How does GR handle metric transition for a spherical mass shell?

[Q-reeus]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime? And with that proviso that the angular departure is larger the larger the triangle? I will take it there is unanimously a yes and yes to the above. So here's the thing. If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 60⁰, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature. Now go the next step to 3D. Instead of flat tiles, build an outer tetrahedron from much smaller ones. Anyone doubt the same issues will manifest as for 2D, only proportionately greater in effect? Now head back to the s-sided cube arrangement between concentric shells. It didn't fit. But then it was supposed to be a perfect cube. Still wouldn't exactly fit given the above, but depending on assumed constraints, that certified in flat spacetime perfectly cubic cube has either puckered like a pin-cushion, or one notices a 'strange' addition to the vertex angles that 'shouldn't' be there. And 'strangely', one can fit more counting spheres inside said cube given the latter constraint (geodesically straight edges and flat faces).

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

 

[pervect]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?

Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.

 

[Q-reeus]

Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.

Fair comment. I'm assuming we are relating this to the case of something similar to the original setup - stationary spherical mass and stationary observers. That would be positively curved spacetime I presume? In which angles add to more than 180⁰. [have since edited and added word 'positively' to curved - thanks]

 

[PAllen]

I'm putting it down to this: both you and Peter 'know' I probably have everything wrong, so what I actually argue tends to go in one ear and out the other, to be replaced with an image of what 'I probably meant', and then proceed to tear down that straw man. Point to anywhere, not just above but any previous entry, where I claimed what you think I did. I'm not so stupid as to imagine that running a hoop over an 'egg' allows one to detect 3-curvature at all. Have endlessly now referred to this as 2D analogue of 3D situation. Having said that, there *is* I think an in principle 2D manifestation of 3-curvature, which will be discussed in another posting.

Then it is not a straw men. Your last sentence is a mathematical falsehood, with no possible valid interpretation. That is, anything you can detect restricted to a 2-surface (or thin 3-analog of it) will tell you nothing about presence of absence of spatial curvature.

 

[PAllen]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰,

No, this is false. As long as you restrict to a 2-surface, and use physical procedures (which naturally pick out the flattest possible interpretation of local reality), you will detect no deviation. To make this plausible, go right back to my argument that you accused of being strawman - 2-sphere embedded in Euclidean 3-space. The 2-surface is curved, the 3-space is flat. As a result, anything you find restricted to a 2-surface cannot be distinguished from an embedding artifact; further, in seeking straightness, you will pick out a flat 2-surface and find no deviation.

The discussion I had with Peter went further than this. That if spacetime curvature is simple enough, then even fully general measures of 3-space curvature (spanning large regions) can show exact flatness. In each case Peter mentioned it is the 'most natural observer' - the inertial one, or the one who sees CMB radiation as isotropic - that detects no spatial at curvature at all, by any means. His K, in SC geometry, only applies to stationary observers, who are non-inertial. One can detect similar things for non-inertial observers in flat spacetime, so even K is related more to non-inertial motion than intrinsic spatial curvature.

 

[PeterDonis]

Now move this composite triangle to a region of 'uniformly' positively curved spacetime.

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?

Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.

Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.

If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 60⁰, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature.

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.

I won't bother commenting on the 3D case because it works out exactly the same.

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

None of what you say is incorrect. It just doesn't mean what you appear to think it means.

Added in edit, after seeing PAllen's post: Here's another way to say what I just said. I say none of what you have said is incorrect; but at the same time, PAllen is right when he says the comment he quoted is false. That's because I am saying you are correct as a matter of abstract geometry only, while he is saying you are incorrect in how you are trying to relate the geometry to the physics. *Both* of our comments are valid. Can you see why?

 

[Q-reeus]

Originally Posted by Q-reeus: "Now move this composite triangle to a region of 'uniformly' positively curved spacetime."

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?

From #151 "In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles."
Miss that bit - tiling, as in tiles? A composite, physical object.

Originally Posted by Q-reeus:
"Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?"

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.

