followed our earlier discussions about
that.

Article 10475 (4 more) in sci.physics:
From: [email-masked] (Brett McInnes)
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 02:04:25 GMT
Organization: National University of Singapore
Lines: 39
Distribution: world
NNTP-Posting-Host: [email-masked]
X-Newsreader: TIN [version 1.2 PL2]

john baez ([email-masked]) wrote:

: Note: I will stop putting quotes "at rest" as soon as it's clear
: everyone understands I really mean "following a worldline for which x,y,
: and z remain constant as t varies."  The point is that "at rest" is a
: coordinate-dependent notion.  There is no "absolute rest"!  When I
: say "at rest" it not meant to have any grand metaphysical significance;
: it's simply a handy quick way of speaking.  

I think that it's easy to get carried away by this "coordinates are 
garbage" philosophy---I know, because I have done it myself. In this
case,particularly, it is rather misleading to say that "at rest"
is a coordinate-dependent notion. In cosmology we begin by assuming
that we have a family of inertial observers who are distinguished by
being the ones who see isotropy. These observers are at rest with
respect to each other in the sense that their "restspaces" [ie,
the orthogonal complements of the tangent vector to the worldline
at each point] integrate up to give you a foliation of the spacetime.
This is of course what we mean by "relatively at rest" in SR. That
is, these observers *agree* as to whether two given events are 
simultaneous, which is a sensible definition of "relatively at rest".
Notice that nothing I have said here depends on a choice of coordinates.
So in this sense it is not true to say that "at rest" is a coordinate-
dependent notion.

Now you might argue that talk of "observers" is just a fancy way
of talking about coordinates. That is not so, however: the foliation
of the spacetime is a geometrical property, quite independent of
coordinates. Again, one might object that we could choose another
set of observers [who would not see isotropy] and that they would
have a different definition of "at rest". That is true, but it
does not change the fact that a *given* field of observers 
regard themselves as being at rest with respect to each other.

In short, the galaxies are at rest, not "at rest"--apart,of course
from the peculiar velocities. [coordinate-independent definition
of peculiar velocity: a galaxy is said to have a peculiar velocity
if its restspace is not tangential to the spacelike slices defined
by the isotropic observers.]

Article 10620 (24 more) in sci.physics:
From: [email-masked] (Emory F. Bunn)
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
Date: 18 Apr 1996 23:30:57 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 85
Distribution: world
NNTP-Posting-Host: physics12.berkeley.edu

In article <[email-masked]>,
Brett McInnes <[email-masked]> wrote:
>
>I think that it's easy to get carried away by this "coordinates are 
>garbage" philosophy---I know, because I have done it myself. In this
>case,particularly, it is rather misleading to say that "at rest"
>is a coordinate-dependent notion. In cosmology we begin by assuming
>that we have a family of inertial observers who are distinguished by
>being the ones who see isotropy. These observers are at rest with
>respect to each other in the sense that their "restspaces" [ie,
>the orthogonal complements of the tangent vector to the worldline
>at each point] integrate up to give you a foliation of the spacetime.
>This is of course what we mean by "relatively at rest" in SR. 

This is definitely *not* what we mean by "relatively at rest" in
special relativity.

Let me explain why.  First I'd like to repeat your argument in my own
words to make sure I've got it right:

1. In Minkowski space, there is a well-defined, coordinate-independent
   notion of two observers being "at rest" with respect to each other.

2. It would be nice to express that notion in a purely geometrical way
   so that we could try to apply it to other spacetimes besides Minkowski.

3. Here's how you do it: A family of observers are defined to be
   "at rest" with respect to each other if (a) each observer sees 
   isotropy and (b) you can foliate spacetime with spacelike slices that
   are orthogonal to the worldlines of all of these observers.

Have I summarized your position fairly?  If not, please let me know.

I completely agree with 1.  I'm pretty much indifferent regarding 2,
but I'll accept it for now.  But item 3 above is definitely,
unambiguously false.  There are infinitely many ways I can fill
Minkowski space with observers who satisfy conditions (a) and (b).  In
only some of those ways are the observers at rest in the usual
Minkowski sense.  Specifically, I'm thinking of all of the different
families of Milne observers.  They satisfy your criteria: their
surroundings look isotropic and you can foliate spacetime with their
"restspaces."  But they're not at rest in the usual sense.

If your goal is to come up with a generalizable geometrical definition
of Minkowskian rest, the first thing you'd better check is that your
definition gives the right answer in Minkowski space.  If it doesn't
-- and yours doesn't -- then it doesn't really make much difference
what it says about other spacetimes.

Here's a more succinct way to put this argument.  According to your
notion of rest, the Milne observers are "really at rest" with respect
to each other.  But I know (from e-mail discussions with you) that you
don't believe that.  You believe that the Milne observers are "really
moving apart."

By the way, here's my choice for how to define the notion of mutual
rest in Minkowski space without reference to coordinates: two
observers are at rest with respect to each other if they can send
light signals back and forth without any redshift or blueshift.
(Actually, I stole this definition from Baez.)  Using this definition,
you can't fill a (non-Minkowski) Friedmann space with observers who
are all at rest with each other.  This may make you sad, depending on
your temperament, but it's the way it is.

That's it for the physics.  I have one more thing to say from a
pedagogical point of view.  I think that item 2 in my summary of your
position is a Bad Thing pedagogically.  If your goal in discussing the
Friedmann models is to impart some intuition for how to think about
geometry in arbitrary curved spacetimes, then trying to generalize the
notion of Minkowskian rest to other spacetimes is at best distracting
and at worst misleading.  From the point of view of general relativity
in general (i.e., beyond the Friedmann models), the rightful heirs to
the Minkowskian inertial frames are the locally inertial frames.  It's
dangerous to get too hung up on the specialness of comoving
coordinates in the Friedmann models.

Let me emphasize again that this is a matter of pedagogy, not
physics.  From a geometrical point of view, the comoving coordinates
of a Friedmann model *are* special in a certain sense: they respect
the spacetime's isometries.  But one of the main reasons for studying
the Friedmann models is to learn how to think about arbitrary
spacetimes, so I think it's best to emphasize tools (like locally
inertial coordinate frames) that work in arbitrary spacetimes.

-Ted

Article 10624 (20 more) in sci.physics:
From: [email-masked] (john baez)
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 16:33:14 -0700
Organization: University of California, Riverside
Lines: 121
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] wri
tes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Hmm.  So you claim 

>>C^0_ii = R(t)R'(t),

>>but I think Esch got something else.  I'm afraid we're going to have to
>>call in an unbiased third party, like Oz, to figure out which of you is
>>right.

>What!

>Ha ha ha ha ha!

>Er, what's a Cri-whatever symbol whatsit?

>I have been waiting for someone, somewhere, to tell me.

Sigh.  You're falling behind... perhaps it's a relativistic time
dilation effect?  First of all, you don't need to have the foggiest clue
as to what the Christoffel symbols MEAN to calculate them.  Quite a
while back I said:

`By the way, Oz was allowed to go behind the curtain, and this is what he
saw:

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

So you can use this to compute the Riemann tensor whenever you want ---
at least, after I tell you want those commas mean in things like
g_{cd,b} and C^a_{bd,c}.  They mean partial differentiation!  So for
example

g_{cd,b} = partial g_{cd} / partial x^b

where "partial" is a feeble ASCII equivalent of one of those curly
d's they use for partial derivatives.  Here x^b is just the coordinate
t, x, y, or z depending on whether the superscript b is 0, 1, 2, or 3.

Similarly,  

C^a_{bd,c} = partial C^a_{bd} / partial x^c

and so on.  

So whenever you have a little spare time you can compute the Riemann
curvature of this big bang metric and see for yourself that it's
nonzero.'

Of course, it's rather demoralizing to compute something without
knowing what it means!  Luckily Kevin and I explained what the
Christoffel symbols mean --- though not yet why they are given by that
crazy formula above.  To repeat:

`I'd put it this way: say we WORK IN LOCAL COORDINATES and take a unit
vector pointing in the d direction and parallel transport it an amount
epsilon in the c direction.  Now it won't be pointing in the "same
direction" anymore!  It'll point a bit more in the a direction than it
had been.  How much?  Well, about

                      - epsilon C^a_{cd} 

much.'

I also expanded on that, saying

`I might as well remind folks of our definition of the Riemann tensor
and the Christoffel symbols:

RIEMANN: 

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number. The vector w comes back a bit changed by its
journey; it is now a new vector w'. We then have

    w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3 

CHRISTOFFEL:

FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the d
direction and parallel transport it an amount epsilon in the c
direction.  See how much its component in the a direction changes.  
(This component started out being 0 unless a = d, in which case it was
1.)  It changes by 

     - epsilon C^a_{cd} + terms of order epsilon^2

Note what's going on here.  On the one hand, the definition of the
Christoffel symbols involves parallel transporting a vector along a
little path.  It thus involves comparing vectors *different points*,
which we can only do by introducing a coordinate system and comparing
their components in these coordinates.  On the other hand, the Riemann
tensor involves parallel transporting a vector around a little loop.  It
thus involves comparing vectors at *the same point*, which we can do
without coordinates.  (The space of tangent vectors at a given point is
a vector space, so we have no trouble making sense of the difference of
two of them.)

For this reason, even though the Riemann tensor is more complicated in
some ways, it doesn't require any coordinate system to define, which
winds up meaning that it transforms very simply under coordinate
transformations.'

So really the only missing link, as far as the Christoffel symbols go,
is a derivation of that crazy formula for them in terms of the metric.  
Not surprisingly, this hasn't stopped some people from racing ahead and
computing the the Christoffel symbols and Riemann tensor of the big bang
metric.  A 2-pronged approach like this is nice: on the one hand, start
calculating away so you can see what the big bang cosmology is like; on
the other hand, get me to explain WHY the formulas are what they are.

Article 10484 (3 more) in sci.physics:
From: [email-masked]    (Norbert Pfannerer)
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 12:12:14 GMT
Organization: IBM ELBU
Lines: 39
NNTP-Posting-Host: pfanner.speech.ibm.com
X-Newsreader: IBM NewsReader/2 v1.9d - NLS

>: In article <[email-masked]> [email-masked] (Kevin Ant
hony Scaldeferri) writes:
>
>: >So, here's what I got.
>
>: >i can be 1,2, or 3 in the following, no Einstein summation notation
>: >
>: >C^0_ii = R(t)R'(t)
>: >C^i_0i = C^i_i0 = R'(t) / R(t)
>: >all others = 0
>
>: >So, based on that, what is the physical meaning of the Christoffel
>: >symbols for this metric
>
>: >C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>: >vector in the x direction.  Then if we move forward in time a little
>: >bit, this vector has 'stretched' in the x direction by a factor of
>: >R'(t)/R(t).  This, I think, just describes the Hubble expansion and

After thinking about this a bit, I have to disagree. Remeber that minus
sign that John told us about!
(Having not done the calculation myself (yet), I don't know where it is
missing here; but I suspect the calculation is correct, but the
interpretation is not)

I think a vector in x direction will *shrink* by a factor of R'/R.
By shrink, I mean relative to the coordinate system. The length of the
vector will stay the same (since it is parallel transported), so the
components with respect to the coordinate system will shrink.
Equivalently I could say the coordinate system is expanding, i.e. 
"space" is expanding.

(All of this under the assumption that R(t) is increasing)

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

From: Geoffrey A. Landis <[email-masked]>
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 13:22:19 GMT
Organization: Ohio Aerospace Institute
Lines: 36
Distribution: world
NNTP-Posting-Host: glandis.lerc.nasa.gov
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Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 8bit
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X-XXMessage-ID: <[email-masked]>
X-XXDate: Thu, 18 Apr 1996 13:23:52 GMT

In article <[email-masked]> Brett McInnes,
[email-masked] writes:
>...In this
>case,particularly, it is rather misleading to say that "at rest"
>is a coordinate-dependent notion. In cosmology we begin by assuming
>that we have a family of inertial observers who are distinguished by
>being the ones who see isotropy. These observers are at rest with
>respect to each other in the sense that their "restspaces" [ie,
>the orthogonal complements of the tangent vector to the worldline
>at each point] integrate up to give you a foliation of the spacetime.
>This is of course what we mean by "relatively at rest" in SR. That
>is, these observers *agree* as to whether two given events are 
>simultaneous, which is a sensible definition of "relatively at rest".

A very broad statement!  You just said that two events at different
coordinates x but the same coordinate t would be measured to occur at the
same time t in any frame of unmoving in the coordinate system (x,y,z)
defining the space part of the Robertson metric.  Could you please
demonstrate that this is true using the metric 
            ds^2 = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)   ?
This is not obvious.

>Notice that nothing I have said here depends on a choice of coordinates.

Again, unclear.  To "agree" as to whether two events at different space
coordinates have the same time coordinates, the two observers need
coordinates to measure with.

>So in this sense it is not true to say that "at rest" is a coordinate-
>dependent notion.
>...

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center
physicist and part-time science fiction writer
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Summary: 
Expires: 
References: <[email-masked]> <[email-masked]> <[email-masked]>
Sender: 
Followup-To: 
Distribution: world
Organization: University of California, Riverside
Keywords: 

Let's see.  Various people have calculated the Christoffel symbols of
the big bang metric

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

using the mysterious formula written on that tome in the Wizard's back
room:

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b}).

Let's see what they say!  First Kevin Scaldeferri weighed in with the
following:

>So, here's what I got.

>i can be 1,2, or 3 in the following, no Einstein summation notation

>C^0_ii = R(t)R'(t)
>C^i_0i = C^i_i0 = R'(t) / R(t)
>all others = 0

Then Patrick van Esch wrote:

>Ok, I tried to get to the Riemann tensor over lunch, didn't 
>really get there, only the Christoffel symbols:

>Apart from arithmetic errors, this is what I find:

>C^0_ii = 1/2 . R'
>C^i_0i = 1/2.R'/R
>C^i_i0 = 1/2.R'/R

: >i = 1, 2 and 3

: >all the rest = 0.

Different!  But then he noted that he was using a metric g with R in it
where it should have been R^2.  If you replace his R by R^2, you get the
same thing Kevin did.

That's promising.  But of course we couldn't know for sure until Oz
worked it out.  A model of professionalism as always, he writes:

>Oh dear, I've blown it haven't I?
>The relationships were given by HRH Baez at the beginning.
>Oh dear, no excuse. Boo hoo.
>I'll have to think godammit.
>
>Well, having made a first stab (er, like writing a bit out) I of course
>come to the first question. 
>
>I can see where the results come from (I think), and bearing in mind
>that g^{ab} has diagonal (I sincerely hope) 
>
>(-1, 1/R(t)^2, 1/R(t)^2, 1/R(t)^2) then
>
>C^0_ii comes from the (1/2) g^{ab}.-g_{bd,c) term
>
>so       (-1/2) (-1) @r(t)^2/@t = R(t) R'(t)
>
>and
>
>C^i_0i  from  (1/2) g^{ab}. g_{bc,d} (and C^i_i0 from the remaining)
>
>so      (1/2) 1/(R(t)^2). @r(t)^2/@t = R'(t)/R(t)

So it looks like everyone agrees, though Oz didn't calculate *all* the
components.  I get the same answers, too.  

By the way, I'm not sure why Oz writes:

>Trouble is that when I go through what I think is the mechanics I get 
>zero for everything every time for every element. I therefore deduce
>that I don't know how to do the mechanics correctly. I presumed that it
>worked just like tensors, but I've probably (zap, frazzle, kerpow!) not
>remembered that correctly or misinterpreted how I should do this
>slightly different version. Er, um, ahem, so how do you evaluate these
>expressions?? Cringe.

because he DIDN'T get zero for everything.  I suppose he is just being
too modest to admit he actually got the right answer.

Okay.  Now why don't we try to understand what they mean!  If the
answers make sense, we'll get a little confidence in that mysterious
formula for the Christoffel symbols, even though we won't really know
why it works.  

Remember the definition of the Christoffel symbols:

FIXING A LOCAL COORDINATE SYSTEM, take a tangent vector pointing in the d
direction and extending one unit in these coordinates, and parallel
transport it an amount epsilon in the c direction.  See how much its
component in the a direction changes.  (This component started out being
0 unless a = d, in which case it was 1.)  It changes by

     - epsilon C^a_{cd} + terms of order epsilon^2

So for example, what does C^1_{01} mean?  First, we start with a
tangent vector that extends one coordinate unit in the x direction,
since "1" means "x" in this game.  

[Obnoxious Note: it needn't have LENGTH one!  It is defined using
coordinates in a manner that doesn't ever mention its length.  This is
crucial, but was obscure in my earlier attempts to describe the
Christoffel symbols. Also, it's just a tangent vector, not an actual
stick in spacetime.  You can regard it as an infinitesimal stick, if you
want, even though it extends one unit in the x direction.  Does this make
sense?  Probably not.  Well, we actually discussed these issues before
when we were first learning about tangent vectors, but can talk about
them again if necessary.]

Then we take that tangent vector and parallel transport it an amount
epsilon in the time direction, since "0" means "t".  

Then we see how much it extends in the x direction now.  It may have
changed!  It will change by 

     - epsilon C^1_{01} + terms of order epsilon^2

Let's see if our answer

               C^i_{0i} = R'(t)/R(t)

makes sense.  Well, it makes some rough sense, because C^i_{0i} has to
do with how much more or less a tiny vector pointing in the x direction
extends in that direction as time passes, and R'(t)/R(t) is the "Hubble
constant" that measures the "rate at which the universe is expanding".

The problem is to see if we got the sign right!  Suppose R'(t) > 0 so
the universe is expanding.  Well, we are parallel transporting our
tangent vector forwards in time, so its length stays the same.  But the
distance between two points (x0,y0,z0) and (x1,y1,z1) at time t gets
BIGGER as time passes if R'(t) > 0.   So it must be that as time passes
our ruler will extend LESS in the x direction.  But there is a minus
sign in the definition of the Christoffel symbols, so this implies 
C^i_{0i} should be POSITIVE!

And it is!

Whew!

As you can see, this is like setting the clock forwards to set the sun
back so as to get up earlier relative to the sunrise... if we don't pay
careful attention we'll miss one of those minus signs and get everything
backwards!

I think this may clarify what Norbert Pfannerer said:

In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:
>>Patrick van Esch writes:
>>So, based on that, what is the physical meaning of the Christoffel
>>symbols for this metric

>>C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>>vector in the x direction.  Then if we move forward in time a little
>>bit, this vector has 'stretched' in the x direction by a factor of
>>R'(t)/R(t).  This, I think, just describes the Hubble expansion [....]

>After thinking about this a bit, I have to disagree. Remember that minus
>sign that John told us about!

>(Having not done the calculation myself (yet), I don't know where it is
>missing here; but I suspect the calculation is correct, but the
>interpretation is not)

>I think a vector in x direction will *shrink* by a factor of R'/R.
>By shrink, I mean relative to the coordinate system. The length of the
>vector will stay the same (since it is parallel transported), so the
>components with respect to the coordinate system will shrink.
>Equivalently I could say the coordinate system is expanding, i.e. 
>"space" is expanding.
>
>(All of this under the assumption that R(t) is increasing)
From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Fri Apr 19 10:54:43 PDT 1996
Article: 10644 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 17:26:59 -0700
Organization: University of California, Riverside
Lines: 17
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:

>Is it possible to suffer mental damage from a surfeit of super-
>subscripts?

Yes.  Elie Cartan, the wonderful mathematician who (among other things)
occaisionally helped out Einstein with his geometry, spoke
disparagingly of the "debauch of indices".  Like any other sort of
debauch, it can be debilitating.  But we need to do a bit of index
juggling to get a feel for what the workday of an actual relativity
theorist is like.  One shouldn't over-romanticize it.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Fri Apr 19 10:56:32 PDT 1996
Article: 10652 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 18 Apr 1996 18:18:24 -0700
Organization: University of California, Riverside
Lines: 173
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

Let's see.  Various people have calculated the Christoffel symbols of
the big bang metric

              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)

using the mysterious formula written on that tome in the Wizard's back
room:

         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b}).

Let's see what they say!  First Kevin Scaldeferri weighed in with the
following:

>So, here's what I got.

>i can be 1,2, or 3 in the following, no Einstein summation notation

>C^0_ii = R(t)R'(t)
>C^i_0i = C^i_i0 = R'(t) / R(t)
>all others = 0

Then Patrick van Esch wrote:

>Ok, I tried to get to the Riemann tensor over lunch, didn't 
>really get there, only the Christoffel symbols:

>Apart from arithmetic errors, this is what I find:

>C^0_ii = 1/2 . R'
>C^i_0i = 1/2.R'/R
>C^i_i0 = 1/2.R'/R

: >i = 1, 2 and 3

: >all the rest = 0.

Different!  But then he noted that he was using a metric g with R in it
where it should have been R^2.  If you replace his R by R^2, you get the
same thing Kevin did.

That's promising.  But of course we couldn't know for sure until Oz
worked it out.  A model of professionalism as always, he writes:

>Oh dear, I've blown it haven't I?
>The relationships were given by HRH Baez at the beginning.
>Oh dear, no excuse. Boo hoo.
>I'll have to think godammit.
>
>Well, having made a first stab (er, like writing a bit out) I of course
>come to the first question. 
>
>I can see where the results come from (I think), and bearing in mind
>that g^{ab} has diagonal (I sincerely hope) 
>
>(-1, 1/R(t)^2, 1/R(t)^2, 1/R(t)^2) then
>
>C^0_ii comes from the (1/2) g^{ab}.-g_{bd,c) term
>
>so       (-1/2) (-1) @r(t)^2/@t = R(t) R'(t)
>
>and
>
>C^i_0i  from  (1/2) g^{ab}. g_{bc,d} (and C^i_i0 from the remaining)
>
>so      (1/2) 1/(R(t)^2). @r(t)^2/@t = R'(t)/R(t)

So it looks like everyone agrees, though Oz didn't calculate *all* the
components.  I get the same answers, too.  

By the way, I'm not sure why Oz writes:

>Trouble is that when I go through what I think is the mechanics I get 
>zero for everything every time for every element. I therefore deduce
>that I don't know how to do the mechanics correctly. I presumed that it
>worked just like tensors, but I've probably (zap, frazzle, kerpow!) not
>remembered that correctly or misinterpreted how I should do this
>slightly different version. Er, um, ahem, so how do you evaluate these
>expressions?? Cringe.

because he DIDN'T get zero for everything.  I suppose he is just being
too modest to admit he actually got the right answer.

Okay.  Now why don't we try to understand what they mean!  If the
answers make sense, we'll get a little confidence in that mysterious
formula for the Christoffel symbols, even though we won't really know
why it works.  

Remember the definition of the Christoffel symbols:

FIXING A LOCAL COORDINATE SYSTEM, take a tangent vector pointing in the d
direction and extending one unit in these coordinates, and parallel
transport it an amount epsilon in the c direction.  See how much its
component in the a direction changes.  (This component started out being
0 unless a = d, in which case it was 1.)  It changes by

     - epsilon C^a_{cd} + terms of order epsilon^2

So for example, what does C^1_{01} mean?  First, we start with a
tangent vector that extends one coordinate unit in the x direction,
since "1" means "x" in this game.  

[Obnoxious Note: it needn't have LENGTH one!  It is defined using
coordinates in a manner that doesn't ever mention its length.  This is
crucial, but was obscure in my earlier attempts to describe the
Christoffel symbols. Also, it's just a tangent vector, not an actual
stick in spacetime.  You can regard it as an infinitesimal stick, if you
want, even though it extends one unit in the x direction.  Does this make
sense?  Probably not.  Well, we actually discussed these issues before
when we were first learning about tangent vectors, but can talk about
them again if necessary.]

Then we take that tangent vector and parallel transport it an amount
epsilon in the time direction, since "0" means "t".  

Then we see how much it extends in the x direction now.  It may have
changed!  It will change by 

     - epsilon C^1_{01} + terms of order epsilon^2

Let's see if our answer

               C^i_{0i} = R'(t)/R(t)

makes sense.  Well, it makes some rough sense, because C^i_{0i} has to
do with how much more or less a tiny vector pointing in the x direction
extends in that direction as time passes, and R'(t)/R(t) is the "Hubble
constant" that measures the "rate at which the universe is expanding".

The problem is to see if we got the sign right!  Suppose R'(t) > 0 so
the universe is expanding.  Well, we are parallel transporting our
tangent vector forwards in time, so its length stays the same.  But the
distance between two points (x0,y0,z0) and (x1,y1,z1) at time t gets
BIGGER as time passes if R'(t) > 0.   So it must be that as time passes
our ruler will extend LESS in the x direction.  But there is a minus
sign in the definition of the Christoffel symbols, so this implies 
C^i_{0i} should be POSITIVE!

And it is!

Whew!

As you can see, this is like setting the clock forwards to set the sun
back so as to get up earlier relative to the sunrise... if we don't pay
careful attention we'll miss one of those minus signs and get everything
backwards!

I think this may clarify what Norbert Pfannerer said:

In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:
>>Patrick van Esch writes:
>>So, based on that, what is the physical meaning of the Christoffel
>>symbols for this metric

>>C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>>vector in the x direction.  Then if we move forward in time a little
>>bit, this vector has 'stretched' in the x direction by a factor of
>>R'(t)/R(t).  This, I think, just describes the Hubble expansion [....]

>After thinking about this a bit, I have to disagree. Remember that minus
>sign that John told us about!

>(Having not done the calculation myself (yet), I don't know where it is
>missing here; but I suspect the calculation is correct, but the
>interpretation is not)

>I think a vector in x direction will *shrink* by a factor of R'/R.
>By shrink, I mean relative to the coordinate system. The length of the
>vector will stay the same (since it is parallel transported), so the
>components with respect to the coordinate system will shrink.
>Equivalently I could say the coordinate system is expanding, i.e. 
>"space" is expanding.
>
>(All of this under the assumption that R(t) is increasing)

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From: [email-masked]    (Norbert Pfannerer)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 13:29:35 GMT
Organization: IBM ELBU
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In <[email-masked]>, [email-masked] (john baez) writes:

... (stuff deleted)
>The problem is to see if we got the sign right!  Suppose R'(t) > 0 so

>the universe is expanding.  Well, we are parallel transporting our
>tangent vector forwards in time, so its length stays the same.  But the
>distance between two points (x0,y0,z0) and (x1,y1,z1) at time t gets
>BIGGER as time passes if R'(t) > 0.   So it must be that as time passes
>our ruler will extend LESS in the x direction.  But there is a minus
>sign in the definition of the Christoffel symbols, so this implies 
>C^i_{0i} should be POSITIVE!
>
>And it is!
>
>Whew!
>
>As you can see, this is like setting the clock forwards to set the sun
>back so as to get up earlier relative to the sunrise... if we don't pay
>careful attention we'll miss one of those minus signs and get everything
>backwards!
>
>I think this may clarify what Norbert Pfannerer said:
>
>In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:
>>After thinking about this a bit, I have to disagree. Remember that minus
>>sign that John told us about!
>
>>I think a vector in x direction will *shrink* by a factor of R'/R.
>>By shrink, I mean relative to the coordinate system. The length of the
>>vector will stay the same (since it is parallel transported), so the
>>components with respect to the coordinate system will shrink.
>>Equivalently I could say the coordinate system is expanding, i.e. 
>>"space" is expanding.
>>
>>(All of this under the assumption that R(t) is increasing)
>

Wow! No thunderbolts! The teacher even agrees with what I said!
Of course his words explain this much more clearly than I could.

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 19 Apr 1996 09:33:49 +0100
Organization: Oz
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, Kevin Anthony Scaldeferri
<[email-masked]> writes
>
>After checking mine, I fine that I should of had these too.  So, I
>think we are in agreement on the Riemann tensor with:
>
>R^i_i00 = - R^i_0i0 = R"/R
>R^0_i0i = - R^0_0ii = RR"
>R^i_jij = - R^i_ijj = (R')^2
>
>And, once again I post my appeal to those who understand the
>symmetries of the Riemann tensor.  Please explain.  I see antisymmetry
>in b and c, but no others are obvious.  And speaking of MTW, I don't
>know what they mean when they state the symmetries:
>
>R^a_bcd = R^a_[bc]d            apparently denotes the b/c antisymmetry
>
>R^a_[bcd] = 0                  means what?

OK, OK, Kevin. Now please could you explain what b/c antisymmetry
actually means. Actually, what this nice (!) bunch of solutions says to
us physically. Well, 'us' means those who are huffing and puffing on
behind, you must not leave us so far behind that we are out of sight.

Now let's see if I have this right. We started with a metric that
allowed us to model some time dependant spacial dimensions. Then we
Cristoffeled the metric apparently to find out how the length of vectors
changed as we moved around in spacetime, which has a certain logic to it
in this context even if it brought in the bete noir of co-ordinate
dependance. Then this allowed (some people) to produce the appropriate
Riemann tensor, which is slightly odd since this is supposed to be co-
ordinate independant but since the Cristoffel thingies are co-ordinate
dependant one might imagine that this Riemann tensor is too.

Finally one might imagine that one ought to fill space up with something
and see how it all links together. One hopes that some sort of
(hopefully simple ha ha) expression(s) for R(t) that makes sense might
float up from the murky depths.

Er, or something.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 10813 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 19 Apr 1996 09:41:45 +0100
Organization: Oz
Lines: 40
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, Kevin Anthony Scaldeferri
<[email-masked]> writes
>
>>So the covariant derivative is not a tensor.
>
>Ahh, of course.  In fact we were just talking about this type of thing
>in my topology course.  I should have realized that.  Oh well, live
>and learn.

Oh, cool. I always wanted to know a bit about how you worked things out 
mathematically in topology, it always seemed rather impossible. Now, for 
some reason, it doesn't. It just looks absurdly difficult!

>>But what about Kevin's claim that C^0_{11} and the like are nonzero?
>>
>>>Yeah, I'd really like to understand this one.  It has me a little
>>>worried.  It suggests that a straight line in space is not
>>>autoparallel and thus not a geodesic. 
>>
>>It definitely would imply that.
>>
>>>But elsewhere on the thread, we
>>>are talking about the distance between two points at the same time
>>>just being the familiar Euclidean straight line distance.  And
>>>shouldn't this mean that that straight line is a geodesic?  
>>
>>Well, let me use "space at time T" to stand for the hypersurface t = T,
>>where T is some number.  This is what folks call a "spacelike slice" of
>>spacetime.
>>
>>Okay: there is certainly no paradox in what you suggest.  It's perfectly
>>possible for a curve in space at time T to be a geodesic in space at
>>time T, but not be a geodesic in spacetime!  

Ah. What, like an orbit? No, this is presumably the reverse.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 10783 of sci.physics
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From: [email-masked]    (Norbert Pfannerer)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 15 Apr 1996 09:12:36 GMT
Organization: IBM ELBU
Lines: 49
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In <[email-masked]>, Oz <[email-masked]> writes:
>
>We need two lightlike paths in this spacetime. Hmm what might that mean.
>Well locally we ought still to be Minkowskian so light travels at 45deg.
>I guess that it would mean that dx/dt=1 and we can set dy,dz=0. 
>(Shaky ground feeling)
>
>so      dS^2= -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)  
>becomes dS^2= -dt^2 + R(t)^2 dx^2               and substituting dx=dt
>        dS^2= -dt^2 + R(t)^2 dt^2 = (R(t)-1)dt^2        or
>        dS=(R(t)-1)^0.5 dt                      which looks promising

I don't think this is the correct way. I think we have to assume dS=0
to get a lightlike path. I'll also assume dy=dz=0, so our lightrays move
out in the x direction.

    dS^2=-dt^2 + R(t)^2 dx^2
    dt^2 = R(t)^2 dx^2
    dt = R(t)dx

Now assuming R(t) is increasing, the lightrays in an (x,t) coordinate
system are curves that get steeper as time passes.
           
In the crude ascii diagram one can easily see that more time passes at
x1 where the lightrays arrive.

                         |
                         |
                         |
                         |
          t|         .   |      .
           |        .    |     .
           |       .     |    .
           |     .       |  .
           |   .         |.
           |.          . |
           |        .    |
           |    .        |
           |.            |
           +----------------------------  x
           x0            x1                

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 19 Apr 1996 09:33:49 +0100
Organization: Oz
Lines: 44
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 <[email-masked]>
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In article <[email-masked]>, Kevin Anthony Scaldeferri
<[email-masked]> writes
>
>After checking mine, I fine that I should of had these too.  So, I
>think we are in agreement on the Riemann tensor with:
>
>R^i_i00 = - R^i_0i0 = R"/R
>R^0_i0i = - R^0_0ii = RR"
>R^i_jij = - R^i_ijj = (R')^2
>
>And, once again I post my appeal to those who understand the
>symmetries of the Riemann tensor.  Please explain.  I see antisymmetry
>in b and c, but no others are obvious.  And speaking of MTW, I don't
>know what they mean when they state the symmetries:
>
>R^a_bcd = R^a_[bc]d            apparently denotes the b/c antisymmetry
>
>R^a_[bcd] = 0                  means what?

OK, OK, Kevin. Now please could you explain what b/c antisymmetry
actually means. Actually, what this nice (!) bunch of solutions says to
us physically. Well, 'us' means those who are huffing and puffing on
behind, you must not leave us so far behind that we are out of sight.

Now let's see if I have this right. We started with a metric that
allowed us to model some time dependant spacial dimensions. Then we
Cristoffeled the metric apparently to find out how the length of vectors
changed as we moved around in spacetime, which has a certain logic to it
in this context even if it brought in the bete noir of co-ordinate
dependance. Then this allowed (some people) to produce the appropriate
Riemann tensor, which is slightly odd since this is supposed to be co-
ordinate independant but since the Cristoffel thingies are co-ordinate
dependant one might imagine that this Riemann tensor is too.

Finally one might imagine that one ought to fill space up with something
and see how it all links together. One hopes that some sort of
(hopefully simple ha ha) expression(s) for R(t) that makes sense might
float up from the murky depths.

Er, or something.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 10922 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 09:44:58 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
Lines: 55
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References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]>,
Norbert Pfannerer <[email-masked]> wrote:
>>: In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
>>
>>: >So, here's what I got.
>>
>>: >i can be 1,2, or 3 in the following, no Einstein summation notation
>>: >
>>: >C^0_ii = R(t)R'(t)
>>: >C^i_0i = C^i_i0 = R'(t) / R(t)
>>: >all others = 0
>>
>>: >So, based on that, what is the physical meaning of the Christoffel
>>: >symbols for this metric
>>
>>: >C^i_0i = R'(t) / R(t) I think I understand. Say we look at a unit
>>: >vector in the x direction.  Then if we move forward in time a little
>>: >bit, this vector has 'stretched' in the x direction by a factor of
>>: >R'(t)/R(t).  This, I think, just describes the Hubble expansion and
>
>After thinking about this a bit, I have to disagree. Remeber that minus
>sign that John told us about!
>(Having not done the calculation myself (yet), I don't know where it is
>missing here; but I suspect the calculation is correct, but the
>interpretation is not)
>
>I think a vector in x direction will *shrink* by a factor of R'/R.
>By shrink, I mean relative to the coordinate system. The length of the
>vector will stay the same (since it is parallel transported), so the
>components with respect to the coordinate system will shrink.
>Equivalently I could say the coordinate system is expanding, i.e. 
>"space" is expanding.
>

Yes, you're actually right here.  I realized this after posting it.
However, this still shows the Hubble expansion, because I should
really have said something like this.  When we move the unit x-vector
forward in time a little, it appears to 'shrink' by a factor of R'/R
(or 1-R'/R in the sense that VanEsch prefers).  Equivalently, we can
say that the rest of the universe appears to have expanded relative to
this still unit-length vector.  This is like what someone else was
discussing of how we could view the expansion of the universe as a
shrinking of our rulers, but people would probably look at us funny.
So, this symbol does indeed point out to us the expansion of the
universe.

BTW, I'm not just trying to save my argument here, I actually realized
this and thought about it myself previously.

Kevin Scaldeferri

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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 10:14:03 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
Lines: 87
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In article <[email-masked]>, john baez <[email-masked]> wrote:
>Let's see.  Various people have calculated the Christoffel symbols of
>the big bang metric
>
>              g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2)
>
>using the mysterious formula written on that tome in the Wizard's back
>room:
>
>         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b}).
>
...
>
>That's promising.  But of course we couldn't know for sure until Oz
>worked it out.  A model of professionalism as always, he writes:
>
>>Oh dear, I've blown it haven't I?
>>The relationships were given by HRH Baez at the beginning.
>>Oh dear, no excuse. Boo hoo.
>>I'll have to think godammit.
>>
>>Well, having made a first stab (er, like writing a bit out) I of course
>>come to the first question. 
>>
>>I can see where the results come from (I think), and bearing in mind
>>that g^{ab} has diagonal (I sincerely hope) 
>>
>>(-1, 1/R(t)^2, 1/R(t)^2, 1/R(t)^2) then
>>
>>C^0_ii comes from the (1/2) g^{ab}.-g_{bd,c) term
>>
>>so       (-1/2) (-1) @r(t)^2/@t = R(t) R'(t)
>>
>>and
>>
>>C^i_0i  from  (1/2) g^{ab}. g_{bc,d} (and C^i_i0 from the remaining)
>>
>>so      (1/2) 1/(R(t)^2). @r(t)^2/@t = R'(t)/R(t)
>

Uh oh...I think I see where some of Oz's confusion lies.  I think we
need to back up a little before running off to new and more
complicated things.  (Forgive me if any of this ends up sounding
condescending, that's not my intention, I just want to do things step
by step.  And if you haven't made a mistake and I just misinterpreted
things, mea culpa)

OK, so we have our equation

>         C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b}).

Now, remember this is already a little misleading.  This is not one
equation giving us all the symbols, instead it is many equations each
giving one symbol.  So, for example

C^0_11 = (1/2) g^0b (g_{b1,1} + g_{b1,1} - g_{11,b})

(remembering that we have an implied summation over all b's.)
But, g^ab is diagonal, so luckily there's only one non-zero piece in
this summation, the one multiplied by g^00

C^0_11 = (1/2) g^00 (g_{01,1} + g_{01,1} - g_{11,0})

Now, we notice that the only variable in the metric is t, so any
partial derivatives with respect to other coordinates are zero.  So,
the first two terms drop out and, substituting in the pieces of the
metric, we are left with

C^0_11 = (1/2) (-1) (-@(R^2(t))/@t)
       = (1/2) (2 R'(t) R(t))
       = R'(t) R(t)

Whew, that seems like a lot of work for just one of 4x4x4=64 symbols.
But, wait, the Christoffel symbols are symmetric wrt the lower
indices, so there are only 40.  And this universe is homogeneous and
isotropic, so there's a lot more that are the same.  All told, for
this metric, there's only 4 or 5 calculations like this that you have
to do.

That's enough for now

Kevin

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat Apr 20 12:22:14 PDT 1996
Article: 10936 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 21:18:55 -0700
Organization: University of California, Riverside
Lines: 18
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References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Edward Green) writes:
>[email-masked] (john baez)' wrote: 

>>>: I think I should pound my point in once more: the Christoffel symbols 
>>>: are BLATANTLY COORDINATE-DEPENDENT.   We cannot define them 
>>>: in a coordinate-free way as we could with the Riemann tensor! 

>Is this mortal sin the reason they are forbidden to even appear in the
>index of Misner,  Thorne and Wheeler? 

Heh.  No.  Strangely, it seems those guys never speak of "Christoffel
symbols".  They call 'em "connection coefficients".  See page 209 and
all over the place after that.  Like everyone else they are written with
a capital Greek letter gamma.  I denote them as C in this thread as part
of my "Greek in ASCII sucks" policy.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat Apr 20 12:22:26 PDT 1996
Article: 10950 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 19 Apr 1996 22:01:40 -0700
Organization: University of California, Riverside
Lines: 24
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In article <[email-masked]> "Paul G. White" <[email-masked]> writes:

>But by the same token, that seems to put a kink in the explanation of
>expansion from the Friedmann solution. If, in order to match
>observations, we have to graft non-expanding patches onto our expanding
>balloon, it seems unsatisfactory, somehow.  Why are these patches
>exempted?

They aren't "exempted" as if by some arbitrary decree.  Essentially, you
solve Einstein's equation and *show* that gravitationally bound lumps of
matter like galaxies don't expand in a big bang universe.  Of course it
ain't so easy to solve Einstein's equation in practice, but Ted said you
can look at the simplified case of a spherically symmetric galaxy and
work out that it doesn't expand.  The symmetry is used to make the
math easier.  I suspect that there's a way to show this sort of thing
more generally without explicit computations, but I admit I don't know
how I'd do it.  All I can say is I'm not at all surprised that
gravitationally bound lumps of stuff don't expand!

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Article: 10967 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 19 Apr 1996 20:34:36 GMT
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In article <[email-masked]>, Paul G. White <[email-masked]> wrote:

>I'm sorry, I've been taking inanity exercises to get up to the proper
>level. I'll try again. 

Maybe you just don't have the knack for it.

>I read the FAQ quote, twice, and I must admit
>to being less that satisfied by the explanation. In fact, it's not
>really an explanation as much as an apology. I can accept the fact
>that interatomic distances are set by quantum mechanics, specifically
>Planck's constant, which doesn't seem to be necessarily connected to
>Hubble's constant. OK, so material objects don't take part in the
>universal expansion, but objects following a geodesic through
>spacetime should, one would think. 

This is not necessarily true, for two completely separate reasons.

Reason 1. Unlike an idealized Robertson-Walker spacetime, the spacetime we
actually live in is not expanding uniformly.  In certain high-density
regions, like the region we live in, spacetime just doesn't look
locally like an expanding Robertson-Walker spacetime.  Our spacetime
is expanding (in the Robertson-Walker sense) when you average it over
large scales, but lots of individual pockets of it are not expanding.

I gather this explanation doesn't satisfy you.  Can you tell me why
not?  Do you (a) not understand what it means, or (b) not believe that
it's true, or (c) something else?

