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 Message-ID: <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 2-dimensional Riemannian manifold", but it sits conveniently in (more or less) flat 3-dimensional Euclidean space, so we can think of a tangent vector to it as being an arrow whose base is at one point of the pumpkin, and which sticks out tangent to the pumpkin. We say it's a "tangent vector at a point" of the pumpkin. Now we have to abstract things a bit! First, remove the 3d ambient Euclidean space and think only of the surface of the pumpkin! We can still define a "tangent vector"... the actual definition being rather mathematical... but one way to visualize it is as a teeny-weeny itsy-bitsy little arrow drawn on the surface of the pumpkin, with its base at the specified point. We make it small --- in fact, infinitesimal --- just in order to avoid worrying about the fact that the pumpkin is curved. After all, if we had an ambient 3d space as before, we could ignore the difference between a vector tangent to the pumpkin, and a vector actually drawn *on* the pumpkin, in the limit where the arrow became very small. This is the sort of thing mathematicians can make precise; don't worry about it too much for now. Now what's a tensor? Well, there are a million ways to think of it, but a good way is to think of it as a machine that eats a list of say, 3 tangent vectors, and spits out a number, for example, or maybe a tangent vector. (This isn't quite the most general sort of tensor but it's good enough for starters.) We require that the output depend in a linear way on each of the inputs. So for example a while back I discussed the Riemann tensor R^a_{bcd}. This was a thing that ate 3 tangent vectors and spit out a tangent vector... which is why there are 3 subscripts and one superscript. Namely, if we parallel translate a tangent vector u around the little parallelogram of size epsilon whose edges point in the directions of the tangent vectors v and w, it changes by a little bit. Namely, it changes by the tangent vector whose component in the a direction is - epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3 Here we sum over b, c, and d. The thing "v^b" is the component of the vector v in the b direction... in whatever the hell coordinate system we happen to be using. And remember, indices like a,b,c,d range from 0 to 3 if we are working in 4d spacetime. From: 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 Message-ID: <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 horizontally-pointing javelin from the north pole to the equator along >)a line of constant longitude, then around the equator a bit, then back >)to the north pole. It's very good to work this out in detail. >Unfortunately, from this description I imagine a gyroscope (allowing >arrow to pivot vertically so it is always parallel to surface) attached >to the arrow, and the thing winding up pointing in the original >direction :-(. Why you imagine a gyroscope when I say "javelin" is beyond me. :-) >Also I imagine being at/near the north pole, going down to the >equator, back up, and being very confused about the forth leg! >(I can only go south from the north pole - what do I do _now_?). There ain't no fourth leg. Where'd I say there was a fourth leg? True, in my later description of the Riemann tensor I talked about carrying a tangent vector around a little square. But here we are carrying it around a big "triangle". In fact, one can do this parallel translation game for any sort of loop that ends where it started, or even any path. >Can you (while retaining this degree of informality) also achieve >the precision to avoid the above (presumably unintentional) interpretation >of what you actually meant by this? Well, I really meant just what I said. Say you were a Roman gladiator up at the north pole (historical accuracy not being my strong point) and you were handed a javelin. "Hold it horizontally, pointing that way," says your commander, pointing at a lump of ice at the horizon. You do so smartly, an exemplar of military precision. It points off to your left. "Now march forwards! Go straight ahead, and never rotate the javelin in the least, under penalty of death! Stop when you reach the equator!" And so on. You march along, never letting the javelin sway or rotate in the least. (If you buy Gauge Fields, Knots and Gravity by Baez and Muniain, you will see the result on pp. 232-233, although this loop goes from the north pole all the way to the south pole along a line of longitude, and then back up along another one.) From: 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 Message-ID: <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 >co-ordinates is also somewhat meaningless if they are >conventional ones since these are usually 'distance' >co-ordinates (taking time as a distance). I don't know what "distance coordinates" are. But I think I vaguely understand your discomfiture. You are getting used to the fact that the world doesn't come with a grid of lines painted on it. Of course, even in the old Newtonian days folks knew that any rotated or translated Cartesian coordinate system was "just as good" as any other. So it's not as if they thought there was a particular coordinate system handed down by God that was the "best" one. Instead, one had a manageable family of "good" coordinates, any one of which was related to any other by an easily understandable sort of transformation: rotation or translation. The same applied back in the days of special relativity. It may freak some people out that in addition to rotations and translations one has "Lorentz transformations" in which the new t' coordinate depends on the old t and x coordinates (say), but there are still a manageable set of "best" coordinates, corresponding physically to the inertial frames. General relativity takes a completely different approach to spacetime. In general relativity, spacetime is wiggly in a fairly arbitrary way (though it must satisfy Einstein's equation), so there is in general no manageable set of "best" coordinates. Think in terms of curved 2d space if you have trouble visualizing curved 4d spacetime. (In the long run you should learn how to visualize curved 4d spacetime, but visualizing curved 2d space is the best way I know to get to that point.) Say someone hands you a flat piece of paper. Then you can draw a Cartesian coordinate system on it in which every line is straight and all intersecting lines meet at right angles. There are different ways to do it, but they all differ only by rotations and translations. (And reflections, if one wants to nitpick.) Now say someone hands you a pumpkin. It's wiggly and bumpy... so there's no obvious best way to coordinatize its surface, not even any obvious manageable set of better-than-average ways. Say you try to draw a straight line on it. The best you can do is to draw a "geodesic": a curve that goes "locally as straight as possible". This is the curve a very tiny ant might follow if it was doing its best to follow its nose and walk as straight as possible. (Subconsiously, this should remind you of the fact that free fall in curved spacetime is motion along a geodesic, "locally as straight as possible", i.e. feeling no acceleration. I'm secretly indoctrinating you in a new worldview: physics as geometry.) Okay, so you don't have "straight lines" but you do have "geodesics" on a pumpkin. So say you try to draw a grid on the pumpkin such that each curve in it is a geodesic and whenever two geodesics intersect, they do so at right angles. If you could do that, it would be a good stab at Cartesian coordinates. But you can't. That's because the surface of the pumpkin is a "curved 2-dimensional Riemannian manifold" --- with the emphasis here on *curved*. So what do we do in general relativity, where spacetime is as bumpy as the surface of a pumpkin. We give up all attempts to pick "best" or "good" coordinates ---- except in working on certain very special problems with lots of symmetry! ---- and decide to do things in such a way that ANY choice of coordinates will work as smoothly as ANY OTHER. We don't exactly abandon coordinates; we just relegate them to the status of completely arbitrary tools. >I wonder if a lot of my confusion is in the intermix of >cosmology and GR. Too much to follow all in one shot, too >many new ways to look at things at once. Certainly this is a big part of it. Personally I can't teach you cosmology without teaching you GR, any more than I could teach you celestial mechanics without teaching you F = ma. There might be some way to study cosmology without GR, but to me that seems to miss the whole grandeur and strangeness of the subject. For example, it's true that one could not bother learning GR and only try to understand one solution of Einstein's equation, the Robertson-Friedman-Walker metric which describes the standard big bang cosmology. This might allow one a certain tempting conceptual sloppiness: one could write this metric down in the "standard coordinates" which take advantage of the symmetry of that solution, and ignore my remarks above about the arbitrariness of coordinates. But one would really be missing a lot of the fun! For example, you've already seen that in the Milne cosmology, different coordinates can give one very different pictures of what's going on. This mental flexibility is crucial if one gets into questions like "why is there a redshift: is it really due to the expansion of space, or is it a gravitational effect?" Without understanding physics as geometry and the arbitrary nature of ones choice of coordinates, these riddles can really throw one for a loop. *With* this understanding one can, so to speak, deconstruct the question and figure out the *right* question and the right answer. >Perhaps a simple GR example might help. I have assumed that >the moon orbits the earth in a GR explanation because it >simply travels in a straight line, but space (3D) is curved >(ie orbital sized curving). However I am getting the >impression that the GR curvature is very very tiny. Yes indeed. >I also >remember Baez commenting that in spacetime the geodesic of a >stone tossed up into the air is very very long and only very >slightly curved since the time distance is ct. Presumably >then the spacetime curvature that bends the moons orbit >round the earth is similarly only curved in a minuscule and >almost undetectable amount, but because of the 20 lightday >distance in the time direction we see it as taking a very >curved path in our *almost flat 3D slice* of spacetime. I >hope this has a small element of correctness in it because I >am finding it difficult piecing all the various questions >and answers together to form some coherent model that >doesn't fall foul of your devastatingly correct objections. Yes, this is right. Visualize the x and y directions as lying in the horizontal plane and the t direction as pointing "up". Then the geodesic path traced out by the moon is a very very stretched-out helix which goes "up" in time 28 lightdays each time it goes around, while its radius is less than a measly lightminute. This is very close to what a geodesic would be in flat spacetime (namely, a straight line). That's as expected, because the curvature is so small. Article 95270 (3700 more) in sci.physics: From: Michael Weiss (SAME) Subject: Re: Feeling stressed out? Date: 22 Jan 1996 22:20:37 GMT Organization: OSF Research Institute Lines: 22 NNTP-Posting-Host: pleides.osf.org In-reply-to: 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 x-axis, but you can't interchange the x-axis with the t-axis. This distinction can be expressed mathematically in a number of ways. For example, in the Minkowski formula for the spacetime "interval": ds^2 = dx^2 + dy^2 + dz^2 - dt^2 the dt^2 is the only term that gets a minus sign. Something called Sylvester's Theorem on Signatures implies that we'll get the same pattern of +'s and -'s no matter what frame of reference we use. Article 95116 (3694 more) in sci.physics: From: john baez (SAME) Subject: Re: Feeling stressed out? Date: 22 Jan 1996 12:40:51 -0800 Organization: University of California, Riverside Lines: 93 NNTP-Posting-Host: guitar.ucr.edu In article <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 n-space, real or complex, these tensors actually live on n >"timelike" dimensions, n "spacelike" dimesions, and possibly, k "k-like" >dimensions, l "l-like" dimensions, and so forth. When we transform >coordinates they are all mixed together, but it always turns out we can >identify the same number of coordinates of the same flavors we started >with. These tensors are really defined on direct sum spaces of the form ( >n | m | l | k | ...). Their theory is due to [fill in eastern European >sounding name]... and they are a treated in the branch of mathematics call >[fill in any suitable sounding name]... :-) " Well, what we really say is that there's a theory of n-dimensional semi-Riemannian manifolds of signature (p,q), where p + q = n. This means that at any point we can find an orthonormal basis of tangent vectors, p of which are spacelike and q of which are timelike. (Or the other way round, depending on your conventions!) This has been intensively studied, but the most deeply studied cases are the case where q = 0 --- the "Riemannian" case, which one might think of physically as the geometry of "space" --- and the case where q = 1 --- the "Lorentzian" case, which one might think of physically as the geometry of "spacetime". One can study worlds with 2 timelike and 2 spacelike directions... and in fact people do; the symmetry makes things very interesting... but from the viewpoint of physics this is rather decadent. To get started, read Semi-Riemannian geometry : with applications to relativity / Barrett O'Neill. New York : Academic Press, 1983. Article 95097 (3693 more) in sci.physics: From: john baez (SAME) Subject: Re: Feeling stressed out? Date: 22 Jan 1996 18:39:48 GMT Organization: University of California, Riverside Lines: 85 NNTP-Posting-Host: guitar.ucr.edu In article <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 stress-energy tensor is. Then, on the NYC subway, a dark thought hit >me: John Baez was pulling my leg! No, I wasn't. >Yes. No. >He claims that this object is a "tensor". Now, what little I know >about tensors leads me to believe they represent some linearization of a >local property of space. Their transformation properties are a guarantee >that they express a physical property of space, not an artifact of >coordinates. It's a guarantee that they express a physical property of space(time) or *stuff in it*. Here by stuff I mean fields and that sort of thing. >So what's so special about the top row??? >In three space the top row of a two index object may be called "x", but >conceptually it's not any different from "y". There is no special "x-ness" >about this position. So how can we say the top row of T keeps tract of >flow wrt *time*. Sounds to me like we are trying to shoe-horn time into >some calculational scheme, and it doesn't quite fit. Even if different >observers have different t's, there is still something intrinsically >"timelike" about this row sticking out like a sore thumb. We can work with tensors using whatever coordinates we want, and their transformation properties are a guarantee that if we watch our step, our results will not be a mere artifact of a coordinate choice. Pick any local coordinates x^0, x^1, x^2, x^3 on spacetime whatsoever. Use these to define the notion of a tangent vector in the ith direction (i = 0,1,2,3). Use them also to define the notion of momentum in the jth direction (0,1,2,3). The entry T_{ij} in the ith row and jth column of the tensor T then keeps track of the flow in the ith direction of the jth component of momentum. Nothing special about the top row of T_{ij} here! It is merely a matter of *convention* that, if we have a metric on spacetime, and thus a way of determining whether a given vector is timelike or spacelike, that we often use coordinates such that the tangent vector in the 0th direction is timelike, while those in the 1st, 2nd, and 3rd directions are spacelike. If we follow this convention, it is then common to indulge in a quaint anachronism and call momentum in the 0th direction "energy". It is also common to call flow in the 0th direction "density". That's what I was doing. But I'm not sure that's what you're worrying about. Maybe you're worrying about the fact that in spacetime, as opposed to space, not all directions were created equal? Some are spacelike and some are timelike! A tangent vector v is spacelike if its dot product with itself --- which we write as g(v,v) --- is positive, and timelike if g(v,v) is negative. (The gadget g is called the metric; we'll probably talk about that a lot more later if I wind up teaching a sci.physics course on general relativity. Often people use the opposite sign conventions from those above.) Now if you ask me why one can find a basis of orthogonal tangent vectors with 3 spacelike ones and 1 timelike one --- i.e., why there is only one dimension's worth of space and 3 of time --- I have to say I don't know. Nobody knows. Not yet, anyway. But don't mistake that puzzle for anything funny about *coordinates*. It's a completely coordinate-independent fact --- unlike the fact that folks like to use coordinates in which the x^0 coordinate corresponds to a time-like direction. The statement "there exists a basis of orthogonal tangent vectors with 3 spacelike ones and 1 timelike one" is a completely coordinate-independent statement. Article 95066 (3692 more) in sci.physics: From: Edward Green (SAME) Subject: Re: Feeling stressed out? Date: 21 Jan 1996 21:15:09 -0500 Organization: The Pipeline Lines: 53 NNTP-Posting-Host: pipe11.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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 0-component of >the energy-momentum 4-vector Thank you for your answers! I still feel there must be something special about the top-row, though where it falls between the mathematics and the physics I couldn't say. I gave this my best shot last time, but let me try one more analogy. Suppose somebody plopped us down at rest in a strange coordinate system. Now, none of us would have any trouble identifying which of the four dimesions in this frame was X_0, aka time, unless we had been chewing a few too many peyote buttons. But if somebody dropped a set of orthogonal spacial axes in our lap, and asked us "which one is X_1", we could rightly answer "how the hell should I know, jack? which ever one you want it to be, ok?" (I guess we are a little testy from being plopped in a strange coordinate frame). The point is, even though our transformations mix time with space, when we stop and consider a particular one there is always a special dimension sticking out, a dimension physically different from the others, which we call "time". I had hoped somebody with an abstract mathematical bent would say something like this: "Ok, while we loosely call both the objects relating to crystals in three space and the objects living in four dimensional spacetime "tensors", one is actually a more general concept. In ordinary three space all the coordinates have the same flavor, but in spacetime one of them has a different flavor, and no matter how we mix them, one of them always comes out tasting more of time than the others." He continues... "We capture this mathematically by saying that instead of living on a single n-space, real or complex, these tensors actually live on n "timelike" dimensions, n "spacelike" dimesions, and possibly, k "k-like" dimensions, l "l-like" dimensions, and so forth. When we transform coordinates they are all mixed together, but it always turns out we can identify the same number of coordinates of the same flavors we started with. These tensors are really defined on direct sum spaces of the form ( n | m | l | k | ...). Their theory is due to [fill in eastern European sounding name]... and they are a treated in the branch of mathematics call [fill in any suitable sounding name]... :-) " I hope this doesn't seem too sophmoric or argumentative coming from someone who started with so many questions! Call it mathematical fiction if you like. Does life imitate art? -- Ed Green 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 NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk Lines: 61 >In article <4dl1fq$19e@agate.berkeley.edu> ted@physics.berkeley.edu writes: >>Cosmologists often talk about the curvature of *space* (meaning >>ordinary three-dimensional space) rather than spacetime. This is >>perhaps unfortunate, since it confuses people like Oz, but that's the >>way it is. >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 3-dimensional >Euclidean space --- nice and flat --- and slice it up into a bunch of >concentric spheres --- which are curved. I am a little loath to ask a question, since I seem to have irritated the supervisors (apologies). However I am left in a rather unsatisfactory situation. Obviously I need a new paradigm, my quasi-classical one having been comprehensively demolished. For example. I can see than 'distance' is not a description of anything, cosmologically. As a result, plonking down a set of co-ordinates is also somewhat meaningless if they are conventional ones since these are usually 'distance' co-ordinates (taking time as a distance). Another measure or mechanism is needed. Even the initially non-physical co-moving co-ordinate system now looks to be more attractive although I am beginning to suspect that it sort of linearises the time direction only, dunno. I can also see why Baez insists that we can only know what is observed at a point. This cannot be ambiguous. I wonder if a lot of my confusion is in the intermix of cosmology and GR. Too much to follow all in one shot, too many new ways to look at things at once. Perhaps a simple GR example might help. I have assumed that the moon orbits the earth in a GR explanation because it simply travels in a straight line, but space (3D) is curved (ie orbital sized curving). However I am getting the impression that the GR curvature is very very tiny. I also remember Baez commenting that in spacetime the geodesic of a stone tossed up into the air is very very long and only very slightly curved since the time distance is ct. Presumably then the spacetime curvature that bends the moons orbit round the earth is similarly only curved in a minuscule and almost undetectable amount, but because of the 20 lightday distance in the time direction we see it as taking a very curved path in our *almost flat 3D slice* of spacetime. I hope this has a small element of correctness in it because I am finding it difficult piecing all the various questions and answers together to form some coherent model that doesn't fall foul of your devastatingly correct objections. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 95018 (3683 more) in sci.physics: From: john baez (SAME) Subject: Re: Red Shift {Was: Center of Universe?} Date: 21 Jan 1996 13:09:30 -0800 Organization: University of California, Riverside Lines: 37 NNTP-Posting-Host: guitar.ucr.edu In article <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 2-dimensional surface of the earth, and the javelin is the >>"little arrow" or "tangent vector", which must remain tangent to >>"space".) After marching down to the equator, march 90 degrees around >>the equator, and then march back up to the north pole, always keeping >>the javelin pointing horizontally and "in as same a direction as >>possible". >>By the time you get back to the north pole, the javelin is pointing a >>different direction! >>That's because the surface of the earth is curved. >I'm sorry, but I have a really obvious question here, so obvious you >didn't address it. What do we do with the javelin at the corners? Do we >try to hold it "pointing in the same direction" as near as possible, while >we turn, or do do we rotate it with us by the angle we turn through? Hold the javelin pointing in the same direction even at the corners! The idea is to do your damnedest to never rotate that sucker, even at corners. We want to study how, even when we do our best not to rotate a tangent vector as we carry it around, it can come back rotated due to the curvature of spacetime (or space, or the earth). Also, since corners are just a limiting case of sharp turns without corners, it would be a bad rule to do something different at corners than at non-corners. Article 95016 (3682 more) in sci.physics: From: john baez (SAME) Subject: Re: Red Shift {Was: Center of Universe?} Date: 21 Jan 1996 13:03:04 -0800 Organization: University of California, Riverside Lines: 73 NNTP-Posting-Host: guitar.ucr.edu In article <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 three-dimensional space) rather than spacetime. This is >perhaps unfortunate, since it confuses people like Oz, but that's the >way it is. Oh, so that's what the confusion is about. Sorry. Yes indeed, if you choose to slice up spacetime into slices called "space", the curvature of the slices depends on how you do the slicing. Indeed, the slices can be curved even if the spacetime itself is flat. Here's an example that starts with 3d space rather than 4d spacetime: take 3-dimensional Euclidean space --- nice and flat --- and slice it up into a bunch of concentric spheres --- which are curved. Whenever you do this slicing stuff, say with spacetime, there are nice relationships between: 1. the curvature of the spacetime 2. the curvature of the slices themselves --- their so-called "intrinsic curvature" and 3. the curvedness of how the slices sit inside the spacetime --- their so-called "extrinsic curvature". These are called the Gauss-Codazzi relations, and they are very important whenever we want to take spacetime, slice it, and study general relativity that way. I won't go in to them here, but for starters it's worthwhile trying to understand that 1, 2, and 3 are different concepts. > When someone asks whether space is curved, what they mean >is this. Choose a particular instant of time, and consider the slice >through four-dimensional spacetime that represents space at that one >instant. This is a perfectly good three-dimensional manifold, and >it's perfectly reasonable to ask whether or not this manifold is >curved. Yes, this is the so-called "intrinsic curvature" of the slice we're calling "space". >Worse, the >question of whether space is curved depends on what coordinates you >use. Specifically, choosing a particular "instant of time" everywhere >in space is a coordinate-dependent act, because there's no >coordinate-independent way of deciding whether two distant events are >at the same time. So your answer to the question of whether space (as >opposed to spacetime) is curved will in general depend on how you >choose to slice spacetime up. I'd prefer to say, not that the curvature of the slice depends on what coordinates you use, but that it depends on the slice! In other words, I like the last sentence in the above bit more than the first one. Of course, if you define your slice by taking a "time coordinate" t and letting your slice be described by the equation t = constant, the slice, hence its curvature, depends on the coordinate t. This indeed comes up a lot, since slicing spacetime into slices of "space" is a very practical tool. Still, it's worth keeping in mind Eddington's remark, that if we're trying to study the anatomy of a pig it can be rather confusing to study bacon. Article 95340 (3681 more) in sci.physics: From: john baez Subject: Re: Red Shift {Was: Center of Universe?} Date: 23 Jan 1996 13:03:24 -0800 Organization: University of California, Riverside Lines: 66 NNTP-Posting-Host: guitar.ucr.edu In article <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 2-dimensional Riemannian manifold", but it sits conveniently in (more or less) flat 3-dimensional Euclidean space, so we can think of a tangent vector to it as being an arrow whose base is at one point of the pumpkin, and which sticks out tangent to the pumpkin. We say it's a "tangent vector at a point" of the pumpkin. Now we have to abstract things a bit! First, remove the 3d ambient Euclidean space and think only of the surface of the pumpkin! We can still define a "tangent vector"... the actual definition being rather mathematical... but one way to visualize it is as a teeny-weeny itsy-bitsy little arrow drawn on the surface of the pumpkin, with its base at the specified point. We make it small --- in fact, infinitesimal --- just in order to avoid worrying about the fact that the pumpkin is curved. After all, if we had an ambient 3d space as before, we could ignore the difference between a vector tangent to the pumpkin, and a vector actually drawn *on* the pumpkin, in the limit where the arrow became very small. This is the sort of thing mathematicians can make precise; don't worry about it too much for now. Now what's a tensor? Well, there are a million ways to think of it, but a good way is to think of it as a machine that eats a list of say, 3 tangent vectors, and spits out a number, for example, or maybe a tangent vector. (This isn't quite the most general sort of tensor but it's good enough for starters.) We require that the output depend in a linear way on each of the inputs. So for example a while back I discussed the Riemann tensor R^a_{bcd}. This was a thing that ate 3 tangent vectors and spit out a tangent vector... which is why there are 3 subscripts and one superscript. Namely, if we parallel translate a tangent vector u around the little parallelogram of size epsilon whose edges point in the directions of the tangent vectors v and w, it changes by a little bit. Namely, it changes by the tangent vector whose component in the a direction is - epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3 Here we sum over b, c, and d. The thing "v^b" is the component of the vector v in the b direction... in whatever the hell coordinate system we happen to be using. And remember, indices like a,b,c,d range from 0 to 3 if we are working in 4d spacetime. Article 95253 (1 more) in sci.physics: From: Bruce Scott TOK (SAME) Subject: Re: Feeling stressed out? Date: 22 Jan 1996 19:19:27 GMT Organization: Rechenzentrum der Max-Planck-Gesellschaft in Garching Lines: 61 NNTP-Posting-Host: s4bds.aug.ipp-garching.mpg.de X-Newsreader: TIN [version 1.2 PL2] Edward Green (egreen@nyc.pipeline.com) wrote: : Thank you for your answers! I still feel there must be something special : about the top-row, though where it falls between the mathematics and the : physics I couldn't say. I gave this my best shot last time, but let me : try one more analogy. [...] The way you make the top row unique is to look at it in the rest frame of the thing whose stress tensor you're studying. For example, a dissipationless fluid has a stress tensor of T^ab = n e u^a u^b + p (g^ab + u^a u^b) (n is the particle density in the fluid's rest frame, e is the thermal energy per particle, and p is the pressure), which looks a bit complicated. In fact, there is _nothing_ special about the top row of this if g^ab has time-space components (example: Kruskal coordinates for the black hole problem) and u is relativistic (especially if it has significant variation in any of the four coordinates). Note that the only physical requirement on u is that it must be a timelike _unit_ vector: u_a u^a = -1 (in units with c = 1). But if you contract this with the fluid's velocity you get a peek at the fluid's rest frame: - T^ab u_b = n e u^a. Now, that looks a lot more like an energy flux, doesn't it? It is in fact a four-vector. But what about the pressure? You can find the fluid's rest frame by transforming such that u^a = (1,0,0,0). Note you have to do a local Lorentz transformation: one that is defined for the small neighborhood about a single point. This is because u for a fluid is itself a field variable. But in the local rest frame you find T^ab = diag ( ne, p, p, p ), where "diag" denotes a diagonal matrix. The 00-component is the thermal energy density and the ii-components are the pressure (one assumes an isotropic pressure for a ideal fluid). Note that "thermal energy" here includes rest energy. Your question was actually aimed at distinguishing among the ii-components. This is, as you surmise, not possible without arbitrariness, unless there is a physical phenomenon (such as a local magnetic field; gee, isn't plasma physics nice :-] ) to show you the way. There is a popular book I saw once where the question of explaining left and right to an extraterrestrial was discussed, to show the arbitrariness in our convention. A complicated set of mental gymnastics was offered by which one could use symmetry violation in kaon decays to do this. No, I don't remember the details :-) -- Mach's gut! Bruce Scott The deadliest bullshit is Max-Planck-Institut fuer Plasmaphysik odorless and transparent bds@ipp-garching.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 NNTP-Posting-Host: pleides.osf.org In-reply-to: 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 "time-like" separations and the t-axis, and Bruce Scott's nice explanation of the stress-energy tensor for the perfect fluid, I don't know that there's much to add. However, I do remember one thing about T_ij that puzzled me when I first encountered it, and the "aha" mental lightbulb that cleared things up. Here's the puzzle: if T_ij measures the amount of i-momentum being transferred in the j-direction, then how come T_ij isn't identically zero in the rest frame of the fluid, when the momentum vanishes? Let's polish off a couple of points quickly. As Bruce Scott points out, in general a fluid doesn't have a global rest frame. We can say we're just talking about an infinitesimal neighborhood of a point, or the special case of a motionless fluid. Next, T_00 is a special case. In the rest frame of the fluid, the velocity 4-vector of a fluid element is (E,0,0,0) -- you can make the v_i components vanish for i=1,2,3 by picking the right frame of reference, but not the v_0 component, aka the energy. (I'm being sloppy about the difference between tensors and tensor-densities, but let's ignore that. What's a determinant among friends?) So it's not surprising that T_00 doesn't vanish, but why doesn't T_ii vanish for i=1,2,3? (Once we've plopped ourselves down in the rest frame of the fluid, that is.) Answer: think microscopically, and remember that T depends *quadratically* on v. Consider the yz-plane, for yucks. Particles of fluid constantly zip through this plane in all directions. If a particle with 4-vector (E, v_x, v_y, v_z) passes across the plane, headed in the +x direction (i.e., v_x>0), then it transfers a bit of v_x momentum in that direction, and likewise v_y momentum and v_z momentum. OK, now here's the kicker. If we look at the *average* transfer of v_y and v_z across the yz-plane, we get zero. That's because v_x is uncorrelated with v_y and v_z, so the average of (v_x.v_y) is zero, ditto (v_x.v_z). (If we weren't sitting in the rest-frame of the fluid--- if we felt a breeze--- then this wouldn't necessarily be true.) But the average of (v_x.v_x) is positive, of course; it doesn't take much imagination to see why this represents the pressure. From Oz@upthorpe.demon.co.uk Wed Jan 24 10:54 PST 1996 Received: from relay-4.mail.demon.net (relay-4.mail.demon.net [158.152.1.108]) by math.ucr.edu (8.6.12/8.6.12) with SMTP id KAA29235 for ; Wed, 24 Jan 1996 10:50:32 -0800 Received: from post.demon.co.uk ([158.152.1.72]) by relay-4.mail.demon.net id ae17044; 24 Jan 96 14:24 GMT Received: from upthorpe.demon.co.uk ([158.152.116.64]) by relay-3.mail.demon.net id ab13561; 24 Jan 96 13:33 GMT From: Oz To: john baez Newsgroups: sci.physics Subject: Re: General relativity tutorial Date: Wed, 24 Jan 1996 13:32:13 GMT Reply-To: Oz@upthorpe.demon.co.uk Message-Id: <3105f1bb.96882427@post.demon.co.uk> References: <30f6c596.25015591@news.demon.co.uk> <4damsr$pon@pipe9.nyc.pipeline.com> <4dp0pf$bu8@guitar.ucr.edu> In-Reply-To: <4dp0pf$bu8@guitar.ucr.edu> X-Mailer: Forte Agent .99c/16.141 Content-Type: text Content-Length: 3093 Status: OR Maybe this one? On 19 Jan 1996 13:00:31 -0800, you wrote: >In article <4damsr$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 horizontally-pointing javelin from the north pole to the equator along >a line of constant longitude, then around the equator a bit, then back >to the north pole. It's very good to work this out in detail. > >Okay, now for some more jargon: by "arrow" we really mean "tangent >vector", which we can think of roughly as an "infinitesimal arrow" based >some point of space. And this process of carrying a tangent vector >around while doing our best not to rotate or stretch it is called >"parallel translation". > >Now for the definition of the Riemann curvature tensor R^a_{bcd}. This >gadget knows all about the curvature of space and we will use it to cook >up the Einstein tensor G_{ab} that sits on the left hand side of >Einstein's equation > >G_{ab} = T_{ab}. > >Here I will use letters at the beginning of the alphabet to denote the >numbers 0,1,2,3. The 0,1,2 and 3 components of something are the t,x,y, >and z components of something... where we use *completely arbitrary* >coordinates t,x,y, and z. (In setting up general relativity, everything >should work nicely NO MATTER WHAT COORDINATES we use.) I have already >said what T_{ab} is: it's the flow in the a direction of momentum in the >b direction. (E.g., T_{00} is the flow in the time direction of >momentum in the time direction, i.e. the density of energy.) So now I >gotta say what G_{ab} is.... but the crucial thing is to define the >Riemann curvature tensor R^a_{bcd}. > >Remember, the letters a,b,c and d stand for anything like 0,1,2 or 3, >or in other words, t,x,y and z. > >Say we take a tangent vector pointing in the d direction and carry it >around a little square in the b-c plane. We go in the b direction until >the b coordinate has changed by epsilon, then we go in the c direction >until the c coordinate has changed by epsilon, then we go back in the b >direction until the b coordinate is what it started out as, and then we >go back in the c direction until the c coordinate is what it started out >as. Our tangent vector may have rotated a little bit since space is >curved. Its component in the a direction has changed a bit, say > >-epsilon^2 R^a_{bcd} > >plus terms of order epsilon^3. > >This defines R^a_{bcd}. The minus sign is just an annoying convention. > >That's the glorious Riemann curvature tensor. > > ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 95429 (10 more) in sci.physics: From: john baez Subject: Getting tenser? Date: 24 Jan 1996 10:58:54 -0800 Organization: University of California, Riverside Lines: 76 NNTP-Posting-Host: guitar.ucr.edu In article <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 higher-dimensional Euclidean space: it's easy to visualize what we mean by a vector *tangent* to a point of that manifold. But in fact, even if we do not think of our spacetime as embedded in some higher-dimensional space, one may define the notion of a "tangent vector" at a point of spacetime. One could just as well drop the "tangent" business and call it a "vector", but hotshot physicists use so many different kinds of vectors they like to keep track of things this way, and besides, the "tangent vector" terminology encourages the useful kind of geometrical thinking that I was trying to cultivate in folks with that pumpkin metaphor. Think of a tangent vector at a point of spacetime, if you like, as a wee arrow whose tail is pinned to that point. >Now I never really needed to do anything >with tensors in anger, or even at all. However I did read a >little about them many years ago and I vaguely remember >deciding they were basically vectors with position. Hmm, that's not what tensors are... "vectors with position" is actually a pretty decent way of saying what *tangent vectors* are. >So I >would imagine them defining a 4D vector at every point in >spacetime. Presumably one does some sort of path integral >incorporating one's 'original' direction and the Riemann >curvature tensor. The Riemann curvature tensor can >presumably be expressed as some function over all points on >the path you are interested in. It sounds the sort of >operation you hope you only have to do with conveniently >tractable functions, say 4D conic sections or whatever. Hmm. I can't understand a word of this, and I'm afraid that if I keep on trying I will. >Ah-ha. I presume your '4-tangents' are related to geodesics. >So I stuff a 'rectangular' co-ordinate system over space (I >hope that's allowed) then fire off some object from some >point in spacetime. This Riemann curvature tensor will tell >me the change in direction at every point and in this way I >can plot it's path wrt the chosen co-ordinate system. Hmm. There is a certain vague truth to this, but I would be reluctant to say it's right. There are many things we have to get to before we can fully understand how the following things are related: 1. the metric 2. parallel translation 3. the Riemann curvature tensor 4. geodesics 5. the Einstein tensor 6. the stress-energy tensor I've described how to compute 3 in terms of 2. (Did someone out there keep a copy of my definition of the Riemann tensor in terms of carrying a vector around a little square? I want to save a copy! Oz and Ed Green have not coughed one up.) I've also said that Einstein's equation says 5 and 6 are proportional. I've also said that freely falling objects move along 4. But I need to say what 4 has to do with 2. And, to really understand GR, I need to tell you how you compute 5 in terms of 3 --- that's pretty easy --- and how you compute 2 in terms of 1 --- that's a bit hard. Then everything will be all hooked together. From Oz@upthorpe.demon.co.uk Wed Jan 24 16:23 PST 1996 Received: from relay-4.mail.demon.net (relay-4.mail.demon.net [158.152.1.108]) by math.ucr.edu (8.6.12/8.6.12) with SMTP id QAA06531 for ; Wed, 24 Jan 1996 16:23:09 -0800 Received: from post.demon.co.uk ([158.152.1.72]) by relay-4.mail.demon.net id aa17751; 24 Jan 96 14:30 GMT Received: from upthorpe.demon.co.uk ([158.152.116.64]) by relay-3.mail.demon.net id ad13561; 24 Jan 96 13:33 GMT 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 Reply-To: Oz@upthorpe.demon.co.uk Message-Id: <3105f2b9.97136125@post.demon.co.uk> References: <4du5tb$96i@agate.berkeley.edu> <4e0vru$8p3@status.gen.nz> <4e3f69$dte@guitar.ucr.edu> In-Reply-To: <4e3f69$dte@guitar.ucr.edu> X-Mailer: Forte Agent .99c/16.141 Content-Type: text Content-Length: 1998 Status: OR Ahh, is this it? Oz baez@guitar.ucr.edu (john baez) wrote: >In article <4df7gg$jd9@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward Green) writes: > >>What stress tensor? > >The stress-energy tensor, aka energy-momentum tensor, T_{ij}, where >i,j go from 0 to 3. This gadget is the thing that appears on the right >side of Einstein's equation for general relativity: > >G_{ij} = T_{ij} > >(in nice units). The thing on the left is the Einstein tensor, that >summarizes some information about spacetime curvature. > >> What top row? > >The top row, T_{0j}, of this 4x4 matrix, keeps track of the *density* of >energy --- that's T_{00} --- and the density of momentum in the x,y, and >z directions --- that's T_{01}, T_{02}, and T_{03} respectively. > >This should make sense if you remember that i=0,1,2,3 correspond to >t,x,y, and z respectively. And that energy is just the same as momentum >in the time direction. > >The other entries