Like gaps appearing for instance. I gave two restraint examples - gaps appear, or sides curve. Not physics? Assuming some kind of elastic glue joining the triangle tiles, naturally, forcing the outer triangle's to be straight requires imposing stresses - evidence of living in curved spacetime. Gaps or curved sides or stresses. Not there, or not needed in flat spacetime. Can't see any physics at work; no makings of an in principle 'detector'?

Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.

Fine, but in present scenario, whether triangles have perfectly 'straight' sides is just a convenience - we are only really interested in detecting change. Straight sides just makes the job easier. Point is, there is detectable change - choose a restraint - see the effect.

Originally Posted by Q-reeus:
"If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 600, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density."

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.

Gone over that above. But another angle on it - what assumptions are you making in saying one must force the outer triangle to have straight sides? That assumes, as I guessed in answering above, that the tiles are glued together somehow. That's one configuration. Another could be tiles just sitting together snuggly but loosely on a plane. Moved en masse to curved space, why would they not simply drift apart slightly owing to curvature? There are any number of possible constraints - everyone I have considered results in detectable change - my idea of physics at work.

Originally Posted by Q-reeus:
"Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature."

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.

Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

 

[PeterDonis]

Warning: long post!

Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

Yes, things change. And the change means you can't use the things as standards of distance.

Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.

At the risk of piling on scenario after scenario, consider the following:

Take a piece of paper and try to wrap it around a globe. You can't; the paper will buckle. That's a physical realization of one scenario you were describing, the one where the sides of the triangles remain straight in the Euclidean sense; if we imagine the paper tesselated with tiny triangles, the paper behaves in such a way as to keep the triangles as flat Euclidean triangles. The paper may buckle along the edges between triangles, but the triangles themselves maintain their shape. So the paper can't possibly conform to any curved surface.

Now take a piece of flat rubber and wrap it around the same globe. You can do it, but only by exerting force on the rubber to deform it into the shape of the globe. This is a physical realization of another scenario you described, where we allow the triangles to adjust to the curvature of the surface; if we imagine the rubber tesselated by tiny triangles, the deformation of the rubber will deform the triangles too. But the deformation is not somehow magically caused by the globe's curvature; it's caused by us, exerting external force on the rubber to change its shape.

We can use the rubber to measure the K factor as follows: suppose that, when sitting on a flat Euclidean plane, the paper and the rubber have identical shape and area. We cut the paper into little tiny squares, and use them to tile the rubber. They fit exactly. Now we take the rubber and stretch it over the globe, and it deforms; we can see the deformation by watching the little squares and seeing that they no longer exactly tile the rubber; we might, as you say, see little gaps open up between the squares, assuming that the rubber is being deformed appropriately. But in order to know exactly how the squares will tile the rubber once it's wrapped around the globe, we have to specify *how*, exactly, the rubber is being wrapped.

Here's one way of specifying the wrapping. Take a circular disk of rubber instead of a square, and a circular piece of paper that, on a flat Euclidean plane, is exactly the same size. We cut the paper up into tiny shapes (triangles, squares, whatever you like) and verify that on a flat Euclidean plane, they tile the rubber exactly. Now take a sphere and draw a circle on it, centered on the North Pole, such that the circumference of the circle is exactly the same as the circumference of the rubber circle when the latter is sitting on the flat Euclidean plane. If we wrap the rubber circle onto the sphere in such a way that its circumference lies exactly on the circle we drew on the sphere, we will need to stretch the rubber; its area will now be larger than it was on the flat plane, and we can verify this by trying to tile it with the little pieces we cut out of the paper and seeing that there is still area left over. This is a manifestation of the K factor being greater than 1. But again, we had to physically stretch the rubber--exert external force on it--to get it to fit the designated area of the sphere.

(As I've noted before, the K factor will vary over the portion of the sphere being used for this experiment, but the average will be greater than 1; if we wanted to limit ourselves to a single value for K, we could cut a small annular ring of rubber on the flat plane, draw two circles on the sphere that matched its outer and inner circumferences, and stretch the rubber between them, and verify that the area increased by tiling with little pieces cut out of a paper annulus that exactly overlapped the rubber on a flat plane. We'll need this version for the 3-D case, as we'll see in a moment.)