Reason 2. Even in the idealized Robertson-Walker spacetime (in which
spacetime is expanding uniformly on all scales), not all geodesics
move apart from each other.  All of the geodesics that correspond to
comoving observers move apart from each other, but it's easy to cook
up two world lines for two objects in an expanding Robertson-Walker
spacetime that approach each other.

It's really just a matter of initial conditions.  Suppose I put two
test objects in my expanding Universe.  I give one of them a peculiar
velocity towards the other one, so that at some initial time they're
not moving towards each other or away from each other.  (That is,
object 2 has a peculiar velocity that just cancels its Hubble velocity
with respect to object 1.)  Then I let the two objects follow their
respective geodesics through spacetime.  These geodesics don't move
apart from each other; in fact, they converge toward each other and
eventually collide!  This is true even though "spacetime is expanding"
all around them.

Here's one way to think about it.  This is very heuristic and vague,
so take it however you want.  The key point is to drop the notion that
there's some "expansion force" pushing everything apart.  In an
expanding Robertson-Walker model, everything is moving apart from
everything else, essentially because that's what they've always been
doing.  There's nothing pushing everything apart -- things just move
apart because of inertia.  So if you have a system in which, for
whatever reason, two objects are not moving apart at one time, they'll
never start moving apart.  That's why the Earth and Sun aren't moving
apart -- simply because they never have been, and there's no force
pushing them apart.

>What's hard to accept are the constant distances in the solar system,
>for example.  The distances at which the planets orbit are set by
>angular momenta and gravitational attraction, to borrow Newtonian
>terms. If these orbits were really expanding over time then this would
>either imply that either the angular momenta were increasing, which
>hardly seems reasonable, or that the gravitational attractions were
>decreasing. Since this is not supported by observation, we conclude
>that the gravitational constant is not decreasing.

That's right.  The orbits aren't expanding, and G is not decreasing.

>But by the same token, that seems to put a kink in the explanation of
>expansion from the Friedmann solution. If, in order to match
>observations, we have to graft non-expanding patches onto our
>expanding balloon, it seems unsatisfactory, somehow.  Why are these
>patches exempted? Granted, it may match observations pretty well, but
>it leaves the door open, in my mind, for a better solution which would
>apply universally, not just in selected areas.

Let me emphasize that you don't just graft non-expanding patches ad
hoc onto a Friedmann solution.  That's just what you get when you
solve the equations of general relativity.  Specifically, start with
initial conditions at some early time in which the Universe is
expanding pretty much uniformly everywhere.  Suppose that the density
is almost the same everywhere, but not quite.  There are slight
overdensities and slight underdensities.  Now solve the Einstein
equation to see how this Universe evolves with time.  You find that,
averaged over large scales, the Universe continues to expand like a
Robertson-Walker spacetime, but on small scales the overdense regions
stop expanding and collapse to form pretty much static, dense lumps.

That's just what the equations tell you.

>An analogy which comes to mind is the relative velocities from
>fragments from an explosion. From the standpoint of any randomly
>selected fragment, all the others would be receding. The further other
>fragments were, the faster the recession. I'm certainly not a
>cosmologist and I wouldn't want this to be taken as a proposed model
>(it's not original, I read it somewhere), but to me, it seems more
>satisfactory than Friedmann.I am reading the references in MTW, but
>this would never get answered if I waited until I understood that!

OK.  Now take that model and throw in the additional assumption that
the fragments are massive enough that you can't neglect their mutual
gravitational attraction.  Then a clump of nearby fragments will stop
moving apart from each other and collapse into a single, non-expanding
lump.  That's what we see.

-Ted

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat Apr 20 19:27:36 PDT 1996
Article: 11059 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 20 Apr 1996 18:27:33 -0700
Organization: University of California, Riverside
Lines: 44
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (LBsys) writes:
>Oz <[email-masked]> schreibt:

>>john baez <[email-masked]>

>>>In fact, the extrinsic curvature of a spacelike slice precisely measures
>>>the failure of geodesics in that slice to be geodesics in the spacetime
>>>itself.  

>>Another quotable quote here. 
>>I read it five times to be sure I had it right.

>Great! Be proud! I read it fifty times and couldn't make anything out of
>it. 

Geez, was it that murky?  Basically, we say a surface A in some
spacetime B has "extrinsic curvature" if a straight as possible line
on A isn't a straight as possible line in B.  Since "straight as
possible line" sounds stupid, people usually say "geodesic".  

For example (yet again), let A be the surface of a balloon, and B be the
surrounding 3-dimensional space.  The straightest posssible lines in A
are great circles, like the equator.  But these aren't the straightest
possible lines in B.   So we say A has "extrinsic curvature" in B.  

In short: the imaginary 2-dimensional people on A have a different notion of
"straight" than us 3-dimensional people in B.  

But what about the 4-dimensional people?

Well, in the big bang universe that we're currently toying with in this
tutorial, the 3-dimensional space B is just a surface in a 4-dimensional
spacetime C!  And what we were just noticing is that the straight as
possible lines in B, which us 3-dimensional folks call simply "lines",
aren't straight as possible in C!  So B has "extrinsic curvature" in C.

More technically, there is a gadget called the "extrinsic curvature
tensor" which makes this all very precise.  It says precisely how much a
geodesic in some surface fails to be a geodesic in the surrounding space
or spacetime.  When you study general relativity in detail, this tensor
plays a big role.

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Article: 11142 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 17:52:39 -0700
Organization: University of California, Riverside
Lines: 22
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>>JB wrote:
>>>Okay: there is certainly no paradox in what you suggest.  It's perfectly
>>>possible for a curve in space at time T to be a geodesic in space at
>>>time T, but not be a geodesic in spacetime!  

>Ah. What, like an orbit? No, this is presumably the reverse.

A geodesic in Schwarzschild spacetime looks like a roughly elliptical
thingie, certainly no geodesic, when we project it down to "space".  
But as you note, this is the related reverse problem.  A geodesic in a given
spacelike slice needn't be a geodesic in spacetime.  Again, think of the
simple example of the sphere in R^3: a geodesic on the sphere isn't a
geodesic in R^3, because the sphere sits inside R^3 in a curved way,
i.e., it has "extrinsic curvature".  

In fact, the extrinsic curvature of a spacelike slice precisely measures
the failure of geodesics in that slice to be geodesics in the spacetime
itself.  

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Article: 11182 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 19 Apr 1996 17:48:10 -0700
Organization: University of California, Riverside
Lines: 51
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] writes:

>Now let's see if I have this right. We started with a metric that
>allowed us to model some time dependant spacial dimensions. Then we
>Cristoffeled the metric apparently to find out how the length of vectors
>changed as we moved around in spacetime, 

Beware: the LENGTH of a vector never changes when we parallel
translate it.  Parallel translation is defined to be "metric-
compatible", meaning that as you carry your javelin about it doesn't get
longer or shorter.  What the Christoffel symbols tell us is how the
COMPONENTS of a vector (in some coordinate system!) change as we
parallel translate it.

That's what you meant, right?

>which has a certain logic to it
>in this context even if it brought in the bete noir of co-ordinate
>dependance. 

Yes.

>Then this allowed (some people) to produce the appropriate
>Riemann tensor, which is slightly odd since this is supposed to be co-
>ordinate independant but since the Cristoffel thingies are co-ordinate
>dependant one might imagine that this Riemann tensor is too.

Slightly odd but not too odd.  First, one can often reach a
coordinate-independent goal by coordinate-dependent methods.  Second,
since our metric was given to us in a certain coordinate system, we are
DOOMED to soil our hands with coordinates in this problem.

>Finally one might imagine that one ought to fill space up with something
>and see how it all links together. One hopes that some sort of
>(hopefully simple ha ha) expression(s) for R(t) that makes sense might
>float up from the murky depths.

Well, show us how you computed a component of the Riemann tensor and got
something 0 when you felt you shouldn't have.  Then we'll straighten you
out and we'll compute the whole darn Riemann tensor.  Then we'll compute
the Ricci tensor --- which you ALREADY know a lot about, thanks to that valor
test.  Then we'll compute the Einstein tensor.  And then we will solve
for R(t).

A note to van Esch, Scaldeferri and Pfannerer: Please don't give away
all the answers to the computations before Oz does them!  He needs the
practice.  Besides, I enjoy torturing him.

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Article: 11203 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sun, 21 Apr 1996 11:51:37 +0100
Organization: Oz
Lines: 49
Distribution: world
Message-ID: <[email-masked]>
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] (LBsys) writes:
>>Oz <[email-masked]> schreibt:
>
>>>john baez <[email-masked]>
>
>>>>In fact, the extrinsic curvature of a spacelike slice precisely measures
>>>>the failure of geodesics in that slice to be geodesics in the spacetime
>>>>itself.  
>
>>>Another quotable quote here. 
>>>I read it five times to be sure I had it right.
>
>>Great! Be proud! I read it fifty times and couldn't make anything out of
>>it. 
>
>Geez, was it that murky?  

Well, lets say it was, er, succinct.  :-)

And just to keep it really down to earth, 
spring appears to have arrived in England. 

The sky is blue with fleecy clouds
and buds are burst with start of spring,
and rooks are building nests on high
to bring some joy to everything.

The daffs are waving heads of gold
all pointing to the warming sun
and skylarks soar into the sky
and lambs across the fields run.

Winter's grip's forgotten now
belonging to the distant past
and everything's renewed again,
another year's begun at last.

Ugh! :-{

Well, not as poetic as Baez's quotable quote, but just to let you know
there is a world out there this morning. Shock tactics.  :-)

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!sunsite.doc.ic.ac.uk!aixssc.uk.ibm.com!hursley.ibm.com!lisa.speech.ibm.com!usenet Sun Apr 21 18:51:03 PDT 1996
Article: 10976 of sci.physics
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From: [email-masked]    (Norbert Pfannerer)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 16 Apr 1996 13:15:27 GMT
Organization: IBM ELBU
Lines: 39
Message-ID: <[email-masked]>
References: <[email-masked]>
Reply-To: [email-masked] (Norbert Pfannerer)
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Since I came late to his thread, I assume I have to pass another test.  
So lets do some more calculations (this is going to be the second
calculation that I do on my own in GR, the first one was the lightlike path
yesterday).

Some days ago you asked how much a volume will change. So I'll assume
8 coffee grounds on the corners of a cube, at rest (I know this has
a meaning only relative to the coordinate system). The volume of a cube is
(length of edge)^3, since distances scale with R(t), the volume
scales with R(t)^3.

When the wizard talked to Oz, the sound waves (vibrating air molecules)
produced some gravitational waves that were recorded by a device called
UFAEAG (Usenet Feed by Ascii Extraction from Amplification of Gravitons),  
and so all course notes eventually made it to my computer screen.

A few minutes ago I wanted to use the geodetic deviation equation to 
compute the Ricci or Riemann tensor, but now I can't see how I can apply
the formulas to the given situation. The question I have: 
When I take the velocity vector of one coffee cround ( (1,0,0,0) at
(t0,x0,y0,z0)) and parellel translate it to another corner of the
cube, what will I get? The other coffee ground's velocity vector
is (1,0,0,0) at the point (t0,x1,y0,z0). Is this the same as the
translated v of the 1st coffee ground? 

Someone calculated the Christoffel symbols, probably I could use that
to answer my question. I'll try later.

BTW, no thunderbolts arrived yet in my mailbox or my newsreader. Either
the news is slow, or I talk so much blunder that the wizard doesn't want
to waste his thunderbolts ... 

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

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Article: 10996 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 20 Apr 1996 19:37:33 +0100
Organization: Oz
Lines: 62
Distribution: world
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 <[email-masked]> <[email-masked]>
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In article <[email-masked]>, Patrick van Esch
<[email-masked]> writes
>Oz ([email-masked]) wrote:
>: >: >
>: >: >This is different from what Kevin had, but then I calculated it very 
>: >: >hastedly, so I might very well be wrong...
>
>: Well, I have learned more from Patrick's mistakes than I ever would have 
>: picked up had he simply agreed. Not least because that nasty old wizard 
>: made me have a go, which in the end is essential.
>
>Ah !  A true environmentalist, even recycles my mistakes :-)
>

Having worked through van Esch's derivation of the Riemann tensor and 
more or less followed what he did, and we can add the ones he found 
later that I haven't a clue how he came by 'cos the look like they 
should be zero to me, but then I'm dim, we can now substitute the 
relevant C-symbols in to get the following dogs dinner that really 
requires you to stand on your head and hum ommmmmm to understand. Wow, 
that was a long sentence. Shoof I'm getting i's in front of my eyes.

R^0_i0i =  R(t)"        which is nice for a change.
R^0_0ii = -R(t)"        not unexpected.

R^i_i00 =  R(t)"/R(t)   not too bad I suppose
R^i_0i0 = -R(t)"/R(t)

And now for the ones that look like the name of some Dutch town, no
wonder Van Esch has no trouble with these super&suffixes, the symbols
probably refer to a dutch word he already knows. Stick a few 'n's here
and there and we could probably make a sentence in dutch! :-)

R^i_ijj = -R(t)'^2
R^i_jij =  R(t)'^2

I have probably made lots of mistakes. It's worked out too easily. I 
know this is 'O' level stuff but hey, that was decades ago. I positively 
refuse to integrate any of these. 

Well, it's turned out not so bad. Now, what was the definition of the 
Riemann tensor again? Indeed is it still really a proper tensor after 
all these co-ordinate shinnanigans? Perhaps we had better stick to co-
ordinate representations.

Lets look at R^0_i0i =  R(t)" . Does this give us the change in the time 
direction of a vector in the x-direction as it's moved round a little 
parallelogram in the time and x-direction? Almost feels reasonable.

What to do next? Do we calculate the Ricci tensor from the Riemann 
tensor, shove some stuff in here and there and OhMyGod are we back to 
the original course notes with a weird metric and T_{00} to find out? Oh 
no, no, no, no, no, no! I'm off down the village, lets leave Patrick to 
carry the heavy load piled on by the demonically enthusiastic wizard! 
Hmm, perhaps if I hide behind this curtain I can spy what they're at and 
they won't see me. Hmm sounds a good idea that. Yup, they can't see me 
here so I'm safe. Ho, ho, ho!

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 11274 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 21 Apr 1996 14:14:25 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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In article <[email-masked]>,
Norbert Pfannerer <[email-masked]> wrote:
>Since I came late to his thread, I assume I have to pass another test.  
>So lets do some more calculations (this is going to be the second
>calculation that I do on my own in GR, the first one was the lightlike path
>yesterday).
>
>Some days ago you asked how much a volume will change. So I'll assume
>8 coffee grounds on the corners of a cube, at rest (I know this has
>a meaning only relative to the coordinate system). The volume of a cube is
>(length of edge)^3, since distances scale with R(t), the volume
>scales with R(t)^3.
>
>When the wizard talked to Oz, the sound waves (vibrating air molecules)
>produced some gravitational waves that were recorded by a device called
>UFAEAG (Usenet Feed by Ascii Extraction from Amplification of Gravitons),  
>and so all course notes eventually made it to my computer screen.
>
>A few minutes ago I wanted to use the geodetic deviation equation to 
>compute the Ricci or Riemann tensor, but now I can't see how I can apply
>the formulas to the given situation. 

Well, I haven't gotten to the geodesic deviation equation in my
studying yet, but isn't it unnecessary for this task?  John has handed
down the knowledge (or at least the formula) for computing the Riemann
tensor from the Christoffel symbols (in fact, elsewhere on this thread
are some tentative results of this calculation), and from the Riemann
tensor, the Ricci tensor is cake.

>The question I have: 
>When I take the velocity vector of one coffee cround ( (1,0,0,0) at
>(t0,x0,y0,z0)) and parellel translate it to another corner of the
>cube, what will I get? The other coffee ground's velocity vector
>is (1,0,0,0) at the point (t0,x1,y0,z0). Is this the same as the
>translated v of the 1st coffee ground? 
>

No, the velocity of another ground is definitely not the parallel
transport of the first.  If they were, the paths they are following
would be parallel, and there would be no change in the volume of the cube.

>Someone calculated the Christoffel symbols, probably I could use that
>to answer my question. I'll try later.
>

The Christoffel symbols will definitely shed some light on this.  In
fact, one of them basically is the representation of this problem.

Kevin Scaldeferri

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Article: 11283 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 21 Apr 1996 14:32:26 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>In article <[email-masked]>, Kevin Anthony Scaldeferri
><[email-masked]> writes
>>
>>After checking mine, I fine that I should of had these too.  So, I
>>think we are in agreement on the Riemann tensor with:
>>
>>R^i_i00 = - R^i_0i0 = R"/R
>>R^0_i0i = - R^0_0ii = RR"
>>R^i_jij = - R^i_ijj = (R')^2
>>
>>And, once again I post my appeal to those who understand the
>>symmetries of the Riemann tensor.  Please explain.  I see antisymmetry
>>in b and c, but no others are obvious.  And speaking of MTW, I don't
>>know what they mean when they state the symmetries:
>>
>>R^a_bcd = R^a_[bc]d            apparently denotes the b/c antisymmetry
>>
>>R^a_[bcd] = 0                  means what?
>
>OK, OK, Kevin. Now please could you explain what b/c antisymmetry
>actually means. Actually, what this nice (!) bunch of solutions says to
>us physically. Well, 'us' means those who are huffing and puffing on
>behind, you must not leave us so far behind that we are out of sight.
>

OK, by the b/c antisymmetry we just mean that

R^a_bcd = - R^a_cbd

If you find the formula we used and stare at it for a moment, it will
be obvious (mathematically, at least) that this is true.  Now as for
the physical interpretation, remember that our picture is a little
parallelogram that we parallel transport some vector around.  Now, if
we switch b and c, we will have the same parallelogram, but we go
around it in the other direction.  If you try to picture the whole
procedure in your mind, I hope you will that we should just get a
negative sign tacked on.  I guess the thing to realize is that on each
leg of the trip, reversing the direction reverses the effect (at least
in the differential limit).  Let us know if you have difficulty seeing
why this works, maybe someone can come up with a good visualization
trick.

As for describing the physical significance of each of the non-zero
term in the riemann tensor, I'm not quite there yet.  All I can say is
that since not everything is zero, our spacetime is curved, which is
reassuring because otherwise this wouldn't be a very interesting
problem.

Kevin

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!physics12.Berkeley.EDU!ted Sun Apr 21 19:02:42 PDT 1996
Article: 11285 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Followup-To: sci.physics
Date: 21 Apr 1996 20:06:07 GMT
Organization: Physics Department, U.C. Berkeley
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In article <[email-masked]>, ale2 <[email-masked]> wrote:
>i'm confused, say our universe is closed and in the future will
>collapse to have a radius of one cm, won't a galaxy have to at some
>point get a little smaller as measured by the same person who measures
>the radius of the universe to be 1 cm?

Oh, sure.  If the big crunch comes, the galaxies will get crunched
too.  Is that what you're saying?

I'll try to be more specific.  As I said before, it's tricky to solve
in detail for what an inhomogeneous expanding spacetime looks like.
However, there are a couple of simple cases that are tractable.
Consider a Universe that's filled with some ideal gas.  At some
particular early time, most of the Universe has a density rho0, but
there's a small spherical pocket that has a larger density rho1.  At
that early time, the whole shebang is expanding uniformly according to
Hubble's law.

Let's suppose that rho0 is greater than the critical density, so the
whole Universe is eventually headed for a big crunch.  The spherical
pocket has a higher density, though, and as a result its expansion
slows down more rapidly.  You can solve exactly for what it does.  The
overdense pocket expands for a while, slows down, stops, and starts to
recollapse.  If it were true pressureless dust, then the pocket would
recollapse back down to a big-crunch singularity, but let's say
instead that it's a gas with some temperature and pressure.  Then
it'll recollapse for a while, but eventually it winds up being
pressure-supported.  It becomes a more or less static sphere of gas.
Gravity is holding it together, but pressure is keeping it from
collapsing further.  (It's not 100% completely static, but it's
very close.  Its density evolves on a time scale that's much, much
longer than the expansion time scale.)

That's what the overdense pocket does.  Meanwhile, the rest of the
Universe is expanding on its merry way.  Far away from the overdense
region, things look just like an ordinary Robertson-Walker model.
Near the boundary of the spherical pocket, things are more
complicated, but you can solve the equations and find out what
happens.  You find that that static sphere of pressure-supported gas
joins smoothly onto the expanding background.

Given that the overall density is greater than the critical density,
the Universe as a whole is going to recollapse eventually.  When it
does, all of the rest of the stuff out there collapses in on our
pressure-supported sphere, and it's not pressure-supported anymore.
The whole enchilada collapses in a big crunch.

The key point here is that there's a very long period of time during
which the sphere is pretty much static.  That's what galaxies are
thought to be like right now: more or less static collections of stuff
that live in an expanding background.  But if the big crunch comes,
they won't be static anymore!

(In case it's not clear, I mean "static" here in a sort of statistical
sense.  The galaxies as a whole aren't expanding or contracting, but
the individual bits of them -- stars, dust, gas, and whatever dark
matter there is -- are moving around.  It's just like saying that the
air is "static" if there's no wind around: the individual molecules
are still whizzing every which way, but the system as a whole is
sitting still.)

-Ted

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sun Apr 21 19:03:03 PDT 1996
Article: 11286 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics,sci.math
Subject: Re: General Relativity Tutorial
Date: 21 Apr 1996 13:12:53 -0700
Organization: University of California, Riverside
Lines: 84
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu
Xref: noise.ucr.edu sci.physics:11286 sci.math:6494

In article <[email-masked]> [email-masked] (LBsys) writes:
>Im Artikel <[email-masked]>, [email-masked] (john baez)
>schreibt:
>>>>john baez <[email-masked]> wrote:
>>>>>In fact, the extrinsic curvature of a spacelike slice precisely 
>>>>>measures the failure of geodesics in that slice to be geodesics in the
>>>>>spacetime itself.  

>>Geez, was it that murky?  

>No, it wasn't 'murky'. I just didn't get the right notion of "precisely
>measures the failure". If "failure" had been "deviation", may be 499 times
>reading would have been enough :-)

Oh, okay.  I'm afraid I lapsed into "math-ese".  When some equation you
might at first think is true, and sometimes is true, turns out not to
always be true, mathematicians often like to cook up some gadget that
quantitatively measures exactly how much the two sides aren't equal.  In
these situations they say the gadget "measures the failure" of the
equation.

For example, you might have thought at first that if you parallel
transported a vector around a little loop, it would come back just the
same.  But it doesn't unless space is flat.  The Riemann curvature tensor
"measures the failure" of parallel transport around a loop to leave a
vector unchanged.

The torsion tensor measures the failure of some other thing to be true:

Say we have coordinates x^a.  Take a little vector of size epsilon
pointing in the a direction, and a little vector of size epsilon
pointing in the b direction.  Parallel translate the vector pointing in
the a direction by an amount epsilon in the b direction.  Similarly,
parallel translate the vector pointing in the b direction by an amount
epsilon in the a direction.  (Draw the resulting two vectors.)  If the
tips touch, up to terms of epsilon^3, there's no torsion!  Otherwise
take the difference of the tips and divide by epsilon^2.  Taking the
limit as epsilon -> 0 we get the torsion t_{ab}.

You might have thought the tips would touch (up to order epsilon^3), but
the torsion measures the failure of this to be true.

>A stupid question: Would that 'tensor' being applied to 4-space be the
>same that takes us from 2-space to 3-space? 

Yes, the same math applies whenever you are studying the extrinsic
curvature of an n-dimensional space sitting in an (n+1)-dimensional one.
(More precisely, whenever you have an n-dimensional submanifold of an
(n+1)-dimensional Riemannian or Lorentzian manifold.... students in the
GR tutorial will all know what I mean by that.)

>In other words, could I apply
>the same maths, that took my from Euclids 2-space geometrie to
>Who-ever-invented-it 3-space geometrie? (could it be then that the sum of
>angles in a triangle with equal angles is 360 degrees?, oops, in case of
>starting at the 'equator' and  moving right out to the 'pole', of course,
>one of 'my beloved' special cases... Can it be that easy?).

I read that 500 times and I still don't quite get it.  Let me just say,
in case it's relevant, that extrinsic curvature is different from
intrinsic curvature.  The 2-dimensional surface of a globe is
intrinsically curved: little 2-dimensional people on it could tell it
was curved, just by seeing how the angles of "spherical triangles" on it
don't add up to 180 degrees.  (Or, if they were more clever, they
could measure the Riemann curvature!)  So when we have the surface of a
globe --- which us mathematicians call simply a sphere --- sitting
inside good old Euclidean 3-dimensional space, the sphere is both
intrinsically curved and extrinsically curved.  

But extrinsic curvature is not to be confused with intrinsic curvature!

If you take a piece of paper and roll it up into a cylinder, that
will be extrinsically curved but not intrinsically curved.  For example,
the angles of "triangles" drawn with geodesics on the cylinder always
add up to 180 degrees.  

The spacelike slice of the big bang universe we are considering in the
GR tutorial is also extrinsically but not intrinsically curved.  

I leave it as a puzzle to find a 2-dimensional submanifold of a
3-dimensional Riemannian manifold that's intrinsically but not
extrinsically curved.

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Article: 11307 of sci.physics
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From: [email-masked] (Keith Ramsay)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: Sun, 21 Apr 1996 18:18:18 -0400
Organization: Boston College
Lines: 26
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
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X-Newsreader: Yet Another NewsWatcher 2.2.0b6

In article <[email-masked]>, [email-masked] (Kevin
Anthony Scaldeferri) wrote:
| If, in order to match observations, we have to graft 
|>non-expanding patches onto our expanding balloon, it seems 
unsatisfactory, somehow. 
|>Why are these patches exempted?

It illustrates the limitations of the "expanding balloon" 
metaphor. Balloons have air in them pushing the sides in
all directions. Our GR solutions have no such "exterior"
force, of course. The parts that are "expanding" are 
expanding autonomously, as it were. Likewise, any piece
which begins to act differently, collapsing into a black
hole perhaps, will continue to "do its own thing" excepting 
the influence neighboring regions have on it. Presumedly
there is some sort of adjustment to the orbits of a system 
of gravitationally bound bodies, due to their being in a
curved surrounding space. It seems from what is being said
here, however, that it doesn't amount to the surrounding
space saying, "hey, come expand with us!". It's more 
complicated, and I wouldn't bet upon being able to figure
it out without pulling out the mathematical tool kit and
doing some calculations.

Keith Ramsay

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sun Apr 21 19:24:02 PDT 1996
Article: 11317 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 21 Apr 1996 18:50:36 -0700
Organization: University of California, Riverside
Lines: 75
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:

>Some days ago you asked how much a volume will change. So I'll assume
>8 coffee grounds on the corners of a cube, at rest (I know this has
>a meaning only relative to the coordinate system). The volume of a cube is
>(length of edge)^3, since distances scale with R(t), the volume
>scales with R(t)^3.

That's right.  I'm sure Oz could have gotten this one, but he was
probably busy fighting Christoffel symbols.

>A few minutes ago I wanted to use the geodetic deviation equation to 
>compute the Ricci or Riemann tensor, but now I can't see how I can apply
>the formulas to the given situation. The question I have: 
>When I take the velocity vector of one coffee cround ( (1,0,0,0) at
>(t0,x0,y0,z0)) and parallel translate it to another corner of the
>cube, what will I get? 

Let me tell you how to solve these problems in general.  Let me use
slightly different notion, though.

Say you have coordinates around, as we do here.  Say you have a vector v
at the point x = (x0,x1,x2,x3), and you want to parallel translate it
along the curve C given by

C(s) = (x0,x1,x2,x3) + s(w0,w1,w2,w3)

where w = (w0,w1,w2,w3) is any old list of 4 numbers and s is a real
number.  More precisely, say you parallel translate v from x to
C(s) along the curve C.  You get some vector v(s).  

You want to know how the components v^a(s) of the vector v(s).  

I'll tell you how v^a(s) changes as you change s, and you can do an
integral to figure out v^a(s) if you know v^a to begin with.

Here's how:

(d/ds) v^a(s) = - C^a_{bc} w^b v^c(s).

I said this before, but not in quite the same way.  This way of saying
it should help you solve your puzzle.

What I said before was:

FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the d
direction and parallel transport it an amount epsilon in the c
direction.  See how much its component in the a direction changes.  
(This component started out being 0 unless a = d, in which case it was
1.)  It changes by 

     - epsilon C^a_{cd} + terms of order epsilon^2

This is really saying the same thing more tersely.

>Someone calculated the Christoffel symbols, probably I could use that
>to answer my question. 

Yes indeed, see above.

>BTW, no thunderbolts arrived yet in my mailbox or my newsreader. Either
>the news is slow, or I talk so much blunder that the wizard doesn't want
>to waste his thunderbolts ... 

Actually he doesn't go hurling thunderbolts at random people who ask him
questions, only his official apprentices.  Right now he just has one.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sun Apr 21 19:25:02 PDT 1996
Article: 11319 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 21 Apr 1996 19:01:11 -0700
Organization: University of California, Riverside
Lines: 73
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] writes:

>R^0_i0i =  R(t)"        which is nice for a change.
>R^0_0ii = -R(t)"        not unexpected.
>
>R^i_i00 =  R(t)"/R(t)   not too bad I suppose
>R^i_0i0 = -R(t)"/R(t)

>R^i_ijj = -R(t)'^2
>R^i_jij =  R(t)'^2

>I have probably made lots of mistakes. It's worked out too easily. I 
>know this is 'O' level stuff but hey, that was decades ago. I positively 
>refuse to integrate any of these. 

The wizard growls: "You're not supposed to integrate them, you buffoon.
They look approximately correct.  I forget if they are exactly right.  
Why don't you work out the Ricci tensor using them.  I know what that's
supposed to look like.  And you do too, a least a little bit!  Remember
what you worked out about the Ricci tensor?  In your test of valor?
That will serve as a check on whether you are doing things right!"

>Well, it's turned out not so bad. Now, what was the definition of the 
>Riemann tensor again? Indeed is it still really a proper tensor after 
>all these co-ordinate shinnanigans? Perhaps we had better stick to co-
>ordinate representations.

"What?!  You forget the definition of the Riemann tensor?!"  The wizard 
waves his staff threateningly at Oz, and then turns and pulls out a
yellowed, decaying piece of parchment from the enormous stack on his
desk.

"Here it is:

The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point
of spacetime.  Thus it takes three tangent vectors, say u, v, and w as
inputs, and outputs one tangent vector, say R(u,v,w).  The Riemann tensor
is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number.  The vector w comes back a bit changed by its
journey; it is now a new vector w'.  We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Or in coordinates," he adds, and all of a sudden a swarm of superscripts
and subscripts fly down from some cracks in the ceiling and land on the
parchment, "it looks like this:

Take the vector w^a, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u^b and epsilon v^c , where
epsilon is a small number.  The vector w^a comes back a bit changed by its
journey; it is now a new vector w'^a.  We then have

w'^a - w^a = -epsilon^2 R^a_{bcd} u^b v^c w^d + terms of order epsilon^3

These are two ways of saying the same thing."

>Lets look at R^0_i0i =  R(t)" . Does this give us the change in the time 
>direction of a vector in the x-direction as it's moved round a little 
>parallelogram in the time and x-direction? Almost feels reasonable.

"Oh, so you DID remember the definition.  Hmmph.  Playing dumb again."

>What to do next? Do we calculate the Ricci tensor from the Riemann 
>tensor, shove some stuff in here and there and OhMyGod are we back to 
>the original course notes with a weird metric and T_{00} to find out? Oh 
>no, no, no, no, no, no!

"Yes!"

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Article: 11412 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics,sci.math
Subject: Re: General Relativity Tutorial
Date: 22 Apr 1996 09:52:22 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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Xref: noise.ucr.edu sci.physics:11412 sci.math:6559

In article <[email-masked]>, john baez <[email-masked]> wrote:
>
>I leave it as a puzzle to find a 2-dimensional submanifold of a
>3-dimensional Riemannian manifold that's intrinsically but not
>extrinsically curved.
>
>

Note: another one to skip and save for later if you want to try and
figure this out yourself.

So we want a manifold which is intrinsically but not extrinsically
curved.  Well, the first time we are going to have to do is find a
good choice of a 3-manifold.  Being Euclidean beasts ourself, of
course the one we are most familiar with is R^3.  But this is a flat
space, so no matter what we do, any intrinsically curve submanifold
will have extrinsic curvature.  To see this, look at is this way...any
intrinsically curved space has an angle surplus or deficiency in
triangles.  This surplus (or deficiency, I'm going to stop writing
both) will be preserved when we look at it in the ambient space.  But,
true triangles in R^3 have no angle surplus.

So...we are going to have to pick a curved 3-manifold and a
2-submanifold whose curvature 'matches' that of the larger space.  Ok,
simplest curved 3-manifold is the 3-sphere.  Now we just need to take
a nice slice of the 3-sphere.  But this is just a 2-sphere.  To
confirm, we know that the 2-sphere has intrinsic curvature,
independent of what space it is embedded in because that's what we
mean by intrinsic.  But geodesics on the 2-sphere will be geodesics on
the 3-sphere, so the 2-sphere is not extinsically curved _wrt_ the
3-sphere.  (It is ext. curved as a submanifold of other 3-manifolds,
like say R^3, but to someone living in S^3, S^2 appears to have no
more or less curvature than S^3)

Kevin Scaldeferri

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue Apr 23 12:15:25 PDT 1996
Article: 11560 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 22 Apr 1996 22:46:15 -0700
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Okay, I'm wondering if everyone who's following this course understands
what I said about the Christoffel symbols.  If you really understand it,
you should be in the position to figure out some cool and very useful
stuff.

So remember my basic definition of them:

CHRISTOFFEL SYMBOLS: 

FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the c
direction and parallel transport it an amount s in the b
direction.  See how much its component in the a direction changes.  
It changes by 

     - s C^a_{bc} + terms of order s^2

Note: here "unit vector in the c direction" doesn't mean vector of
length one, it means a vector pointing out one unit in the c direction
in our coordinate system.  Remember, subscripts like c can be
0,1,2, or 3, corresponding to the 4 dimensions of spacetime, so for
example if c = 2 the "unit vector in the c direction" is just (0,0,1,0).  

Now remember that parallel translation was defined to be linear.  So if
you parallel translate a couple of vectors and then add them, or
multiply them by numbers, you could just as well have first added them
or multiplied them by numbers, and THEN parallel translated them.  No
difference!  

So we can take our definition of the Christoffel symbols and see what it
implies for vectors that are linear combinations of the basis vectors. 
(Here by "basis vectors" I mean the vectors pointing in the coordinate
directions like the c direction and so on).  And we get:

FIXING A LOCAL COORDINATE SYSTEM, take the vector v and parallel
transport it an amount s in the u direction.  See how much its
component in the a direction changes.  It changes by

     - s C^a_{bc} u^b v^c + terms of order s^2

Does everyone see how I got that?  We took the vector v and wrote it as

(v^0,v^1,v^2,v^3) = 

v^0 (1,0,0,0) + v^1 (0,1,0,0) + v^2 (0,0,1,0) + v^3 (0,0,0,1) 

and then used linearity and the previous definition of the Christoffel
symbols.  And ditto for u.

Thank god for linearity, eh?

Now let's consider parallel translating v along a curve that starts
out pointing in the u direction.  I.e. we have some curve K(s) with
K(0) being the point we're starting at, and the derivative K'(0), that is
the tangent vector of the curve K(s) at s = 0, happens to equal u.  
We get:

Take the vector v and parallel transport it along the curve K to the
point K(s), getting the new vector v(s).  Now FIXING A COORDINATE
SYSTEM, let v^a(s) be the components of v(s).  Differentiating these
components we get

(d/ds) v^a(s) |       =  - C^a_{bc} K'^b(0) v^c 
               s=0

See how we got that?  We just took the earlier formula

     - s C^a_{bc} u^b v^c + terms of order s^2

and differentiated it with respect to s.  Also, we used K' = u.  

We are just saying the same thing in different ways, but the last way we
said should make you realize we have a handy method of parallel
translating vectors along curves.  

Okay, one last thing.  Say the curve C(s) is a geodesic.  That means
ITS OWN TANGENT VECTOR IS PARALLEL TRANSLATED ALONG ITSELF: it "follows
its nose", going as straight as possible in the wiggly spacetime we
inhabit.

In other words, we can take v above to be the tangent vector K'(s) of
the curve.  So we get:

(d/ds) K'(s)^a |       =  - C^a_{bc} K'(0)^b K'(0)^c 
                s=0

or just

K''(0)^a =  - C^a_{bc} K'(0)^b K'(0)^c 

But since whatever's going on at s = 0 is just like what's going on at
any old s, we more generally get

K''(s)^a =  - C^a_{bc} K'(s)^b K'(s)^c 

See?  The cool self-referential definition of geodesic as a curve whose
own tangent vector is parallel translated along itself, gives us an
equation where the curve's second derivative K'' --- closely related to
acceleration --- is determined by its velocity K' in a *quadratic* sort
of way.  

Anyway, this is the equation for geodesics.  

Okay, folks, now use it to show that the galaxies "at rest" in our
big-bang universe are following geodesics.  Remember that "at rest"
means their x, y, and z coordinates don't change as t changes.  And our
hardy band of students agreed the Christoffel symbols were

C^0_ii = R(t)R'(t)

C^i_0i = C^i_i0 = R'(t) / R(t)

all others = 0

where i = 1,2,3 as usual, and we aren't doing Einstein summation.

It's good to know that being "at rest" in the coordinate sense implies
following a geodesic, that is, an unaccelerated path!

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Article: 11565 of sci.physics
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From: [email-masked] (Edward Green)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 22 Apr 1996 22:13:10 -0400
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[email-masked] (LBsys)' wrote: 
 
>>>In fact, the extrinsic curvature of a spacelike slice precisely measures

>>>the failure of geodesics in that slice to be geodesics in the spacetime 
>>>itself.   
>> 
>>Another quotable quote here.  
>>I read it five times to be sure I had it right. 
> 
>Great! Be proud! I read it fifty times and couldn't make anything out of 
>it. And I'm sure I could read it 500 times and still would have it wrong 
>(as long as having the feeling of *not_understanding_it_at_all* is equal 
>to have it wrong :-). 
 
You are too modest.   You should pound on the screen and tell everybody
else it must be wrong.   :-)  
 
Sans any calculational machinery to give a precise meaning to "precisely", 
I think we can just think of this in terms of a sphere,  and the fact that
the geodesics on the sphere (great circles) are *curved* in three space. 
In fact,  they are curved precisely as much as the surface of the sphere!!!
 Ohmigod. 
 
-- 
 
Ed Green /  [email-masked] 
 
Occam used a double-edged razor.

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Article: 11590 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 22 Apr 1996 20:15:55 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>In article <[email-masked]>, Patrick van Esch
><[email-masked]> writes
>>Oz ([email-masked]) wrote:
>>: >: >
>>: >: >This is different from what Kevin had, but then I calculated it very 
>>: >: >hastedly, so I might very well be wrong...
>>
>>: Well, I have learned more from Patrick's mistakes than I ever would have 
>>: picked up had he simply agreed. Not least because that nasty old wizard 
>>: made me have a go, which in the end is essential.
>>
>>Ah !  A true environmentalist, even recycles my mistakes :-)
>>
>
>Having worked through van Esch's derivation of the Riemann tensor and 
>more or less followed what he did, and we can add the ones he found 
>later that I haven't a clue how he came by 'cos the look like they 
>should be zero to me, but then I'm dim, we can now substitute the 
>relevant C-symbols in to get the following dogs dinner that really 
>requires you to stand on your head and hum ommmmmm to understand. Wow, 
>that was a long sentence. Shoof I'm getting i's in front of my eyes.
>

Ummm...I won't ask.

>R^0_i0i =  R(t)"        which is nice for a change.
>R^0_0ii = -R(t)"        not unexpected.
>
>R^i_i00 =  R(t)"/R(t)   not too bad I suppose
>R^i_0i0 = -R(t)"/R(t)
>
>And now for the ones that look like the name of some Dutch town, no
>wonder Van Esch has no trouble with these super&suffixes, the symbols
>probably refer to a dutch word he already knows. Stick a few 'n's here
>and there and we could probably make a sentence in dutch! :-)
>
>R^i_ijj = -R(t)'^2
>R^i_jij =  R(t)'^2
>
>I have probably made lots of mistakes. It's worked out too easily. I 
>know this is 'O' level stuff but hey, that was decades ago. I positively 
>refuse to integrate any of these. 
>

These look close, but not quite what van Esch and I got.  They do have
the proper b/c antisymmetry.  I'd double check all the math though to
make sure they're dead on.

>
>What to do next? Do we calculate the Ricci tensor from the Riemann 
>tensor, shove some stuff in here and there and OhMyGod are we back to 
>the original course notes with a weird metric and T_{00} to find out? Oh 
>no, no, no, no, no, no! I'm off down the village, lets leave Patrick to 
>carry the heavy load piled on by the demonically enthusiastic wizard! 
>Hmm, perhaps if I hide behind this curtain I can spy what they're at and 
>they won't see me. Hmm sounds a good idea that. Yup, they can't see me 
>here so I'm safe. Ho, ho, ho!
>

Oh come now, it's all downhill once you've got the Riemann tensor.

Kevin

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 23 Apr 1996 11:27:04 +0100
Organization: Oz
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Hmmm. Nobody will speak to me and tell me where I went wrong.

In article <[email-masked]>, Oz
<[email-masked]> writes
>
>What to do next? Do we calculate the Ricci tensor from the Riemann 
>tensor, shove some stuff in here and there and OhMyGod are we back to 
>the original course notes with a weird metric and T_{00} to find out? Oh 
>no, no, no, no, no, no! I'm off down the village, lets leave Patrick to 
>carry the heavy load piled on by the demonically enthusiastic wizard! 
>Hmm, perhaps if I hide behind this curtain I can spy what they're at and 
>they won't see me. Hmm sounds a good idea that. Yup, they can't see me 
>here so I'm safe. Ho, ho, ho!

Well, they are being a long time, so perhaps I could make a start.