of the stress-energy keep track of the *flow* of >energy and momentum in various spatial directions. > >In brief, T_{ij} is the flow in the i direction of momentum in the j >direction. > >This gadget tells you everything about what energy and momentum are >doing at your given point of spacetime. > >>What language are you talking? > >Physics, ca 20th century. > >(I sure can be snide. Sorry.) > >>You wouldn't care to give me a crash descriptive course in GR, would you? >>(That is the language you are speaking, isn't it?). > >Well, the stress-energy tensor is a basic gadget throughout physics, >since we all want to know where energy and momentum are going and how >much there is sitting around, right? But it's only in general >relativity where the stress-energy tensor is sitting proudly on the >right side of an equation, telling spacetime how to curve. > > > > > ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" From KMULDREW@jiarg.mccaig.ucalgary.ca Fri Jan 26 12:30 PST 1996 Received: from fsa.cpsc.ucalgary.ca (fsa.cpsc.ucalgary.ca [136.159.2.1]) by math.ucr.edu (8.6.12/8.6.12) with ESMTP id MAA07096 for ; Fri, 26 Jan 1996 12:30:46 -0800 Received: from mccaig.ucalgary.ca (epicenter.mccaig.ucalgary.ca [136.159.150.10]) by fsa.cpsc.ucalgary.ca (1.8) id ; Fri, 26 Jan 1996 13:30:51 -0700 Received: from by mccaig.ucalgary.ca (4.1/SMI-EPI-4.1) id AB00698; Fri, 26 Jan 96 13:30:17 MST Received: from NOVELL1/SpoolDir by jiarg.mccaig.ucalgary.ca (Mercury 1.21); 26 Jan 96 13:37:48 -0700 Received: from SpoolDir by NOVELL1 (Mercury 1.21); 26 Jan 96 13:37:36 -0700 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 Reply-To: kmuldrew@acs.ucalgary.ca Priority: normal X-Mailer: Pegasus Mail for Windows (v2.01) Message-Id: <1F4F4D625CE@jiarg.mccaig.ucalgary.ca> Content-Type: text Content-Length: 11093 Status: OR ======== 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 1-element tensor in this case). A standard matrix multiplication is then Y = AX (matrix notation) b b a Y = A a X (tensor notation). Because the 'a' is repeated, it us understood that one sums over it - one sums over any repeated indeex. In this case, one sums over the row of A (row, because it's a subscript) of a and the column (column because it's a superscript) of X. The convention is that one always sums over one row and one column, never a row and a row or a column or a column. The difference between row and column vectors becomes very important and is called "tangent and cotangent spaces" or "covariant and contravariant" components. The distinction doesn't matter when one has a nice, uniform, Cartesian metric, (i.e. s= x^2+y^2+z^2) but becomes important when one has to deal with other metrics, such as with either non-cartesian co-ordinate systems or the x^2+y^2+z^2 -t^2 metric of GR. Actually the difference isn't fundamentally important as long as one keeps tract of what's what. If one misplaces an index, one is off by the value of the metric, which is no longer the identity matrix, which it used to be. in the simplest case. So one has to keep tract of it, but once one is aware of the need to keep tract of it one relegates this to an unpleasant but necessary task that one has to do to get the right answers). Dunno if the metric stuff made any sense - do you recall having seen at least some stuff about matrices and "quadratic forms"? If you have, the metric stuff should make more sense - the quadratic form for x^2+y^2+z^2 is the identity matrix, which is why it's possible to not to worry about covariance vs. contravariance. This isn't possible when the quadratic form of the metric isn't equivalent to an identity metric. Anyway, a matrix is just a two-element tensor, one can think of a matrix as either generating a column vector from a column vector (the usual way), or one can consider it as taking in a column vector, and a row vector (aka - one form), and generating a scalar result. A three element tensor can be thought of as taking in two row/column vectors and producing a r/c vector, or it can be thought of as taking in three r/c vectors and producing a scalar. The last definition is what most mathematicians use, a general tensor of degree m+n takes in m row vectors and n column vectors and spits out a scalar. (This may seem odd, but it's an easy way to generalize). As I recall, the Rienmann tensor is a 4-element tensor - it takes in 3 things and spits out another. This is what defines the "curvature" of space in GR. One other point I should make explicitly - it's all linear. Hope this isn't two confused - that's my take on the topic, anyway, I'm not as familiar with all this yet as I'd really like myself. ps - "Gravitation" does have an intro to this stuff, but it moves quickly, like it's supposed to be a review (IMO). ======== Newsgroups: sci.physics Subject: Getting tenser? From: 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 higher-dimensional Euclidean space: it's easy to visualize what we mean by a vector *tangent* to a point of that manifold. But in fact, even if we do not think of our spacetime as embedded in some higher-dimensional space, one may define the notion of a "tangent vector" at a point of spacetime. One could just as well drop the "tangent" business and call it a "vector", but hotshot physicists use so many different kinds of vectors they like to keep track of things this way, and besides, the "tangent vector" terminology encourages the useful kind of geometrical thinking that I was trying to cultivate in folks with that pumpkin metaphor. Think of a tangent vector at a point of spacetime, if you like, as a wee arrow whose tail is pinned to that point. >Now I never really needed to do anything >with tensors in anger, or even at all. However I did read a >little about them many years ago and I vaguely remember >deciding they were basically vectors with position. Hmm, that's not what tensors are... "vectors with position" is actually a pretty decent way of saying what *tangent vectors* are. >So I >would imagine them defining a 4D vector at every point in >spacetime. Presumably one does some sort of path integral >incorporating one's 'original' direction and the Riemann >curvature tensor. The Riemann curvature tensor can >presumably be expressed as some function over all points on >the path you are interested in. It sounds the sort of >operation you hope you only have to do with conveniently >tractable functions, say 4D conic sections or whatever. Hmm. I can't understand a word of this, and I'm afraid that if I keep on trying I will. >Ah-ha. I presume your '4-tangents' are related to geodesics. >So I stuff a 'rectangular' co-ordinate system over space (I >hope that's allowed) then fire off some object from some >point in spacetime. This Riemann curvature tensor will tell >me the change in direction at every point and in this way I >can plot it's path wrt the chosen co-ordinate system. Hmm. There is a certain vague truth to this, but I would be reluctant to say it's right. There are many things we have to get to before we can fully understand how the following things are related: 1. the metric 2. parallel translation 3. the Riemann curvature tensor 4. geodesics 5. the Einstein tensor 6. the stress-energy tensor I've described how to compute 3 in terms of 2. (Did someone out there keep a copy of my definition of the Riemann tensor in terms of carrying a vector around a little square? I want to save a copy! Oz and Ed Green have not coughed one up.) I've also said that Einstein's equation says 5 and 6 are proportional. I've also said that freely falling objects move along 4. But I need to say what 4 has to do with 2. And, to really understand GR, I need to tell you how you compute 5 in terms of 3 --- that's pretty easy --- and how you compute 2 in terms of 1 --- that's a bit hard. Then everything will be all hooked together. Article 95557 (379 more) in sci.physics: From: Oz Subject: Re: Red Shift {Was: Center of Universe?} Date: Wed, 24 Jan 1996 13:31:55 GMT Lines: 133 NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk X-Newsreader: Forte Agent .99c/16.141 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) co-ordinate system and 'another view', or at least be able to chose one that suited the problem and could be induced to give 'sensible' answers. Here we seem to be in the situation that the problem, which is non trivial, is to ask a clearly defined question first. Indeed many reasonable questions seem to have become undefined, this (luckily for you) seriously inhibits question asking unless you know the subject very well indeed at a deep level. For this (dammit) you need to have studied it in it's technical depth. A somewhat depressing conclusion. However I appreciate the small scent of the subject I have gleaned. Not really satisfactory, but I suppose that's life. >The same applied back in the days of special relativity. It may freak >some people out that in addition to rotations and translations one has >"Lorentz transformations" in which the new t' coordinate depends on the >old t and x coordinates (say), but there are still a manageable set of >"best" coordinates, corresponding physically to the inertial frames. Yes, this mislead me for a very long while. >General relativity takes a completely different approach to spacetime. >In general relativity, spacetime is wiggly in a fairly arbitrary way >(though it must satisfy Einstein's equation), so there is in general no >manageable set of "best" coordinates. >Okay, so you don't have "straight lines" but you do have "geodesics" on >a pumpkin. So say you try to draw a grid on the pumpkin such that each >curve in it is a geodesic and whenever two geodesics intersect, they do >so at right angles. If you could do that, it would be a good stab at >Cartesian coordinates. But you can't. That's because the surface of >the pumpkin is a "curved 2-dimensional Riemannian manifold" --- with the >emphasis here on *curved*. > >So what do we do in general relativity, where spacetime is as bumpy as >the surface of a pumpkin. We give up all attempts to pick "best" or >"good" coordinates ---- except in working on certain very special >problems with lots of symmetry! ---- and decide to do things in such a >way that ANY choice of coordinates will work as smoothly as ANY OTHER. Comment: Most of the structures that seem to be bandied about seem to be exactly these highly symmetrical structures. In fact precious little else. Should I deduce that in general these are the only ones that we (I mean 'you') can properly handle in an unambiguous enough way to make a reasonable and general statement of both the problem and it's solution? >We don't exactly abandon coordinates; we just relegate them to the >status of completely arbitrary tools. Hmmm. Some esoteric mathematical construct that takes a near genius several years to begin to master. Gloom. >>I wonder if a lot of my confusion is in the intermix of >>cosmology and GR. Too much to follow all in one shot, too >>many new ways to look at things at once. > >Certainly this is a big part of it. Personally I can't teach you >cosmology without teaching you GR, any more than I could teach you >celestial mechanics without teaching you F = ma. There might be some >way to study cosmology without GR, but to me that seems to miss the whole >grandeur and strangeness of the subject. This is, of course, true. However I was rather hoping that a basic understanding, but below the level of being able to properly manipulate real problems, might be enough to follow the reasoning well enough. For example understanding F=ma will not allow you to calculate most dynamic problems, but will allow you to follow someone elses explanation. Anyway I still live in some small hope. >For example, it's true that one could not bother learning GR and only >try to understand one solution of Einstein's equation, the >Robertson-Friedman-Walker metric which describes the standard big bang >cosmology. This might allow one a certain tempting conceptual >sloppiness: one could write this metric down in the "standard >coordinates" which take advantage of the symmetry of that solution, and >ignore my remarks above about the arbitrariness of coordinates. But one >would really be missing a lot of the fun! For example, you've already >seen that in the Milne cosmology, different coordinates can give one >very different pictures of what's going on. This mental flexibility is >crucial if one gets into questions like "why is there a redshift: is it >really due to the expansion of space, or is it a gravitational effect?" >Without understanding physics as geometry and the arbitrary nature of >ones choice of coordinates, these riddles can really throw one for a >loop. *With* this understanding one can, so to speak, deconstruct the >question and figure out the *right* question and the right answer. Exactly, of course, why it interests me. I rather like the unintuitive behaviour of the cosmos, it stretches the brain. I do not have the time, or enough brain cells any more, to take a prostgrad (UK postgrad that is) course on the subject. I must make do with crumbs from the table. >>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 NNTP-Posting-Host: 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 teeny-weeny >>itsy-bitsy little arrow drawn on the surface of the pumpkin, with its >>base at the specified point. We make it small --- in fact, >>infinitesimal --- just in order to avoid worrying about the fact that >>the pumpkin is curved. After all, if we had an ambient 3d space as >>before, we could ignore the difference between a vector tangent to the >>pumpkin, and a vector actually drawn *on* the pumpkin, in the limit >>where the arrow became very small. >OK, perhaps, just perhaps, I get the idea. 'Normal' geometry >is obsessed with orthogonality in some way. We abandon this >and go for tangents. Now it's tempting to say that any small >bit of pumpkin is flat, but that loses the curvaceousness of >it. On the other hand we can't really define a big long >tangent line on a 2D pumpkin since it cruises off into >another non-existant dimension. We can however, always >define an infinitesimally short one, possibly in a >similarish way to Newton, at least in concept. That's the basic idea. Mathematicians have had many years to make Newton's "infinitesimals" precise... in quite a variety of different ways which are pretty much equivalent for practical purposes. Thus we may unabashedly imagine a tangent vector to a pumpkin as an vector tangent to the pumpkin, but infinitesimal, so that it doesn't cruise off into the 3d space which is, after, quite nonexistent to the Pumpkin People, the 2-dimensional beings who inhabit the surface of the pumpkin. >Incidentally I presume your warty old pumpkin is properly >2D, ie it's really completely smooth but has some strange >spacial properties where different paths give different >distances in "whatever co-ordinate system we define". Since >we find it hard to visualise this we bend it up into a >non-existent 3rd dimension. Precisely. For us, of course, the 3rd dimension is real and the surface of the pumpkin is a mere "submanifold", but for the Pumpkin People the pumpkin is all of space and the 3rd dimension would simply be a mathematical fiction. Luckily, there is no need to argue. Mathematicians have mastered both "extrinsic geometry," in which a curved space is treated as a "submanifold" of some other, perhaps flat, space, and "intrinsic geometry", where you treat curved space in its own right and don't imagine it sitting in some higher-dimensional space. The intrinsic viewpoint, developed by Gauss and Riemann in the late 1800s, is harder to get good at but it's often better. Why bother with extra dimensions you never really see or use anyway, if you don't need to? >>a good way is to think of it as a machine that eats a list of say, 3 >>tangent vectors, and spits out a number, for example, or maybe a tangent >>vector. (This isn't quite the most general sort of tensor but it's good >>enough for starters.) We require that the output depend in a linear way >>on each of the inputs. >OK, I bet in reality it's not quite as simple as this >however. Well, that was a pretty precise definition a large class of tensors, but not of the most general kind. Here it is again, more formally so you will feel the suffering normally associated with education: A tensor of "rank (0,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a number. The output must depend linearly on each input. A tensor of "rank (1,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a tangent vector at the point p. The output must depend linearly on each input. I am avoiding defining tensors of rank (n,k) for other values of n, because there is actually a fair amount of physics I can do without dragging them in. Eventually I would need to explain them, and you'd see they weren't much worse. >I take it that "linear way" is a fundamental >property of a Tensor. Indeed!!!!!!!!!! It's a branch of Linear Algebra. >>So for example a while back I discussed the Riemann tensor R^a_{bcd}. >>This was a thing that ate 3 tangent vectors and spit out a tangent >>vector... which is why there are 3 subscripts and one superscript. >> >>Namely, if we parallel translate a tangent vector u around the little >>parallelogram of size epsilon whose edges point in the directions of the >>tangent vectors v and w, it changes by a little bit. Namely, it changes >>by the tangent vector whose component in the a direction is >> >>- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3 >> >>Here we sum over b, c, and d. The thing "v^b" is the component of the >>vector v in the b direction... in whatever the hell coordinate system we >>happen to be using. And remember, indices like a,b,c,d range >>from 0 to 3 if we are working in 4d spacetime. >1) In general I assume there are an infinity of possible >tangent vector directions like v,w above defining some >parallelogram of size epsilon (which some nasty person will >presumably tend to zero). I presume this is operational for >a well behaved space over which tangent vectors are >definable. Sure, there are infinitely many tangent vectors at a point. This is not so bad. For example, note that the output R^a_{bcd} v^b w^c u^d (let's ignore that epsilon junk and the niggly minus sign) depends linearly on the inputs u, v, and w, so we don't need to know it for *infinitely* many choices of u, v, and w to figure out what it will be for all possible choices. Linearity keeps life simple. >2) Somebody has sneakily brought in some co-ordinates whilst >nobody was looking (a,b,c ....). Now telling me they are >local just won't do and nor will telling me they are >'whatever co-ordinates you desire'. Well, certainly they are local coordinates, because you would be hard pressed to flatten out a whole pumpkin and impose *global* coordinates on it, rendering it a mere plane. And certainly the coordinates above are indeed "whatever coordinates you desire". But you are starting to sound like a mathematician --- high praise in my book --- with your complaint about the unpleasant appearance of coordinates. So to reward you, I will explain how it works without coordinates. You'll see it's much simpler. The Riemann tensor is a tensor of rank (1,3) at each point of spacetime. Thus it takes three tangent vectors, say u, v, and w as inputs, and outputs 1 tangent vector, say R(u,v,w). As usual, the output depends linearly on each input. The Riemann tensor is defined like this: Take the vector w, and parallel transport it around a wee parallelogram whose two edges are the vectors epsilon u and epsilon v , where epsilon is a tiny number longing to approach zero. The vector w comes back a bit changed by its journey; it is now a new vector w'. We then have w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3 Note: I am now saying just what I said before, but without those yucky coordinates! If you insist on using coordinates, I will say "Go ahead! Pick any that you like! I don't care which!" The only effect will be to turn the above elegant equation into the grungier but sometimes more practical one: w'^a - w^a = - epsilon^2 R^a_{bcd} v^b w^c u^d + terms of order epsilon^3 >I have had (in GR >context) been bludgeoned into realising that I need to >reconsider the concept of 'co-ordinates' altogether. Good. There's nothing like a good bludgeoning now and then. Article 95805 (376 more) in sci.physics: From: Keith Ramsay (SAME) Subject: Re: Red Shift {Was: Center of Universe?} Date: 25 Jan 1996 04:43:34 GMT Organization: Boston College Lines: 89 NNTP-Posting-Host: mt14.bc.edu In article <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 space-time pumpkin. You can draw parametrized curves on it. You can have functions varying from point to point on it. Assume we're dealing in both cases with curves and functions which are differentiable (no "corners"). Given a parametrized curve passing through a point, and a function, you can consider the rate of change of the function as you follow the point. If you like: d f(v(t)) _________ dt where v(t) parametrizes the curve, and f is the function varying on the pumpkin. So if, as I tell my students, the parametrized curve describes the path of a fly, and f describes the temperature at each point, then we're talking about how fast the fly is getting hotter when it is passing through the point. Now various parametrized curves passing through the same point will be "equivalent" in the sense that the rate of increase is the same for any given f. Likewise, various functions will be "equivalent" in the sense that for any given parametrized curve (did I say it had to be differentiable?) passing through the point, the functions are both increasing at the same rate. The set of all the curves which are equivalent in this way to a given one are said to define a "tangent vector" at that point. (If it were in space, it would be a "velocity".) It's the tangent vector for those curves. The set of all functions which are equivalent in this respect to a given one are said to define a "cotangent vector" at the point. It is essentially the gradient vector of the function. |Now what's a tensor? Well, there are a million ways to think of it, but |a good way is to think of it as a machine that eats a list of say, 3 |tangent vectors, and spits out a number, for example, or maybe a tangent |vector. (This isn't quite the most general sort of tensor but it's good |enough for starters.) We require that the output depend in a linear way |on each of the inputs. What John is skirting around here is simply the bit about cotangent vectors. Some tensors you have to think of as taking in cotangent vectors, if you want to think of them this way. Taking in a tangent vector is a lot like spitting out a cotangent vector, and vice versa. |So for example a while back I discussed the Riemann tensor R^a_{bcd}. |This was a thing that ate 3 tangent vectors and spit out a tangent |vector... which is why there are 3 subscripts and one superscript. It's equivalent to think of it as taking in three tangent vectors and a cotangent vector (at the same point), and spitting out a number in a multi-linear way. (Let Baez's gizmo take the three vectors and give back one. Then combine it with the given cotangent vector and output the result.) Up to a point, people are happy to blur the distinction between tangent vectors and cotangent vectors. They both work out as coordinate vectors if you have a fixed coordinate system. But to identify them with each other, you need a metric. Concretely, you associate a tangent vector with a covector by imagining that your fly is going directly toward the hotter air, and moves at a speed in proportion to how quickly it is warming up with distance. This requires, however, having a notion of "distance". That's typically okay. In GR, though, the metric is something you have to figure out. It's not given. So one is more careful when converting vectors to covectors and vice-versa. In terms of notation, it's called "raising and lowering indices". Given a metric, different types of tensors of the same rank can be put in correspondence with the others. I guess I should stop. Keith Ramsay Article 95871 (375 more) in sci.physics: From: Keith Ramsay (SAME) Subject: Re: Red Shift {Was: Center of Universe?} Date: 25 Jan 1996 18:32:58 GMT Organization: Boston College Lines: 105 NNTP-Posting-Host: mt14.bc.edu In article <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 co-ordinates whilst |nobody was looking (a,b,c ....). Now telling me they are |local just won't do and nor will telling me they are |'whatever co-ordinates you desire'. The point is that there are some things which are hard to describe without using coordinates, but can be described in a coordinate- invariant way. If you change the coordinates, you have to change the numbers (R^0_{000}, R^1_{230}, v^2 etc.) appearing in the formula. But the result you get will be the same tangent vector, but expressed in the new coordinates. Here's a simple example. T_a would be a tensor which takes a tangent vector and gives back a number (i.e., a cotangent vector). v^a would be a tangent vector. If we have coordinates at a point, then the tangent vector v there has coordinates (v^0, v^1, v^2, v^3). The coordinates of T are (T_0, T_1, T_2, T_3). They are defined as the answers T gives when fed the vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), and (0,0,0,1) respectively. By the linearity of T, the answer you get when you plug in v is v^0 times T applied to (1,0,0,0) + v^1 times T applied to (0,1,0,0) + etc., i.e. T_0 v^0 + T_1 v^1 + T_2 v^2 + T_3 v^3. Now, to keep things managable we have summation notation: sum{a} T_a v^a. Then, to make it even simpler to write, we have the "Einstein summation convention": if an index appears as a subscript and as a superscript, then sum over it. So we just have to write T_a v^a. So you see, it doesn't depend on which coordinate system you have. I like the notation Wald uses in his book, General Relativity. For formulas which are like the ones we have here, in which any coordinate system will do, but it's helpful to write it out suggesting how we would calculate the result if we had a particular one in mind, one has an "abstract index notation". One way to think of it is as a way of labelling the inputs and outputs of your tensors. In some products, wires are labelled red, green, blue for which socket matches which plug. In calculations with tensors, it's convenient sometimes to label which "slots" in tensors match up with which other ones. So the formula above is just a way to describe plugging the vectors v,w,u into R and getting a tangent vector out. Here's an odd thing one can do. (Those with dirty minds may be excused.) Take a tensor T^a_b, the kind which takes a tangent vector and gives back one. It turns out T^a_a (plugging one end into the other) makes sense as a number, independent of coordinate system. The components T^i_j of T in a given coordinate system can be thought of as the entries of a matrix giving the linear transformation on tangent vectors which T is. T^a_a represents the sum T^0_0+T^1_1+T^2_2+T^3_3 of the diagonal entries of T's matrix. This is called the "trace" of the matrix. Interestingly enough, if you change the coordinate system, the matrix in the new coordinate system has the same trace. Since also sometimes you want to calculate with the components of tensors in a given coordinate system, the notation allows for that. Switch the index letters to Greek! This is just a way to say, "No, this formula is not guaranteed to be correct for all coordinate systems." Keith Ramsay Article 96015 (374 more) in sci.physics: From: john baez (SAME) Subject: Re: Red Shift {Was: Center of Universe?} Date: 26 Jan 1996 13:27:13 -0800 Organization: University of California, Riverside Lines: 113 NNTP-Posting-Host: guitar.ucr.edu In article <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) co-ordinate >system and 'another view', or at least be able to chose one >that suited the problem and could be induced to give >'sensible' answers. Here we seem to be in the situation that >the problem, which is non trivial, is to ask a clearly >defined question first. Indeed many reasonable questions ^seemingly >seem to have become undefined, this (luckily for you) >seriously inhibits question asking unless you know the >subject very well indeed at a deep level. For some reason your asking of questions seems not to have been inhibited one whit. >For this (dammit) >you need to have studied it in it's technical depth. Perhaps. Or else you need to learn the art of asking operational questions --- questions that come along with an experimental procedure for determining their answer. Those are the two ways to stay reasonable in modern physics: either learn the mathematical formalism so you can ask questions about that, or stick with operational questions. >>So what do we do in general relativity, where spacetime is as bumpy as >>the surface of a pumpkin. We give up all attempts to pick "best" or >>"good" coordinates ---- except in working on certain very special >>problems with lots of symmetry! ---- and decide to do things in such a >>way that ANY choice of coordinates will work as smoothly as ANY OTHER. >Most of the structures that seem to be bandied about seem to >be exactly these highly symmetrical structures. In fact >precious little else. Should I deduce that in general these >are the only ones that we (I mean 'you') can properly handle >in an unambiguous enough way to make a reasonable and >general statement of both the problem and it's solution? It's true that things get trickier when there are no symmetries around, hence no "specially nice" coordinates. There are two aspects to this trickiness. The first and simpler one is that in the absence of symmetries, the math gets tough: there are very few "exactly solvable" problems in GR without symmetry. In these cases one typically needs a computer to solve things numerically. So, for example, the problem of two in-spiralling black holes is one of the National Science Foundation's "grand challenge problems", and lots of people are writing code to solve it. The second one is the deeper issue of what questions make sense. For a while now I have been attempting to force you to only ask questions that make sense in the *general case*, i.e. that make sense when spacetime is wiggly and bumpy in an arbitrary way. While you continue to kick and scream and ask questions that don't make sense, I think you are getting the idea of why they don't make sense. But I hope you also get an idea of what questions DO make sense! There are lots of them out there... as there had damn well better be! Basically, IF YOU TELL ME EXACTLY HOW TO PERFORM AN EXPERIMENT, I CAN TELL YOU WHAT WOULD HAPPEN. I write this in capital letters for three reasons: one, because it's the fundamental criterion for a theory to be complete, two, because there's never been any reason to expect any *more* from a theory, and three, because only a crackpot would say that they personally could meet *any* challenge of this kind: some physics problems like this may be too hard in practice, but *in principle* the theory should tell us what would happen. Of course this is a bit sneaky, because when you start describing an experiment and say "measure the distance from A to B", I will say "how?" And you will have to tell me a procedure. And if that procedure involves coordinates, I will say "which coordinates?" And so on. This question-and-answer game may take a while to play, but if you play it well, it *does* end with you giving a exact specification of an experiment, which I can then in principle describe the results of. (Of course, in quantum mechanics the results would be probabilistic in nature. But we're not talking about that.) >>We don't exactly abandon coordinates; we just relegate them to the >>status of completely arbitrary tools. >Hmmm. Some esoteric mathematical construct that takes a near >genius several years to begin to master. Gloom. Huh? It's not some "esoteric mathematical construct" you need to master, it's an *attitude*. You have already mastered the esoteric mathematical construct known as coordinates. Now you just need to master the attitude that the coordinates are arbitrary: that any calculation you want to do, you can do in any coordinates whatsoever. (Sometimes it's easier in some coordinates than others, of course.) This means that you don't get attached to any particular choice of coordinates; you don't attribute too much "physical reality" (whatever that is) to it. >This is, of course, true. However I was rather hoping that a >basic understanding, but below the level of being able to >properly manipulate real problems, might be enough to follow >the reasoning well enough. For example understanding F=ma >will not allow you to calculate most dynamic problems, but >will allow you to follow someone elses explanation. Anyway I >still live in some small hope. Oh, certainly you can get this basic understanding without being able to do any calculations. That's what I'm shooting for. You need to learn about tensors, metrics, parallel transport, and curvature, but you don't need to know how to calculate your way out of a paper bag. Luckily, they are all geometrical concepts so they are very easy to understand without any long and complicated formulas. Article 96159 (373 more) in sci.physics: From: Oz Subject: Re: Red Shift {Was: Center of Universe?} Date: Fri, 26 Jan 1996 13:43:08 GMT Lines: 48 NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk X-Newsreader: Forte Agent .99c/16.141 Ramsay-MT@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 NNTP-Posting-Host: guitar.ucr.edu In article <4e43j8$6db@pipe10.nyc.pipeline.com> egreen@nyc.pipeline.com (Edward Green) writes: >'bds@ipp-garching.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 NNTP-Posting-Host: guitar.ucr.edu In article <3105f6bd.98164211@news.demon.co.uk> Oz@upthorpe.demon.co.uk writes: >Uh-huh. So as expected there is some 'definition' of >parallel transportation. Is it a straightforward one or do I >not want to know? In what I've said so far in this mini-course on general relativity, I have taken parallel translation as a (fairly) easy-to-grasp starting point. It is simply a function that, given a point p, a curve from p to q, and a tangent vector v at p, spits out a tangent vector at q. One can do this in the nitty-gritty mathematical theory, as well. One then requires this function to satisfy a few obvious axioms: e.g., if you have a curve from p to q, and another curve from q to r, you can glom them together to get a curve from p to r, and then we can parallel translate a tangent vector at p over to r either in two stages or in one fell swoop, and we had better get the same answer. (There are a few more axioms.) But what we need to get to eventually is the marvelous fact that the *metric* on spacetime determines a "best" recipe for parallel translation. Remember, the "metric" g is a tensor such that if you feed in two tangent vectors v and w, g(v,w) is the "dot product" or "inner product" of v and w. This is the gadget that lets us calculate angles between vectors, lengths of curves, and all that. Our spacetime has a metric on it, a "Lorentzian" metric g (meaning *roughly* that the dot product of a vector with itself can be negative... for details see the stuff I posted in response to Ed). By a certain amount hard work we can get from the metric g_{ab} to parallel translation to <--------------------- I already described this step! the Riemann curvature tensor R^a_{bcd} to the Einstein tensor G_{ab} (I stuck in indices just so you know how many inputs there are to each of the tensors listed. Parallel translation is not a tensor because it involves two points and a curve between them, not just one point.) Once we have done this, when we look at Einstein's equation G_{ab} = T_{ab} we will know how the left hand side is cooked up from the metric, and what it has to do with the curvature of spacetime. We already know about the right hand side, the stress-energy (or energy-momentum) tensor. So we will understand how the presence of energy and momentum curve spacetime! I hope this course outline inspires you and reassures you that there's not a vast indefinite number of things you need to learn. Article 96392 (33 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 28 Jan 1996 11:46:40 -0800 Organization: University of California, Riverside Lines: 119 NNTP-Posting-Host: guitar.ucr.edu In article <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 co-ordinates since it points >from here to there, and there can be as close as we want and >we could (probably) co-ordinatise here and there if we >wanted to in any co-ordinate system we chose. >Am I in the ball park? Same orbit? Yes, this is a good informal summary of what tangent vectors are like. In particular, while we can describe them using any coordinates we like, we don't need any coordinates to get our hands on them. This is true both at the intuitive level --- what does a teeny infinitesimal arrow care about coordinates?? --- and at the mathematically rigorous level. >Riemann curvature tensor derivation. This little >parallelogram that results from 'parallel transport' really >does need to be looked at a teeny bit more deeply. It's >John's own fault really, he has battered my preconceived >ideas into abject submission, so I am hypercautious. In fact >I am trying to cultivate an outlook that makes Ted look >positively reckless. A praiseworthy goal, though you'd be hard put to make Ted look reckless. >A) Our loop round local space. >1) Am I right in assuming that we always arrive back at our >starting point after our little local excursion? Well, we can parallel transport a tangent vector at the point P along any path from P to the point Q and get a tangent vector at Q. E.g., our Roman can carry his javelin from the north pole to the equator and leave it there. This is important. But for the present purposes, let's concentrate on what happens when we parallel translate a tangent vector around a teeny loop that ends where it started. >2) For n dimensions I presume we have n tangent vectors to >follow so we enclose an n dimensional volume. Yikes!!!!!!!!! No, I like to say what I mean, and you must respect that quirk of mine. I gave the definition of the Riemann tensor in any number of dimensions a while back; I'll quote a post of mine where I did it without coordinates: The Riemann tensor takes three tangent vectors, say u, v, and w, as inputs, and outputs one tangent vector, say R(u,v,w). It's defined like this: Take the vector w, and parallel transport it around a wee parallelogram whose two edges go in the directions epsilon u and epsilon v , where epsilon is a tiny number longing to approach zero. The vector w comes back a bit changed by its journey; it is now a new vector w'. We then have w' - w = - epsilon^2 R(u,v,w) + terms of order epsilon^3 The point is this. The Riemann tensor is supposed to tell us in a "local" or "infinitesimal" way how space (or spacetime) is curved. What does that mean? Well, space being curved means that when you parallel transport a tangent vector around a loop, it comes back changed. The idea of the Riemann tensor is that we can take any big loop, span it with a surface, and then chop up that surface into tons of wee parallelograms, thus reducing the problem of "parallel translation around a big loop" to lots of problems of "parallel translation around a wee parallelogram". This works in any dimension. So we never need to worry about "parallel translating around a hyperquadrilateral." (Note: if our space has a funky topology, there may be no surface spanning a given big loop! Let's not worry about that now, though there is a whole branch of math devoted to it. There are always lots of loops that *can* be spanned by a surface.) >Same ball-park? Same solar system? Well, you gotta let that parallelogram thing sink in. Get rid of the hyperquadrilaterals. >Typical prof. Hang the carrot out and the donkies just gotta >follow for the payoff. Hmmm, looks really nice and juicy, if >only I could reach it! Sure you can; if it was that hard to understand general relativity there wouldn't be thousands of people who understood it. Especially since you just want a rough idea of it, it's really just a matter of putting some time into thinking about what I say, and keeping on asking questions. >Due to the unreal time element of usenet there are active >threads all over on, or associated, with this subject from >the same posters. In order to allow John the chance to be >coherent I propose that we all post to one thread, and this >one is as good as any. I will try to start collecting the various threads along these lines and renaming them Re: General relativity tutorial which is what one of them is already called. Article 96402 (25 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 28 Jan 1996 12:45:21 -0800 Organization: University of California, Riverside Lines: 89 NNTP-Posting-Host: guitar.ucr.edu In article <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 well-defined nature of >"trying very hard not to rotate the arrow". The point of the exercise is >after all, try as hard as we like and the damn thing will have rotated >when we get back! But just how "hard" is "very hard"... The key thing is the *local* nature of parallel transport. I think I noted this in my first explanation: at *each step of the way* you do your best not to rotate the vector. It's cheating to know your route ahead of time and sneakily diddle with your vector so that it comes out pointing the same way at the end of the journey. I think you know this and are just trying to cause trouble. Let's go back to the example I originally gave. Say our Roman starts at the north pole with his javelin. Say his javelin points directly in front of him as he begins his journey down the meridian to the equator. Assuming the earth is a perfect sphere and he doesn't gratuitously swing his javelin from side to side, he will end up at the equator with the javelin pointing due south. Now he turns 90 degrees --- NOT rotating the javelin, of course!!! --- and walks along the equator due west. The javelin continues to point due south, to his left. He goes 90 degrees along the equator and stops. He does NOT sneakily rotate his vector because he guesses what's coming. He's a Roman soldier, after all, trained to follow orders. He turns 90 degrees again until he's facing north. He does NOT rotate the javelin --- jeez, how many times do I need to say this: he doesn't EVER rotate that javelin --- so it is still facing due south, directly behind him. He now marches up the meridian to the north pole, carrying the javelin pointing directly behind him, not rotating it even a teeny weeny little bit. When he returns to the north pole the javelin is pointing a different direction than when he started. In fact, it's rotated by an angle of 90 degrees from his initial position. For fun, notice that if we think of the earth as a unit sphere, it's area is 4 pi, and our Roman has just travelled around a region of land having area one eighth of that, hence pi/2. His javelin has rotated by an angle of pi/2! This is no coincidence: on the unit sphere, whenever you go around a simple closed curve enclosing an area A, parallel translation gives a rotation of angle A. >It's like SR... when we speak of length contraction, that means we have >tried "very hard" to measure the correct length... but not too hard! If >we tried "hard enough", ie, allowing for the Lorentz transfomation, we >would measure the correct rest length! And if we tried hard enough here, >maybe doing a bit of local surveying, perhaps we *could* triumphantly >come back to the same orientation, even in curved space! Of course you can always cleverly correct for things, but that's not the point. DON'T cleverly correct for things. In the case of Lorentz contractions, just read what the damn ruler says. In the case of parallel translations, just follow orders like a Roman soldier and don't ever swing that javelin around. >[Other perverse objections deleted.] >In fact, let me take a wild educated guess: If we get around to stating >some STOKES like theorem, to the effect the total discrepency accumulated >in walking a path is equal to some surface integral, it will turn out that >we first state it for a path made of segments of geodesics, and then >"arbitrarily well approximate" our arbitrary path by little pieces of >geodesic, in an appropriate sense, of course. Well, you can see in the paragraph beginning "for fun" that there is a Stokes-like theorem operating here. But there's no need to prove it first for piecewise geodesic paths; in fact it applies just as well to situations where parallel translation is well-defined but geodesics are not --- situations that are irrelevant to general relativity, but very important in discussing the strong, weak, and electromagnetic forces, which are ALSO all about parallel translation. Article 96411 (21 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 28 Jan 1996 13:28:44 -0800 Organization: University of California, Riverside Lines: 154 NNTP-Posting-Host: guitar.ucr.edu To keep this course rolling, let me just state where we've gotten so far, and rapidly finish explaining general relativity!! Well, I won't really explain all of general relativity here. But if you read this and understand it somewhat, you will know what Einstein's equation, the basic equation of general relativity, says. 1. A TANGENT VECTOR at the point p of spacetime may be visualized as an infinitesimal arrow with tail at the point p. 2. A TENSOR of "rank (0,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a number. The output must depend linearly on each input. A TENSOR of "rank (1,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a tangent vector at the point p. The output must depend linearly on each input. 