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 Li