From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: Red Shift {Was: Center of Universe?}
Date: 23 Jan 1996 13:03:24 0800
Organization: University of California, Riverside
MessageID: <4e3iet$e0j@guitar.ucr.edu>
References: <4dgrig$ae1@guitar.ucr.edu> <30fd365b.18293265@news.demon.co.uk> <1996Jan22.213139.23783@schbbs.mot.com>
In article <1996Jan22.213139.23783@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bronis Vidugiris) writes:
>In article <30fd365b.18293265@news.demon.co.uk>,
>Oz 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 2dimensional Riemannian manifold", but
it sits conveniently in (more or less) flat 3dimensional 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 teenyweeny
itsybitsy 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: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 23 Jan 1996 13:13:32 0800
Organization: University of California, Riverside
MessageID: <4e3j1s$e1s@guitar.ucr.edu>
References: <4damsr$pon@pipe9.nyc.pipeline.com> <4dp0pf$bu8@guitar.ucr.edu> <1996Jan22.214958.24698@schbbs.mot.com>
In article <1996Jan22.214958.24698@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bronis Vidugiris) writes:
>In article <4dp0pf$bu8@guitar.ucr.edu>, john baez 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 horizontallypointing 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. 232233, 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: baez@guitar.ucr.edu (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
MessageID: <4e3h8t$dv9@guitar.ucr.edu>
References: <4dl1fq$19e@agate.berkeley.edu> <4du9m8$cf7@guitar.ucr.edu> <31033b1d.34440117@news.demon.co.uk>
In article <31033b1d.34440117@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>...plonking down a set of
>coordinates is also somewhat meaningless if they are
>conventional ones since these are usually 'distance'
>coordinates (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 betterthanaverage 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 2dimensional 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
RobertsonFriedmanWalker 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 stretchedout 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
NNTPPostingHost: pleides.osf.org
Inreplyto: egreen@nyc.pipeline.com's message of 21 Jan 1996 21:15:09 0500
cc: egreen@nyc.pipeline.com
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
xaxis, but you can't interchange the xaxis with the taxis.
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
NNTPPostingHost: guitar.ucr.edu
In article <4durvd$11m@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (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 nspace, real or complex, these tensors actually live on n
>"timelike" dimensions, n "spacelike" dimesions, and possibly, k "klike"
>dimensions, l "llike" 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 ndimensional
semiRiemannian 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
SemiRiemannian 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
NNTPPostingHost: guitar.ucr.edu
In article <4dk7v8$fqn@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward G
reen) writes:
>Yet more.
>I left my house today with a warm glow of understanding. Now I knew what
>the stressenergy 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 "xness"
>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 shoehorn 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 coordinateindependent fact  unlike the fact that
folks like to use coordinates in which the x^0 coordinate corresponds to
a timelike direction. The statement "there exists a basis of
orthogonal tangent vectors with 3 spacelike ones and 1 timelike one" is
a completely coordinateindependent 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
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XPipeUser: egreen
XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: The Pipeline v3.4.0
'columbus@pleides.osf.org (Michael Weiss)' wrote:
>
>4. So what's so special about the top row???
>
>Nothing. No more is there anything special about the 0component of
>the energymomentum 4vector
Thank you for your answers! I still feel there must be something special
about the toprow, 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 nspace, real or complex, these tensors actually live on n
"timelike" dimensions, n "spacelike" dimesions, and possibly, k "klike"
dimensions, l "llike" 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
egreen@nyc.pipeline.com
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
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
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>In article <4dl1fq$19e@agate.berkeley.edu> ted@physics.berkeley.edu writes:
>>Cosmologists often talk about the curvature of *space* (meaning
>>ordinary threedimensional space) rather than spacetime. This is
>>perhaps unfortunate, since it confuses people like Oz, but that's the
>>way it is.
>baez@guitar.ucr.edu (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 3dimensional
>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 quasiclassical 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
coordinates is also somewhat meaningless if they are
conventional ones since these are usually 'distance'
coordinates (taking time as a distance). Another measure or
mechanism is needed. Even the initially nonphysical
comoving coordinate 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
NNTPPostingHost: guitar.ucr.edu
In article <4do512$4ov@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'baez@guitar.ucr.edu (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 2dimensional 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 noncorners.
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
NNTPPostingHost: guitar.ucr.edu
In article <4dl1fq$19e@agate.berkeley.edu> ted@physics.berkeley.edu writes:
>In article <4dgrig$ae1@guitar.ucr.edu>, john baez 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 threedimensional 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 3dimensional
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 socalled "intrinsic
curvature"
and
3. the curvedness of how the slices sit inside the spacetime  their
socalled "extrinsic curvature".
These are called the GaussCodazzi 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 fourdimensional spacetime that represents space at that one
>instant. This is a perfectly good threedimensional manifold, and
>it's perfectly reasonable to ask whether or not this manifold is
>curved.
Yes, this is the socalled "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 coordinatedependent act, because there's no
>coordinateindependent 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
NNTPPostingHost: guitar.ucr.edu
In article <1996Jan22.213139.23783@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Br
onis Vidugiris) writes:
>In article <30fd365b.18293265@news.demon.co.uk>,
>Oz 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 2dimensional Riemannian manifold", but
it sits conveniently in (more or less) flat 3dimensional 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 teenyweeny
itsybitsy 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 MaxPlanckGesellschaft in Garching
Lines: 61
NNTPPostingHost: s4bds.aug.ippgarching.mpg.de
XNewsreader: TIN [version 1.2 PL2]
Edward Green (egreen@nyc.pipeline.com) wrote:
: Thank you for your answers! I still feel there must be something special
: about the toprow, 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 timespace 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 fourvector. 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 00component is the thermal
energy density and the iicomponents 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
iicomponents. 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
MaxPlanckInstitut fuer Plasmaphysik odorless and transparent
bds@ippgarching.mpg.de  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
NNTPPostingHost: pleides.osf.org
Inreplyto: columbus@pleides.osf.org's message of 22 Jan 1996 22:20:37 GMT
cc: egreen@nyc.pipeline.com
Between John Baez's beautiful job on the difference between
"timelike" separations and the taxis, and Bruce Scott's nice
explanation of the stressenergy 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 imomentum being
transferred in the jdirection, 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 4vector 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 tensordensities, 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 yzplane, for yucks. Particles of
fluid constantly zip through this plane in all directions. If a
particle with 4vector (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 yzplane, 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 restframe 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 Oz@upthorpe.demon.co.uk Wed Jan 24 10:54 PST 1996
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From: Oz
To: john baez
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: Wed, 24 Jan 1996 13:32:13 GMT
ReplyTo: Oz@upthorpe.demon.co.uk
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Maybe this one?
On 19 Jan 1996 13:00:31 0800, you wrote:
>In article <4damsr$pon@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (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 horizontallypointing 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 bc 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
NNTPPostingHost: guitar.ucr.edu
In article <30fd365b.18293265@news.demon.co.uk> Oz@upthorpe.demon.co.uk
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 higherdimensional 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 higherdimensional 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.
>Ahha. I presume your '4tangents' are related to geodesics.
>So I stuff a 'rectangular' coordinate 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 coordinate 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 stressenergy 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 Oz@upthorpe.demon.co.uk Wed Jan 24 16:23 PST 1996
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From: Oz
To: john baez
Newsgroups: sci.physics,sci.astro,sci.philosophy.meta
Subject: Re: Stressed out?