3. The METRIC g is a tensor of rank (0,2). It eats two tangent vectors v,w and spits out a number g(v,w), which we think of as the "dot product" or "inner product" of the vectors v and w. This lets us compute the length of any tangent vector, or the angle between two tangent vectors. Since we are talking about spacetime, the metric need not satisfy g(v,v) > 0 for all nonzero v. A vector v is SPACELIKE if g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0. 4. PARALLEL TRANSLATION is an operation which, given a curve from p to q and a tangent vector v at p, spits out a tangent vector v' at q. We think of this as the result of dragging v from p to q while at each step of the way not rotating or stretching it. There's an important theorem saying that if we have a metric g, there is a unique way to do parallel translation which is: a. Linear: the output v' depends linearly on v. b. Compatible with the metric: if we parallel translate two vectors v and w from p to q, and get two vectors v' and w', then g(v',w') = g(v,w). This means that parallel translation preserves lengths and angles. This is what we mean by "no stretching". c. Torsion-free: this is a way of making precise the notion of "no rotating". I don't think I want to go into the math of "torsion" just yet. Let's see the overall picture first. 5. The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point of spacetime. Thus it takes three tangent vectors, say u, v, and w as inputs, and outputs one tangent vector, say R(u,v,w). The Riemann tensor is defined like this: Take the vector w, and parallel transport it around a wee parallelogram whose two edges point in the directions epsilon u and epsilon v , where epsilon is a small number. The vector w comes back a bit changed by its journey; it is now a new vector w'. We then have w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3 Thus the Riemann tensor keeps track of how much parallel translation around a wee parallelogram changes the vector w. 6. Introducing COORDINATES. Now say we choose coordinates on some patch of spacetime near the point p. Call these coordinates x^a (where a = 0,1,2,3). Then given any tangent vector v at p, we may speak of its components v^a in this basis. The inner product g(v,w) of two tangent vectors is given by g(v,w) = g_{ab} v^a w^b for some matrix of numbers g_{ab}, where as usual we sum over the repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION. Another way to think of it is that our coordinates give us a basis of tangent vectors at p, and g_{ab} is the inner product of the basis vector pointing in the x^a direction and the basis vector pointing in the x^b direction. Similarly, the vector R(u,v,w) has components R(u,v,w)^a = R^a_{bcd} u^b v^c w^d where we sum over the indices b,c,d. 7. The EINSTEIN TENSOR. The matrix g_{ab} is invertible and we write its inverse as g^{ab}. We use this to cook up some tensors starting from the Riemann curvature tensor and leading to the Einstein tensor, which appears on the left side of Einstein's marvelous equation for general relativity. We will do this using coordinates to save time... though later we should do this over again without coordinates. This part is the only profound and mysterious part, at least to me. Okay, starting from the Riemann tensor, which has components R^a_{bcd}, we now define the RICCI TENSOR to have components R_{bd} = R^c_{bcd} where as usual we sum over the repeated index c. Then we "raise an index" and define R^a_d = g^{ab} R_{bd}, and then we define the RICCI SCALAR by R = R^a_a The Riemann tensor knows everything about spacetime curvature, but these gadgets distill certain aspects of that information which turn out to be important in physics. Finally, we define the Einstein tensor by G_{ab} = R_{ab} - (1/2)R g_{ab}. You should not feel you understand why I am defining it this way!! Don't worry! That will take quite a bit longer to explain. But we are almost at Einstein's equation; all we need is 8. The STRESS-ENERGY TENSOR. The stress-energy is what appears on the right side of Einstein's equation. It is a tensor of rank (0,2), and it defined as follows: given any two tangent vectors u and v at a point p, the number T(u,v) says how much momentum-in-the-v-direction is flowing through the point p in the u direction. Writing it out in terms of components in any coordinates, we have T(u,v) = T_{ab} u^a v^b In coordinates where x^0 is the time direction t while x^1, x^2, x^3 are the space directions (x,y,z), we have the following physical interpretation of the components T_{ab}: The top row of this 4x4 matrix, keeps track of the density of energy --- that's T_{00} --- and the density of momentum in the x,y, and z directions --- those are T_{01}, T_{02}, and T_{03} respectively. This should make sense if you remember that "density" is the same as "flow in the time direction" and "energy" is the same as "momentum in the time direction". The other components of the stress-energy tensor keep track of the flow of energy and momentum in various spatial directions. 9. EINSTEIN'S EQUATION: This is what general relativity is based on. It says that G = T or if you like coordinates and more standard units, G_{ab} = 8 pi k T_{ab} where k is Newton's gravitational constant. So it says how the flow of energy and momentum through a given point of spacetime affect the curvature of spacetime there. That's it! Article 96587 (29 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 29 Jan 1996 12:57:52 -0800 Organization: University of California, Riverside Lines: 243 NNTP-Posting-Host: guitar.ucr.edu In article <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 present-day theories of forces: Maxwell's equations for electromagnetism, Einstein's theory of general relativity, and the Yang-Mills equations for the electroweak and strong forces. We say they are all "gauge theories". What this means is that the basic field involved in any of these forces is a "connection" which describes what happens to particles when you move them along a path. Various internal degrees of freedom get "parallel transported". When you take them around a loop they don't come back as they were. When you study this infinitesimally, you get a "curvature tensor" describing parallel translation around infinitesimal parallelograms --- of which the Riemann curvature is an example. There is a formula for this curvature tensor as a kind of "curl" of the connection. In the case of magnetostatics, this takes the simple form: B = curl A That is, the magnetic field is the curl of the vector potential. The vector potential is the connection in this case, and the magnetic field is the curvature. In the case of electromagnetism in 4d spacetime we have the same sort of thing: F = dA where the electromagnetic field F is the curvature and the (4d) vector potential A is the connection. Here d is a 4d analog of the curl, called the "exterior derivative". In Yang-Mills theory and gravity we have the same sort of thing, only somewhat fancier. Not too surprising, in a way, since Einstein, Yang and Mills were all deliberately trying to copy the shining example of Maxwell's equations. Now you might ask: in general relativity, when a parallel translate a tangent vector around a loop, its actual direction changes. But in electromagnetism, if I carry a charged particle around a loop, what changes? It's phase! (Here quantum theory rears its ugly head.) Article 96602 (18 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 29 Jan 1996 14:01:27 -0800 Organization: University of California, Riverside Lines: 71 NNTP-Posting-Host: guitar.ucr.edu In article <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 four-dimensional >>spacetime). What's F_{abc}, you ask? Any old batch of numbers!!! So typographically speaking, the above equation just means a b c F(u,v,w) = F u v w abc If you're wondering what THAT means, reread the above explanation and then ask question... or maybe someone will be so kind as to go over an example of how this tensor stuff works. I gotta go teach class. >>It just works out that way: the amount the vector w changes when you >>carry it around a little parallelogram with sides epsilon u and epsilon >>v is roughly proportional to epsilon^2. >Hmmm, sounds plausible. One might expect the amount of >curvature to relate to area. Yup. Article 96646 (58 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 29 Jan 1996 17:23:09 -0800 Organization: University of California, Riverside Lines: 115 NNTP-Posting-Host: guitar.ucr.edu In article <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 fancy-ass mathematical physics like "topological quantum field theories". >A topological space >with no metric? Do such things actually have physical applications? Well, we don't need to consider topological spaces with no metric, we can simply consider smooth manifolds with no metric, but with some other geometrical structure, to get situations where parallel transport of some *other* sort of vectors (not tangent vectors) might be well-defined, but not geodesics. It's sort of amusing: to the outsider these possibilities must seem quite bizarre, but in fact they have been under intense scrutiny ever since the early 20th century, and are "standard material" that every mathematician or theoretical physicist should be acquainted with. They call it simply "geometry" or "differential geometry". Just to reminisce slightly... when I learned general relativity in college, I realized I needed to learn more geometry to really understand what was going on. The assigned text was Weinberg's Gravitation and Cosmology, but in a notorious passage in that book Weinberg writes `the passage of time has taught us not to expect that the strong, weak, and electromagnetic interactions can be understood in geometrical terms, and too great an emphasis on geometry can only obscure the deep connections between gravitation and the rest of physics.' Many theoretical physicists these days would laugh their heads off if someone said that now. Anyway, I naturally turned to Misner Thorne and Wheeler's Gravitation, which *does* emphasize the geometry, and I liked this much better, but I realized I needed a good solid mathematical treatment of geometry, so I eventually bumped into Analysis, Manifolds, and Physics, by Yvonne Choquet-Bruhat, Cecile DeWitt-Morette, and Margaret Dillard-Bleick, North Holland, New York, 1982. This covers a lot of the geometry and other math physicists need, and I fell in love with it; I just kept reading and rereading it 'til I knew all that stuff. Later in grad school I studied differential geometry with Guillemin (famous for his work on symplectic geometry, which is the geometry of *phase space* rather than physical space or spacetime), but I didn't go too deep in the subject. When I started working in earnest on quantum gravity, I found I needed to dig deeper into geometry. So I've been learning a bit more about it. Not a whole lot more, mind you: it's a vast and deep field, and if I got too carried away with it I'd never get to all the other stuff I need to know, and do, in studying quantum gravity. Okay, here's a reminder of how you get from the metric to parallel transport: PARALLEL TRANSLATION is an operation which, given a curve from p to q and a tangent vector v at p, spits out a tangent vector v' at q. We think of this as the result of dragging v from p to q while at each step of the way not rotating or stretching it. There's an important theorem saying that if we have a metric g, there is a unique way to do parallel translation which is: a. Linear: the output v' depends linearly on v. b. Compatible with the metric: if we parallel translate two vectors v and w from p to q, and get two vectors v' and w', then g(v',w') = g(v,w). This means that parallel translation preserves lengths and angles. This is what we mean by "no stretching". c. Torsion-free: this is a way of making precise the notion of "no rotating". Article 96675 (57 more) in sci.physics: From: john baez (SAME) Subject: Re: general relativity tutorial Date: 29 Jan 1996 19:16:01 -0800 Organization: University of California, Riverside Lines: 60 NNTP-Posting-Host: guitar.ucr.edu In article <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 one-form? I always >thought those wore considered linear (?). Or are they called >anti-linear? Or what? Let's see. I was thinking of u, v, and w as three different vectors. In my example, I considered the function which eats three vectors u, v, and w, and outputs the vector 2u + v + 7w. You might naively think this is a tensor of rank (1,3), since there are 3 vector inputs and 1 vector output, and it looks sorta linear. My point was that it's not, since it's not separately linear in each argument. But you seem to be thinking of u, v, and w as the 3 components of a single vector! Perfectly understandable, since I didn't make myself terribly clear! So let's think about this. Now we are saying F is a function which eats one vector (u,v,w) and spits out a number 2u + v + 7w. Now F turns out to be a tensor of rank (0,1), since there is one vector input and one number output, and it now *is* linear in its one input. You are right; this kind of tensor is also called a one-form or cotangent vector. So with sufficiently ambiguous notation it's completely impossible to tell whether something is a tensor or not. :-) I'll try to be clearer about notation in the future. Article 96864 (58 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 30 Jan 1996 11:25:16 -0800 Organization: University of California, Riverside Lines: 162 NNTP-Posting-Host: guitar.ucr.edu In article <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 co-ordinate invariant. Yes, "geodesic" is a coordinate-invariant notion on any Riemannian manifold. To move along a geodesic is, so to speak, simply to "follow your nose" while keeping your nose pointing straight forwards. (By the latter I secretly mean that your nose is being parallel transported.) Thus the geodesics on the sphere may be defined in a coordinate-independent way, and they work out to be simply the great circles. >OTOH if a constant path was a line of constant >lattitude, it becomes something totally different (probably co-ordinate >dependent, too). "Lines of latitude" are, of course, coordinate-dependent. The only line of latitude that's a geodesic is the equator. If you go on any other line of latitude, you are constantly "swerving" slightly. This is easiest to understand if you consider the line of latitude which is a circle one foot in diameter going around the north pole! So it's bad to think of lines of latitude as having any special privileged role as "constant paths". Parallel translation along lines of latitude is complicated. If you parallel transport a vector all around a line of latitude, it does not come back to where it started, unless the line of latitude happened to be the equator. (For smart alecks, I suppose I should add that the north pole and south pole could be considered degenerate "lines of latitude", and parallel transport along these... i.e., just sitting there... doesn't change a tangent vector.) >)But if you know what it means for the tangent vector to your path not to >)rotate, how come you don't know what it means for the javelin --- >)another vector --- not to rotate??? > >I can accept as a primitive operation "move in the dirction of the >tangent vector", but I have difficulty accepting "don't rotate" >as a primitive operation - it just doesn't click for me. Well, believe it or not, it's well-defined in a coordinate-independent way. It is, however, rather subtle! Recall what I said: PARALLEL TRANSLATION is an operation which, given a curve from p to q and a tangent vector v at p, spits out a tangent vector v' at q. We think of this as the result of dragging v from p to q while at each step of the way not rotating or stretching it. There's an important theorem saying that if we have a metric g, there is a unique way to do parallel translation which is: a. Linear: the output v' depends linearly on v. b. Compatible with the metric: if we parallel translate two vectors v and w from p to q, and get two vectors v' and w', then g(v',w') = g(v,w). This means that parallel translation preserves lengths and angles. This is what we mean by "no stretching". c. Torsion-free: this is a way of making precise the notion of "no rotating". The notion of "torsion-free" parallel transport is what you are worrying about. I am beginning to fear that I am going to have to give you the mathematical definition to make you happy. :-) Luckily I know a way to do this without writing down any big equations. But I won't do it just yet. You see, I had hoped that the notion of "carrying a vector along while not rotating it" would be geometrically obvious in a rough intuitive sort of way. Certainly, for example, if two guys in a column of soldiers are marching along carrying javelins (not necessarily pointing forwards), and one is not rotating it, while the other is sassily swinging it back and forth, you can tell who is following orders! You don't need to have coordinates around to see who is rotating something and who is not. You might think this is only true in flat space, but actually it's still true in curved space. That's the point of "torsion-freeness". >)I think the more primitive concept is that of parallel translation --- >)"carrying a vector along while never rotating or stretching it". The >)concept of geodesic may then be defined as a path whose tangent >)vector is parallel translated along that path. In this approach, one >)doesn't want to go back and define parallel translation using the notion >)of geodesic. However, if the notion of geodesic seems more basic to >)you, then your approach is okay. At least it suffices to define >)parallel translation along curves that are piecewise geodesic: >Hmmm - well, I saw some stuff in "Gravitation" that may give me a clue >as to your motivations the other day more or less by chance >(you probably want to avoid having to define a metric), but >it seems kind of strange and backwards to me as of yet. I defined a metric a few times by now... The METRIC g is a tensor of rank (0,2). It eats two tangent vectors v,w and spits out a number g(v,w), which we think of as the "dot product" or "inner product" of the vectors v and w. This lets us compute the length of any tangent vector, or the angle between two tangent vectors. Since we are talking about spacetime, the metric need not satisfy g(v,v) > 0 for all nonzero v. A vector v is SPACELIKE if g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0. If my pedagogy seems perverse, it's because I'm secretly trying to make clear the following facts: 1. A metric determines a unique notion of parallel transport satisfying a-c as above. 2. Not every notion of parallel transport comes from a metric in this way! It's often good to think of parallel transport as a fundamental notion. 3. Given a notion of parallel transport we may define geodesics. 4. Just given geodesics we *cannot* easily define parallel transport. Two different notions of parallel transport may happen to give the same geodesics. It's true that one may take a shortcut and define geodesics directly from the metric, by saying: A geodesic is a path that's locally a path of minimum (or maximum) length. But I wasn't wanting to take this shortcut, because the notion of parallel transport is needed for all sorts of things in GR, like defining the Riemann curvature, so we must face up to it. As you can see, there is a fairly intricate web of concepts here, and I was attempting to cut my way through it with maximum efficiency, but now you have gone and made me explain more of the mathematics. :-) Article 96867 (57 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 30 Jan 1996 11:55:21 -0800 Organization: University of California, Riverside Lines: 147 NNTP-Posting-Host: guitar.ucr.edu In article <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 itsy-bitsy step of the way the legionary carries his javelin, "not rotating or stretching it", according to a recipe that depends only on what his path is like and what space is like in the immediate vicinity. If we make the vicinity immediate enough, we can get away with pretending it's flat, for these purposes. There is, however, a certain care one must exercise in this way of thinking, which is why I haven't brought it up sooner. It has to do with whether one is keeping track of terms of order epsilon, or epsilon^2, or what. We can pretend space is flat locally to a good enough degree to define parallel transport, but it's still true that if we parallel transport a vector around a rectangle whose edges have length epsilon, it changes by an amount proportional to epsilon^2... so it's not as if space really *is* flat. >Baez intends to discard terms in epsilon^3 in the time-honoured Newtonian >way. Something like that. If this all seems confusing, remember the freshman calculus mistake: "if I change x a teeny bit epsilon, then f(x) changes by a teeny bit proportional to epsilon. But then as I take the limit epsilon -> 0 that means f doesn't change at all. So the derivative of f is zero." Or for example (d/dx) x^2 | = (d/dx) 2^2 = (d/dx) 4 = 0. x = 2 >I take it that Baez is going to show us the next steps with minimal maths ..... ? Sure, I am not going to get into these epsilons and deltas more than absolutely necessary. I'll prefer to send them to zero and say to hell with them. I have laid out a road map that'll take us to Einstein's equation, and now I will wait for people to ask me questions about it. To repeat it, leaving out the boring stuff we have already been over thoroughly (??), it goes like this: 1. A TANGENT VECTOR is.... 2. A TENSOR of "rank (0,k)" is.... A TENSOR of "rank (1,k)" is... 3. The METRIC g is... 4. PARALLEL TRANSLATION is... 5. The RIEMANN CURVATURE TENSOR is... 6. Introducing COORDINATES. Now say we choose coordinates on some patch of spacetime near the point p. Call these coordinates x^a (where a = 0,1,2,3). Then given any tangent vector v at p, we may speak of its components v^a in this basis. The inner product g(v,w) of two tangent vectors is given by g(v,w) = g_{ab} v^a v^b for some matrix of numbers g_{ab}, where as usual we sum over the repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION. Another way to think of it is that our coordinates give us a basis of tangent vectors at p, and g_{ab} is the inner product of the basis vector pointing in the x^a direction and the basis vector pointing in the x^b direction. Similarly, if R is the Riemann tensor, the vector R(u,v,w) has components R(u,v,w)^a = R^a_{bcd} u^b v^c w^d where we sum over the indices b,c,d. 7. The EINSTEIN TENSOR. The matrix g_{ab} is invertible and we write its inverse as g^{ab}. We use this to cook up some tensors starting from the Riemann curvature tensor and leading to the Einstein tensor, which appears on the left side of Einstein's marvelous equation for general relativity. We will do this using coordinates to save time... though later we should do this over again without coordinates. This part is the only profound and mysterious part, at least to me. Okay, starting from the Riemann tensor, which has components R^a_{bcd}, we now define the RICCI TENSOR to have components R_{bd} = R^c_{bcd} where as usual we sum over the repeated index c. Then we "raise an index" and define R^a_d = g^{ab} R_{bd}, and then we define the RICCI SCALAR by R = R^a_a The Riemann tensor knows everything about spacetime curvature, but these gadgets distill certain aspects of that information which turn out to be important in physics. Finally, we define the Einstein tensor by G_{ab} = R_{ab} - (1/2)R g_{ab}. You should not feel you understand why I am defining it this way!! Don't worry! That will take quite a bit longer to explain. But we are almost at Einstein's equation; all we need is 8. The STRESS-ENERGY TENSOR is... 9. EINSTEIN'S EQUATION: This is what general relativity is based on. It says that G = T or if you like coordinates and more standard units, G_{ab} = 8 pi k T_{ab} where k is Newton's gravitational constant. So it says how the flow of energy and momentum through a given point of spacetime affect the curvature of spacetime there. That's it! Article 96869 (56 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 30 Jan 1996 12:02:56 -0800 Organization: University of California, Riverside Lines: 32 NNTP-Posting-Host: guitar.ucr.edu In article <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 NNTP-Posting-Host: sagi.wistar.upenn.edu In-reply-to: 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 3-space. Perhaps the best introduction is a neat new book, John McCleary GEOMETRY FROM A DIFFERENTIABLE VIEWPOINT, Cambridge, 1994. In particular, he explains well that it shows up rather naturally via lim_{r->0} (2.pi.r - circum(radius r))/r^3: the second derivative, so to speak, of the variation from Euclidean geometry. At any point on a smooth manifold, you can pick local coordinates whose origin is the point in question, and which has zero first derivatives of each coordinate with respect to each other. In general, that is the best you can do. Curvature shows up via non-zero second derivatives. This is vague, but it's enough to motivate what happens next. Given a tangent vector at the starting point, one can interpolate a geodesic whose starting velocity is that tangent vector. Now push off like this for a plane P's worth of tangent vectors in the tangent space. Near the point in question, these describe a 2-dimensional submanifold P* of the original manifold. Let u,v be an orthonormal basis for the tangent plane P. Then Riemann(u,v,u,v)=Gaussian curvature(P*). The full tensor can be recovered from this by a simple polarization. This explains R_ijkl. One would also to understand R^i_jkl just as concretely. That's where holonomy comes in. Given P,u,v as above, one can push off on these geodesics in an epsilon rhombus as follows. First to u.eps, then u.eps+v.eps, then backwards to v.eps, then back to 0. You don't quite hit home again, but the error shrinks like eps^2, so ignore it. Now, in addition to running around, carry an arbitrary vector w with you along for the ride. Carry it in a parallel way the whole ride. When you get home, the vector, now w', could be pointing anywhere, thanks to curvature. What's very interesting is that (w'-w)/eps^2 approaches a limit as eps->0, and this limit is linear in w. So for each u,v, we have R(u,v):w|->e, a kind of second derivative that measures holonomic deviation. What's very very very interesting is that R itself is linear in u and v! And so we have R(u_j,u_k)u_l=Sum R^i_jkl.u_i, the Riemann curvature tensor. The Ricci tensor, summing over i and l, can be thought of as an "average" measurement of R(u,v)'s absolute scaling without twisting. That is, cut to just the positive orthant, and ignore any holonomic deviation in other directions by projecting back on to the original vector each time, and see how big it is. That's the part that contributes to gravity. (Which makes a bit of rough sense, even: no sidewise gravity!) (Note: signs and factors have been gleefully ignored. For one thing, there are different conventions out there.) -- -Matthew P Wiener (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 NNTP-Posting-Host: 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 widely-used definitions of geodesic *relies upon* the concept of "parallel transport without rotation". Folks define a geodesic as a path whose tangent vector is parallel translated along that very path. So if you are scared that you don't know how to carry a javelin along without rotating it, you should be scared that you don't know how to walk along in a straight line. Personally I have no trouble doing either. >This is why I tried to take refuge in the idea of geodesics, like >Bronis. They make some sense to me as a primitive notion.... One is allowed to take whatever notions one wants as primitive and then try to develop others on that basis. It's all a matter of elegance. You guys like geodesics, so you might prefer an approach based on those. But see below for a problem with trying to develop an approach like that. >...and in that >case the idea of holding the javelin at a constant angle w.r.t. the >path also makes sense (I assume I have that much right... that if we >*do* walk a geodesic, holding a constant angle w.r.t. it is the right >thing to do). Let me just point out one problem which I failed to notice before. Say you know what a geodesic is. Then in TWO dimensions you are right: when we are walking along a geodesic, you can parallel translate a vector by holding it at a constant angle from the tangent vector to the geodesic. But in three or more dimensions that's not good enough: even though the vector remained at a constant angle from the tangent vector to the geodesic, it could "swing around". It's only parallel transported when it doesn't rotate at all! So there is trouble seeing how to reconstruct the concept of geodesics from the concept of parallel transport. And indeed, now that I think about it, this is obvious; two different connections can have the same geodesics, and to pick out the "good" one, the one to use in Einstein's equation, we need to pick the one with no "torsion". >Too late. Look, say I am strolling along the surface of a sphere with my >javelin (let's keep this simple): can you give me an *operational* >definition of how to keep the javelin pointing in the same direction as I >move around? And please don't tell me "just don't rotate it". I am >rotationally challenged, and that would be discriminating against me. I >would like you to tell me how to do it in terms of surveying instruments, >or rulers and protractors, and other such crutchs for the differently >torsioned. Okay, I'll do this after I come back from class. Article 97064 (40 more) in sci.physics: From: Edward Green (SAME) Subject: Re: General relativity tutorial Date: 31 Jan 1996 07:33:10 -0500 Organization: The Pipeline Lines: 47 NNTP-Posting-Host: pipe11.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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 epsilon-tisimo bits of path to form the path of dimensions or order epsilon. On the scale of epsilon, the legionary is a microbe! I am not sure this view is totally coherent, but it has the strong enough flavor for me of something on its way from speculation into rigor that I think I am about ready to accept it and move on. I assume this insight could be made rigorous. And I am beginning to think you understood this all along... By the way, did legionaries carry javelins? That's the real CD problem here... he was pointing with his short sword. Now, all is clear :-) . -- Ed Green 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 NNTP-Posting-Host: 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 heavy-handed spiel about torsion just yet. >I think the *two* >infinitesimal distance scales are really the key to thinking about this >sensibly. > >On the one hand, we let epsilon get small enough that *something* is >effectively constant over the dimensions of our little path, that >something presumably being the "curvature" or the Riemann tensor. On the >other hand, we then let distances shrink still further, an "infinitesimal >of an infinitesimal", in considering the legionary's path, so that space >is effectively flat right in front of his face. Then presumably there is >no ambiguity, even for drunken wanderings on this scale, in what we mean >by "holding the javelin at a constant angle". We piece together all these >epsilon-tisimo bits of path to form the path of dimensions or order >epsilon. >I am not sure this view is totally coherent, but it has the strong enough >flavor for me of something on its way from speculation into rigor that I >think I am about ready to accept it and move on. I assume this insight >could be made rigorous. Yes, it can. Let me try to restate it a little more mathematically, just for fun. Let's take any old n-dimensional space with a Riemannian metric on it. Say we want to parallel transport a vector along a curve. Since this is a "local" process, it suffices to parallel transport it from point P along an itsy-bitsy stretch of the curve, of length epsilon, to the nearby point Q. So: consider a little patch of spacetime, whose length, width, etc. are all about epsilon, and which contains the curve from P to Q. It's not flat, but it's almost so. Flatten it out, i.e., take a flat metric that closely approximates our original metric in the patch. Now that we're in flat space, parallel transport is unambiguous! So do parallel transport that way. But wait --- this "flattening out" process was ill-defined: there might be different flat metrics that approximate our original metric pretty well in the patch. That's true, but if two people do this flattening out in different but still reasonable ways, it will only affect the answer by an amount of order at most espilon^2. This means that if we take a long curve of length L and chop it into L/epsilon pieces of length epsilon, and do the above trick repeatedly, the accumulated error will be at most of order L epsilon, when epsilon is sufficiently small. So chopping ever finer we can get as good an approximation as we want. [If any of you ask how the "flattening out" business works I'll challenge you to a duel with javelins at 50 paces. If your manifold is embedded in a metric-preserving way in some higher-dimensional space, this is easy: just project onto the plane tangent to the manifold.] Article 97164 (69 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 31 Jan 1996 18:35:24 -0800 Organization: University of California, Riverside Lines: 89 NNTP-Posting-Host: guitar.ucr.edu In article <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 "torsion-free" --- whereas the other two identities didn't need that. Naturally, real experts in general relativity know these identities by heart and constantly use them to pull all sorts of fiendish tricks. Don't bother remembering them, since they won't be on the final. But some of you folks might enjoy seeing how you can use them to boil the Riemann tensor down to 20 independent components when spacetime has dimension 4. Here is something cool. In 1 dimension, these identities mean that the Riemann tensor has NO nonzero components --- there ain't no intrinsic curvature in a 1-dimensional universe. In 2 dimensions, these identities mean it has ONE linearly independent component --- the Gaussian curvature of a surface. In 3 dimensions, there are 6 independent components, and in 4 dimensions, there are 20. The Einstein tensor G, which appears in Einstein's equation G = T, has exactly as many independent components as the Riemann tensor in dimensions 1, 2, and 3, but not in 4 or more! This means that in the vacuum, where G = 0, the Riemann tensor must vanish in dimensions 1, 2, and 3. So empty space is flat unless you have at least 4 dimensions. In 4 dimensions it needn't be --- which is why gravitational waves exist. Article 97321 (32 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 1 Feb 1996 13:33:21 -0800 Organization: University of California, Riverside Lines: 33 NNTP-Posting-Host: guitar.ucr.edu In article <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 1-dimensional universe. >Interesting. I see this, and yet, if we embed a 1-dimensional universe >in higher dimensional space, we *can* have some kind of curvature, which >is evidently not captured by the Riemann tensor. >Oh, maybe that is why you said "intrinisic" curvature... Yes, the curve in space only has extrinsic curvature. I mentioned this intrinsic/extrinsic curvaturre distinction before. Poetically speaking, the little 1-dimensionsal Linelanders can't do any experiments crawling back forth along their line that enables them to detect any curvature. Their universe is flat as far as they are concerned. Similarly, in GR we don't care about any sort of extrinsic curvature our universe might have if embedded in some 47-dimensional spacetime, even if the folks in 47 dimensions are snickering at us as we crawl about. Here's a good example. The metric on a flat sheet of paper is "flat" in the technical sense: parallel translation around a little parallelogram has no effect. Now take the sheet of paper and roll it into a cylinder. It has extrinsic curvature now but still no intrinsic curvature. It's metric is still the same (if you haven't stretched or ripped the paper), so it's still flat. Ponder that and ye shall become enlightened. Article 97363 (85 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 1 Feb 1996 17:35:25 -0800 Organization: University of California, Riverside Lines: 43 NNTP-Posting-Host: guitar.ucr.edu In article <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 coordinate-independent definition of torsion to make you feel happier, but it's a bit more technical. Take my word for it that t_{ab} is actually a tensor which could be defined in a coordinate-independent way.) From: john baez (SAME) Subject: Re: General relativity tutorial Date: 1 Feb 1996 17:39:29 -0800 Organization: University of California, Riverside Lines: 21 NNTP-Posting-Host: guitar.ucr.edu In article <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 high-school geometry, except that geodesics replace straight-edges. That's nice; I didn't know this construction. >[construction deleted] >This construction parallel-transports vector AB to the point C. Of >course, when I say "draw a line between points X and Y", I mean, >construct a geodesic, so this construction requires geodesics. It >also requires some notion of the distance along a geodesic (otherwise, >we couldn't take half to get a midpoint). I should add that we are supposed to think of A and B as being "infinitesimally far apart", so that AB is really a tangent vector. Article 97524 (27 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 2 Feb 1996 10:32:51 -0800 Organization: University of California, Riverside Lines: 33 NNTP-Posting-Host: guitar.ucr.edu In article <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 NNTP-Posting-Host: 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 agreed-upon scale for intensity of smell. One can present a cotangent vector by giving a function of which it is the gradient at the point in question. In coordinates, the gradient of f has coordinates (@f/@x,@f/@y,@f/@z,@f/@t). Again, we need not get stuck on any one coordinate system; if two functions have the same gradient in one coordinate system, they do in all. They are often treated in elementary math courses as being essentially the same type of thing ("vector"). Part of the reason is that if one has a *metric* (q.v.) one can identify the two. You can associate to the cotangent vector the tangent vector which suggests moving in the direction of fastest increase of the function, and whose length is the rate of increase. This only makes sense, however, because we can compare lengths in different directions. Taking one step to the south, say, increases the smell more than one step in any other direction. Without such a measure of distance as "steps", though, there's no direct comparison between the rate at which the function increases in different directions. Also, tangent and cotangent vectors transform differently when you change coordinates. I mentioned already a difference in how they thansform under a change of units. If you multiply the coordinates of all the points by 10, then the coordinates of a tangent vector also get multiplied by 10, but the coordinates of a cotangent vector are reduced by a factor of 10: the amount by which the function increases per "unit" change in a coordinate is less, not greater. If you like, a more mathematical example. Let u=x+y and v=y be new coordinates for the x-y plane. The parametrized line x=t, y=0 defines a tangent vectorat the origin; call it a. The parametrized line x=0, y=t defines a tangent vector; call it b. In the original (x,y) coordinates these are (1,0) and (0,1). In the (u,v) coordinates these lines become u=t,v=0 and u=t,v=t. Thus in the new coordinates they are (1,0) and (1,1). Now consider the cotangent vectors. The gradient of the function x is often denoted by dx. The gradient of y is dy. They are of course represented by coordinates (1,0) and (0,1). Now consider how they are represented in (u,v) coordinates. We have to take partial derivatives with respect to u and v. We carefully avoid the pitfall of supposing that taking the partial derivative with respect to v and with respect to y are the same, because v=y. In the one case, one is leaving u the same, in the other x. Since x=u-v and y=v, dx=du-dv and dy=dv. Hence in the new coordinates, dx=(1,-1) and dy=(0,1). Thus the matrix for transforming tangent vectors is (1 1) (0 1) but for cotangent vectors it is the inverse, (1 -1) (0 1). The reason for being extra careful in relativity theory is that the metric is part of what one is trying to figure out, and even with a known metric, the correspondence between tangent and cotangent vectors requires calculation as a function of the metric. They call it "raising and lowering indices", because one uses upper indices for one aspect and lower for the other. Keith Ramsay Article 97714 (29 more) in sci.physics: From: Keith Ramsay Subject: Re: General relativity tutorial Date: 3 Feb 1996 03:04:24 GMT Organization: Boston College Lines: 78 NNTP-Posting-Host: mt14.bc.edu In article <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 2-space, but not in 3-space. I think |this may be where I'm going slightly ........ wrong :-(. | |2) a notion of "moving in the direction of the tangent vector". |(There may be some problems with this notion, but if so I don't |quite get it, though some of the stuff I've read indicates there |may be some issues with this seemingly simple idea). Well, it's possible to set out in the same direction with two paths that start to diverge, and you need something besides just a manifold and a tangent direction to tell you which one is "straighter". John has avoided defining geodesics independently of parallel transport, but I think there may be some merit in stating how one can do so. Geodesics are locally extremes of length. In ordinary geometry, paths which take the shortest route between points which are close together on the path. Take a strong cord and stretch it tight.... In general relativity, it is sometimes a maximum of elapsed time. Feynman has a cute illustration in a book of his. Suppose you want to arrive back where you are now in one hour of local time, but with a maximum of time having elapsed for you. Note that going uphill takes you to a place where, informally speaking, time goes faster. But moving fast causes time to go "slower" (informally speaking). What is the tradeoff between the two which leads to an optimum of wasted time? Geodesics in space-time are the *free-fall* paths of objects. So the right thing to do is to shoot yourself out of a cannon so that in free fall, you return to the same spot on the ground. If you transport the "forward" direction on a geodesic along the geodesic, it really ought to remain the "forward" direction. If our Roman turns out to be a clutz, we make him follow a taut cord, and point his spear along it to keep it from wobbling. The more difficult problem is to deal with carrying vectors which are pointing off to one side. One does also require that parallel transport preserve the metric. So the spear shouldn't lengthen, say, as one goes. And the angles between different vectors should remain the same. As you've pointed out, on a two-dimensional manifold, this is enough. But in 3 or more, spears pointed to the side can wobble around while keeping the same angle to the taut cord. Let the Roman take not just a spear, but a set of three which are at right angles. One is pointing along the cord; the others remain at right angles to it and to each other. So far, our instructions don't prevent the spears pointed perpendicular to the path from turning around in the plane perpendicular to the path. "Torsion", as it were. Someone should correct me if I'm mistaken, but I believe that this can be corrected if we require that the tips of the spears take the shortest path compatible with our prior requirements. Spinning them around makes them take a longer path. This is less than absolutely precise, since in fact we want to consider the spears infinitesimal arrows, and for the distance to be an issue, they have to have some small finite length. But I believe that if we make them very small, in the limit the way to carry them which moves them the least will be to parallel transport them as tangent vectors. And this is enough to show how to transport any other tangent vector. Keith Ramsay Article 97745 (28 more) in sci.physics: From: Keith Ramsay Subject: Re: General relativity tutorial Date: 3 Feb 1996 02:36:08 GMT Organization: Boston College Lines: 100 NNTP-Posting-Host: mt14.bc.edu In article <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 space-time consists of the information describing lengths and times. (For those who have time for a book, Geroch's "Relativity from A to B" gives what seems a good explanation of the physical significance of the metric.) In Minkowski space, one has a metric dt^2-dx^2-dy^2-dz^2. This takes two vectors (x1,y1,z1,t1) and (x2,y2,z2,t2) and gives back the number, t1t2-x1x2-y1y2-z1z2. Here I'm setting the units so that the speed c of light is 1. I'm also adopting the "+---" convention. The "-+++" convention (change all the signs) is also used. Suppose you are in Minkowski space-time, and that your path can be described by the parametrized curve (x(s),y(s),z(s),t(s)). The elapsed time on your clock between s=0 and whenever you are at a given value of s is a function of s. Let's call it f(s). f'(s) is the rate at which time is passing per unit of s (whatever s may be). From special relativity, one knows that in this model, the rate of elapsed time for you, relative to t, is sqrt(1-v^2/c^2). Since I'm setting c=1, this is just sqrt(1-v^2). To find the rate of elapsed time w.r.t. s, multiply by dt/ds: sqrt(1-v^2)dt/ds. Then of course v^2=(dx/dt)^2+(dy/dt)^2+(dz/dt)^2, so one gets sqrt((dt/ds)^2 (1-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2)) =sqrt((dt/ds)^2 - (dx/ds)^2 - (dy/ds)^2 - (dz/ds)^2) using the chain rule from calculus. Now the square of this (dt/ds)^2 - (dx/ds)^2 - (dy/ds)^2 - (dz/ds)^2 is what you get by applying the metric, above, to the vector (dx/ds,dy/ds,dz/ds,dt/ds), the tangent vector to your parametrization of your path (world-line) through space-time. This is directly analogous to the way that one can compute the length of a parametrized curve (x(s),y(s),z(s)) in Euclidean space by integrating sqrt((dx/ds)^2+(dy/ds)^2+(dz/ds)^2) over the portion of the curve in question. The metric in space-time "measures" elapsed time, roughly the way the metric in Euclidean space "measures" distance. It also measures distances, although the interpretation seems a little less direct. If you have a tiny rigid rod moving along a path in space-time, the points in space-time occupied by its ends, which are simultaneous in the locally inertial frame for the rod, are a distance apart equal to the length of the rod. Here, you can compute distance using the metric with the sign changed: sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2-(t1-t2)^2), where (x1,y1,z1,t1) and (x2,y2,z2,t2) are the coordinates of the points of space-time in question. Now, all that was special relativity (!). In general relativity, one deals with other space-times. The metric, however, will do the same for you. Some astronaut has had an amazing career in which he's flown close to event horizons of black holes and things like that. Once he's retired and died of old age, it's up to you to write an obituary. One thing you want to work out: how old was he, after all that? Sure he was born in 2025 and died only in 2211, but what with all the time-dilation he certainly wasn't 186 years old. So draw his world-line in space-time, and parametrize it somehow. Then use the metric to determine the rate of elapsed personal time for him, per unit of parameter. (It's sometimes nice to make elapsed time itself the parameter.) Apply the metric tensor to the tangent vector v at a point and v. (It's a 2-tensor, which takes two tangent vectors and gives back a number. Put the same vector in both arguments.) Take the square root of the result. That's the rate of elapsed time per unit of parameter. Angles come out of the metric in a way similar to the way angles can be figured out from distances in ordinary geometry. The metric is directly analogous to the "dot product", v.w=|v||w|cos(angle). The metric is assumed to be one which at any one point can be transformed by a change of coordinates into a Minkowski metric. So there is always an implicit "+---" (or "-+++") "signature". This makes the geometry used in general relativity different from much of metric differential geometry, where the signature is positive. One manifestation of this feature is that in some tangent directions to space-time, the metric is 0. These are "null" directions, and massless particles (such as light, presumedly) travel along them. A common tactic is to employ some mathematical tool to convert the problem into one which involves a positive metric (using "imaginary time"). Keith Ramsay Article 97823 (12 more) in sci.physics: From: john baez Subject: Re: general relativity tutorial Date: 3 Feb 1996 13:01:39 -0800 Organization: University of California, Riverside Lines: 64 NNTP-Posting-Host: guitar.ucr.edu In article <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 coordinate-dependent statement here, or an operational one; if it's an operational one, you should make it operational by making it precise. Let's do that in this example. There are lots of different ways, but let's take a simple one. Say you have two clocks next to each other in your house, at spacetime point A. Then you move one up to a mountain for a while, and then move it back down and compare the two clocks at spacetime point B. You see the one you took up the mountain is ahead of the other. Note that since I am only comparing clocks when they are right next to each other, I don't need to worry about how the information was passed from one to the other, and whether extra time lags were introduced in the process. Note also that I don't need any coordinate system! I just need two clocks. In terms of the mathematics of GR, what's going on? Well, we have two paths from spacetime point A to spacetime point B, and the "length" (or more precisely, "proper time") along one path is more than the other. That's all. The gravitational potential can be a useful tool in *certain* problems, but I prefer this other way of thinking of it, since it tells you something relevant to quite general GR problems. Also, it fits with the idea that the *metric on spacetime* is what really matters: that's what you'd use to compute the proper time along a curve. Also, it seems pretty simple and unmysterious. Article 97828 (9 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 3 Feb 1996 13:27:40 -0800 Organization: University of California, Riverside Lines: 43 NNTP-Posting-Host: guitar.ucr.edu In article <1996Feb3.001629.17341@schbbs.mot.com> bhv@areaplg2.corp.mot.com (Bron is Vidugiris) writes: >Ouch! Sorry, I was being a bit sharp-tongued there. I gotta mellow out a bit here... >It's fine to prove that the stress energy tensor is a contraction >of the Rienmann tensor, but this seems incomplete unless >either it's possible to recover the full Rienmann tensor from the >contraction, *or* if it's possible to calculate what we really want to >know given only the contraction and not the full tensor. It's definitely NOT true that you can recover the Riemann tensor from the stress-energy tensor in 4 dimensions. You can do it dimensions 1, 2, or 3. This is why there are gravitational waves in 4 dimensions but not in lower dimensions. I sorta said this a while back. To expand: In 4 dimensions, the Riemann tensor R^a_{bcd} has 20 independent components. The Ricci tensor R_{ab} --- or for that matter the Einstein tensor G_{ab} --- has 10. So if you know the stress-energy T_{ab}, Einstein's equation G_{ab} = T_{ab} gives you the Einstein tensor. But you can't get the Riemann tensor back... even using whatever sneaky tricks you might attempt to dream up. Indeed, there are lots of different solutions of Einstein's equation with T_{ab} = 0 --- so-called "vacuum solutions", since there's no energy or momentum anywhere. There are lots of such solution with R^a_{bcd} nonzero, in addition to the obvious one with R^a_{bcd} = 0, namely flat Minkowski spacetime. One of them is the Schwarzschild solution. I.e., in the solar system, spacetime is curved even in the vacuum, due to the gravity of the sun. Another sort is the gravitational wave solution, in which spacetime has ripples of curvature moving through the vacuum. So the business of actually solving Einstein's equation proceeds a bit differently than you might have suspected. Article 97841 (7 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 3 Feb 1996 14:56:44 -0800 Organization: University of California, Riverside Lines: 258 NNTP-Posting-Host: guitar.ucr.edu Okay, on today's episode of GENERAL RELATIVITY TUTORIAL we are going to step outside of the studio and see how a typical fan out there is doing at learning general relativity: In article <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 co-ordinates were defined as 'a' where 'a'=0,1,2,3) Oh, I see! Sorry. We feel free to use, not just a, but all letters, particularly those near the beginning of the alphabet, to stand for any of the numbers 0,1,2,3. So when I write g_{ab}, I am quickly referring to g_{00} g_{01} g_{02} g_{03} g_{10} g_{11} g_{12} g_{13} g_{20} g_{21} g_{22} g_{23} g_{30} g_{31} g_{32} g_{33} You can see why I prefer to say g_{ab}. Indeed, if we were in 26 dimensions, the way some nuts believe, the indices a,b,c, etc. would go from 0 to 25. It's much better just to say g_{ab}. >>for some matrix of numbers g_{ab}, where as usual we >'as usual we'? What a compliment. Unfortunately ...... >>sum over the >>repeated indices a,b, following the EINSTEIN SUMMATION CONVENTION. Well, you'll soon get used to it. The way you tell the difference between a physicist and a mathematician these days, by the way, is that physicists instinctively know to sum over repeated indices. So I was just flaunting the fact that, while a mathematician, I can play the physicist. You can too. >>6. The EINSTEIN TENSOR. The matrix g_{ab} is invertible > >Another g(,). Presumably another tensor of rank (0,2) It's not another one, it's the same bloody one every time. The metric. >>7. The STRESS-ENERGY TENSOR. The stress-energy is what appears on the >>right side of Einstein's equation. It is a tensor of rank (0,2), and it >>defined as follows: given any two tangent vectors u and v at a point p, >>the number T(u,v) says how much momentum-in-the-u-direction is flowing >>through the point p in the v direction. > >At last, something that even *sounds* like a physical >system. Hmmm. Momentum, time as a dimension. OhMyGod shades >of VV, yuk! What does momentum in the time dimension mean? >Oh, do a vaguely remember Baez calling it energy. No I am >probably confused. This needs sorting out. Heeeeeellllpppp! Oh-oh, our ratings our dropping!! What's the problem, dear viewer? Energy is momentum in the time direction. Consider a rock "just sitting there". It has lots of velocity in the time direction --- one second per second --- hence lots of momentum in the time direction. That's what we call it's "rest energy". You know, E = mc^2 and all that stuff they taught you in grade school.... >>The top row of this 4x4 matrix, keeps track of the density of energy --- >>that's T_{00} --- and the density of momentum in the x,y, and z >>directions --- those are T_{01}, T_{02}, and T_{03} respectively. This >>should make sense if you remember that "density" is the same as "flow in >>the time direction" and "energy" is the same as "momentum in the time >>direction". The other components of the stress-energy tensor keep track >>of the flow of energy and momentum in various spatial directions. >Oh. Is it just a generalised expression of the conservation >of momentum in spacetime? We haven't said anything about *conservation* of energy-momentum here. But perhaps you are peeking at the next chapter, where I explain that the law D^a T_{ab} = 0 expresses local conservation of energy and momentum, and that it's an automatic consequence of Einstein's equation T_{ab} = G_{ab}. >Hmmm, quite a philosophical point >worthy of some deep consideration. I guess you have all >considered this, I wonder what sort of conclusions/observations you came to? Well, it all boils down to this: local energy-momentum conservation is really just an expression of the fact that the laws of physics are the same in any coordinate system. Cool, huh? It's a special case of "Noether's theorem" relating symmetries and conservation laws. But really, this is quite a digression. >>8. EINSTEIN'S EQUATION: This is what general relativity is based on. >>It says that >> >> G = T >> >>or if you like coordinates and more standard units, >> >> G_{ab} = 8 pi k T_{ab} >> >>where k is Newton's gravitational constant. So it says how the flow of >>energy and momentum through a given point of spacetime affect the >>curvature of spacetime there. >[Cryptic allusions deleted.] >So, this says [...] that the flow of >energy and momentum through a given point of spacetime >affect the curvature of spacetime there. Yes. Precisely. >(That can't be wrong). What's more, it's precisely right! >Hang on a tick. Does this refer to one tiny point in >spacetime? In fact, each and every tiny little point! >Surely not, what if that point is in a vacuum?. Then T_{ab} = 0, of course, 'cause there's no energy and momentum there. What's the big deal???? It just means G_{ab} = 0. >BANG! THUD! Hmm, well, our viewer has died. Or possibly just collapsed in a faint? Perhaps he needs a little review course in index juggling. It seems that he started losing it right around when we formed the Einstein tensor by contracting some indices on the Riemann curvature tensor. Or maybe a more geometric explanation of the Ricci and Einstein tensors would help. We'll be back next week! Article 97842 (6 more) in sci.physics: From: Marco de Innocentis Subject: Re: General relativity tutorial Date: 3 Feb 1996 10:53:20 GMT Organization: The Mathematical Institute, Oxford Lines: 61 NNTP-Posting-Host: yukon.maths.ox.ac.uk Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-Mailer: Mozilla 1.1N (X11; I; SunOS 4.1.3_U1 sun4c) X-URL: news:4etlcj$l4v@guitar.ucr.edu baez@guitar.ucr.edu (john baez) wrote: >In article <4ent4a$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 space-times and see which kinds of tidal effects cannot happen in 2,1 - they are due to Weyl only. For example when you have a 'beam' of null geodesics with circular cross section which gets deformed into an ellipse - this is due to Weyl, because it can't happen in a 2D space. >>I don't know. But going back to my crystal physics book to check on the >>stiffness tensor, I see that symmetry reduces its 81 components to no more >>than 36 independent entries. So I wonder how much information we do >>discard here in contracting the Riemann tensor? > >I don't know if you saw it yet, but in one of my posts I noted that the >Riemann tensor has 20 independent entries (in 4d spacetime), while the >Ricci satisfies > >R_{ab} = R_{ba}, > >so it has 10. So we lose 10 juicy facts about the curvature of spacetime >when we contract the Riemann to get the Ricci. Article 97856 (5 more) in sci.physics: From: Emory F. Bunn Subject: Re: General relativity tutorial Followup-To: sci.physics Date: 3 Feb 1996 23:16:22 GMT Organization: Physics Department, U.C. Berkeley Lines: 39 Distribution: world NNTP-Posting-Host: physics12.berkeley.edu In article <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+1-dimensional spacetime? If so, I'm enormously surprised. Doesn't the Einstein equation say that an empty patch of 2+1-dimensional spacetime is completely flat? Maybe you mean that you can have tidal forces in a region of 2+1-dimensional spacetime that's filled with matter. In that case, I'm still not sure what you're getting at. Can you give a precise definition of a "tidal force"? It seems like it needn't mean anything more than "geodesic deviation", which pretty much just means "curvature." So then you'd be saying that wherever there's matter, there's curvature. That's true, but I'm missing the part where insight into the nature of the Ricci and Weyl tensors takes place. >My understanding (after having discussed this with >my supervisor briefly a few weeks ago) is that Weyl is only associated with >some tidal forces, while Ricci is associated both with compression and >(other) tidal forces. I think it would help me a lot if you could explain precisely why compression doesn't count as a tidal force. >To see which kinds of tidal forces are due to Weyl >and which ones to Ricci, you have to compare 2,1 and 3,1 space-times and >see which kinds of tidal effects cannot happen in 2,1 - they are due to >Weyl only. For example when you have a 'beam' of null geodesics with circular >cross section which gets deformed into an ellipse - this is due to Weyl, >because it can't happen in a 2D space. This is definitely true, and it's why gravitational radiation doesn't exist in fewer than 3+1 dimensions. Circular beams get deformed into ellipses when a gravitational wave passes by. -Ted Article 97865 (4 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 3 Feb 1996 16:40:27 -0800 Organization: University of California, Riverside Lines: 26 NNTP-Posting-Host: guitar.ucr.edu In article Ramsay-MT@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 NNTP-Posting-Host: 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 vice-versa (unfortunately?). |Well, I mean in the sense that you seem to imply that there is only one |parallelly transported path (no direction changes) from a to b, but |there may be more than one geodesic from a to b. No, it's not that. Parallel tranport is the general process of what happens when you carry a vector around on some path, without "turning" it as you go. Any sort of path, carrying any vector you feel like starting out with. A geodesic is a particular kind of path-- the kind with "no direction changes". The two ideas are related: You can tell the yellow brick road is a geodesic by carrying a vector pointing "forward" along it, and seeing that as you ease it down the road, "parallelly", it keeps pointing along the yellow brick road. It is true, by the way, that there is sometimes more than one geodesic between a and b. For instance, returning to our surface-of-the-Earth example, if the Earth were simply a sphere, then antipodal points would be the ones which are joined by more than one geodesic. Any geodesic from the north pole goes to the south pole. If two distinct points on a sphere are not antipodal, then there is just one "great circle" (i.e., geodesic) route between them. |Hmmmm, yes, I could go |with that even in curved 3D space. However I am probably wrong (and |getting used to it). Good. :-) |2) Baez: |>two different connections can have |>the same geodesics, and to pick out the "good" one, the one to use in |>Einstein's equation, we need to pick the one with no "torsion". | |Uh-oh. Can we forget about this and use parallelly transported, or do we |have to figure out the concept 'geodesic' in all it's ramifications? You can forget about the concept of "connection" and just think about "parallel transport" which is one specific kind of connection. If you are happy with the idea of parallel transport, then you need not think about geodesics. (Geodesics are nice to know, though; they tell you what space-time paths correspond to the motions of objects in free-fall.) | Lots of new words like 'torsion', It's purely to distinguish parallel transport (which you want to think about) from certain other sorts of things one could do with vectors as you carry them around (a.k.a. other connections). In our highschool marching band, on parades we would have people twirling mock "rifles". They would be held perpendicular to the way they were facing, but twirled around. This is *not* what we have in mind for parallel transport. Such a way of carrying a vector about involves torsion ("twisting"), even though they may be keeping the angle between their rifle and the path they are taking constant. In the plane, there isn't this problem, because there is no way to turn a vector in place while keeping it at a fixed angle to another one. |'right one to use in Einstein's equation'. I hope none of these words is actually new to you. :-) |Nobody's even mentioned (I think: he says quickly) where |geodesics even come into Einstein's equation yet. I expect it's |'obvious'. Ah, well. Geodesics don't. It's the Ricci tensor R_ij, which can be computed from the curvature tensor, which can be worked out by parallel transport of vectors, which appears in Einstein's equation. Look at the "course outline" again. Keith Ramsay Article 97992 (44 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 4 Feb 1996 20:17:03 -0800 Organization: University of California, Riverside Lines: 174 NNTP-Posting-Host: guitar.ucr.edu In article <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 Followup-To: sci.physics Date: 4 Feb 1996 20:09:54 GMT Organization: Physics Department, U.C. Berkeley Lines: 56 Distribution: world NNTP-Posting-Host: physics12.berkeley.edu In article <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 stress-energy tensor T) isn't enough to tell you all about the curvature. In both cases, you can supplement the source information with some extra initial conditions to get a unique solution. (For example, in electromagnetism you can specify that no electromagnetic waves are zooming in from infinity. That's enough to give you a unique solution to the fields given the sources. For general relativity, you can perform similar feats, although it's technically trickier.) Anyway, I hope that makes it a little bit clearer why people say that the Weyl part of the curvature has to do with gravitational radiation: the Weyl tensor carries information about the kind of curvature that's independent of the source distribution, sort of like electromagnetic waves are fields that propagate independently of whatever sources are around. >3) Misreading what messrs Ted & Marco discuss, suggests that at least >some of the Weyl stuff may describe non-physical thingies (2+1D >spacetime is completely flat so the Weyl bit describing how it curves >would be redundant perhaps). But exactly what the Weyl tensor *does* >represent is not trivial. Let me correct one thing: *empty* 2+1-dimensional spacetime is flat. There certainly are curved 2+1-dimensional spacetimes, but they all have matter in them. Specifically, in 2+1 dimensions, the Einstein equation implies that the curvature is zero wherever T is zero. (That's because in 2+1 dimensions the Ricci and Riemann tensors contain the same amount of information.) -Ted Article 98021 (38 more) in sci.physics: From: Oz Subject: Re: General relativity tutorial Date: Sun, 4 Feb 1996 09:37:51 +0000 Organization: Oz Lines: 56 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 Ramsay-MT@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 free-fall geodesic is (locally: obviously not flatily, so what does it mean - parallelly transportedly?) the longest path in spacetime. At last, something concrete to hang on to. Yup, I follow that. One also wonders what a thingy that wiggled so much that it got dilated to zero might look like, hmmm one is tempted to say 'light' as it does a LOT of wiggling. Presumably this thingy would cover more time in space than it does in time and so would be described by analogy as 'spacelike'. Hmmm. I have a bad feeling about this. >In space, a geodesic is the path a string takes when you pull it taut. >It's spacelike, of course, and it's locally the *shortest* curve from >one point to another. OK, have I got it right. In space (3D) a taut string is the shortest path in space. OK, not unexpected perhaps. In spacetime though, one might dare to wonder because the string will be curved and the path would depend on how fast you went down the string. Baez has thrashed it in to me to treat these sorts of things with great caution. >In both cases there is an "extremal principle" involved. Particles in >free fall travel the path that takes the most proper time because they are >trying to minimize the "action". A stretched string follows the >shortest path because it's trying to minimize the "energy". Hang on. Free-fall expressed above in 4D, taut string in 3D. I follow the idea of course, but we keep being told to stay in 4D. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 98014 (37 more) in sci.physics: From: Matthew P Wiener Subject: Re: Riemann tensor (was: Re: General relativity tutorial) Date: 4 Feb 1996 14:24:49 GMT Organization: The Wistar Institute of Anatomy and Biology Lines: 47 NNTP-Posting-Host: sagi.wistar.upenn.edu In-reply-to: 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 right--ask me if interested and >>I will give you a reference. To give an elementary example, if all you >>know about an inner product is the set of values, then as >> = +2c+c^2, you can solve for (and >>prove the Cauchy-Schwartz inequality to boot). Something similar, but >>algebraically more messy and less ASCIIble, applies here. >Notice that this only works because we know = , 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 NNTP-Posting-Host: 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 NNTP-Posting-Host: guitar.ucr.edu In article Ramsay-MT@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. Free-fall expressed above in 4D, taut string in 3D. I follow >|the idea of course, but we keep being told to stay in 4D. Come come, unless you are already perfectly comfortable with 4d spacetime geometry, the ANALOGY of 3d space geometry is very handy. They are not the same thing; merely nice mathematical analogs. The serious physics happens in 4d spacetime, but there is a lot of nice fun toying around to be had in 3d space. Article 98105 (33 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 5 Feb 1996 12:10:15 -0800 Organization: University of California, Riverside Lines: 72 NNTP-Posting-Host: guitar.ucr.edu A couple of corrections: In article <4f40bv$ngv@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes: >Rather than trying to calculate out how this works in an example, I want >to use these ideas to explain the Ricci tensor. >So, suppose an astronaut taking a space walk accidentally spills a can >of ground coffee. [etc.] > | | > v^ ^v' > | w | > P->-----Q > | | > | | > | | [etc.] >This may seem like a lot of work to say that two coffee grounds are >moving in the same direction at the same speed, but when spacetime is >curved we gotta be very careful. Note that everything I've done is >based on parallel translation! (I defined geodesics using parallel >translation.) Actually, I also used the metric in a direct way at one place: when I said that w was orthogonal to v. But this is not really all that big a deal, since in GR we get parallel translation from the metric in the first place. >A little round ball of coffee >grounds in free fall through outer space! As time passes this ball will >change shape and size depending on how the paths of the coffee gournds >are deflected by the spacetime curvature. Since everything in the >universe is linear to first order, we can imagine shrinking or expanding, >and also getting deformed to an ellipsoid. There is a lot of >information about spacetime curvature encoded in the rate at which this >ball changes shape and size. But let's only keep track of the rate of >change of its volume! This rate is basically the Ricci tensor. >More precisely, the rate of change of volume of this little ball is >equal to >r(v,v) Sorry, the instantaneous rate of change of volume is its original volume *TIMES* r(v,v)! Article 98174 (22 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 5 Feb 1996 17:45:09 -0800 Organization: University of California, Riverside Lines: 52 NNTP-Posting-Host: guitar.ucr.edu In article <4f33qi$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 Lines: 101 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 stress-energy tensor T) isn't enough to tell you all >about the curvature. In both cases, you can supplement the source >information with some extra initial conditions to get a unique >solution. (For example, in electromagnetism you can specify that no >electromagnetic waves are zooming in from infinity. That's enough to >give you a unique solution to the fields given the sources. For >general relativity, you can perform similar feats, although it's >technically trickier.) > Well, since everybody knows I don't know anything, and this bit definitely, I feel I am allowed to speculate pending authoritative pronouncements. If the Ricci bit allos us to calculate the curvature effect of everything we *do* know about (or care to consider), then the Weyl bit must (sort of by definition) be the curvature of everything else. This makes some sort of sense really, when you^H^H^H I think about it. So, as you indicated before, Weyl might be the boundary condition bit. I am not quite sure if this is exactly the best way to think of it since one would tend to imagine (in 4D) that it would leave curvature within spacetime if you sort of removed the Ricci bit. Anyway it's what's left. End of mindless speculation. More mindless speculation (a little later): I have just read Baez's explanation of the Ricci tensor. Fancy using coffee grounds, it should be tealeaves. Tealeaves can travel in the time direction as everyone in Britain knows. That's why you can read the future in tea leaves. Anyway the Ricci tensor seems to be something that operates in a small volume. One might even suspect another infinitesimal one. In fact infinitesimal enough that curvature due to external thingies can be neglected. On top of that it could be infinitesimal enough that it contains, or one can neglect, anisotropy within the volume. In other words the curvature due to a speck of momentum flow within the volume. Inevitably, then, we must lose curvature due to external thingies; indeed we want to. Right ball park? Not even wrong? This mindless stuff is sure to bring on a Baez Bashing: Teacher: (Striding down the classroom, his face a contorted red mask, lightning bolts coming from his eyes) "You infinitesimal moron. Haven't you understood a word I said". (Picks Oz up by the hair until the poor little mite's feet are three feet off the ground. His eyes are an 'O' of terror.) "The dummest clut in the class, and he tries guessing what I meant. I am the teacher here, you learn what I say, when I say it, and don't start thinking about it. Moronic twit". (With a casual flick of his hand he flings Oz across the room by the hair. Oz strikes the wall near the ceiling and he slides down the wall to a dejected heap on the floor. Stars flick in and out of existance round his head.) Teacher: Now, Ed, perhaps you can explain to the class about the Ricci tensor, since you have been doing so brilliantly of late. Ed: (Ed rises confidently. He gives a scornful look to the bedraggled Oz crawling painfully along the floor and back to his seat). "Now the Ricci tensor is a simple construct ........" says Ed. [Apologies to Ed, at least I think so]. :-) ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 98219 (44 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 5 Feb 1996 20:32:54 -0800 Organization: University of California, Riverside Lines: 114 NNTP-Posting-Host: guitar.ucr.edu Let me just comment on a few of many things. I deliberately avoid commenting on those remarks that give me the urge to hurl a thunderbolt, because it's getting late and I will be able to hurl bigger, badder thunderbolts tomorrow when I am all nice and rested. So: In article <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 stress-energy tensor is divergence-free. Using some lingo I haven't explained yet, one writes this as follows: D^a T_{ab} = 0. So if we are going to write down Einstein's equation G_{ab} = T_{ab} we had damn well better have D^a G_{ab} = 0! Now there is a wonderful identity, the Bianchi identity, which says that D^a G_{ab} = 0, no matter what the metric is! It's not true that D^a R_{ab} = 0. So we really do much better to use G_{ab}, since then: IN GENERAL RELATIVITY, LOCAL CONSERVATION OF ENERGY AND MOMENTUM IS AN AUTOMATIC CONSEQUENCE OF TAUTOLOGOUS FACTS ABOUT SPACETIME CURVATURE!!!!!! By "tautologous" I mean, "relying on mathematical identities which hold no matter what the metric is". This is a truly wondrous thing. I leave you to ponder it, and to remember what is the analogous wonderful thing about Maxwell's equations. >1) It is not really the Ricci tensor that has the dilational >interpretation, but the Einstein tensor itself. The second term is a >"correction" occasioned by the use of a local metric that in general >distorts the underlying *physical* properties of space time {I sense a >thunderbolt brewing...} ZAP! Whoops, and here I said I wasn't going to hurl any thunderbolts. I must have forgotten to disable my automatic bullshit detection system. >We notice that this term is the only one involving our metric, whose >physical significance we have been rather quiet about so far... No, the metric is what underlies everything: we used the metric to get parallel transport!!!! Then we used that to get the Riemann tensor, etc.. Everything on the left hand side of Einstein's equation is built out of the metric. We could write it out all explicitly in terms of g_{ab} if we wanted. (Take my word for it: we don't want to.) >2) It really *physically* modifies the Ricci tensor, in the sense that we >are discarding yet more information. In this connection, note what >happens in > > R^a_d = g^{ab} R_{bd}, > > R = R^a_a > > G_{ab} = R_{ab} - (1/2)R g_{ab}. We are not discarding information in passing from Ricci tensor to Einstein tensor: we are rearranging it. We can go backwards. Let's do some index gymnastics.... those who aren't in shape, please go to the back of the room and watch. Okay, class. Raise an index! G^a_b = R^a_b - (1/2)R g^a_b Contract! G^a_a = R^a_a - (1/2)R g^a_a Note that R^a_a = R and that g^a_a = 4 in 4 dimensions! G^a_a = R - 2R = -R Substitute back in the definition of G_{ab}! G_{ab} = R_{ab} + (1/2) G^c_c g_{ab} (Whispers from the back of the room: "How'd the G^a_a turn into G^c_c?" "It's a dummy index, you dummy!") Solve for the Ricci tensor! R_{ab} = G_{ab} - (1/2) G^c_c g_{ab} So you see that the formula for Einstein in terms of Ricci tensor is just like the formula for Ricci in terms of Einstein... at least in dimension 4. Article 98394 (43 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 6 Feb 1996 13:57:41 -0800 Organization: University of California, Riverside Lines: 91 NNTP-Posting-Host: guitar.ucr.edu In article 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 volume-preserving way. Alternatively, the Ricci tensor is the part of the curvature that's determined by the presence of energy and momentum flowing through that very point. The Weyl tensor is the part that's not. So in the vacuum, there's no Ricci, just Weyl. If you've ever thought about "tidal forces", this should make sense. A bunch of coffee grounds floating through the vacuum will experience "tidal forces", meaning that an initially round ball will get stretched in some directions and squashed in others. (Or sheared, or rotated. I need to ponder more deeply the relation here between stretch/squashings, shears, and rotations.) >So, as you indicated before, Weyl might be the boundary condition >bit. I am not quite sure if this is exactly the best way to think >of it since one would tend to imagine (in 4D) that it would leave >curvature within spacetime if you sort of removed the Ricci bit. Yes, you can have the universe be a complete vacuum, but still have gravitational waves rippling through it. These waves have no Ricci curvature, only Weyl. They should be determined by the "boundary conditions at infinity". (The math of boundary conditions for nonlinear equations like Einstein's equations is more tricky than that for linear equations like Maxwell's. Still, folks know a lot about it.) >Anyway it's what's left. Yes. >Anyway the Ricci tensor seems to be something that operates in a >small volume. One might even suspect another infinitesimal one. Yes, all my "small", "itsy-bitsy", and "tiny" things are really meant to be infinitesimal. You know, take that epsilon and send it scurrying on home to zero. >In >fact infinitesimal enough that curvature due to external thingies >can be neglected. On top of that it could be infinitesimal enough >that it contains, or one can neglect, anisotropy within the >volume. In other words the curvature due to a speck of momentum >flow within the volume. Inevitably, then, we must lose curvature >due to external thingies; indeed we want to. Hmm, well, the Ricci curvature is a tensor which takes a value at each point of spacetime, and its value at any point is determined solely by the flow of energy and momentum through that very point. That's what Einstein's equation says. So if I understand you correctly, it sounds right. >Right ball park? Not even wrong? Definitely right ballpark. >This mindless stuff is sure to bring on a Baez Bashing: > >Teacher: >(Striding down the classroom, his face a contorted red mask, >lightning bolts coming from his eyes) >"You infinitesimal moron. Haven't you understood a word I said". > >(Picks Oz up by the hair until the poor little mite's feet are >three feet off the ground. His eyes are an 'O' of terror.) > >"The dummest clut in the class, and he tries guessing what I >meant. I am the teacher here, you learn what I say, when I say it, >and don't start thinking about it. Moronic twit". > >(With a casual flick of his hand he fings Oz across the room by >the hair. Oz strikes the wall near the ceiling and he slides down >the wall to a dejected heap on the floor. Stars flick in and out >of existance round his head.) What I really like about teaching is the affection and respect my students feel for me. Article 98395 (42 more) in sci.physics: From: john baez (SAME) Subject: Re: Ricci Tensor (was: Re: General relativity tutorial) Date: 6 Feb 1996 13:59:52 -0800 Organization: University of California, Riverside Lines: 27 NNTP-Posting-Host: guitar.ucr.edu In article <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 multi-linear machine. You input vectors, or even other tensors, in one side, turn the crank, and out pops a new vector, tensor, or even a scalar on the other side. If your underlying coordinate basis is NOT orthornormal, then the concepts of contravariant and covariant come into play. If your underlying vectorspace IS orthonormal, then the concepts of covariant and contravariant coordinates are not relevant. The components of a tensor are usually specified with respect to the dual basis of the vectors it operates on, unless the metric tensor is included as part of the equation. A tensor can have both contravariant and covariant components. The position of the indices (vertically) denote whether a particular set of components is contravariant or covariant. A defining characteristic of a tensor is the way its components transform with a change of basis. In general, when transforming to a new coordinate system (where T is the transformation matrix), one must multiply by T once for each covariant index, and by (T)^-1 once for each contravariant index. The difference between a rank 2 tensor and a matrix is the difference between a formula and an OBJECT! A tensor is a description of a real physical phenomena, object, etc. such as the strain in a rigid solid. Such objects exist independently of any coordinate system. For an excellent set of examples of real world tensors in the physical sciences (restricted to an orthonormal basis), check out The Feynman Lectures on Physics, vol II, chapter 31. The whole chapter is virtually a list of various tensors, including the polarization tensor of a crystal, the conductivity tensor of a crystal, the Moment of Inertia tensor of an asymmetric solid, the stress tensor of a solid, the strain tensor of a solid, as well as others. -Bruce 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 NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4f644q$o4a@guitar.ucr.edu>, john baez writes >In article Oz@upthorpe.demon.co.uk >writes: 1) In modern notation vectors are/tend-to-be 'column' vectors. OK, that's just fine. Er, one little teeny thought. Matrix operations don't commute. So a 'column' vector transforms differently to a 'row' vector. eg [a1,...,a3][b1]=/=[b1][a1,..,a3], one scalar, other [3x3] [c1] [c1] [d1] [d1] Reading between the lines in your notation vectors: column, covectors: row?? Of course it would help if I had the faintest idea what a covector was. If as above, however, I can see a difference. >Well, let's see. You are using a 4x4 matrix to turn row vectors into row >vectors. Mind if I use one to turn column vectors to column vectors? No, go right ahead. >So let me say: > >[b1,b2,b3,b4] [a1] = [b.a] >[c1.......c4] [a2] [c.a] >[d1.......d4] [a3] [d.a] >[e1.......e4] [a4] [e.a] > >Then our matrix is serving as a machine that turns tangent vectors to >tangent vectors in a linear way, so it's a (1,1) tensor. > >Now, you protest, but doesn't it ALSO serve to turn cotangent vectors >into cotangent vectors? (I.e., row vectors into row vectors, like you >had done.) The answer is yes. (1,1) tensors are multipurpose gadgets. A very simple and obvious question. Do we have to be very caerful WHAT our tensor is used for. Ie just row>row not col>col, or does the *same* tensor do 'double duty' in some way. I am confident this is a misinterpretation, but something along the lines of: Tensor T transforms a vector V to say a rotated vector V', and the *same* tensor T transforms (if I knew what a covector *was* it would help) a covector C (lets say it's some angle to some other vector) to a covector C' which is the angle V' makes with the 'other' vector. If you have understood what I am trying to say, you are a genius. >Now we are getting into some of the subtleties I had wanted to avoid. Hey, I don't want to be able to manipulate these things. Just get an idea of tha salient points, and some idea of the mechanism. Otherwise I can't properly follow what's going on. >Oh well, education occurs even when one least wants it. If I didn't want it, I would have given up weeks ago. Have you any idea how many concepts I have tried to take in over the last few weeks? Basically all the information I have is all here in the thread. No books (ahem) no barroom chats, no exercises, no supervision. This doesn't make it easier, but harder. IMHO you are all doing a surprisingly effective job. I am surprised it's possible to do it at all. >>What would a tensor of rank (0,2) be? > >>Well it turns TWO vectors into a scalar. Oh, dear. >>Ah, well Baez drew g(u,v) out. It was a square 4x4 matrix as above. > >There's something like that which works, but I don't want to explain it >in PreCambrian. How about Jurassic? Say I have a tangent vector v and >another one w. Then a (0,2) tensor like the metric g can be thought of >as a matrix of numbers g_{ab}, and the way we get the scalar g(v,w) is >by doing this: > >g(v,w) = g_{ab} v^a v^b > >Here I am summing over the indices a and b. This is the problem. I have done too much (mis)reading between the lines in the past (rubs ear). There are a whole lot of ways of defining 'summing over the indices a and b'. YOU mean in a particular way. *I* don't know which way you mean. Since we are getting close to the notation you are using in GR, it would help if I knew, I think. >Actually I am afraid I am opening an enormous can of worms here, by >trying to translate PreCambrian into more recent dialects. PreCambrian >is so vague that there are limitless possibilities for argument. Well, I am perfectly happy to be brought up to pre-modern. I am (would prefer) to use your notation, if only I knew what it was! > 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 NNTP-Posting-Host: sagi.wistar.upenn.edu In-reply-to: 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. Off-hand, I have not seen it spelled out anywhere else as this, but I assume it was once standard knowledge. One form of it is the MTW "bongs of a bell" description of "covectors" aka 1-forms. A vector is just an arrow pointing somewhere. A covector (in R^3) is a dissection of R^3, consisting of parallel planes, such that they are labelled "linearly", with 0 for the plane passing through the origin, one of the other planes labelled 1, the one twice as far labelled 2, etc. These may be described with 3 coordinates (a,b,c) by letting ax+by+cz=1 be the plane labelled 1. (0,0,0) is the plane at infinity, so to speak. So covectors form a 3-dimensional space. There is a natural duality between covectors and vectors, namely < (a,b,c) | (x,y,z) > = ax+by+cz. In terms of the starting R^3 dissection, this just identifies the label of the plane that vector (x,y,z) reaches. The unit covectors are commonly denoted dx,dy,dz. Although you were taught dx is "differential x", here its "dual x". But the notation was chosen this way deliberately. 