So now we need the Ricci Tensor. Defined by his wizardship to be

R_{bd} = R^c_{bcd}   this should be more manageable with 16 terms.

Oh dear, I'm too old for this. Lets try to get a term or two to show
willing. 

R_{bd} = R^0_{b0d} + R^1_{b1d} + R^2_{b2d} + R^3_{b3d}

R_{00} = R^0_{000} + R^1_{010} + R^2_{020} + R^3_{030} I hope.

= -3 R"/R       Oooh! Looks promising.

R_{11} = R" + 3R'^2 = R_{22} = R_{33) 

Odd, don't like the R'^2 term. Still.
<Panic> I wish I followed VanEsch's derivation of these.

That seems to be four of them done.

R_{01} = 0 + 0 +0 +0 = 0 = R_{02} = R_{03}
R_{10} = 0 + 0 +0 +0 = 0 = R_{20} = R_{30}
hmmm, OK.

R_{12} = 0 + 0 +0 +0 = 0 = R_{13} = R_{23} etc. alrightish.

<Panic> I ought to go back and check that R^0_ii0 and R^i_00i are
actually zero, but I don't have time.<Damn>

So this gives us a diagonalised 

R_{bd} = -3R"/R, R"+3R'^2, R"+3R'^2, R"+3R'^2

However I feel uncomfortable somehow.

I suppose they will work out the Ricci scalar next.
Oops, we have a surfeit of R's. Ho ho 'R 'R, ahem.

No time. Oh well, quickly then.

Well, that's R^a_d = g^(ab) R_{bd} oh how does it go? er

g^(00) R_{00} + g^(11) R_{11} etc, other terms zero.

Ricci scalar= +3R"/R + 3(R"+3R'^2)/R    yetch!

It feels like it ought to be no more complex than 6R"/R and even that
doesn't feel nice, 9(R'^2)/R would be nicer. Somethings wrong.
Maybe a sign error <Oz checks all over> but I can't see one.<Checks
again>. 
van Esch hasn't been collaborating with the nasty wizard to see if I'm
doing my homework is he? Nah, he is a nice guy, he wouldn't dream of it.

Hmmmmm, let's check out the R^i_jij term as soon as I have time.

Nah, more likely I have made an entire pigs ear of the whole thing, I
mean, what can you expect from a broom pushing apprentice after all.

Oh. We are getting close to the nitty gritty. We now have G_{ab} which
is pretty amazing. So if we specify T_{ab} we have an equation to solve.
Oh that's rather nice, in concept anyway. So what we put in our universe
will determine T, and thus G and thus what's allowd for R(t). Oh, I like
it.

Gosh they are being a long time getting back. It's getting dark.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue Apr 23 14:32:18 PDT 1996
Article: 11627 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 23 Apr 1996 12:38:35 -0700
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In article <[email-masked]> [email-masked] writes:

>So now we need the Ricci Tensor. Defined by his wizardship to be

>R_{bd} = R^c_{bcd}   this should be more manageable with 16 terms.

>Oh dear, I'm too old for this. Lets try to get a term or two to show
>willing. 

>R_{bd} = R^0_{b0d} + R^1_{b1d} + R^2_{b2d} + R^3_{b3d}

>R_{00} = R^0_{000} + R^1_{010} + R^2_{020} + R^3_{030} I hope.

That's right.

>= -3 R"/R       Oooh! Looks promising.

Yes, that looks right to me!

>R_{11} = R" + 3R'^2 = R_{22} = R_{33) 

>Odd, don't like the R'^2 term. Still.

Hmm, here I get something different, though similar in flavor.  It
definitely has two terms, one with an R" and one with an R'^2 in it, if
that's what you're worrying about.  But a few little things are
different.  We'd better check that one.  Maybe someone else in the class
can say what they get, and we can thrash it out.

>That seems to be four of them done.

Yes, those four diagonal entries are the important ones here....

>R_{01} = 0 + 0 +0 +0 = 0 = R_{02} = R_{03}
>R_{10} = 0 + 0 +0 +0 = 0 = R_{20} = R_{30}
>hmmm, OK.
>
>R_{12} = 0 + 0 +0 +0 = 0 = R_{13} = R_{23} etc. alrightish.

Yes, these off-diagonal ones are all zero.  But we knew that already,
right?  You proved that using just the ispotory of space!  (Or
however you spell it.)  So this is mainly just a check on whether
you got your Riemann tensor right.  

>So this gives us a diagonalised 
>
>R_{bd} = -3R"/R, R"+3R'^2, R"+3R'^2, R"+3R'^2
>
>However I feel uncomfortable somehow.

Well, it's darn close.

>I suppose they will work out the Ricci scalar next.

>Ricci scalar= +3R"/R + 3(R"+3R'^2)/R    yetch!

Hmm.

>Oh. We are getting close to the nitty gritty. We now have G_{ab} which
>is pretty amazing. So if we specify T_{ab} we have an equation to solve.
>Oh that's rather nice, in concept anyway. So what we put in our universe
>will determine T, and thus G and thus what's allowed for R(t). Oh, I like
>it.

Yes, it's cool, isn't it, how after having sweated over 4-dimensional
curved spacetime geometry for a couple of months you are now in the
position to work out what the big bang is all about?

Just gotta take care of those annoying little computational errors.  We
don't want you accidentally destroying the universe with a big crunch
just because of sign error.  The stakes are high.

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 23 Apr 1996 17:47:24 +0100
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>
>>Now let's see if I have this right. We started with a metric that
>>allowed us to model some time dependant spacial dimensions. Then we
>>Cristoffeled the metric apparently to find out how the length of vectors
>>changed as we moved around in spacetime, 
>
>Beware: the LENGTH of a vector never changes when we parallel
>translate it.  Parallel translation is defined to be "metric-
>compatible", meaning that as you carry your javelin about it doesn't get
>longer or shorter.  What the Christoffel symbols tell us is how the
>COMPONENTS of a vector (in some coordinate system!) change as we
>parallel translate it.
>
>That's what you meant, right?

To be honest I never considered it in any detail, however if I had I
might well have come to that conclusion.

>>Then this allowed (some people) to produce the appropriate
>>Riemann tensor, which is slightly odd since this is supposed to be co-
>>ordinate independant but since the Cristoffel thingies are co-ordinate
>>dependant one might imagine that this Riemann tensor is too.
>
>Slightly odd but not too odd.  First, one can often reach a
>coordinate-independent goal by coordinate-dependent methods.  Second,
>since our metric was given to us in a certain coordinate system, we are
>DOOMED to soil our hands with coordinates in this problem.

That's the way it felt to me. Nice to have confirmation.

>A note to van Esch, Scaldeferri and Pfannerer: Please don't give away
>all the answers to the computations before Oz does them!  He needs the
>practice.  Besides, I enjoy torturing him.

Well, waddyaknow! Posted on Saturday, too (it's Tuesday 18.00 hrs).
Wot a beastly wizard! Some many moons ago I once asked if he would teach
the group the basics of GR. I really should have said 'tell us the
basics of GR', but then he wouldn't have done it would he.

I hid behind my curtain till I got bored, and they were all sniggering
outside. I woz 'ad, I wuz!

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue Apr 23 14:35:31 PDT 1996
Article: 11635 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Visualizing curved spacetime
Date: 23 Apr 1996 13:15:05 -0700
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In article <[email-masked]> [email-masked] writes:
>Personally, I find it a bit confusing; if you think in terms
>of euclidean 4-space, then the paths of orbiting objects look
>distinctly helical instead of straight.  IMO, the toughest part about
>GR is understanding how such 'bent' paths are really the
>'straightest', since the relevant 4-geometry doesn't really fit nicely
>inside your head (which is probably why it's easier to deal with
>things by diving into the mathematics).

Well, I find it easiest to start by visualizing curved 2-dimensional
surfaces in 3d space.  (This should be a snap, since we see them all
over the place, but when I try to teach vector calculus to undergrads I
find they are frightfully weak when it comes to, say, imagining what
happens when you take two hyperboloids and see where they intersect.  So
for those who don't already have the knack, a crash course in
visualizing surfaces in 3d space would be good.) 

Then I would start visualizing curved 3-dimensional surfaces in
4-dimensional space.  There are lots of ways to work at this.  One is to
pretend they are 2-dimensional surfaces in 3-dimensional space, i.e., to
fake it.  Don't laugh!  This is surprisingly effective.  I've seen 
experts on 4-dimensional topology talking to each other, drawing
pictures of 4-dimensional manifolds in the air with their hands, clearly
communicating quite efficiently by this method.  

Another way is to use the method of slices.  We can slice flat
4-dimensional space into lots of flat 3-dimensional slices, and if we
have a curved 3-dimensional surface in 4-dimensional space, each slice
will contain a curved 2-dimensional slice of that surface.  For starters
visualize the 3-sphere

                w^2 + x^2 + y^2 + z^2 = 1

using this method.  

A closely related way is the method of movies.  Flip through the slices
in your minds eye as if they were frames of a movie; i.e., think of the
slices as moments in time.  This is especially easy for dealing with
physics in flat 4-dimensional spacetime.  The sphere above will look
like a 2-sphere that pops into existence at the origin, expands to
radius one, and then shrinks down and winks out of existence.  

Personally I blend the method of slices, the method of movies, the
method of faking, and various other methods so much that I can't say
any more which one I use when I visualize something like the 3-sphere in
flat 4d spacetime.  I just "picture it". 

Oh, and another method, noted by Hollebeeck, is to "dive into the
equations".  This complements the visual methods.  When solving the
equations, it usually helps to visualize, so you can "see" when you are
making a mistake.  Similarly, when visualizing, it usually helps to
break out some equations to make sure you're not screwing up.

Now how about curved 4-dimensional spacetime?  Well, we can use the
method of slices, thinking of each slice as a curved 3-dimensional space
(which in turn we visualize by the above methods).  We can also use the
method of movies --- "animating" the changing shape of a curved
3-dimensional space.  This is where the "living, pulsing" metaphor comes
in.  Of course, there is the ever-popular method of faking it:
finding a lower-dimensional analog to the problem you are thinking
about.  This is never to underestimated.  Finally, you've got to do the
math, to keep honest.    

This may seem like a rather baroque, complicated exercise, demanding of
endless time and patience, like one of those Tibetan visualization
meditations.  Well, it is!  But it's worth it if you really want to
understand general relativity.  

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From: [email-masked]    (Norbert Pfannerer)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 23 Apr 1996 07:55:48 GMT
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In <[email-masked]>, [email-masked] (john baez) writes:
:
:A note to van Esch, Scaldeferri and Pfannerer: Please don't give away
:all the answers to the computations before Oz does them!  He needs the
:practice.  Besides, I enjoy torturing him.

I'll try very hard. (When I've finally computed it!) 

Also my posts and the answers that I got helped me to better understand
that stuff. (The first time I read about Christoffel symbols was two weeks
ago when I found this tutorial).

I now have computed the Christoffel symbols, and got the same results
that are already posted. By thinking about them and all I've read here,
I think I know what C^0_{ii} not equal 0 (what Kevin was wondering about) 
means: A straight line in a spacelike slice t=constant is not a geodesic
in spacetime, i.e. the spacelike slice is extrinsically curved.
(This was already posted, so I don't think I give avay too much secret
info here).

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

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Article: 11664 of sci.physics
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From: Geoffrey A. Landis <[email-masked]>
Newsgroups: sci.physics
Subject: Christoffel symbol (Re: General Relativity Tutorial)
Date: 23 Apr 1996 15:01:56 GMT
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In article <[email-masked]> john baez, [email-masked]
writes:
>What I said before was:
>
>FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the d
>direction and parallel transport it an amount epsilon in the c
>direction.  See how much its component in the a direction changes.  
>(This component started out being 0 unless a = d, in which case it was
>1.)  It changes by 
>
>     - epsilon C^a_{cd} + terms of order epsilon^2

This seems to imply that the Christoffel symbol components with a=d must
always be unity, right?

 C^a_{ca} = 1            [no summation!]

My logic is: since it's a unit vector parallel transported, the magnitide
doesn't change, only the direction.  The vector thus rotates without
stretching.  If it rotates by theta, the component in the original (d)
direction changes by [cos theta].  For small transport epsilon, theta is
linear with distance.  But the cosine has no linear term in the expansion
to order epsilon, so to first order, cos theta is unity.  Thus the
^a_{ca} term must be unity. 

True?  Or am I misunderstanding something?

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center
physicist and part-time science fiction writer

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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Christoffel symbol (Re: General Relativity Tutorial)
Date: 23 Apr 1996 17:04:12 -0700
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In article <[email-masked]> Geoffrey A. Landis <[email-masked]> writes:
>In article <[email-masked]> john baez, [email-masked]
>writes:
>>What I said before was:

>>FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the d
>>direction and parallel transport it an amount epsilon in the c
>>direction.  See how much its component in the a direction changes.  
>>(This component started out being 0 unless a = d, in which case it was
>>1.)  It changes by 

>>     - epsilon C^a_{cd} + terms of order epsilon^2

>This seems to imply that the Christoffel symbol components with a=d must
>always be unity, right?

> C^a_{ca} = 1            [no summation!]

>My logic is: since it's a unit vector parallel transported, the magnitide
>doesn't change, only the direction.  The vector thus rotates without
>stretching.  If it rotates by theta, the component in the original (d)
>direction changes by [cos theta].  For small transport epsilon, theta is
>linear with distance.  But the cosine has no linear term in the expansion
>to order epsilon, so to first order, cos theta is unity.  Thus the
>C^a_{ca} term must be unity. 

>True?  Or am I misunderstanding something?

Let me stick to 2-dimensional examples below to keep things simple.

I think you are making a couple of mistakes.  Yes, parallel transport
preserves the length of a vector, but no, that doesn't mean that as we
parallel translate a vector like (1,0) that it can only become a vector
like (cos theta, sin theta).  The reason is that we may be working in
coordinates where the metric isn't

1 0 
0 1

In fact, we typically AREN'T working in such coordinates if we're
studying curved spacetime.   So we can easily parallel translate a
vector like (1,0) and have it become (999,0) after a while.

This is why I frequently apologize for using the term "unit vector" in
the definition of Christoffel symbols above.  But I should apologize
again!!  I just mean a vector having one component equal to 1 in our
coordinates, and the rest zero.  This has nothing to do with unit
length.  For example, in a recent post I said:

"FIXING A LOCAL COORDINATE SYSTEM, take a tangent vector pointing in the d
direction and extending one unit in these coordinates, and parallel
transport it an amount epsilon in the c direction.  See how much its
component in the a direction changes.  (This component started out being
0 unless a = d, in which case it was 1.)  It changes by

     - epsilon C^a_{cd} + terms of order epsilon^2

[Obnoxious Note: it needn't have LENGTH one!  It is defined using
coordinates in a manner that doesn't ever mention its length.  This is
crucial, but was obscure in my earlier attempts to describe the
Christoffel symbols. Also, it's just a tangent vector, not an actual
stick in spacetime.  You can regard it as an infinitesimal stick, if you
want, even though it extends one unit in the x direction.  Does this make
sense?  Probably not.  Well, we actually discussed these issues before
when we were first learning about tangent vectors, but can talk about
them again if necessary.]"

Okay.  That's one problem with what you said, but there's another.
Suppose we are in a situation where parallel translation DOES change a
vector like (1,0) into a vector like (cos theta, sin theta).  Say we
parallel transport the vector by a wee bit epsilon.  When we compute how
its first component *changes*, it will *change* by an amount
proportional to epsilon^2, since the derivative of cos theta is 0 at
theta = 0.  So

epsilon^2 ~ the change = - epsilon C^a_{ca} + terms of order epsilon^2

so C^a_{ca} = 0.

In short, I think you are mixing up the fact that cos(0) = 1 with the
more relevant fact that cos'(0) = 0.  

Does this make sense?

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 23 Apr 1996 20:30:18 +0100
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AAAARRRRrrrrrggggggghhhhhhhh. I made a teeny error.

In article <[email-masked]>, Oz
<[email-masked]> writes
>
>Having worked through van Esch's derivation of the Riemann tensor and 
>more or less followed what he did, and we can add the ones he found 
>later that I haven't a clue how he came by 'cos the look like they 
>should be zero to me, but then I'm dim, we can now substitute the 
>relevant C-symbols in to get the following dogs dinner that really 
>requires you to stand on your head and hum ommmmmm to understand. Wow, 
>that was a long sentence. Shoof I'm getting i's in front of my eyes.
>
>R^0_i0i =  R(t)"        which is nice for a change.

Too damn nice. It should be R(t)" R(t) ***********

>R^0_0ii = -R(t)"        not unexpected.

But wrong. It should be -R(t)" R(t)  ************

>
>R^i_i00 =  R(t)"/R(t)   not too bad I suppose
>R^i_0i0 = -R(t)"/R(t)
>
>And now for the ones that look like the name of some Dutch town, no
>wonder Van Esch has no trouble with these super&suffixes, the symbols
>probably refer to a dutch word he already knows. Stick a few 'n's here
>and there and we could probably make a sentence in dutch! :-)
>
>R^i_ijj = -R(t)'^2
>R^i_jij =  R(t)'^2
>
>I have probably made lots of mistakes. 

Well, one at least.

Now lets see about the Ricci.

So now we need the Ricci Tensor. Defined by his wizardship to be

R_{bd} = R^c_{bcd}   this should be more manageable with 16 terms.

Oh dear, I'm too old for this. Lets try to get a term or two to show
willing. 

R_{bd} = R^0_{b0d} + R^1_{b1d} + R^2_{b2d} + R^3_{b3d}

R_{00} = R^0_{000} + R^1_{010} + R^2_{020} + R^3_{030} I hope.

= -3 R"/R       Oooh! Looks promising. 

Oh but rushing the next <shame, shame, sackcloth and ashes>

R_{11} = R^0_{101} + R^1_{111} + R^2_{121} + R^3_{131}

Humpf. Lets just check R^1_{111} it equals ... 0!
Better do R^2_{121} it equals ..... 0 too

van Esch you are a bastard conspirator with the wonky wiz  :-)
No, actually I have been HAD!! :-{ argh! 
And me only a poor old apprentice too. Sadists!
Mutter, mutter, mutter, moan.

R_{11} = R"R = R_{22} = R_{33) 

That seems to be four of them done.
No time to check the rest. Family calls. Press on regardless!

R_{01} = 0 + 0 +0 +0 = 0 = R_{02} = R_{03}
R_{10} = 0 + 0 +0 +0 = 0 = R_{20} = R_{30}
hmmm, OK. <At least I hope so>

R_{12} = 0 + 0 +0 +0 = 0 = R_{13} = R_{23} etc. <I hope>

<Panic> I ought to go back and check that R^0_ii0 and R^i_00i are
actually zero, but I don't have time.<Damn> 
NB Still no time.

So this gives us a diagonalised 

R_{bd} = -3R"/R, R"R, R"R, R"R

That feels better.

I suppose they will work out the Ricci scalar next.
Oops, we have a surfeit of R's. Ho ho 'R 'R, ahem.

No time. Oh well, quickly then.

Well, that's R^a_d = g^(ab) R_{bd} oh how does it go? er

g^(00) R_{00} + g^(11) R_{11} etc, other terms zero.

Ricci scalar= +3R"/R + 3(R"R)/R^2  =  6R"/R (forgot metric R^2)

This is crying out to be zero. Don't like this factor of six.
<Quick sign check. Can't find the mistake. No time, family calls>

Oh. We are getting close to the nitty gritty. We now have G_{ab} which
is pretty amazing. So if we specify T_{ab} we have an equation to solve.
Oh that's rather nice, in concept anyway. So what we put in our universe
will determine T, and thus G and thus what's allowd for R(t). Oh, I like
it.

Gosh they are being a long time getting back. It's getting dark.
And when they come in they can have the back of my broom too!

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 23 Apr 1996 14:41:43 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] (Norbert Pfannerer) writes:
>In <[email-masked]>, [email-masked] (john baez) writes:

>:A note to van Esch, Scaldeferri and Pfannerer: Please don't give away
>:all the answers to the computations before Oz does them!  He needs the
>:practice.  Besides, I enjoy torturing him.

>I'll try very hard. (When I've finally computed it!) 

Well, by now Oz has computed the Riemann tensor, the Ricci tensor, and the
Ricci scalar.  So if you happen to compute it yourself, please post it.
We are in a little mixup about what the components R_{ii} should be.

>I now have computed the Christoffel symbols, and got the same results
>that are already posted. By thinking about them and all I've read here,
>I think I know what C^0_{ii} not equal 0 (what Kevin was wondering about) 
>means: A straight line in a spacelike slice t=constant is not a geodesic
>in spacetime, i.e. the spacelike slice is extrinsically curved.
>(This was already posted, so I don't think I give avay too much secret
>info here).

That's right.  Here's a nice puzzle that requires a convincing VERBAL
answer.  Can you see why "space is expanding" implies that even though
each slice at time t is flat, it has *extrinsic* curvature?  As you say,
it has to do with how geodesics on the slice at time t aren't geodesics
in spacetime.  But can someone explain in simple words why this follows from
the fact that space is expanding?

We don't want to get so caught up in computations that we forget what
we're doing.  So it's good to see how much we can understand verbally.

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Wed, 24 Apr 1996 13:11:47 +0100
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In article <[email-masked]>, Oz
<[email-masked]> writes

06.00 in the morning and all is quiet. Let's have another look.

Um. Ooo. Bother.

Ooops. One or two minor mistakes.
Like taking R^i_iii non zero. And noting that R^1_212 isn't.
More haste means less speed!

So I now have the Ricci as diagonalised as

(-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)

and the Ricci scalar as

6(RR"+R'2)/R^2, which still doesn't feel brilliant yet.

We can also say (I think) that the second time derivative of a free-
falling ball in this metric in this co-ordinate system is R_{00}=-3R"/R 
um, that sign doesn't look quite right, oh dunno.

Hang on a minute, we have (Ricci scalar=Rs)

T_{ab} = R_{ab} - (1/2)Rs g_{ab} so we ought to be able to get T

T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
       = 3R'^2/R^2

T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
       = -2RR" - R'^2

And the rest are zero.

Hmmm. Well

T_{00} is the density of the energy.
T_{0i} is the density of momentum, which we have set equal to zero. OK
T_{ii} is to do with pressure. No, I'm not very clear about it.
T_{i0} is to do with the flow of energy in the i-direction. Zero too.

So the stuff in the universe affects the metric and the metric affects
how the stuff in the universe behaves. Hmmm, that could cause
complications.

We are doing a universe, as against a BH, because it's simple as the
metric is everywhere the same at least in any spacelike slice.
Presumably in a BH situation this is very, very, far from being the case
although you could probably find some 2D surface where it is.

Let's see what the wizard has to say.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Wed Apr 24 16:55:06 PDT 1996
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
Date: 24 Apr 1996 23:00:12 GMT
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In article <[email-masked]>,
Kevin Anthony Scaldeferri <[email-masked]> wrote:
>>So I now have the Ricci as diagonalised as
>>
>>(-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)
>>
>
>This agrees with my calculations
>
>>and the Ricci scalar as
>>
>>6(RR"+R'2)/R^2, which still doesn't feel brilliant yet.
>>
>
>I also get this.  Can't say I'm wild about it, you'd think this metric
>would give us something simpler.

I just checked too.  (Actually, I cheated and had Mathematica
check for me.)  I get the same thing.  I think you guys are 
done with this part of the calculation.

>>Hmmm. Well
>>
>>T_{00} is the density of the energy.
>>T_{0i} is the density of momentum, which we have set equal to zero. OK
>>T_{ii} is to do with pressure. No, I'm not very clear about it.
>>T_{i0} is to do with the flow of energy in the i-direction. Zero too.

This is all correct.  In fact, T_{ii} isn't just something to do with
the pressure -- it *is* the pressure!  And the other components --
T_{ij} with i not equal to j -- are zero too.

Maybe Baez would prefer me to stay out of this, but I'm going to give
you a suggestion anyway.  As a first attempt, why don't you assume
that the Universe is full of pressureless matter (which general
relativists usually call "dust").  Then only one component of T is
nonzero.  Write down the Einstein equations G_{ab} = T_{ab}, and see
what you get.  (There'll be two distinct nontrivial equations.)  Can
you solve them for R(t)?  (There's one or two wrinkles you'll have to
iron out to do that, but I'll let you find them for yourselves.)

After that, if you're feeling ambitions, you might want to throw in
pressure and see how that affects things.

-Ted

From galaxy.ucr.edu!not-for-mail Wed Apr 24 18:29:48 PDT 1996
Article: 113867 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 24 Apr 1996 16:54:56 -0700
Organization: University of California, Riverside
Lines: 80
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>,
>Kevin Anthony Scaldeferri <[email-masked]> wrote:
>>>So I now have the Ricci as diagonalised as

>>>(-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)

>>This agrees with my calculations

>>>and the Ricci scalar as

>>>6(RR"+R'2)/R^2, which still doesn't feel brilliant yet.

>>I also get this.  Can't say I'm wild about it, you'd think this metric
>>would give us something simpler.

One thing I have learned in my years of mathematical physics is that
sometimes you take enough derivatives of something simple and keep
multiplying it by a bunch of other simple stuff and it starts getting
complicated.

>I just checked too.  (Actually, I cheated and had Mathematica
>check for me.)  I get the same thing.  I think you guys are 
>done with this part of the calculation.

Yes, this looks like what I got too.  Good.

>>>Hmmm. Well

>>>T_{00} is the density of the energy.
>>>T_{0i} is the density of momentum, which we have set equal to zero. OK
>>>T_{ii} is to do with pressure. No, I'm not very clear about it.
>>>T_{i0} is to do with the flow of energy in the i-direction. Zero too.
>
>This is all correct.  In fact, T_{ii} isn't just something to do with
>the pressure -- it *is* the pressure!  And the other components --
>T_{ij} with i not equal to j -- are zero too.

T_{ii} is the pressure in this situation.  Some caveats: we need to be
working in the local rest frame of the fluid we're studying for this to
be true, and we need things to be isotropic for the pressure in every
direction to be the same.  

A while back I explained the reason why; it's important and simple.  By
definition it's the flow of i-momentum in the i-direction.  Imagine a
bunch of little balls bouncing all around in an isotropic way.  Then
imagine you have a wall perpendicular to the i direction.  The flow of
i-momentum in the i-direction tells you how much momentum the little
balls moving in the direction of the wall transfer to the wall, per unit
area.  This is nothing but the pressure.

>Maybe Baez would prefer me to stay out of this, but I'm going to give
>you a suggestion anyway.  

Boy, as soon as I say I'm leaving for a long weekend, Ted Bunn sneaks in
and starts teaching you stuff behind my back!  Good thing I peeked back
in here.  By the time I came back you would all know more GR than I do,
and then where would I be?

Seriously, I have no objection to Ted or anyone else helping out; in
fact, it's great --- as long as they don't give away the answers to
computations.  Explaining ideas and helping people over conceptual or
mathematical roadblocks is fine.  But I want these students to sweat
their way honestly to solving the big-bang cosmology.  It's more fun for
everyone that way, after all.  

>As a first attempt, why don't you assume
>that the Universe is full of pressureless matter (which general
>relativists usually call "dust").  Then only one component of T is
>nonzero.  Write down the Einstein equations G_{ab} = T_{ab}, and see
>what you get.  (There'll be two distinct nontrivial equations.)  Can
>you solve them for R(t)?  (There's one or two wrinkles you'll have to
>iron out to do that, but I'll let you find them for yourselves.)

That sounds good.  I was going to have them work out the case where the
universe is full of dust, and then the case where it's full of
radiation.  The "matter-dominated" and "radiation-dominated" extremes.  

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Article: 11974 of sci.physics
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From: [email-masked] (Jan Bielawski)
Subject: Re: General Relativity Tutorial
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	Unless indicated otherwise, everything in this note is personal
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In article <[email-masked]> [email-masked] (Kevin Anthony Scaldeferri) writes:
<
<OK, we're going to have to back up here.  I think I'm blurring the
<Christoffel symbols and the covariant derivative, but...the C. symbols
<describe how the covarient derivative acts on a basis, yes?  And the
<covariant derivative is multi-linear.  So why doesn't all this act
<tensorial?  

Because it's linear on vector *fields*, not on vector field *values
at a point*.
-- 
Jan Bielawski
Molecular Simulations, Inc.   )\._.,--....,'``.         | http://www.msi.com
San Diego, CA                /,   _.. \   _\  ;`._ ,.   | ph.: (619) 458-9990
[email-masked]              fL `._.-(,_..'--(,_..'`-.;.'   | fax: (619) 458-0136
-- 
 
#****************************************************************************#
+DISCLAIMER: Unless indicated otherwise, everything in this note is personal +
+opinion, and not an official statement of Molecular Simulations Inc.        +
#****************************************************************************#

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Article: 12080 of sci.physics
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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 25 Apr 1996 11:39:21 GMT
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Emory F. Bunn ([email-masked]) wrote:
: In article <[email-masked]>,
: Kevin Anthony Scaldeferri <[email-masked]> wrote:
: >>So I now have the Ricci as diagonalised as
: >>
: >>(-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)
: >>
: >
: >This agrees with my calculations
: >
: >>and the Ricci scalar as
: >>
: >>6(RR"+R'2)/R^2, which still doesn't feel brilliant yet.
: >>
: >
: >I also get this.  Can't say I'm wild about it, you'd think this metric
: >would give us something simpler.

: I just checked too.  (Actually, I cheated and had Mathematica
: check for me.)  I get the same thing.  I think you guys are 
: done with this part of the calculation.

Ok, I also get it:
After a lot of napkins (I don't have access to Mathematica during lunch:)
I get (just for reference):

introducing the Christoffel symbols,

A(t) = R'(t).R(t)

B(t) = R'(t)/R(t)

(where we have agreement on now)
See an old post of mine ...

we find:

R^0_i0i = R".R

R^0_0ii = - R".R

R^i_0ii = - R".R

R^i_i00 = R"/R

R^i_0i0 = - R"/R

R^i_ijj = - R'^2

R^i_jij = R'^2

---------------------

Ricci:

R_kl = R^e_kel

non-zero contributions:

no R_ij contributions for i not equal to j.

R_kk = R^0_k0k + R^i_kik = R".R + 2.R'^2

R_00 = R^0_000 + R^i_0i0 = 0 + 3.(- R"/R) = - 3.R"/R

So:

The non-zero components of the Ricci tensor are:

R_kk = R".R + 2.R'^2    for k = 1,2,3

R_00 = -3.R"/R

---------------------

Ricci scalar:

Ricci = g^ef.R_ef

since R_ef is non-zero only if diagonal,
we have:

Ricci = g^00.R_00 + g^11.R_11 + g^22.R_22 + g^33.R_33

with:

       g^00 = -1

       g^ii = 1/R^2

(since it is the inverse of the matrix g_ab)

This gives:

Ricci = 3.R"/R + 3.(R".R+2.R'^2)/R^2

      = 3.R"/R + 3.R"/R + 6.R'^2/R^2 = 6.(R".R + R'^2)/R^2

So:

Ricci = 6.(R".R + R'^2)/R^2

Cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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From: [email-masked]    (Norbert Pfannerer)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 25 Apr 1996 12:23:01 GMT
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In <[email-masked]>, [email-masked] (john baez) writes:
>
>That's right.  Here's a nice puzzle that requires a convincing VERBAL
>answer.  Can you see why "space is expanding" implies that even though
>each slice at time t is flat, it has *extrinsic* curvature?  As you say,
>it has to do with how geodesics on the slice at time t aren't geodesics
>in spacetime.  But can someone explain in simple words why this follows from
>the fact that space is expanding?

I've thought about this a bit, already started last week when I thought
about the meaning of the Chr.-symbols. I don't have much time now
explaining my thoughts, but I have a nice geometrical picture that I
want to share.

It is the (t,x) plane of our spacetime, with R(t)'>0 and R(t)''=0. 
It turns out, this (t,x)-slice is not curved, so I can draw it in the
euclidean plane. t-lines (lines with constant x) are rays starting at
a common point, x-lines are circles with this point as the center.
The line is not intrinsically curved, but of course has extrinsinc
curvature. A geodesic in space is just this circle, a geodesic in
spacetime is a tangent to the circle.

I realize that this is just one of many examples, and it might well not
even satisfy the Einstein-equation (I'll check that later). Also
in this example space is finite, and I don't know yet what our metric
has to say about that.

Also my post is already much longer than I wanted...

-------------------------------------------------------------------
All opinions are my own, not my employer's

Norbert Pfannerer     [email-masked]
Vienna, Austria, Europe

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From: Oz <[email-masked]>
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Subject: Re: General Relativity Tutorial
Date: Thu, 25 Apr 1996 16:08:41 +0100
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In article <[email-masked]>, Norbert Pfannerer
<[email-masked]> writes
>
>
>Yesterday I did all the calculations, Riemann tensor, Ricci tensor,
>Ricci scalar (Oz: be careful. The formula talks about g^ab, not g_ab there.
>It's defined in the course notes.)
>Then the Einstein tensor. I made a reasonable assumption about T_ab
>(which is the only thing from all the above that I'm sure I've done
>correctly, but I won't tell yet). Then I solved the Einstein equation
>and got R(t)''=0 or R(t)=-3, which is obviously false.

OK, since El Grando Wizardo is off for the weekend we gotta do a bit
ourselves. Now I think that between Van Esch, Kevin and Norbert (with
manic confusion and errors contributed by Oz) we ought to be able to get
a bit further.

What procedure do you use to solve the Einstein equation?

We have some results for T, I think, which I made as:

T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
       = 3R'^2/R^2

T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
       = -2RR" - R'^2

And the rest are zero. However, given my record the chances of this
being correct are slim.

And I believe

T_{00} is the density of the energy.
T_{ii} is to do with pressure. No, I'm not very clear about it.

And we have probably defined a universe filled with co-moving particles
from the very start, or anyway that ought to be the easiest to solve. I
presume this is a universe filled with cold dust so T_{00} is the mass
density. So naively, and I stress without thinking (hey I'm only the
apprentice) one might head towards:

T_{00} = mass density of the universe so (shakey shakey) in our co-
ordinate system the mass within a spacelike cube, dx,dy,dz so the volume
is er, um, ds^3 um ooh er, hang on ds^2=R(t)^2(dx^2 + dy^2 + dz^2) I
think for a space like manifold. The mass within this volume will remain
constant so we could describe the mass density as =B/R(t)^3, with B a
constant. Er I think. Incidentally readers should be aware that I am
thrashing around like a fish out of water, really I haven't a clue.
Anyway does this suggest that:

            T = G
        B/R^3 = 3R'^2/R^2    ??????????????????
or          B= 3(R'^2)R

Hmmm. So if the density is zero then R or R' must be zero, yes R' might
be plausibly zero and we have something like the Milne universe. Well,
probably the Milne universe, actually. 

Hmmmm. 
The only solution to this I can dream up is 

        R(t) = (3B/4)^(1/2) t^(2/3)

Then we have T_{11} which is supposed to be the pressure. Surely the
pressure in this cold dusty universe filled with geodesic particles
should be zero? This would give us

       T_{ii} = -2RR" - R'^2 
or          0 = -2RR" - R'^2
or       2RR" = -R'^2

Oh dear. Proper maths. Lets try the little solution above and see.

Good greif, it IS a solution.

Oz faints.

And dreams he has got it right ........

What would a metric of form (-1, kt^(4/3), kt^(4/3), kt^(4/3)) look
like? Well the Ricci tensor tells us how a volume evolves in time.

R_{ab} diagonalises at (-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)

R_{00} = (-2/3)t^-2 which means I have it wrong.

The evolution of the volume of a ball in *this* universe does not depend
on the density.

Sod it!

Hang on though. I said the volume scaled as 1/R^3. Hmm, well it does.

Now I'm really confused. It's something to do with the co-ordinates or I
should have considered presssure to be non-zero. Either that or I am
missing something simple.

>I'll go over my calculations again today, and also calculate a few
>geodesics.

It's nice to know that someone knows what he is doing.

We await some indication of your results.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 26 Apr 1996 13:05:52 GMT
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Oz ([email-masked]) wrote:

: What procedure do you use to solve the Einstein equation?

: We have some results for T, I think, which I made as:

: T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
:        = 3R'^2/R^2

: T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
:        = -2RR" - R'^2

: And the rest are zero. However, given my record the chances of this
: being correct are slim.

: And I believe

: T_{00} is the density of the energy.
: T_{ii} is to do with pressure. No, I'm not very clear about it.

[...]

Well, I approached it from the other side: I said: no pressure.
I don't really know if that is really what is meant by a dusty universe,
but I think so.

So here is what I found:

---------------------------

Dusty Universe.

We suppose there is NO PRESSURE.

So T_ii = 0 for our dusty universe (i = 1,2,3)

This means: G_ii = 0, or :

         - 2.R".R - R'^2 = 0

This is a non-linear homogenious differential equation
of second order.  So let's guess solutions :)

Let us rewrite the equation, dividing by R^2 :

          2.(R"/R) + (R'/R)^2 = 0

suggestion: u = R'/R  (so u = d(log R)/dt).

Then u' = (R".R - R'^2)/R^2 = R"/R - u^2.

This means: R"/R = u' + u^2

            R'/R = u

So our equation becomes:

          2.u' + 2.u^2 + u^2 = 0

or:
 
          2.u' + 3.u^2 = 0

or:
          du/u^2 = -3/2.dt  which has as solution:

          -1/u = - 3/2.t + C.

or          u = 1/(3/2.(t-t0))

Now we return to R:

         d(log R)/dt = 1/(3/2(t-t0))

integrating:

           log R = 2/3.log(t-t0) + C

or:

        ===========================
        |                         |
        |   R(t) = A.(t-t0)^(2/3) |
        |                         |
        ===========================

with A and t0 integration constants.

A less formal derivitation:  GUESS !

         R(t) = t^a

 ===> R' = a.t^(a-1)
      R" = a.(a-1).t^(a-2)

substituting, we find: 2.a.(a-1).t^(2a-2) + a^2.t^(2a-2) = 0

or 2.a^2 - 2a + a^2 = 0 or 3.a^2 - 2.a = 0 or a.(3.a - 2) = 0

so 2 solutions: a = 0 and a = 2/3.

(this isn't the general solution of this equation, just
2 particular solutions...)

a = 0 is boring, let's look at a = 2/3.

                So R(t) = C.t^(2/3), C an arbitrary constant.

Noticing that the substitution t -> t - t0 doesn't change
the equation, one should presume that the same substitution
can be applied to the solution: hence:

             R(t) = C.(t-t0)^(2/3)

However, our guessing has introduced a particular solution that
isn't part of the general solution: 

               ==============
               |            | 
               | R(t) = Cte |
               |            |
               ==============

(is this a static Minkowski universe then ?)

What does this imply for G_00 ?

We will use the general solution:

                   G_00 = 3.R'^2/R^2

this gives us:

     G_00 = 3.(2/3.(t-t0)^(-1/3))^2/(t-t0)^(4/3)

          = 4/3.(t-t0)^(-2) 

                   ===================== 
                  |                     |
                  |             4       | 
   So we have:    | G_00 = ------------ |
                  |         3.(t-t0)^2  |
                  |                     |
                   =====================

           for the general solution ;

   While we have:

                G_00 = 0 

           for the particular solution R(t) = C

-------------------

: The only solution to this I can dream up is 

:         R(t) = (3B/4)^(1/2) t^(2/3)

Well, indeed, that's the same solution I have !

: pressure in this cold dusty universe filled with geodesic particles
: should be zero? This would give us

I started from that one...

:        T_{ii} = -2RR" - R'^2 
: or          0 = -2RR" - R'^2
: or       2RR" = -R'^2

: Oh dear. Proper maths. Lets try the little solution above and see.

: Good greif, it IS a solution.

: Oz faints.

: And dreams he has got it right ........

I guess so :-)

: Hang on though. I said the volume scaled as 1/R^3. Hmm, well it does.

: Now I'm really confused. It's something to do with the co-ordinates or I
: should have considered presssure to be non-zero. Either that or I am
: missing something simple.

Why the confusion ???

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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From: [email-masked] (R Shaw)
Subject: Re: General Relativity Tutorial
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Oz wrote:
> 
> In article <[email-masked]>, Norbert Pfannerer
> <[email-masked]> writes
> >
> >
> >Yesterday I did all the calculations, Riemann tensor, Ricci tensor,
> >Ricci scalar (Oz: be careful. The formula talks about g^ab, not g_ab there.
> >It's defined in the course notes.)
> >Then the Einstein tensor. I made a reasonable assumption about T_ab
> >(which is the only thing from all the above that I'm sure I've done
> >correctly, but I won't tell yet). Then I solved the Einstein equation
> >and got R(t)''=0 or R(t)=-3, which is obviously false.
> 
> OK, since El Grando Wizardo is off for the weekend we gotta do a bit
> ourselves. Now I think that between Van Esch, Kevin and Norbert (with
> manic confusion and errors contributed by Oz) we ought to be able to get
> a bit further.
> 
> What procedure do you use to solve the Einstein equation?
> 
> We have some results for T, I think, which I made as:
> 
> T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
>        = 3R'^2/R^2
> 
> T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
>        = -2RR" - R'^2
> 
> And the rest are zero. However, given my record the chances of this
> being correct are slim.
> 
> And I believe
> 
> T_{00} is the density of the energy.
> T_{ii} is to do with pressure. No, I'm not very clear about it.
> 
> And we have probably defined a universe filled with co-moving particles
> from the very start, or anyway that ought to be the easiest to solve. I
> presume this is a universe filled with cold dust so T_{00} is the mass
> density. So naively, and I stress without thinking (hey I'm only the
> apprentice) one might head towards:
> 
> T_{00} = mass density of the universe so (shakey shakey) in our co-
> ordinate system the mass within a spacelike cube, dx,dy,dz so the volume
> is er, um, ds^3 um ooh er, hang on ds^2=R(t)^2(dx^2 + dy^2 + dz^2) I
> think for a space like manifold. The mass within this volume will remain
> constant so we could describe the mass density as =B/R(t)^3, with B a
> constant. Er I think. Incidentally readers should be aware that I am
> thrashing around like a fish out of water, really I haven't a clue.
>

You don't need to assume anything about the density,
it all comes out in the wash.
There are two ODE's so we can find two functions of time.
Just call the density D(t). The pressure given as P(D)
the equation of state.
You want T_{00}=D(t) T_{ii}=P(D(t))=0
Solve the 2nd first
2RR"+R'R'=0
Homogenous so just try At^n.
I get  2n(n-1)+n^2=0
Then sub into T_{00} to find what the density does.