Date: Wed, 24 Jan 1996 13:32:22 GMT
ReplyTo: Oz@upthorpe.demon.co.uk
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Ahh, is this it? Oz
baez@guitar.ucr.edu (john baez) wrote:
>In article <4df7gg$jd9@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward Green) writes:
>
>>What stress tensor?
>
>The stressenergy tensor, aka energymomentum 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 stressenergy 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 stressenergy 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 stressenergy 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 KMULDREW@jiarg.mccaig.ucalgary.ca Fri Jan 26 12:30 PST 1996
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From: "Ken Muldrew"
Organization: McCaig Centre
To: baez@guitar.ucr.edu
Date: Fri, 26 Jan 1996 13:37:35 MDT
Subject: parallel transport around a parallellogram
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========
Newsgroups: sci.physics
Subject: Re: Red Shift {Was: Center of Universe?}
From: bhv@areaplg2.corp.mot.com (Bronis Vidugiris)
Date: Mon, 22 Jan 1996 21:31:39 GMT
In article <30fd365b.18293265@news.demon.co.uk>,
Oz 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 1element
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 noncartesian coordinate
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 twoelement 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 4element 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: baez@guitar.ucr.edu (john baez)
Date: 24 Jan 1996 10:58:54 0800
In article <30fd365b.18293265@news.demon.co.uk> Oz@upthorpe.demon.co.uk
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 higherdimensional 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 higherdimensional 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.
>Ahha. I presume your '4tangents' are related to geodesics.
>So I stuff a 'rectangular' coordinate 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 coordinate 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 stressenergy 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
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baez@guitar.ucr.edu (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) coordinate
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.
>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 2dimensional 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
>RobertsonFriedmanWalker 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.
>>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.
...........
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
NNTPPostingHost: guitar.ucr.edu
In article <3105d4fe.89525388@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>baez@guitar.ucr.edu (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 teenyweeny
>>itsybitsy 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 nonexistant 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 2dimensional 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 coordinate system we define". Since
>we find it hard to visualise this we bend it up into a
>nonexistent 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 higherdimensional 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 coordinates 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 coordinates 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 'coordinates' 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
NNTPPostingHost: mt14.bc.edu
In article <4e3iet$e0j@guitar.ucr.edu>, baez@guitar.ucr.edu (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 spacetime
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 multilinear 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 viceversa.
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
NNTPPostingHost: mt14.bc.edu
In article <3105d4fe.89525388@news.demon.co.uk>, Oz@upthorpe.demon.co.uk wrote:
baez@guitar.ucr.edu (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 coordinates 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 coordinates 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
NNTPPostingHost: guitar.ucr.edu
In article <31055266.56094078@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>Yes, got it in one. I was imagining that it was reasonable
>to transform between a rectangular (or whatever) coordinate
>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 inspiralling 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
questionandanswer 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
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
XNewsreader: Forte Agent .99c/16.141
RamsayMT@hermes.bc.edu (Keith Ramsay) wrote:
>In article <3105d4fe.89525388@news.demon.co.uk>, Oz@upthorpe.demon.co.uk wrote:
>baez@guitar.ucr.edu (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
NNTPPostingHost: guitar.ucr.edu
In article <4e43j8$6db@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'bds@ippgarching.mpg.de (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
NNTPPostingHost: guitar.ucr.edu
In article <3105f6bd.98164211@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>Uhhuh. 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 minicourse on general relativity, I
have taken parallel translation as a (fairly) easytograsp 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 nittygritty 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 stressenergy (or energymomentum)
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
NNTPPostingHost: guitar.ucr.edu
In article <31077515.76320374@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>baez@guitar.ucr.edu (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 coordinates since it points
>from here to there, and there can be as close as we want and
>we could (probably) coordinatise here and there if we
>wanted to in any coordinate 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 ballpark? 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
NNTPPostingHost: guitar.ucr.edu
In article <4egah7$nrc@pipe8.nyc.pipeline.com> egreen@nyc.pipeline.com (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 welldefined 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
Stokeslike 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 welldefined 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
NNTPPostingHost: 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. Torsionfree: 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 STRESSENERGY TENSOR. The stressenergy 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 momentuminthevdirection 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 stressenergy 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
NNTPPostingHost: guitar.ucr.edu
In article <4egckn$s8a@pipe8.nyc.pipeline.com> egreen@nyc.pipeline.com (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 presentday theories of
forces: Maxwell's equations for electromagnetism, Einstein's theory of
general relativity, and the YangMills 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 YangMills 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
NNTPPostingHost: guitar.ucr.edu
In article <310cc83e.51368111@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>baez@guitar.ucr.edu (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 fourdimensional
>>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
NNTPPostingHost: guitar.ucr.edu
In article <4eitri$9i1@pipe8.nyc.pipeline.com> egreen@nyc.pipeline.com (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 fancyass 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
welldefined, 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 ChoquetBruhat, Cecile
DeWittMorette, and Margaret DillardBleick, 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. Torsionfree: 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
NNTPPostingHost: guitar.ucr.edu
In article <1996Jan29.230030.3000@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bro
nis Vidugiris) writes:
>In article <4eh7uj$hmv@guitar.ucr.edu>, john baez wrote:
>)In article <310bb57a.6566448@news.demon.co.uk> Oz@upthorpe.demon.co.uk 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 oneform? I always
>thought those wore considered linear (?). Or are they called
>antilinear? 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 oneform 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
NNTPPostingHost: guitar.ucr.edu
In article <1996Jan29.222251.1702@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bron
is Vidugiris) writes:
>In article <4ee250$fpp@guitar.ucr.edu>, john baez wrote:
>)In article <4e8lp0$pnu@brokaw.comm.mot.com> bhv@areaplg2.corp.mot.com (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 coordinate invariant.
Yes, "geodesic" is a coordinateinvariant 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
coordinateindependent 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 coordinate
>dependent, too).
"Lines of latitude" are, of course, coordinatedependent. 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 welldefined in a coordinateindependent
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. Torsionfree: this is a way of making precise the notion of
"no rotating".
The notion of "torsionfree" 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 "torsionfreeness".
>)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
ac 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
NNTPPostingHost: guitar.ucr.edu
In article <310dbd78.1404322@news.demon.co.uk> Oz@upthorpe.demon.co.uk 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 (<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 itsybitsy 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 timehonoured 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 STRESSENERGY 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
NNTPPostingHost: guitar.ucr.edu
In article <4elcr6$7qv@sulawesi.lerc.nasa.gov> Geoffrey A. Landis 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
NNTPPostingHost: sagi.wistar.upenn.edu
Inreplyto: Geoffrey A. Landis
In article <4elcr6$7qv@sulawesi.lerc.nasa.gov>, Geoffrey A. Landis 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 3space. 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 nonzero 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 2dimensional 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 (weemba@sagi.wistar.upenn.edu)
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
NNTPPostingHost: guitar.ucr.edu
In article <4emjju$adj@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward G
reen) writes:
>'baez@guitar.ucr.edu (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 widelyused
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
NNTPPostingHost: pipe11.nyc.pipeline.com
XPipeUser: egreen
XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: The Pipeline v3.4.0
'Oz@upthorpe.demon.co.uk (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 (<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
epsilontisimo 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
egreen@nyc.pipeline.com
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
NNTPPostingHost: guitar.ucr.edu
In article <4enni6$9cp@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward G
reen) writes:
>'Oz@upthorpe.demon.co.uk (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 (<>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 heavyhanded 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
>epsilontisimo 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 ndimensional 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
itsybitsy 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 illdefined: 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
metricpreserving way in some higherdimensional 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
NNTPPostingHost: guitar.ucr.edu
In article <4eo6hp$dcj@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward Gr
een) writes:
>'egreen@nyc.pipeline.com (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
"torsionfree"  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 1dimensional 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
NNTPPostingHost: guitar.ucr.edu
In article <4eqnbe$np0@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward G
reen) writes:
>'baez@guitar.ucr.edu (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 1dimensional universe.