1-forms Adx+Bdy+Cdz are naturally understood as covectors, not vectors. For example, given a scalar function f on R^3 (ie, f(x,y,z)=some number), we can take its gradient. You probably learned grad f = (@f/@x).i + ... What piffle. "Really" grad f is df = (@f/@x).dx + .... It forms a covector field, not a vector field. So what does this really look like? It's ultimately just a contour map of f, custom linearized at each point. This is easier to imagine in 2D. Take a contour map of a scalar function "height", assumed differentiable. Pick a point. Kind of curvy, right? Blow up the neighborhood 1000 times, notice how the nearby contours are a lot straighter. So just _redraw_ them as straight, extend to a covector worth of lines dissecting R^2, and scale the line labelled 1 down by 1000. This is the dual version of (F(x+.001)-F(x))/.001 we all know and love. That this is so direct, the moral equivalent of drawing tangent lines, is all the proof anyone really needs that gradients are covectors. By pushing the MTW pictures to the limit, it is actually possible to work out a pretty good visualization of k-forms. For 2-forms, one has two sets of parallel dissections, together forming pipes. For 3-forms, one has three sets of them dissection. In n-dimensions, one works with hyperplanar dissections instead of planar dissections. The differential that maps k-forms to k+1-forms becomes obvious: just more contouring. Given a k-form field, U(p) at each point p, we want to compare the k hyperplanar dissections of U(p) with U(p+h) for small h's. Just draw a new hyperplane perpendicular to the direction of change, and scale by 1/h. As h->0, this converges to a k+1-form. -- Matthew P Wiener (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 NNTP-Posting-Host: 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 old-fashioned linear algebra books. You can think of a row vector as a guy that's dying to eat column vectors and spit out numbers: [a b][c] = ac + bd [d] but if you change your point of view you can think of it the other way around: the column vector eats the row vector. >A very simple and obvious question. Do we have to be very caerful WHAT >our tensor is used for. Ie just row>row not col>col, or does the *same* >tensor do 'double duty' in some way. There are two levels to this question, because you are reading an old book that talks about "row vectors" and "column vectors", and I am struggling to translate it into the modern lingo of "vectors" and "covectors", but the translation is a tricky business, in the way that translating foreign languages always is. In fact, I urge you to not pay much attention to that book, and just let me explain everything. I will wind up doing LESS work that way. Anyway, I can confidently say: yes, a (1,1) tensor does double duty: by definition it serves to turn a vector into a vector, but if we are clever we can use it to go turn a covector into a covector. On the other hand, something like a matrix is even more multipurpose: we can use a matrix to stand for either a (1,1) tensor, or a (0,2) tensor (like the metric!) or a (2,0) tensor. This is because the PreCambrian dialect of "row vectors", "column vectors" and "matrices" fails to make some of the finer-grained distinctions that the modern lingo does. Why? Because it's deceptively easy to turn a column vector into a row vector: just flop it over! E.g. [2] [3] becomes [2 3] The modern approach is set up to make this impossible to do UNLESS you use the metric!!!!!!!!! You can't just "flop over" a vector and get a covector. You do this using the metric (in a way I'll eventually explain.) Again, file this stuff away in your brain, or somewhere, but don't let it get you down now. >Hey, I don't want to be able to manipulate these things. Just get an >idea of the salient points, and some idea of the mechanism. Otherwise I >can't properly follow what's going on. Yes, but well, I am now being forced to explain not just tensors, which is bad enough, but the difference between the way your textbook explains tensors and the way modern mathematical physicists think about tensors. I don't mind doing it, but you are likely to become very confused for a while before it all clears up. It's as if, rather than just learning German, you wanted to learn German out of a book written in Old Norse. >>Oh well, education occurs even when one least wants it. >If I didn't want it, I would have given up weeks ago. Have you any idea >how many concepts I have tried to take in over the last few weeks? Come come, I didn't say you didn't want some education. I was making a cryptic joke about how you are now busily learning about the history of different mathematical notations for tensors! This means you are going to learn a whole bunch of extra concepts which you don't even need to know to understand general relativity. They are good to know, but will perhaps be the straw that breaks the camel's back. >>Say I have a tangent vector v and >>another one w. Then a (0,2) tensor like the metric g can be thought of >>as a matrix of numbers g_{ab}, and the way we get the scalar g(v,w) is >>by doing this: >>> >>g(v,w) = g_{ab} v^a v^b >> >>Here I am summing over the indices a and b. >This is the problem. I have done too much (mis)reading between the lines >in the past (rubs ear). There are a whole lot of ways of defining >'summing over the indices a and b'. YOU mean in a particular way. *I* >don't know which way you mean. Since we are getting close to the >notation you are using in GR, it would help if I knew, I think. Okay, let me spell it out. I was really hoping someone else would do it, since this involves typing long dull rows of symbols. But okay. Let's say we're in 4d spacetime. Then indices like a,b etc. stand for 0,1,2, or 3. When we sum over them, we sum from 0 to 3. So for example, when I speak of "the metric g_{ab}", what I mean is a particular batch of numbers, namely: g_{00} g_{01} g_{02} g_{03} g_{10} g_{11} g_{12} g_{13} g_{20} g_{21} g_{22} g_{23} g_{30} g_{31} g_{32} g_{33} Note how both a and b go from 0 to 3. Each thing like g_{21} above is just a number. So for example g_{ab} might be the matrix -1 0 0 0 0 1 0 0 (*) 0 0 1 0 0 0 0 1 in which g_{22} = 1 and so on. Similarly, when I refer to "a vector v^a", I mean a column vector like v^0 v^1 v^2 v^3 A random example would be the vector 1 2 (**) 0 9 In this example we'd have v^3 = 9. Similarly, when I refer to "the vector w^a" I mean another such thing, like w^0 w^1 w^2 w^3 An example would be, say 2 2 (***) 0 0 Now when I write something like g_{ab} v^a w^b, I mean that I multiply the number g_{ab} by the number v^a and by the number w^b, and then SUM UP letting a and b range from 0 to 3. So this is short for g_{00}v^0w^0 + g_{01}v^0w^1 + g_{02}v^0w^2 + g_{03}v^0w^3 + .... 4 more of these bloody things .... + .... 4 more of them .... + g_{30}v^3w^0 + g_{31}v^3w^1 + g_{32}v^3w^2 + g_{33}v^3w^3 Okay. Here's a test!!! Take the example of the tensor g_{ab} that I gave, back at (*), and the example of v^a I gave, back at (**), and the example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is. The answer is some specific number; what is it? This is how you actually work out the inner product of two vectors v and w: g(v,w) = g_{ab} v^a w^b. Article 98464 (58 more) in sci.physics: From: john baez (SAME) Subject: Re: Ricci Tensor (was: Re: General relativity tutorial) Date: 6 Feb 1996 18:50:42 -0800 Organization: University of California, Riverside Lines: 113 NNTP-Posting-Host: guitar.ucr.edu In article <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 4-space, we will assume that it >is equally straightforward to define the t axis as forward in time along >the geodesic]. Looking at the infinitesimal change in shape from sphere >to ellipse over an infinitesimal d time, we can define a >second-derivative-of-the-radius tensor, which is by definition the >tidal-acceleration tensor, which is by definition a curvature-of-space >tensor. In our nice diagonal coordinate system, the diagonal elements >are simply d^2r_x/dt^2, d^2r_y/dt^2, and d^2r_z/dt^2, plus a 0,0 term >relating to differential time dilation. For simplicity, we can normalize >by dividing by the initial radius of the coffee sphere. >Let me simply assert, without proof, that this >second-derivative-of-the-radius-of-the-coffee-sphere tensor is the same >as the Ricci tensor R_mn. Sorry. The "tidal-acceleration tensor" you describe is not the Ricci tensor, but rather R^m_{anb} v^a v^b where v is the 4-velocity of the coffee ground at the center of the ball, and R^m_{anb} is the Riemann tensor. I could explain WHY it's this: it follows from the "geodesic deviation equation" which relates the deviation of nearby geodesics to the curvature of spacetime. Instead, let's see how this result jibes with my earlier claims. >The volume of this ellipsoid is, to within factors of pi, the product V = >r_x r_y r_z. The (normalized) second derivative of the volume will be >the sum of the second derivative diagonal elements (keeping only first >order terms each time we do a derivative, and ignoring factors of pi >again); that is, the *trace*, or, in the Einstein summing notation, the >tensor contracted to a scalar by summing indices. Right. So to get the second derivative of volume we take the *trace* of the tensor R^m_{anb} v^a v^b on the two free indices m and n, getting R^m_{amb} v^a v^b. Using the definition of the Ricci tensor, R_{ab} = R^m_{amb}, this is the same as R_{ab} v^a v^b In coordinate-free notation this is just r(v,v), where now I write the Ricci tensor with a lowercase r to keep it straight from the Riemann tensor, because the indices don't keep things straight for us. This agrees with what I was saying earlier: "A little round ball of coffee grounds in free fall through outer space! As time passes this ball will change shape and size depending on how the paths of the coffee gournds are deflected by the spacetime curvature. Since everything in the universe is linear to first order, we can imagine it shrinking or expanding, and also getting deformed to an ellipsoid. There is a lot of information about spacetime curvature encoded in the rate at which this ball changes shape and size. But let's only keep track of the rate of change of its volume! This rate is basically the Ricci tensor. More precisely, the second time derivative of the volume of this little ball is equal to r(v,v) times its original volume, where r is a rank (0,2) tensor called the Ricci tensor, and v is as above. (In general r eats two vectors and spits out a number. Here it is eating two copies of v.)" >Then, if we want a tensor which represents the change in *shape* of the >initial sphere, without change in *volume*, we can subtract the change in >volume from the tensor representing the entire change in the sphere. To >subtract a scalar from a tensor, we have to multiply it by the metric. A >factor of 2 comes in from some differentiation I lost track of. Thus, we >get the divergenceless Ricci tensor [typically called the Einstein tensor] >G^ab = R^ab - (1/2) g^ab R > >I think that this is correct. Time to check the homework with the >teacher-- Sorry... nice guess, but the gadget that represents the change of shape of the initial sphere discounting change in volume is built from the Weyl tensor, not the Einstein tensor. As explained in other posts, the Einstein tensor and the Ricci tensor are two ways of repackaging the same information about spacetime curvature. The Weyl tensor expresses the information about curvature that's not in the Ricci tensor (or Einstein tensor). I think you'll find all this clearer if I explain the geodesic deviation equation. Article 98491 (34 more) in sci.physics: From: Edward Green Subject: Re: general relativity tutorial Date: 6 Feb 1996 21:16:17 -0500 Organization: The Pipeline Lines: 73 NNTP-Posting-Host: pipe9.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 I don't want to deviate too much on this question, but since you have moved it here, one more try. Ok? '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 multi-generational experiment), lower *one* of the two clocks on the tower, B, and note the time discrepancy. In yet another century, lower the second clock, C, in the same way as B. Now what as happened? Let the time gains relative to the ground be t_B and t_C. Without knowing what GR predicts, I surmise either: 1) t_C = t_B, or 2) t_C = 2 t_B In other words, we have left the damn things at the top of tower so long, that any effects attributable to the trip there have become negligible, unless that's the only effect in town! These results show either that 1) the discrepancy is attributable only to the relative motion in the gravitational field, a one shot deal, or 2) there is an accumulating discrepancy attributable to their separation. In the second case I would say _operationally_ that "time is moving faster at the top of the tower". If that's not what we mean by such a statement, I don't know what is! Now if you tell me 2) is the predicted outcome, then in GR, in cases where the gravitational potential predicts the time discrepancy, this potential must correspond to some local property of space, unlike in newtonian gravity, where the potential is just a summary of the global field. Space really is more sluggish down there in the graviational well. >The gravitational potential can be a useful tool in *certain* problems, >but I prefer this other way of thinking of it, since it tells you >something relevant to quite general GR problems. Also, it fits with the >idea that the *metric on spacetime* is what really matters: that's what >you'd use to compute the proper time along a curve. Also, it seems >pretty simple and unmysterious. Granted. So where is the metric doing its thing? Just going up and down the towers, or while sitting on the top? -- Ed Green / 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 NNTP-Posting-Host: guitar.ucr.edu While some of the stuff Ed said was wrong, it has helped me understand the simple geometrical essence of Einstein's equation, and that's good. In article <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 off-diagonals, shear. >However, we could always diagonalize it by a coordinate transformation to >principle axes. Then we would find all information in it summarized in *4* >numbers, giving expansion or contraction along the 4 principle axes, >transforming our little hyperspherical test region into a hyper-ellipsoid. >Up to scale factors these axes give the locally *best* coordinate system >for curved spacetime, one that captures the local behavior most >succinctly. Maybe that's true in some sense but I don't know what sense. All I know is that if one lets a little 3d ball of initially comoving test particles freely fall along in the direction v, the ball starts to expand at a rate proportional to r(v,v), where r is the Ricci tensor: r(v,v) = R_{ab} v^a v^b When I say "starts to expand, I'm talking about SECOND derivatives along the geodesic whose tangent vector was initially v, which is what one would expect from the fact that v appears in a QUADRATIC way in that formula. So we have a quadratic form in 4 variables, r(v,v), and thus we can find 4 principal axis for it. Typically, one of these is the axis along which the ball of test particles expands the MOST, one is the axis along which it expands, the LEAST, and then there are two "middle" ones. >The remaining information in the Riemann tensor, which we have thrown out, > captures the *rotation* of our little test sphere under translation. In >the same rank as our Ricci tensor, which is symmetric, we could add >another antisymmetric tensor, which would contain the pure rotations. >Apparently, these rotations are *not* determined by the local energy >density, but may be globally determined by boundry conditions. Dilation >is an intrinsic property, determined locally by the stress tensor, while >rotation is an extrinsic property, determined by the behavior of >nieghboring bits of space. Now I guess you're no longer talking about your "little 4d hyperspherical test region" but my "little 3d spherical ball of comoving coffee grounds," right? I hope so. The Ricci tensor describes how, as the ball freefalls, it expands and contracts. I am struggling to figure out whether it can start to rotate... I have some vague reasons to think it can't. This leaves volume-preserving transformations that stretch out some axes and squash the others. That sounds very much like what gravitational radiation and tidal forces do to stuff... so I think that's the Weyl tensor! >It seems reasonable from the definition R_{bd} = R^c_{bcd} that each >component of R_{bd} is measuring how much space "oriented in the d >direction" is spreading out as we "move in the "b direction", since we sum >over all the side excursions the test vector might make, and in particular >just sum the "spreading" terms (the discrepancy in the same direction as >the side excursion. Just waving my hands.... Something sorta vaguely like this is true... you can figure it all out from the definition of the Riemann tensor I gave. But I'll do it for you, eventually: it's the "geodesic deviation equation" you're after. >But I have no idea what the second term in the Einstein tensor is doing >yet. > G_{ab} = R_{ab} - (1/2)R g_{ab}. Well, here's a good way to think about. Let's do the same index gymnastics we did in the last class, only starting with Einstein's equation. Okay, everyone! Time for index gymnastics. Stand with your feet slightly apart and hands loosely at your sides. Now, assume the Einstein equation! G_{ab} = T_{ab} Substitute the definition of Einstein tensor! R_{ab} - (1/2)R g_{ab} = T_{ab} Raise an index! R^a_b - (1/2)R g^a_b = T^a_b Contract! R^a_a - (1/2)R g^a_a = T^a_a Remember the definition of Ricci scalar, and g^a_a = 4 in 4d! R - 2R = T^a_a Solve! R = - T^a_a Okay. That's already a bit interesting. It says that when Einstein's equation is true, the Ricci scalar R is the sum of the diagonal terms of T^a_a. What are those terms, anyway? Well, they involve energy density and pressure. But let's wait a bit on that... let's put this formula for R back into Einstein's equation: R_{ab} + (1/2) T^c_c g_{ab} = T_{ab} or R_{ab} = T_{ab} - (1/2) T^c_c g_{ab} In other words: you might naively have hoped that since the Ricci tensor has such a nice interpretation in terms of convergence of geodesics, it must be equal to the stress-energy tensor. In fact, I think Einstein made exactly this guess, before realizing that it'd ruin local conservation of energy and momentum. To get the right Einstein equation, you need this "extra term" around, coming from the trace of the stress-energy tensor, T^c_c. If we work this out a bit more, we'll see that Einstein's equation has a simple meaning in terms of a little ball of initially comoving particles in free fall: it says that in the rest frame of the ball, the second time derivative of the volume of the ball is given by a simple formula in terms of the energy density and pressure at that point! What's the simple formula? Hmm, let me think... Article 98676 (15 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 7 Feb 1996 18:42:50 -0800 Organization: University of California, Riverside Lines: 219 NNTP-Posting-Host: guitar.ucr.edu Okay, I claimed I was going to describe the "simple geometrical essence of Einstein's equation, so let me do that. I left off last time after rerwriting Einstein's equation that it looks like this: R_{ab} = T_{ab} - (1/2) T^c_c g_{ab} This is nice because we have a simple geometrical way of understanding the Ricci tensor R_{ab}. Namely, if v is the velocity vector of the particle in the middle of a little ball of initially comoving test particles in free fall, and the ball starts out having volume V, the second time derivative of the volume of the ball is R_{ab} v^a v^b times V. Here "time" means proper time. If we know the above quantity for all velocities v (even all timelike velocities, which are the physically achievable ones), we can reconstruct the Ricci tensor R_{ab}. But we might as well work in the local rest frame of the particle in the middle of the little ball, and use coordinates that make things look just like Minkowski spacetime right near that point. Then g_{ab} = -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 and v^a = 1 0 0 0 So then --- here's a good little computation for you budding tensor jocks --- we get R_{ab} v^a v^b = -R_{00} So in this coordinate system we can say the 2nd time derivative of the volume of the little ball of test particles is just -R_{00}. On the other hand, check out the right side of the equation: R_{ab} = T_{ab} - (1/2) T^c_c g_{ab} Take a = b = 0 and get R_{00} = T_{00} + (1/2) T^c_c Note: demanding this to be true at every point of spacetime, in every local rest frame, is the same as demanding that the whole Einstein equation be true! So we just need to figure out what it MEANS! What's T_{00}? It's just the energy density at the center of our little ball. How about T^c_c? Well, remember this is just g^{ca} T_{ac}, where we sum over a and c. So --- have a go at it, tensor jocks and jockettes! --- it equals -T_{00} + T_{11} + T_{22} + T_{33}. So if I haven't made a mistake (and I darn well could've) we get R_{00} = (1/2) T_{00} + T_{11} + T_{22} + T_{33}. What about T_{11}, T_{22}, and T_{33}? In general these are the flow of x-momentum in the x direction, and so on. So the "simple geometrical essence of Einstein's equation" is this: Take any small ball of initially comoving test particles in free fall. Work in the local rest frame of this ball. As time passes the ball changes volume; calculate its second derivative at time zero and divide by the original volume. The negative of this equals 1/2 the energy density at the center of the ball, plus the flow of x-momentum in the x direction there, plus the flow of y-momentum in the y direction, plus the flow of z-momentum in the z direction. Note: all of general relativity can in principle be recovered from the above paragraph! I'm glad I'm "wasting time" talking about general relativity on sci.physics, because I'd never known this formulation of general relativity until you folks forced me to explain the Ricci tensor. Note that the minus sign in that paragraph is good, since it says if you have POSITIVE energy density, the ball of test particles SHRINKS. I.e., gravity is attractive. Now you might wonder about what all that "x-momentum in the x direction" stuff really means. I don't have anything thrillingly insightful to say about this except that, for a perfect fluid, each of these terms is just the pressure! I will quit here for now, leaving Michael Weiss to explain why that's true. Next time (when I'm not too busy swatting down questions) I will explain the "geodesic deviation equation" which relates the change of shape of this little ball of freely falling particles to the Riemann curvature. Okay, here's Michael and Bruce: [etc.] Article 98586 (52 more) in sci.physics: From: john baez Subject: Re: general relativity tutorial Date: 7 Feb 1996 12:47:42 -0800 Organization: University of California, Riverside Lines: 102 NNTP-Posting-Host: guitar.ucr.edu In article <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 multi-generational experiment), lower >*one* of the two clocks on the tower, B, and note the time discrepancy. >In yet another century, lower the second clock, C, in the same way as B. >Now what as happened? Let the time gains relative to the ground be t_B >and t_C. Without knowing what GR predicts, I surmise either: >1) t_C = t_B, or > >2) t_C = 2 t_B And the answer is.... 2)! >In other words, we have left the damn things at the top of tower so long, >that any effects attributable to the trip there have become negligible, >unless that's the only effect in town! These results show either that 1) >the discrepancy is attributable only to the relative motion in the >gravitational field, a one shot deal, or 2) there is an accumulating >discrepancy attributable to their separation. In the second case I would >say _operationally_ that "time is moving faster at the top of the tower". > If that's not what we mean by such a statement, I don't know what is! I sympathize.... But as you begin to get more rhapsodic, I get more nervous: >Now if you tell me 2) is the predicted outcome, then in GR, in cases >where the gravitational potential predicts the time discrepancy, this >potential must correspond to some local property of space, unlike in >newtonian gravity, where the potential is just a summary of the global >field. Space really is more sluggish down there in the graviational well. "Space is more sluggish"... hmm, now we are really waxing rhapsodic, aren't we! What makes me really nervous is your talk of a "local property of space" being different. This can be fatally ambiguous: 1. The equivalence principle says that to an object in free fall, every tiny patch of spacetime looks, to first order, just like every other. What do I mean by that? I mean that if you work in the local rest frame of a freely falling object at a given moment and do an experiment in its immediate locale, the result will always be the same, as long as we ignore higher-order effects --- namely CURVATURE --- which show up when the experiment is not infinitesimal. This is one sense in which spacetime has no local properties. 2. But of course your house and the tower are NOT in free fall, and we are NOT doing an infinitesimal experiment in the immediate locale of one spacetime event. So to some extent 1 is not relevant. I just thought I'd mention it. 3. If we do take into account second-order effects, we may say that spacetime at any given point DOES have local properties, namely curvature. This curvature is higher in your house than up on the tower. But the way the "time runs faster at the top of the tower" is NOT simply proportional to the curvature or anything like that; the relation is subtler. So curvature is not a "local property of spacetime which makes time run slower or faster". A better way to think of the role of curvature is this. Think about a 2d surface rather than 4d spacetime. If we know the surface is flat we know how to compute the length of any path on it. If we THINK it's flat but it's NOT, our computations will be wrong. Some paths will be mysteriously shorter or longer than we expect. That's what's happening in the tale of two clocks: two worldlines in spacetime which commonsense suggests should have the same length, do not, because spacetime is curved. 4. In a sense, one might say the *metric* on spacetime is the "local property" that makes paths long or short --- perhaps longer or shorter than naive commonsense suggested --- hence makes a clock run "faster" or "slower" than expected. This has a certain truth to it, despite 1. As I said: >>... the *metric on spacetime* is what really matters: that's what >>you'd use to compute the proper time along a curve. Also, it seems >>pretty simple and unmysterious. >Granted. So where is the metric doing its thing? Just going up and down >the towers, or while sitting on the top? Mainly while sitting on top, if the clock sits on top for a long time. I almost feel like showing you the formulas, but I don't like typing formulas too much. Article 98666 (54 more) in sci.physics: From: Emory F. Bunn (SAME) Subject: Re: General relativity tutorial Followup-To: sci.physics Date: 8 Feb 1996 00:40:02 GMT Organization: Physics Department, U.C. Berkeley Lines: 31 Distribution: world NNTP-Posting-Host: physics12.berkeley.edu In article , Keith Ramsay wrote: >In article <4f6tq6$13v8@hearst.cac.psu.edu>, ale2@psu.edu (ale2) wrote: >|For any spacetime point around, say the sun, is there an infinite >|number of geodesics through that point? > >Yes, one for each tangent direction through it. Lest anyone get confused, I just want to point out that when Keith says "each direction" he means each direction through *spacetime*, not just space. As long as you've gotten into the proper relativity spirit, that should be obvious. So if you're sitting at a particular point in spacetime, you can't just choose a particular direction through space (say, North) and expect to find a unique geodesic in that direction through that point. That's because North is just a direction in space, not a direction through spacetime. (Of course, before a word like "North" even makes sense, you have to have laid down coordinates on your patch of spacetime.) If you want to specify a unique geodesic through a particular spacetime point, one way to do it is to specify the *velocity* that a particle would have if it passed through that point following that geodesic. That's because the velocity is a 4-dimensional vector (a tangent vector) in spacetime, so specifying the velocity is a good way to specify a direction through spacetime. -Ted Article 98602 (39 more) in sci.physics: From: john baez Subject: Re: Tensors for twits please. Date: 7 Feb 1996 13:27:02 -0800 Organization: University of California, Riverside Lines: 117 NNTP-Posting-Host: guitar.ucr.edu In article Steven_Hall@mit.edu ( Steven Hall) writes: >In article <4f90ct$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 "old-fashioned" matrix/vector notation >dominates.) The connections you are making between the two notations are >certainly helping me. That's great! It's really a relief to know that typing in those enormous matrices is doing *somebody* some good. In Oz's case, I was afraid I might simply be confusing the heck out of him unnecessarily. >OK, let's see if I have it: > >> Okay, let me spell it out. I was really hoping someone else would do >> it, since this involves typing long dull rows of symbols. But okay. >> Let's say we're in 4d spacetime. Then indices like a,b etc. stand for >> 0,1,2, or 3. When we sum over them, we sum from 0 to 3. So for >> example, when I speak of "the metric g_{ab}", what I mean is a >> particular batch of numbers, namely: >> >> g_{00} g_{01} g_{02} g_{03} >> g_{10} g_{11} g_{12} g_{13} >> g_{20} g_{21} g_{22} g_{23} >> g_{30} g_{31} g_{32} g_{33} >> >> Note how both a and b go from 0 to 3. Each thing like g_{21} above is >> just a number. So for example g_{ab} might be the matrix >> >> -1 0 0 0 >> 0 1 0 0 (*) >> 0 0 1 0 >> 0 0 0 1 > >[more deletia] >So to an engineer, we would just have the matrix G, with which we could do >any number of things. We could multiply by a column vector x, yielding a >column vector y = G x, or we could multiply by a row vector u, to get v = >u G, are take an inner product, p G q. Exactly. >In tensor notation, we can do only the last, because the tensor is g_{ab}, >and the inner product would be g_{ab} u^a v^b. The first would be written >y^a = g^a_b x^b, and the second would be v_a = g_a^b u_b. Exactly!!!!! >If I have that right, then the interpretation is that with a matrix, you >have to know the allowable operations from the context of the problem, >whereas with tensor notation, the tensor tells you what to do. Right. Now one extra comment. Why do engineers feel so free to treat matrices in this multipurpose way? Or conversely, why are general relativists so fussy, that they DON'T want to do so? The key thing is that to an engineer, it's perfectly acceptable to turn a column vector into a row vector. It's no big deal. But to a relativist, one should be careful turning vectors (aka tensors of rank (1,0)) into covectors (aka tensors of rank (0,1)). It's not against the law, but one does it using the METRIC. This is a tensor g_{ab}. We define the tensor g^{ab} in such a way that g_{ab}g^{bc} = delta_a^c where delta_a^c is 0 if a is not c, and 1 if a = c. In other words, in matrix lingo g^{bc} is just the inverse of the matrix g_{bc}. We can then turn a vector v^a into the corresponding covector --- written simply v_a --- using the metric: v_a = g_{ab}v^b Or conversely, we can turn a covector w_a into the corresponding vector --- written simply w^a --- as follows: w^a = g^{ab}w_b These are inverse processes since g_{ab} and g^{ab} are inverse matrices (thinking of them as mere matrices). Now: most engineers act like they live in ordinary Euclidean 3-dimensional space. The metric for that is just 1 0 0 0 1 0 0 0 1 so this process of "raising and lowering indices using the metric" --- i.e. turning vectors into covectors and vice versa --- is completely invisible! It just amounts to multiplying a column vector by the identity matrix and then flopping it over to being a row vector. But in special relativity, the metric is -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 so one has to be more careful... and in general relativity, the metric g_{ab} can be ANY invertible symmetric matrix, and it depends on where you are in spacetime, too! So you gotta be *really* careful not to blithely mix up vectors and covectors. See some stuff Matthew Wiener wrote recently, for a more geometrical explanation of the difference between vectors and covectors. Article 98610 (34 more) in sci.physics: From: john baez Subject: Re: Gravitational Red Shifts - Real or Apparent ? Date: 7 Feb 1996 14:04:07 -0800 Organization: University of California, Riverside Lines: 60 NNTP-Posting-Host: guitar.ucr.edu In article <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? Oh-oh. Look. You talked about one very specific problem involving two clocks at different elevations in a spherically symmetric static gravitational field (the earth's), and came up with an experiment, and you said if the answer was answer number "2", you would say time ran slower on mountains. And I warned you that everything you said after that would be your own fault. And now you are acting as if it was some generally true sort of fact, not just an artifact of this one very special situation, that there is something called the "gravitational potential" and that clocks "run slower" when it's lower. And that's completely wrong, or at least it would be wrong if it were even meaningful. >Now that I >understand the framework a little better, I would say: According to GR, >the local conditions are *not* the same. The potential must have a local >physical significance. I'd prefer to put it this way: the potential has no bloody physical significance. A slight exaggeration, but quite a useful principle to keep in mind for starters. Article 98590 (23 more) in sci.physics: From: Oz Subject: Re: GR Tutorial Water Cooler Date: Wed, 7 Feb 1996 19:43:42 +0000 Organization: Oz Lines: 49 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 >>tongue-lashing for some things I overheard from that last post. I agree >>that the students need to be able to talk things over freely. There are >>all sorts of things that the "experts" are too familiar with to be able >>to explain very well. > >Tongue-lashing? Oh, oh. I don't think I am going to feel good when I see >the grade for that assignment. Oh wow. I just caught up with the postings, at least Demon has. I must look very slow at replying. This has just arrived. Must catch up with the reams of stuff that's just come in so I'll be back for a chat later. 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 NNTP-Posting-Host: 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 NNTP-Posting-Host: guitar.ucr.edu In article <4f6tq6$13v8@hearst.cac.psu.edu> ale2@psu.edu (ale2) writes: >Student interupts (many classmates give him a look of disgust)... >For any spacetime point around, say the sun, is there an infinite >number of geodesics through that point? Yes indeed, basically one for each unit vector through that point. Remember a spacetime point is a moment and a place. At any given moment, at any given place, there are many ways a freely falling rock could zip through that place at that moment. The velocity of the rock (a 4d vector!) could be any future-pointing timelike vector. As it continues to whiz along its worldline traces out a geodesic. Then there are the lightlike and spacelike geodesics.... >Is the geodesic of a particle >at some point in the spacetime around the sun determined once we know >the velocity of the particle at that spacetime point? Yes, if we are told the geometry of spacetime, we can completely work out a geodesic if we know a point it goes through and its tangent vector (velocity) at that point. It satisfies a second-order differential equation, which I have avoided writing down, and you just solve them. Article 98699 (48 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 7 Feb 1996 20:12:40 -0800 Organization: University of California, Riverside Lines: 75 NNTP-Posting-Host: guitar.ucr.edu In article <0M++SOAJTGGxEwLK@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk writ es: >In article <4f8r0u$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 non-sarcastic information-rich understandable >sentences would suffice. Enough to place it in a footnote of the brain. Non-sarcastic, eh? You must think I'm really mean and nasty, when actually I am the sweetest, nicest guy you could ever meet. Relatively few people understand why in GR we assume the connection --- the gadget we use to do parallel translation --- is torsion-free. It's often presented as a technical assumption not worth trying to understand. But here's a nice way understanding what torsion-free-ness means (in conjunction with our other assumptions: that parallel translation be linear and metric-compatible). Take a tangent vector v at P. Parallel translate it along a very short curve from P to Q, a curve of length epsilon. We get a new tangent vector w at Q. Now let two particles free-fall with velocities v and w starting at the points P and Q. They trace out two geodesics. Let me try to draw this: | | | | | | ^v ^w | | P------Q Remember, this is a picture in spaceTIME. Here I've drawn what it might look like in flat Minkowski spacetime, where the geodesics are boring old straight lines, and I've drawn everything very rectilinearly, since ASCII is so bad at drawing curves. Okay. Now, let's call our two geodesics C(t) and D(t), respectively. Here we use as the parameter t the proper time: the time ticked out by stopwatches falling along the geodesics. (We set the stopwatches to zero at the points P and Q, respectively.) Now we ask: what's the time derivative of the distance between C(t) and D(t)? Note this "distance" makes sense because C(t) and D(t) are really close, so we can define the distance between them to be the arclength along the shortest geodesic between them. If, no matter how we choose P and Q and v, the time derivative of the distance between C(t) and D(t) at t = 0 is ZERO, up to terms proportional to epsilon^2, then the torsion is zero! And conversely! If v got "rotated" a bit when we dragged it over to Q, and things looked like this: | / | / | / ^v ^w | / P------Q then the time derivative of the distance would not be zero (it'd be proportional to epsilon). In this case the torsion would not be zero. In GR we assume this kind of "rotation" effect doesn't happen. In some other theories of gravity there is torsion. But there's no experimental evidence for torsion, so most people stick with GR. Article 98736 (58 more) in sci.physics: From: Oz Subject: Re: General relativity tutorial Date: Thu, 8 Feb 1996 06:57:37 +0000 Organization: Oz Lines: 56 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article , Doug Merritt writes >In article <4fbt7o$puu@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes: >>Non-sarcastic, eh? You must think I'm really mean and nasty, when >>actually I am the sweetest, nicest guy you could ever meet. > >Having actually met John in the flesh, I figure I should now take >bids from opposing sides as to his nature. High bid wins. > Ah ....... Well I would guess he has a smile on his face quite a bit of the time. I also guess he likes a fast thought out response to a carelessly framed comment, that may be cutting at times if you were stupid enough to take it that way. Obviously he is considerate, you can see that from his apologies when he inadvertntly steps on some delicate ego. He also does it on occasion when he has stepped on a non-delicate ego so I guess that must be right. Among his peers I would guess he delights in utterly confusing them at every opportunity, probably with a real big grin. If he has something wrong the smile vanishes and his brow puckers as he tries to work out how an earth he could have made such a booboo. He is worried he is getting a reputation as a grumpy and irascible lecturer. I cannot imagine why that should be (not guilty y'r 'onnor). He is doing this GR course primarily to make sure he has it absolutely right for his students. He also hoped to chide them on with a comment like "Good grief, even Oz has this figured out", but has been woefully let down. He also hopes to avoid getting caught out by any smart-assed students who try to do what he did in their position and catch out the prof. I suppose in short probably best described as a mischevous sprite. At least from the subset of Baez as indicated by his sci.physics persona. I bet this will cause hernias amongst his friends. >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 $5 right or wrong. Sheesh, the risk of posting this on the group. I must find myself a shrink. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 98747 (57 more) in sci.physics: From: Oz Subject: Re: General relativity tutorial Date: Thu, 8 Feb 1996 12:56:43 +0000 Organization: Oz Lines: 118 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4fbp1l$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 NNTP-Posting-Host: 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 pre-primer, because we are avoiding all the nasty equations that you'd see in a typical introduction to general relativity. As a result, you may be in for a shock when you finally see, for example, the standard formula for the Riemann curvature in a textbook. On the other hand, this pre-primer is more thorough than many textbooks in certain ways, in that we are getting into the "inner meaning" of a lot of concepts that books typically gloss over. Sure, you can extract this inner meaning from the nasty equations with enough effort, but I don't think everyone gets around to it. For example, my attempt to gloss over the concept of torsion failed miserably: Ed and Bronis kept asking what the hell it meant to parallel transport a vector without rotating it, then Oz came up with a nice way of describing, then Darryl described the "Schild's ladder" construction for torsion-free parallel translation, and finally Oz forced an operational definition of "torsion" out of me. This operational definition makes it clear (I hope) exactly what physical assumption is built into the "torsion-free" condition on parallel transport in general relativity. On the contrary, if you look in Wald's General Relativity, which is in many respects an excellent book, the only thing you will see about this is: "5. Torsion free: for all f, D_a D_b f = D_b D_a f This condition is sometimes dropped, and indeed there are theories of gravity where it is not imposed. However, in general relativity the derivative operator is assumed to satisfy condition 5, and, unless otherwise stated, all derivative operators in this book will be assumed to be torsion free." This leaves reader completely in the dark as to the meaning of torsion, and why we assume it vanishes. So while we won't get very far in some respects, we will be way ahead of the crowd in terms of really understanding what we've covered. "Petrified knowledge" is a constant danger in science: someone figures something out that's very important and interesting, and then it becomes reduced in the textbooks to a brief remark without any clue to the reader about what's really going on, or turned into an equation whose inner meaning can only be unpacked with a lot of effort. There's no way to get the "inner meaning" out of textbooks without making that effort! For example, it was only in this "tutorial" that I finally understood the geometrical essence of the Ricci tensor; I always knew I should do it someday, but I never saw it spelled out anywhere, and I was always too lazy to do it. That's pretty terrible. The inner meaning --- or *one* inner meaning, anyway, because there are usually many layers of inner meaning --- lay lurking in Raychaudhuri's equation, which only a fairly dedicated reader of Wald would ever get to. But once it's unearthed it's not all that complicated. By the way, thanks go out to Dan McGuirk for catching a factor-of-two error. Here's the new improved "inner meaning of Einstein's equation": Take any small ball of initially comoving test particles in free fall. Work in the local rest frame of this ball. As time passes the ball changes volume; calculate its second derivative at time zero and divide by the original volume. The negative of this equals 1/2 of: the energy density at the center of the ball, plus the flow of x-momentum in the x direction there, plus the flow of y-momentum in the y direction, plus the flow of z-momentum in the z direction. Or: R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] Article 98782 (55 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 8 Feb 1996 10:38:55 -0800 Organization: University of California, Riverside Lines: 106 NNTP-Posting-Host: guitar.ucr.edu In article 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 b-c plane. We go in the b direction until the b coordinate has changed by epsilon, then we go in the c direction until the c coordinate has changed by epsilon, then we go back in the b direction until the b coordinate is what it started out as, and then we go back in the c direction until the c coordinate is what it started out as. Our tangent vector may have rotated a little bit since space is curved. Its component in the a direction has changed a bit, say -epsilon^2 R^a_{bcd} " So R^2_{121} tells us how much a vector pointing in the theta direction swings over towards the phi direction when we parallel transport it around a little square. If you visualize it... [little vector pointing east appears near the equator on the hovering globe; moves east, then south, then west, and north back to where it was, but has rotated a wee bit southwards in the process]... you'll see this makes sense. Unless I got a sign wrong.... [North pole on globe flips over to south pole, east flips over to west, globe lets out a snicker and disappears in a puff of smoke.] Article 98817 (53 more) in sci.physics: From: Daryl McCullough (SAME) Subject: Re: General relativity tutorial Date: 8 Feb 1996 10:44:51 -0500 Organization: Odyssey Research Associates, Inc., Ithaca NY Lines: 23 Distribution: world NNTP-Posting-Host: www.oracorp.com I have a question or comment about the Ricci tensor and the missing information needed to go from it to the full Riemannian tensor. If you are in vaccuum nearby a large mass, then the stress-energy tensor, and therefore the Ricci tensor, will be zero. However, it seems to me that the presence of mass nearby is indicated by geodesic deviation (falling particles tend to get closer together as they get closer to the center of the mass). This geodesic deviation is determined by the non-Ricci part of the Riemannian curvature tensor (the Weyl tensor). So information about mass-energy is really present in the Weyl tensor as well as the Ricci tensor. Maybe it is like Gauss's theorem for electrostatics. Although the divergence of E is zero everywhere outside of a charged particle, knowledge of E on a surface can tell us that there is charge inside a surface (and therefore nonzero divergence inside). Similarly, maybe knowledge of the Weyl tensor on a surface can tell us that there is mass-energy inside the surface (and therefore a nonzero Ricci tensor) even though the stress-energy tensor is zero everywhere on the surface. Daryl McCullough ORA Corp. Ithaca, NY Article 98772 (52 more) in sci.physics: From: Oz Subject: Re: Tensors for twits please. Date: Wed, 7 Feb 1996 18:08:01 +0000 Organization: Oz Lines: 236 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 re-read my reply. Just a note to point out that I follow that in f(v)=x one can look at it that *either* v eats f, or f eats v) Yes, but I feel that covectors should do slightly different things to vectors. For one thing if you take your example above then f(v) is a quite different beastie to v(f) and indeed one or the other may not even exist. Er, I suspect. (Prays hard) >But it's lurking in the "row vector" vs "column vector" stuff you always >see in old-fashioned linear algebra books. You can think of a row >vector as a guy that's dying to eat column vectors and spit out numbers: > >[a b][c] = ac + bd > [d] > >but if you change your point of view you can think of it the other way >around: the column vector eats the row vector. Agreed. How about [c][a b]=[2x2] it's different. [d] >>A very simple and obvious question. Do we have to be very caerful WHAT >>our tensor is used for. Ie just row>row not col>col, or does the *same* >>tensor do 'double duty' in some way. > >There are two levels to this question, because you are reading an old >book that talks about "row vectors" and "column vectors", and I am >struggling to translate it into the modern lingo of "vectors" and >"covectors", but the translation is a tricky business, in the way that >translating foreign languages always is. In fact, I urge you to not pay >much attention to that book, and just let me explain everything. I will >wind up doing LESS work that way. OK, boss. Consider it banned. Well just a *little* peek. >Anyway, I can confidently say: yes, a (1,1) tensor does double duty: by >definition it serves to turn a vector into a vector, but if we are >clever we can use it to go turn a covector into a covector. > >On the other hand, something like a matrix is even more >multipurpose: we can use a matrix to stand for either a (1,1) tensor, or >a (0,2) tensor (like the metric!) or a (2,0) tensor. Ugh. But these are *different*! You'll get mixed up terribly if you let this go on. Need a new notation - and fast, Batman! >This is because >the PreCambrian dialect of "row vectors", "column vectors" and >"matrices" fails to make some of the finer-grained distinctions that the >modern lingo does. Why? Because it's deceptively easy to turn a column >vector into a row vector: just flop it over! E.g. > >[2] >[3] > >becomes > >[2 3] > >The modern approach is set up to make this impossible to do UNLESS you >use the metric!!!!!!!!! You can't just "flop over" a vector and get a >covector. You do this using the metric (in a way I'll eventually >explain.) > >Again, file this stuff away in your brain, or somewhere, but don't let >it get you down now. No, I like this. I didn't bother with matrices years ago because they didn't 'feel' right, and anyway I could the things I needed to do with vanilla flavoured vectors. This looks better, at least vectors stay as vectors. >>>Oh well, education occurs even when one least wants it. > >>If I didn't want it, I would have given up weeks ago. Have you any idea >>how many concepts I have tried to take in over the last few weeks? > >Come come, I didn't say you didn't want some education. I was making a >cryptic joke about how you are now busily learning about the history of >different mathematical notations for tensors! This means you are going >to learn a whole bunch of extra concepts which you don't even need to >know to understand general relativity. They are good to know, but will >perhaps be the straw that breaks the camel's back. Hey, it's a nice break from GR for a while, let some of those conceots sink in a bit. Anyway it's rather fun, I am just a little disappointed that in a few months I won't remember it well enough and get told off! Anyway, at least there's no exam at the end! >Okay, let me spell it out. I was really hoping someone else would do >it, since this involves typing long dull rows of symbols. But okay. >Let's say we're in 4d spacetime. Then indices like a,b etc. stand for >0,1,2, or 3. When we sum over them, we sum from 0 to 3. So for >example, when I speak of "the metric g_{ab}", what I mean is a >particular batch of numbers, namely: > >g_{00} g_{01} g_{02} g_{03} >g_{10} g_{11} g_{12} g_{13} >g_{20} g_{21} g_{22} g_{23} >g_{30} g_{31} g_{32} g_{33} > >Note how both a and b go from 0 to 3. Each thing like g_{21} above is >just a number. So for example g_{ab} might be the matrix > > -1 0 0 0 > 0 1 0 0 (*) > 0 0 1 0 > 0 0 0 1 > >in which g_{22} = 1 and so on. Similarly, when I refer to "a vector >v^a", I mean a column vector like > > v^0 > v^1 > v^2 > v^3 > >A random example would be the vector > > 1 > 2 (**) > 0 > 9 > >In this example we'd have v^3 = 9. Similarly, when I refer to "the >vector w^a" I mean another such thing, like > > w^0 > w^1 > w^2 > w^3 > >An example would be, say > > 2 > 2 (***) > 0 > 0 > >Now when I write something like g_{ab} v^a w^b, I mean that I multiply >the number g_{ab} by the number v^a and by the number w^b, and then >SUM UP letting a and b range from 0 to 3. So this is short for > >g_{00}v^0w^0 + g_{01}v^0w^1 + g_{02}v^0w^2 + g_{03}v^0w^3 + >.... 4 more of these bloody things .... + >.... 4 more of them .... + >g_{30}v^3w^0 + g_{31}v^3w^1 + g_{32}v^3w^2 + g_{33}v^3w^3 > >Okay. Here's a test!!! Take the example of the tensor g_{ab} that I >gave, back at (*), and the example of v^a I gave, back at (**), and the >example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is. >The answer is some specific number; what is it? Oooh. I didn't expect this. No exam indeed, humpf. OK, this looks to be fun. Now I see it out longhand I can see that its: -- a b \ g v w oh dear. You have said this before, / ab but now I follow the notation. -- so g_{ab} above is mostly 0's. Only the diagonal has non-zero indices so we can simplify it as: g_{00}v^0w^0 + g_{11}v^1w^1 + g_{22}v^2w^2 + g_{33}v^3w^3 Hmmmm, only the diagonal is non-zero. So no nasty cross terms. Hmmm this wouldn't be flat space (I suppose in spacetime this would be Minkowski space) would it? Must check later. Anyway g_{ab}v^aw^b = -1x1x2 + 1x2x2 + 1x0x0 + 1x9x0 = -2 + 4 = 2 Hope to god this is right. Hmmm also get some idea why you are so pernickity about your indices and whatsits. Like g(uv) is different to g_{ab} and to a vector g^a, I presume. Anyway lets just check using (for the last time, promise) the old fashioned matrix formulation. [1,2,0,9][2]= whaddayaknow=1x2+2x2+0x0+9x0=2+4=6 BUT no g_{00}=-1 [2] ^time thingy should be negative [0] so works out as +2 as before. [0] so it looks like flat spacetimey thingy. Phew, at least he stared with an easy one . >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 NNTP-Posting-Host: 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 re-read my reply. Just a note to point out that I follow that in >f(v)=x one can look at it that *either* v eats f, or f eats v) >Yes, but I feel that covectors should do slightly different things to >vectors. For one thing if you take your example above then f(v) is a >quite different beastie to v(f) and indeed one or the other may not even >exist. Er, I suspect. (Prays hard) Oh yes, vectors and covectors are very very different in character, this is a big theme in modern mathematics. When you have a metric around you can freely convert one to the other, by lowering or raising indices. But they have a different geometrical significance... and the process of raising and lowering indices has a very geometrical description as well. If you think the "index gymnastics" I'm describing seems somewhat mechanical and removed from the physics and geometry, that's just because I was focussing on getting you going on the mechanics. There is also a lot to say about the inner meaning of it all. Matthew Wiener said a bunch so I quote him... this may or may not make sense to you just yet: "A vector is just an arrow pointing somewhere. A covector (in R^3) is a dissection of R^3, consisting of parallel planes, such that they are labelled "linearly", with 0 for the plane passing through the origin, one of the other planes labelled 1, the one twice as far labelled 2, etc. These may be described with 3 coordinates (a,b,c) by letting ax+by+cz=1 be the plane labelled 1. (0,0,0) is the plane at infinity, so to speak. So covectors form a 3-dimensional space. There is a natural duality between covectors and vectors, namely < (a,b,c) | (x,y,z) > = ax+by+cz. In terms of the starting R^3 dissection, this just identifies the label of the plane that vector (x,y,z) reaches. The unit covectors are commonly denoted dx,dy,dz. Although you were taught dx is "differential x", here its "dual x". But the notation was chosen this way deliberately. 1-forms Adx+Bdy+Cdz are naturally understood as covectors, not vectors. For example, given a scalar function f on R^3 (ie, f(x,y,z)=some number), we can take its gradient. You probably learned grad f = (@f/@x).i + ... What piffle. "Really" grad f is df = (@f/@x).dx + .... It forms a covector field, not a vector field. So what does this really look like? It's ultimately just a contour map of f, custom linearized at each point. This is easier to imagine in 2D. Take a contour map of a scalar function "height", assumed differentiable. Pick a point. Kind of curvy, right? Blow up the neighborhood 1000 times, notice how the nearby contours are a lot straighter. So just _redraw_ them as straight, extend to a covector worth of lines dissecting R^2, and scale the line labelled 1 down by 1000. This is the dual version of (F(x+.001)-F(x))/.001 we all know and love. That this is so direct, the moral equivalent of drawing tangent lines, is all the proof anyone really needs that gradients are covectors." To continue..... >>On the other hand, something like a matrix is even more >>multipurpose: we can use a matrix to stand for either a (1,1) tensor, or >>a (0,2) tensor (like the metric!) or a (2,0) tensor. > >Ugh. But these are *different*! You'll get mixed up terribly if you let >this go on. Need a new notation - and fast, Batman! Yes, well, that's why we write (1,1) tensors as things like g^a_b, (0,2) tensors as things like g_{ab}, and (2,0) tensors as things like g^{ab}. >Hey, it's a nice break from GR for a while, let some of those conceots >sink in a bit. Anyway it's rather fun, I am just a little disappointed >that in a few months I won't remember it well enough and get told off! What, you mean you're planning on forgetting all this stuff? >Anyway, at least there's no exam at the end! That's what YOU think. In fact... >>So for example g_{ab} might be the matrix >> >> -1 0 0 0 >> 0 1 0 0 (*) >> 0 0 1 0 >> 0 0 0 1 >>[v^a might be the vector] >> >> 1 >> 2 (**) >> 0 >> 9 >>[and w^b might be the vector] >> >> 2 >> 2 (***) >> 0 >> 0 >>Okay. Here's a test!!! Take the example of the tensor g_{ab} that I >>gave, back at (*), and the example of v^a I gave, back at (**), and the >>example of w^a I gave, back at (***), and work out what g_{ab}v^aw^b is. >>The answer is some specific number; what is it? >Oooh. I didn't expect this. No exam indeed, humpf. OK, this looks to be >fun. Now I see it out longhand I can see that its: > >-- a b >\ g v w oh dear. You have said this before, >/ ab but now I follow the notation. >-- > >so g_{ab} above is mostly 0's. Only the diagonal has non-zero indices so >we can simplify it as: > >g_{00}v^0w^0 + g_{11}v^1w^1 + g_{22}v^2w^2 + g_{33}v^3w^3 > >Hmmmm, only the diagonal is non-zero. So no nasty cross terms. Hmmm this >wouldn't be flat space (I suppose in spacetime this would be Minkowski >space) would it? Must check later. Yes, that metric g^{ab} I wrote down is the metric for Minkowski spacetime. I was subtly trying to get this metric to seem familiar. >Anyway > >g_{ab}v^aw^b = -1x1x2 + 1x2x2 + 1x0x0 + 1x9x0 = -2 + 4 = 2 Yes indeed. Correct!! You are picking up the art of index juggling. >Hope to god this is right. Hmmm also get some idea why you are so >pernickity about your indices and whatsits. Like > >g(uv) is different to g_{ab} and to a vector g^a, I presume. Aheh, yes, by the way we write g(u,v) with a comma in there when we think of the metric g as a guy waiting to swallow two hapless vectors v and w and spit out their inner product g(u,v). >Er, ahem, cough, cough, er, "Prof, prof, er, prof sir?" > >I don't like to ask, but how do I handle a tensor that turns two vectors >into another vector, and what is the notation, please sir? > You have a strong masochistic streak, Oz. This may actually come in handy when learning about tensors. The notation (using indices) for a tensor that turns two vectors into one is --- well, say our tensor is called H --- H^c_{ab}. One index downstairs for each input vector, one upstairs for each output vector! If we were in 4d spacetime this thing could be viewed in a lowbrow way as a 4 x 4 x 4 array of numbers. How does it act on two vectors u and v to give the vector H(u,v)? Note: now H(u,v), the output, is a vector! So it has components H(u,v)^c where c goes from 0 to 3. And what are they? Here: H(u,v)^c = H^c_{ab} u^a v_b where as usual we sum on repeated indices (and, to keep us from screwing up, repeated indices only come in pairs, one up and one down). Article 98845 (3 more) in sci.physics: From: Emory F. Bunn Subject: Re: GR Tutorial Water Cooler Followup-To: sci.physics Date: 8 Feb 1996 20:47:26 GMT Organization: Physics Department, U.C. Berkeley Lines: 43 Distribution: world NNTP-Posting-Host: physics12.berkeley.edu In article <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, index-based definition. >I picked up on the claim that it captures the *other* ten independent >components of the Riemann, the ones we didn't use in the Einstein. > >But what's its rank? Same as the Einstein? Is it antisymmetric? If you only knew about four-dimensional geometry, you would be tempted to guess that it was rank 2 like the Ricci tensor, since in four dimensions Ricci and Weyl have the same number of pieces of information. But things aren't that simple, as you can see if you look at how many pieces of information the various tensors contain in different numbers of dimensions: Dimensions Riemann Ricci Weyl 1 0 0 0 2 1 1 0 3 6 6 0 4 20 10 10 and so on. (Do I have all these numbers right? Something doesn't look right to me. Anybody want to confirm them?) It turns out that Weyl is a rank-4 tensor like Riemann, but it has some extra conditions it has to satisfy, which is why it has fewer bits of information hidden in it. Basically, Weyl is "trace-free," which means that if you try to contract Weyl to a rank-2 tensor in the same way you contract Riemann to get Ricci, you always get zero. (Ricci is R^a_{bad}. Try to perform the same trick on Weyl, and you get C^a_{bad}, which turns out to be identically zero.) -Ted Article 98867 (7 more) in sci.physics: From: john baez Subject: Re: GR Tutorial Water Cooler Date: 8 Feb 1996 18:00:55 -0800 Organization: University of California, Riverside Lines: 78 NNTP-Posting-Host: guitar.ucr.edu In article <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, index-based definition. I don't yet know a beautiful geometrical definition, though I'm working on it (in 4d). So far I think of it as just "the stuff about curvature that's not in the Ricci tensor" --- as Ed recalls: >>it captures the *other* ten independent >>components of the Riemann, the ones we didn't use in the Einstein. But another way to think of it is that we take R^a_{bcd} and subtract off stuff to make it trace-free on all its indices, while still keeping all the symmetries the Riemann tensor has. We get a gadget they call C^a_{bcd}. >>But what's its rank? Same as the Einstein? Is it antisymmetric? Same rank as the Einstein, same symmetries, only now it's "trace-free on all the indices", meaning that C^a_{bad} = 0, and C^a_{acd} = 0, and C^a_{bca} = 0. The Einstein tensor satisfied the last of these properties already, by virtue of its symmetries, but R^a_{bad} = R_{bd} was the Ricci tensor, and R^a_{bca} was just minus the Ricci tensor. So what we have in effect done in forming the Weyl tensor is to take the Riemann tensor and subtract out the stuff that gives the Ricci. >If you only knew about four-dimensional geometry, you would be tempted >to guess that it was rank 2 like the Ricci tensor, since in four >dimensions Ricci and Weyl have the same number of pieces of >information. But things aren't that simple, as you can see if you >look at how many pieces of information the various tensors contain in >different numbers of dimensions: >Dimensions Riemann Ricci Weyl > 1 0 0 0 > 2 1 1 0 > 3 6 6 0 > 4 20 10 10 > >and so on. (Do I have all these numbers right? Something doesn't >look right to me. Anybody want to confirm them?) It looks right to me. For dimension n, the Riemann has n^2(n^2 - 1)/12 independent components. (I wrote my book on general relativity in part so I could easily look up this kind of thing!) The Ricci, being an n x n symmetric guy, has n(n+1)/2 --- except in dimensions 1 and 2, amusingly!! Perhaps those exceptions in dimension 1 and 2 were what was bugging Ted. In dimension 1 there is no curvature at all. In dimension 2 it turns out that R_{ab} = (1/2)R g_{ab}, so there is just one degree of freedom in the Ricci tensor. (By the way, this identity wreaks havoc for GR in 2 dimensions --- guess why!) I think there is something mystically important about Weyl and Ricci having the same number of independent entries in dimension 4, maybe something to do with the "duality" symmetry present in 4d, but I dunno. >It turns out that Weyl is a rank-4 tensor like Riemann, but it has >some extra conditions it has to satisfy, which is why it has fewer >bits of information hidden in it. Basically, Weyl is "trace-free," >which means that if you try to contract Weyl to a rank-2 tensor in the >same way you contract Riemann to get Ricci, you always get zero. >(Ricci is R^a_{bad}. Try to perform the same trick on Weyl, and you >get C^a_{bad}, which turns out to be identically zero.) Oh, whoops, Ted already said this stuff. Well --- it's like he said! Article 98861 (82 more) in sci.physics: From: john baez Subject: Re: General relativity tutorial Date: 8 Feb 1996 17:32:19 -0800 Organization: University of California, Riverside Lines: 62 NNTP-Posting-Host: guitar.ucr.edu In article 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 NNTP-Posting-Host: 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 "x-momentum in the x direction" |stuff really means. I don't have anything thrillingly insightful to |say about this except that, for a perfect fluid, each of these terms is |just the pressure! I will quit here for now, leaving Michael Weiss to |explain why that's true. I have a prosaic example to add to the explanation. You know those little toys they have, with a row of metal balls hung on threads? You pull the first little ball to the side, and let it go. The mechanics of the toy are such that the ball you let go comes to a stop in its resting position, and the other balls remain where they were (more or less), excepting the last one, which swings up on the other side. This is an example of momentum in a given direction being "transported" in that direction. It is of course by means of pressure of the balls in the series upon each other that the momentum is "transported" from one end to the other. When you have pressure, you can think of it as the collective effect of a lot of little particles being bounced back and forth. When you switch to other coordinate systems, the energy of the system is different, in a way which depends upon the distribution of velocities of the particles, even if the average is fixed. Keith Ramsay Article 99051 (39 more) in sci.physics: From: john baez Subject: Re: Tensors for twits please. Date: 9 Feb 1996 13:40:06 -0800 Organization: University of California, Riverside Lines: 181 NNTP-Posting-Host: guitar.ucr.edu [Oz knocks on the wizard's door, and then opens it. He sees the wizard juggling dozens of glowing Greek and Roman letters in complicated, everchanging patterns. The wizard catches sight of Oz, frowns, and the letters explode in a blinding white flash. He motions for Oz to come in.] So, how did you do on your homework? You were learning about tensors... right? What did you learn? In article 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 double-bladed axe and polishes it fondly.] >Hmm, lets just do H(0). This seems to imply c=0 to me. True, but folks usually call it H(u,v)^0 or H^0, since that c was a superscript. Anyway: >H(0)= >K^0(0,0)u^0v^0 + K^0(0,1)u^0v^1 + K^0(0,2)u^0v^2 + K^0(0,3)u^0v^3 + >K^0(1,0)u^1v^0 + K^0(1,1)u^1v^1 + K^0(1,2)u^1v^2 + K^0(1,3)u^1v^3 + >K^0(2,0)u^2v^0 + etc >K^0(3,0)u^3v^0 + ............................... + K^0(3,3)u^3v^3 Very good! You got it, despite my typo, which was a deliberate error thrown in to confuse you (just kidding). I would have said: H(u,v)^0 = H^0_{00}u^0v^0 + H^0_{01}u^0v^1 + H^0_{02} u^0v^2 + H^0_{03}u^0v^3 + H^0_{10}u^1v^0 + H^0_{11}u^1v^1 + H^0_{12} u^1v^2 + H^0_{13}u^1v^3 + ....... H^0_{30}u^3v^0 + H^0_{31}u^3v^1 + H^0_{32} u^3v^2 + H^0_{33}u^3v^3 + but the difference is purely notational, no big deal. >Well, I have to say I vastly prefer this notation to a 'do what you like >even if it's wrong' matrix method. I would probably like it even more if >I knew what I was doing, too. Although I have no wish to do it, it would >appear that this notion would allow things that produced tensors of any >rank from other tensors of any rank. I think a computer program would, >ahem, do it with less effort. Indeed the terse notation lets you immediately know what sums you would have to do to compute, say R^a_{bcd} u^a v^b w^d or R^a_{bcd}R_a^{bcd} or any such thing that might come up in general relativity. The beauty of the notation is that it tells you what sum you would have to do, in a very terse way, without actually making you do the sum. Actually DOING the sums is often best left to a computer, as you note. So now when I write stuff using tensor notation you will, at least in some sense, know what it means. At least you'll know what sum it describes. That leaves the more interesting aspect, namely what the hell we're doing it for. Article 98956 (32 more) in sci.physics: From: Oz Subject: Re: General relativity tutorial Date: Fri, 9 Feb 1996 08:56:26 +0000 Organization: Oz Lines: 82 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 second-order differential >equation, which I have avoided writing down, and you just solve that. Honestly, a red flag to a bull, "which I have avoided writing down". Any chance of seeing the form of this? After all one of the 'things' about GR seems to be the ability to calculate geodesics. We don't really have any idea as yet. Probably, on reflection never will, but still. [NB Cattle are colourblind so it doesn't matter what colour flag you wave. Bull is probably an appropriate word. Most bulls are more interested in 'other things' than chasing people.] A sort of moronic thought occured to me. (I never have problems with having these). If 4D space is curved in a GR-like fashion what do we really mean by this? Rather basic, I know, and we have lots of mathematical thingies to describe it but what are they actually describing? Maybe it would be a good idea to be clear about it? How about some thoughts like: 1) An unaccelerated body follows it's tangent vector. Since we have been told this is a geodesic, this would seem right. 2) The Riemann tensor 'derivation' we were given involves rotation of a tangent vector when parallely transported to another point in spacetime on a path that is not entirely on its geodesic. It would seem to me that we cannot say that anything is parallel to something else unless they start from the same point in spacetime because any other description will be path dependent (although I expect geodesics have some 'nice' properties I have excluded them above). The only way to get to another part of spacetime (as defined above) is to accelerate, no? This strikes me as meaning something, but I'm not sure exactly what. 3) It is valid to describe distances measured along an (arbitarily long) path. Of course you won't get the same distance if you choose a different path, but a distance as measured along a path is a valid distance. I think this is valid even if the path has accelerations along it. Indeed I suspect that this is the ONLY valid distance you can define. Hmmm, or is it simply the definition of a 'distance'. Note that this path need not be a geodesic. Is this right??? 3) In a torsion-free GR then an infinitesimal sphere of massless test points surrounding an (even more) infinitesimal speck of momentum-flow will behave with spherical symmetry under 'control' of the Ricci tensor. We note that this essentially says that the *local* spacetime in this infinitesimal volume can be considered flat. We also note that this is also the case for a torsion-free parallel transportation: it is in locally flat spacetime. This sort of implies that GR models a world where a suitably small piece of spacetime can be considered locally flat. On a philosophoical note one wonders if a black hole horizon is such a place for example, and if any extension of GR will remove this simplification resulting in a truly complex piece of mathematics that will require new tools to manage it. 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 NNTP-Posting-Host: guitar.ucr.edu [The wizard and Oz are talking.... the wizard says:] >>Yes, if we are told the geometry of spacetime, we can completely work >>out a geodesic if we know a point it goes through and its tangent vector >>(velocity) at that point. It satisfies a second-order differential >>equation, which I have avoided writing down, and you just solve that. [Excited, Oz replies:] >Honestly, a red flag to a bull, "which I have avoided writing down". >Any chance of seeing the form of this? WHAT?!?! [The wizard's eyebrows bristle and daggers of light shoot out from them.] You want to see THAT?? NO!!! I keep that equation in the room back there, behind that curtain, where I do my work. That's where I keep all my heavy-duty equipment. I can't let you go back there, because you don't know how to use that machinery and it's VERY DANGEROUS. There's really no telling what trouble you might get into if I let you back there! So, just remember this: never, NEVER go back there, especially when I'm not around. Now get out of here! The impudence!! [Sheepishly, Oz skulks off, not without a peek to see what's back behind the curtain.] Article 99005 (137 more) in sci.physics: From: Matthew P Wiener Subject: Applied Covectors and Symmetric Tensors Date: 9 Feb 1996 16:49:32 GMT Organization: The Wistar Institute of Anatomy and Biology Lines: 97 NNTP-Posting-Host: sagi.wistar.upenn.edu The story so far.... A nonzero covector (a,b,c) can be visualized as the plane ax+by+cz=1. This plane can be thought of as part of a dissection of R^3 into all the parallel planes, each labelled by the value ax+by+cz. Covectors and vectors are dual, with < (a,b,c) | (x,y,z) > = ax+by+cz, the label of the plane in the (a,b,c) dissection that the vector (x,y,z) reaches out to. One thing I didn't spell out before is that scalar multiples of (a,b,c) are found simply by multiplying the labels by the scalar in question. If you scale by 10, the 1-labeled plane is now 10-labeled, with the new 1-labeled plane .1 the way down. And I mentioned that gradients are really covectors, with grad f properly thought of as df = @f/@x dx + ..., where dx = "dual x" = (1,0,0) etc are the unit covectors. So let's start using them. The most common use of gradient is for a potential: force = - grad U. So this says that force is a covector all along. And if force is a covector, and force = d/dt (momentum), momentum too is a covector. You thought position/momentum duality was a quantum thing? No, it was part of classical mechanics, all along. You can see this in relativity too, where the 4-momentum has the inverse Lorentz transformation from the 4-position. What's amazing is that the covector dissection, in the right units, is just a de Broglie wave. Momentum is the inverse of wavelength, so a covector for very high momentum corresponds, as per the scaling above, to a dissection with the integral labelled planes packed very close together. The de Broglie p=h/l relationship just means that the momentum push is packed in the positive covector direction, and its magnitude is a count %%%of the covector planes with integral labels from here to here + h%%%. Simple, right? For p in the x direction, what happens to a wave function? If the position operator is multiplication by x, how about the momentum operator just being multiplication--of a covector 1-off push--by p: W(x+l)-W(x) @W p.W(x+l)-W(x) = h ----------- = h -- ? l @x This is one standard heuristic derivation in disguise, by the way, but in covector language it goes over much better. Now that you're convinced that quantum mechanics could have been figured out much faster had people back then known that momentum was a covector, let's see more about those classical covectors. I started out with a conservative force being the gradient of a potential: F=-dU. What if there's a constraint, some surface H(r)=0? The equilibrium can be found as follows. The constraint force, which you thought was the normal vector sticking out, is actually a tangent covector. Right away, you know this *has* to be, a priori, the only proper invariant description: perpendicularity is _not_ invariant in relativity, but parallelism _is_. The actual constraint force is thus l.dH, some scalar l. At equilibrium, the net force is zero: so F+l.dH=0 or equivalently dU=l.dH. Look familiar? Yes! It's a Lagrange multiplier! So with covectors firmly established, it is natural to expect physics to occur relating vectors with covectors. A simple example is Hooke's law: F=kx, some k. This is _not_ a vector equation. It only works out that way in the simplest of cases. In full generality, it is F=K.x, where F is the force covector, x is the displacement vector, and K is the _strain tensor_. So what's a tensor? In this case, just a linear map vectors to covectors. Mathematically, covectors are duals to vectors: they work by taking vectors to numbers. (See above: (a,b,c)--our R^3 planar dissection--sits and waits for (x,y,z) so as to get ax+by+cz.) So what we have here is something that takes a vector, and then sits and waits for another vector to show up, so as to get a number. In other words, we have a bilinear map K, sending pairs of vectors to numbers. That just takes a matrix to describe. In the Hooke case, K is symmetric and positive definite, and thus can be described as an ellipsoid. Geometrically, if you are given the displacement, starting at the ellipsoid's center and pointing out somewhere, find the cone with vertex the vector's far end and which is tangent to the ellipsoid. The tangency occurs in a single plane, which is the 1-labelled plane for the covector F=K.v. If the vector is short, scale it up first, and then compensate. Note that one can reverse the process. The usual Euclidean metric (where the bilinear map is the dot product) is just the unit sphere. And in the Minkowski case, one can do this using a hyperboloid. -- -Matthew P Wiener (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 NNTP-Posting-Host: pipe9.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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 off-diagonals, shear. However, we could always diagonalize it by a coordinate transformation to principle axes. Then we would find all information in it summarized in *4* numbers, giving expansion or contraction along the 4 principle axes, transforming our little hyperspherical test region into a hyper-ellipsoid. Up to scale factors these axes give the locally *best* coordinate system for curved spacetime, one that captures the local behavior most succinctly." "All coordinate systems are equal, but some are more equal than others". Article 99174 (134 more) in sci.physics: From: Oz (SAME) Subject: Re: General relativity tutorial Date: Sat, 10 Feb 1996 00:17:35 +0000 Organization: Oz Lines: 32 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4fgc3b$r3h@guitar.ucr.edu>, john baez writes >In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk >writes: > >>4) We don't seem to have any relativistic effects modelled into our >>version of GR yet. > >HUH? > Oh, dear. I must have made a *serious* booboo. Anyway it's past midnight and my apology for a brain isn't functioning very well. I will go away and commit hari-kari. Have a nice weekend. I will speak again when I'm sober. Hoc. I mean Hic. Oh, dear. Have a good weekend, from Oz. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 99133 (133 more) in sci.physics: From: Oz (SAME) Subject: Re: General relativity tutorial Date: Sat, 10 Feb 1996 00:23:58 +0000 Organization: Oz Lines: 40 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4fgff2$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$r3h@guitar.ucr.edu> john baez, baez@guitar.ucr.edu >replies: >>HUH? > >In this case, I think that the professor's answer, while correct, is a >little to concise. > >What I think he means to say here is, by using a metric with a diagonal >[-1,1,1,1], known as the "Lorentz metric", *all* of the special >relativistic effects are already included in the geometry. (Message timed at 12.15 at night) What!!!!!!! That does it all!!! Wow. Oh dear, a whole lotta questions. Whoever thought this up was, indisputably, a genius! Why didn't they tell us before? Why do they even *bother* with SR!!! This is *brilliant*!!! Frabjabjous! I will post again when sober. Many thanks for your post. Many thanks. I think Baez, atypically, has missed something ever so slightly important out of his description. Obvious to some, maybe. But important. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" (SAME) Subject: Re: General relativity tutorial Date: Sat, 10 Feb 1996 08:07:15 +0000 Organization: Oz Lines: 46 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4fgff2$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$r3h@guitar.ucr.edu> john baez, baez@guitar.ucr.edu >replies: >>HUH? > >In this case, I think that the professor's answer, while correct, is a >little to concise. Hey, I have managed to get Baez to do a post ENTIRELY in capitals. It's also completely devoid of content. It must carry a record Baez crank index at least. Er, unless of course it's a strange notation he has that encompasses the entire universe. Well I guess that "HUH?" is as good an explanation for it as any. It is at least in keeping with 99.99% of the population's explanation of the entire universe, so it's got that 'common touch' to it. Makes for a mighty short publication though. :-( > >What I think he means to say here is, by using a metric with a diagonal >[-1,1,1,1], known as the "Lorentz metric", *all* of the special >relativistic effects are already included in the geometry. Yeah. Braindead, that's me. I even bet that there could be a little 'c' somewhere that has been downgraded to a '1', I think that was mentioned waaaaaay up the thread near when it started, some seconds after the big bang. I *knew* the metric had been glossed over. What I don't quite see is how this tensor produces relativistic effects. I am sure it's quite simple and obvious to you tensor tyros, you have done a full and proper course. Can anyone spell it out in words of one syllable, oh well Ok then three syllables, how this happens? ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 99202 (131 more) in sci.physics: From: Oz Subject: Re: General relativity tutorial Date: Sat, 10 Feb 1996 08:46:16 +0000 Organization: Oz Lines: 81 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 In article <4fgg6e$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 second-order differential >>>equation, which I have avoided writing down, and you just solve that. > >[Excited, Oz replies:] > >>Honestly, a red flag to a bull, "which I have avoided writing down". >>Any chance of seeing the form of this? > >WHAT?!?! > >[The wizard's eyebrows bristle and daggers of light shoot out >from them.] > >You want to see THAT?? NO!!! I keep that equation in the room back >there, behind that curtain, where I do my work. That's where I keep all my >heavy-duty equipment. I can't let you go back there, because you don't >know how to use that machinery and it's VERY DANGEROUS. There's really >no telling what trouble you might get into if I let you back there! >So, just remember this: never, NEVER go back there, especially when I'm >not around. > >Now get out of here! The impudence!! > >[Sheepishly, Oz skulks off, not without a peek to see what's back behind >the curtain.] > In another dimension, far, far away ........ [Quiet, doleful music plays in the background.] [From Ducas "The Sorcerer's Apprentice of course: what else?] The aged retainer shuffles back to his hole in the wall, dragging his broom behind him over the solid milk quartz floor. From time to time he peers wistfully over his shoulder at the Great Wizard who ignores him and is busy curving space with a wave of his arms and producing black holes with a flick of his fingers. He arrives at his cramped, cold and damp cave in the corner, pulls the tattered curtain across the doorway and sits on his straw filled bunk. It's ichy. He is not so sure that getting sorcery lessons and food (food: ha!) for keeping the Great Wizards cave tidy is such a good deal as it seemed originally, but determines to soldier on. The Great Wizard watches his aged retainer close the curtain. "At last," he thinks, "he has gone, time for some proper work. Time to cut this simple child's stuff. Got to get up the wizard rankings.". [Music changes to Borodin: Night on a Bare Mountain: fortissimo.] The yelps from the Great Wizard and occasional fireballs wake the aged retainer from his reverie. He peers cautiously through one of the many holes in his curtain. The Great Wizard seems to be working on something terribly small that keeps slipping from his fingers. He just can't hold it in one place, every time he does it zooms away painfully taking a chunk of wizard with it. Fireballs erupt from the clear shielding the Great Wizard has thrown up around himself. "Oh well." thinks the aged retainer "It's not always that much fun being a Great Wizard, either.". [Music changes to Brahms lullaby] The aged retainer falls asleep on his itchy bunk. He dreams of creating black holes with a flick of his finger, but always ends up being dragged in and painfully turned into sorcerer's spaghetti. He dreams that he goes behind the curtain and is immediately filled with knowledge without any effort at all. If only he could remember it all in the morning. [Music fades] Dream on. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 99235 (129 more) in sci.physics: From: Jerry Freedman (SAME) Subject: Re: General relativity tutorial Date: 9 Feb 1996 17:06:03 GMT Organization: GTE Government Systems Corporation Lines: 8 NNTP-Posting-Host: 155.95.68.41 I am just a poor software type whose mathematical education was cut short and I don't really read this group much but I LIKE this thread, particularly Mr. Baez' contributions. If this, or something like it were captured, organized and printed Id buy it. Its the first time I have had any chance of comprehending this stuff so at the least, continue gentle people while I go back to polite lurking Jerry Freedman,Jr Article 99318 (127 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 10 Feb 1996 23:01:25 GMT Organization: University of California, Riverside Lines: 23 NNTP-Posting-Host: guitar.ucr.edu In article <4fgc3b$r3h@guitar.ucr.edu> baez@guitar.ucr.edu (john baez) writes: >In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk wri tes: > >>4) We don't seem to have any relativistic effects modelled into our >>version of GR yet. > >HUH? To clarify: we never did anything that WASN'T perfectly "relativistic". Not only did we never introduce any yucky "absolute rest", or any yucky split of spacetime into space and time, we didn't even introduce any yucky "global inertial frames"! So not only we have avoided the errors of Aristotelian and Newtonian mechanics, we have avoided the errors of special relativity. So we are doing general relativity. Perhaps you mean that I haven't said any cool stuff about black holes or the big bang? Actually we're almost ready for some of that fun stuff. Article 99319 (126 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 10 Feb 1996 23:03:12 GMT Organization: University of California, Riverside Lines: 92 NNTP-Posting-Host: guitar.ucr.edu In article <4fcvok$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 fiddle-faddle. No, I mean how local conservation of charge is an automatic consequence of Maxwell's equations! Start with div E = rho div B = 0 curl B = j + dE/dt curl E = -dB/dt and derive the equation for local conservation of charge, d rho/dt = - div j. That's a prototype of how you derive local conservation of energy- momentum from Einstein's equation. Einstein was not as dumb as most people think. He knew that one of the coolest features of Maxwell's equations is how conservation of charge is an automatic spinoff, and made damn sure something like this worked in general relativity. In fact, in late 1915 he first wrote down a wrong version of his equation which included only a piece of the Ricci tensor, and then ANOTHER wrong version that went (in modern notation) R_{ab} = T_{ab}. But by November 25th he corrected this to (in modern notation) R_{ab} - (1/2) R g_{ab} = T_{ab}, saying "With this step, general relativity is finally completed as a logical structure." >I still have a hunch the distortions of the test sphere into the ellipsoid >are captured by the Einstein part, not the leftovers, a la some kind of >principle axis construction. ^principal >In other words, I think you can't know the >volume dilation without knowing this also. Please take one more look at >the first paragraph of this B.S. (appended) and tell me if it rings a bell? It rings a bell; I know what you're thinking and I read what you wrote with interest the first time. I have been trying to better understand the separation of transformations of a sphere into dilations, stretching-squishings, and rotations, so as to see how it corresponds to the separation of the curvature into Ricci and Weyl. E.g., I thought, perhaps we don't need to worry about one of the 3 kinds of transformations of the sphere too much for some reason? But the stretching-squishings seem possible even in the absence of Ricci curvature --- that's what gravitational waves do when they go through empty space! So I don't hope to calculate THOSE using the Ricci tensor as you suggest. Instead, I hope to see why, even though it seems there are more sorts of stretching-squishings and rotations than dilations, the Weyl tensor has just as many degrees of freedom as the Ricci. And stuff like that. >How many numbers do you need to capture a general rotation of a body in >four space? Hmm, I didn't know we were talking about that, but the answer is (4 x 3)/2 = 6. To capture a general linear transformation of a body in 3-space it takes 3 x 3 = 9 numbers. 1 of these keeps track of dilation. 3 keep track of rotation. The remaining 5 keep track of stretching-squishings. (That's what I was worrying about.) For the mathematically macho, it's easiest to understand this using the Lie algebra gl(3). This splits up into: scalar multiples of the identiy (1 dim), skew-symmetric matrices (3 dim), and symmetric traceless matrices (5 dim). Article 99320 (125 more) in sci.physics: From: john baez (SAME) Subject: Re: General relativity tutorial Date: 10 Feb 1996 23:03:49 GMT Organization: University of California, Riverside Lines: 42 NNTP-Posting-Host: guitar.ucr.edu In article <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 NNTP-Posting-Host: guitar.ucr.edu In article <3bCWTNA6wwGxEwLh@upthorpe.demon.co.uk> Oz@upthorpe.demon.co.uk writ es: >If 4D space is curved in a GR-like fashion what do we >really mean by this? I told you this once a while ago, but you'll probably understand it a lot better now. Here's an operational definition of the sentence "spacetime is curved": we say spacetime is curved at the point p if, in a very small neighborhood of p, initially comoving geodesics accelerate relative to one another. I could make that more precise but actually I already did. Remember those coffee grounds? We set them up very carefully to follow "initially comoving geodesics", and then a little ball of them started rotating and changing shape. If, regardless of the velocity v of the central coffee ground at the point P, the ball does NOT start rotating or change shape, then space is flat at P. Moreover, every bit of information about the Riemann curvature tensor at P can be recovered from knowing how these balls change shape (for all possible velocities v). So we can quite rightly say: "Operationally, curvature of spacetime is just a way of talking about the relative acceleration of initially comoving nearby geodesics." And that's good, right, because curvature of spacetime is all about GRAVITY, and gravity is what you need to understand about WHAT HAPPENS IN FREE FALL. >1) An unaccelerated body follows it's tangent vector. Since we have been >told this is a geodesic, this would seem right. Yeah. More precisely: it follows a geodesic, which is a curve whose own tangent vector is parallel translated along itself. >2) The Riemann tensor 'derivation' we were given involves rotation of a >tangent vector when parallely transported to another point in spacetime >on a path that is not entirely on its geodesic. It would seem to me that >we cannot say that anything is parallel to something else unless they >start from the same point in spacetime because any other description >will be path dependent (although I expect geodesics have some 'nice' >properties I have excluded them above). The only way to get to another >part of spacetime (as defined above) is to accelerate, no? This strikes >me as meaning something, but I'm not sure exactly what. I really don't follow this. >3) It is valid to describe distances measured along an (arbitarily long) >path. Of course you won't get the same distance if you choose a >different path, but a distance as measured along a path is a valid >distance. I think this is valid even if the path has accelerations along >it. Indeed I suspect that this is the ONLY valid >distance you can define. Hmmm, or is it simply the definition of a >'distance'. Note that this path need not be a geodesic. Is this right??? WOW! YES! You seem to be seeing exactly why I would always scold you for talking about the distance between P and Q without specifying a path between P and Q! YES! The "arclength" or "proper time" along a path is well-defined both operationally and in our mathematical formalism for GR. Operationally: if the path is spacelike you can use rulers; if it is timelike you can use watches. Mathematically: the metric lets us compute the length of a tangent vector, and we integrate this length along a given path from P to Q to compute the length of the path. And YES! Distance between points is ONLY defined given a path between them; this is what distance means in GR. >3) In a torsion-free GR then an infinitesimal sphere of massless test >points surrounding an (even more) infinitesimal speck of momentum-flow >will behave with spherical symmetry under 'control' of the Ricci tensor. >We note that this essentially says that the *local* spacetime in this >infinitesimal volume can be considered flat. We also note that this is >also the case for a torsion-free parallel transportation: it is in >locally flat spacetime. This sort of implies that GR models a world >where a suitably small piece of spacetime can be considered locally >flat. If I understand you aright, yes. This is what they call the "equivalence principle". >On a philosophical note one wonders if a black hole horizon is >such a place for example, and if any extension of GR will remove this >simplification resulting in a truly complex piece of mathematics that >will require new tools to manage it. 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 NNTP-Posting-Host: pipe10.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: 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 batter-dipped deep-fried ball of eggplant? That's right, little deep-fried balls of green chilies. Then, I recently stopped in a little place here while I was working called the "Punjabi Deli", a hole in the wall whose chief product lines are Indian audio cassettes, Indian Newspapers, Indian food, and motor oil (I think most of their clientele is Punjabi cabbies). Their food is only mildly spicy (possibly cooked in the motor oil), but they had a little bowl of whole green chilies out on the counter. Appetizers? Candy? I didn't ask what they were for. Maybe toothpicks. You are right, Baez is a pretty nice guy. I seem to keep testing it. I am getting overwhelmed keeping up with the volume here... I'm behind to about the middle of last week right now, if you want to know the truth. I guess I'll just have to give up sifting through sci-physics to find threads to make snide comments in, like "haven't you ever heard of a little thing called a correlation, buddy?", and getting flamed by the irate; enjoyable as that activity is. Next I'll have to give up Monday night wrestling! About that cabana... The prices remind me of a famous tourist destination in the north of India, where beautiful houseboats on the lake can be rented for a song, most especially after the guerillas kidnapped and killed that American tourist... Still, he was walking in the mountains. The guerillas *never* come down to the houseboats. So they say. -- Ed Green / 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 NNTP-Posting-Host: guitar.ucr.edu In article <4ff2d8$1g1u@info.estec.esa.nl> Grai w rites: >baez@guitar.ucr.edu (john baez) wrote: >> >>See some stuff Matthew Wiener wrote recently, for a more geometrical >>explanation of the difference between vectors and covectors. >Just a question on nomenclature. I always used the terms vector and >1-form. Is covector a new, more general entity applicable to other >vector spaces, or just exactly the same thing as a 1-form. If you hand me any vector space I could call its elements vectors and call the elements of the dual vector space "covectors". It's certainly interesting to think about this. But I wasn't really trying to be so general here... I was mainly talking about the tangent space, whose elements I call "tangent vectors" or (for short) "vectors", and its dual space, the cotangent space, whose elements I call "cotangent vectors" or (for short) "covectors". If I have a tangent vector at each point of spacetime I say I have a "vector field", while if I have a cotangent vector at each point of spacetime I might say I have a "1-form". Article 99130 (17 more + 1 Marked to return) in sci.physics: From: Steve Carlip (SAME) Subject: Re: General relativity tutorial Date: 9 Feb 1996 22:28:01 GMT Organization: University of California, Davis Lines: 69 NNTP-Posting-Host: dirac.ucdavis.edu X-Newsreader: TIN [version 1.2 PL2] john baez (baez@guitar.ucr.edu) wrote: : In article <4f33qi$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 understand---I'll explain below.) Start with your ball of coffee grounds, with an initial four-velocity u^a. (In the rest frame, u^0=1 and the rest of the components are zero.) If you look at the configuration slightly later, it will typically have changed in three ways. First, it may have expanded or contracted (its volume may have changed). Second, it may have twisted. Third, it may have sheared---that is, become distorted from a sphere to an ellipsoid without changing volume. The Weyl tensor (in empty space) gives the rate of change of the shear. Specifically, you can describe shear by a three by three symmetric matrix, usually denoted by a lower-case sigma (or \sigma to LaTeX users). The three eigenvectors of \sigma give three spatial axes, and the eigenvalues give the rate of expansion along each axis. The fact that \sigma is traceless means that the sum of the rates of expansion is zero, i.e., the total volume is remaining constant. In empty space, if your ball of coffee grounds has no initial shear, rotation, or expansion, the rate of change of \sigma is given by the Weyl tensor contracted twice with u. (If you want that in symbols, I mean C^a_bcd u^bu^d, or in index-free notation, C(*,u,*,u).) This means that the Weyl tensor is the thing gravitational radiation detectors are designed to measure. A laser interferometer detector like LIGO or VIRGO (now under construction) is just a very large, very accurate interferometer with two perpendicular arms. If the Weyl tensor changes, one arm will expand while the other one contracts, changing the relative lengths and thus the interference pattern. Now, the part I don't understand. The Weyl tensor really splits into two parts, and "electric" part (with five components) and a "magnetic" part (also with five components). The contraction C(*,u,*,u) that I described above is the electric part, in the rest frame of the coffee grounds. Like ordinary electric and magnetic fields, the electric and magnetic components of the Weyl tensor mix under coordinate changes. But it would be nice to have a direct geometrical picture of the magnetic part. Does anybody know one? (Presumably it's hidden in what's called the Newman-Penrose formalism, but I've never learned that very well.) (The obvious guess is that it has to do with rotation, but this is wrong---rotation isn't produced directly by geometry, but only indirectly by expansion and shear.) Steve Carlip 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 NNTP-Posting-Host: 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 index-free notation, C(*,u,*,u).) I see, so it's really that simple! My problem was that I didn't see why this gives us a 3x3 *symmetric* traceless matrix, describing only the *shears* and not the rotations of our little ball of coffee grounds. Now I see. Let me just say it rather quickly so the experts like Steve can see it's finally penetrated my thick skull; later I may attempt more of a "pop" exposition of this stuff. Working in the rest frame of the center of the little ball of coffee, so that u = (1,0,0,0) and the metric is the Minkowski one (at that point), the 2nd derivative of the shape of the ball is given by R^i_{0j0}, and when we subtract out the part of this that describes the 2nd derivative of the volume we get C^i_{0j0}. (Here i,j = 1,2,3.) We may harmlessly lower an index and get C_{0j0i}, using the convention that the top index goes to the back. The point is: this is *symmetric* since the Weyl tensor, like the Riemann, has C_{abcd} = C_{cdab}. >Now, the part I don't understand. The Weyl tensor really splits into >two parts, and "electric" part (with five components) and a "magnetic" >part (also with five components). The contraction C(*,u,*,u) that I >described above is the electric part, in the rest frame of the coffee >grounds. Like ordinary electric and magnetic fields, the electric and >magnetic components of the Weyl tensor mix under coordinate changes. >But it would be nice to have a direct geometrical picture of the magnetic >part. Does anybody know one? (Presumably it's hidden in what's called >the Newman-Penrose formalism, but I've never learned that very well.) Perhaps the magnetic part could be understood in terms of the shear of a ball of tachyonic coffee grounds, but I bet that's not quite the answer you're looking for. >(The obvious guess is that it has to do with rotation, but this is >wrong---rotation isn't produced directly by geometry, but only >indirectly by expansion and shear.) That's what I hadn't been able to see, though I suspected it. Article 99406 (16 more + 1 Marked to return) in sci.physics: From: Oz (SAME) Subject: Re: General relativity tutorial Date: Sun, 11 Feb 1996 09:13:33 +0000 Organization: Oz Lines: 42 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 relativistic-like effects. Other posters have pointed out it is due to the metric. (-,+,+,+). Here we arrive at the situation where I should have already done 10,000 examples of tensors so I would immediately see this. I bet there is a lecture just on this bit. It's all to do with it defining a hyperbolic geometry, posters have said. I just *knew* that a "metric" meant more than a bundle of signs. But it doesn't say this to me. It's just a tensor. Could someone walk me through (real easy) how I get to see this metric as representing a hyperboloid space. I think this is VERY VERY important to follow this clearly. It musn't be glossed over. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 99414 (15 more + 1 Marked to return) in sci.physics: From: Oz (SAME) Subject: Re: General relativity tutorial Date: Sun, 11 Feb 1996 13:12:20 +0000 Organization: Oz Lines: 78 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 co-ordinates and a length of 1, or .000001 for that matter. Or looking at it the other way round, if one co-ordinate is eversobig and the length is 1, the other has to be eversobig positive or negative. Needs a bit more thought, dunnit? ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" From: john baez Subject: Re: General relativity tutorial Date: 11 Feb 1996 19:41:57 GMT Organization: University of California, Riverside Lines: 69 Distribution: world NNTP-Posting-Host: guitar.ucr.edu In article 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 energy-momentum curves spacetime. Remember, if we stop hiding the gravitational constant k, Einstein's equation is G_{ab} = 8 pi k T_{ab} (Of course c is still 1 here.) And, the cool part, while the geometry of Minkowski spacetime is boringly flat, it still gives you all the usual stuff about special relativity. That's how special relativity should be taught: not as a melange of "relativistic effects", but as the natural outgrowth of Lorentzian geometry. >This is *brilliant*!!! Frabjabjous! Yes. >I will post again when sober. Many thanks for your post. Many thanks. I >think Baez, atypically, has missed something ever so slightly important >out of his description. Obvious to some, maybe. But important. Yes, very important; for some reason I assumed you knew it. Article 99483 (10 more) in sci.physics: From: john baez Subject: Re: Minkowski Metrics for Tensor Twits Date: 11 Feb 1996 12:40:19 -0800 Organization: University of California, Riverside Lines: 37 NNTP-Posting-Host: guitar.ucr.edu In article 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 NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 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 co-ordinate system is not required, so we have to distinguish between a co-ordinate system and a metric even though they do look very similar. So what's the difference between a metric(theta,phi) and a co-ordinate system(theta,phi)? Well the only thing I can see is that metric(theta,phi) is a relative thingy. A co-ordinate(theta,phi) is wrt a set of fixed axes. Never properly discussed, but there you go. >This means that its length squared is r^2 sin^2(phi), or in other words, >its length is r sin(phi). That's supposed to make sense. You do >remember your spherical coordinates, right? [Pained, worried look. >Helpful review of trig formulas flashes by on the sphere.] Hey, spherical geometry at this level is kid's stuff. Even I remember it. I have only regressed to age 17ish, this stuff you learn at 14!! I have to say that it's a much easier formalism than the old fashioned way. If there are any schoolteachers listening in (particularly UK ones) could they comment if this stuff (tensors) is done at school nowadays? Even if they aren't called tensors. Presumably we could pick some other strange metric that isn't orthogonal, and all that would happen is that we would get some cross terms coming into the metric and probably they would all be nasty expressions. But hey, who cares, plug it into the computer and out comes the answer. >But you wanted to know the Riemann tensor of the sphere, not the metric >or the Ricci scalar! Well, okay, so we take the metric and feed it into >this machine... [scurries behind a curtain; loud banging noises ensue, >followed by a deafening explosion and a puff of smoke; returns somewhat >blackened but smiling]... and it computes the Riemann tensor for us. >Don't worry about that machine in the other room just yet, someday you >too may learn to use it... or maybe not. > >So, the Riemann tensor has lots of components, namely 2 x 2 x 2 x 2 of >them, but it also has lots of symmetries, so let me tell just tell you >one: > >R^2_{121} = sin^2(phi) >Its component in the a direction has changed a bit, say > > -epsilon^2 R^a_{bcd} " > 2 rotation in this direction (only) -epsilon R {starting direction,second direction,tangent direction} Ok, so now we see why it has 2^4 components. so R^2_{121}, 1=theta, 2=phi 2 phi -epsilon R = sin^2(phi) {theta, phi, theta} ahhhhh No, no, no, no,no. This really will not do. How do you expect us to get a proper grasp, nay even a basic vague concept, if we can't even see how to work out one element of what is likely the simplest non-trivial Riemann tensor. I know, I don't like it either, but it's no good fudging it. It's just gotta be done. You just have to put guards on the machine, hand out hard hats, dark goggles, and make sure we all stay clear, keep our fingers out of the way, and just watch. I have to say, in my heart of hearts, that I don't really like this definition very much. Not really. For one thing each leg is not epsilon long. I would rather prefer to see it as going in a little square and finding I was *not* back where I started. Then I would have a little path back to where I *had* started from which would be a vector I could 'easily calculate' and would (I think) give me a measure of curvature. I also rather suspect that it would give me the same measure of curvature. Not the actual vector itself, of course, you would have to fiddle with it a bit to get the curvature. >Let me describe the Ricci scalar, R, in 2d. This is positive at a given >point if the surface looks locally like a sphere or ellipsoid there, and >negative if it looks like a hyperboloid --- or "saddle". If the R is >positive at a point, the angles of a small triangle there made out of >geodesics add up to a bit more than 180 degrees. If R is negative, they >add up to a bit less. > >For example, a round sphere of radius r has Ricci scalar curvature R = >2/r^2 at every point. [With a click of his fingers, a sphere of radius >r appears on it, with a small triangle drawn on it, edges bulging >slightly.] Well, I don't like to mention this, but nobody has mentioned a Ricci SCALAR before. I take it we are still talking about a spherical shell too. Anyway, not wanting to quibble too much, lets see what we can find out from this. Well, the first thing that strikes you is that if r is very small then R becomes very large. As r tends to infinity, R rends to zero so one expects R of an infinite plane to be zero, which is not unexpected. One has the irresistable urge to throw down a tiny circle on the spherical shell and measure it's area, and feel that the Ricci scalar reflects the ratio of the (difference between the area of the cap and the area of the circle) and the circle. Hmmm. 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 Summary: Expires: References: <4fdfvv$q2k@guitar.ucr.edu> Sender: Followup-To: Distribution: world Organization: University of California, Riverside Keywords: 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. Torsion-free: this is a way of making precise the notion of "no rotating". >Now G.Wiz insists at this stage that a >co-ordinate system is not required, so we have to distinguish between a >co-ordinate system and a metric even though they do look very similar. WHAT????? A metric is just a gadget g that eats two tangent vectors v and w and spits out a number g(v,w)! How does that look like a coordinate system, which is a gadget that assigns coordinates to every point?? Look: if I hand you a perfectly round sphere of a certain size, you know the metric: you know how to compute dots products of tangent vectors, you know how to compute lengths of tangent vectors and angles between them, etc.. This is very different than giving you an abstract sphere with a coordinate system, e.g. a longitude-latitude grid or some other weird coordinate system! Knowing a metric does NOT tell you a coordinate system. Knowing the coordinate system does NOT tell you a metric. The metric is a very physical aspect of spacetime: in GR, it completely describes the gravitational field. A coordinate system is a very unphysical thing that we lay down on a patch of spacetime to help us do calculations. Given a metric, we can work out its components in ANY coordinate system. Here I have taken the round sphere of radius r and worked out the components of the metric in ONE PARTICULAR coordinate system, namely, spherical coordinates. Remember, FIRST my sphere with its metric appeared: "With a click of his fingers, a sphere of radius r appears on it, with a small triangle drawn on it, edges bulging slightly." And THEN I decided to start computing things... "Well, for this we need some coordinates. Let's use the usual spherical coordinates theta, phi.... [Coordinate lines appear on the sphere, which floats back and forth playfully.] When we write something like g_{ab} or R^a_{bcd}, the indices will go from 1 to 2, with "1" corresponding to theta and "2" corresponding to phi. For example, the components g_{ab} of the metric in this coordinate system are: g_{11} g_{12} = r^2 sin^2(phi) 0 g_{21} g_{22} 0 r^2 " So hopefully you see how confused you were when you wrote: >So what's the difference between a metric(theta,phi) and a co-ordinate >system(theta,phi)? Well the only thing I can see is that >metric(theta,phi) is a relative thingy. A co-ordinate(theta,phi) is wrt >a set of fixed axes. Never properly discussed, but there you go. >Presumably we could pick some other strange metric that isn't >orthogonal, and all that would happen is that we would get some cross >terms coming into the metric and probably they would all be nasty >expressions. But hey, who cares, plug it into the computer and out comes >the answer. Note, it's not the metric itself which is or is not "orthogonal" --- which I guess is your new terminology for "diagonal" --- instead, its the COMPONENTS of the metric IN A PARTICULAR COORDINATE SYSTEM which are, or aren't, diagonal. We could take the usual round metric on the sphere and work in some screwy wiggly coordinate system, and then its components in that coordinate system would not be diagonal. >>But you wanted to know the Riemann tensor of the sphere, not the metric >>or the Ricci scalar! Well, okay, so we take the metric and feed it into >>this machine... [scurries behind a curtain; loud banging noises ensue, >>followed by a deafening explosion and a puff of smoke; returns somewhat >>blackened but smiling]... and it computes the Riemann tensor for us. >>Don't worry about that machine in the other room just yet, someday you >>too may learn to use it... or maybe not. >>So, the Riemann tensor has lots of components, namely 2 x 2 x 2 x 2 of >>them, but it also has lots of symmetries, so let me tell just tell you >>one: >> >>R^2_{121} = sin^2(phi) By the way, the above is only good for a unit sphere. For a sphere of radius r, which is what I really had, we have R^2_{121} = sin^2(phi)/r^2 A small sphere is more curved! >No, no, no, no,no. This really will not do. How do you expect us to get >a proper grasp, nay even a basic vague concept, if we can't even see how >to work out one element of what is likely the simplest non-trivial >Riemann tensor. I know, I don't like it either, but it's no good fudging >it. It's just gotta be done. You just have to put guards on the machine, >hand out hard hats, dark goggles, and make sure we all stay clear, keep >our fingers out of the way, and just watch. NO!!!! You must NEVER, NEVER go back into that room where I actually compute things. The machines are VERY dangerous. If they could only cut off your fingers, frankly I wouldn't mind; I'd let you go in. But they can destroy your very soul! I was AFRAID you'd start wanting to go back there. [Wizard pauses, frowns, fingers his beard fretfully as he ponders what to do.] Let me show you what I mean. Hold on a second. [He walks over to the black curtain and slips behind it, motioning for Oz to keep his distance. A click is heard and then a long, high creaking noise as of a rusty door opening. Then Oz some thumping around and a loud clang as of a door slamming shut. After some more noise, the wizard lifts the curtain and rolls out a stretcher. Oz is shocked to find on the stretcher a human figure completely covered with white crystals of ice.] This, my friend, is what I meant. See this poor fellow? He is frozen stiff. The worst thing is, he's still alive under all that ice! Do you want to know how this fellow got that way? [Oz nods, getting over his shock and gradually moving closer towards the stretcher.] This was once a student of mine, like you, eager to learn general relativity. He was an excellent student, much better than SOME, and he had progressed to the point where he was writing code to do numerical simulations of Einstein's equation. He was trying to figure out what happened when two black holes collide --- a problem, by the way, that is still not fully understood. Anyway, he noticed a lot of problems. When he reduced the mesh size --- never mind, this is just some jargon --- sometimes his answers seemed to converge, other times not. He asked me about it so I suggested that he read a bit on numerical analysis. Oh, had I only known! [The wizard pauses sadly a moment.] To understand the numerical analysis he realized he needed to learn a bit of analysis. After all, how could you compute the answer to something if you weren't sure the solution existed in the first place?? Pretty soon he was quite an expert on existence and uniqueness for nonlinear hyperbolic PDE. He studied Sobolev spaces, and energy bounds, and the work of Choquet-Bruhat.... But as he did he noticed something funny happening. Occaisionally he would feel a slight chill. He disregarded it and kept on working, delving ever more deep into nonlinear analysis. He lost interest in his original goal of simulating black hole collisions. Proving existence of solutions to equations seemed much more interesting than actually solving them. After a while he noticed frost forming on his glasses. He just wiped it off and kept on proving theorems. Unfortunately he failed to notice the icicles growing on his desk.... By the time we found him, it was too late. He was frozen solid, but still thinking about existence and uniqueness of solutions of nonlinear PDE.... And here he is, still that way, in a condition of... rigor mortis. [Oz leaned forwards and touched the frozen figure with his finger. He felt a strange longing, but also a chill, which seemed to seep up his finger and into his heart.] DON'T DO THAT!!!!! [The wizard raised his staff and aimed it at Oz, sending a fireball at him, and wheeled the stretcher away from Oz.] BACK!!!! Little do you know the dangers!!!! I am protected from the infectious chill by many magic spells, but you are not. Oh, you fool! Let me put this back into the vault. Stay there. [He wheels the stretcher into the back room again, and Oz again hears the clanging of a great metal door being slammed shut. The wizard then reappeared....] So, you may think it a little thing, a trifling thing, to learn how I calculated the Riemann tensor of a sphere, but I assure you it is not. Very few know that secret. And if you learn that, you may be irresistably drawn to mathematics, and thence towards RIGOR, and you too may, like my poor student, perish in the icy splendor thereof. [Oz, abashed, decides not to press the point. But after a long pause, he starts thinking about the Riemann tensor of the sphere again... and says:] >I have to say, in my heart of hearts, that I don't really like this >definition very much. Not really. For one thing each leg is not epsilon >long. I would rather prefer to see it as going in a little square and >finding I was *not* back where I started. Then I would have a little >path back to where I *had* started from which would be a vector I could >'easily calculate' and would (I think) give me a measure of curvature. I >also rather suspect that it would give me the same measure of curvature. >Not the actual vector itself, of course, you would have to fiddle with >it a bit to get the curvature. Hmm. I'm not sure what you are saying. First of all, as I said, the LENGTH of the edges of the "little square" plays no role in the definition of the curvature. To compute R^2_{121} 1 = theta, 2 = phi we simply take the vector in the theta direction at a point P, parallel translate it from P over to the point whose theta coordinate is epsilon more, then over to the point whose phi coordinate is epsilon more, then over to the point whose theta coordinate is epsilon less, and then back where we started, and we see how much it now points in the phi direction. Then we divide by epsilon^2, throw in a minus sign for good luck, and take the limit as epsilon goes to zero. There are various things about the above paragraph which are a wee bit subtle and may confuse you... but anyway, I'm just parrotting the recipe for curvature, and if this is confusing, that means perhaps you didn't quite understand the original definition of curvature. Which is nothing to be ashamed of, because often things look simple in the abstract and then mysteriously become confusing in any concrete special case. [Oz picks up some old notes and reads them:] >>Let me describe the Ricci scalar, R, in 2d. This is positive at a given >>point if the surface looks locally like a sphere or ellipsoid there, and >>negative if it looks like a hyperboloid --- or "saddle". If the R is >>positive at a point, the angles of a small triangle there made out of >>geodesics add up to a bit more than 180 degrees. If R is negative, they >>add up to a bit less. >> >>For example, a round sphere of radius r has Ricci scalar curvature R = >>2/r^2 at every point. [With a click of his fingers, a sphere of radius >>r appears on it, with a small triangle drawn on it, edges bulging >>slightly.] >Well, I don't like to mention this, but nobody has mentioned a Ricci >SCALAR before. WHAT?????????? [The wizard, obviously still stressed from the previous incident, goes ballistic. He waves his staff about and shoots fireballs in all four directions of the compass, cursing with anger.] Listen here, Oz! I keep careful notes on EVERYTHING I EVER TELL YOU, so don't say I never mentiond the Ricci scalar. Here's what I said, and I quote.... [He ruffles around on his desk through enormous stacks of yellowing papers, pauses, scratches his head, and then yanks out one from the middle of the very tallest pile.] Ahem: "7. The EINSTEIN TENSOR. The matrix g_{ab} is invertible and we write its inverse as g^{ab}. We use this to cook up some tensors starting from the Riemann curvature tensor and leading to the Einstein tensor, which appears on the left side of Einstein's marvelous equation for general relativity. We will do this using coordinates to save time... though later we should do this over again without coordinates. This part is the only profound and mysterious part, at least to me. Okay, starting from the Riemann tensor, which has components R^a_{bcd}, we now define the RICCI TENSOR to have components R_{bd} = R^c_{bcd} where as usual we sum over the repeated index c. Then we "raise an index" and define R^a_d = g^{ab} R_{bd}, and then we define the RICCI SCALAR by R = R^a_a The Riemann tensor knows everything about spacetime curvature, but these gadgets distill certain aspects of that information which turn out to be important in physics. Finally, we define the Einstein tensor by G_{ab} = R_{ab} - (1/2)R g_{ab}." Now GET OUT! How do you expect to become a sorcerer at this rate? [Oz skulks out, thinking the old fellow must have drank too much coffee today... "What an old fart," he mutters.] Article 99503 (40 more) in sci.physics: From: Edward Green Subject: Re: Tensors for twits please. Date: 11 Feb 1996Article 99503 (40 more) in sci.physics: From: Edward Green Subject: Re: Tensors for twits please. Date: 11 Feb 1996 18:07:05 -0500 Organization: The Pipeline Lines: 55 NNTP-Posting-Host: pipe11.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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?). Organization: The Pipeline Lines: 55 NNTP-Posting-Host: pipe11.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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 "1-form". Oh... so that's what a 1-form is... so much nomenclature, more than concepts even! I've got a little list: tangent/cotangent, direct/dual, covariant/contravariant... And don't forget the 1-forms... [Q]: Am I right that through an historical quirk we say tangent vectors are "covariant" under coordinate transformations while cotangent vector are "contravariant"? Are we that fortunate? :-) Because, if Y is a (coordinate representation of a) tangent vector and Y' = TY our transformed representation, and if X a (coordinate representation of a) covector, then we must have: X' = XT^(-1) . That way; X'.Y' = X'T^(-1)TY = X.Y i.e., scalars are invariant (I am thinking in the bad old evil matrix notation... I will get this index stuff down presently) So it seems reasonable that since the inverse transformation appears next to the cotangent vector, it "contravaries"... Or have I stated this backwards and created confusion? (As usual). -- Ed Green / 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 Organization: The Pipeline Lines: 72 NNTP-Posting-Host: pipe10.