    I only did GR for 3 months in my undergrad days.
This thread has been an interesting reminder.

-- 
They both savoured the strange warm glow of being
much more ignorant than ordinary people, who were
only ignorant of ordinary things.
Terry Pratchett, 'Equal Rites'

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Article: 12358 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 26 Apr 1996 09:13:03 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>In article <[email-masked]>, Norbert Pfannerer
><[email-masked]> writes
>>
>>
>>Yesterday I did all the calculations, Riemann tensor, Ricci tensor,
>>Ricci scalar (Oz: be careful. The formula talks about g^ab, not g_ab there.
>>It's defined in the course notes.)
>>Then the Einstein tensor. I made a reasonable assumption about T_ab
>>(which is the only thing from all the above that I'm sure I've done
>>correctly, but I won't tell yet). Then I solved the Einstein equation
>>and got R(t)''=0 or R(t)=-3, which is obviously false.
>
>OK, since El Grando Wizardo is off for the weekend we gotta do a bit
>ourselves. Now I think that between Van Esch, Kevin and Norbert (with
>manic confusion and errors contributed by Oz) we ought to be able to get
>a bit further.
>

Well, unfortunately, I too will be away for the weekend, however I may
be able to check my email occasionally

>What procedure do you use to solve the Einstein equation?
>
>We have some results for T, I think, which I made as:
>
>T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
>       = 3R'^2/R^2
>
>T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
>       = -2RR" - R'^2
>
>And the rest are zero. However, given my record the chances of this
>being correct are slim.
>
>And I believe
>
>T_{00} is the density of the energy.
>T_{ii} is to do with pressure. No, I'm not very clear about it.
>
>And we have probably defined a universe filled with co-moving particles
>from the very start, or anyway that ought to be the easiest to solve. I
>presume this is a universe filled with cold dust so T_{00} is the mass
>density. So naively, and I stress without thinking (hey I'm only the
>apprentice) one might head towards:
>
>T_{00} = mass density of the universe so (shakey shakey) in our co-
>ordinate system the mass within a spacelike cube, dx,dy,dz so the volume
>is er, um, ds^3 um ooh er, hang on ds^2=R(t)^2(dx^2 + dy^2 + dz^2) I
>think for a space like manifold. The mass within this volume will remain
>constant so we could describe the mass density as =B/R(t)^3, with B a
>constant. Er I think. Incidentally readers should be aware that I am
>thrashing around like a fish out of water, really I haven't a clue.
>Anyway does this suggest that:
>
>            T = G
>        B/R^3 = 3R'^2/R^2    ??????????????????
>or          B= 3(R'^2)R
>
>Hmmm. So if the density is zero then R or R' must be zero, yes R' might
>be plausibly zero and we have something like the Milne universe. Well,
>probably the Milne universe, actually. 
>
>Hmmmm. 
>The only solution to this I can dream up is 
>
>        R(t) = (3B/4)^(1/2) t^(2/3)
>

I think you meant to type R(t) = (3B/4)^(1/3) t^(2/3)

I checked this and, by jove, it works.

Hooray!

>
>What would a metric of form (-1, kt^(4/3), kt^(4/3), kt^(4/3)) look
>like? Well the Ricci tensor tells us how a volume evolves in time.
>
>R_{ab} diagonalises at (-3R"/R, RR"+2R'2, RR"+2R'2, RR"+2R'2)
>
>R_{00} = (-2/3)t^-2 which means I have it wrong.
>

(+2/3)t^-2 is what I get.

Why were you upset with this.  Was is because it was negative, or
something else I'm missing.  Well, as I pointed out, I think you last
a minus sign somewhere.

Is there still cause for concern?

Kevin 

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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
Date: 26 Apr 1996 21:29:35 GMT
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In article <[email-masked]>,
Oz  <[email-masked]> wrote:
>In article <[email-masked]>, Patrick van Esch
><[email-masked]> writes
>>
>>
>>However, our guessing has introduced a particular solution that
>>isn't part of the general solution: 
>>
>>               ==============
>>               |            | 
>>               | R(t) = Cte |
>>               |            |
>>               ==============
>>
>>(is this a static Minkowski universe then ?)

If Cte means "constant" ("constante" in some European language, perhaps?)
then the answer is yes.

[...]
>>: The only solution to this I can dream up is 
>>
>>:         R(t) = (3B/4)^(1/2) t^(2/3)
>>
>>Well, indeed, that's the same solution I have !

And it's right.  In an expanding dust-filled flat Robertson-Walker
spacetime, the expansion goes like the two-thirds power of time.
Well done.

>I expected to find that a volume would evolve two ways. With the
>density=0, we would have a Milne universe that expands forever, OK I
>*think* we have this. 

When the density is zero, you get just ordinary flat spacetime
R(t)=constant.  This is Minkowski space, described in ordinary
standard Minkowski coordinates.

You did not find the solution in Milne coordinates, i.e., the coordinates
that make Minkowski space "look like it's expanding."  For those
who don't recall, I'll just remind you that Milne coordinates
are just coordinates for ordinary Minkowski spacetime that
make the metric look like this:

ds^2 = -dt^2 + t^2 dspace^2,

Here "dspace" is my shorthand for an ordinary three-dimensional spatial
line element.  This is a Robertson-Walker solution with R(t)=t.

You might wonder why you didn't find this particular solution when you
solved the Einstein equation.  You found R(t) going like t to the
two-thirds and R(t) constant, but not R(t) going like t.  Why not?
(The R(t)=constant solution is geometrically the same as Milne, just
in different coordinates, but you might wonder why you didn't find the
Milne solution in Milne coordinates.)

The answer lies in a fact about Milne coordinates that you might
have forgotten.  The "dspace" that goes into them is not flat
three-space.  That is, it's not dx^2+dy^2+dz^2.  It's negatively
curved (hyperbolic) three-space.  When Baez started you working on
the Robertson-Walker spacetimes, he explicitly told you to worry
only about the spatially flat case, and so Milne space (in Milne
coordinates) was excluded from consideration from step 1.

I'll repeat this business about Milne coordinates again, since it
might confuse people.  The spacetime described by this metric:

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

is ordinary flat Minkowski spacetime, the spacetime of special relativity.
The spacetime described by this metric:

ds^2 = -dt^2 + t^2 dspace^2,

where dspace is a *negatively curved* three-dimensional spatial
metric, is the same spacetime as Minkowski space, just written in
different coordinates.  In these Milne coordinates, the constant-time
spatial slices (that is, the three-dimensional surfaces t=constant) do
have curvature, even though the spacetime as a whole is flat.  When
you look at Minkowski space in Milne coordinates, you're just slicing
spacetime up into spatial sections in a different way that makes
things look funny.

>However as the density increases we ought to have a universe where a
>volume evolves initially by increasing in volume, and then decreasing. I
>can't see how to show this. This irritates me.

This doesn't happen for spatially flat dust-filled Robertson-Walker
spacetimes.  They just expand forever like t to the two-thirds power.
It turns out that in order to get a big-crunch Universe that expands
and then contracts, you need to construct a Robertson-Walker model
that has spatial curvature.  To keep the math relatively simple, Baez
has had you considering only spatially flat models.

Specifically, the metric you all have been using is

ds^2 = -dt^2 + R(t)^2 (dx^2+dy^2+dz^2).

To get a Universe that undergoes a big crunch, you need to replace
that spatial part with a piece whose geometry is that of a
three-dimensional sphere.  Specifically, you'd want to consider
metrics like this:

ds^2 = -dt^2 + R(t)^2 [dr^2+ sin(r)^2 (dtheta^2+ sin(theta)^2 dphi^2)].

You'd have to go back to step one and calculate the Christoffel
symbols and Riemann curvature all over again.  If you feel like doing
that, go right ahead, although I think that Baez would rather have you
work out some more things with the flat models first.

Specifically, I think he mentioned something about working out R(t)
in the case where the Universe was filled with radiation instead of
dust.  The difference is that radiation has pressure, so those
other three diagonal elements of T_{ab} aren't zero.  It's a little
bit tricky to figure out what to do about that, but I'll let you
think about it rather than giving any hints right now.

>Basically I think it's time for an expert to wax lyrical and talk around
>the subject. Now Baez is away, but I don't think Ted is, so how about
>some talking around the subject, please Ted.

That's as lyrical as I get.

-Ted

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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 26 Apr 1996 09:42:57 GMT
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Oz ([email-masked]) wrote:

: OK, since El Grando Wizardo is off for the weekend we gotta do a bit
: ourselves. Now I think that between Van Esch, Kevin and Norbert (with
: manic confusion and errors contributed by Oz) we ought to be able to get
: a bit further.

: What procedure do you use to solve the Einstein equation?

: We have some results for T, I think, which I made as:

: T_{00} = -3R"/R -(-1)3(RR"+R'^2)/R^2= 3R"/R +3R'^2/R^2 -3R"/R
:        = 3R'^2/R^2

: T_{ii} = RR"+2R'^2 -3R^2(RR"+R'^2)/R^2= RR"+2R'^2 -3RR" -3R'^2
:        = -2RR" - R'^2

: And the rest are zero. However, given my record the chances of this
: being correct are slim.

Well, given my record, they are even slimmer, because I find the
same thing :-)
I prefer ito call it still the Einstein tensor...

G_ab = R_ab - (1/2).Ricci.g_ab

Clearly, non-diagonal elements are 0.

G_00 = R_00 - 1/2 . 6.(R".R + R'^2)/R^2.(-1) 
    
     = -3.R"/R + 3.R"/R + 3.R'^2/R^2 = 3.R'^2/R^2

for i = 1,2,3:

G_ii = R_ii - 1/2. 6.(R".R + R'^2)/R^2 . R^2

     = R".R + 2.R'^2 - 3.(R".R + R'^2)

     = - 2.R".R - R'^2

So:

G_00 = 3.R'^2/R^2

G_ii = - 2.R".R - R'^2

Cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 27 Apr 1996 16:13:50 GMT
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Oz ([email-masked]) wrote:

: I expected to find that a volume would evolve two ways. With the
: density=0, we would have a Milne universe that expands forever, OK I
: *think* we have this. 

No, we don't !  R(t) = constant, so no t dependence, hence no 
expansion, but flat, infinite, boring Minkowski space.

: However as the density increases we ought to have a universe where a
: volume evolves initially by increasing in volume, and then decreasing. I
: can't see how to show this. This irritates me.

Well, this is not possible with our flat space part.  We actually
put a lot of constraints on the universe by assuming an Euclidean
space part.   So what actually happens is that *in order to get
a flat space part* the rate of expansion (given by R') is proportional
to the density of the universe.  So if you put more matter in
the universe (higher density) our particular solution will be
such that we will automatically get a higher expansion rate, in
such a way that we will never witness collapse.  This is no magic,
it is because we *started* with the condition of a flat spacial
part.

I don't know JB's plans, probably he wanted us to play a bit with
this simple case to get used to turning the handles of the maths
machinery.
What you want, is to consider a "closed" universe, whose metric
is given by:

     ds^2 = -dt^2 + a^2(t) [dChi^2 + sin^2Chi.{d theta^2 + ...
               ...+ sin^2theta.dphi^2}]

where Chi, theta and phi  are the 3 angular variables describing
a 3-sphere: Chi and theta vary between 0 and pi, while
phi varies between 0 and 2pi.

As you can see, the space part isn't a flat, Euclidean
metric anymore but curved the way a sphere would do.  But 
this is tedious to start with 8-)

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 27 Apr 1996 07:18:51 +0100
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In article <[email-masked]>, "Emory F. Bunn"
<[email-masked]> writes
>
>To get a Universe that undergoes a big crunch, you need to replace
>that spatial part with a piece whose geometry is that of a
>three-dimensional sphere.  Specifically, you'd want to consider
>metrics like this:
>
>ds^2 = -dt^2 + R(t)^2 [dr^2+ sin(r)^2 (dtheta^2+ sin(theta)^2 dphi^2)].
>
>You'd have to go back to step one and calculate the Christoffel
>symbols and Riemann curvature all over again.  

I just KNEW it was too easy. I should have sussed from the start that if
the spacial slices are flat, then we would end up with, well, a universe
with flat spacial slices. Duh.

NOW we have turned another page in the wizards book, we get to see the
slightly heavier mathematical machinery we are going to need, and it's
*frightening* to neophytes.

Someone of my limited maths ain't going to solve this lot without some
serious hand holding. Well,not so much hand-holding as dragging
violently through the mathematical undergrowth backwards whilst being
force fed on answers. Sad.

I dunno, I suppose if I had a spare week ..... <smack!>

>If you feel like doing
>that, go right ahead, although I think that Baez would rather have you
>work out some more things with the flat models first.

What a good idea!

>Specifically, I think he mentioned something about working out R(t)
>in the case where the Universe was filled with radiation instead of
>dust.  The difference is that radiation has pressure, so those
>other three diagonal elements of T_{ab} aren't zero.  It's a little
>bit tricky to figure out what to do about that, but I'll let you
>think about it rather than giving any hints right now.

OOoh. A hint! Did I want to know?

>>Basically I think it's time for an expert to wax lyrical and talk around
>>the subject. Now Baez is away, but I don't think Ted is, so how about
>>some talking around the subject, please Ted.
>
>That's as lyrical as I get.

Nah. I've seen you more lyrical than this, but it's a real good post.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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From: [email-masked] (Matthew J. McIrvin)
Subject: Re: General Relativity Tutorial
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In article <[email-masked]>, [email-masked] wrote:

> The answer lies in a fact about Milne coordinates that you might
> have forgotten.  The "dspace" that goes into them is not flat
> three-space.  That is, it's not dx^2+dy^2+dz^2.  It's negatively
> curved (hyperbolic) three-space.  When Baez started you working on
> the Robertson-Walker spacetimes, he explicitly told you to worry
> only about the spatially flat case, and so Milne space (in Milne
> coordinates) was excluded from consideration from step 1.

Incidentally, this provides those familiar with special relativity with a
wonderfully easy way to visualize hyperbolic space. The three-spaces in
Milne coordinates are hyperbolic, that is, they have constant negative
curvature. These are surfaces of constant "cosmic time," that is,
constant proper time of a comoving object (as is the case in the
coordinates usually used for all Robertson-Walker solutions).

But if you go into ordinary Minkowski coordinates, then the comoving test
particles in a Milne cosmology just look like they explode out from a
particular point in space at a particular time, and go flying off in all
directions at different, constant velocities, filling up the interior of
the "Big Bang"'s future light cone. If you sprinkle test particles with a
constant density on the constant cosmic-time surfaces in Milne
coordinates, then in Minkowski coordinates they'll be sparsest in the
middle of the cone, and scrunched together divergingly tightly by Lorentz
contraction (as well as increasingly time-dilated) near the edges. You
can fit an infinite number of them into the cone! Outside the cone there
is empty spacetime.

(A digression from my digression: Note that this violates the dogma that
the Big Bang does not happen at a particular point in space. But, of
course, this Milne cosmology is an idealized case where the "galaxies" are
assumed to have no energy or mass--they're not really there [or,
equivalently, there is no gravity]. Can you see what will happen if you
give them a little bit of mass?)

So the hyperbolic spaces are just the hypersurfaces of constant proper
time for objects that fly out from the origin at different constant
velocities. That is, they are surfaces of *equal spacetime interval*
from the origin along straight paths, or solutions to the equation

tau^2 = t^2 - x^2 - y^2 - z^2

for constant values of tau. In a spacetime diagram, these surfaces look
like hyperboloidal bowls. In fact, they are exactly the same shape as the
hyperboloids

m^2 = E^2 - p_x^2 - p_y^2 - p_z^2

that describe the possible energies and momenta of a particle of mass
m.

So what is the shape of a space of constant negative curvature? It's the
same shape as a mass hyperboloid! You might think that a bowl-shaped
surface would have positive curvature, but in Minkowski spacetime that's
not necessarily true. I find this a big help when attempting to imagine
the "pseudosphere," which is the space of constant negative curvature in
two dimensions. It can't be embedded into three-dimensional Euclidean
space, but it can be embedded into 2+1 dimensional *Minkowski* spacetime,
as a hyperboloid of exactly this variety. And I deal with SR enough that
I have fair intuition about Minkowski spacetime.

-- 
Matt McIrvin   http://world.std.com/~mmcirvin/

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!info.ucla.edu!agate!howland.reston.ans.net!vixen.cso.uiuc.edu!uwm.edu!fnnews.fnal.gov!lakesis.fapesp.br!bee.uspnet.usp.br!not-for-mail Mon Apr 29 15:00:15 PDT 1996
Article: 12814 of sci.physics
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From: [email-masked] (Henrique Fleming {F})
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 29 Apr 1996 10:39:20 GMT
Organization: Universidade de Sao Paulo / Brasil
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john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] writes:
: 
: >Is it possible to suffer mental damage from a surfeit of super-
: >subscripts?
: 
: Yes.  Elie Cartan, the wonderful mathematician who (among other things)
: occaisionally helped out Einstein with his geometry, spoke
: disparagingly of the "debauch of indices".  Like any other sort of
: debauch, it can be debilitating.  But we need to do a bit of index
: juggling to get a feel for what the workday of an actual relativity
: theorist is like.  One shouldn't over-romanticize it.
: 
Yes. I use the first chapters of Cartan's "Lecons sur la Geometrie
des Espaces de Riemann" im my introductory course of tensor analysis
for physicists. It is beautifully done, uses a lot of indices
and prepares the mind for the more abstract, intrinsic study of
tensors and forms. Actually,the index juggling technique has its
own charm, especially when combined with the wise choice of
some special coordinate frame. On the other hand, try to understand
Hawking-Penrose singularity theorems without the intrinsic methods!

-- 
---------------------------------------------------------------
Henrique Fleming                    Doubt, one of the forms of
University of Sao Paulo, Brazil     intelligence... 
[email-masked]                    J.L. Borges 

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Mon Apr 29 17:00:18 PDT 1996
Article: 12874 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 29 Apr 1996 14:59:23 -0700
Organization: University of California, Riverside
Lines: 59
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, "Emory F. Bunn"
><[email-masked]> writes

>>To get a Universe that undergoes a big crunch, you need to replace
>>that spatial part with a piece whose geometry is that of a
>>three-dimensional sphere.  Specifically, you'd want to consider
>>metrics like this:

>>ds^2 = -dt^2 + R(t)^2 [dr^2+ sin(r)^2 (dtheta^2+ sin(theta)^2 dphi^2)].

>>You'd have to go back to step one and calculate the Christoffel
>>symbols and Riemann curvature all over again.  

Ugh.  What a bummer.  

>NOW we have turned another page in the wizards book, we get to see the
>slightly heavier mathematical machinery we are going to need, and it's
>*frightening* to neophytes.

Calculating the Riemann tensor when space is S^3 doesn't involve heavier
mathematical machinery.  It involves trigonometry: you gotta take lots
of derivatives of sine and cosine, multiply 'em, differentiate 'em some
more, and stuff like that.  This is frightening even to paleophytes!
Let's not mess with it!

>>If you feel like doing
>>that, go right ahead, although I think that Baez would rather have you
>>work out some more things with the flat models first.

>What a good idea!

Indeed.  That, and try to understand where the heck those formulas for
the Christoffel symbol and Riemann tensor came from.  No, they did not
spring full-formed by sheer inspiration from Riemann's domed head.  

>>Specifically, I think he mentioned something about working out R(t)
>>in the case where the Universe was filled with radiation instead of
>>dust.  The difference is that radiation has pressure, so those
>>other three diagonal elements of T_{ab} aren't zero.  It's a little
>>bit tricky to figure out what to do about that, but I'll let you
>>think about it rather than giving any hints right now.

>OOoh. A hint! Did I want to know?

Well, let's see.  The main thing you need to know, I think, is how the
pressure and the energy density are related for electromagnetic radiation.  
I don't think this can be too hard to work out if we pretend the
radiation is a bunch of photons moving every which way at the speed of
light.  By the way, the speed of light is 1 in this course.  

That's all the hint I'm going to give right now.  

Well... 

Don't forget the mass of the photon, either.  

From noise.ucr.edu!guitar!baez Mon Apr 29 17:13:15 PDT 1996
Article: 12889 of sci.physics
Path: noise.ucr.edu!guitar!baez
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 29 Apr 1996 23:57:55 GMT
Organization: University of California, Riverside
Lines: 133
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In article <[email-masked]> [email-masked] writes:
>Some comments. I'm lost!

Good!  You get the point!  On the one hand, you now see how to use the
equations I gave to figure out how the big bang works --- at least for a
spatially flat homogeneous universe full of dust.  Other cosmologies 
work more or less the same way.  The algebra gets messier, but the 
basic ideas are the same.

On the other hand, when it comes to the formulas for the 
Christoffel symbols and Riemann tensor:
 
        C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

many mysteries remain.  You have no idea why these are true!  Being smart, 
this makes you feel uneasy.  So we should go back and try to understand the
relation between the metric, the Christoffel symbols, and the Riemann
tensor more deeply.  As you note:

>The tensor notation is one of the neatest things I have come across. 
>However, I have to admit that understanding what I am doing is rather 
>harder than the mechanics. Doubtless if I did 10,000 examples it would 
>be clearer, but being realistic this ain't going to happen. Associated 
>with that is a basic lack of understanding of the interactions of the 
>metric, the Riemann tensor and the Ricci with what we are doing. Now 
>this would definitely respond to the 10,000 examples treatment but no 
>thank you.

10,000 examples would NOT be enough for you to truly understand these
formulas: 

        C^a_{cd} = (1/2) g^{ab} (g_{bc,d} + g_{bd,c} - g_{cd,b})

R^a_{bcd} = C^a_{bd,c} - C^a_{cd,b} + C^e_{bd}C^a_{ec} - C^e_{cd}C^a_{eb}

We really need to sit down and think about geometry a bit more!  Okay?

>How do they inter relate?
>
>1) The metric gives us distances between co-ordinates for the system we 
>are modelling.

I don't know what "distances between coordinates" are.  Presumably you
are attempting to say something like this: the metric gives us the
lengths of tangent vectors, and thus the lengths of curves in spacetime.
The components of the metric in any particular coordinate system let us
easily calculate these lengths in that coordinate system.

>2) The Riemann gives us a measure of how vectors change DIRECTION as we 
>move them round space.

That sounds exactly correct.

>3) The Ricci gives us a measure of how space is curved due to 
>gravitational effects in a small volume. It is a subset of the Riemann.

That sounds right too.  I'd say the Ricci tensor is a "contraction" of
the Riemann tensor, but since you are a physicist, I will simply grit my
teeth, force a smile, and nod when you call it a "subset" of the Riemann
tensor.  It contains only some of the information contained in the
Riemann tensor, and this is your way of saying that.  Right?

>4) The connection would seem to tell us how we can link the expression 
>of curved space offered by the Riemann tensor to the actual co-ordinate 
>system we are using, or possibly vice-versa. It also tells us how a 
>vector will change in LENGTH as we move it through spacetime AS SEEN IN 
>OUR CO-ORDINATE SYSTEM. I think.

Hey!  I thought you agreed already that parallel translation does not
actually change the length of a vector, as measured by the metric.  The
LENGTH of a vector is the same no matter what our COORDINATE SYSTEM
happens to be, but the COMPONENTS of the vector may differ drastically
depending on our coordinates.  The Christoffel symbols, or connection,
tell us how the COMPONENTS of a vector change as we parallel transport
it.  

(If you are going to use capital letters, I am going to demand a higher
level of precision of you, or else I'm going to announce to the WHOLE WORLD 
that you are a CRACKPOT who does NOT understand the TRUE MEANING of
EINSTEIN'S THEORY.)

>Well that's how I see it anyway.

I think you basically get it, though you're speak in a way that
makes me wince a bit at times.  I hope you see why speaking about these 
things very precisely is worthwhile, at least when you're just learning.
Once you are an expert you can slack off.

>A problem though.
>
>As I understand it we would need to use both the Riemann (or the 
>connection) AND the metric to model a real system, say the volume of a 
>cube (which is co-ordinate dependant), unless the metric is simply 
>included within the representation of the Riemann and/or the
>connection. 

Here I am mystified.  Let me remind you of two facts.  You should repeat
these statements every night just before bed until you understand them
deeply.  

1) THE GEOMETRY OF SPACETIME IS DETERMINED BY THE METRIC.  Everything
else, like the Christoffel symbols, the Riemann tensor, and the Ricci
tensor, can be computed from the metric.  They are basically just
different ways of analyzing the geometry of a spacetime with the given
metric.  

2) EVERY COMPUTATION IN GENERAL RELATIVITY CAN BE DONE IN ANY
COORDINATE SYSTEM; THE PHYSICS REMAINS THE SAME NO MATTER WHAT
COORDINATE SYSTEM WE USE.  If Fred knows the metric in
coordinate system 1, while Betty knows it in coordinate system 2, both
of them can happily compute whatever they want, and there are
well-understood rules for transforming any of Fred's results over to
Betty's coordinate system, to see if their results agree.  

Perhaps at some point I will make you do some computations in two
different coordinate systems, to help you understand point 2).
I have trouble understanding your problem above, because it sounds
as if you aren't taking points 1) and 2) into account.

>In which case the length of things in our *co-ordinate system* would not 
>be directly calculated using the metric, but via the Riemann (already 
>co-ordinatified via the connection and the metric). Or something.

Yikes.  "Or something" is right!

From noise.ucr.edu!guitar!baez Mon Apr 29 17:13:25 PDT 1996
Article: 12891 of sci.physics
Path: noise.ucr.edu!guitar!baez
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 30 Apr 1996 00:02:25 GMT
Organization: University of California, Riverside
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Oz ([email-masked]) wrote:

> However as the density increases we ought to have a universe where a
> volume evolves initially by increasing in volume, and then decreasing. I
> can't see how to show this. This irritates me.

Don't be irritated!  Be pleased!  You have discovered a surprising fact:

When the universe is spatially flat, homogeneous, and full of dust,
there cannot be a big crunch!

In fact, a big crunch typically occurs when each spacelike slice looks
not like R^3 (Euclidean space) but like S^3 (the 3-sphere).  Layfolk
call the former an "open" universe, and the latter a "closed" universe or a 
"finite but unbounded" universe.  It's typical for "closed" universes to 
have a big crunch, but not "open" ones.  Here you are beginning to see that!

As Patrick Esch notes:

>We actually
>put a lot of constraints on the universe by assuming an Euclidean
>space part.   So what actually happens is that *in order to get
>a flat space part* the rate of expansion (given by R') is proportional
>to the density of the universe.  So if you put more matter in
>the universe (higher density) our particular solution will be
>such that we will automatically get a higher expansion rate, in
>such a way that we will never witness collapse.  This is no magic,
>it is because we *started* with the condition of a flat spacial
>part.

Right.  But it's still a bit wonderful that it works this way.  As it
is, there's a fascinating relationship between the universe being
open, and it lasting forever with no big crunch.  This is not obvious,
but simple, and therefore cool.

>I don't know JB's plans, probably he wanted us to play a bit with
>this simple case to get used to turning the handles of the maths
>machinery.

Yes.  And I'm not even going to get into the more complicated cases,
like the S^3 case, where the metric is:

>     ds^2 = -dt^2 + a^2(t) [dChi^2 + sin^2Chi.{d theta^2 + ...
>               ...+ sin^2theta.dphi^2}]

I'm just teaching a tutorial on general relativity.  So I'm focusing on
the fundamental ideas and keeping the calculations as simple as possible.  

For a full course on cosmology, you gotta hire a cosmologist.  :-) 

From noise.ucr.edu!guitar!baez Mon Apr 29 17:14:49 PDT 1996
Article: 12893 of sci.physics
Path: noise.ucr.edu!guitar!baez
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 30 Apr 1996 00:05:13 GMT
Organization: University of California, Riverside
Lines: 59
Distribution: world
Message-ID: <[email-masked]>
References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, "Emory F. Bunn"
><[email-masked]> writes

>>To get a Universe that undergoes a big crunch, you need to replace
>>that spatial part with a piece whose geometry is that of a
>>three-dimensional sphere.  Specifically, you'd want to consider
>>metrics like this:

>>ds^2 = -dt^2 + R(t)^2 [dr^2+ sin(r)^2 (dtheta^2+ sin(theta)^2 dphi^2)].

>>You'd have to go back to step one and calculate the Christoffel
>>symbols and Riemann curvature all over again.  

What a bummer.  

>NOW we have turned another page in the wizards book, we get to see the
>slightly heavier mathematical machinery we are going to need, and it's
>*frightening* to neophytes.

Calculating the Riemann tensor when space is S^3 doesn't involve heavier
mathematical machinery.  It involves trigonometry: you gotta take lots
of derivatives of sine and cosine, multiply them, differentiate them some
more, and stuff like that.  This is frightening even to paleophytes!
Let's not mess with it!

>>If you feel like doing
>>that, go right ahead, although I think that Baez would rather have you
>>work out some more things with the flat models first.

>What a good idea!

Indeed.  That, and try to understand where the heck those formulas for
the Christoffel symbol and Riemann tensor came from.  Hint: they did not
spring forth by divine decree from Riemann's domed head.  

>>Specifically, I think he mentioned something about working out R(t)
>>in the case where the Universe was filled with radiation instead of
>>dust.  The difference is that radiation has pressure, so those
>>other three diagonal elements of T_{ab} aren't zero.  It's a little
>>bit tricky to figure out what to do about that, but I'll let you
>>think about it rather than giving any hints right now.

>OOoh. A hint! Did I want to know?

Well, let's see.  The main thing you need to know, I think, is how the
pressure and the energy density are related for electromagnetic radiation.  
I don't think this can be too hard to work out if we pretend the
radiation is a bunch of photons moving every which way at the speed of
light.  By the way, the speed of light is 1 in this course.  

That's all the hint I'm going to give right now.  

Well... 

Don't forget the mass of the photon, either.  

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!howland.reston.ans.net!newsfeed.internetmci.com!btnet!zetnet.co.uk!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Wed May  1 22:34:38 PDT 1996
Article: 13079 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 30 Apr 1996 20:02:50 +0100
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>
>Well, let's see.  The main thing you need to know, I think, is how the
>pressure and the energy density are related for electromagnetic radiation.  
>I don't think this can be too hard to work out if we pretend the
>radiation is a bunch of photons moving every which way at the speed of
>light.  By the way, the speed of light is 1 in this course.  
>
>That's all the hint I'm going to give right now.  
>
>Well... 
>
>Don't forget the mass of the photon, either.  

Well, it was bound to come out. My guilty secret. I can't remember a 
thing about pressures, light and so on. The only thing I can remember is 
some derivation that involved bouncing photons off a plate and thus 
deriving a force, hence a pressure. Following the age old choice I can 
either blather on pseudoconvincingly or go back to first principles. I 
think I prefer the latter even though I am pretty sure that this will 
expose more layers of ignorance to the wizard's piercing gaze.

Like a lamb to the slaughter. Oh, well.

Right. To start at the end, so to speak.

We have a couple of expressions.

G_{00} = 3R'^2/R^2 and G_{ii} = -2RR" - R'^2 

For a cold dusty universe we have solutions 
R(t)= a constant and R(t) = At^(2/3) since here
G_{00} = constant/R^3 and G_{ii} = 0.

Now we want to sort out the case when G_{ii} is not zero. One is tempted 
to consider a universe filled only with photons as an initial case, just 
to get some clues, you understand.

The first thing to consider is what G_{ii} really is. Now if I say my 
naive view, then it can be corrected and move on to the correct view. 
After all it just means I have to expose my ignorance and take a few GW 
of incandescent firebolts. What you have to do for knowledge, shoof!

Well T_{ii} should be the density of momentum in the i-direction and
T_{00} is the density of the energy. Now for light, with zero mass, the 
energy equals the momentum. Now I think we should be rather cautious in 
assuming that the energy of a photon in an expanding universe is 
constant since there is evidence to the contrary. We ought to be able to 
derive an expression for the change in energy of a photon as this 
universe expands, but I hope it won't come to that (wishful thinking). 
Er, hang on, with a bit of luck it will be inversely proportional to a 
spatial length and we know how that varies, it varies as R(t). The 
number of photons must surely remain constant, but their energy varies 
as 1/R(t). So the (ooh, should have tried this on Sunday) energy density 
varies as 1/(R^4). Yuk! Oh, I don't know

G_{00} = k/(R^4) = 3R'^2/R^2 => k = 3R'^2 R^2 which has solutions

R = (4k/3)^(1/4)  t^(1/2) = At^(1/2) and t = constant.

Hmm, lets see how this fits into G_{ii}. Oh dear, what's the pressure of 
a photonic gas? Dunno. I suppose I could try to work it out. Hmm.

How about a cheat then. Yes, why not. Let's shove our solution above 
into G_{ii}.

G_(ii) = -2RR" - R'^2 =<chunter chunter>= (1/4) A^2 t(-1) sod it.

How do I plausibly argue that the 'pressure' varies like this? Of course 
it would help a lot if I had a clear idea of what exactly is meant by 
pressure in this context. Hmmm. Well pressure is a force/area thingy. 
Hmm. Well the number of impacts will depend on the momentum of each 
photon, which I think should vary inversely as R. The number of impacts 
per unit time should depend inversely on the linear distance and since 
the ......

I'm not thinking straight!

Perhaps if someone could give a little definition of pressure?

Oh dear. What few credits I might have collected have all been blown.

<Oz trogs despondently to the back of the class.>

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!guitar!baez Wed May  1 22:37:58 PDT 1996
Article: 13156 of sci.physics
Path: noise.ucr.edu!guitar!baez
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 2 May 1996 04:15:25 GMT
Organization: University of California, Riverside
Lines: 172
Distribution: world
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NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>Well, let's see.  The main thing you need to know, I think, is how the
>>pressure and the energy density are related for electromagnetic radiation.  
>>I don't think this can be too hard to work out if we pretend the
>>radiation is a bunch of photons moving every which way at the speed of
>>light.  By the way, the speed of light is 1 in this course.  

>Well, it was bound to come out. My guilty secret. I can't remember a 
>thing about pressures, light and so on. The only thing I can remember is 
>some derivation that involved bouncing photons off a plate and thus 
>deriving a force, hence a pressure. Following the age old choice I can 
>either blather on pseudoconvincingly or go back to first principles. I 
>think I prefer the latter even though I am pretty sure that this will 
>expose more layers of ignorance to the wizard's piercing gaze.

More layers of ignorance, eh?  As ones sphere of knowledge increases,
ones circumference of ignorance also grows.  The only consolation is
that the former grows in proportion to a higher power of the radius, 
and even this is only true for a space that is flat or positively
curved.  

>We have a couple of expressions.
>
>G_{00} = 3R'^2/R^2 and G_{ii} = -2RR" - R'^2 
>
>Now we want to sort out the case when G_{ii} is not zero. One is tempted 
>to consider a universe filled only with photons as an initial case, just 
>to get some clues, you understand.

Yes.  You know the famous line: "Let there be light! --- just to simplify
the math at first". 

>The first thing to consider is what G_{ii} really is. Now if I say my 
>naive view, then it can be corrected and move on to the correct view. 
>After all it just means I have to expose my ignorance and take a few GW 
>of incandescent firebolts. What you have to do for knowledge, shoof!

>Well T_{ii} should be the density of momentum in the i-direction...

ZAP!!!!!!!!!!!!!

Remember, in coordinates where the metric looks like the Minkowski one,
T_{ij} is the flow in the i direction of momentum in the j-direction.
So the "density of momentum in the i-direction" is T_{0i}, since flow
in the 0 or time direction is what apprentices like you call "density".

But if you're trying to understand T_{ii} --- and you are --- that's
gonna be the flow in the i-direction of i-momentum.  And that is just
the *pressure* in the i direction.  That's what you were getting at with
the business of photons bouncing off a plate, right?

>T_{00} is the density of the energy. Now for light, with zero mass, the 
>energy equals the momentum. 

Right.  From this you could derive a relation between the energy density and
pressure for isotropic radiation.  That would be one way to solve this
problem.  Actually you figured out a different way, which is also good. 

But just for your own benefit, I think you should figure out the
relation between the pressure P and the density rho for isotropic
radiation.  After all, doesn't that sound like an interesting thing to
know?  It's probably pretty simple, too.

>Now I think we should be rather cautious in 
>assuming that the energy of a photon in an expanding universe is 
>constant since there is evidence to the contrary. We ought to be able to 
>derive an expression for the change in energy of a photon as this 
>universe expands, but I hope it won't come to that (wishful thinking). 
>Er, hang on, with a bit of luck it will be inversely proportional to a 
>spatial length and we know how that varies, it varies as R(t). The 
>number of photons must surely remain constant, but their energy varies 
>as 1/R(t). So the (ooh, should have tried this on Sunday) energy density 
>varies as 1/(R^4). Yuk! 

Yuk?  You mean "Hurray!"  You weaseled your way through this one quite
nicely.  You never needed to understand pressure after all, you sneaky
devil.  

>Oh, I don't know
>
>G_{00} = k/(R^4) = 3R'^2/R^2 => k = 3R'^2 R^2 which has solutions
>
>R = (4k/3)^(1/4)  t^(1/2) = At^(1/2) and t = constant.

Yes!  Correct.

Good work.

>I'm not thinking straight!

You're not supposed to.  This is curved spacetime.

>Perhaps if someone could give a little definition of pressure?

Ahem, we've done it MANY TIMES now.  Let me dig them up from earlier
posts, just to emphasize the advantages of keeping your notes from
earlier lectures.  

Article 98908 (52 more) in sci.physics:
From: Keith Ramsay
Subject: Re: General relativity tutorial
Date: 8 Feb 1996 19:49:25 GMT
Organization: Boston College
Lines: 28
NNTP-Posting-Host: mt14.bc.edu

In article <[email-masked]>, [email-masked] (john baez) wrote:

|Now you might wonder about what all that "x-momentum in the x direction"
|stuff really means.   I don't have anything thrillingly insightful to
|say about this except that, for a perfect fluid, each of these terms is
|just the pressure!  I will quit here for now, leaving Michael Weiss to
|explain why that's true.  

I have a prosaic example to add to the explanation. You know those little
toys they have, with a row of metal balls hung on threads? You pull the first
little ball to the side, and let it go. The mechanics of the toy are such
that the ball you let go comes to a stop in its resting position, and the
other balls remain where they were (more or less), excepting the last one, 
which swings up on the other side.

This is an example of momentum in a given direction being "transported"
in that direction. It is of course by means of pressure of the balls
in the series upon each other that the momentum is "transported" from
one end to the other.

When you have pressure, you can think of it as the collective effect
of a lot of little particles being bounced back and forth. When you
switch to other coordinate systems, the energy of the system is
different, in a way which depends upon the distribution of velocities 
of the particles, even if the average is fixed.

Keith Ramsay

>From galaxy.ucr.edu!not-for-mail Sun Mar  3 20:00:34 PST 1996
Article: 103267 of sci.physics
Path: galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 3 Mar 1996 18:40:16 -0800
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Oz writes: 

>Let's see. T_{ij} must be
>the flow in the i direction of momentum in the j direction. Now we gotta
>be careful here or we will convince ourselves that T_{ii} equals zero
>too. I have the distinct feeling that this will result in a period of
>time spent as a colonic parasite, which is even worse than an intestinal
>one. Well, T_{ii} is a pressure. It's a little hard to see how T_{ij}
>(j=/=i) could be a pressure as long as we keep our hands waving at all
>times since the flow is perpendicular to the action, so to speak. Of
>course if we knew the derivation of T_{ii} being a pressure then it would
>almost certainly be perfectly clear.

The derivation is quite simple and I think I said it before.  For
example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
for visual vividness that this flow is carried by little ball-shaped
atoms.  Then if you put a wall in their way they push on the wall in
the x direction, with a certain pressure.  This pressure is a force per
unit area, which is just the same as "momentum per time per unit area".
This is given by the flow of x-momentum in the x-direction!

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Thu May  2 15:01:35 PDT 1996
Article: 13241 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 1 May 1996 20:16:10 -0700
Organization: University of California, Riverside
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In article <[email-masked]> "Alexander E. Meier" <[email-masked]> writes:
>I might be completely off, but I seem to recall that in SR you have an
>energy-momentum tensor for an electromagnetic field.. isn't it  possible
>to derive some kind of pressure from it...?
>That's the farthest I can go since I'm still trying to grasp what GR is
>all about.

That's definitely one interesting way to tackle this problem.  Actually
it was Michael Faraday who noticed that electric and magnetic field
lines have tension and pressure!  Tension is simply negative pressure.
To be precise, if you have an electric field, there is a tension of
E^2/2 along the lines of force, and a pressure of E^2/2 in both
directions perpendicular to the lines of force, where E is the electric
field.  Similarly, there is a tension of B^2/2 along magnetic force
lines, and a pressure of B^2/2 in both directions perpendicular to the
magnetic force lines.

Amazing what they never tell you in school, eh?  If I weren't so
respectable you might think I was making that stuff up.  It can be found
in any decent book on electromagnetism, buried in the equations.

There is also an energy density of (E^2 + B^2)/2 due to an
electromagnetic field, and also a flux of energy equal to E x B.  This
last thing is often called the Poynting vector.

So yes, knowing this stuff, and knowing the definition of the
stress-energy tensor, and the fact that it's symmetric (T_{ab} =
T_{ba}), one can work out the pressure and energy density of a bath of
isotropic radiation.  But somehow I think it's easier to treat the light
as a bunch of massless particles zipping isotropically in all directions
at the speed of light.  Can someone guess how the pressure and energy
density will be related?  That's all we need, to figure out how the big
bang works in a universe full of radiation.