>Interesting. I see this, and yet, if we embed a 1dimensional 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 1dimensionsal 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 47dimensional 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
NNTPPostingHost: guitar.ucr.edu
In article <4ep87s$8er@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (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 coordinateindependent
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 coordinateindependent 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
NNTPPostingHost: guitar.ucr.edu
In article <4enufs$alp@www.oracorp.com> daryl@oracorp.com (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 highschool geometry, except that geodesics replace straightedges.
That's nice; I didn't know this construction.
>[construction deleted]
>This construction paralleltransports 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
NNTPPostingHost: guitar.ucr.edu
In article <4ent4a$hql@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (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
NNTPPostingHost: mt14.bc.edu
In article <4ep8cb$8ur@pipe10.nyc.pipeline.com>,
egreen@nyc.pipeline.com (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 agreedupon 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 xy 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=uv and y=v, dx=dudv 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
NNTPPostingHost: mt14.bc.edu
In article <1996Feb1.204744.17506@schbbs.mot.com>,
bhv@areaplg2.corp.mot.com (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 2space, but not in 3space. 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 spacetime are the *freefall* 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 twodimensional 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
NNTPPostingHost: mt14.bc.edu
In article <3110c6a8.14983979@news.demon.co.uk>, Oz@upthorpe.demon.co.uk 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 spacetime 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^2dx^2dy^2dz^2.
This takes two vectors (x1,y1,z1,t1) and (x2,y2,z2,t2) and gives
back the number, t1t2x1x2y1y2z1z2. 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 spacetime, 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(1v^2/c^2).
Since I'm setting c=1, this is just sqrt(1v^2). To find the
rate of elapsed time w.r.t. s, multiply by dt/ds:
sqrt(1v^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 (worldline) through spacetime.
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 spacetime
"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
spacetime, the points in spacetime 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((x1x2)^2+(y1y2)^2+(z1z2)^2(t1t2)^2),
where (x1,y1,z1,t1) and (x2,y2,z2,t2) are the coordinates of the
points of spacetime in question.
Now, all that was special relativity (!). In general relativity, one
deals with other spacetimes. 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 timedilation
he certainly wasn't 186 years old. So draw his worldline in spacetime,
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 2tensor, 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=vwcos(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 spacetime, 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
NNTPPostingHost: guitar.ucr.edu
In article <4eue1b$c78@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward Gr
een) writes:
>'daryl@oracorp.com (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 coordinatedependent
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
NNTPPostingHost: guitar.ucr.edu
In article <1996Feb3.001629.17341@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bron
is Vidugiris) writes:
>Ouch!
Sorry, I was being a bit sharptongued 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 stressenergy 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 stressenergy 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  socalled "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
NNTPPostingHost: 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 <3110c6a8.14983979@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes:
>baez@guitar.ucr.edu (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 coordinates 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 STRESSENERGY TENSOR. The stressenergy 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 momentumintheudirection 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!
Ohoh, 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 stressenergy 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 energymomentum 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 energymomentum 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
Subject: Re: General relativity tutorial
Date: 3 Feb 1996 10:53:20 GMT
Organization: The Mathematical Institute, Oxford
Lines: 61
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baez@guitar.ucr.edu (john baez) wrote:
>In article <4ent4a$hql@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (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 spacetimes 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
FollowupTo: sci.physics
Date: 3 Feb 1996 23:16:22 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 39
Distribution: world
NNTPPostingHost: physics12.berkeley.edu
In article <4ever0$nou@news.ox.ac.uk>,
Marco de Innocentis 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+1dimensional spacetime? If
so, I'm enormously surprised. Doesn't the Einstein equation say that
an empty patch of 2+1dimensional spacetime is completely flat?
Maybe you mean that you can have tidal forces in a region of
2+1dimensional 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 spacetimes 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
NNTPPostingHost: guitar.ucr.edu
In article RamsayMT@hermes.bc.edu (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
NNTPPostingHost: mt14.bc.edu
In article ,
Oz@upthorpe.demon.co.uk 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 viceversa (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 surfaceoftheEarth
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".

Uhoh. 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 spacetime paths correspond to the motions of objects in
freefall.)
 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
NNTPPostingHost: guitar.ucr.edu
In article <4f1age$adv@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (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
FollowupTo: sci.physics
Date: 4 Feb 1996 20:09:54 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 56
Distribution: world
NNTPPostingHost: physics12.berkeley.edu
In article <5btFSGAF8FFxEwDh@upthorpe.demon.co.uk>,
Oz 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 stressenergy 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 nonphysical 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+1dimensional spacetime is flat.
There certainly are curved 2+1dimensional 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
Subject: Re: General relativity tutorial
Date: Sun, 4 Feb 1996 09:37:51 +0000
Organization: Oz
Lines: 56
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
Gosh, this is getting involved. Can't we just stay in 4D spacetime for a
while? No? Ok, you're the boss.
In article <4f0v9r$m97@guitar.ucr.edu>, john baez
writes
>In article RamsayMT@hermes.bc.edu
>(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 freefall 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. Freefall 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
NNTPPostingHost: sagi.wistar.upenn.edu
Inreplyto: toby@ugcs.caltech.edu (Toby Bartels)
In article <4evipr$1ak@gap.cco.caltech.edu>, toby@ugcs (Toby Bartels) writes:
>Matthew P Wiener (weemba@sagi.wistar.upenn.edu) wrote:
>>I'd have to look this one up to get it rightask me if interested and
>>I will give you a reference. To give an elementary example, if all you
>>know about an inner product is the set of values, then as
>> = +2c+c^2, you can solve for (and
>>prove the CauchySchwartz inequality to boot). Something similar, but
>>algebraically more messy and less ASCIIble, applies here.
>Notice that this only works because we know = , a symmetry.
>This symmetry of the inner product is reflected geometrically
>in that there's a definition that only involves the lengths .
>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 (weemba@sagi.wistar.upenn.edu)
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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk writ
es:
>In article <4f0p7c$m7t@guitar.ucr.edu>, john baez
> 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
NNTPPostingHost: guitar.ucr.edu
In article RamsayMT@hermes.bc.edu (Ke
ith Ramsay) writes:
>In article <453l6UAv5HFxEwBB@upthorpe.demon.co.uk>,
> Oz@upthorpe.demon.co.uk 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. Freefall 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
NNTPPostingHost: guitar.ucr.edu
A couple of corrections:
In article <4f40bv$ngv@guitar.ucr.edu> baez@guitar.ucr.edu (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
NNTPPostingHost: guitar.ucr.edu
In article <4f33qi$htk@agate.berkeley.edu> ted@physics.berkeley.edu 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
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In article <4f33qi$htk@agate.berkeley.edu>, "Emory F. Bunn"
writes
>In article <5btFSGAF8FFxEwDh@upthorpe.demon.co.uk>,
>Oz 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.
>
>
>So in electromagnetism, knowing all about the sources isn't enough to
>specify the fields. In general relativity, knowing all about the
>sources (the stressenergy 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
NNTPPostingHost: 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 <4f5dvt$22g@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (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 stressenergy tensor is divergencefree. 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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk 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 volumepreserving
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", "itsybitsy", 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
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In article <4f6jvo$obf@guitar.ucr.edu> baez@guitar.ucr.edu (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 , by Oz :
> 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 multilinear 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 bbowen@megatest.com

Bruce Bowen Take only meat and hide, leave only guts
bbowen@megatest.com "Just say F.O."