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: The Pipeline v3.4.0 '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 ! >Off-hand, I have not seen it spelled out anywhere else >as this, but I assume it was once standard knowledge. Some relatively useful ways of looking at things seem to have fallen off the standard technical curriculum at some point. There used to be such a subject as "spherical trigonometry" once upon a time, did there not? Well I never met it, and though I do not think it is the ne-plus-ultra of sophistication, it probably could be useful to know about. For example, the kind of Stokes theorem we noticed for parallel transporting a vector around the surface of a sphere; rotation is proportional to the enclosed area (I assume the ambiguity in the enclosed area gives us the compliment w.r.t. pi); must be a factoid from this area. -- Ed Green / 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 Lines: 80 NNTP-Posting-Host: sagi.wistar.upenn.edu In-reply-to: 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 "1-form". >Oh... so that's what a 1-form is... so much nomenclature, more than >concepts even! I've got a little list: > tangent/cotangent, direct/dual, covariant/contravariant... >And don't forget the 1-forms... The standard examples of a vector field on a manifold (for example, spacetime) are sums of the form A.@/@x + B.@/@y + .... The standard examples of a covector field are sums P.dx + Q.dy + .... Are these different? Most assuredly! If you change coordinates, you get different results. Consider X=X(x,y,...), Y=Y(x,y,...), .... Then for any function f on the manifold (meaning f(x,y,...)=f(X,Y,...) and so on in coordinates): @f @f @X @f @Y @f @X @f @Y @f A.-- + B.-- + ... = A.(--.-- + --.-- +...) + B.(--.-- + --.-- +...) + ... @x @y @x @X @x @Y @y @X @y @Y @X @X @f @Y @Y @f = (A.-- + B.-- + ...).-- + (A.-- + B.-- + ...).-- + ... @x @y @X @x @y @Y In contrast, @x @x @y @y P.dx + Q.dy + ... = P.( -- dX + -- dY + ...) + Q.( -- dX + -- dY + ...) + ... @X @Y @X @Y @x @y @x @y = (P.-- + Q.-- + ...).dX + (P.-- + Q.-- + ...).dY + .... @X @X @Y @Y The assignment of n-tuples, one for each coordinate system, such that they relate under a change of coordinates as in the first expression above, is called a "contravariant" vector field. If they relate under a change of coordinates as in the second expression above, they are called a "covariant" vector field. The @/@x, @/@y, ... form a basis for the contravariant vectors, while dx, dy, ... form a basis for the covariant vectors. This is old-style talk. New style talk is to call the former vectors, and the latter covectors. Covector fields are also called 1-forms. >[Q]: Am I right that through an historical quirk we say tangent vectors >are "covariant" under coordinate transformations while cotangent vector are >"contravariant"? Exactly backwards. (Although, as you could have guessed, there are people who use the backwards convention.) @/@x strikes out a tangent direction, along the direction of increasing x coordinate. Note that I am not assuming that x is *the* x coordinate. Rather, I am assuming that we have coordinates on our manifold, meaning a map from some portion of our manifold to a vector of R^n values, and the first coordinate is x=x(p), the second coordinate is y=y(p), etc. If I fix a point p0 on the manifold, with coordates x0,y0,..., and then I freeze the y0,z0,... coordinates but let the x vary, I get a curve that passes through p0. @?/@x, to be evaluated at x0, perfectly captures the notion of tangent in the x direction at p0. In contrast dx pushes, not in the x direction, but perpendicular to the other tangent directions @/@y, @/@z, .... The picture of covectors as a hyperplane dissection still applies here: the 0-labelled hyperplace is the one containing @/@y, @/@z, ...., and the 1-labelled hyperplane is parallel to this but an infinitesimal distance dx over. -- -Matthew P Wiener (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 Lines: 83 NNTP-Posting-Host: pipe10.nyc.pipeline.com X-PipeUser: egreen X-PipeHub: nyc.pipeline.com X-PipeGCOS: (Edward Green) X-Newsreader: 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 stress-energy tensor is non-zero. G = T tells us how the drumhead has to respond there (well partially... ignore this qualification for a moment). But... what tells us the shape of the rest of the drumhead? G = 0. Well, that's part of the story. I suppose any value for the Riemann tensor consistent with G = 0 is ok then? But how much does this limit us? Someone said there were "gravitational wave solutions". I assume these are analogous to electromagnetic waves... source-free, we can take any superposition of them, and have a source-free solution. So are these the only ambiguity? How far can we push the drum head analogy? We are also tempted to notice that the drumhead stretched by the pencil can vibrate, and equate these with the gravitational waves (except that in space time, nothing ever really happens... it's timeless, changeless, etern... whoops... I hear a rhapsody... anyway space vibrates, spacetime just is, but ignore this. So on the drumhead, when we specify "there are no waves" we then have a unique solution for the shape of the head poked from underneath by the pencil. Is the same sort of thing true in GR? Is there a "base" solution corresponding to the undistrubed drumhead, shaped only by nearby mass distributions? What *is* the equation determing the "propagation" of the Riemann in empty space time (T = 0)... is "G = 0" a sufficient description? Does this tie the Riemann down enough so that "the answer is specified up to a linear superposition of gravity waves", whatever that might mean? The Weyl is supposed to have 10 independent components. So another way of asking this is, are all ten of these tied down by picking an element out of the gravity wave Hibert space... (zap!... sorry, I have no idea if they form a Hilbert space...) Actually, Oz (are you still there?), maybe this is what you were getting at... excuse me, what "Bert" was gettin at by "Oh, it might be like a taut string as wot sumone av plucked sumwere far away. Its dead flat until the triangular wave comes flying past." But what is this dern "tidal force" thing some of these guys are talking about... there seems to seems to be some feeling this should be associated with the Weyl too... seems to mess up this picture a bit... this is due to local mass distribution... Hey! Here comes a passing expert! Let's ask HIM!!! (Volunteers?) -- Ed Green / 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 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 >stress-energy tensor is non-zero. G = T tells us how the drumhead has to >respond there (well partially... ignore this qualification for a moment). >But... what tells us the shape of the rest of the drumhead? I think we just gotta measure it. Keeps the experimentalists in work. >G = 0. Well, that's part of the story. I suppose any value for the >Riemann tensor consistent with G = 0 is ok then? But how much does this >limit us? Someone said there were "gravitational wave solutions". I >assume these are analogous to electromagnetic waves... source-free, we can >take any superposition of them, and have a source-free solution. G=0 would surely correspond to a situation where space bending was due to another cause, and anyway outside of our (observable?) volume. Unless it's just a local G=0. >So are these the only ambiguity? How far can we push the drum head >analogy? We are also tempted to notice that the drumhead stretched by the >pencil can vibrate, and equate these with the gravitational waves (except >that in space time, nothing ever really happens... it's timeless, >changeless, etern... whoops... I hear a rhapsody... anyway space >vibrates, spacetime just is, but ignore this. Yes, but we can still describe dx/dt, dx/dy and so on. So it doesn't stop us working up some messy maths to keep the mathematicians in work. And that's before we start projecting things onto planes, onto curves even, and oh dear, do we really want to go through with this? >So on the drumhead, when we specify "there are no waves" we then have a >unique solution for the shape of the head poked from underneath by the >pencil. Is the same sort of thing true in GR? Is there a "base" solution >corresponding to the undistrubed drumhead, shaped only by nearby mass >distributions? What *is* the equation determing the "propagation" of the >Riemann in empty space time (T = 0)... is "G = 0" a sufficient >description? Does this tie the Riemann down enough so that "the answer is >specified up to a linear superposition of gravity waves", whatever that >might mean? Wow Ed, you have lost me here. I dunno, you are way past me. I just hope all this stuff is linear so we can just add curvatures or whatever in a simple way. Hahahaha, ...... ha? >The Weyl is supposed to have 10 independent components. So another way of >asking this is, are all ten of these tied down by picking an element out >of the gravity wave Hibert space... (zap!... sorry, I have no idea if they >form a Hilbert space...) Actually, Oz (are you still there?), maybe >this is what you were getting at... excuse me, what "Bert" was gettin at >by "Oh, it might be like a taut string as wot sumone av plucked sumwere far >away. Its dead flat until the triangular wave comes flying past." Me I just thought these 10 independent components were the 'external' forms of the Ricci. One describes whatsits due to the momentum flow inside, the other the curvature coming from the outside. >But what is this dern "tidal force" thing some of these guys are talking >about... there seems to seems to be some feeling this should be >associated with the Weyl too... seems to mess up this picture a bit... >this is due to local mass distribution... Hey, I am not too sure that the Ricci IS due to AN infinitesimal speck of momentum flow. It feels like it should explain how to calculate groups of bits of momentum flow due to gravitational curvature. In this case tidal and wave expressions all have an effect. We ought to ask prof in the lecture and see if he can explain. Maybe, if we get to working stuff out, it will become clear? >Hey! Here comes a passing expert! Let's ask HIM!!! (Volunteers?) Dunno about that Ed. There's experts, and then there's experts. Particularly in this college. Some of them float 6" above the floor, and they are STRANGE. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 99422 (34 more) in sci.physics: From: Matthew P Wiener Subject: Minkowski Metrics for Tensor Twits Date: 11 Feb 1996 15:24:59 GMT Organization: The Wistar Institute of Anatomy and Biology Lines: 96 Distribution: world NNTP-Posting-Host: sagi.wistar.upenn.edu In-reply-to: Oz 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)=10c-ct/2, y=z=0 for t=10 to 20. That is, travel at half the speed of light in the x direction for 10 years, and then back home for the next 10 years. (Note that "10c" in the formula above is really "10 light-(time-unit)s".) dx/dt=c/2 going out, and -c/2 coming back. (dx/dt)^2=c^2/4 either way. So ds^2=dt^2-dx^2/c^2=(1-1/4)dt^2, and ds=sqrt(1-1/4)dt. Integrating from t=0 to 10 gives sqrt(3/4).10=8.66 years. This is sometimes called the "twin paradox". If you want to replace x=x(t) with a more realistic path, be my guest. The integral is much messier, of course, but the idea is the exact same. The sqrt(1-1/4) factor is just the world famous sqrt(1-v^2/c^2) factor, as you've no doubt noticed already. > I am sure it's quite simple >and obvious to you tensor tyros, you have done a full and proper course. >Can anyone spell it out in words of one syllable, oh well Ok then three >syllables, how this happens? This was Minkowski's great contribution. The mathematics of relativity existed before Einstein. Einstein rederived it, but more importantly, he realized that it was secondary to the relative principle, and not, as others were thinking, a conspiracy of mathematics to force relativistic invariance. (So backwards was the thinking pre-Einstein, that no one really noticed that they _had_ invariance.) Nevertheless, Einstein's version of the mathematics was piecemeal and not particularly conceptual on its own. It fit the physics, and that was that. Minkowski realized relativity was a certain skewed but still very clear form of geometric thinking. That is what the metric encapsulates. For example, to compute arc length of a curve, one merely has to write down ds^2=du^2+dv^2 and identify the appropriate u and v and integrate. Advanced calculus stuff, very Pythagorean. Any beginning book on the differential geometry of curves and surfaces should contain this in gory detail somewhere. What you _generally_ get, though, is ds^2 = A.du^2 + 2B.du.dv + C.dv^2. Think of ds^2=du^2+dv^2 as special geometry and the above as general geometry. The kinematic passage from SR to GR is pretty much the same. As Einstein realized from the equivalence principle, relativistic gravity is simply not compatible with Euclidean geometry. In his writings, he liked to give the example of the relativistic merry-go-round. But the mere business of light--which follows a geodesic--bending because of gravity is already non-Euclidean. So the first step on the path to GR is doing physics on a metric. So long as the metric is [1,-1,-1,-1], you are only doing SR. But you are doing SR in a form ready to jump to GR. -- I don't know how often he told me, "You're stupid" and suchlike. That helped me a lot. --Heisenberg on Pauli -Matthew P Wiener (weemba@sagi.wistar.upenn.edu) Newsgroups: sci.physics Subject: Re: general relativity tutorial Summary: Expires: References: <1996Feb19.175727.16773@schbbs.mot.com> <4gcuc6$lig@pipe12.nyc.pipeline.com> Sender: Followup-To: Distribution: Organization: University of California, Riverside Keywords: 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 quasi-stationarily about our >system, provided we make the additional physical assumption that space is >flat at infinity. >Okey doke? Have I dotted the t's and crossed the i's so far? :-) Pretty good, though instead of "for a stationary distribution of masses" you should say "for a static spacetime geometry", since that's what counts. After all, it's really the fact that the geometry of spacetime is not rollicking around, and behaves like flat Minkowski spacetime at infinity, which lets you do your clock experiment and define a "gravitational potential" with the properties you desire. The distribution of masses, or even the whole stress-energy tensor, could in principle be static without the spacetime geometry being static. Remember, the stress-energy tensor does NOT completely determine the geometry of spacetime, since gravitational waves with no Ricci curvature but plenty of Weyl curvature could be zipping past --- even in an asymptotically flat spacetime. Of course, properly defining a "static, asymptotically flat spacetime geometry" is a wee bit subtle, but let's not worry about that. So: I'm back, Ed seems to have got cleared up on the "gravitational potential" business, and Oz has relearnt his special relativity from the modern viewpoint, in which everything is based on the metric. Moreover, everybody has spent the weekend boning up on their tensors and is ready to go. That means it's time for a new improved course outline. It looks like the old one, but mysteriously it has gotten a bit longer in various places.... 1. A TANGENT VECTOR or simply VECTOR at the point p of spacetime may be visualized as an infinitesimal arrow with tail at the point p. The tangent vectors at p form a vector space called the TANGENT SPACE; in other words, we can add them and multiply them by real numbers. Suppose we work in a local coordinate system with coordinates (x^0,x^1,x^2,x^3). (Since we are working in 4d spacetime there are 4 coordinates; we may think of x^0 as the time coordinate t and the other 3 as x, y, and z, but we don't need to think of them this way, since we're using an utterly arbitrary coordinate system.) Then we can describe a tangent vector v by listing its components (v^0,v^1,v^2,v^3) in this coordinate system. For short we write these components as v^a, where the superscript a, like all of our superscripts and subscripts, goes from 0 to 3. 2. A COTANGENT VECTOR or simply COVECTOR at the point p is a function f that eats a tangent vector v and spits out a real number f(v) in a linear way. Cotangent vectors can be viewed as ordered stacks of parallel planes in the tangent space at p. They don't "point" like tangent vectors do; instead, they "copoint". Working in local coordinates, we define the components of a covector f to be the numbers (f_0,f_1,f_2,f_3) you get you get when you evaluate f on the basis vectors: f_0 = f(1,0,0,0) f_1 = f(0,1,0,0) f_2 = f(0,0,1,0) f_3 = f(0,0,0,1) 3. A TENSOR of "rank (0,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a number. The output must depend linearly on each input. A TENSOR of "rank (1,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a tangent vector at the point p. The output must depend linearly on each input. More generally, a TENSOR of "rank (j,k) at a point p of spacetime is a function that takes as input a list of j cotangent vectors and k tangent vectors and returns as output a number. The output must depend linearly on each input. Note that this definition is compatible with the previous ones! This is obvious for the rank (0,k) tensors, but for the rank (1,k) ones we need to check that a function that eats k vectors and spits out a vector v can be reinterpreted as a function that eats k vectors and one covector f and spits out a number. We just let the covector f eat the vector v and spit out f(v)! Similarly, note that a vector can be reinterpreted as a tensor of rank (1,0), and a covector can be reinterpreted as a tensor of rank (0,1). In local coordinates we write the components of a tensor T of rank (j,k) as a monstrous array T^{ab....c}_{de....f} with j superscripts and k subscripts. Again, all superscripts and subscripts range from 0 to 3; each number T^{ab....c}_{de....f} is simply the number the tensor spits out when fed a suitable wad of basis vectors and covectors. I will describe this in more detail in the following example: 4. The METRIC g is a tensor of rank (0,2). It eats two tangent vectors v,w and spits out a number g(v,w), which we think of as the "dot product" or "inner product" of the vectors v and w. This lets us compute the length of any tangent vector, or the angle between two tangent vectors. Since we are talking about spacetime, the metric need not satisfy g(v,v) > 0 for all nonzero v. A vector v is SPACELIKE if g(v,v) > 0, TIMELIKE if g(v,v) < 0, and LIGHTLIKE if g(v,v) = 0. The inner product g(v,w) of two tangent vectors is given by g(v,w) = g_{ab} v^a w^b for some matrix of numbers g_{ab}, where we sum over the repeated indices a,b (this being the so-called EINSTEIN SUMMATION CONVENTION). Another way to think of it is that our coordinates give us a basis of tangent vectors at p, and g_{ab} is the inner product of the basis vector pointing in the x^a direction and the basis vector pointing in the x^b direction. The metric is the star of general relativity. It describes everything about the geometry of spacetime, since it lets us measure angles and distances. Einstein's equation describes how the flow of energy and momentum through spacetime affects the metric. What it actually affects is something about the metric called the "curvature". The biggest job in learning general relativity is learning to understand curvature. 5. PARALLEL TRANSLATION is an operation which, given a curve from p to q and a tangent vector v at p, spits out a tangent vector v' at q. We think of this as the result of dragging v from p to q while at each step of the way not rotating or stretching it. There's an important theorem saying that if we have a metric g, there is a unique way to do parallel translation which is: a. Linear: the output v' depends linearly on v. b. Compatible with the metric: if we parallel translate two vectors v and w from p to q, and get two vectors v' and w', then g(v',w') = g(v,w). This means that parallel translation preserves lengths and angles. This is what we mean by "no stretching". c. Torsion-free: this is a way of making precise the notion of "no rotating". We can define the "torsion tensor", with components t_{ab}, as follows. Take a little vector of size epsilon pointing in the a direction, and a little vector of size epsilon pointing in the b direction. Parallel translate the vector pointing in the a direction by an amount epsilon in the b direction. Similarly, parallel translate the vector pointing in the b direction by an amount epsilon in the a direction. (Draw the resulting two vectors.) If the tips touch, up to terms of epsilon^3, there's no torsion! Otherwise take the difference of the tips and divide by epsilon^2. Taking the limit as epsilon -> 0 we get the torsion t_{ab}. We say that parallel translation is "torsion-free" if t_{ab} = 0. A GEODESIC is a curve whose tangent vector is parallel transported along itself. I.e., to follow a geodesic is to follow ones nose while never turning ones nose. A particle in free fall follows a geodesic in spacetime. 6. The CONNECTION is a mathematical gadget that describes "parallel translation along an infinitesimal curve in a given direction". In local coordinates the connection may be described using the components of the CHRISTOFFEL SYMBOL Gamma_{ab}^c. There is an explicit formula for these components in terms of components g_{ab} of the metric, which may be derived from the assumptions a-c above. However, this formula is very frightening, so I will only describe it to those who have passed certain tests of courage and valor. 7. The RIEMANN CURVATURE TENSOR is a tensor of rank (1,3) at each point of spacetime. Thus it takes three tangent vectors, say u, v, and w as inputs, and outputs one tangent vector, say R(u,v,w). The Riemann tensor is defined like this: Take the vector w, and parallel transport it around a wee parallelogram whose two edges point in the directions epsilon u and epsilon v , where epsilon is a small number. The vector w comes back a bit changed by its journey; it is now a new vector w'. We then have w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3 Thus the Riemann tensor keeps track of how much parallel translation around a wee parallelogram changes the vector w. In addition to this simple coordinate-free definition of the Riemann tensor, we may describe its components R^a_{bcd} using coordinates. Namely, the vector R(u,v,w) has components R(u,v,w)^a = R^a_{bcd} u^b v^c w^d where we sum over the indices b,c,d. Another way to think of this is that if we feed the Riemann tensor 3 basis vectors in the x^b, x^c, x^d directions, respectively, it spits out a vector whose component in the x^a direction is R^a_{bcd}. There is an explicit formula for the components R^a_{bcd} in terms of the Christoffel symbols. Together with the aforementioned formula for the Christoffel symbols in terms of the metric, this lets us compute the Riemann tensor of any metric! Thus to do computations in general relativity, these formulas are quite important. However, they are not for the faint of heart, so I will not describe them here! 7. The RICCI TENSOR. The matrix g_{ab} is invertible and we write its inverse as g^{ab}. We use this to cook up some tensors starting from the Riemann curvature tensor and leading to the Einstein tensor, which appears on the left side of Einstein's marvelous equation for general relativity. We will do this using coordinates to save time... though later we should do this over again without coordinates. Okay, starting from the Riemann tensor, which has components R^a_{bcd}, we now define the RICCI TENSOR to have components R_{bd} = R^c_{bcd} where as usual we sum over the repeated index c. The physical significance of the Ricci tensor is best explained by an example. So, suppose an astronaut taking a space walk accidentally spills a can of ground coffee. Consider one coffee ground. Say that a given moment it's at the point P of spacetime, and its velocity vector is the tangent vector v. Note: since we are doing relativity, its velocity is defined to be the tangent vector to its path in *spacetime*, so if we used coordinates v would have 4 components, not 3. The path the coffee ground traces out in spacetime is called its "worldline". Let's draw a little bit of its worldline near P: | v^ | P | | | The vector v is an arrow with tail P, pointing straight up. I've tried to draw it in, using crappy ASCII graphics. Now imagine a bunch of comoving coffee grounds right near our original one. What does this mean? Well, it means that for any tangent vector w at P which is orthogonal to v, if we follow a geodesic along w for a certain while, we find ourselves at a point Q where there's another coffee ground. Let me draw the worldline of this other coffee ground. | | v^ ^v' | w | P->-----Q | | | | | | I've drawn w so you can see how it is orthogonal to the worldline of our first coffee ground. The horizontal path is a geodesic from P to Q, which has tangent vector w at Q. I have also drawn the worldline of the coffee ground which goes through the point Q of spacetime, and I've also drawn the velocity vector v' of this other coffee ground. What does it mean to say the coffee grounds are comoving? It means simply that if we take v and parallel translate it over to Q along the horizontal path, we get v'. This may seem like a lot of work to say that two coffee grounds are moving in the same direction at the same speed, but when spacetime is curved we gotta be very careful. Note that everything I've done is based on parallel translation! (I defined geodesics using parallel translation.) Now consider, not just two coffee grounds, but a whole swarm of comoving coffee grounds near P. If spacetime were flat, these coffee grounds would *stay* comoving as time passed. But if there is a gravitational field around (and there is, even in space), spacetime is not flat. So what happens? Well, basically the coffee grounds will tend to be deflected, relative to one another. It's not hard to figure out exactly how much they will be deflected! We just use the definition of the Riemann curvature! We get an equation called the "geodesic deviation equation". But let me not do that just yet. Instead, let me say what the Ricci tensor has to do with all this. Then, when we use the "geodesic deviation equation" to work out the deflection of the coffee grounds using the Riemann curvature, we will see what this has to do with the usual definition of the Ricci tensor in terms of the Riemann curvature. Imagine a bunch of coffee grounds near the coffee ground that went through the point P. Consider, for example, all the coffee grounds that were within a given distance at time zero (in the local rest frame of the coffee ground that went through P). A little round ball of coffee grounds in free fall through outer space! As time passes this ball will change shape and size depending on how the paths of the coffee gournds are deflected by the spacetime curvature. Since everything in the universe is linear to first order, we can imagine shrinking or expanding, and also getting deformed to an ellipsoid. There is a lot of information about spacetime curvature encoded in the rate at which this ball changes shape and size. But let's only keep track of the rate of change of its volume! This rate is basically the Ricci tensor. More precisely, the second time derivative of the volume of this little ball is approximately R_{ab} v^a v^b times the original volume of the ball. This approximation becomes better and better in the limit as the ball gets smaller and smaller. 8. The RICCI SCALAR. Starting from the Ricci tensor, we define R^a_d = g^{ab} R_{bd}. As always, we follow the Einstein summation convention and sum over repeated indices when one is up and the other is down. This process, which turned one subscript on the Ricci tensor into a superscript, is called RAISING AN INDEX. Similarly we can LOWER AN INDEX, turning any superscript into a subscript, using g_{ab}. Then we define the RICCI SCALAR by R = R^a_a. This process, whereby we get rid of a superscript and a subscript in a tensor by summing over them a la Einstein, is called CONTRACTING. 9. The EINSTEIN TENSOR. Finally, we define the Einstein tensor by G_{ab} = R_{ab} - (1/2)R g_{ab}. You should not feel you understand why I am defining it this way!! Don't worry! That will take quite a bit longer to explain; the point is that with this definition, local conservation of energy and momentum will be an automatic consequence of Einstein's equation. To understand this, we need to know Einstein's equation, so we need to know about: 10. The STRESS-ENERGY TENSOR. The stress-energy is what appears on the right side of Einstein's equation. It is a tensor of rank (0,2), and it defined as follows: given any two tangent vectors u and v at a point p, the number T(u,v) says how much momentum-in-the-u-direction is flowing through the point p in the v direction. Writing it out in terms of components in any coordinates, we have T(u,v) = T_{ab} u^a v^b In coordinates where x^0 is the time direction t while x^1, x^2, x^3 are the space directions (x,y,z), we have the following physical interpretation of the components T_{ab}: The top row of this 4x4 matrix, keeps track of the density of energy --- that's T_{00} --- and the density of momentum in the x,y, and z directions --- those are T_{01}, T_{02}, and T_{03} respectively. This should make sense if you remember that "density" is the same as "flow in the time direction" and "energy" is the same as "momentum in the time direction". The other components of the stress-energy tensor keep track of the flow of energy and momentum in various spatial directions. 11. EINSTEIN'S EQUATION: This is what general relativity is all about. It says that G = T or if you like coordinates and more standard units, G_{ab} = 8 pi k/c^2 T_{ab} where k is Newton's gravitational constant and c is the speed of light. So it says how the flow of energy and momentum through a given point of spacetime affect the curvature of spacetime there. But what does it mean? To see this, let's do some "index gymnastics". Stand with your feet slightly apart and hands loosely at your sides. Now, assume the Einstein equation! G_{ab} = T_{ab} Substitute the definition of Einstein tensor! R_{ab} - (1/2)R g_{ab} = T_{ab} Raise an index! R^a_b - (1/2)R g^a_b = T^a_b Contract! R^a_a - (1/2)R g^a_a = T^a_a Remember the definition of Ricci scalar, and note that g^a_a = 4 in 4d! R - 2R = T^a_a Solve! R = - T^a_a Okay. That's already a bit interesting. It says that when Einstein's equation is true, the Ricci scalar R is the sum of the diagonal terms of T^a_a. What are those terms, anyway? Well, they involve energy density and pressure. But let's wait a bit on that... let's put this formula for R back into Einstein's equation: R_{ab} + (1/2) T^c_c g_{ab} = T_{ab} or R_{ab} = T_{ab} - (1/2) T^c_c g_{ab}. This equation is equivalent to Einstein's equation. What does it mean? Well, first of all, it's nice because we have a simple geometrical way of understanding the Ricci tensor R_{ab} in terms of convergence of geodesics. Remember, if v is the velocity vector of the particle in the middle of a little ball of initially comoving test particles in free fall, and the ball starts out having volume V, the second time derivative of the volume of the ball is -R_{ab} v^a v^b times V. If we know the above quantity for all velocities v (even all timelike velocities, which are the physically achievable ones), we can reconstruct the Ricci tensor R_{ab}. But we might as well work in the local rest frame of the particle in the middle of the little ball, and use coordinates that make things look just like Minkowski spacetime right near that point. Then g_{ab} = -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 and v^a = 1 0 0 0 So then --- here's a good little computation for you budding tensor jocks --- we get R_{ab} v^a v^b = R_{00} So in this coordinate system we can say the 2nd time derivative of the volume of the little ball of test particles is just -R_{00}. On the other hand, check out the right side of the equation: R_{ab} = T_{ab} - (1/2) T^c_c g_{ab} Take a = b = 0 and get R_{00} = T_{00} + (1/2) T^c_c Note: demanding this to be true at every point of spacetime, in every local rest frame, is the same as demanding that the whole Einstein equation be true! So we just need to figure out what it MEANS! What's T_{00}? It's just the energy density at the center of our little ball. How about T^c_c? Well, remember this is just g^{ca} T_{ac}, where we sum over a and c. So --- have a go at it, tensor jocks and jockettes! --- it equals -T_{00} + T_{11} + T_{22} + T_{33}. So we get R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] What about T_{11}, T_{22}, and T_{33}? In general these are the flow of x-momentum in the x direction, and so on. In a typical fluid at rest, these are all equal to the pressure. So the "simple geometrical essence of Einstein's equation" is this: -------------------------------------------------------------------------- Take any small ball of initially comoving test particles in free fall. Work in the local rest frame of this ball. As time passes the ball changes volume; calculate its second derivative at time zero and divide by the original volume. The negative of this equals 1/2 the energy density at the center of the ball, plus the flow of x-momentum in the x direction there, plus the flow of y-momentum in the y direction, plus the flow of z-momentum in the z direction. -------------------------------------------------------------------------- Or, if you want a less precise but more catchy version: -------------------------------------------------------------------------- Take any small ball of initially comoving test particles in free fall. Work in the local rest frame of this ball. As time passes, the rate at which the ball begins to shrink in volume is proportional to the energy density at the center of the ball plus the flow of x-momentum in the x direction there plus the flow of y-momentum in the y direction plus the flow of z-momentum in the z direction. -------------------------------------------------------------------------- Note: all of general relativity can in principle be recovered from the above paragraph! Also note that the minus sign in that paragraph is good, since it says if you have POSITIVE energy density, the ball of test particles SHRINKS. I.e., gravity is attractive. Article 101363 (14 more) in sci.physics: From: john baez Subject: Re: Tensors for twits please. Date: 21 Feb 1996 17:23:18 -0800 Organization: University of California, Riverside Lines: 58 NNTP-Posting-Host: guitar.ucr.edu In article <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 2-sphere of radius r, for example, where we use coordinates theta and phi, the metric is g_{11} g_{12} = r^2 sin^2(phi) 0 g_{21} g_{22} 0 r^2 with "1" corresponding to theta and "2" corresponding to phi. Now --- here's the sneaky part --- g is a (0,2) tensor so we can alternatively write just g = r^2 sin^2(phi) (d theta)^2 + r^2 (d phi)^2. Here we are using the mysterious fact that d theta and d phi are cotangent vectors, and the mysterious fact that some sort of product of two cotangent vectors is a (0,2) tensor, to express g in terms of d theta and d phi. Alternatively, we can call the metric "ds^2" --- this is the "line element" thingie you know and love, hopefully --- and then we get ds^2 = r^2 sin^2(phi) (d theta)^2 + r^2 (d phi)^2. However, ds^2 is really just an older-fashioned way of talking about the metric g, so this is no big deal. >Now I am getting the distinct flavor that to really follow your >description, I would actually have to seek a text, and work some >problems, and all would become clear... Oh, hopefully the above makes enough sense so that you won't have to resort to such drastic measures. Article 101295 (1 more) in sci.physics: From: Oz Subject: Re: Gravitational Red Shifts - Real or Apparent ? Date: Wed, 21 Feb 1996 18:19:25 +0000 Organization: Oz Lines: 63 Distribution: world NNTP-Posting-Host: upthorpe.demon.co.uk X-NNTP-Posting-Host: upthorpe.demon.co.uk MIME-Version: 1.0 X-Newsreader: Turnpike Version 1.11 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 over-generalized conclusions from a very >special case is always dangerous. Suppose we just learned how to solve >the problem of the falling body in classical mechanics. The height h is >the following function of time: > >h = h_0 - mgt^2/2 > >We start to philosophize: > >"Wow, so height always decreases with time!" > >"In fact, time is a just a way of measuring height, because >t = sqrt(2(h_0 - h)/mg)!" > >"Cool, so time is really just basically the same thing as height!" > >Etc. etc.. Ooo. Straight to the heart. I bet I am the worst culprit for this. However your point is very well made. If viciously well observed :-( The whole point as far as I was concerned in starting out following this GR course for idiots was to have some rather better understanding of what GR was, and what it predicted. Do we have nearly enough math yet to follow it at an ever-so-slightly deeper level. The simplest form would seem to be an isolated mass in otherwise empty space with minkowski space at infinity. Presumably there are some expressions that we can dissect. Say for an observer falling from infinity, and one hovering at a fixed r from the mass at the origin? Hopefully the expressions will not be excessively complex such that they can reasonably be written down in a posting (somehow I have a bad feeling about this). I have to admit that I did try to make a start on this, but my pathetic efforts simply weren't replied to. I suspect that they were either so pathetic or in the 'not even wrong' category that kind people decided not to reply. It is nice to know that people here are actually a considerate bunch. It is clear from Baez's posting that it is easier to get the wrong end of things than the right end, so hopefully he, Wiener et al can pull the rabbits from the hat and not ask us to pull them out wrongly. At least not too often, that is. ------------------------------- 'Oz "When I knew little, all was certain. The more I learnt, the less sure I was. Is this the uncertainty principle?" Article 101406 (4 more) in sci.physics: From: john baez Subject: Re: Gravitational Red Shifts - Real or Apparent ? Date: 21 Feb 1996 22:34:24 -0800 Organization: University of California, Riverside Lines: 105 NNTP-Posting-Host: guitar.ucr.edu In article 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 over-generalized conclusions from a very >>special case is always dangerous. Suppose we just learned how to solve >>the problem of the falling body in classical mechanics. The height h is >>the following function of time: >> >>h = h_0 - mgt^2/2 >> >>We start to philosophize: >> >>"Wow, so height always decreases with time!" >> >>"In fact, time is a just a way of measuring height, because >>t = sqrt(2(h_0 - h)/mg)!" >> >>"Cool, so time is really just basically the same thing as height!" >> >>Etc. etc.. >Ooo. Straight to the heart. I bet I am the worst culprit for this. >However your point is very well made. If viciously well observed :-( I just thought that a little viciousness would make my point clearer. Of course, science is all about making generalizations, most of which are false. It's all a matter of *levels* of naivete. When studying general relativity, however, it's good to start with the presumption that nothing need work like it did in Newtonian mechanics or special relativity unless there's a damn good reason. That's because it's really a revolutionary theory. >The whole point as far as I was concerned in starting out following this >GR course for idiots was to have some rather better understanding of >what GR was, and what it predicted. Do we have nearly enough math yet to >follow it at an ever-so-slightly deeper level. We're getting there. Check out the new course outline. >The simplest form would >seem to be an isolated mass in otherwise empty space with minkowski >space at infinity. Presumably there are some expressions that we can >dissect. Say for an observer falling from infinity, and one hovering at >a fixed r from the mass at the origin? Hopefully the expressions will >not be excessively complex such that they can reasonably be written down >in a posting (somehow I have a bad feeling about this). Well, the simplest interesting solution of Einstein's equation is the "static point mass" solution you describe, better known as the Schwarzschild solution. Actually, though, there's not really a point mass! The solution describes a black hole with the stress-energy tensor equal to zero everywhere. I could write down the metric for this solution, and then I could write down what the geodesics look like in this solution, and so on. Of course, to know that it *is* a solution, you need to be able to compute the Riemann tensor from the metric! And this, of course, means we need to be able to compute parallel transport (or more precisely, the "connection") starting from the metric. Similarly, to know why the geodesics of a given metric *are* geodesics, we need to know how to compute parallel transport (or more precisely, the connection) starting from the metric. I have been avoiding telling you how do do these things, since I didn't want to scare you silly until I had already lulled you into the false impression that learning GR was just a bowl of cherries. In short: yes, your bad feeling was a premonition: it all has to do with the mathematics going on in the back room, that the wizard suspects you don't really want to see. More on that wizard business later.... Anyway, you have two alternatives. Either take a peek at the Schwarzschild metric and start to see how some of the weird time-warping, space-bending properties of black holes work, taking my word for it that all the math actually works out like I say. Or get a bit deeper understanding of what we skimmed over in the earlier version of the course outline, so you'll at least know what sort of calculations one must do to figure these things out.... even if we don't actually DO the calculations. (It's too yucky to do these calculations in ASCII.) Article 101630 (32 more) in sci.physics: From: john baez Subject: Re: GR Tutorial Water Cooler Date: 22 Feb 1996 21:50:07 -0800 Organization: University of California, Riverside Lines: 144 NNTP-Posting-Host: guitar.ucr.edu In article <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, Dirac-delta-like in its jaggedness and its suddenness. This is not the case with space, where elliptic regularity reigns. But that's always what time was supposed to do: to allow surprises to occur, to make the future something one would have to wait for, rather than merely predict from local observations! One triumph of special relativity is thus the fact that when spacetime is a Lorentzian manifold, the laws of physics are hyperbolic PDE, elliptic regularity fails, and the future constantly brings new information crashing in upon us. >Maybe that's enough for one post. Now somebody ask me how the >Cauchy-Riemann equations fit into this. Come on, ask me. I dare you. The Cauchy-Riemann equations are another primordial elliptic PDE. Article 101754 (15 more) in sci.physics: From: john baez Subject: Re: Tensors for twits please. Date: 23 Feb 1996 13:45:37 -0800 Organization: University of California, Riverside Lines: 182 NNTP-Posting-Host: guitar.ucr.edu In article <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." "Right-oh, your magicness sir." said Oz, pretty pi**sed off. "No sooner said than done y'r wizardness." "OK, see you at sunset." said G. Wiz as he tottered out of the door. "Hours of work here," thought Oz "No sooner said than done, indeed! Humph, he's never even held a broom!" Oz swept up, put the debris into the black hole recycler and cleaned the larger scorch marks from the white quartzite floor. Th