All of the above is consistent with what you've been saying about how curvature of a surface manifests itself, and nobody has been disputing it. Now carry the analogy forward to the 3-D case. Here, since we're going to have a gravitating body in the center, we need to use the "annular" version of the scenario, as I just noted. We go to a region far away from all gravitating bodies, so spacetime in the vicinity is flat, and we cut ourselves a hollow sphere of rubber, with inner surface area A and outer surface area A + dA. We also cut a hollow sphere of metal (we use this instead of paper as our standard for material that will maintain its shape instead of deforming) with the same inner and outer surface area. Then we cut the metal up into little tiny pieces and verify that the pieces exactly fill the volume of the rubber. (This is a thought experiment, so assume that we can do this "tiling" in the 3-D case as we did in the 2-D case.)

Now take the hollow rubber sphere and place it around a gravitating body, in such a way that its inner and outer surface area are exactly the same as they were in the flat spacetime region. (Again, this is a thought experiment, so assume we can actually do this as we did in the 2-D case.) We will find that we need to physically stretch the rubber in order for the inner and outer surface areas to match up; there is "more volume" between the two surfaces than there was in the flat spacetime region. Again, we can verify this by trying to tile the volume of the rubber with the little metal pieces, and finding that we run short; there is volume left over when all the pieces have been used up. This excess volume is a physical measure of the K factor. (If dA is small enough compared to A, K is basically constant in this scenario, as it was in the "annular" 2-D case.)

The key fact, physically, is that we have to apply an external force to the rubber, stretch it, to make it fit when it's wrapped around a gravitating body, just as we had to apply an external force to the 2-D rubber circle to make it fit in the specified way around the sphere. So we *cannot* say that the rubber was stretched "by the K factor"--the K factor wasn't what applied the force. We did. Remember that we are assuming that tidal forces, and any other "forces", are negligible; the *only* effect we are considering is the K factor. (Again, this is easier to do because we are now dealing with a "local" scenario, where K is constant; if we covered a wider area, so K varied, we would also find it impossible, or at least a lot harder, to ignore or factor out the effects of tidal gravity. In the "local" case it's easy to set up the scenario so that tidal gravity is negligible, while the K factor itself is not; we just choose the mass M of the gravitating body and the radial coordinate r at which we evaluate K appropriately.)

You may ask, well, what about if we *don't* exert any force on the rubber? What then? In the 2-D case, the answer is that if we exert no force on the rubber, we can't make it conform to the curved surface at all. We specified one way of fitting the rubber to the sphere--so that its circumference remained the same (we checked that by drawing in advance a circle on the sphere centered on the North Pole, to use as a guide). We could specify another way--keep the total area of the rubber constant. But if we do that, then the rubber's circumference will be smaller, so again the rubber has to deform. There is *no* way to make the flat piece of rubber exactly fit to the curved surface without deforming it somehow.

For the spacetime case, it's a little more complicated, because if we take the hollow sphere of rubber we made in a flat spacetime region, and move it around a gravitating body, *something* will happen to it if we exert no force on it (other than the force needed to move it into place). But the general point still applies: there is no way to "fit" the object into a curved spacetime region without deforming it somehow. Since we can't actually wrap a full hollow sphere around a gravitating body without breaking it, consider a portion of the sphere covering a given solid angle, as PAllen proposed in a previous post. Suppose we have done all the work described earlier in a flat spacetime region: we have the inner surface area A and outer surface area A + dA measured (now not covering the total solid angle 4 pi of the sphere, but something less), and we have a piece of metal of the exact same shape and size in flat spacetime cut up into little pieces, which tile the volume of the rubber exactly in flat spacetime.