Current theories say the universe was "radiation-dominated" until
sometime between a thousand and ten million years after the big bang.
In other words, most of the energy density was in the form of radiation, and
there was lots of pressure.  Later, most of the energy density wound up
in the form of matter, and the pressure dropped way down.  So it's for
studying the early history of our world that it's nice to know how a
universe full of radiation would expand.  

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Fri May  3 09:07:02 PDT 1996
Article: 13334 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Gravitons & Doppler
Date: 2 May 1996 11:40:46 -0700
Organization: University of California, Riverside
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Keywords: graviton doppler effect

In article <[email-masked]> [email-masked] (ThInK) writes:
>	Hello, everyone. This is my first post in this group. You seem like a
>group of friendly and very knowledgeable people,

Flattery will get you nowhere.  We are a bunch of ignorant,
argumentative jerks.   

>so let me pose a question
>that I came up with on my own: if gravitons are wave-like, would it be
>possible to observe some sort of doppler effect on particles travelling
>rapidly towards or away from a graviton source (e.g. the Earth)? As you moved
>rapidly away from a source, the force of gravity at your position would be
>less than the force of gravity at your position if you were standing
>still.

Don't mix up gravitational radiation with gravitational attraction.

Gravitational radiation is wavelike, and mathematically somewhat similar
to light or other forms of electromagnetic radiation.  You are right
that a moving observer should see gravitational radiation being
"redshifted" or "blueshifted" depending on whether they are moving away
from or towards the source.  It's a good point which had never occurred
to me.  Unfortunately, we are currently unable to detect gravitational
radiation.  (See the end of this post for something about an experiment
to detect it.)

On the other hand, gravitational attraction is mathematically somewhat
similar to the "Coulomb force", or electrostatic force, exerted by a
charged static particle.  No fields are "waving", they are just sitting
there.  One does expect interesting effects when one moves rapidly
through such a gravitational or electric field, but not a simple
increase or decrease in the strength of the field as you suggest.  For
example, when you move rapidly through an electric field you see some
magnetic field as well!  A similar thing should be true for gravity,
which also can be decomposed into so-called "gravitoelectric" and
"gravitomagnetic" components.  (Technically speaking, this is a way to
decompose the 10-component Weyl tensor into 2 5-component objects, in a
way which depends on your frame of reference.)

>I
>am considering pursuing this idea to equations and then an actual project,
>but in the meantime could any of you physics gurus tell me whether or not
>this has actually been attempted yet? 

I'm sure this stuff has been worked out somewhere, but that shouldn't
prevent you from having the fun of doing it yourself!  Usually the only
way to understand something is to work it out oneself.  I'd suggest that
you start by working out in detail what an observer moving through the
electrostatic potential produced by a charged particle will detect.
You'll need to figure out how to Lorentz transform the electromagnetic
field.  Like I say, the gravitational case is similar but more
complicated.

>From "week80":

4) Ignazio Ciufolini and John Archibald Wheeler, Gravitation and
Inertia, Princeton University Press, 1995.

and 

5) Kip Thorne, Richard Price and Douglas Macdonald, eds., Black Holes:
The Membrane Paradigm, 1986.

The book by Ciufolini and Wheeler is full of interesting stuff, but it
concentrates on "gravitomagnetism": the tendency, predicted by general
relativity, for a massive spinning body to apply a torque to nearby
objects.  This is related to Mach's old idea that just as spinning a
bucket pulls the water in it up to the edges, thanks to the centrifugal
force, the same thing should happen if instead we make lots of stars
rotate around the bucket!  Einstein's theory of general relativity was
inspired by Mach, but there has been a long-running debate over whether
general relativity is "truly Machian" --- in part because nobody knows
what "truly Machian" means.  In any event, Ciufolini and Wheeler argue
that gravitomagnetism exhibits the Machian nature of general relativity,
and they give a very nice tour of gravitomagnetic effects.

That is fine in theory.  However, the gravitomagnetic effect has never
yet been observed!   It was supposed to be tested by Gravity Probe B, a
satellite flying at an altitude of about 650 kilometers, containing a
superconducting gyroscope that should precess at a rate of 42
milliarcseconds per year thanks to gravitomagnetism.  I don't know what
ever happened with this, though: the following web page says "Gravity
Probe B is expected to fly in 1995", but now it's 1996, right?  Maybe
someone can clue me in to the latest news.... I seem to remember some
arguments about funding the program.

6) The story of Gravity Probe B, http://www-leland.stanford.edu/~michman/RELATIVITYmosaic/GPBmosaic/GPB.html#GPBstory

Kip Thorne's name comes up a lot in conjuction with black holes and the
LIGO --- or Laser-Interferometer Gravitational-Wave Observatory --- project.
As pairs of black holes or neutron stars spiral emit gravitational
radiation, they should spiral in towards each other.  In their final
moments, as they merge, they should emit a "chirp" of gravitational
radiation, increasing in frequency and amplitude until their ecstatic
union is complete.  The LIGO project aims to observe these chirps, and
any other sufficiently strong gravitational radiation that happens to be
passing by our way.  LIGO aims to do this by using laser interferometry
to measure the distance between two points about 4 kilometers apart to
an accuracy of about 10^{-18} meters, thus detecting tiny ripples in the
spaceteim metric.  For more on LIGO, try

7) LIGO project home page, http://www.ligo.caltech.edu/

Thorne helped develop a nice way to think of black holes by envisioning
their event horizon as a kind of "membrane" with well-defined
mechanical, electrical and magnetic properties.  This is called the
membrane paradigm, and is useful for calculations and understanding what
black holes are really like.  The book "Black Holes: The Membrane
Paradigm" is a good place to learn about this.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Fri May  3 09:07:11 PDT 1996
Article: 13340 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: solutions of GR? very small wave packets of high curvature?
Date: 2 May 1996 12:09:41 -0700
Organization: University of California, Riverside
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References: <[email-masked]>
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In article <[email-masked]> [email-masked] (ale2) writes:
>please stop me when i'm wrong.

>1) GR has plane-wave type solutions,

Sorta, yeah.

>2) like other wave phenomena these plane-wave solutions can be
>superimposed to give a localized wave-packet (with some finite
>lifetime?).

It's not a linear equation so you can't really "superpose" things as
with a linear equation, but if the waves are small they are
approximately linear, so the superposition principle works
approximately.  Folks trying to detect gravitational radiation are happy
to work with the linearized Einstein equation, because the ripples are
so small.

>3) then we can construct two such wave-packets which are highly
>localized, have equal and opposite momentum, and which would pass each
>other at a distance L if there were no interactions 

Yes.

>3b) (guess we have to specify the polarization also?). 

Something like that, though unlike photons, gravitons are spin-2.  Don't
take the particle-like language too seriously here; all I really mean is
that more information is needed to specify the "polarization" of a
gravitational wave than of an electromagnetic wave.  

>4) then for some proper initial conditions bound states of two such
>wave-packets will form and,

This certainly doesn't happen for linear equations; it might happen for
Einstein's equation due to the nonlinearity.  In 1962 Wheeler
constructed solutions of the vacuum Einstein equation called "geons"
which are basically regions of curved spacetime that hang together due
to their own gravity.  Unfortunately these are never stable.  See
_Gravitation_ for references.

>5) one might guess that some gravitational radiation would be given off
>when this happened?

Sure, probably.

>5b) Such a solution can be considered a massive particle made out of
>massless (massless in the sense of rest mass) wave-packets? 

I'd guess geons act like massive particles.

>6) The nonlinear aspect of GR will make talking about the proceeding
>very difficult for very large curvatures.

Yes, but not impossible.  Anyway, when in the nonlinear regime, it is
often easier to bite the bullet and solve the nasty nonlinear equations
than it is to try to construct solutions via "superposition".  There are
some equations, like the KdV equation for shallow water waves, which
admit a "nonlinear superposition principle".  This is so cool --- like
having your cake and eating it too --- that it's been all the rage in
mathematical physics for the last decade or so.  You can find out about
it in books on soliton theory, like Rajaraman's _Solitons and
Instantons_.  But Einstein's equation isn't that simple.  The self-dual
Einstein equation, a special case, can be studied using instanton
methods.  But a widely-used rule of thumb is that any equation which you can
exactly solve is probably not complicated enough to describe the real
world.

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Article: 13497 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 3 May 1996 20:33:57 +0100
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I have been chastised by the wizard for not doing as I have been told.

Sob, sob.

However, since he has been right in the past, I had better do what he
says before everyone kills this thread. If they haven't already that is.

In article <[email-masked]>, john baez <[email-masked]>
writes

NB How long has Baez been classifying this thread as @noise.ucr.edu???

>
>But if you're trying to understand T_{ii} --- and you are --- that's
>gonna be the flow in the i-direction of i-momentum.  And that is just
>the *pressure* in the i direction.  That's what you were getting at with
>the business of photons bouncing off a plate, right?
>
>>T_{00} is the density of the energy. Now for light, with zero mass, the 
>>energy equals the momentum. 
>
>Right.  From this you could derive a relation between the energy density and
>pressure for isotropic radiation.  That would be one way to solve this
>problem.  Actually you figured out a different way, which is also good. 
>
>But just for your own benefit, I think you should figure out the
>relation between the pressure P and the density rho for isotropic
>radiation.  After all, doesn't that sound like an interesting thing to
>know?  It's probably pretty simple, too.

OK. Lets consider a plate in a sea of photons. Let's have the energy 
density as D and the plate of area A. For a photon the energy equals the 
momentum. 

Just to keep the model clear and simple for me lets consider this 
density to be comprised of N photons of momentum Q each per unit volume.

In a short time dt there will be dn collisions with the plate. Now if 
ALL the photons were (laserlike) heading towards my plate then in time 
dt all the photons in a column of volume Acdt would strike my plate, and 
no others. So in this case a volume Acdt worth of photons since they 
travel at c. However in Baezworld c=1 so

dn = NAdt 

Since each photon has momentum Q this will give me a total momentum 
transferred to the plate of

dp = QNAdt => dp/dt = QNA

and we note that the energy density is QN = D

and the force on the plate is dp/dt so the force on the plate due to 
incoming photons will be

DA and thus the pressure will be P = DA/A = D

No I have an irresistible urge to have a reflecting plate here, just to 
avoid the photons accumulating on my plate. In this case they will 
'rebound' and give me double this momentum transfer. I expect I can 
offer more plausible reasons for wanting to chose this.

So P = 2D for a laserbeam intersecting my nice shiny plate.

Now we have been told to get the pressure for an isotropic sea of 
photons. Well, we must take a third of the figure derived above because 
the resolved components perpendicular to the plate (ie the laser is in 
the x-direction, so the y & z directions will transfer no momentum). 
Also half the photons are heading in the -x direction so won't hit the 
plate, so we had better divide by two as well. 

So my supercrude derivation comes out to

P = D/3

So the pressure is 1/3 of the energy density.

Of course in real life there ought to be few 'c's about too.
Dunno, this feels right. I just wish I could remember if it is.

>The derivation is quite simple and I think I said it before.  For
>example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
>for visual vividness that this flow is carried by little ball-shaped
>atoms.  Then if you put a wall in their way they push on the wall in
>the x direction, with a certain pressure.  This pressure is a force per
>unit area, which is just the same as "momentum per time per unit area".
>This is given by the flow of x-momentum in the x-direction!

I have an irresistible urge to imagine a flow of momentum through a 
surface (instead of a plate). The only problem I have with this is that 
the momentum flowing from right to left is cancelled by the momentum 
flowing from left to right. Now you are going to tell me that if I stick 
my reflective plate instead of the surface (an infinitely thin one of 
course) the distribution of photons will remain quite isotropic. Thus 
the pressure is everywhere uniform. I still find this difficult to match 
unless you are going to start talking about the flow of momentum in the 
-x direction too.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 13545 of sci.physics
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From: [email-masked] (Patrick van Esch)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 11:45:11 GMT
Organization: DESY
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Oz ([email-masked]) wrote:

: >
: >But just for your own benefit, I think you should figure out the
: >relation between the pressure P and the density rho for isotropic
: >radiation.  After all, doesn't that sound like an interesting thing to
: >know?  It's probably pretty simple, too.

: OK. Lets consider a plate in a sea of photons. Let's have the energy 
: density as D and the plate of area A. For a photon the energy equals the 
: momentum. 

: Just to keep the model clear and simple for me lets consider this 
: density to be comprised of N photons of momentum Q each per unit volume.

: In a short time dt there will be dn collisions with the plate. Now if 
: ALL the photons were (laserlike) heading towards my plate then in time 
: dt all the photons in a column of volume Acdt would strike my plate, and 
: no others. So in this case a volume Acdt worth of photons since they 
: travel at c. However in Baezworld c=1 so

: dn = NAdt 

: Since each photon has momentum Q this will give me a total momentum 
: transferred to the plate of

: dp = QNAdt => dp/dt = QNA

: and we note that the energy density is QN = D

: and the force on the plate is dp/dt so the force on the plate due to 
: incoming photons will be

: DA and thus the pressure will be P = DA/A = D

: No I have an irresistible urge to have a reflecting plate here, just to 
: avoid the photons accumulating on my plate. In this case they will 
: 'rebound' and give me double this momentum transfer. I expect I can 
: offer more plausible reasons for wanting to chose this.

: So P = 2D for a laserbeam intersecting my nice shiny plate.

: Now we have been told to get the pressure for an isotropic sea of 
: photons. Well, we must take a third of the figure derived above because 
: the resolved components perpendicular to the plate (ie the laser is in 
: the x-direction, so the y & z directions will transfer no momentum). 
: Also half the photons are heading in the -x direction so won't hit the 
: plate, so we had better divide by two as well. 

: So my supercrude derivation comes out to

: P = D/3

: So the pressure is 1/3 of the energy density.

: Of course in real life there ought to be few 'c's about too.
: Dunno, this feels right. I just wish I could remember if it is.

Well, this very intuitive reasoning goes beyond my capacity.
However, I tried to work out the above formally, here it goes.
I haven't got much more time, unfortunately.

-------------------

radiation pressure.

there are N photons per unit of volume, say.
The probability that a photon at a distance r and an angle theta
hits a surface of area dA, is given by:
 
            ____|dA
         theta /|
              /
             /r 
            /
           *
         photon

                         dA.cos(theta)
                      ------------------
                           4.pi.r^2

(this is the fraction of the area seen by the photon over the
area of the sphere with radius r, because due to isotropy, every
bit of area of that sphere is equally likely to be hit by the photon)

Now consider the half-sphere on the left side of dA.
We cut it into different volume elements:
   
                  dV = r^2.dr.sin(theta).dtheta.dphi

We will count the number of photons that will make it through our
dA in a time dt.  In a time dt, a photon will travel c.dt, so 
r will go from 0 to c.dt.
phi is the angle around the symmetry axis and will go from 0 to 2pi.
theta will go from 0 to pi/2, since we want half a sphere.

Actually, we don't want to *count* the photons, but measure their
momentum component perpendicular to dA.  That one is:

             p_dA =  p_photon * cos(theta)

so, we have the following integral:

                                         dA.cos(theta)
Integral over r,theta,phi   {     p_dA.------------------. N.dV   }
                                           4.pi.r^2
                                                          number of photons
                               momentum 
                             contribution     
                                        probability to make it
                                             to dA 

                           dA.cos(theta)
 = p_photon * cos(theta).----------------- .N.r^2.dr.sin(theta).dtheta.dphi
                             4.pi.r^2

                  (integrated over)

then:
 * r disappears from the integrand, so integration over r yields c.dt
 * integration over phi yields 2.pi
 * reordering factors, we get:

                  1
 = p_photon.dA.N.---.c.dt.  ...
                  2

  ... .integral(theta:0-->pi/2) {cos^2(theta).sin(theta).dtheta}
                  
 substitution x = cos(theta) in this integral yields 1/3 for this one.

So we find: the momentum flux through a surface dA in a time dt is 
equal to:

            momflux = p_photon.N.(1/6).c.dt.dA

The momentum flux per unit of surface and time = the PRESSURE.

         so the pressure P = p_photon.N.(1/6).c

----------------

cheers,
Patrick.

--
Patrick Van Esch
http://www.iihe.ac.be/hep/pp/vanesch
mail:   [email-masked]
for PGP public key: finger [email-masked]

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Article: 13632 of sci.physics
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From: [email-masked] (Vanesch P.)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 22:59:04 GMT
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(my newsserver in DESY is down again...)

Oz ([email-masked]) wrote:
: In article <[email-masked]>, Patrick van Esch
: <[email-masked]> writes
: >
: >Well, this very intuitive reasoning goes beyond my capacity.

: Ho ho. ROFL. Beyond your capacity indeed!

You'll see why I say this...

: However it's nice to see it done properly with lots of integrations and
: stuff.

Yes, that's why I love physics :)

: >So we find: the momentum flux through a surface dA in a time dt is 
: >equal to:
: >
: >            momflux = p_photon.N.(1/6).c.dt.dA
: >
: >The momentum flux per unit of surface and time = the PRESSURE.
: >
: >         so the pressure P = p_photon.N.(1/6).c

: Oh dear. This isn't what I got.

: Perhaps Baez can help us here.

It is very close to what you had.
You had: P = D/3 with D = p_photon.N

Now, the c is no problem, because you put it to 1.

the only difference then is I have 1/6 and you have 1/3.

But if I understood you, you were counting pressure from both
sides of the surface, so you introduced a factor 2.  I looked
only at one side.  I think this is the way the pressure is really
defined, but I could be wrong.  So essentially you DID get the
same thing, but then you multiplied by 2, something I didn't do.
I think one shouldn't but I could be wrong.

So I had to do all this nasty math to get your solution, while
you just invented a story to tell... see what I meant with
"beyond my capacity" ? %-)

cheers,
Patrick.

: ------------------------------- 
: 'Oz     "When I knew little, all was certain. The more I learnt,
:         the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!howland.reston.ans.net!nntp.coast.net!zombie.ncsc.mil!news.mathworks.com!newsfeed.internetmci.com!btnet!zetnet.co.uk!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sat May  4 19:39:18 PDT 1996
Article: 13497 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 3 May 1996 20:33:57 +0100
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I have been chastised by the wizard for not doing as I have been told.

Sob, sob.

However, since he has been right in the past, I had better do what he
says before everyone kills this thread. If they haven't already that is.

In article <[email-masked]>, john baez <[email-masked]>
writes

NB How long has Baez been classifying this thread as @noise.ucr.edu???

>
>But if you're trying to understand T_{ii} --- and you are --- that's
>gonna be the flow in the i-direction of i-momentum.  And that is just
>the *pressure* in the i direction.  That's what you were getting at with
>the business of photons bouncing off a plate, right?
>
>>T_{00} is the density of the energy. Now for light, with zero mass, the 
>>energy equals the momentum. 
>
>Right.  From this you could derive a relation between the energy density and
>pressure for isotropic radiation.  That would be one way to solve this
>problem.  Actually you figured out a different way, which is also good. 
>
>But just for your own benefit, I think you should figure out the
>relation between the pressure P and the density rho for isotropic
>radiation.  After all, doesn't that sound like an interesting thing to
>know?  It's probably pretty simple, too.

OK. Lets consider a plate in a sea of photons. Let's have the energy 
density as D and the plate of area A. For a photon the energy equals the 
momentum. 

Just to keep the model clear and simple for me lets consider this 
density to be comprised of N photons of momentum Q each per unit volume.

In a short time dt there will be dn collisions with the plate. Now if 
ALL the photons were (laserlike) heading towards my plate then in time 
dt all the photons in a column of volume Acdt would strike my plate, and 
no others. So in this case a volume Acdt worth of photons since they 
travel at c. However in Baezworld c=1 so

dn = NAdt 

Since each photon has momentum Q this will give me a total momentum 
transferred to the plate of

dp = QNAdt => dp/dt = QNA

and we note that the energy density is QN = D

and the force on the plate is dp/dt so the force on the plate due to 
incoming photons will be

DA and thus the pressure will be P = DA/A = D

No I have an irresistible urge to have a reflecting plate here, just to 
avoid the photons accumulating on my plate. In this case they will 
'rebound' and give me double this momentum transfer. I expect I can 
offer more plausible reasons for wanting to chose this.

So P = 2D for a laserbeam intersecting my nice shiny plate.

Now we have been told to get the pressure for an isotropic sea of 
photons. Well, we must take a third of the figure derived above because 
the resolved components perpendicular to the plate (ie the laser is in 
the x-direction, so the y & z directions will transfer no momentum). 
Also half the photons are heading in the -x direction so won't hit the 
plate, so we had better divide by two as well. 

So my supercrude derivation comes out to

P = D/3

So the pressure is 1/3 of the energy density.

Of course in real life there ought to be few 'c's about too.
Dunno, this feels right. I just wish I could remember if it is.

>The derivation is quite simple and I think I said it before.  For
>example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
>for visual vividness that this flow is carried by little ball-shaped
>atoms.  Then if you put a wall in their way they push on the wall in
>the x direction, with a certain pressure.  This pressure is a force per
>unit area, which is just the same as "momentum per time per unit area".
>This is given by the flow of x-momentum in the x-direction!

I have an irresistible urge to imagine a flow of momentum through a 
surface (instead of a plate). The only problem I have with this is that 
the momentum flowing from right to left is cancelled by the momentum 
flowing from left to right. Now you are going to tell me that if I stick 
my reflective plate instead of the surface (an infinitely thin one of 
course) the distribution of photons will remain quite isotropic. Thus 
the pressure is everywhere uniform. I still find this difficult to match 
unless you are going to start talking about the flow of momentum in the 
-x direction too.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May  4 20:53:31 PDT 1996
Article: 13675 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 18:57:15 -0700
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>NB How long has Baez been classifying this thread as @noise.ucr.edu???

Our newsreader is now called "noise", and aptly so.

>>... I think you should figure out the
>>relation between the pressure P and the density rho for isotropic
>>radiation.  After all, doesn't that sound like an interesting thing to
>>know?  It's probably pretty simple, too.

>OK. 

[Nice simple derivation deleted]

>So the pressure is 1/3 of the energy density.

Right.  Briefly: if all the photons were moving back and forth in the x
direction (50% moving back and 50% moving forth), the pressure in the x
direction would equal the energy density, since the momentum of a photon
equals its energy.  But if we consider photons moving in the x, y, and z
directions, only those moving in the x direction will contribute to the
pressure in the x direction, while all of them will contribute to the
energy density.  Thus the pressure in the x direction --- or any
direction --- will be 1/3 the energy density.

This may seem unrealistic, because of course photons don't just move in
the x, y, and z directions; they can go every which way.  However, the
stress-energy tensor we have just computed is

E  0   0   0
0 E/3  0   0
0  0  E/3  0
0  0   0  E/3

and if we apply a rotation to this tensor it doesn't change.  Thus
if we take our batch of photons moving along the x, y, and z axes, and
throw in other batches of photons moving along the x', y', and z' axes,
for various rotated coordinate systems (x',y',z'), we will continue to
get a stress-energy tensor of the above form.  So even when the photons
have an equal chance of going in *any* direction, the stress-energy
tensor will be of the above form.

Sneaky, huh?  Of course we could also have done the calculation in a
brutally straightfoward manner, assuming from the start that we had a
bunch of photons with equal probability going in every direction.  I
think someone has posted a derivation like that.  The above way seems
easier, though.

>Of course in real life there ought to be few 'c's about too.

Sure, add them to taste. 

>>The derivation is quite simple and I think I said it before.  For
>>example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
>>for visual vividness that this flow is carried by little ball-shaped
>>atoms.  Then if you put a wall in their way they push on the wall in
>>the x direction, with a certain pressure.  This pressure is a force per
>>unit area, which is just the same as "momentum per time per unit area".
>>This is given by the flow of x-momentum in the x-direction!

>I have an irresistible urge to imagine a flow of momentum through a 
>surface (instead of a plate). The only problem I have with this is that 
>the momentum flowing from right to left is cancelled by the momentum 
>flowing from left to right. 

You have to be careful with your minus signs in this game: x-momentum
flowing in the x-direction works out to be the SAME as (-x)-momentum
flowing in the (-x)-direction, so even though there are atoms going both
ways, their contributions to T_{xx} don't cancel out: they add up.  I'm
not sure if that's relevant to your worries here, but it's important.

Why does it work out this way?  Well, let me just say that it has a lot
to do with the fact that T_{xx} is not a component of a vector.  It's a
component of a rank 2 tensor.  When you perform a coordinate
transformation like x -> -x and work out how T_{xx} changes, you get not
one minus sign but two, so T_{xx} doesn't change at all.

On the other hand, something like T_{tx}, which is the density of
x-momentum, does change sign when we do x -> -x.  Basically because
there's just one x in "T_{tx}".  And that's related to how when you've
got an equal flow of atoms going both ways, their contributions to
T_{tx} can cancel out.

I hope you're not mixing up the flow of x-momentum in the x-direction,
T_{xx}, with the density of x-momentum, T_{tx}.  They are very different
things.  In the rest frame of a gas, the first one is the pressure and
the second one is zero.

>Now you are going to tell me that if I stick 
>my reflective plate instead of the surface (an infinitely thin one of 
>course) the distribution of photons will remain quite isotropic. Thus 
>the pressure is everywhere uniform. I still find this difficult to match 
>unless you are going to start talking about the flow of momentum in the 
>-x direction too.

I hope the above stuff helps.  You didn't make it very precise what you
thought the problem was, so I just shot off random remarks in every
direction and hoped a few hit.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May  4 20:57:08 PDT 1996
Article: 13683 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 19:25:09 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

>Well, this very intuitive reasoning goes beyond my capacity.
>However, I tried to work out the above formally, here it goes.

For some people the intuitive approach is easier.  For others it's
easier to calculate things out and see what happens!  It's fun to see
both approaches.

>So we find: the momentum flux through a surface dA in a time dt is 
>equal to:
>
>            momflux = p_photon.N.(1/6).c.dt.dA

Looks good.

>The momentum flux per unit of surface and time = the PRESSURE.
>
>         so the pressure P = p_photon.N.(1/6).c
>

Hmm.  Here I worry a bit.  Using the fact that the momentum of a photon
equals its energy, together with the fact that c = 1, you are claiming
the momentum flux per unit of surface and time is 1/6 the energy
density.

How does this jibe with Oz's calculation that the pressure is 1/3 the
energy density?  

Well, maybe the pressure is not equal to the momentum flux you
calculated, but twice that!  

I can think of two arguments for this.  The down-to-earth way is to
imagine you have photons bouncing off the wall of a silvered box.  When
they bounce off, the component of their momentum perpendicular to the
wall changes sign.  So the change of momentum in this direction is twice
the original momentum in that direction.  So the pressure is twice the
momentum flux you calculated.  

Another way is more abstract.  You imagine you are in the middle of the
gas, and you realize that pressure in the x direction is the flow of
x-momentum in the x-direction.  But you are very careful and you notice
that computing the flow of x-momentum in the x-direction requires you to
keep track of p_x v_x for particles having v_x negative as well as
positive.  So you do the calculation you just did, but doing the
integral from theta = 0 to theta = 2 pi.  All the way around.   This
doubles the answer.  

This is related to a point I just made in a post to Oz, that
"x-momentum in the x-direction" is also "(-x)-momentum in the
(-x)-direction".  Rank 2 tensors are a little counter-intuitive since we 
are more used to vectors.  This makes concepts like pressure, stress,
strain, and the like seem rather sneaky.  

I used to wonder: if I pull on both ends of a rope, the forces cancel
out, so what exactly is this "tension" the rope feels, anyway?  Well,
it's a flow of (-x)-momentum in the x-direction, and an equal flow of
x-momentum in the (-x)-direction.  Both contribute to making T_{xx} be
negative; that's tension.  

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May  4 20:59:45 PDT 1996
Article: 13702 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 20:40:05 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>
>writes

>>So we should go back and try to understand the
>>relation between the metric, the Christoffel symbols, and the Riemann
>>tensor more deeply.

>We should. I get the feeling that we will too.

Well, basically, that's the only thing left for me to do.  Once you
understand that stuff you'll understand the main ideas of general
relativity.  

But first I want to say a couple more things about this big bang
business.  I never quite got around to saying how I would work this
problem.  

We started by assuming a metric of the form

g = -dt^2 + R(t)^2 (dx^2 + dy^2 + dz^2).

and computing its Einstein tensor.  With a fair amount of pain and
suffering we arrived at:

 G_{00} = 3 (R')^2 / R^2          G_{ii} = - 2 R" R - (R')^2     (*)

with the rest of the components being zero.  

Since Einstein's equation says G_{ab} = T_{ab}, we know G_{00} and
G_{ii} should be related to the energy density E and the pressure P,
respectively.

In fact, I said that T_{00} = E and T_{ii} = P when we are working in
coordinates at a point that make the metric look like the Minkowski
metric.  But we're not.  To get our metric to look like the Minkowski
metric AT A GIVEN POINT OF SPACETIME (t,x,y,z), we can use new
coordinates t' = t, x' = x/R(t), y' = y/R(t), z' = z/R(t).  In THOSE
coordinates we'll have T_{00} = E, T_{ii} = P.  But that means --- doing
a quick little calculation --- that in our original coordinates we
have

 T_{00} = E                         T_{ii} = P R^2               (**)

Okay, so using (*) and (**) we get

 3 (R')^2 = E R^2      - 2 R" R - (R')^2 = P R^2             

The second equation is a bit disgusting, so go ahead and use the first
to replace the (R')^2 term by something more palatable.   We thus have

 3 (R')^2 = E R^2       R" = (-1/6) (E + 3P) R                  (***)

Now to do anything with these we need to assume something about the
energy density E and the pressure P.  If we assume the world is full of
"dust", we have P = 0.  We can then solve the equations --- Oz and Esch
and maybe some other people did --- and see that R(t) is proportional to
t^{2/3}.  So the universe expands forever, but ever more slowly.  If we
assume the world is full of "radiation", we have P = E/3.  One can again
solve the equations and see that R(t) is proportional to t^{1/2}.  Again
the universe expands forever.  

Note that the presence of the radiation pressure *slows* the expansion
of the universe, after a while, since t^{1/2} grows more slowly than
t^{2/3}.  This is a typical example of how the gravity created by
pressure makes things contract --- despite the usual way pressure makes
things expand.  Of course, we can also see that pressure slows the
expansion of the universe directly by looking at the second equation in
(***).  

I didn't show you how to solve the equations above, because it's more
fun to do that in combination with some other new ideas.   I'll post
this separately.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May  4 21:00:30 PDT 1996
Article: 13704 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 4 May 1996 20:50:39 -0700
Organization: University of California, Riverside
Lines: 77
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In my last post on this thread I looked at something we have been
studying for a few weeks now: a spatially flat homogeneous universe.  
This is a special case of the big bang model.  At each time space is 
Euclidean R^3.  Let's call this the "k = 0" case, for reasons to become
clear soon enough.  

We got the following equations:

 3(R')^2 = E R^2       R" = (-1/6) (E + 3P) R    

where E is the energy density, P is the pressure, and R measures the
"size of the universe", or more precisely, the distance between two
nearby points in space.  

It's also interesting to study what happens when space is positively
curved and homogeneous --- the "k = 1" case.  In this case, space is a
3-sphere, or S^3.  And it's also fun to study what happens when space is
negatively curved and homogeneous --- the "k = -1" case.  In this case,
space is hyperbolic 3-space, or H^3.  Working out the Einstein tensor in
these other cases is even more of a pain than in the k = 0 case.  But
the final result is strikingly similar for all three cases.  We can
handle all 3 at once if we take the equations above and stick in a new
term depending on k: 

 3(R')^2 = E R^2 - 3k      R" = (-1/6) (E + 3P) R             (*)

Here as before R measures the distance between nearby points with fixed
coordinates in space at time t.  

Let's play with these and see what we get.  Let me do the case where the
universe is full of dust so P = 0.  We get

 3(R')^2 = E R^2 - 3k      R" = - ER/6                        (**)

Now, you might wonder whether we need another equation expressing
"conservation of dust".  We don't: it follows from the equations above
(which are really a special case of Einstein's equation).  To see this,
differentiate the first equation with respect to t and get

 6 R' R" = E' R^2 + 2 E R R'

and plug the second equation in to this getting

- E R R' = E' R^2 + 2 E R R'

or
       
       0 = E' R^2 + 3 E R R'
       
         = E' R^3 + 3 E R^2 R'
  
         = (d/dt) (E R^3)

Don't ask me how I knew to do that, or I will tell you, and then I'll
have to kill you, because it's top secret.  The point is that the
density of dust, E, times the volume of a patch of space, R^3, stays
constant.  So we have conservation of dust.

Let's write 

      D = E R^3

for this conserved quantity, and use this to simplify the equations (**).
We get

 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

Okay.  Now I claim these equations are incredibly familiar to anybody
who has ever studied gravity at all.  Not general relativity, classical
Newtonian gravity!  What are they????   

Once you figure out that puzzle it should be easy to understand the big
bang in all three cases: flat, positively curved, and negatively curved.
At least for a universe with zero pressure.

From noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!gatech!swrinde!newsfeed.internetmci.com!news.dacom.co.kr!arclight.uoregon.edu!dispatch.news.demon.net!demon!upthorpe.demon.co.uk!Oz Sun May  5 14:25:28 PDT 1996
Article: 13747 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Thu, 2 May 1996 18:12:25 +0100
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>
>>
>>G_{00} = k/(R^4) = 3R'^2/R^2 => k = 3R'^2 R^2 which has solutions
>>
>>R = (4k/3)^(1/4)  t^(1/2) = At^(1/2) and t = constant.
>
>Yes!  Correct.
>
>Good work.

Good grief!

>
>>I'm not thinking straight!
>
>You're not supposed to.  This is curved spacetime.
>
>>Perhaps if someone could give a little definition of pressure?
>
>The derivation is quite simple and I think I said it before.  For
>example, T_{xx} is the flow in the x direction of x-momentum.  Suppose
>for visual vividness that this flow is carried by little ball-shaped
>atoms.  Then if you put a wall in their way they push on the wall in
>the x direction, with a certain pressure.  This pressure is a force per
>unit area, which is just the same as "momentum per time per unit area".
>This is given by the flow of x-momentum in the x-direction!

Poxy Wizard. He just doesn't let you get away with a damned thing.
Mutter, mutter, grumble, grumble, gripe, gripe, moan, moan.

Hokay. We do it the simple minded numskulls method for peasants.

Lets erect a plate in space and consider the change of momentum of these 
pesky photons. Let us have it of area A. Now let's see, hmmmmmmm, well 
since we have an isotopic thingy then half the phots will be heading 
away, so we have a factor of 1/2. Hmmm, two thirds will be non 
interacting as they are parallel to the plate (resolving here and there) 
so we should only count a third. So we should only count a sixth of the 
phots.

I said this was physics for peasants, didn't I? Well it's going to get 
worse, and cruder, I promise. Now I intend to have my phots travelling 
at c for now, so I can keep some sort of track.

Right. Now let's see. In time t we will have had a column of phots ct 
long bash into our plate transferring 2hn (n=nu) each (on account of 
these are reflecting phots). So the number of phots is the number in a 
volume A ct. The momentum is thus h(nu) of that, lets make h=1 to give 
us a momentum of Act(nu). Well, the volume is varying as R^3 but the 
number of phots stays the same so the momentum is proportional to 
Act(nu), say kAct(nu) (probably a different k, maybe check later)

So pressure = F/A = (1/A) dp/dt = (1/A) d(kAct(nu))/dt = kc(nu) and c=1

so the pressure varies as k(nu). Oh sugar, how did I blow this when I 
used the back of that envelope? Me superthick.

Now nu varies as 1/R so the pressure varies as 
k(some constant or other)/R or K R(-1).

Which is what we had as the form of our solution for G_{00}.

Apart from a bit of tidying up that's done. Phew!
Wot a mess you get when you try to work things out straight
onto the screen.

So the solution for a cold dusty universe is R = A t^(2/3)
For a universe comprised of light only it's R = B t^(1/2)

Hmmmm. I would guess that with a universe filled with a bit of light and 
a bit of warmth would be er, um probably somewhere in between.

Ahem.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 13880 of sci.physics
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From: [email-masked] (Brett McInnes)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 6 May 1996 02:24:52 GMT
Organization: National University of Singapore
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john baez ([email-masked]) wrote:

:  3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

: Okay.  Now I claim these equations are incredibly familiar to anybody
: who has ever studied gravity at all.  Not general relativity, classical
: Newtonian gravity!  What are they????   

They are formally analogous to the equations describing the motion
of a stone thrown into the air. Which is utterly irrelevant to the
present case, because the reason stones fall down is because of the
fact that the situation is NOT isotropic, whereas the above 
equations describe a situation which IS isotropic. So the formal
analogy may be useful for guessing the solution of the differential
equations, but its physical significance is nil. 

The point is an interesting one because it leads to the fundamental
reason for the fact that GR cosmology works so much better than
Newtonian cosmology. The reason is this. The basic object in 
Newtonian gravity is the force vector. The basic object in GR
is the metric tensor. An isotropic vector has to be trivial, ie
the zero vector. That is why Newtonian gravity cannot handle
isotropy. But a spatially isotropic *tensor* on space-time need
not be trivial. And that's why GR cosmology works. This is why
you *can't* explain,eg, the deceleration parameter by saying that
a gravitational pull "slows everything down". In an isotropic
universe, any "gravitational pull" would have to be the same in
all directions, and hence would cancel. So Newtonian analogies
work particularly badly in cosmology. Note that the usual way
to do Newtonian cosmology is to start with a ball of galaxies
floating in empty space: this is not isotropic except around
one point, and this anisotropy is precisely what causes the 
deceleration.

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Article: 13896 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 3 May 1996 08:03:00 +0100
Organization: Oz
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Ok, it's early morning. No phones. No kids. All is peace.

Time to get one's thoughts straight. 

To get the appropriate diff. equ. I need to have the pressure varying as 
1/R^2. However I really cannot get this, no matter how hard I try. I 
need someone to point out my mistake(s).

Now because c is a constant it is clear that the number of photons 
crossing a surface whose co-ordinates are fixed in our co-ordinate 
system is going to be inversely proportional to R. This is quite 
obvious. As a cube containing the same number of photons expands as R, 
the number of photons 'within reach' of one face in a short time 
interval t is those within a small slice ct from the face. As each side 
increases as R the proportion that can reach the face declines as 1/R. 
Alternatively you can take the density as varying as 1/R^3, the face 
increasing as R^2 and the small slice is of length ct, which is a 
constant, to get the same result. This 'feels' correct.

The force on the side will be proportional to the number of photons 
striking the face in a given time, and the momentum of the photons. 
Unless I've missed something gross, I can't see anything wrong here.

The momentum of the photons declines as 1/R due to 'stretching' of their 
wavelength. Really, I would prefer to have this fall out of the solution 
rather than put it in by hand. However this still 'feels' right.

So the force on a face of constant co-ordinate varies as 1/R^2.

However the area of this face increases as R^2 so the pressure decreases 
as 1/R^4.

What am I doing wrong?

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sun May  5 21:43:43 PDT 1996
Article: 13902 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 21:19:17 -0700
Organization: University of California, Riverside
Lines: 48
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NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> [email-masked] (Brett McInnes) writes:
>john baez ([email-masked]) wrote:
>
>:  3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)
>
>: Okay.  Now I claim these equations are incredibly familiar to anybody
>: who has ever studied gravity at all.  Not general relativity, classical
>: Newtonian gravity!  What are they????   

>They are formally analogous to the equations describing the motion
>of a stone thrown into the air.

Right.  

>Which is utterly irrelevant to the
>present case, because the reason stones fall down is because of the
>fact that the situation is NOT isotropic, whereas the above 
>equations describe a situation which IS isotropic. So the formal
>analogy may be useful for guessing the solution of the differential
>equations, but its physical significance is nil. 

Please, don't get carried away with enthusiasm now just because you saw
the analogy!  We don't want anyone bubbling over with delight here: this
is physics.

To me, any analogy that lets you guess the solution of a physics problem
is "physically significant".  After all, part of the business of physics is
to solve problems.  Armed with the analogy you mention, the students
should easily be able to guess the behavior of the solutions of (***).

Nonetheless, your sober warnings should serve as a useful corrective to
any dangerous urge to read too much into this analogy and confuse
general relativity with Newtonian gravity.  I certainly didn't mean to
lead them into this temptation.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sun May  5 22:29:24 PDT 1996
Article: 13915 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 21:39:19 -0700
Organization: University of California, Riverside
Lines: 39
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References: <[email-masked]> <[email-masked]> <[email-masked]>
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In article <[email-masked]> [email-masked] writes:
>
>Ok, it's early morning. No phones. No kids. All is peace.
>
>Time to get one's thoughts straight. 
>
>To get the appropriate diff. equ. I need to have the pressure varying as 
>1/R^2. However I really cannot get this, no matter how hard I try. I 
>need someone to point out my mistake(s).

Well, in a recent post I showed that for the spatially flat homogeneous
universe full of dust, we can get use the basic equations

 3(R')^2 = E R^2       R" = - ER/6                 

to show that the energy density E varies as 1/R^3.  Maybe you can
pull a similar trick to get a relation between P and R from the
equations for a spatially flat homogeneous universe full of radiation:

 (R')^2 = P R^2       R" = -P R    

Remember, you and the rest of the GR gang showed that

 3 (R')^2 = E R^2      - 2 R" R - (R')^2 = P R^2             

for any spatially flat homogeneous cosmology.  I brought these equations
into the following more pleasant form:

3 (R')^2 = E R^2       R" = (-1/6) (E + 3P) R     

Since you showed E = 3P for radiation, you know

 (R')^2 = P R^2       R" = -P R    

So, play around with this and see what you get.