Article 98357 (39 more) in sci.physics:
From: Oz
(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 <4f644q$o4a@guitar.ucr.edu>, john baez
writes
>In article Oz@upthorpe.demon.co.uk
>writes:
1) In modern notation vectors are/tendtobe '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
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 premodern. I am (would
prefer) to use your notation, if only I knew what it was!
> 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. .
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
NNTPPostingHost: sagi.wistar.upenn.edu
Inreplyto: Oz
In article , Oz 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. Offhand, 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 1forms. 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 3dimensional 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. 1forms 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 kforms. For 2forms, one has two sets
of parallel dissections, together forming pipes. For 3forms, one has
three sets of them dissection. In ndimensions, one works with hyperplanar
dissections instead of planar dissections.
The differential that maps kforms to k+1forms becomes obvious: just
more contouring. Given a kform 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+1form.

Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk writ
es:
>In article <4f644q$o4a@guitar.ucr.edu>, john baez
>writes
>>In article Oz@upthorpe.demon.co.uk
>>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 oldfashioned 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 finergrained 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
NNTPPostingHost: guitar.ucr.edu
In article <4f7tvu$caa@sulawesi.lerc.nasa.gov> Geoffrey A. Landis 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 4space, 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
>secondderivativeoftheradius tensor, which is by definition the
>tidalacceleration tensor, which is by definition a curvatureofspace
>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
>secondderivativeoftheradiusofthecoffeesphere tensor is the same
>as the Ricci tensor R_mn.
Sorry. The "tidalacceleration tensor" you describe is not the Ricci
tensor, but rather
R^m_{anb} v^a v^b
where v is the 4velocity 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 coordinatefree 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
NNTPPostingHost: pipe9.nyc.pipeline.com
XPipeUser: egreen
XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: 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?
'baez@guitar.ucr.edu (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 multigenerational 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 / egreen@nyc.pipeline.com
"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
NNTPPostingHost: 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 <4f5dvt$22g@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (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 offdiagonals, 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 hyperellipsoid.
>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 volumepreserving 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 stressenergy 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 stressenergy 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
NNTPPostingHost: 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
xmomentum 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 xmomentum in the x
direction there, plus the flow of ymomentum in the y direction, plus
the flow of zmomentum 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 "xmomentum 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
NNTPPostingHost: guitar.ucr.edu
In article <4f921h$mg@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (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 multigenerational 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 higherorder 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 secondorder 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
FollowupTo: sci.physics
Date: 8 Feb 1996 00:40:02 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 31
Distribution: world
NNTPPostingHost: physics12.berkeley.edu
In article ,
Keith Ramsay wrote:
>In article <4f6tq6$13v8@hearst.cac.psu.edu>, ale2@psu.edu (ale2) wrote:
>For any spacetime point around, say the sun, is there an infinite
>number of geodesics through that point?
>
>Yes, one for each tangent direction through it.
Lest anyone get confused, I just want to point out that when Keith
says "each direction" he means each direction through *spacetime*, not
just space. As long as you've gotten into the proper relativity
spirit, that should be obvious.
So if you're sitting at a particular point in spacetime, you can't just
choose a particular direction through space (say, North) and expect
to find a unique geodesic in that direction through that point.
That's because North is just a direction in space, not a direction
through spacetime.
(Of course, before a word like "North" even makes sense, you have to
have laid down coordinates on your patch of spacetime.)
If you want to specify a unique geodesic through a particular
spacetime point, one way to do it is to specify the *velocity* that a
particle would have if it passed through that point following that
geodesic. That's because the velocity is a 4dimensional vector (a
tangent vector) in spacetime, so specifying the velocity is a good way
to specify a direction through spacetime.
Ted
Article 98602 (39 more) in sci.physics:
From: john baez
Subject: Re: Tensors for twits please.
Date: 7 Feb 1996 13:27:02 0800
Organization: University of California, Riverside
Lines: 117
NNTPPostingHost: guitar.ucr.edu
In article Steven_Hall@mit.edu (
Steven Hall) writes:
>In article <4f90ct$ovl@guitar.ucr.edu>, baez@guitar.ucr.edu (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 "oldfashioned" 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
3dimensional 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
NNTPPostingHost: guitar.ucr.edu
In article <4fac3s$1p4@pipe12.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'jonathan_scott@vnet.ibm.com (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?
Ohoh. 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
Subject: Re: GR Tutorial Water Cooler
Date: Wed, 7 Feb 1996 19:43:42 +0000
Organization: Oz
Lines: 49
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4faapl$sgd@pipe12.nyc.pipeline.com>, Edward Green
writes
>'baez@guitar.ucr.edu (john baez)' wrote:
>
>Oh, hi there, professor.
>
>>I promise to restrain myself in the future, though I gave Ed quite a
>>tonguelashing 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.
>
>Tonguelashing? 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.
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 why don't YOU ask him about it. You always get good grades.
Brrrrriiiiiing
Oh dear, back we go.
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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk writ
es:
>In article <4egpqc$hi1@guitar.ucr.edu>, john baez
>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 , so one might think this has something
>to do with 'length'. .
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
NNTPPostingHost: guitar.ucr.edu
In article <4f6tq6$13v8@hearst.cac.psu.edu> ale2@psu.edu (ale2) writes:
>Student interupts (many classmates give him a look of disgust)...
>For any spacetime point around, say the sun, is there an infinite
>number of geodesics through that point?
Yes indeed, basically one for each unit vector through that point.
Remember a spacetime point is a moment and a place. At any given
moment, at any given place, there are many ways a freely falling rock
could zip through that place at that moment. The velocity of the rock
(a 4d vector!) could be any futurepointing timelike vector. As it
continues to whiz along its worldline traces out a geodesic. Then
there are the lightlike and spacelike geodesics....
>Is the geodesic of a particle
>at some point in the spacetime around the sun determined once we know
>the velocity of the particle at that spacetime point?
Yes, if we are told the geometry of spacetime, we can completely work
out a geodesic if we know a point it goes through and its tangent vector
(velocity) at that point. It satisfies a secondorder differential
equation, which I have avoided writing down, and you just solve them.
Article 98699 (48 more) in sci.physics:
From: john baez
Subject: Re: General relativity tutorial
Date: 7 Feb 1996 20:12:40 0800
Organization: University of California, Riverside
Lines: 75
NNTPPostingHost: guitar.ucr.edu
In article <0M++SOAJTGGxEwLK@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk writ
es:
>In article <4f8r0u$oov@guitar.ucr.edu>, john baez
>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 nonsarcastic informationrich understandable
>sentences would suffice. Enough to place it in a footnote of the brain.
Nonsarcastic, 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 torsionfree. It's
often presented as a technical assumption not worth trying to understand.
But here's a nice way understanding what torsionfreeness means (in
conjunction with our other assumptions: that parallel translation be
linear and metriccompatible).
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 freefall 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
 
PQ
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
 /
PQ
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
Subject: Re: General relativity tutorial
Date: Thu, 8 Feb 1996 06:57:37 +0000
Organization: Oz
Lines: 56
Distribution: world
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In article , Doug Merritt
writes
>In article <4fbt7o$puu@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes:
>>Nonsarcastic, 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.
>
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 nondelicate 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 smartassed
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.