Now we slowly lower the rubber into place "hovering" over a gravitating body. We have to specify *some* constraint for where it ends up. Suppose we specify that the inner surface lies in a portion of a 2-sphere such that the solid angle it covers is exactly the same as what it covered in flat spacetime--i.e., the ratio of the inner surface area A of the rubber piece to the total area of the 2-sphere is the same as it was for the 2-sphere that the inner surface was cut out of in flat spacetime. That tells us that the inner surface of the 2-sphere is not deformed; but now we have to decide what to do with the outer surface. We have the same problem as we did in the 2-D case: if we specify that the outer surface area remains the same, relative to the inner surface area (i.e., the outer surface "fits" into a corresponding 2-sphere the same way the inner surface does--the outer surface is not deformed), then the rubber will have to stretch to match up, and there will be extra volume in the rubber, according to the K factor (which we can verify by trying to tile with the little pieces of metal and seeing that there is volume left over). Or, if we specify that we want the volume of the rubber to remain the same (i.e., we want to be able to tile it exactly with the little pieces of metal), then the outer surface area will "fit" to a smaller 2-sphere, so the shape of the outer surface will deform. And, of course, if we relax the constraint on the inner surface, there are even more ways we could deform the rubber; but there is *no* way to fit it without *some* kind of deformation.

As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.

Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow. That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.

 

[Q-reeus]

Warning: long post!
You are I deduce a speed typist - massive effort!

...Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.

Understand and agree with that almost completely. Sure the curvature is not exerting any direct forces, so in that sense yes is not physics. Had thought though we could use the 'deformity correction' externally applied forces as a gauge of 'geometry'. More on that below.

As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.

This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before. If small triangular gussetts were initially glued to the apexes, owing to their reduced susceptibility to angular change, ought to partially restrain angular expansion of the much larger triangle tube joints. In other words, there should be some internal stresses - miniscule but in principle measurable. I think.

Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow.

Many thanks Peter for going to all that fuss - I do now appreciate more just why K factor, relating to a global geometry restraint, requires an implied stretching to make sense. So an isolated, stress-free container (or sheet in 2D case) cannot exhibit that unless forced into an equivalent geometry - like the solid angle segment mentioned. OK then agree that my earlier subdividing of concentric shells argument is not appropriate. And the hoop/marbles thing is presumably analogous. To explain a null result, we must assume there is no effective sampling of curvature, which is still hard for me to fathom re curved vs flat surface. Or is it a fundamental difference betwen 2D vs 3D , which may have been talked about elsewhere.

That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.

So how to interpret goings on in the case of say a very rigid tubular frame in the shape of a perfect tetrahedron in flat spacetime. It is the simplest polyhedron that is fully constrained re shape when all sides are connected via freely hinging ball joints. A 3D arrangement of four triangles. If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime. Can we say just how that would not be true? S'pose it's down to fierce math proof - yes? Scratching head over that one.
Thanks again, Cheers :zzz:

 

[PeterDonis]

You are I deduce a speed typist

Thanks to my Dad forcing me to learn to type one summer in high school, yes. I learned on the old Olympic typewriter that he had used to type his master's thesis, which was older than I was. So my fingers learned how to squeeze a decent word rate out of those old, half-rusted keys that took about ten pounds of effort each to strike, and now they just laugh at a typical computer keyboard, which takes practically no effort by comparison. (It probably helps that I play keyboards too.)

This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before.

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform. We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).

Given that these conditions are met, yes, we can in principle use the triangle as described (or the tetrahedron in the 3-D case) to measure differences between the space we are interested in and flat space; but there is still the question of how we decide to constrain the deformation. In the case quoted above, you are essentially constraining the side lengths of the tetrahedron (or triangle) to be constant, and letting the angles vary. Consider the 2-D case first (the triangle), and note that to physically realize this, we not only need to let the joint angles expand; we also need to bend the sides of the triangle since they are no longer Euclidean straight lines, but geodesics of the sphere, i.e., segments of great circles. (At least, I assume this was your intention in specifying side lengths constant and letting angles vary.) For thought experiment purposes, we can stipulate that this can be done while keeping the length of the sides constant, and similarly, in 3-D, the sides of the tetrahedron will have to bend slightly, but we can stipulate that their lengths are still held constant. In both cases, however, the sides will clearly undergo deformation, and we will have to exert external force on the tetrahedron to effect this deformation.

If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime.

With the constraint as described above, yes; I would expect the volume inside the tetrahedron (or the area of the triangle, in the analogous 2-D case) to be larger if the side lengths are stipulated to be held constant. But as I just noted, to realize this case we have to exert external force to deform the tetrahedron (or triangle), and this external force is what causes the volume to expand.