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Article: 13974 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 3 May 1996 09:41:07 -0700
Organization: University of California, Riverside
Lines: 45
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References: <[email-masked]> <[email-masked]>
NNTP-Posting-Host: guitar.ucr.edu

In article <[email-masked]> "Alexander E. Meier" <[email-masked]> writes:
>I got a thought..
>somewhere in those Einstein equations, you have the tensor T which is
>the energy momentum tensor right? why not use the one SR gives for
>electromagnetic waves?

Indeed, why not?  Very good idea.  If you supplement the resulting
version of Einstein's equation by Maxwell's equation, you get the theory
of gravity coupled to electromagnetism!  The so-called "Einstein-
Maxwell" theory.  A very pretty theory indeed.  A world of only gravity
and light --- what could be more romantic?  For example, it's possible
to get solutions corresponding to a ball of electromagnetic fields held
together by its own gravity.

I described the tensor T for electromagnetism in a recent reply to a
post of yours.  Not in a slick covariant form, but component by
component in an arbitrary orthonormal coordinate system.  It's sort of
easier to understand it's physical meaning that way.  Let me repeat it,
just for fun:

It was Michael Faraday who noticed that electric and magnetic field
lines have tension and pressure!  Tension is simply negative pressure.
To be precise, if you have an electric field, there is a tension of
E^2/2 along the lines of force, and a pressure of E^2/2 in both
directions perpendicular to the lines of force, where E is the electric
field.  Similarly, there is a tension of B^2/2 along magnetic force
lines, and a pressure of B^2/2 in both directions perpendicular to the
magnetic force lines.

Amazing what they never tell you in school, eh?  If I weren't so
respectable you might think I was making that stuff up.  It can be found
in any decent book on electromagnetism, buried in the equations.

There is also an energy density of (E^2 + B^2)/2 due to an
electromagnetic field, and also a flux of energy equal to E x B.  This
last thing is often called the Poynting vector.

Knowing this stuff, and knowing the definition of the stress-energy
tensor, and the fact that it's symmetric (T_{ab} = T_{ba}), one can work
out the stress-energy tensor for electromagnetism.  Just rememember that
in orthonormal coordinates, T_{ab} is the flow in the a direction of
momentum in the b direction.

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Article: 14017 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 4 May 1996 18:21:49 +0100
Organization: Oz
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In article <[email-masked]>, Patrick van Esch
<[email-masked]> writes
>
>Well, this very intuitive reasoning goes beyond my capacity.

Ho ho. ROFL. Beyond your capacity indeed!

I am the dim and ignorant one! Oh mighty postgrad.

However it's nice to see it done properly with lots of integrations and
stuff.

>However, I tried to work out the above formally, here it goes.
>I haven't got much more time, unfortunately.
>
>-------------------
>
>radiation pressure.
>
>there are N photons per unit of volume, say.
>The probability that a photon at a distance r and an angle theta
>hits a surface of area dA, is given by:
> 
>            ____|dA
>         theta /|
>              /
>             /r 
>            /
>           *
>         photon
>
>                         dA.cos(theta)
>                      ------------------
>                           4.pi.r^2

<Snip of nasty maths>

>So we find: the momentum flux through a surface dA in a time dt is 
>equal to:
>
>            momflux = p_photon.N.(1/6).c.dt.dA
>
>The momentum flux per unit of surface and time = the PRESSURE.
>
>         so the pressure P = p_photon.N.(1/6).c

Oh dear. This isn't what I got.

Perhaps Baez can help us here.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 14043 of sci.physics
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From: [email-masked] (Bill Rowe)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 03 May 1996 17:09:33 -0600
Organization: none apparent
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In article <[email-masked]>,
[email-masked] wrote:

>To get the appropriate diff. equ. I need to have the pressure varying as 
>1/R^2. However I really cannot get this, no matter how hard I try. I 
>need someone to point out my mistake(s).

>Now because c is a constant it is clear that the number of photons 
>crossing a surface whose co-ordinates are fixed in our co-ordinate 
>system is going to be inversely proportional to R. This is quite 
>obvious. As a cube containing the same number of photons expands as R, 
>the number of photons 'within reach' of one face in a short time 
>interval t is those within a small slice ct from the face. As each side 
>increases as R the proportion that can reach the face declines as 1/R. 
>Alternatively you can take the density as varying as 1/R^3, the face 
>increasing as R^2 and the small slice is of length ct, which is a 
>constant, to get the same result. This 'feels' correct.

Hmmm. I would note that pressure normally thought of as force per unit
area is the same as energy per unit volume. So it would seem the energy
density ought to be something like

            N h nu
            ------
             R^3

which also ought to be the pressure against the "walls". At least this is
dimensionally correct even though it does't have a 1/R^2 term.

>The force on the side will be proportional to the number of photons 
>striking the face in a given time, and the momentum of the photons. 
>Unless I've missed something gross, I can't see anything wrong here.

>The momentum of the photons declines as 1/R due to 'stretching' of their 
>wavelength. Really, I would prefer to have this fall out of the solution 
>rather than put it in by hand. However this still 'feels' right.

I am guessing the idea of momentum declining as 1/R comes from the
redshift that is expected. While I can't argue with the redshift, I
definitely don't care for the idea of momentum declining. There shouldn't
be anything about the metric we are discussion which would produce a
violation of conservation of momentum or energy. So if the momentum
decreases, where is the corresponding increase?

>So the force on a face of constant co-ordinate varies as 1/R^2.

>However the area of this face increases as R^2 so the pressure decreases 
>as 1/R^4.

>What am I doing wrong?

I am not actually sure what is wrong. In terms of momentum, pressure would
be change in momentum per unit time per unit area. I suspect the time
needs to be expressed in terms of R. Once this is done, I suspect you get
the same result as I've given above for energy density.

-- 
"Against supidity, the Gods themselves contend in vain"

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Article: 14043 of sci.physics
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From: [email-masked] (Bill Rowe)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 03 May 1996 17:09:33 -0600
Organization: none apparent
Lines: 58
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In article <[email-masked]>,
[email-masked] wrote:

>To get the appropriate diff. equ. I need to have the pressure varying as 
>1/R^2. However I really cannot get this, no matter how hard I try. I 
>need someone to point out my mistake(s).

>Now because c is a constant it is clear that the number of photons 
>crossing a surface whose co-ordinates are fixed in our co-ordinate 
>system is going to be inversely proportional to R. This is quite 
>obvious. As a cube containing the same number of photons expands as R, 
>the number of photons 'within reach' of one face in a short time 
>interval t is those within a small slice ct from the face. As each side 
>increases as R the proportion that can reach the face declines as 1/R. 
>Alternatively you can take the density as varying as 1/R^3, the face 
>increasing as R^2 and the small slice is of length ct, which is a 
>constant, to get the same result. This 'feels' correct.

Hmmm. I would note that pressure normally thought of as force per unit
area is the same as energy per unit volume. So it would seem the energy
density ought to be something like

            N h nu
            ------
             R^3

which also ought to be the pressure against the "walls". At least this is
dimensionally correct even though it does't have a 1/R^2 term.

>The force on the side will be proportional to the number of photons 
>striking the face in a given time, and the momentum of the photons. 
>Unless I've missed something gross, I can't see anything wrong here.

>The momentum of the photons declines as 1/R due to 'stretching' of their 
>wavelength. Really, I would prefer to have this fall out of the solution 
>rather than put it in by hand. However this still 'feels' right.

I am guessing the idea of momentum declining as 1/R comes from the
redshift that is expected. While I can't argue with the redshift, I
definitely don't care for the idea of momentum declining. There shouldn't
be anything about the metric we are discussion which would produce a
violation of conservation of momentum or energy. So if the momentum
decreases, where is the corresponding increase?

>So the force on a face of constant co-ordinate varies as 1/R^2.

>However the area of this face increases as R^2 so the pressure decreases 
>as 1/R^4.

>What am I doing wrong?

I am not actually sure what is wrong. In terms of momentum, pressure would
be change in momentum per unit time per unit area. I suspect the time
needs to be expressed in terms of R. Once this is done, I suspect you get
the same result as I've given above for energy density.

-- 
"Against supidity, the Gods themselves contend in vain"

From noise.ucr.edu!news.service.uci.edu!usc!howland.reston.ans.net!surfnet.nl!swsbe6.switch.ch!news.belnet.be!news.vub.ac.be!orca!pvanesch Mon May  6 23:03:03 PDT 1996
Article: 14089 of sci.physics
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From: [email-masked] (Vanesch P.)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 09:43:00 GMT
Organization: Brussels Free Universities VUB/ULB
Lines: 37
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References: <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: orca.vub.ac.be

Vanesch P. ([email-masked]) wrote:
: (my newsserver in DESY is down again...)

: : >            momflux = p_photon.N.(1/6).c.dt.dA
: : >
: : >The momentum flux per unit of surface and time = the PRESSURE.
: : >
: : >         so the pressure P = p_photon.N.(1/6).c

: : Oh dear. This isn't what I got.

: : Perhaps Baez can help us here.

: It is very close to what you had.
: You had: P = D/3 with D = p_photon.N

: Now, the c is no problem, because you put it to 1.

: the only difference then is I have 1/6 and you have 1/3.

: But if I understood you, you were counting pressure from both
: sides of the surface, so you introduced a factor 2.  I looked
: only at one side.  I think this is the way the pressure is really
: defined, but I could be wrong.  So essentially you DID get the
: same thing, but then you multiplied by 2, something I didn't do.
: I think one shouldn't but I could be wrong.

Well, as John points out further, I was wrong.
The reflective stuff you used was my problem, but I had forgotten
indeed to take into account the -px flux with negative "flux"...

I integrated over half a sphere and should have done it over
the whole sphere.

cheers,
Patrick.

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Article: 14088 of sci.physics
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From: [email-masked] (Vanesch P.)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 09:37:47 GMT
Organization: Brussels Free Universities VUB/ULB
Lines: 105
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References: <[email-masked]> <[email-masked]> <[email-masked]> <[email-masked]>
NNTP-Posting-Host: orca.vub.ac.be

john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

: >Well, this very intuitive reasoning goes beyond my capacity.
: >However, I tried to work out the above formally, here it goes.

: For some people the intuitive approach is easier.  For others it's
: easier to calculate things out and see what happens!  It's fun to see
: both approaches.

: >So we find: the momentum flux through a surface dA in a time dt is 
: >equal to:
: >
: >            momflux = p_photon.N.(1/6).c.dt.dA

: Looks good.

: >The momentum flux per unit of surface and time = the PRESSURE.
: >
: >         so the pressure P = p_photon.N.(1/6).c
: >

: Hmm.  Here I worry a bit.  Using the fact that the momentum of a photon
: equals its energy, together with the fact that c = 1, you are claiming
: the momentum flux per unit of surface and time is 1/6 the energy
: density.

: How does this jibe with Oz's calculation that the pressure is 1/3 the
: energy density?  

: Well, maybe the pressure is not equal to the momentum flux you
: calculated, but twice that!  

: I can think of two arguments for this.  The down-to-earth way is to
: imagine you have photons bouncing off the wall of a silvered box.  When

Yes, this is what Oz did and that was what I was objecting to.  After all,
there isn't a silvered box.  But, as you point out below, I should indeed
have taken into account the -px flux coming from the other side.
When you look to my little calculation I posted, you'll see I integrated
only over half a sphere (theta : 0 --> pi/2).  It should have been a full
sphere and that will give me an extra factor of 2.

: they bounce off, the component of their momentum perpendicular to the
: wall changes sign.  So the change of momentum in this direction is twice
: the original momentum in that direction.  So the pressure is twice the
: momentum flux you calculated.  

: Another way is more abstract.  You imagine you are in the middle of the
: gas, and you realize that pressure in the x direction is the flow of
: x-momentum in the x-direction.  But you are very careful and you notice
: that computing the flow of x-momentum in the x-direction requires you to
: keep track of p_x v_x for particles having v_x negative as well as
: positive.  So you do the calculation you just did, but doing the
: integral from theta = 0 to theta = 2 pi.  All the way around.   This
: doubles the answer.  

Right.
I should have checked my answer before posting it but I was in a hurry.

: This is related to a point I just made in a post to Oz, that
: "x-momentum in the x-direction" is also "(-x)-momentum in the
: (-x)-direction".  Rank 2 tensors are a little counter-intuitive since we 
: are more used to vectors.  This makes concepts like pressure, stress,
: strain, and the like seem rather sneaky.  

: I used to wonder: if I pull on both ends of a rope, the forces cancel
: out, so what exactly is this "tension" the rope feels, anyway?  Well,
: it's a flow of (-x)-momentum in the x-direction, and an equal flow of
: x-momentum in the (-x)-direction.  Both contribute to making T_{xx} be
: negative; that's tension.  

Well, that's the most tricky part !  In material science, what you
are supposed to do is to make a small surface cut (perpendicular to 
direction j) and the force needed to make up for that cut has a component
i, then the material stress tensor component sigma_ij is exactly that.
So here you take only the force coming "from one side".  That was what
confused me and made me integrate over only half a sphere.
So in your example (I'm happy you take it because it illustrates exactly
what's tricky), you should cut the rope in 2 pieces, see what force you
need to apply on ONE of the pieces to compensate for the loss of the
other one,... which is ONLY ONE TIMES THE FORCE.
So in your rope example you exactly don't have the problem that is
showing up with the photons !
(not that I want to hide behind it, I was plainly wrong there but it
illustrates how nasty these things are).

cheers,
Patrick.

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Article: 14093 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sun, 5 May 1996 09:55:49 +0100
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>
>I used to wonder: if I pull on both ends of a rope, the forces cancel
>out, so what exactly is this "tension" the rope feels, anyway?  Well,
>it's a flow of (-x)-momentum in the x-direction, and an equal flow of
>x-momentum in the (-x)-direction.  Both contribute to making T_{xx} be
>negative; that's tension.  

Oooh. I like that!

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 14088 of sci.physics
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From: [email-masked] (Vanesch P.)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 09:37:47 GMT
Organization: Brussels Free Universities VUB/ULB
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john baez ([email-masked]) wrote:
: In article <[email-masked]> [email-masked] (Patrick van Esch) writes:

: >Well, this very intuitive reasoning goes beyond my capacity.
: >However, I tried to work out the above formally, here it goes.

: For some people the intuitive approach is easier.  For others it's
: easier to calculate things out and see what happens!  It's fun to see
: both approaches.

: >So we find: the momentum flux through a surface dA in a time dt is 
: >equal to:
: >
: >            momflux = p_photon.N.(1/6).c.dt.dA

: Looks good.

: >The momentum flux per unit of surface and time = the PRESSURE.
: >
: >         so the pressure P = p_photon.N.(1/6).c
: >

: Hmm.  Here I worry a bit.  Using the fact that the momentum of a photon
: equals its energy, together with the fact that c = 1, you are claiming
: the momentum flux per unit of surface and time is 1/6 the energy
: density.

: How does this jibe with Oz's calculation that the pressure is 1/3 the
: energy density?  

: Well, maybe the pressure is not equal to the momentum flux you
: calculated, but twice that!  

: I can think of two arguments for this.  The down-to-earth way is to
: imagine you have photons bouncing off the wall of a silvered box.  When

Yes, this is what Oz did and that was what I was objecting to.  After all,
there isn't a silvered box.  But, as you point out below, I should indeed
have taken into account the -px flux coming from the other side.
When you look to my little calculation I posted, you'll see I integrated
only over half a sphere (theta : 0 --> pi/2).  It should have been a full
sphere and that will give me an extra factor of 2.

: they bounce off, the component of their momentum perpendicular to the
: wall changes sign.  So the change of momentum in this direction is twice
: the original momentum in that direction.  So the pressure is twice the
: momentum flux you calculated.  

: Another way is more abstract.  You imagine you are in the middle of the
: gas, and you realize that pressure in the x direction is the flow of
: x-momentum in the x-direction.  But you are very careful and you notice
: that computing the flow of x-momentum in the x-direction requires you to
: keep track of p_x v_x for particles having v_x negative as well as
: positive.  So you do the calculation you just did, but doing the
: integral from theta = 0 to theta = 2 pi.  All the way around.   This
: doubles the answer.  

Right.
I should have checked my answer before posting it but I was in a hurry.

: This is related to a point I just made in a post to Oz, that
: "x-momentum in the x-direction" is also "(-x)-momentum in the
: (-x)-direction".  Rank 2 tensors are a little counter-intuitive since we 
: are more used to vectors.  This makes concepts like pressure, stress,
: strain, and the like seem rather sneaky.  

: I used to wonder: if I pull on both ends of a rope, the forces cancel
: out, so what exactly is this "tension" the rope feels, anyway?  Well,
: it's a flow of (-x)-momentum in the x-direction, and an equal flow of
: x-momentum in the (-x)-direction.  Both contribute to making T_{xx} be
: negative; that's tension.  

Well, that's the most tricky part !  In material science, what you
are supposed to do is to make a small surface cut (perpendicular to 
direction j) and the force needed to make up for that cut has a component
i, then the material stress tensor component sigma_ij is exactly that.
So here you take only the force coming "from one side".  That was what
confused me and made me integrate over only half a sphere.
So in your example (I'm happy you take it because it illustrates exactly
what's tricky), you should cut the rope in 2 pieces, see what force you
need to apply on ONE of the pieces to compensate for the loss of the
other one,... which is ONLY ONE TIMES THE FORCE.
So in your rope example you exactly don't have the problem that is
showing up with the photons !
(not that I want to hide behind it, I was plainly wrong there but it
illustrates how nasty these things are).

cheers,
Patrick.

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue May  7 12:51:21 PDT 1996
Article: 14107 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 6 May 1996 23:35:00 -0700
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In article <[email-masked]> [email-masked] (Vanesch P.) writes:
>john baez ([email-masked]) wrote:

>: Well, maybe the pressure is not equal to the momentum flux you
>: calculated, but twice that!  

>: I can think of two arguments for this.  The down-to-earth way is to
>: imagine you have photons bouncing off the wall of a silvered box.  

>Yes, this is what Oz did and that was what I was objecting to.  After all,
>there isn't a silvered box.

"Assume there is a silvered box."  

"But there isn't a silvered box!"  

That reminds me of the story about when the math professor said, "Assume
f(x) = sin x," and the student raised his hand and interjected, "But
sir, what if it's not?"

Sorry, I couldn't resist.  The thing is, a down-to-earth definition of
the pressure of a gas is the force per unit area with which its
molecules push against a wall as they bounce off.  So it's not utterly
insane to try to define the pressure of radiation as the force per unit
area with which the photons push against a wall as they bounce off.  Of
course, for them to bounce off, the wall has to be silvered.  (We don't
want a bunch of photons sticking to the wall; they're a real nuisance to
scrape off.)

However, it's also important to have an abstract definition of pressure
that doesn't require Gedankenequipment like imaginary walls, silvered
boxes, and the like.  And this is simply the total flux of x-momentum in
the x-direction.  Since this is the same as (-x)-momentum in the
(-x)-direction, you have to do your integral from 0 to 2pi to catch all
the pressure.  On that we agree.

>: I used to wonder: if I pull on both ends of a rope, the forces cancel
>: out, so what exactly is this "tension" the rope feels, anyway?  Well,
>: it's a flow of (-x)-momentum in the x-direction, and an equal flow of
>: x-momentum in the (-x)-direction.  Both contribute to making T_{xx} be
>: negative; that's tension.  

>Well, that's the most tricky part !  In material science, what you
>are supposed to do is to make a small surface cut (perpendicular to 
>direction j) and the force needed to make up for that cut has a component
>i, then the material stress tensor component sigma_ij is exactly that.
>So here you take only the force coming "from one side".  That was what
>confused me and made me integrate over only half a sphere.

So now you're going to get your revenge by confusing me, eh?  :-)
I suppose it's only fair.

>So in your example (I'm happy you take it because it illustrates exactly
>what's tricky), you should cut the rope in 2 pieces, see what force you
>need to apply on ONE of the pieces to compensate for the loss of the
>other one,... which is ONLY ONE TIMES THE FORCE.
>So in your rope example you exactly don't have the problem that is
>showing up with the photons !
>(not that I want to hide behind it, I was plainly wrong there but it
>illustrates how nasty these things are).

Yes... and just when I thought I understood that damn rope!  I can't
take the pressure, the tension, the stress, the STRAIN of these darn
rank-2 tensors.  Take me back to the days of vectors, when I knew which
end was up!   

I actually think it all should make sense if one thinks about it.  
The tension on the rope is the force you'd need to apply to ONE piece to
hold it still after cutting it in half --- I'll take you word for that.
And the pressure in a box of gas is the force that you'd need to apply
to ONE wall if you split the box in two and held the gas in with two new
walls.  It seems consistent.

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Article: 14141 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 7 May 1996 06:54:41 +0100
Organization: Oz
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>
>Hmmm.  It seems safer to me to just solve Einstein's equation.  I
>sketched how to start doing this in a recent post.  All we need to know
>is the relationship for pressure and energy density of photons, and we
>have that: E = 3P.
>
>I'm sure one can fight ones way to the correct physics through the above
>heuristics, but it seems a bit too much work.  Often it's easier to work
>out the simpler heuristic explanation after you've done the calculation
>using insightless grunge-work.

I wouldn't quite describe it like that. The insightless grunge work is
hardly insightless, and it is clear, careful and cautious. It is also
essential. The handwavingly so-called 'intuitive' method is only
appropriate for describing to unskilled denizens of the group (me) when
you know the correct answer. Alternatively for the use of unskilled
denizens when there is somebody who knows the *correct* answer who can
point out the errors when the inevitable error occurs.

The unskilled denizens have to use it because they can't do the maths
any more. (sob).

Anyway, I think the reason it doesn't work is because T_{ii} = P R^2 and
not T_{ii} = P in this metric. It's not clear to me from the original
course notes why this is so, but I can see that it is not unlikely.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 14146 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sun, 5 May 1996 21:04:46 +0100
Organization: Oz
Lines: 40
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In article <[email-masked]>, john baez <[email-masked]>
writes
>  The point is that the
>density of dust, E, times the volume of a patch of space, R^3, stays
>constant.  So we have conservation of dust.
>
>Let's write 
>
>      D = E R^3
>
>for this conserved quantity, and use this to simplify the equations (**).
>We get
>
> 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)
>
>Okay.  Now I claim these equations are incredibly familiar to anybody
>who has ever studied gravity at all.  Not general relativity, classical
>Newtonian gravity!  What are they????   

Sorry, John, I'm out of this one (sigh). I definitely didn't do enough 
Newtonian gravity anyway. Not hardly enough. The only thing that springs 
to mind, with no evidence or anything, just a sort of association for no 
reason is Kepler's laws. I am confident that this will be wrong, 
however. I can't even see a connection.

I suppose if we felt really 'blue skylike' R" = -D/(6R^2) does look a 
little like F=Gm1m2/(r^2) or r" = -Gm1/(r^2) if you felt so inclined. 
Simply Newton's law.

I suppose if you want to be even more absurd you could look at 
3(R')^2 = D/R as similar in for to an orbital thingy where 
v^2 = m1m2G/(2pi r).

Mind you , you could dream hundreds up if you tried, I'm sure.

vanEsch where are you when you are needed?

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 14153 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 6 May 1996 23:07:13 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
Lines: 56
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No, I haven't dropped off the face of the planet.  Just keeping myself
a little too busy.  So, here's a little bit I had some time to do just
now.

In article <[email-masked]>, john baez <[email-masked]> wrote:
>
>Well, in a recent post I showed that for the spatially flat homogeneous
>universe full of dust, we can get use the basic equations
>
> 3(R')^2 = E R^2       R" = - ER/6                 
>
>to show that the energy density E varies as 1/R^3.  Maybe you can
>pull a similar trick to get a relation between P and R from the
>equations for a spatially flat homogeneous universe full of radiation:
>
> (R')^2 = P R^2       R" = -P R    
>

So, let's differentiate the first to get,

2 R' R" = 2 P R R' + P' R^2

Subsititute the second

-2 P R R' = 2 P R R' + P' R^2
0 = 4 P R R' + P' R^2
  = 4 P R^3 R' + P' R^4
  = d/dt (P R^4)

So, (P R^4) is a constant which, for lack of anything better I will
call Q.  Going back to the original equations.

R" = - Q / R^3
R" R^3 = -Q

Let's try R = A t^1/2
	  R' = 1/2 A t^-1/2
	  R" = -1/4 A t^-3/2

R" R^3 = -1/4 A^4 = -Q

A = (4 Q)^1/4

R = (4 Q)^1/4 t^1/2

Hmmm...later I'll think about the mixed dust and radiation universe.
However that has me a little worried as to whether we can finesse
things in quite the same way, as we will have the mass energy density,
the radiation pressure and our expansion function R to figure and only
two equations with which to do it.  Hmm...

Kevin

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Tue May  7 16:45:07 PDT 1996
Article: 14185 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 7 May 1996 13:12:35 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked] writes:
>In article <[email-masked]>, john baez <[email-masked]>

>> 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

>>Okay.  Now I claim these equations are incredibly familiar to anybody
>>who has ever studied gravity at all.  Not general relativity, classical
>>Newtonian gravity!  What are they????   

>Sorry, John, I'm out of this one (sigh). I definitely didn't do enough 
>Newtonian gravity anyway. Not hardly enough. The only thing that springs 
>to mind, with no evidence or anything, just a sort of association for no 
>reason is Kepler's laws. I am confident that this will be wrong, 
>however. I can't even see a connection.

Oz loves to play dumb like this!  I'm sure he does it just to keep our
expectations low.  First he says "I can't even see a connection."  And
then, immediately afterwards, he explains the connection:

>I suppose if we felt really 'blue skylike' R" = -D/(6R^2) does look a 
>little like F=Gm1m2/(r^2) or r" = -Gm1/(r^2) if you felt so inclined. 
>Simply Newton's law.

Right.  The second equation is mathematically analogous to the 
inverse square force law.   R" is the acceleration, and -D/(6R^2) is the
force.  (Let's not worry too much about what plays the role of the
"masses" or Newton's constant G in this analogy.  Some combination of
them is gonna give us that D/6 stuff.  What matters is the basic
mathematical pattern, not the clutter of constants.)  

Okay, so the first equation is closely related to the inverse square
force law.  What is it, exactly?  What does (R')^2 remind you of?  What
does -1/R remind you of?

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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 5 May 1996 14:39:54 -0700
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In article <[email-masked]> "Alexander E. Meier" <[email-masked]> writes:
>now I try to represent what a cotangent vector looks like and arghhhhhhh
>I just can't have any clear picture of what such a thing looks
>like,......
>Any wizards or apprentices out there who can help me ?????

Ah, finally Oz has a serious competitor when it comes to kicking and
screaming when one doesn't understand something!  Good.  The squeaky
wheel gets the grease.

A tangent vector points.  A cotangent vector "copoints".  You can
roughly visualize it as an cooriented stack of hyperplanes -- at least
if you are careful.  "Cooriented" means that all the hyperplanes have a
"front" and "back", like the even and odd sides of the pages in a book.

The bigger the covector is, the more densely packed these hyperplanes
are.  If you take a vector v and a covector f, you can count how many
hyperplanes the vector pokes through, to get a number f(v).  We use the
coorientation of the hyperplanes to determine whether f(v) is positive
or negative.

The gradient of a function is a cotangent vector field.  The level
surfaces of the function are what give you, infinitesimally, the stack
of hyperplanes. 

All this is infinitely easier to understand with pictures.  Try
_Gravitation_ by Misner Thorne and Wheeler, or _Gauge Fields, Knots and
Gravity_ by Baez and Muniain.  

Cotangent vectors are also called covectors or 1-forms.  

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Article: 14585 of sci.physics
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From: Geoffrey A. Landis <[email-masked]>
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Subject: Momentum (Re: General Relativity Tutorial)
Date: 9 May 1996 18:22:13 GMT
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In article <[email-masked]> john baez, [email-masked]
writes:
>>However, it's also important to have an abstract definition of pressure
>>that doesn't require Gedankenequipment like imaginary walls, silvered
>>boxes, and the like.  And this is simply the total flux of x-momentum in the 
>>x-direction. 

I have to admit that I've never really understood this way of looking at
things.

Consider a gas of uniform pressure P, and consider some unit area A
oriented with normal pointing in the x direction.  The amount of gas
molecules carrying momentum crossing this unit area in the +x direction
exactly equals the amount of gas molecules carrying momentum across the
unit area in the -x direction, so the flux of momentum in the x direction
is zero.

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Thu May  9 23:25:35 PDT 1996
Article: 14659 of sci.physics
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From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: Momentum (Re: General Relativity Tutorial)
Date: 9 May 1996 21:15:06 -0700
Organization: University of California, Riverside
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In article <[email-masked]> Geoffrey A. Landis <[email-masked]> writes:
>In article <[email-masked]> john baez, [email-masked]
>writes:
>>>However, it's also important to have an abstract definition of pressure
>>>that doesn't require Gedankenequipment like imaginary walls, silvered
>>>boxes, and the like.  And this is simply the total flux of x-momentum
>>>in the x-direction. 

>I have to admit that I've never really understood this way of looking at
>things.

>Consider a gas of uniform pressure P, and consider some unit area A
>oriented with normal pointing in the x direction.  The amount of gas
>molecules carrying momentum crossing this unit area in the +x direction
>exactly equals the amount of gas molecules carrying momentum across the
>unit area in the -x direction, so the flux of momentum in the x direction
>is zero.

Right.  But I wasn't talking about what you call the flux of momentum in
the x direction --- which I might call the density of x-momentum.  I was
talking about what I call the flux of x-momentum in the x-direction.
Admittedly my language is a bit tortured.  But let me explain it!

Let's work in plain old Minkowski space using the standard coordinates.
Then the component T_{ab} of the stress-energy tensor is what I'd call
the "flow in the a-direction of b-momentum".  This is easiest to
understand if we have a gas made of point particles.  Then each point
particle has a momentum p and a velocity v.  Both the momentum p and the
velocity are 4-vectors.  The contribution of a particular particle (wait
--- aren't all particles particular?) to the stress-energy tensor is
then, roughly speaking, v_a p_b.  In other words, the rate at which the
particle is carrying momentum-in-the-b-direction in the a-direction is
v_a p_b.

So for example if the particle is moving northeast, it's carrying

x-momentum in the y-direction

and also 

y-momentum in the x-direction

and also for that matter it's carrying

x-momentum in the x-direction

and 

y-momentum in the y-direction.

So this particle is contributing positive amounts to T_{xy}, T_{yx},
T_{xx}, and T_{yy}.  It's also contributing to T_{tx} and so on.  For
example, T_{tx} is the flow in the time direction of x-momentum, which
is what I'd often call the *density* of x-momentum.  This is what you
are calling the "flux" of momentum in the x direction.  It's different
from T_{xx}!

If you get confused you may find it a relief to know the following
nontrivial fact: T_{ab} = T_{ba}. 

Now: what if we have a particle moving in the x-direction and a particle
of the same mass moving at the same speed in the (-x)-direction?  Well,
their contributions to T_{tx} cancel out, as you note.  But their
contributions to T_{xx} add up!  The particle moving in the
(-x)-direction is carrying NEGATIVE x momentum in the NEGATIVE x
direction, which is the same --- by the rules of this game --- as
carrying POSITIVE x momentum in the POSITIVE x direction. 

Who made up the rules of this nutty game, you might ask?  Well, one
answer to that is that it's not nutty at all.  I'm simply saying that

(-v_x)(-p_x) = v_x p_x ; 

the signs cancel out!  Another deeper answer is that the name of this
game is "rank 2 tensors", and rank 2 tensors transform differently under
coordinate transformations than vectors.  If we take a vector v and do a
reflection across the yz plane, the component v_x changes sign.  But if
we take a rank two tensor T and do a reflection across the yz plane, the
component T_{xx} changes sign TWICE, or in other words, not at all!

Above I haven't been worrying too much about the fact that we don't
often talk about T_{ab} of a single particle; usually we talk of T_{ab}
as some kind of continuous field on spacetime.  

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Fri May 10 18:26:44 PDT 1996
Article: 14762 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
Date: 10 May 1996 19:32:50 GMT
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Michael Weiss just reminded me of some e-mail we exchanged a while ago
about dust-filled and radiation-filled Robertson-Walker cosmologies.
He suggested that some of it might be of interest to readers of this
thread.  So here it is.  

(Thanks, Michael, for reminding me about it -- I'd forgotten I'd
written this -- and for saving a copy!)

First, let's talk about the expansion rate of the Universe.  Suppose
that the Universe is homogeneous and isotropic, and is filled with a
perfect fluid characterized by some (time-dependent) density rho and
pressure P.  Then the Einstein equation reduces to a simple ODE, the
Friedmann equation, for the scale factor.  Specifically, if R(t) is
the scale factor and we define the time-dependent Hubble parameter
H=(1/R) dR/dt, then we have

H^2 = 8 pi G rho/3 - k/R^2.

Here k is +1, 0, or -1 depending on the spatial curvature.  Let's set
k=0 for simplicity.

Notice that the pressure doesn't explicitly enter this equation at
all.  All that matters is rho(t).  But the energy density of
relativistic matter has a different time dependence from that of
nonrelativistic matter, because relativistic matter redshifts.
Specifically, rho is proportional to R^{-3} for nonrelativistic matter
and to R^{-4} for relativistic matter.  You can combine this fact with
the Friedmann equation and discover that R is proportional to t^{2/3}
if the Universe is full of nonrelativistic matter and to t^{1/2} if it
is full of relativistic matter.

We can take little steps away from this direction in either the
"intuition" direction or the "formal mathematics" direction.
Intuitively, what's going on is this: we know (if we're willing to
assume spatial flatness and we know the Hubble constant) the density
of matter today.  What was the density at some earlier time, say when
R was 10 times smaller than it is today?  Well, it depends on whether
the matter is relativistic or not.  The density at a particular time
in the past (or rather a particular redshift) was higher if the matter
is relativistic than if it isn't.  If you like you can think of that
past high density as the reason why the deceleration is more
pronounced for relativistic matter than for nonrelativistic.

To make everything a very little more formal, and to see the
connection with pressure, note that we need to know something about
rho(t) before we can solve the Friedmann equation.  So we need one
more equation.  Local energy conservation comes to the rescue.  The
change in energy of a volume is equal to the work done, P dV.  But the
volume is proportional to R^3, so we have

d (rho R^3) = -P d(R^3).

But even with this extra bit of information we still need to know
something about the pressure before we can solve the Friedmann
equation.  Specifically, we can use the equation of state relating P
to rho.  For relativistic matter, P=rho/3, while for dust, P=0.

-Ted

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Article: 14796 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Fri, 10 May 1996 13:51:20 +0100
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In article <[email-masked]>,
[email-masked] writes
>'"Alexander E. Meier" <[email-masked]>' wrote: 
> 
>>Well I'm never offended :) 
> 
>Actually,   I just emailed that note  :) 
> 
>>Actually your answer made me think even hrader about it... 
>>Transporting the second clock back to the location of the first one and 
>>to compare their times is dumb...since parallel transport is changing 
>>the vector...so I can't really tell what it was before I transported 
>>it...(Or maybe I could????? arghhhh I don't wanna think about it...it 
>>might involve tons of nasty Christoffel symbols anyway). 
>>But on reflexion I found another way.  Send a photon to a neighbouring 
>>point, where it bounces of , and let it come back to you...In the 
>>meantime, your 'time' coordinate changed by x^0... I guess you can 
>>safely take the point at x^0/2 to be simultaneous with the instant the 
>>photon bounced of at the neighbouring location.....Now the question is 
>>what is x^0...well I guess it's gonna involve taking ds^2=g_{ab}dx^adx^b 
>>along the path where it is equal to zero, and then use some kind of 
>>magic....arghhh I just can't figure out what magic to use  
>>Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp 
>>:) 

Well this did come up earlier. You only seem to be able to do it in 
certain circumstances. If you start with all your clocks close together 
and at rest to each other then you can synchronise them quite well. The 
start of the big bang would be an excellent choice, but you could do it 
anywhere that space is flat enough. Hmmm, come to think of it the start 
of the BB is hardly flat at any scale. HHeeeeeeeeeeeeeeeellllllllpppp!

Then as they separate due to the curvature of spacetime you can use co-
moving co-ordinates to describe what happens. I think that they will 
define a spacelike manifold in 4-space. This is handy. I have my doubts 
as to whether this is true as they separate at speeds near c, but the 
metric should take this into account 'automatically'.

After the synchronisation, unless you have Minkowski space, then further 
synchronisation doesn't make much sense. Baez will probably come in here 
and say that you cannot synchronise even in Minkowski space unless the 
clocks are 'at rest' wrt each other and 'close'.

I am not as clear about this as I should be either. Mostly because it 
hasn't really been discussed at length because we have only just begun 
to understand the ramifications of curved spacetime. I think we might 
now be in a position to discuss this with a modicum of sense.

But then again, perhaps not.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May 11 00:25:32 PDT 1996
Article: 14915 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 10 May 1996 23:44:22 -0700
Organization: University of California, Riverside
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Oz posted this but it seems to have had some trouble reaching me, and
maybe other folks, so I'll post it again over at this end of the world.
Perhaps when the two copies collide there will be an explosion.

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>>In article <[email-masked]>, john baez <[email-masked]>
>
>>> 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

>Oz loves to play dumb like this!  I'm sure he does it just to keep our
>expectations low.  First he says "I can't even see a connection."  And
>then, immediately afterwards, he explains the connection:

It's embarrassing to expose my thinking processes in print. I ought 
really to go back and edit all the stupidities out so that I look smart. 
Hmm, too much trouble I suppose. The last time I did any Newtonian 
gravity was at school. I think they thought it either too simple for 
undergraduates at one level, or more suitable for astronomers at the 
more complex level. Anyway, the poor old brain cells took an inordinate 
time to put one and one together and make two. Neuronal degeneration, I 
expect.

>Okay, so the first equation is closely related to the inverse square
>force law.  What is it, exactly?  What does (R')^2 remind you of?  What
>does -1/R remind you of?

Oh lets try something so simple and crude and straightforward that it 
can't have anything to do with it.

What does R'^2 remind me of?  
Well how about kinetic energy, like (1/2)mv^2? Kinda elementary really.
The only trouble is that this is as classical as one could get.

What does -1/R remind me of?
Well how about gravitational energy, noooo.
How about gravitational potential energy? Ooh, that was a long time ago. 
Is my memory faulty? Hang on gotta do an (aargh!) integration.
<Scribble, curse, hmmmm>
Yup, that looks plausible.

Oh, it makes 3(R')^2 = D/R - 3k look slightly sensible looked at this 
way. If k=0 then we can do our potential to infinity, if it's positive 
then we can't get to infinity, only to k. If it's negative we can never 
get to infinity, we gotta go past it to -k. Usual caveats for signs.

OK. I plumb for 3(R')^2 = D/R - 3k being a analogue of energly like
thingy whatsits.

I can't get any simpler. (That doesn't sound right.)

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May 11 00:27:54 PDT 1996
Article: 14916 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 11 May 1996 00:25:24 -0700
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Oz wrote:

>In article <[email-masked]>, john baez <[email-masked]>
>writes
>>In article <[email-masked]> [email-masked] 
>>writes:
>>>In article <[email-masked]>, john baez <[email-masked]>

>>>> 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

>>Oz loves to play dumb like this!  I'm sure he does it just to keep our
>>expectations low.  First he says "I can't even see a connection."  And
>>then, immediately afterwards, he explains the connection:

>It's embarrassing to expose my thinking processes in print. 

Hey, look at the bright side: at least you have thinking processes to
expose, unlike some around here.

>>Okay, so the first equation is closely related to the inverse square
>>force law.  What is it, exactly?  What does (R')^2 remind you of?  What
>>does -1/R remind you of?

>Oh lets try something so simple and crude and straightforward that it 
>can't have anything to do with it.

Here you go, playing coy again.

>What does R'^2 remind me of?  
>Well how about kinetic energy, like (1/2)mv^2? Kinda elementary really.
>The only trouble is that this is as classical as one could get.

Sounds good.  As for it being classical, remember I said that (***) was
something everybody should already recognize from Newtonian gravity.

>What does -1/R remind me of?
>Well how about gravitational energy, noooo.
>How about gravitational potential energy? Ooh, that was a long time ago. 
>Is my memory faulty? Hang on gotta do an (aargh!) integration.
><Scribble, curse, hmmmm>
>Yup, that looks plausible.

Yes, an inverse square force law comes as the derivative of a -1/R
potential.  I like how you had to rederive that.  It means you don't
have too many unnecessary facts about Newtonian gravity cluttering up
your brain and making it harder to learn general relativity.  Just the
bare essentials.

>Oh, it makes 3(R')^2 = D/R - 3k look slightly sensible looked at this 
>way. If k=0 then we can do our potential to infinity, if it's positive 
>then we can't get to infinity, only to k. If it's negative we can never 
>get to infinity, we gotta go past it to -k. Usual caveats for signs.

"Can never get to infinity, we gotta go past it" --- well, that's cute.
But basically you mean we CAN and DO go to infinity.  Also you left out
some D's and 3's in your calculation.  But you get the idea.

>OK. I plumb for 3(R')^2 = D/R - 3k being a analogue of energly like
>thingy whatsits.

>I can't get any simpler. (That doesn't sound right.)

"Energly like thingy whatsits" --- I presume this is a quaint Anglicism
for what we prosaic Americans would call "conservation of energy".  

Okay, to wrap it up:  When we consider a universe full of dust, its size
R as function of time satisfies

            3(R')^2 = D/R - 3k      R" = -D/6R^2        

The second equation is just the Newtonian inverse square force law!
The first one contains something like kinetic energy and something like
the Newtonian gravitational potential energy!  Rewrite it like this

3(R')^2 - D/R = -3k

and it's just like the law of conservation of energy.  So in fact our
problem is completely isomorphic to the problem of an inverse square
force law.  All this work developing general relativity and here we are
doing something shockingly similar to Newtonian gravity!

As Brett McInnes forcefully pointed out, one shouldn't take this too
seriously.  If we consider some other problems, like the big bang with
radiation instead of dust in our universe, we'll get very different
answers.  Nonetheless the coincidence we see is useful, and one can also
enjoy oneself immensely figuring out just why it occurs.