>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
Subject: Re: General relativity tutorial
Date: Thu, 8 Feb 1996 12:56:43 +0000
Organization: Oz
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In article <4fbp1l$pmq@guitar.ucr.edu>, john baez
>
>>OK. Doing this on another thread.
>>Urk. g(v,w) is a scalar , so one might think this has something
>>to do with 'length'. .
>
>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.
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.
> 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 ?
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.
>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".
>
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.
>Yeah, but that's it for me. After you guys understand Einstein's
>equation you're on your own.

'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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk 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 preprimer, 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 preprimer 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 torsionfree 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
"torsionfree" 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 factoroftwo
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 xmomentum in the x
direction there, plus the flow of ymomentum in the y direction, plus
the flow of zmomentum 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
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In article Oz@upthorpe.demon.co.uk 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 bc 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
NNTPPostingHost: 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
stressenergy 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 nonRicci
part of the Riemannian curvature tensor (the Weyl tensor). So
information about massenergy 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 massenergy inside the surface (and therefore a nonzero
Ricci tensor) even though the stressenergy tensor is zero
everywhere on the surface.
Daryl McCullough
ORA Corp.
Ithaca, NY
Article 98772 (52 more) in sci.physics:
From: Oz
Subject: Re: Tensors for twits please.
Date: Wed, 7 Feb 1996 18:08:01 +0000
Organization: Oz
Lines: 236
Distribution: world
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In article <4f90ct$ovl@guitar.ucr.edu>, john baez
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 reread 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 oldfashioned 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.
>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 finergrained 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 nonzero 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 nonzero. 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 .
>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?

'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
NNTPPostingHost: guitar.ucr.edu
I answer only a random subset of the vast list of questions posed by Oz:
In article Oz@upthorpe.demon.co.uk 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 reread 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 3dimensional 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. 1forms 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 nonzero 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 nonzero. 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?
>
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
FollowupTo: sci.physics
Date: 8 Feb 1996 20:47:26 GMT
Organization: Physics Department, U.C. Berkeley
Lines: 43
Distribution: world
NNTPPostingHost: physics12.berkeley.edu
In article <4fcvne$j1l@pipe9.nyc.pipeline.com>,
Edward Green 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, indexbased 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 fourdimensional 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 rank4 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 "tracefree,"
which means that if you try to contract Weyl to a rank2 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
NNTPPostingHost: guitar.ucr.edu
In article <4fdngu$tf@agate.berkeley.edu> ted@physics.berkeley.edu writes:
>In article <4fcvne$j1l@pipe9.nyc.pipeline.com>,
>Edward Green 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, indexbased 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 tracefree 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 "tracefree 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 fourdimensional 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 rank4 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 "tracefree,"
>which means that if you try to contract Weyl to a rank2 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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk 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?
>
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
NNTPPostingHost: mt14.bc.edu
In article <4fbnva$pjh@guitar.ucr.edu>, baez@guitar.ucr.edu (john baez) wrote:
Now you might wonder about what all that "xmomentum 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
NNTPPostingHost: 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
Oz@upthorpe.demon.co.uk 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!
>
>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 doublebladed 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
Subject: Re: General relativity tutorial
Date: Fri, 9 Feb 1996 08:56:26 +0000
Organization: Oz
Lines: 82
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4fbp9v$po3@guitar.ucr.edu>, john baez
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 secondorder 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 GRlike 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. 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 torsionfree GR then an infinitesimal sphere of massless test
points surrounding an (even more) infinitesimal speck of momentumflow
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 torsionfree 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.
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. .
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
NNTPPostingHost: 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 secondorder 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
heavyduty 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
NNTPPostingHost: 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 1labeled plane is now 10labeled, with the
new 1labeled 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 4momentum has the inverse Lorentz transformation from
the 4position.
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 multiplicationof a covector 1off pushby 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 dissectionsits 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 1labelled 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 (weemba@sagi.wistar.upenn.edu)
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
NNTPPostingHost: pipe9.nyc.pipeline.com
XPipeUser: egreen
XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: The Pipeline v3.4.0
'baez@guitar.ucr.edu (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.... :)
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 ,
'baez@guitar.ucr.edu (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 , 'egreen@nyc.pipeline.com (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 / egreen@nyc.pipeline.com
"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 offdiagonals, 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 hyperellipsoid.
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
(SAME) Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 00:17:35 +0000
Organization: Oz
Lines: 32
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4fgc3b$r3h@guitar.ucr.edu>, john baez
writes
>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk
>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 harikari.
Have a nice weekend. I will speak again when I'm sober. Hoc. I mean Hic.
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
(SAME) Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 00:23:58 +0000
Organization: Oz
Lines: 40
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4fgff2$fir@sulawesi.lerc.nasa.gov>, Geoffrey A
writes
>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk>
>Oz@upthorpe.demon.co.uk writes:
>>>4) We don't seem to have any relativistic effects modelled into our
>>>version of GR yet.
>
>In article <4fgc3b$r3h@guitar.ucr.edu> john baez, baez@guitar.ucr.edu
>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.
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
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4fgff2$fir@sulawesi.lerc.nasa.gov>, Geoffrey A
writes
>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk>
>Oz@upthorpe.demon.co.uk writes:
>>>4) We don't seem to have any relativistic effects modelled into our
>>>version of GR yet.
>
>In article <4fgc3b$r3h@guitar.ucr.edu> john baez, baez@guitar.ucr.edu
>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
Subject: Re: General relativity tutorial
Date: Sat, 10 Feb 1996 08:46:16 +0000
Organization: Oz
Lines: 81
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
In article <4fgg6e$r8r@guitar.ucr.edu>, john baez
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 secondorder 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
>heavyduty 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
NNTPPostingHost: 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
NNTPPostingHost: guitar.ucr.edu
In article <4fgc3b$r3h@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes:
>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk wri
tes:
>
>>4) We don't seem to have any relativistic effects modelled into our
>>version of GR yet.
>
>HUH?
To clarify: we never did anything that WASN'T perfectly "relativistic".
Not only did we never introduce any yucky "absolute rest", or any yucky
split of spacetime into space and time, we didn't even introduce any
yucky "global inertial frames"! So not only we have avoided the errors
of Aristotelian and Newtonian mechanics, we have avoided the errors of
special relativity. So we are doing general relativity.
Perhaps you mean that I haven't said any cool stuff about black holes or
the big bang? Actually we're almost ready for some of that fun
stuff.
Article 99319 (126 more) in sci.physics:
From: john baez
(SAME) Subject: Re: General relativity tutorial
Date: 10 Feb 1996 23:03:12 GMT
Organization: University of California, Riverside
Lines: 92
NNTPPostingHost: guitar.ucr.edu
In article <4fcvok$j4j@pipe9.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'baez@guitar.ucr.edu (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 fiddlefaddle. 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,
stretchingsquishings, 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
stretchingsquishings 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 stretchingsquishings 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 3space 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 stretchingsquishings. (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), skewsymmetric 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
NNTPPostingHost: guitar.ucr.edu
In article <4ffoog$157@pipe12.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'baez@guitar.ucr.edu (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
NNTPPostingHost: guitar.ucr.edu
In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk writ
es:
>If 4D space is curved in a GRlike 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. 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 welldefined 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 torsionfree GR then an infinitesimal sphere of massless test
>points surrounding an (even more) infinitesimal speck of momentumflow
>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 torsionfree 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.
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. .