OTOH, we could stipulate a different constraint, that the tetrahedron should be "unstressed"--or more precisely, that we should not *impose* any stress on it by exerting external force. In this case, I'm not sure what would happen, other than that I would expect the tetrahedron's volume to remain constant--always assuming that we can neglect tidal forces (as I noted previously, we can always choose the mass M of the central body and the radius r appropriately to make K measurably different from 1 while still having tidal gravity negligible). I *think* that the angles would still expand, but the sides would shorten, keeping the resulting volume constant but changing the shape.

 

[JDoolin]

Hi, I apologize if this seems off topic. You guys are talking about a stationary shell, but another interesting idea to consider is an expanding shell.

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles? It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

 

[PeterDonis]

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

Lorentz-contrated and time-dilated relative to what?

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles?

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

 

[Q-reeus]

Originally Posted by Q-reeus:
"This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before."

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform.

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

 

[PAllen]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have explained this multiple times and you have continued to ignore it.

 

[Q-reeus]

I have explained this multiple times and you have continued to ignore it.

Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have quoted a proof by J.L. Synge that five vertices are the minimum to detect curvature of spacetime (that is, even a tetrahedron can always be constructed to conform to Euclidean expectations).

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry
[Another grab, from: http://en.wikipedia.org/wiki/Curved_space#Open.2C_flat.2C_closed
"Triangles which lie on the surface of an open space will have a sum of angles which is less than 180°. Triangles which lie on the surface of a closed space will have a sum of angles which is greater than 180°. The volume, however, is not (4 / 3)πr³" In context this appears to me to be a generalized statement applicable to higher than 2D curvature. As I say , haven't studied this subject at all.]

 

[PAllen]

Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry

If you look at that link, the context is geometry on a 2-surface. Popular statements don't get into the issue of embedding - they are over-simplifed. I am not just making assertions, I am asking you to think, as I'll do again. Apply your idea to 3-space "if a triangle is non-pythagorean, the *space* is non-euclidean' to flat 3-space. On a 2-sphere in 3-space, you conclude, correctly, that the 2-sphere is curved (triangle is non-pythagorean). What does that tell you about the 3-space: nothing. The 3-space is still flat.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.

 

[Q-reeus]

If you look at that link, the context is geometry on a 2-surface.

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

 

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

The whole passage refers to geometry of a 2-surface. Euclid's Parrallel postulate is a postulate about plane geometry. Spherical geometry is the geometry of a 2-sphere - the *surface* of a sphere. And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.

 

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

I am saying if you try to make Euclidean triangles in a general 4-d semi-riemannian manifold, you will succeed. For 3-space, I'm not sure whether you can *always* do it. There are fewer degrees of freedom, and the simple counting argument I used for 4-manifold does not work. It is certainly true that not all 2-manifolds can be embedded in flat 3-space, so it is plausible that some 3-manifolds will not embed a section of flat 2-surface.

Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.

 

[JDoolin]

Lorentz-contrated and time-dilated relative to what?

If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

This deserves further animations; I'd rather show you, if I can, rather than tell you. But suffice it to say, the animation above could be Lorentz transformed so as to place any particle in the center of the circle. However, because there are only a finite number of particles in the picture, there would be a gravitational asymmetry for every particle, except for one.

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

There are a couple different ways I might specify the initial conditions; (1) starting with an equipartition of rapidity (2) starting with an equipartition of momentum.

In this animation I assumed equal masses, and used equipartition of momentum.

Let

q\equiv \left \|\frac{\vec p}{m c} \right \| = \left \|\frac{(\vec v/c)}{\sqrt{1-v^2/c^2}} \right \|<3\frac{v}{c}=\sqrt{\frac{q^2}{q^2+1}}

Once I calculated the velocities, I animated, finding position, by simply multiplying the velocities by t.
The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

There's no limit to what q might be. If I set q=10^100, then all the particles that are in this animation would be so close to the center that they would experience essentially no net force.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

I have a plan in mind to present the animation from the reference frame of a particle on the edge of the mass by Lorentz Transforming all of the velocities. Definitely, as long as the total mass of the distribution is finite, then there would be particles on the edge that experience extreme accelerations toward the center. If the mass were infinite (and thus no edge), you could invoke symmetry, and there would be no net force in any direction for any particle.