It's useful because we now promptly see, as Oz saw, that whether the
universe recollapses or not depends in a simple way on k:

If k = 1 (positively curved space) we see the "energy" 3(R')^2 - D/R
= -3k is negative and there is not enough energy to get out to R =
infinity.   There's a big bang and the universe expands.  Then R reaches a
maximum when the energy is all in potential form, at R = D/3.  Then R
rolls back down and we get a big crunch.

If k = 0 (flat space) we see the energy is just *barely* enough to get
out to R = infinity.  The universe expands at an ever more lethargic
pace.  In fact, since 3(R')^2 - D/R = 0, the rate of expansion R' is
proportional to R^{-1/2}.  Doing a little calculation we see this means
R is proportional to t^{2/3}.  That's good --- that's what Oz got
earlier.

If k = -1 (negatively curved space) we see the energy is more than
enough to get us out to R = infinity.  The universe expands at a rate
which approaches some constant as t -> infinity.  

So we see the marvelous fact that whether the universe is closed
(positively curved space) or open (flat or negatively curved space)
determines whether or not there will be a big crunch!  At least, that's
what we get in this simplified model which is homogeneous and full of
just dust.

Okay, that does it for cosmology as far as I'm concerned.  

From noise.ucr.edu!news.service.uci.edu!unogate!mvb.saic.com!news.mathworks.com!newsfeed.internetmci.com!in2.uu.net!world!mmcirvin Sat May 11 12:38:17 PDT 1996
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From: [email-masked] (Matthew J. McIrvin)
Subject: Re: Momentum (Re: General Relativity Tutorial)
Message-ID: <[email-masked]>
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In article <[email-masked]>, Geoffrey A. Landis
<[email-masked]> wrote:

> Consider a gas of uniform pressure P, and consider some unit area A
> oriented with normal pointing in the x direction.  The amount of gas
> molecules carrying momentum crossing this unit area in the +x direction
> exactly equals the amount of gas molecules carrying momentum across the
> unit area in the -x direction, so the flux of momentum in the x direction
> is zero.

The molecules are carrying momentum in the +x direction in the +x
direction, and carrying momentum in the -x direction in the -x direction.
So the "xx" component of the momentum-flux tensor is the *sum* of those
two terms; their overall signs are the same.

-- 
Matt McIrvin   http://world.std.com/~mmcirvin/
-- 
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Path: mmcirvin
Date: Sat, 30 Mar 1996 02:41:04 -0500
From: [email-masked] (Matt McIrvin)

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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 11 May 1996 06:52:16 +0100
Organization: Oz
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I posted this on Thursday 9th May, it's not appeared as far as I can see
so I repost. Apologies if there is a double posting.

In article <[email-masked]>, john baez <[email-masked]>
writes
>In article <[email-masked]> [email-masked] 
>writes:
>>In article <[email-masked]>, john baez <[email-masked]>
>
>>> 3(R')^2 = D/R - 3k      R" = -D/6R^2                             (***)

>Oz loves to play dumb like this!  I'm sure he does it just to keep our
>expectations low.  First he says "I can't even see a connection."  And
>then, immediately afterwards, he explains the connection:

It's embarrassing to expose my thinking processes in print. I ought 
really to go back and edit all the stupidities out so that I look smart. 
Hmm, too much trouble I suppose. The last time I did any Newtonian 
gravity was at school. I think they thought it either too simple for 
undergraduates at one level, or more suitable for astronomers at the 
more complex level. Anyway, the poor old brain cells took an inordinate 
time to put one and one together and make two. Neuronal degeneration, I 
expect.

>Okay, so the first equation is closely related to the inverse square
>force law.  What is it, exactly?  What does (R')^2 remind you of?  What
>does -1/R remind you of?

Oh lets try something so simple and crude and straightforward that it 
can't have anything to do with it.

What does R'^2 remind me of?  
Well how about kinetic energy, like (1/2)mv^2? Kinda elementary really.
The only trouble is that this is as classical as one could get.

What does -1/R remind me of?
Well how about gravitational energy, noooo.
How about gravitational potential energy? Ooh, that was a long time ago. 
Is my memory faulty? Hang on gotta do an (aargh!) integration.
<Scribble, curse, hmmmm>
Yup, that looks plausible.

Oh, it makes 3(R')^2 = D/R - 3k look slightly sensible looked at this 
way. If k=0 then we can do our potential to infinity, if it's positive 
then we can't get to infinity, only to k. If it's negative we can never 
get to infinity, we gotta go past it to -k. Usual caveats for signs.

OK. I plumb for 3(R')^2 = D/R - 3k being a analogue of energly like
thingy whatsits.

I can't get any simpler. (That doesn't sound right.)

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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Article: 14985 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 11 May 1996 10:35:07 +0100
Organization: Oz
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In article <[email-masked]>, "Emory F. Bunn"
<[email-masked]> writes
>
>Notice that the pressure doesn't explicitly enter this equation at
>all.  All that matters is rho(t).  But the energy density of
>relativistic matter has a different time dependence from that of
>nonrelativistic matter, because relativistic matter redshifts.
>Specifically, rho is proportional to R^{-3} for nonrelativistic matter
>and to R^{-4} for relativistic matter.  You can combine this fact with
>the Friedmann equation and discover that R is proportional to t^{2/3}
>if the Universe is full of nonrelativistic matter and to t^{1/2} if it
>is full of relativistic matter.

But what about a mixture of the two? Naively there would appear to be
two forms of solution. One where t^n behaves like 1/2 < n < 2/3 and the
other where the solution is a combination of t^(1/2) and t^(2/3). To
save more metal anguish I would be grateful if you could indicate which
is the more correct answer. Instinctively I feel that a solution along
the lines od At^(1/2) + Bt(2/3) ought to be about right but .....

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May 11 19:55:56 PDT 1996
Article: 15044 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 11 May 1996 12:37:52 -0700
Organization: University of California, Riverside
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In article <[email-masked]> [email-masked](Edward Green) writes:
>'"Alexander E. Meier" <[email-masked]>' wrote: 

>>Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp 

>Now,  what was you original question:  "How do we synchronize clocks in
>GR".   Hmm...  I'm not sure what you are asking,  so maybe the heavies
>aren't either. 

Since I play the role of a "heavy" in this grade-B movie, let me
comment that I am not sure what "synchronizing clocks in GR" is or why
one would want to do it.  This is why I hadn't answered earlier.  I
couldn't understand the question.

How can you tell if two clocks are "synchronized" other than by bringing
them together to the same place?  One might try to send a signal from
one clock to the other using light, carrier pigeon, or camel caravan.
Unfortunately, when the signal from one clock reaches the other will
depend drastically on the mode of transportation, the path the signal
takes, and the curvature of the intervening spacetime.  Trying to
specify a "best" route and mode of transportation between each pair of a
large number of clocks is too complicated to be worth bothering with,
unless we have a very specific experiment in mind, where we can work out
all the details.  

The best thing is to realize that this issue is not very important for
how general relativity really works.  Anyone who has followed the
general relativity tutorial has begun to see we can solve lots of
problems without ever getting entangled in this distraction.  

This is a distraction caused by earlier ways of thinking about
spacetime: the Aristotelian, Newtonian and special-relativistic
traditions.

>I know we agree that two clocks initially sychronized at
>an event,  each taking a little tour and reuniting later,  will in general
>show different times.   So are you proposing to "synchronize" clocks on
>nearby geodesics?   I think that's what I hear you saying.   I still think
>the answer is "We don't",  but somehow that seems to be inadequate. 

We don't.  At least I don't.  Each clock falling along its own geodesic,
or travelling any route whatsoever, keeps its own time.  This is a very
important and useful thing.  And it's all one really needs, as far as
time-keeping is concerned.

>Maybe you had some prior issue in mind,  or an experiment.   Pretending I
>am John Baez for a moment ( a very thin illusion ),  I might say "I don't
>know what you mean by 'sychronize' .  The word has no generally accepted
>meaning in GR,  other than to synchronize two clocks *at a single
>event*.  Perhaps you could explain to me exactly what you want to accomplish
>operationally..."  

That's a pretty good imitation.  Took the words right out of my
mouth!  Soon I will be superfluous here!

From noise.ucr.edu!galaxy.ucr.edu!library.ucla.edu!agate!physics12.Berkeley.EDU!ted Sat May 11 21:35:49 PDT 1996
Article: 15100 of sci.physics
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From: [email-masked] (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Followup-To: sci.physics
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In article <[email-masked]>, john baez <[email-masked]> wrote:
>So I'd expect something that looks roughly like t^{1/2} for small t and
>t^{2/3} for big t.  Your guess, At^(1/2) + Bt(2/3) does have that
>property.  So it may well be a decent fit.  But I doubt it's exactly
>right.  

Correct on both counts: it's got the right qualitative behavior, but
it's not exactly right.  

I don't think I've ever seen a simple closed-form solution for the
scale factor in a Universe with both dust and radiation, which makes
me suspect that one doesn't exist.  If you're interested in studying
the behavior of the scale factor during the time when the energy
densities of both species are comparable, you just solve the equation
numerically.

-Ted

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Article: 15097 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Sat, 11 May 1996 23:48:58 +0100
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In article <[email-masked]>, john baez <[email-masked]>
writes
>
>>But what about a mixture of the two? Naively there would appear to be
>>two forms of solution. One where t^n behaves like 1/2 < n < 2/3 and the
>>other where the solution is a combination of t^(1/2) and t^(2/3). To
>>save more metal anguish I would be grateful if you could indicate which
>>is the more correct answer. Instinctively I feel that a solution along
>>the lines of At^(1/2) + Bt(2/3) ought to be about right but .....
>
>As for myself, instinctively I feel that a solution along those lines
>can't be exactly right.  It looks like you're pretending you have a
>linear equation, and are getting two solutions and adding them up with
>coefficients A and B.  Einstein's equation is nastily nonlinear!  Even
>for our simplified flat homogeneous space we get the nonlinear
>equations:
>
>3 (R')^2 = E R^2       R" = (-1/6) (E + 3P) R   
>
>We'd need to solve these given whatever assumptions we make about
>the relationship between E and P, to get the answer.
>
>In reality it appears the universe isn't just a constant mix of
>relativistic and nonrelativistic matter.  It was "radiation-dominated"
>with lots of pressure and hot, highly relativistic stuff flying around
>at early times, and "matter-dominated" with almost no pressure at late
>times --- like now.
>
>So I'd expect something that looks roughly like t^{1/2} for small t and
>t^{2/3} for big t.  Your guess, At^(1/2) + Bt(2/3) does have that
>property.  So it may well be a decent fit.  But I doubt it's exactly
>right.  

Yeah, well I have to say that I rather think what you say is right. To 
do it properly would result in a really nasty set of equations that 
ended up with a solution starting somewhere near t^1/2 and slid towards 
t^2/3 as the radiation/energy balance slipped towards matter dominance. 
A sort of t^(f(t)) sort of nasty. Ugh! It makes you realise why vast and 
powerful parallel processing hypercomputers get used for it. Far beyond 
mere analytical stuff. It's rather what I said back a bit. (Warning, 
heresy coming) Most real life solutions are insoluble, so at the end of 
the day why bother with too much maths, just understand the physics 
enough to model it well enough to stuff it on a computer and take a 
holiday while it works it out. I just wish that in my day practical 
efficient computers existed, say as powerful as a 386 DX. Hey, (this 
will stagger you) in my day there really weren't even any reasonable 
electronic calculators!

Save the maths for elegant and powerful notations like tensors. I hope 
the QED notations are equally elegant and powerful, or I (for one) am 
doomed.

At this final point in the GR cosmology it would be really nice if Ted 
could post a short piece on real up to date cosmology relevant to what 
we have achieved so far. No maths (much) but enough to see how current 
thinking has developed past our elementary model.

Then the cristoffel stuff finishes the job.

I really would like to see how orbits are seen under GR more formally 
than has been done so far. Obviously there are some substantial 
approximations, but it would be nice to tie old man Newton to GR in a 
simple way to end with.

It's been enormous fun.

(NB I make it 232 postings!)

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

From noise.ucr.edu!galaxy.ucr.edu!not-for-mail Sat May 11 21:59:07 PDT 1996
Article: 15130 of sci.physics
Path: noise.ucr.edu!galaxy.ucr.edu!not-for-mail
From: [email-masked] (john baez)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 11 May 1996 21:35:10 -0700
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In article <[email-masked]> [email-masked] writes:

>>3 (R')^2 = E R^2       R" = (-1/6) (E + 3P) R   

>>We'd need to solve these given whatever assumptions we make about
>>the relationship between E and P, to get the answer.

>Yeah, well I have to say that I rather think what you say is right. To 
>do it properly would result in a really nasty set of equations that 
>ended up with a solution starting somewhere near t^1/2 and slid towards 
>t^2/3 as the radiation/energy balance slipped towards matter dominance. 
>A sort of t^(f(t)) sort of nasty. Ugh! It makes you realise why vast and 
>powerful parallel processing hypercomputers get used for it. Far beyond 
>mere analytical stuff. 

Let's not get carried away!  It's just an ordinary differential
equation; you could easily write a wee program to solve it if you happened
to know the relation between P and E.  Most of the engineering students
I see tend to carry around programmable calculators that can do this
sort of thing.

It's for full-fledged GR, without assumptions like "homogeneous,
isotropic, utterly boring," that one needs the heavy-duty equipment.
Say you want to see how two black holes spiral into each other.  That's
a National Science Foundation "grand challenge" supercomputer problem.

>It's rather what I said back a bit. (Warning, 
>heresy coming) Most real life solutions are insoluble, so at the end of 
>the day why bother with too much maths, just understand the physics 
>enough to model it well enough to stuff it on a computer and take a 
>holiday while it works it out. 

Yeah, right.  Just a wee problem: say you happen to be interested in
general relativity.  You want to hand your problem over to a computer
and take a holiday while it works it out.  Okay, well, you need to know
enough math to write an adaptive grid code to solve constrained
nonlinear hyperbolic PDE involving tensor fields!  I.e., writing the
program itself requires a hell of a lot of math.  (Math that I for one
intend never to learn.)

Even for simpler things like fluid dynamics, one can't blindly "stuff
the problem into a computer" and expect to get much of anywhere.  At
least not yet.  

But you might enjoy thinking for a minute or two about how you'd solve
Einstein's equation using a computer.  Hint: you do not write the
equation on a napkin and stuff it into the floppy drive.  :-)

>Save the maths for elegant and powerful notations like tensors. 

Here I agree with you: I like to stick to simple, beautiful stuff.

>I hope the QED notations are equally elegant and powerful, or I (for one) am 
>doomed.

You are doomed.

>At this final point in the GR cosmology it would be really nice if Ted 
>could post a short piece on real up to date cosmology relevant to what 
>we have achieved so far. No maths (much) but enough to see how current 
>thinking has developed past our elementary model.

Sounds cool.

>Then the cristoffel stuff finishes the job.

Yup.  

(Heh.  Little do you know what you are in for here!  I have NO idea how
to explain this stuff in simple terms.)

>I really would like to see how orbits are seen under GR more formally 
>than has been done so far. Obviously there are some substantial 
>approximations, but it would be nice to tie old man Newton to GR in a 
>simple way to end with.

You solve the Einstein equation assuming a spherically symmetric, static
solution.  You get the Schwarzschild solution.  You write down the
equation for geodesics in this spacetime.  (I wrote down the general
equation for geodesics a while back; since you didn't respond I assume
it was too trivial to be interesting to you.)  With much agonizing
grungework, you solve it.  You find precessing orbits.  That's what old
Al did, anyway.

>It's been enormous fun.

Indeed.

>(NB I make it 232 postings!)

Yikes.  Loquacious, ain't we.

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Article: 15486 of sci.physics
Newsgroups: sci.physics
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From: [email-masked] (Matthew J. McIrvin)
Subject: Re: General Relativity Tutorial
Message-ID: <[email-masked]>
Sender: [email-masked] (Mr Usenet Himself)
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Date: Mon, 13 May 1996 23:33:21 GMT
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In article <[email-masked]>, [email-masked] (Ken
Muldrew) wrote:

> [email-masked] (john baez) wrote:

> >Yeah, right.  Just a wee problem: say you happen to be interested in
> >general relativity.  You want to hand your problem over to a computer
> >and take a holiday while it works it out.  Okay, well, you need to know
> >enough math to write an adaptive grid code to solve constrained
> >nonlinear hyperbolic PDE involving tensor fields!  I.e., writing the
> >program itself requires a hell of a lot of math.  (Math that I for one
> >intend never to learn.)
> 
> What assumptions did Einstein make to get the predictions for the
> perihelion of Mercury and the bending of starlight? Were they fierce
> calculations? Did he recruit some helpers (and take a holiday while
> they worked it out)?

In *that* case the math was much simpler, because the situation was one
that could be approximated to high accuracy by assuming spherical
symmetry, which makes it possible to solve the equations in closed form.
Then, using the resulting solution (Karl Schwarzschild's, actually), you
can calculate the perihelion precession and the bending of starlight. 

It's simple enough that it's worked out in full in textbooks on general
relativity.

-- 
Matt McIrvin   http://world.std.com/~mmcirvin/
-- 
Nntp-Posting-Host: world.std.com
Path: mmcirvin
Date: Sat, 30 Mar 1996 02:41:04 -0500
From: [email-masked] (Matt McIrvin)

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Article: 15532 of sci.physics
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From: Oz <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: Tue, 14 May 1996 06:34:43 +0100
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In article <[email-masked]>, Patrick van Esch
<[email-masked]> writes
>Oz ([email-masked]) wrote:
>
>: Well, GR (at my trivial level) looks rather neat and elegant. In fact
>: rather beautiful. I also liked the mental reforming that was required to
>: grasp it at my basic level. Many, many thanks to John, Ted and all who
>: did it. It took some effort, but then all worthwhile things do.
>
>Sounds like kinda goodbye ???  

Not from me. You ain't that lucky!

Maybe a goodbye to the GR tutorial (formal) since JB seems to be having 
serious difficulty in figuring out how to explain cristoffel's simply 
enough for morons like me to follow. I don't know if he will ever have 
time to produce his handwavingly simple exposition. Lets face it he has 
put a whole bundle of work in, and he has other things to do.

------------------------------- 
'Oz     "When I knew little, all was certain. The more I learnt,
        the less sure I was. Is this the uncertainty principle?"

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From: persjohn <[email-masked]>
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 18 May 1996 16:57:21 GMT
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This is my first post to sci.physics, so I hope it works. The GR 
tutorial has been wonderful -- thanks to John Baez and all you 
other active participants, I've picked up some intuitive feel for 
GR, and enough confidence to buy the MTW book and study more. 
Hopefully I'm not the only silent lurker to have been enlightened 
by this epitome pioneering Internet tutorial.

The simple picture of a ball of free-falling coffee grounds is a 
great visualization tool, which I tried to apply to the examples 
in the first chapter of MTW. In free space above the surface of 
the earth, our ball elongates in the z (vertical) direction, but 
shrinks half as much in x and y (MTW p. 29 & 37), so the volume 
remains constant; as we learned, no stress-energy within the 
ball, no volume change, just distortion.

I puzzled over relating MTW's picture (p. 39) of (some) Riemann 
components within a constant-density earth to our ball of coffee 
grounds, until I realized that they were essentially talking 
about a ball of coffee the size of the earth; everything falls 
toward the center, and the volume shrinks at a rate determined by 
the mass density -- another intuitive connection, which made me 
really happy about getting a grip on some of this GR stuff, 
especially when I realized that the result doesn't depend on the 
size of the sphere: even with a soccer-size ball, if it's 
permeated with a (frictionless) mass density of 5.52 g/cm3 then 
it will take 21 minutes for the coffee to collapse to a point, 
then overshoot inside-out and continue to oscillate with a period 
of 84 minutes. One could imagine a desktop Einstein clock 
something like one of those in a bell jar with a slowly twisting 
disk hung on a torsion fiber, but instead for a timing mechanism 
a gold sphere with just a thin horizontal gap cut through the 
equator and prepared as a frictionless surface on which a small 
puck would slide back and forth with a 45 minute period (since 
gold has a density of 19.32 g/cm3, and the period depends on the 
square root of the density). Undoubtedly impractical, but such a 
cool concept -- the local mass density determines the curvature 
of spacetime, and keeps time.

The MTW picture of dropping test particles through a hole bored 
through the center of the earth bothered me however: after all, 
in the hole the particles are moving in empty space, so why 
should they contract? If our vertically-stretching, horizontally-
shrinking (but fixed volume) ball of coffee grounds falls from 
above the earth into the hole bored through it, does it start to 
shrink in all directions (and so volume too) once it passes into 
the interior? Plainly it would if the hole weren't there, by the 
classical spherical shell theorem, but with the hole the mass 
density within the ball is zero, so our GR tutorial notes say no 
volume change. Eager to elaborate what I was seeing as a very 
attractive bridge between classical physics and GR, I set about 
calculating. If I can't keep up with the Wizard or even the 
apprentices, maybe I should do some grunge work as penance. . .

I figured the most doable case would be with our little ball of 
coffee grounds at the center of the earth, on the central axis of 
the manhole which extends out to the surface in both directions. 
Release the grounds -- what happens? Do they collapse uniformly 
to a point, as expected from spherical symmetry, or is the 
missing mass of the cylinder of hole enough to let the ball keep 
its shape? Consider the reverse case: just the cylinder of 
constant mass density, and remove the rest of the earth. We'll 
assume the coffee grounds can move freely through the solid 
cylinder, although it seems ironic that the only reason it's 
there in the first place is to prevent students worrying about 
how test particles can fall freely through the solid earth. Never 
mind, we'll just assume free movement, and move on. It was at 
this point that I had my first COOL REALIZATION, that even if we 
just have a small sphere of mass, the coffee grounds oscillate 
with the same period, 84 minutes for 5.52 g/cm3, whether or not 
this sphere is contained in a larger sphere (like the whole 
earth).

Well, after much calculus that could have been a lot neater in 
retrospect, I showed that a ball of coffee grounds within the 
middle of a massive cylinder will retain it's size in the z 
direction (z the axis of the cylinder), but shrink in x and y at 
1.5 times the rate that all dimensions would shrink in a sphere, 
so the Ricci scalar would be the same:
    (1.5 + 1.5 + 0) = (1 + 1 + 1), in appropriate units, 
and the volume would change at the same rate. And then, if we put 
back the earth and remove the cylinder, we subtract and get:
    (-.5 + -.5 + 1) = 0.
No volume change for our ball of coffee grounds in the manhole at 
the center of the earth, just a flattening in the z direction 
balanced by swelling in x and y, caused by the asymmetry of the 
manhole. And that was the COOLEST thing of all. Groping across 
the bridge of classical concepts of space and time, I'm thrilling 
to the vista of GR-land opening through the mists, and salute the 
merry band of adventurers who have led the way past such reticent 
cyberlurkers as myself. 

And so I add my voice to those asking for more, MORE, not 
necessarily GR, but now emboldened to hope for enlightenment on 
QFT etc.

Daniel Johnson

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Article: 16842 of sci.physics
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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: General Relativity Tutorial
Date: 20 May 1996 14:35:40 -0400
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I'd like to go back to a question I had a while ago.  It's not so much
a conceptual question as a computational one.

It's been said that there are really only 20 independent components of
the Riemann tensor.  Using these symmetries

R^{a}_{[bc]d} = R^{a}_{bcd}
R^{a}_{[bcd]} = 0

I can drop the number of independent components from 256 to 80, quite
an improvement.

I know that there are two other symmetry relations that let us drop
the number to 20.  Could someone state and explain these.  Maybe leave
it as an exercise for us apprentices to show that they leave only 20,
and to figure out which 20 are the minimal set.

Kevin

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From: [email-masked]
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial, Riemann tensor symmetries
Date: 21 May 1996 15:30:31 GMT
Organization: Universitaet Muenchen (Germany)
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[email-masked] (Kevin Anthony Scaldeferri) writes:

>It's been said that there are really only 20 independent components of
>the Riemann tensor.  Using these symmetries

>R^{a}_{[bc]d} = R^{a}_{bcd}
>R^{a}_{[bcd]} = 0

>I can drop the number of independent components from 256 to 80, quite
>an improvement.

 Let R_{bcde} = g_{ae} R^a_{bcd}.

The symmetries for the Riemannian curvature tensor are
(for the metric and symmetric case considered in GR)

 R_{bcde}= -R_{bced}
 R_{bcde}= -R_{cbed}
 R_{bcde}= R_{debc}.

They reduce the number to 21.

Then, R^{a}_{[bcd]} = 0 reduces the number to 20, 
for a given coordinate system.

Considering 6 space-time rotations reduces the number of
independent components to 14.

(Landau -Lifschitz II, $ 91.)

Alexander
http://www.lrz-muenchen.de/~u7f01bf/WWW/homepage.html

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From: [email-masked] (Kevin Anthony Scaldeferri)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial, Riemann tensor symmetries
Date: 22 May 1996 11:53:19 -0400
Organization: Project GLUE, University of Maryland, College Park, MD
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References: <[email-masked]> <[email-masked]>
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In article <[email-masked]>,
 <[email-masked]> wrote:
>[email-masked] (Kevin Anthony Scaldeferri) writes:
>
>
>
>>It's been said that there are really only 20 independent components of
>>the Riemann tensor.  Using these symmetries
>
>>R^{a}_{[bc]d} = R^{a}_{bcd}
>>R^{a}_{[bcd]} = 0
>
>>I can drop the number of independent components from 256 to 80, quite
>>an improvement.
>
> Let R_{bcde} = g_{ae} R^a_{bcd}.
>
>The symmetries for the Riemannian curvature tensor are
>(for the metric and symmetric case considered in GR)
>
> R_{bcde}= -R_{bced}
> R_{bcde}= -R_{cbed}
> R_{bcde}= R_{debc}.
>
>They reduce the number to 21.

Assuming that correction I posted earlier is right, and the second one
should be

R_{bcde} = -R_{cbde}

Then, the b/c and d/e antisymmetries mean that for each one of them we
only need to look at the 6 pairs {10,20,30,21,31,32}.  And 6x6=36
means we are down to 36 independent components.  Now, the bc,de
symmetry lets us drop of (5)(6)/2 = 15 more, which indeed leave just
21 elements, namely

R_{1010},R_{1020},R_{1030},R_{1021},R_{1031},R_{1032},
R_{2020},R_{2030},R_{2021},R_{2031},R_{2032},
R_{3030},R_{3021},R_{3031},R_{3032},
R_{2121},R_{2131},R_{2132},
R_{3131},R_{3132},
R_{3232}

and finding the rest from these is easy.

I haven't worked out yet what this translates to for the R^a_{bcd}
yet.

Kevin

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Article: 17416 of sci.physics
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From: [email-masked] (Daryl McCullough)
Newsgroups: sci.physics
Subject: Re: General Relativity Tutorial
Date: 22 May 1996 09:50:58 -0400
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[email-masked] (john baez) writes:
[a summer rerun about torsion]

>Take a tangent vector v at P.  Parallel translate it along a very short
>curve from P to Q, a curve of length epsilon.  We get a new tangent
>vector w at Q.  Now let two particles free-fall with velocities v and w
>starting at the points P and Q. They trace out two geodesics....

>Okay.  Now, let's call our two geodesics C(t) and D(t), respectively....

>Now we ask: what's the time derivative of the distance between C(t) and
>D(t)?...
>
>If, no matter how we choose P and Q and v, the time derivative of the
>distance between C(t) and D(t) at t = 0 is ZERO, up to terms
>proportional to epsilon^2, then the torsion is zero!  And conversely!

I don't quite understand this. I know that in your GR tutorial,
you made parallel transport primitive and defined geodesics in
terms of that, but it seems to me that it makes just as much 
sense to take "proper time parameterized geodesics" as primitive,
and define vectors v and w are parallel if and only if C(t) and
D(t) maintain constant distance. But if one takes the latter
approach, then torsion would seem to be zero by definition!

In order for torsion as you defined it to be something experimentally
testable, there would have to be an operational definition of when
two vectors are "parallel transports" of each other. But how,
operationally, do you do parallel transport?

Daryl McCullough
ORA Corp.
Ithaca, NY

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Article: 18024 of sci.physics
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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: 27 May 1996 05:08:24 GMT
Organization: Virginia Polytechnic Institute and State University
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In article <[email-masked]>, "J. Matthew Nyman" <[email-masked]> wrote:

> Ok, assuming I'm not crazy here and assuming I understand a little of 
> what you guys are telling me (and thank you very much for the obviously 
> well-thought out responses) Alcubierre is using general relativity to 
> postulate a universe of hypersurfaces - but these hypersurfaces are just 
> areas in space that would normally not be causally connected, hence the 
> spacelike separation.  Now, allow me to quote from his paper: 
 
Yes, that's pretty much it.  Although phrasing it as "using general
relativity to postulate a universe of hypersurfaces" is a little weird.
Spacetime can be foliated into space and time under for just about
any physically meaningful spacetime.  (I'm not sure what the rigorous
conditions for a manifold to admit a foliation are, but they must be pretty
general.)  The 3+1 formalism assumes you can do this.  If I hunted
through Wald, I could probably find out when it's allowed.

And two points in space at the same "time" (residing in the same
spacelike hypersurface) are not causally connected, so you're right
there.

> "Notice that as long as the metric [gamma]_ij is positive definite for 
> all values of t (as it should be in order for it be a spatial metric), 
> the spacetime is guaranteed to be globally hyperbolic."
> 
> What does he mean by "positive definite?"  Does this just mean that time 
> will never be negative, or, in other words, flowing in the opposite 
> direction?  (This would seem to suggest that time travel is ruled out of 
> the Alcubierre warp idea, which is probably a good thing.)  He makes 
> mention that the 3+1 formalism of general relativity will have no closed 
> causal curves so, again, I assume that rules out time travel of any sort 
> and is based on the aforementioned "positive definite" values and I 
> assume that this is, in general, what global hyperbolicity entails.
 
"Positive definite" means that the inner product of two vectors is
always positive.  A positive definite metric on a hypersurface means
that two events which lie in the same hypersurface have a spacelike
separation.  This is good, because they supposedly take place at the
same "time" (at least from the perspective of some observer) -- all
events in one hypersurface occur at the same coordinate time.  A better
way of thinking about spacelike separation is that events which are
spacelike separated cannot influence one another casually.  For a causal
influence to occur, one event has to be "before" the other; in fact, its
time separation has to be larger than its space separation.  This goes
back to the Lorentz metric on Minkowski spacetime:  ds^2 = -dt^2 + dx^2
+ dy^2 + dz^2 (in units where c=1, otherwise there's a c^2 before the
dt^2).  If ds^2 > 0, the separation is spacelike.  For example, in flat
spacetime, the events t=0,x=y=z=0 and t=0,x=y=z=1 meter are spacelike
separated.. a signal from x=y=z=0 wouldn't be able to reach x=y=z=1
until t=however long it takes light to travel that far.

The definition of "globally hyperbolic" is rather technical, but it
means that you can take the state of the universe on one of these
surfaces at one time and predict the entire past and future of the
universe from it.  This means that there can be no closed timelike
loops.  (A closed timelike loop is just a timelike worldline that
doubles back on itself.  Remember, the timelike curves are the ones
that real particles travel on, so that just boils down to saying that
you could travel in such a way as to meet yourself in your past.)

It can be proven that if the 3-metric is positive-definite on all
hypersurfaces in the foliation, then spacetime is globally hyperbolic,
and hence closed timelike loops (CTLs) and time travel don't exist.
Time travel is automatically ruled out anytime you have such a
foliation and anytime you work in the 3+1 formalism.

> [Now, if I could hammer away at that first formula of Alcubierre's that 
> describes the metric he intends on postulating, my notes in class 

(By the way, Alcubierre's first formula is the decomposition of _any_
4D metric into space and time parts, not just the metric he intends on
postulating.)

> indicate that the "Greek indices will take the values (0,1,2,3) and are 
> contravariant tensors and Latin indices take the values (1,2,3), which 
> are covariant tensors."  Is there any way to briefly explain the 
> difference between covariant and contravariant?]

That's not quite right.  Greek indices will take the values (0,1,2,3)
and indicate spacetime vectors (or, in general, tensors).  Latin ones
indicate space vectors/tensors (which reside in a spacelike
hypersurface) which take the values (1,2,3) (no time dimension).

But they both can indicate contravariance or covariance.  Upstairs
indices (superscripts) indicate contravariance, downstairs indices
(subscripts) indicate covariance.  (Or vice versa..  some people reverse
the terminology.)  I don't want to attempt to take you through a whole
tutorial on GR (we just got through doing one on sci.physics!), so I'll
just very briefly say a few things.

The notation b^i refers to a vector, also known as a contravariant
tensor of rank 1.  b_i is a covariant tensor of rank 1, or a "1-form".
It's something that eats a vector and spits out a number, in a linear
way.  You can raise and lower indices to convert between the two.  For
example, b_i = g_ij b^j.  The inner product of two vectors u and v,
which I wrote as g(u,v) in a previous post, is written as g_ij u^i v^j.  
This can also be written as u_j v^j.  (This is because you define the
1-form u which corresponds to the vector u as the one such that u(v) = g(u,v). 
Here, the u on the left is the 1-form u which is eating a vector, and
the u on the right is the vector u which corresponds to it.)  The
metric itself is a covariant tensor of rank 2; it takes _two_ vectors
and linearlly spits out a number, their dot product.

So, for example, when you see b_i b^i, what that really means is g_ij
b^i b^j, or g(b,b).  In other words, the squared magnitude of the
vector b.

If you have chosen a particular basis, you can think of these things
using linear algebra.  The nice thing about the tensor notation is that
it works for any basis, but if you want to choose one and get all icky
and disgusting by working in a particular coordinate system, you can
think of b^i as a column vector, b_i as a row vector, and g_ij as a
matrix.  By the equation b_i = g_ij b^j, we see that this is just a
matrix multiplied by a vector.  (Repeated indiced denote summation, so
this equation actually means the sum over j of g_ij*b^j.  In something
like g_ij b^i b^j, we have a double summation over i and j.)  It
actually works out that b_i = transpose of the column vector b^i
multiplied by the matrix g on the right.  So g(u,v) = u_i v^i = g_ij
u^i v^j = u^t g v.  If g is the 3x3 identity matrix, you get the
Euclidean dot product.  If g is the 4x4 matrix with {-1,1,1,1} on the
diagonal, we have the Lorenztian metric.

But enough of this component, coordinate system based stuff.  It's
very ugly, but I don't have the time to go on about the abstract
version.  Look at the GR tutorial thread for that, or in any GR book.

You may be interested to see the derivation of Alcubierre's formula (1).
I worked it out for fun last night.  You'll have to refer to the other
message I wrote.  I'll work it out abstractly without the indices, since
the index notation isn't really designed to work with 4-vectors and
tensors and 3-vectors and tensors mixed together.  But I'll try to
convince you that the result is the same.  (Actually, I'll throw in a
few indices to help you compare, but things are clearer, I think, if I
avoid them.)

Okay, given a tangent vector dx^a, let's compute its spacetime length
from the metric g_ab.  Its spacetime length is written ds^2, or -dtau^2,
where dtau is the proper time you would experience if you travelled for
a short distance with velocity dx^a.

Let's pick a basis for our vectors, {t,x,y,z}.  (We can do this if the
time vector t does not lie competely in the spacelike hypersurface,
which is what we would expect.  Things which are tangent to the
hypersurface are regarded as "space" vectors with no time component, so
a time vector should stick out of the surface.)  That means that we can
write dx^a in terms of that basis:  dx^a = dt t + dx x + dy y + dz z.
The latter three terms are the spatial part of the vector, and we will
refer to that vector as as dx^i.  So dx^a = dt t + dx^i.

(The index notation isn't really designed to handle equations with Greek
and Latin indices, because things like g_ab dx^i dx^j don't quite make
sense, although by common sense we would interpret that as the dot
product of the spatial 3-vector dx^i with itself, using the
4-dimensional metric g:  g(dx^i,dx^i).  I've opted to bypass these
oddities by bypassing the index notation.)

Now, the spacetime length of dx^a may be calculated from the metric.
The square of the length of the vector is just the dot product of the
vector with itself, which is g(dx^a,dx^a).  (This is written as g_ab
dx^a dx^b in index notation.)  A little calculation reveals:

ds^2 = -dtau^2 = g(dx^a,dx^a)
               = g(dt t + dx^i, dt t + dx^i)
               = g(dt t, dt t) + g(dx^i, dt t) + g(dt t, dx^i) + g(dx^i, dx^i)
               = g(dt t, dt t) + 2g(dt t, dx^i) + g(dx^i, dx^i)
               = dt^2 g(t,t) + 2dt g(t,dx^i) + g(dx^i,dx^i)

I have utilized the linearity property of the metric where g(au+bv,w)
= ag(u,w) + bg(v,w), as well as the symmetry property g(u,v) = g(v,u).
These are familiar properties of the dot product.

Okay, let's work out some of these terms.  In my previous article, I
showed that the time vector t may be broken down into components normal
and tangent to the hypersurface, using the lapse function and shift
vector.  That is, t = an + b.  (This would be written as t^a = an^a +
b^i or some such in index notation, where 'a' is alpha and 'b' is
beta, of course.)

So,

g(t,t) = g(an+b,an+b) = a^2 g(n,n) + 2a g(n,b) + g(b,b),

by exactly the same procedure we did above.  However g(n,n) = -1,
because we defined n to be a unit timelike vector normal to the
hypersurface.  (Unit implies squared mangitude +/- 1, timelike implies
squared magnitude < 0.)  Furthermore, g(n,b) = 0.  This is because b,
the shift vector, is tangent to the hypersurface (lies in it), whereas
n is normal to it.  Thus, they are perpendicular, so their dot product
is zero.  So, we have:

g(t,t) = -a^2 + g(b,b) = -(a^2 - g(b,b))

Now we have one of the three terms for ds^2 worked out.  Let's do the
next one:

g(t,dx^i) = g(an+b,dx^i) = ag(n,dx^i) + g(b,dx^i) = g(b,dx^i).

(g(n,dx^i) = 0, because dx^i is a spatial vector lying in the
hypersurface again, while n is normal to it.)

We're not going to do anything with the last term for ds^2, so let's
rewrite it in terms of what we've worked out:

ds^2 = -(a^2 - g(b,b)) dt^2 + 2g(b,dx^i) dt + g(dx^i,dx^i)

This, I claim, is Alcubierre's equation (1).  Let's try to write it in
index notation.

g(b,b) is just the squared length of the vector b^i, which is
g_ij b^i b^j, or b_i b^i.  So that part looks right.  Next, let's look
at g(dx^i,dx^i).  That's just the squared length of the spatial vector
dx^i.  But remember, we defined the 3-metric y so that it agrees with
the 4-metric g whenever both the vectors it takes are spatial vectors.
This is the case here, so g(dx^i,dx^i) = y(dx^i,dx^i).  In index
notation, this is y_ij dx^i dx^j.  Finally, we have g(b,dx^i).  This
is the dot product of those two vectors, which is g_ij b^i dx^j.  (Or
g_ij b^j dx^i if you are attached to using the 'i' index for the
latter vector; it doesn't matter).  But this just equals b_i dx^i.
(Or equivalently, b^i dx_i ...)

So, finally we obtain:

ds^2 = -(a^2 - b_i b^i) dt^2 + 2 b_i dx^i + y_ij dx^i dx^j

Look familiar?

> One more thing and I promise I'll go away for now :-)
> 
> Alcubierre says that his metric will be described by a metric that has 
> the following property: G=c=1.  Now, I've seen things like this before 
> where certain constants are set to unity so as to make computing values 
> easier by relating, say, time and energy or something like that.  Is 
> this what is happening here?

Unfortunately, most scientists have chosen to adopt silly units.  I
prefer to choose to measure time in seconds and distance in
light-seconds, so that c=1.  We can also arrange things so that G=1
too.  That way, you don't have to carry around all those dumb c's and
G's in all your equations.  For example, the flat spacetime metric is
ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 instead of -c^2 dt^2 + ...
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
--------------------------------------------------------------------------

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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: 27 May 1996 21:56:41 -0400
Organization: Virginia Polytechnic Institute and State University
Lines: 77
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In article <[email-masked]>, "J. Matthew Nyman" <[email-masked]> wrote:

> Now, in Alcubierre's formulas 2, 3, 4, 5 he seems to delineate further 
> the metric he is using.  For example, he seems to set the lapse function 
> to 1.  [How about that, huh?  Notice me picking up on on this 
> terminology stuff!!! :-)]  He then goes through some pains to explain 

"Seems to"?  He _does_ set the lapse function to 1, and does define his
metric.  :)  Further, he defines the shift vector to have only an
x-component (because he wants motion only in the x-direction), and he
sets the 3-metric to the identity matrix.  (This implies that the
spatial hypersurfaces are flat Euclidean 3-space.)

> the shift vector by stating that v_s(t) = dx_s(t) / dt.  Now, I don't 
> pretend to understand what any of that means but then he goes on to 

Alcubierre has defined x_s(t) to be the path the spaceship takes through
space as a function of time; x as a function of t.  (This works because
he has defined a coordinate system, so you can talk about a particular x
coordinate.)  Velocity is merely the rate of change of position with
velocity, which is just dx_s(t)/dt.  What Alcubierre has done is
_define_ his metric so that the spaceship travels along the path x_s(t).
That is, he started with a path through spacetime that he wanted the
spaceship to take, and invented a spacetime curvature which would
produce that motion.

> describe a function or r_s and says that for certain values this 
> function rapidly approaches a "top hat" function.  What the heck is a 
> "top hat" function?  None of my references list anything like this 
> although my professor says that the speed of light is a top hat 
> function, in that it sets the ultimate limit on information propagation. 
>  Is this all a top hat function really is?

Uh, no.  That's not how the term "top hat" is used.  (I don't know what
your professor is thinking.  Perhaps you misinterpreted?  What course
is this for, anyway..?  I missed the beginning of the thread.)