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
NNTPPostingHost: pipe10.nyc.pipeline.com
XPipeUser: egreen
XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: The Pipeline v3.4.0
' 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
batterdipped deepfried ball of eggplant? That's right, little
deepfried 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 sciphysics 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 / egreen@nyc.pipeline.com
"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
NNTPPostingHost: guitar.ucr.edu
In article <4ff2d8$1g1u@info.estec.esa.nl> Grai w
rites:
>baez@guitar.ucr.edu (john baez) wrote:
>>
>>See some stuff Matthew Wiener wrote recently, for a more geometrical
>>explanation of the difference between vectors and covectors.
>Just a question on nomenclature. I always used the terms vector and
>1form. Is covector a new, more general entity applicable to other
>vector spaces, or just exactly the same thing as a 1form.
If you hand me any vector space I could call its elements vectors and
call the elements of the dual vector space "covectors". It's certainly
interesting to think about this. But I wasn't really trying to be so
general here... I was mainly talking about the tangent space, whose
elements I call "tangent vectors" or (for short) "vectors", and its dual
space, the cotangent space, whose elements I call "cotangent vectors" or
(for short) "covectors". If I have a tangent vector at each point of
spacetime I say I have a "vector field", while if I have a cotangent
vector at each point of spacetime I might say I have a "1form".
Article 99130 (17 more + 1 Marked to return) in sci.physics:
From: Steve Carlip
(SAME) Subject: Re: General relativity tutorial
Date: 9 Feb 1996 22:28:01 GMT
Organization: University of California, Davis
Lines: 69
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john baez (baez@guitar.ucr.edu) wrote:
: In article <4f33qi$htk@agate.berkeley.edu> ted@physics.berkeley.edu 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 understandI'll explain below.) Start
with your ball of coffee grounds, with an initial fourvelocity
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 shearedthat 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 lowercase 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 indexfree 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 NewmanPenrose formalism, but I've never learned that very well.)
(The obvious guess is that it has to do with rotation, but this is
wrongrotation isn't produced directly by geometry, but only
indirectly by expansion and shear.)
Steve Carlip
carlip@dirac.ucdavis.edu
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
NNTPPostingHost: guitar.ucr.edu
In article <4fghph$355@mark.ucdavis.edu> carlip@dirac.ucdavis.edu (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 indexfree 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 NewmanPenrose 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
>wrongrotation 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
(SAME) Subject: Re: General relativity tutorial
Date: Sun, 11 Feb 1996 09:13:33 +0000
Organization: Oz
Lines: 42
Distribution: world
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In article <4fj845$6a1@galaxy.ucr.edu>, john baez
writes
>In article <4fgc3b$r3h@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes:
>>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk
>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 relativisticlike 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
(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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XNewsreader: Turnpike Version 1.11
In article , Doug Merritt
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.
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:
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 coordinates and a length of 1, or .000001 for
that matter. Or looking at it the other way round, if one coordinate is
eversobig and the length is 1, the other has to be eversobig positive or
negative.
Needs a bit more thought, dunnit?

'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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk write
s:
>In article <4fgff2$fir@sulawesi.lerc.nasa.gov>, Geoffrey A
>writes
>>In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk>
>>Oz@upthorpe.demon.co.uk writes:
>>>>4) We don't seem to have any relativistic effects modelled into our
>>>>version of GR yet.
>>In article <4fgc3b$r3h@guitar.ucr.edu> john baez, baez@guitar.ucr.edu
>>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 energymomentum 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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk 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
Subject: Re: General relativity tutorial
Date: Mon, 12 Feb 1996 14:06:24 +0000
Organization: Oz
Lines: 179
Distribution: world
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In article <4fdfvv$q2k@guitar.ucr.edu>, john baez
writes
>In article Oz@upthorpe.demon.co.uk
>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
coordinate system is not required, so we have to distinguish between a
coordinate system and a metric even though they do look very similar.
So what's the difference between a metric(theta,phi) and a coordinate
system(theta,phi)? Well the only thing I can see is that
metric(theta,phi) is a relative thingy. A coordinate(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)
>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 nontrivial
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.
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?

'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
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In article Oz@upthorpe.demon.co.uk writes:
>In article <4fdfvv$q2k@guitar.ucr.edu>, john baez
>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. Torsionfree: this is a way of making precise the notion of
"no rotating".
>Now G.Wiz insists at this stage that a
>coordinate system is not required, so we have to distinguish between a
>coordinate 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 longitudelatitude 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 coordinate
>system(theta,phi)? Well the only thing I can see is that
>metric(theta,phi) is a relative thingy. A coordinate(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 nontrivial
>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 ChoquetBruhat....
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
Organization: The Pipeline
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XPipeHub: nyc.pipeline.com
XPipeGCOS: (Edward Green)
XNewsreader: The Pipeline v3.4.0
'baez@guitar.ucr.edu (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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'baez@guitar.ucr.edu (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 "1form".
Oh... so that's what a 1form 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 1forms...
[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 / egreen@nyc.pipeline.com
"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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'weemba@sagi.wistar.upenn.edu (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 !
>Offhand, 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 neplusultra 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 / egreen@nyc.pipeline.com
"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
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NNTPPostingHost: sagi.wistar.upenn.edu
Inreplyto: egreen@nyc.pipeline.com (Edward Green)
In article <4flsqp$29m@pipe11.nyc.pipeline.com>, 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 "1form".
>Oh... so that's what a 1form 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 1forms...
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 ntuples, 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 oldstyle talk. New style talk is to call the former vectors, and
the latter covectors. Covector fields are also called 1forms.
>[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 0labelled hyperplace
is the one containing @/@y, @/@z, ...., and the 1labelled hyperplane
is parallel to this but an infinitesimal distance dx over.

Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Article 99356 (37 more) in sci.physics:
From: Edward Green
Subject: Re: GR Tutorial Water Cooler
Date: 10 Feb 1996 19:51:30 0500
Organization: The Pipeline
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XNewsreader: The Pipeline v3.4.0
' 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
stressenergy tensor is nonzero. 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... sourcefree, we can
take any superposition of them, and have a sourcefree 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 / egreen@nyc.pipeline.com
"All coordinate systems are equal,
but some are more equal than others".
Article 99481 (36 more) in sci.physics:
From: Oz
(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 <4fjeii$l9e@pipe10.nyc.pipeline.com>, Edward Green
writes
>' 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.
>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
>stressenergy tensor is nonzero. 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... sourcefree, we can
>take any superposition of them, and have a sourcefree 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.

'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
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Inreplyto: Oz
In article , Oz In article <4fgff2$fir@sulawesi.lerc.nasa.gov>, Geoffrey A
>>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)=10cct/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(timeunit)s".)
dx/dt=c/2 going out, and c/2 coming back. (dx/dt)^2=c^2/4 either way.
So ds^2=dt^2dx^2/c^2=(11/4)dt^2, and ds=sqrt(11/4)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(11/4) factor is just the world famous sqrt(1v^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 preEinstein, 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 merrygoround. But the
mere business of lightwhich follows a geodesicbending because of
gravity is already nonEuclidean.
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 (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Summary:
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In article <4gcuc6$lig@pipe12.nyc.pipeline.com> egreen@nyc.pipeline.com (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 quasistationarily 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 stressenergy tensor, could in
principle be static without the spacetime geometry being static.
Remember, the stressenergy 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 socalled 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. Torsionfree: 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 "torsionfree" 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 ac 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 coordinatefree 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 STRESSENERGY TENSOR. The stressenergy 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 momentumintheudirection 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 stressenergy 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
xmomentum 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 xmomentum in the x
direction there, plus the flow of ymomentum in the y direction, plus
the flow of zmomentum 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
xmomentum in the x direction there plus the flow of ymomentum in the
y direction plus the flow of zmomentum 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
NNTPPostingHost: guitar.ucr.edu
In article <4gb15v$jb7@pipe12.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward
Green) writes:
>'weemba@sagi.wistar.upenn.edu (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
2sphere 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 olderfashioned 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
Subject: Re: Gravitational Red Shifts  Real or Apparent ?