But then you'd have the problem, being inside a shell of infinite mass, that at any given point inside, you are at an infinite negative gravitational potential

On the other hand, with a finite mass, the particles at the edge could experience, at least for a time, acceleration equivalent to a black hole. I'm not at all sure what theoretical ramifications that would have.

Anyway, I'll work on the other animations, and hope that makes my meaning clearer.

 

[PeterDonis]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent).

If the triangle is *assembled* in the curved space, in the right way, this will be true. But you specified that the triangle (or tetrahedron) was assembled in flat space, and then *moved* to curved space. That means the "natural" configuration of the sides is the flat configuration, Euclidean straight lines, and they must be deformed to get into the curved configuration, geodesics on a curved surface.

If, instead, you assembled the triangle carefully in a curved region, so that the "natural" configuration of its sides was as geodesics on the curved surface, and the "natural" angles at each vertex summed to more than 180 degrees, then the triangle would deform if you tried to make it conform to a flat Euclidean plane. Similarly, if you assembled a tetrahedron in a region of curved space, hovering over a gravitating body, so that it was unstressed in that configuration, and then took it far away from gravitating bodies where space was flat, it would undergo stress and deformation in the course of conforming to the flat space.

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

It depends on what you want to make an analogy to. If we are trying to construct an analogy to what happens when we take an object constructed far away from gravitating bodies and move it into a gravity well, the above is *not* a good analogy for that, because deflating the balloon corresponds to contracting the spacetime as a whole. That would do as an analogy for a collapsing universe, but *not* for moving into a gravity well. A better analogy for that would be to consider a surface like the Flamm paraboloid...

http://en.wikipedia.org/wiki/File:Flamm's_paraboloid.svg

...and think of moving a triangle from the flat region to the curved region.

 

[PeterDonis]

If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.

The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

How are you calculating the force?

 

[JDoolin]

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.
How are you calculating the force?

This is a good question, or more specifically, how should I (or how should we) calculate the force? So far, essentially, I'm not calculating the force. Invoking symmetry, there is no net force. There's a gravitational potential, but no net force on a particle at or near the center of the distribution.

But let's look at the other extreme now, from the very edge of the distribution.

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Let me give you a couple of premises of how I would go about calculating the force, and see if you agree with these premises, or if I am hopelessly naive.

#1 The net gravitational force on a particle is F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i
#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.
#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.
#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

 

[PeterDonis]

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Well, the problem as a whole has spherical symmetry--or at least, you can impose that condition as a reasonable idealization to make the problem tractable. If the problem has spherical symmetry, then the "force" on any particle (a) must point in the radial direction, and (b) must be a function only of its radial coordinate. If the matter is all ordinary matter, as I described in my last post, then (a) can be further refined: the force on any particle must point radially *inward*, i.e., the expansion of the shell must be decelerating. As I note below, viewing this deceleration as caused by a "force" may not be the simplest way to view this problem.

#1 The net gravitational force on a particle is F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.

A better way to state this would be: the 4-force on a particle is a covariant geometric object, it is the particle's rest mass times the particle's 4-acceleration, which is the covariant derivative of its 4-velocity with respect to its proper time.

#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.

A better way to state this would be: the "gravitational field" is determined by the stress-energy tensor of the matter, which is a covariant geometric object. The SET determines the field via the Einstein Field Equation. It also correctly accounts for the fact that "rest mass" is what affects gravity, not momentum due purely to kinematics.

#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

If you are assuming this, then you can't use #1 as your force equation. Consider a simple example: the force on the Earth at a given instant does *not* point towards where the Sun was 8 minutes ago by the Earth's clock. It points towards where the Sun is "now" by the Earth's clock. (Technically, there are some small correction factors, but they can be ignored for this discussion.) So if you use your #1, you have to plug in the position of the Earth relative to the Sun "now", not the "retarded" position, or you'll get the wrong answer. (Steve Carlip wrote an excellent paper some time ago that explains all this: it's at http://arxiv.org/abs/gr-qc/9909087.)