A top hat function is just a function that is constant on some interval
and zero outside of it.  It is exactly what equation (7) says:  the top
hat function is 1 when -R <= r_s <= R, and 0 otherwise.  It looks like a
flat "top hat" sticking up from the r_s-axis, centered at the origin.

The reason he mentions this is because (when sigma is large, at least)
the fact that the shift vector vanishes outside the interval [-R,R]
implies that spacetime is (essentially) flat there.  (From the
definition of the shift vector, you can see that if f(r_s) vanishes, so
does the shift vector.)  Thus, we only need to really concentrate on
the spacetime disturbance in a region of radius R around the
spaceship.

> One other thing of a general nature --- Alcubierre's scheme obviously 
> requires what has been called "exotic matter."  Now, my professor says 
> that this requires negative pressures and/or negative mass.  So is 
> exotic matter then the same thing as negative matter?  I've have always 
> thought the two were separate ideas, with exotic matter being a little 
> more plausible.  Does general relativity really allow for something like 
> exotic matter that is required in Alcubierre's scheme?  (Part of my 
> research paper will certainly be as to whether the idea is even possible 
> in theory or not.)

I don't know if there is a definition of "exotic matter".  I think it is
just generally used to mean "weird stuff we haven't seen yet".

General relativity does not rule out negative matter, but the
strong/weak/dominant energy conditions are postulated to say that that
can't happen.  (Simply because we don't know of anything with negative
mass.)  Alcubierre mentions that the quantum Casimir effect may provide
a way out, but that seems rather far-fetched, especially on a scale
large enough to power a warp drive and move a spaceship between stars.
Don't ask me to explain the Casimir effect; I don't know anything about
how it works.  Other people around here know about it, though, and it
might be in the FAQ.
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
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Article: 18178 of sci.physics
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From: "J. Matthew Nyman" <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: Tue, 28 May 1996 00:17:49 -0500
Organization: Advanced Studies Group
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Nathan Urban wrote:

> "Seems to"?  He _does_ set the lapse function to 1, and does define   
> his metric.  :)  Further, he defines the shift vector to have only an
> x-component (because he wants motion only in the x-direction), and he
> sets the 3-metric to the identity matrix.  (This implies that the
> spatial hypersurfaces are flat Euclidean 3-space.)

GOD!  How do you learn all this stuff????  My head's going buggy.  I 
feel like a moron more and more each day!  :-)  (At least the people 
here don't criticize you - you should hear my professor.)

Now, I do understand lapse functions somewhat from a previous post on 
this thread.  But could you elaborate as to why he sets it to 1.  You 
mention the shift vector which I guess is his equation (3) B^x = -v_s(t) 
f(r_s(t)).  So this is the x motion that you speak of and that I 
somewhat understand (believe it or not!)  Now, equation (4) B^y = B^z = 
0 seems to simply rule out the other spatial directions, right?  I think 
I got that part too.  He basically wanted his model simple, it sounds 
like.  But that last part threw  me off; what do you mean that he set 
the 3-metric to the identity matrix?  I assume this must relate to 
equation (5). What is an "identity matrix?"
 
>What Alcubierre has done is_define_ his metric so that the spaceship   
>travels along the path x_s(t). That is, he started with a path through 
>spacetime that he wanted the spaceship to take, and invented a 
>spacetime curvature which would produce that motion.

Okay, I understand then the v_s(t) = dx_s(t) / dt.  But he also says 
that r_s = [(x - x_s(t)^2 + y^2 + z^2]^1/2.  Now, the r_s appeared 
earlier in the shift vector.  Why would he need to set this parameter 
here again?  (Sorry if I appear overly ignorant here!)  But this 
certainly brings up a good point.  Are you saying that Alcubierre was 
not originally basing his model on anything real, per se.  In other 
words, it's sounding like he had a model of a "warp" engine he wanted so 
he fudges the figures to fit the facts to see if it works out in the 
end, presumably without any violations of physical law.

Basically, I am having a great deal of trouble understanding exactly how 
he set up this metric of his, which seems to be contained in equations 
(1) through (8).  I have barely gotten past that point.  You see, that 
basically is what my research paper has to be on - the development of 
Alcubierre's idea of the warp drive, which of course hinges on the 
metric he has chosen to use, and whether or not it is feasible.  Now, 
the latter I realize will probably be contingent on whether exotic 
and/or negative matter is found to exist so I'm trying to concentrate 
more on the former.

> Uh, no.  That's not how the term "top hat" is used.  (I don't know    
> what your professor is thinking.  Perhaps you misinterpreted?

No, I'm sure that is what he said.  I record some of the class sessions 
as well.  To be sure, I have no clue what he was thinking either.  There 
are many things he has told the class that later turned out to be 
inaccurate, although these things dealt mostly with quantum mechanics 
and quantum gravity.  However, part of the problem I think is that he is 
dead set that Alcubierre's paper is just total trash (his words) and it 
is not worth my time to investigate it.  However, we have free choice so 
I chose it.  My professor is convinced that the math in the paper is 
just faulty, although he will not explain why.  In fact, he has made 
mention to me that the paper will be graded almost solely on whether or 
not I can give some justification for the math, or, if not, that I can 
totally disprove it.

> What course is this for, anyway..?  I missed the beginning of the > > thread.) 

The course is "Quantitative Literacy Series: Theoretical Physics 
Research."

Thanks for the explanation of "top hot."  I will be sure to 
not-so-subtly insinuate in the body of the paper that the prof. was 
quite incorrect in his "assumption."

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Article: 18242 of sci.physics
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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: 28 May 1996 16:05:45 -0400
Organization: Virginia Polytechnic Institute and State University
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In article <[email-masked]>, "J. Matthew Nyman" <[email-masked]> wrote:

> > "Seems to"?  He _does_ set the lapse function to 1, and does define   
> > his metric.  :)  Further, he defines the shift vector to have only an
> > x-component (because he wants motion only in the x-direction), and he
> > sets the 3-metric to the identity matrix.  (This implies that the
> > spatial hypersurfaces are flat Euclidean 3-space.)

> GOD!  How do you learn all this stuff????

You spend several years learning it out of some good books.  :)

>  My head's going buggy.  I 
> feel like a moron more and more each day!  :-)  (At least the people 
> here don't criticize you - you should hear my professor.)

Ah, you haven't been here long enough.  Criticism flies fast and furious
around here.  :)

> Now, I do understand lapse functions somewhat from a previous post on 
> this thread.  But could you elaborate as to why he sets it to 1.

I'm not quite sure about that.  I think the main thing is for it to be
constant.  (That corresponds to the hypersurfaces being evenly spaced in
time, since the shift function can be thought of as how much in the
"time" direction the normal to the hypersurfaces are.)  I guess it could
be any constant, but if you changed the constant, I think you'd have to
readjust the shift vector by the same factor to get the same motion.
(That's because the lapse function can be thought of as a measure of how
fast time flows normal to the hypersurface; it is the normal component
of the time vector.  So, if you made the lapse function 2 instead of 1,
you'd need to divide the shift vector by 2 --  making the lapse function
2 means that you're measuring time twice as fast as before, so you'd
need to travel twice as slow to get the same curve.)

(I could be wrong on some of this.  I'm learning the 3+1 formalism as I
go along.  Good thing I've got all these GR books lying around.)

>  You 
> mention the shift vector which I guess is his equation (3) B^x = -v_s(t) 
> f(r_s(t)).

The shift vector is defined in equations (3) and (4).

>  So this is the x motion that you speak of and that I 
> somewhat understand (believe it or not!)

You can easily see that if there is any motion at all, it has to be
x-motion.  That's because the lapse function does not depend on x,y, or
z, so it doesn't introduce a direction dependence.  Similarly the
3-metric is symmetric in x, y, and z, so it can't either.  But the shift
vector points along the x-axis, so that's where x-motion must come from.

>  Now, equation (4) B^y = B^z = 
> 0 seems to simply rule out the other spatial directions, right?  I think 
> I got that part too.

Yes.  (An asymmetry in the lapse function or 3-metric could make it so
other spatial directions are not ruled out, but that's not the case
here.)

>  He basically wanted his model simple, it sounds 
> like.

Yes.

>  But that last part threw  me off; what do you mean that he set 
> the 3-metric to the identity matrix?  I assume this must relate to 
> equation (5). What is an "identity matrix?"

Hmm, I think I've been talking over your head if you don't know about
matrices.  (Do you know about dot products?  If not, then none of the
stuff I've been explaining about the metric, lapse function, and shift
vector will make any intuitive sense.  Did you follow any of that?)

An identity matrix is one which, when multiplied by any vector, gives
the same vector.  There is only one identity matrix for a given vector
space.  The 3x3 identity matrix is:

     [ 1 0 0 ]
     [ 0 1 0 ]
     [ 0 0 1 ]

This happens to be the metric for 3D Euclidean space.  In a previous
message, I tried to explain how to use the metric to compute the squared
length of a vector, but I guess you didn't understand.  If you have a
vector (dx,dy,dz), its squared length is given by:

                [ 1 0 0 ] [ dx ]
     [dx dy dz] [ 0 1 0 ] [ dy ]
                [ 0 0 1 ] [ dz ]

which, by the rules of matrix multiplication, gives you the number (or
1x1 matrix) dx^2 + dy^2 + dz^2.

Now, for 4D Minkowski space (the Lorentzian flat spacetime of special
relativity), the metric is:

     [-1 0 0 0 ]
     [ 0 1 0 0 ]
     [ 0 0 1 0 ]
     [ 0 0 0 1 ]

If you use it to compute the (squared) length of the spacetime 4-vector
(dt,dx,dy,dz) you'll get -dt^2 + dx^2 + dy^2 + dz^2.

Let me mention that these expressions depend on what coordinate system
you're using.  (The language of tensors was invented so you could do
math without worrying about what coordinate system you're using.)  For
example, in 2D Euclidean space (the plane), the metric is written as 

     [ 1 0 ]
     [ 0 1 ]

in Cartesian coordinates.  (I hope this is obvious to you by now!)
That's because ds^2 = dx^2 + dy^2.  However, in polar coordinates (r,t)
(where 't' is 'theta', not "time"), a simple geometric argument will
tell you that ds^2 = dr^2 + r^2 dt^2.  That means that the metric in
polar coordinates is written as

     [ 1 0 ]
     [ 0 r ]

The catch is, _both_ of the matrices represent the same metric.  But
there are ways of telling if they're really the same.  The most
important test is to see if the metric is flat.  (You can write down
metrics for curved spaces, too, such as the surface of a sphere.)  It
turns out that if you get that the metric is the identity matrix
everywhere, then that metric describes flat Euclidean space.  So we
know that the fact that the 3-metric is the identity matrix really means
that our spacelike hypersurfaces are flat Euclidean 3D space.

> >What Alcubierre has done is_define_ his metric so that the spaceship   
> >travels along the path x_s(t). That is, he started with a path through 
> >spacetime that he wanted the spaceship to take, and invented a 
> >spacetime curvature which would produce that motion.

> Okay, I understand then the v_s(t) = dx_s(t) / dt.  But he also says 
> that r_s = [(x - x_s(t)^2 + y^2 + z^2]^1/2.  Now, the r_s appeared 
> earlier in the shift vector.  Why would he need to set this parameter 
> here again?  (Sorry if I appear overly ignorant here!)

Note that r_s defines a spherically symmetric function, centered around
the point (x_s(t),0,0).  This is the location of the spaceship.  The
v_s(t) term is necessary to give the spaceship the desired velocity.
The f(r_s(t)) term multiplies it by a spherically symmetric
"top-hat"-like function. (Remember, it's only a top hat for large
sigma.  I believe he threw the sigma in so that you can adjust how
top-hat-like the function is by varying sigma.)  This makes it so that
spacetime is flat far away from the spaceship.  Although space is flat,
spacetime is not, because the shift vector is nonzero.  But because of
the top hat function, the shift vector gets very small far away from
the spaceship.  That means that far from the spaceship, the lapse
function is 1, the 3-metric is the identity, _and_ the shift vector is
zero.  Plugging all of those things into equation (1) will tell you
that spacetime is flat.  I'll demonstrate this.

ds^2 = -(a^2 - b_i b^i) dt^2 + 2 b_i dx^i dt + y_ij dx^i dx^j

Now, a^2 = 1.  Since b^i = 0, we have that b_i b^i is zero.  That's
because b_i b^i is just the squared length of the shift vector, which of
course is zero since the vector is zero.  Likewise, the term b_i dx^i is
just the dot product of b^i with dx^i.  But the dot product of any
vector with the zero vector is zero, so that term vanishes as well.
So we have:

ds^2 = -dt^2 + y_ij dx^i dx^j

But the second term is just the squared length of the space vector dx^i.
(We would ordinarily write it as dx_i dx^i, like we did with the shift
vector, but having one index up and one down means compute the length
using the 4-metric instead of the 3-metric.  The strange thing is that
the shift vector b^i is a space vector too, so computing its length with
the 3-metric is defined to be the same as computing it with the
4-metric.  I guess he just wanted to make the fact that he was using the
3-metric more explicit for dx^i.)  Since y_ij is just the identity
matrix, or the Euclidean 3-metric, the squared length of dx^i = (dx,dy,dz)
is, of course, dx^2 + dy^2 + dz^2.  That means

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

Which is just the metric for flat spacetime.

>  But this 
> certainly brings up a good point.  Are you saying that Alcubierre was 
> not originally basing his model on anything real, per se.  In other 
> words, it's sounding like he had a model of a "warp" engine he wanted so 
> he fudges the figures to fit the facts to see if it works out in the 
> end, presumably without any violations of physical law.

Exactly.  He just cooked up a metric to suit the effect he wanted to
produce, and then found out that it requires negative mass to produce
the warp effect.  (We don't know how he came up with the metric, though
by working backward we can justify that it should produce the effect he
claims.  But, like all good scientific papers, he covers up all the
time he spent coming up with the metric and just gives the answer.)

(The way he calculated that negative matter would be required is by
using Einstein's equation.  Using the metric, which tells you
_everything_ about spacetime and its curvature, you can calculate all
sorts of things.  This includes a curvature-related quantity, the
Einstein tensor G_ab.  Einstein's equation relates spacetime curvature
to the matter occupying spacetime, by saying that the Einstein tensor
G_ab is proportional to the stress-energy tensor T_ab.  By calculating
G_ab from the metric, you can thus use Einstein's equation to find T_ab.
The time-time component of T_ab, which is T_00, is just energy density.
That turns out to be a negative number in the matter's rest frame.)

> Basically, I am having a great deal of trouble understanding exactly how 
> he set up this metric of his, which seems to be contained in equations 
> (1) through (8).  I have barely gotten past that point.  You see, that 
> basically is what my research paper has to be on - the development of 
> Alcubierre's idea of the warp drive, which of course hinges on the 
> metric he has chosen to use, and whether or not it is feasible.  Now, 
> the latter I realize will probably be contingent on whether exotic 
> and/or negative matter is found to exist so I'm trying to concentrate 
> more on the former.

Well, you're going to have some trouble then..  he doesn't explicitly
say how he came up with that particular metric.  (Maybe you should ask
him.  He's probably sick of getting e-mail from weirdos asking him how
to build one, though..)  :)  Presumably, there are a lot of metrics you
could cook up which would produce the same effect.  I think we've got
most of it worked out, though.  We've discussed why the lapse function
is 1, and why the shift vector is in the x-direction, and why the
3-metric is the identity matrix..  as well as why r_s and f(r_s) are
defined as they are (to give approximately flat spacetime far from a
spherical region around the spaceship).

(This is desirable because of equation (13) that says that dtau = dt.
tau is the proper time that the obsrever in the spaceship measures.  t
is just some time coordinate which we've decided to use.. it doesn't
necessarily have to correspond to anything physical.  But the fact that
for this metric we have dtau = dt is convenient.  In _flat_ spacetime,
coordinate time corresponds to the proper time of observers.  Since
spacetime is flat far away from the spaceship, that means that the
spaceship's time corresponds to the time of distant observers as well.
So if the trip takes a short time from the spaceship's perspective, it
takes a short time from the distant observers' perspective as well.)

> > Uh, no.  That's not how the term "top hat" is used.  (I don't know    
> > what your professor is thinking.  Perhaps you misinterpreted?

> No, I'm sure that is what he said.  I record some of the class sessions 
> as well.  To be sure, I have no clue what he was thinking either.  There 
> are many things he has told the class that later turned out to be 
> inaccurate, although these things dealt mostly with quantum mechanics 
> and quantum gravity.

What field of physics does your professor work in?

>  However, part of the problem I think is that he is 
> dead set that Alcubierre's paper is just total trash (his words) and it 
> is not worth my time to investigate it.  However, we have free choice so 
> I chose it.  My professor is convinced that the math in the paper is 
> just faulty, although he will not explain why.  In fact, he has made 
> mention to me that the paper will be graded almost solely on whether or 
> not I can give some justification for the math, or, if not, that I can 
> totally disprove it.

Yay.  The deck is kinda stacked against you, huh?

As far as I can tell, Alcubierre's math is correct.  (I also think that
if it were wrong, I would have heard about it here.  It comes up from
time to time on this newsgroup, and other than the exotic matter
problem, I haven't heard a GR expert say the paper has problems.)  I'm
not totally convinced that Alcubiere's metric gives the motion x_s(t)
for the spaceship, but that's just because I haven't worked with the 3+1
formalism before.  I'm sure that it would turn out to be right if I
studied it enough.

> > What course is this for, anyway..?  I missed the beginning of the > > thread.) 

> The course is "Quantitative Literacy Series: Theoretical Physics 
> Research."

I guess the goal is to pick some research paper and explain it?

> Thanks for the explanation of "top hot."  I will be sure to 
> not-so-subtly insinuate in the body of the paper that the prof. was 
> quite incorrect in his "assumption."

Yeah, good way to go..  piss off the professor right from page 1.  :)

It sounds like he doesn't even understand special relativity correctly,
if he though that the top hat function was referring to the maximum
speed of information propagation..
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
--------------------------------------------------------------------------

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From: "J. Matthew Nyman" <[email-masked]>
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: Thu, 30 May 1996 22:42:21 -0500
Organization: Advanced Studies Group
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>>GOD! How do you learn all this stuff?

>You spend several years learning it out of some good books.	.....
>Good thing I've got all these GR books lying around.

What reading materials do you have?  I am looking for good books that 
will explain all of these concepts like you've been doing.  Many of the 
books I find will explain the theory, but forgo the mathematics or, 
conversely, will have reams of math but little or no explanation.  Could 
you possibly recommend some good books, or at least the ones that have 
helped you out?

>Hmm, I think I've been talking over your head if you don't know about matrices.  (Do you know about >dot products?  If not, then none of the 
stuff I've been explaining about the metric, lapse function, and >shift 
vector will make any intuitive sense.  Did you follow any of that?) 

Well, sort of.  You are right that I don't have that intuitive feel for 
what's going on here, at least on a mathematical level.  Mainly I'm 
trying to understand just enough to be able to form a decent opinion 
(and be able to back that opinion up) about the validity of the 
mathematics in  Alcubierre's paper.  Unfortunately, I'm finding that, as 
you've mentioned, I've got to learn about spacetime metrics, tensor 
fields, manifolds, tangent and cotangent vectors, and the kitchen sink. 
 This is really a trial by fire for me since I was not too bright on 
these subjects to begin with.  I've managed to pick up some material 
from our library that is sort of helping me, but the books are not 
geared towards physics and therefore their usefulness is somewhat 
limited.

Do I know what a dot product is?  If I am not sadly mistaken it has to 
do with the scalar (magnitude) properties; something like the cosine of 
the angle (?) between two vectors.  I also know that it somehow hinges 
upon the actual length of the vectors.  However, I'll be the first to 
admit - my knowledge of matrices is shaky at best.  That's something I'm 
currently learning.  But this brings up something: in an earlier post 
you said "we have something called a 'metric,' which is like an inner 
(or 'dot') product."  Does this refer to the fact that a metric will 
describe the curvature of space?  In other words, are you saying that 
the metric acts like the measure of an angle between two vectors and 
that angle is what we take as the curvature? 

Another problem is this whole "metric" thing.  I did know that a metric 
describes the local curvature of spacetime.  Now, I have heard that 
Einstein's metric g is given by the following array:
	g_11	g_12	g_13	g_14
	g_21	g_22	g_23	g_24
	g_31	g_32	g_33	g_34
	g_41	g_42	g_43	g_44

I then hear that g_12 = g_21, g_13 = g_31, g_14 = g_41 so that there are 
only ten components.  But what actually are the values of these 
components?  What is it that the numbers are actually describing?  What, 
in essence, is g?

Basically, you were correct in an earlier post when you said I had to 
learn more about tangent vectors, manifolds, tensors, etc, etc.  Know 
any good web sites or any place at all where this stuff is pretty 
clearly delineated for a beginner?

>What field of physics does your professor work in?

He says that his speciality is something called Global General 
Relativity.

>Yay.  The deck is kinda stacked against you, huh?

Yes, in some ways it is.  The problem is that my professor really is 
smart - at least he seems so to me, although I suppose, when you 
consider the source, that may not be saying much.  But he strikes me as 
very arrogant.  And when he disagrees with something he just says he 
does and will not back up his reasoning with anything solid and that is 
something I can't stand.

>I'm not totally convinced that Alcubiere's metric gives the motion x_s(t) for the spaceship, but that's just >because I haven't worked with 
the 3+1 formalism before.  I'm sure that it would turn out to be right 
if I >studied it enough.

Interesting.  Why are you not convinced, at this juncture, as to whether 
Alcubierre's metric works?

>I guess the goal is to pick some research paper and explain it?

You got it.  And this paper seemed like a good one for two reasons.  
Number one: I like the idea of the papre.  Number two: my professor 
hates the idea of the paper.  It's a match made in heaven.

>Yeah, good way to go..  piss off the professor right from page 1.  :)

Yes, I guess on second thought that might not be a good idea.  I'd just 
like to rub his face in something.  Incidentally, I did ask him again 
about top-hat functions.  He says that they are anything that has an 
ultimate limit of infinity.  He said that the speed of light is just one 
example since it would take an infinite amount of energy for something 
with mass to attain the speed of light.  The "top" of  the "hat" (if you 
imagine it as a real top hat) corresponds to something that is infinite 
if that something concerns some form of information propagation.

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Article: 18748 of sci.physics
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From: [email-masked] (Nathan Urban)
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: 31 May 1996 02:14:48 -0400
Organization: Virginia Polytechnic Institute and State University
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In article <[email-masked]>, "J. Matthew Nyman" <[email-masked]> wrote:

> What reading materials do you have?  I am looking for good books that 
> will explain all of these concepts like you've been doing.  Many of the 
> books I find will explain the theory, but forgo the mathematics or, 
> conversely, will have reams of math but little or no explanation.  Could 
> you possibly recommend some good books, or at least the ones that have 
> helped you out?

Okay, here are some recommendations:

_Flat and Curved Space-Times_, Ellis and Williams
This is a pretty good intro book.  It spends a bunch of time on special
relativity, and then moves on to some of the simpler aspects of curved
spacetimes.  No tensors involved though, except for the metric.  Very
light on the math -- elementary calculus at worst.

_Gravitation_, Misner, Thorne and Wheeler
This is the standard first-year grad text at many schools.  HUGE book.
It's got a lot of good stuff in it, and they give a pretty good shot at
trying to help you visualize things, but the book is just so big that
it's too easy to get lost in it.  They also use a nonstandard notation.

_General Relativity_, Wald
This is the most concise intro to full GR I've seen.  Another standard
first-year grad text, more modern than MTW.  Very heavy on the math,
probably too hard for you right now.  But it discusses a lot of useful
stuff.  Good reference work.

There are some other intro books out there that are more at your level,
but I haven't looked at any of them.  To get familiar with some special
relativity concepts in flat spacetime, Taylor and Wheeler's _Spacetime
Physics_ is the standard book.  I hear they're supposed to be coming out
with an intro general relativity book pretty soon, I think.  That should
be good.

(Also, the calculations I was doing with the 3+1 formalism were based
out of a good explanation that may be found towards the back of Baez
and Muniain's _Gauge Fields, Knots, And Gravity_.  Not a GR book, but
it has lots of neat stuff in it and plenty of pictures.  Heavy on the
abstract math, though.)

> Well, sort of.  You are right that I don't have that intuitive feel for 
> what's going on here, at least on a mathematical level.

It would help immensely if you had some diagrams to look at.  Those
don't come across in ASCII too well.  There are some good ones in the
aforementioned book by Baez and Muniain for the 3+1 formalism.

> Do I know what a dot product is?  If I am not sadly mistaken it has to 
> do with the scalar (magnitude) properties; something like the cosine of 
> the angle (?) between two vectors.  I also know that it somehow hinges 
> upon the actual length of the vectors.

The dot product equals u.v = |u||v|cos(theta).

The 3D Euclidean dot product also equals u1*v1 + u2*v2 + u3*v3.
Amazingly enough, that turns out to be the first formula in terms of
lengths and angles.

The dot product we use in relativity is a little different.  The first
formula still holds, but the angle theta isn't quite what you would
think it would be.  But we use this to _define_ the angle theta
anyway..  We define the dot product in special relativity (in the
standard Cartesian coordinates) to be -u1*v1 + u2*v2 + u3*v3 + u4*v4.
Then by saying that this equals |u||v|cos(theta), we can solve for
theta.  Note that the dot product can be less than zero with this dot
product.

Two things to note about the dot product:  u.u = |u|^2, so the dot
product of a vector with itself is the squared magnitude of that vector.
(With the SR dot product, which can be less than zero, we usually
multiply this by -1 before taking the square root to get the magnitude
if it's negative, to avoid getting an imaginary length.)  The other
thing to note is that if u.v = 0, we say the two vectors are perpendicular,
or "orthogonal".
 
> However, I'll be the first to 
> admit - my knowledge of matrices is shaky at best.  That's something I'm 
> currently learning.  But this brings up something: in an earlier post 
> you said "we have something called a 'metric,' which is like an inner 
> (or 'dot') product."

Actually, a metric _is_ a dot product.  Actually, it's a dot product
defined at every point in spacetime; the way you take a dot product will
vary from point to point.  It's a "dot product field".  (More precisely,
a type-(0,2) tensor field.)

>  Does this refer to the fact that a metric will 
> describe the curvature of space?  In other words, are you saying that 
> the metric acts like the measure of an angle between two vectors and 
> that angle is what we take as the curvature? 

Not quite.  It is true that the metric describes the curvature of space,
in some sense.  More precisely, it tells us how to measure the lengths
of vectors, and hence curves and things.  However, it is not what we
think of as curvature.  There is a tensor, called the Riemann curvature
tensor, which tells you everything about the curvature of spacetime. 
Given the metric, you can calculate the Riemann tensor, so the metric is
a more useful thing.  _Everything_ about the shape of spacetime is
contained in the metric.

Think of a surface approximated by a mesh of rubber bands tied
together.  It's floppy and deformable, all you can tell about it is its
topology.  But replace the rubber bands with toothpicks, and it becomes
rigid.  By measuring the lengths of the toothpicks, you can tell
everything about the shape of this rigid mesh.  This is what the metric
does for a space.

> Another problem is this whole "metric" thing.  I did know that a metric 
> describes the local curvature of spacetime.  Now, I have heard that 
> Einstein's metric g is given by the following array:
> 	g_11	g_12	g_13	g_14
> 	g_21	g_22	g_23	g_24
> 	g_31	g_32	g_33	g_34
> 	g_41	g_42	g_43	g_44

> I then hear that g_12 = g_21, g_13 = g_31, g_14 = g_41 so that there are 
> only ten components.  But what actually are the values of these 
> components?  What is it that the numbers are actually describing?  What, 
> in essence, is g?

g is the dot product.  To use the language of MTW, it is a machine with
two slots.  Each slot accepts a vector, and the machine spits out a
number, the dot product of the two vectors.  It satisfies the following
properties:

g(au,v) = ag(u,v)
g(u+v,w) = g(u,w)+g(v,w)
[the linearity properties]

g(u,v) = g(v,u)
[the symmetry property]

(If you don't like the notation, replace g(u,v) with u.v wherever you
see it.)

In fact, this is how we define the metric, abstractly -- it's anything
that satifies those properties.

Now, I'm sure you're lacking in abstract linear algebra.  When you read
about matrices, _be sure to also read about abstract vector spaces and
linear transformations_.  Linear algebra is worthless if you only know
about matrices, and nothing of the abstract approach.

Let me just tell you a tiny bit about this.  Given a vector space of
dimension n, you can pick out n vectors which form a basis for that
space.  That is, you can write any vector in that space as a unique
linear combination of basis vectors.  For example, the standard basis of
the x,y,z unit vectors.  However, bases are not unique; for example, you
could choose a basis x',y',z' which is rotated with respect to the first
basis.

Now, a linear transformation is a linear machine with one slot, that
takes a vector and spits out another vector.  Since it's linear, it
satisfies:

T(au) = aT(u), T(u+v) = T(u)+T(v)

Let's suppose our vector space is R^3, and we've chosen a basis x,y,z.
Then any vector v can be written as v = ax + by + cz.  Calculate T(v).
You'll get (by linearity) T(v) = aT(x) + bT(y) + cT(z).  T(x), etc. are
vectors, remember.  (Let's assume that T maps R^3 into itself, so T(v)
is in R^3 too.)  Now, we can decompose T(x) in terms of our basis too:
T(x) = T_xx x + T_xy y + T_xz z, where T_xx, etc. are just what I'm
choosing to call the coefficients.  Do the same for T(y) and T(z).
Then we see, by rearranging terms, that 

T(v) = a(T_xx x + T_xy y + T_xz z) + b(T_yx x + T_yy y + T_yz z)
      +c(T_zx x + T_zy y + T_zz z)
     = (aT_xx + bT_yx + cT_zx) x + (aT_xy + bT_yy + cT_zy) y
      +(aT_xz + bT_yz + cT_zz) z

We defined v so that v=(a,b,c) in this basis.  If w = T(v), then
we get w = (T_xx*a + T_yx*b + T_zx*c, ..., ...).

Or, we could write it as:

[ wx ]       [ T_xx T_xy T_xz ][ vx ]
[ wy ]   =   [ T_yx T_yy T_yz ][ vy ]
[ wz ]       [ T_zx T_zy T_zz ][ vz ] 

(Where I changed notation so vx=a, vy=b, vz=c.)

This is precisely where matrices and matrix multiplication comes from.
Note that the first column of the T matrix is just the components of
T(x), the second column is the components of T(y), etc.  A matrix is
just a representation of a linear transformation once you've chosen a
particular basis.  Change the basis and you'll change the matrix, but it
will still represent the same linear transformation.  Matrices which
represent the same linear transformation, but in different basis, are
said to be related by a "similarity transformation".

In general, given n basis vectors {e1,e2,...,en}, and a linear
transformation T, the component T_ij of the matrix corresponding to T in
that basis is just the ith component of the vector T(ej).

Now finally, on to the metric.  It turns out that this is just like a
linear transformation -- something that takes a vector and returns a
vector can be reinterpreted as something that takes _two_ vectors and
returns a _number_.  I don't want to get into the full details, but I'll
just say that the component g_ij of the metric is just g(ei,ej), the dot
product of the ith and jth basis vectors.  Of course, the components of
the metric's matrix change if you use a different basis (coordinate
system), which is why we like to take the abstract approach and write
equations which don't depend on any particular basis.

Also, since the metric is actually a _field_ of dot products, its
elements aren't numbers, but _functions_.  Refer to a previous article
for an example of the Euclidean metric in 2D written in Cartesian (x,y)
coordinates and polar (r,t) coordinates.  Same metric, different matrix
representations.

Hope this clears things up a bit.  (Note that because of the symmetry of
the metric, its matrix is symmetric across the diagonal, which reduces
it to 10 independent components instead of 16 in 4D, as you noted.)

> Basically, you were correct in an earlier post when you said I had to 
> learn more about tangent vectors, manifolds, tensors, etc, etc.  Know 
> any good web sites or any place at all where this stuff is pretty 
> clearly delineated for a beginner?

As a matter of fact, a multi-month general relativity tutorial on
sci.physics just finished up a little while ago.  Much of it is archived
at:

http://math.ucr.edu/home/baez/gr/gr.html

> Interesting.  Why are you not convinced, at this juncture, as to whether 
> Alcubierre's metric works?

Well, maybe that was poor wording.  I have nothing to suggest it's
wrong, it's just that I'm unable to verify it completely.  Specifically,
I don't know enough about the 3+1 formalism to show that given a
particular initial velocity, the spaceship will follow the advertised
trajectory over time.

> Incidentally, I did ask him again 
> about top-hat functions.  He says that they are anything that has an 
> ultimate limit of infinity.  He said that the speed of light is just one 
> example since it would take an infinite amount of energy for something 
> with mass to attain the speed of light.  The "top" of  the "hat" (if you 
> imagine it as a real top hat) corresponds to something that is infinite 
> if that something concerns some form of information propagation.

Well, that's certainly not the way mathematicians use the term "top hat
function".  Maybe global general relativists use the term differently.  :)

Now, given your newfound knowledge of what the metric is, go back
through my old articles and see if they make sense yet.  They're pretty
much just simple vector algebra, so it's not really that hard if you
know what the key concepts are. 

Oh, let me just mention that in spacetime, the vectors you're interested
in are usually tangent vectors to some curve (worldline).  For a massive
particle, the tangent vectors are always timelike, which means that
their squared length is negative.  The tangent vectors for massless
particles like photons are always null, which means their squared length
is zero.
-- 
--------------------------------------------------------------------------
Nathan Urban  | Undergrad {CS,Physics,Math} |        Virginia Tech
[email-masked] |  {NeXT,MIME} mail welcome   | http://nurban.campus.vt.edu/
--------------------------------------------------------------------------

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Article: 18831 of sci.physics
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From: [email-masked] (Steve Carlip)
Newsgroups: sci.physics
Subject: Re: Hypersurfaces (Re: General Relativity Tutorial)
Date: 31 May 1996 16:28:44 GMT
Organization: University of California, Davis
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Nathan Urban ([email-masked]) wrote:
: In article <[email-masked]>, "J. Matthew Nyman" <[email-masked]> wrote:

: > What reading materials do you have?  

[...]
: There are some other intro books out there that are more at your level,
: but I haven't looked at any of them.  To get familiar with some special
: relativity concepts in flat spacetime, Taylor and Wheeler's _Spacetime
: Physics_ is the standard book.  I hear they're supposed to be coming out
: with an intro general relativity book pretty soon, I think.  That should
: be good.

I think it's called "Scouting Black Holes," and is due out soon.

Here are a few more suggestions:

Geroch, _General Relativity from A to B_, is a very nice conceptual
introduction to the way spacetime is viewed in general relativity.
The book is almost entirely nonmathematical, but takes a lot of
thought about assumptions you might be taking for granted.

Schutz, _A First Course in General Relativity_, is a good introductory
textbook at the upper division undergraduate/first year graduate level.

Einstein, _The Meaning of Relativity_, is a classic.  Some of the
notation is a bit outdated, but it's a beautiful book.

I'd second Nathan's suggestion of Wald, _General Relativity_, which is
rather terse, but is a good way of seeing whether you really understand
what's going on.  

Steve Carlip
[email-masked]

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Subject: Re: Torsion and the Equivalence Principle
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In article <[email-masked]> [email-masked] writes:

>Just in passing the connection in the GR tutorial was never really 
>explained, due to the difficulty of explaining it adequately to morons. 
>Now I can readily see why this should be true.

>However, is it derived from the metric and the curvature or is it an 
>independent thingy? In other words (and I mean words) what route is
>taken to produce it?

I talked about the connection a huge amount, and so did you... only, not
calling it that!  First I talked about it in a "global" way, calling it
"parallel transport".  You remember all that stuff about Roman soldiers
carrying javelins around, and how there is a unique best way to do this
that is torsion-free and metric-compatible.  We saw that the connection
(in the guise of parallel transport) is not an independent thingy; it's
derived from the metric. 

Then I talked about the connection in a "local" way, calling it the
"Christoffel symbols".  At this point you learned even more explicitly
how to calculate it from the metric.  Remember that horrible formula?
Unfortunately I never got around to explaining THE REASON for that
horrible formula for it in terms of the metric.  

By "global" above I mean that parallel transport is all about dragging a
tangent vector along a path of finite length.  By "local" above I mean
that the Christoffel symbols are all about dragging a tangent vector
along a path of infinitesimal length.  I started with the notion of
parallel transport and developed the notion of Christoffel symbols by a
"global to local" process, namely:

"CHRISTOFFEL SYMBOLS: 

FIXING A LOCAL COORDINATE SYSTEM, take a unit vector pointing in the c
direction and parallel transport it an amount s in the b
direction.  See how much its component in the a direction changes.  
It changes by 

     - s C^a_{bc} + terms of order s^2

where C^a_{bc} are the Christoffel symbols."

Many textbooks take the reverse approach, first introducing the
Christoffel symbols and developing the notion of parallel transport by a
"local to global process", which involves integrating rather than (as
above) differentiating.

>Or did I miss a post in the tutorial.

Perhaps you missed the post where I gave an outline of the course?  I
reprint a section of it below.  It emphasizes that one can compute the
connection from the metric, and the Riemann curvature from the
connection.

I didn't talk too much about the term "connection".  People usually use
this term when they are studying this concept in a local way, not using
the Christoffel symbols ---- which depend on a choice of coordinate
system in a rather ugly way --- but in a more coordinate-invariant
approach.  Unfortunately I never got around to this slick approach.

Here's the relevant part of the course outline:

4.  The METRIC g is a tensor of rank (0,2).  It eats two tangent vectors
v,w and spits out a number g(v,w), which we think of as the "dot
product" or "inner product" of the vectors v and w.  This lets us
compute the length of any tangent vector, or the angle between two
tangent vectors.  Since we are talking about spacetime, the metric need
not satisfy g(v,v) > 0 for all nonzero v.  A vector v is SPACELIKE if 
g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0.

The inner product g(v,w) of two tangent vectors is given by

        g(v,w) = g_{ab} v^a w^b

for some matrix of numbers g_{ab}, where we sum over the repeated
indices a,b (this being the so-called EINSTEIN SUMMATION CONVENTION).
Another way to think of it is that our coordinates give us a basis of
tangent vectors at p, and g_{ab} is the inner product of the basis
vector pointing in the x^a direction and the basis vector pointing in
the x^b direction.

The metric is the star of general relativity.  It describes everything
about the geometry of spacetime, since it lets us measure angles and
sitances.  Einstein's equation describes how the flow of energy and
momentum through spacetime affects the metric.  What it actually affects
is something about the metric called the "curvature".  The biggest job
in learning general relativity is learning to understand curvature.

5.  PARALLEL TRANSLATION is an operation which, given a curve from p to
q and a tangent vector v at p, spits out a tangent vector v' at q.  We think
of this as the result of dragging v from p to q while at each step of
the way not rotating or stretching it.  There's an important theorem
saying that if we have a metric g, there is a unique way to do parallel
translation which is: 

        a. Linear: the output v' depends linearly on v.

        b. Compatible with the metric: if we parallel translate two 
           vectors v and w from p to q, and get two vectors v' and w',
           then g(v',w') = g(v,w).  This means that parallel translation
           preserves lengths and angles.  This is what we mean by "no
           stretching".  

        c. Torsion-free: this is a way of making precise the notion of
           "no rotating".  We can define the "torsion tensor", with
           components t_{ab}, as follows.  Take a little vector of size
           epsilon pointing in the a direction, and a little vector of
           size epsilon pointing in the b direction.  Parallel translate
           the vector pointing in the a direction by an amount epsilon
           in the b direction.  Similarly, parallel translate the vector
           pointing in the b direction by an amount epsilon in the a
           direction.  (Draw the resulting two vectors.)  If the tips
           touch, up to terms of epsilon^3, there's no torsion!
           Otherwise take the difference of the tips and divide by epsilon^2.
           Taking the limit as epsilon -> 0 we get the torsion t_{ab}.
           We say that parallel translation is "torsion-free" if t_{ab}
           = 0.

A GEODESIC is a curve whose tangent vector is parallel transported along
itself.  I.e., to follow a geodesic is to follow ones nose while never
turning ones nose.  A particle in free fall follows a geodesic in
spacetime.

6.  The CONNECTION is a mathematical gadget that describes "parallel
translation along an infinitesimal curve in a given direction".  In
local coordinates the connection may be described using the components
of the CHRISTOFFEL SYMBOL Gamma_{ab}^c.  There is an explicit formula
for these components in terms of components g_{ab} of the metric, which
may be derived from the assumptions a-c above.  However, this formula is
very frightening, so I will only describe it to those who have passed
certain tests of courage and valor.

7.  The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point
of spacetime.  Thus it takes three tangent vectors, say u, v, and w as
inputs, and outputs one tangent vector, say R(u,v,w).  The Riemann tensor
is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram
whose two edges point in the directions epsilon u and epsilon v , where
epsilon is a small number.  The vector w comes back a bit changed by its
journey; it is now a new vector w'.  We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Thus the Riemann tensor keeps track of how much parallel translation
around a wee parallelogram changes the vector w.             

In addition to this simple coordinate-free definition of the Riemann
tensor, we may describe its components R^a_{bcd} using coordinates.  Namely, 
the vector R(u,v,w) has components  

        R(u,v,w)^a = R^a_{bcd} u^b v^c w^d

where we sum over the indices b,c,d.  Another way to think of this is
that if we feed the Riemann tensor 3 basis vectors in the x^b, x^c, x^d
directions, respectively, it spits out a vector whose component in the
x^a direction is R^a_{bcd}. 

There is an explicit formula for the components R^a_{bcd} in terms of
the Christoffel symbols.  Together with the aforementioned formula for
the Christoffel symbols in terms of the metric, this lets us compute the
Riemann tensor of any metric!  Thus to do computations in general
relativity, these formulas are quite important.  However, they are not
for the faint of heart, so I will not describe them here!"

For more, folks should go to

http://math.ucr.edu/home/baez/gr/gr.html