Date: Wed, 21 Feb 1996 18:19:25 +0000
Organization: Oz
Lines: 63
Distribution: world
NNTPPostingHost: upthorpe.demon.co.uk
XNNTPPostingHost: upthorpe.demon.co.uk
MIMEVersion: 1.0
XNewsreader: Turnpike Version 1.11
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 <4ftln7$1n6@guitar.ucr.edu>, john baez
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 overgeneralized 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 eversoslightly 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
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk write
s:
>In article <4ftln7$1n6@guitar.ucr.edu>, john baez
>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 overgeneralized 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 eversoslightly 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 stressenergy 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 timewarping,
spacebending 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
NNTPPostingHost: guitar.ucr.edu
In article <4gi8q7$69s@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (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, Diracdeltalike 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
>CauchyRiemann equations fit into this. Come on, ask me. I dare you.
The CauchyRiemann 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
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In article <4gj3sc$7ar@agate.berkeley.edu> doug@remarque.berkeley.edu (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."
"Rightoh, 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 3dimensional 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 strawfilled 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
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
MessageID:
References: <1996Feb12.182018.18379@schbbs.mot.com>
<4ftln7$1n6@guitar.ucr.edu>
<4gh2pg$2ma@guitar.ucr.edu>
ReplyTo: Oz@upthorpe.demon.co.uk
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In article <4gh2pg$2ma@guitar.ucr.edu>, john baez
writes
>
>Anyway, you have two alternatives. Either take a peek at the Schwarzschild
>metric and start to see how some of the weird timewarping,
>spacebending 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!notformail Fri Feb 23 19:26:47 PST 1996
Article: 101774 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (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
MessageID: <4gluc6$4n3@guitar.ucr.edu>
References: <4gh2pg$2ma@guitar.ucr.edu>
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk writes:
>In article <4gh2pg$2ma@guitar.ucr.edu>, john baez
>writes
>>Anyway, you have two alternatives. Either take a peek at the Schwarzschild
>>metric and start to see how some of the weird timewarping,
>>spacebending 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!notformail Sat Feb 24 21:15:23 PST 1996
Article: 101513 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (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
MessageID: <4gifh3$35a@guitar.ucr.edu>
References: <4gcts5$dm0@sulawesi.lerc.nasa.gov> <4gg2tg$27l@guitar.ucr.edu>
NNTPPostingHost: guitar.ucr.edu
In article Oz@upthorpe.demon.co.uk writes:
>In article <4gg2tg$27l@guitar.ucr.edu>, john baez
>>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 stressenergy tensor, Einstein
tensor and the like. If you use the usual recipes for calculating
stressenergy tensors you find that that stressenergy 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
stressenergy 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
stressenergy 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 finitesized 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 pseudotensor. (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. Pseudotensors 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 pseudotensors have some rather strange properties. If you
choose the "wrong" coordinates, they are nonzero 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 pseudotensors 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 pseudotensors 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
socalled stressenergy 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 nonlinear; 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
*energymomentum 4vector*. To learn the whole scoop on 4vectors, read a
text on SR, for example Taylor and Wheeler (see refs.) For our purposes,
it's enough to remark that 4vectors 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 quasiNewtonian 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 wellestablished, both theoretically and observationally.
(4) Expansion of the universe leading to cosmological redshift.
The Cosmic Background Radiation (CBR) has redshifted over billions
of years. Each photon gets redder and redder. What happens to this
energy? Cosmologists model the expanding universe with
FriedmannRobertsonWalker (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 pseudotensors 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 catchphrase "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
pseudotensors slide around this difficulty?
We've seen already that we should be talking about the
energymomentum 4vector, not just its timelike 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 energymomentum
conservation says that the energymomentum inside V changes only because of
energymomentum 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 = energymomentum 4vector)
T is called the stressenergy tensor. You don't need to know what
that means! just that you can integrate T, as shown, to get
4vectors. 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
4dimensional frame of mind, you'll discover it really says that the
flux across the boundary of a certain 4dimensional 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 4dimensional 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
energymomentum 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 nonMinkowskian 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 energymomentum 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 LeviCivita 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 nontrivial
mathematics lurks behind the phrase in quotation marks. But even
popscience 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 welldefined, because we're dealing with an
infinitesimal hypervolume. But to add up fluxes all over a finitesized
hypervolume (as in the contemplated extension of Gauss's theorem) runs
smack into the dependence on transportation path. So the flux integral is
not welldefined, 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
pseudotensors 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 semitechnical discussion of the
controversy over gravitational radiation.
Wheeler, "A Journey into Gravity and Spacetime". Wheeler's try at
a "popscience" treatment of GR. Chapters 6 and 7 are a
tourdeforce: Wheeler tries for a nontechnical 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 energyloss 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 BondiSachs
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
BondiSachs 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: <2dTq8JAEAvLxEwTJ@upthorpe.demon.co.uk>
Sender:
FollowupTo:
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 grumpylooking 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 ncategories 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 4vector 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!notformail Thu Feb 29 17:05:26 PST 1996
Article: 102534 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 29 Feb 1996 13:35:57 0800
Organization: University of California, Riverside
Lines: 142
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Recall one of the previous episodes...
Oz awoke to find himself in his cave, lying on the strawfilled 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
ncategories, 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 highpitched 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 LateTwentieth 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
laborsaving 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.ohiostate.edu!math.ohiostate.edu!uwm.edu!newsfeed.internetmci.com!in2.uu.net!psinntp!psinntp!psinntp!pipeline!notformail Thu Feb 29 18:58:42 PST 1996
Article: 102571 of sci.physics
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From: egreen@nyc.pipeline.com (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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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 / egreen@nyc.pipeline.com
"All coordinate systems are equal,
but some are more equal than others".
From galaxy.ucr.edu!notformail Thu Feb 29 18:58:59 PST 1996
Article: 102375 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (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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NNTPPostingHost: guitar.ucr.edu
In article <4gv12u$fqn@pipe11.nyc.pipeline.com> egreen@nyc.pipeline.com (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: columbus@pleides.osf.org (Michael Weiss)
Newsgroups: sci.physics
Subject: Re: Is the equivalence principle true in homo. grav. fields?
Date: 27 Feb 1996 20:40:04 GMT
Organization: OSF Research Institute
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ReplyTo: columbus@osf.org (Michael Weiss)
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Inreplyto: egreen@nyc.pipeline.com's message of 27 Feb 1996 08:31:10 0500
cc: egreen@nyc.pipeline.com
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 lowtech
pencil and paper and follow my directions. First draw your good old
horizontal xaxis and your vertical taxis. (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 "10forward" (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 chalkboard, 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?
Uhoh, 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
spacestation 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 DS9measured 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 lightpulses 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
lightpulses 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 lightsignals 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
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 iddybiddy 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 taxis 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 .
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 spacespace 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 coordinates 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?"
From galaxy.ucr.edu!ihnp4.ucsd.edu!munnari.OZ.AU!news.hawaii.edu!ames!uhog.mit.edu!news.mathworks.com!newsfeed.internetmci.com!in1.uu.net!psinntp!psinntp!psinntp!pipeline!notformail Thu Feb 29 18:59:23 PST 1996
Article: 102544 of sci.physics
Path: galaxy.ucr.edu!ihnp4.ucsd.edu!munnari.OZ.AU!news.hawaii.edu!ames!uhog.mit.edu!news.mathworks.com!newsfeed.internetmci.com!in1.uu.net!psinntp!psinntp!psinntp!pipeline!notformail
From: egreen@nyc.pipeline.com (Edward Green)
Newsgroups: sci.physics
Subject: Re: general relativity tutorial
Date: 28 Feb 1996 09:02:58 0500
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'carlip@dirac.ucdavis.edu (Steve Carlip)' wrote:
>: Suppose we followed a timelike 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 hereit'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 / egreen@nyc.pipeline.com
"All coordinate systems are equal,
but some are more equal than others".
From galaxy.ucr.edu!notformail Thu Feb 29 19:32:16 PST 1996
Article: 102576 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: General relativity tutorial
Date: 29 Feb 1996 17:14:04 0800
Organization: University of California, Riverside
Lines: 27
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In article <4h1huj$d6c@pipe12.nyc.pipeline.com> egreen@nyc.pipeline.com (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!notformail Fri Mar 1 17:09:55 PST 1996
Article: 102753 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (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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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 mindreading 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 iddybiddy 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 taxis 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 4dimensional 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 skewsymmetric 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!notformail Sat Mar 2 12:04:38 PST 1996
Article: 102965 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (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 <4h692o$t5p@csugrad.cs.vt.edu> nurban@vt.edu 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 indexgymnastic
>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.ohiostate.edu!pacific.mps.ohiostate.edu!freenet.columbus.oh.us!magnus.acs.ohiostate.edu!csn!news1.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: rmiller@landau.ucdavis.edu (Roger Miller)
Newsgroups: sci.astro.amateur,sci.astro,alt.journalism,sci.physics
Subject: Re: Today is the last Feb 29th of the millenium
FollowupTo: sci.astro.amateur,sci.astro,alt.journalism,sci.physics
Date: 1 Mar 1996 23:11:33 GMT
Organization: University of California, Davis
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Lori Eichelberger (LKEich@gslis.utexas.edu) wrote:
: ez049941@boris.ucdavis.edu (Jedidiah Whitten) wrote:
: >Emory F. Bunn (ted@physics12.Berkeley.EDU) 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 heremany, 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 goodYear 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
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Article: 103183 of sci.physics
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From: Oz
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
Organization: Oz
Lines: 163
Distribution: world
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In article <4h86ka$8uk@guitar.ucr.edu>, john baez
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 Wihiz, woohoo, 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 skewsymmetric in the
>first two slots."
>
>"Cool," says Oz.
Wondering what *exactly* "skewsymmetric 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 tdirection. 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 Stressenergy 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 spacetime 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!notformail Sun Mar 3 14:09:44 PST 1996
Article: 103139 of sci.physics
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From: egreen@nyc.pipeline.com (Edward Green)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 1 Mar 1996 12:26:22 0500
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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
mindbending special effects thingies involving swirling colors and the end
of the popular song by the dungbeetles, '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
alterego 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 toedents,
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
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Fri, 1 Mar 1996 21:59:55 +0000
Organization: Oz
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In article <4h7bvu$3ck@pipe9.nyc.pipeline.com>, Edward Green
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 smartass
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!notformail Sun Mar 3 15:44:35 PST 1996
Article: 102999 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 2 Mar 1996 15:48:37 0800
Organization: University of California, Riverside
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In article Oz@upthorpe.demon.co.uk 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 "radiationdominated 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 "matterdominated 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 offdiagonal 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 bmomentum. So the
component T_{0i} represents the density of imomentum (i.e., xmomentum,
ymomentum, or zmomentum), 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.
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 wellfounded.
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 bigbang universe as
a sphere (a 3sphere) 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
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 17:10:31 +0000
Organization: Oz
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In article <4h9nfv$fps@pipe12.nyc.pipeline.com>, Edward Green
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 reappears 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 Wiztime over in the United Plates. A whole lotta demons
are using the lodelines over there for their own nefarious purposes,
and the lodemasters have gone on short time working. "
"Gosh" said Ed "you had better be careful. You definitely wouldn't want
to crosslink 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?"
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Article: 103183 of sci.physics
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From: Oz
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
Organization: Oz
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In article <4h86ka$8uk@guitar.ucr.edu>, john baez
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 Wihiz, woohoo, 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 skewsymmetric in the
>first two slots."
>
>"Cool," says Oz.
Wondering what *exactly* "skewsymmetric 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 tdirection. 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 Stressenergy 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 spacetime 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
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sat, 2 Mar 1996 20:59:12 +0000
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In article <4h86ka$8uk@guitar.ucr.edu>, john baez
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 Wihiz, woohoo, 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 skewsymmetric in the
>first two slots."
>
>"Cool," says Oz.
Wondering what *exactly* "skewsymmetric 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 tdirection. 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 Stressenergy 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 spacetime 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!pipexsa.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
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 3 Mar 1996 10:52:07 +0000
Organization: Oz
Lines: 128
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In article <4hamol$9i2@guitar.ucr.edu>, john baez
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 offdiagonal 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 bmomentum. So the
>component T_{0i} represents the density of imomentum (i.e., xmomentum,
>ymomentum, or zmomentum), 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 imomentum. 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 idirection, +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 tdirection?
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 (hohoho) 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 xdirection???????
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!notformail Sun Mar 3 20:00:34 PST 1996
Article: 103267 of sci.physics
Path: galaxy.ucr.edu!notformail
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 3 Mar 1996 18:40:16 0800
Organization: University of California, Riverside
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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 stressenergy 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 bmomentum. So the
>>component T_{0i} represents the density of imomentum (i.e., xmomentum,
>>ymomentum, or zmomentum), 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 imomentum. 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 idirection, +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 idirection 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 xmomentum. Suppose
for visual vividness that this flow is carried by little ballshaped
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 xmomentum in the xdirection!
>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 offdiagonal
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 tdirection?
>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
Newsgroups: sci.physics
Subject: Re: Friedmann Models
Date: Thu, 29 Feb 1996 06:36:06 +0000
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In article <4h2jjg$oqv@agate.berkeley.edu>, "Emory F. Bunn"
writes
>dr^2+r^2 dphi^2.
>
>A uniform curved twodimensional surface is either a sphere or
>twodimensional hyperbolic space. If we label the sphere with
>polar coordinates (theta,phi), the line element is
>
>dtheta^2+(sin theta)^2 dphi^2.
>
>
>The line element for twodimensional 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 threedimensional
>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 threesphere has line element
>
>dr^2 + (sin r)^2 (dtheta^2 + (sin theta)^2 dphi^2),
>
>and you get the answer for negatively curved threespace 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 treekiller 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 fourdimensional spacetime is curved, the
>threedimensional 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
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From: ted@physics12.Berkeley.EDU (Emory F. Bunn)
Newsgroups: sci.physics
Subject: Re: Friedmann Models
FollowupTo: sci.physics
Date: 3 Mar 1996 23:10:31 GMT
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In article <+jcG$yAt+xNxEwBs@upthorpe.demon.co.uk>,
Oz 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 energymomentum 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
paralleltransported 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!notformail Mon Mar 4 17:01:50 PST 1996
Article: 103494 of sci.physics
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From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: 4 Mar 1996 15:43:38 0800
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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.
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Article: 103335 of sci.physics
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From: Oz
Newsgroups: sci.physics
Subject: Re: Riemann Awakes
Date: Sun, 3 Mar 1996 14:59:31 +0000
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In article <4h86ka$8uk@guitar.ucr.edu>, john baez