As I noted above, using #1 requires the problem to be non-relativistic, and you don't seem to be imposing that limitation. For the relativistic case, you can't really use a "force" equation for this problem, or at least it does not seem to be the easiest way to approach it. A better model would be an expanding matter-dominated FRW-type model, as I mentioned in a previous post, especially since it doesn't seem like your particles are a "shell", since there's no interior vacuum region, as far as I can see. This type of model does not view gravity as a "force"; it just solves for the dynamics of a curved spacetime using the EFE and an expression for the stress-energy tensor of the matter. You can still view individual particles as being subject to a "force", but that force can be more easily calculated *after* you have constructed the model from the EFE.

If the spatial extent of the expanding "shell" is limited, then at the surface of the shell, the FRW-type solution would be matched to an exterior Schwarzschild vacuum solution; this would basically be the time reverse of the Oppenheimer-Snyder solution for the gravitational collapse of a star.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

Huh? I don't understand what you're getting at here at all. How can you accelerate toward a past event? For that matter, how can you accelerate toward a future event? You accelerate toward a point in space, not an event in spacetime. Also, what kind of "reference frames" are you talking about? If you're talking about ordinary Lorentz frames, they are only valid locally; you can't use them to correctly evaluate distances to faraway objects.

It looks like this discussion might be better moved to a new thread, since it appears to be getting further from the topic of this one.

 

[JDoolin]

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

Okay, I see where I had one mistake.

http://en.wikipedia.org/wiki/Gauss's_law#Relation_to_Coulomb.27s_law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically-symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

I was thinking that coulomb's law and gauss law were equivalent. But they are only equivalent when the charge is not moving. I'll try to read more of the article to see what I've missed. May take me a bit of time to catch up so I don't think I'm ready to start another thread at this time.

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means. Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?

 

[PeterDonis]

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

The answer to this will be easier to understand if I answer your other question first. See below.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means.

It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.

Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?

The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.

 

[pervect]

There's a well known solution for an expanding (or collapsing) sphere of pressureless dust. Which isn't quite what was asked for, but you might be able to graft onto it to get the solution. The expanding pressureles dust sphere solution is just the FRW solution of cosmology, by the way.

As far as forces go, they're pretty much not used in GR. You can calculate them as an afterthought when you have the metric by evaluating u^a \nabla_a u^a, where u^a is your velocity vector.

[add]Conceptually, what the above expression does is calculate what an accelerometer following the worldline would read.

The reason forces aren't used much in GR that they transform in a complex manner - i.e. they don't transform as a tensor.

In SR, you can still deal with 4-forces as a tensor under Lorentz boosts. In GR, with accelerated coordinate systems, forces obviously cannot transform as a tensor. For instance, if you have an unaccelerated system with zero force, an accelerated system will have a nonzero force, but a tensor that is zero in one coordinate system must be zero in all.

 

[Q-reeus]

And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.

No problem in agreeing with that situation.

Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.

In that more generaized context I see the point, no problems.

 

[Q-reeus]

A better analogy for that would be to consider a surface like the Flamm paraboloid...
http://en.wikipedia.org/wiki/File:Fl...paraboloid.svg

Handy reminder; while I think DrGreg had posted on it much earlier, upon finding the original article at http://en.wikipedia.org/wiki/Flamm's_paraboloid#Flamm.27s_paraboloid , studied that piece with some more attention. Had never bothered to understand the significance of the w ordinate, but now appreciate how it gives a useful handle on visualising spatial part of SM. 'Inverting' w ordinate gives the sense of radial contraction as I had envisaged. Matching that to shell transition is the interesting exercise. Brings it back to on topic, and having come to appreciate the general view of the difficulties of applying a simple reading of SC's, conclude the original posting was an inappropriate vehicle for finding problems with SM. Thanks for all the inputs.

 

[JDoolin]

The answer to this will be easier to understand if I answer your other question first. See below.



It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.



The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

Although, depending on what you mean by "set up a Lorentz frame using its worldline as the t axis," I may have an even greater objection:

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames (in which case, I cannot agree; curved paths are not straight lines) or are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame?

In any case, this no longer has anything to do with shells, so I can invite you to read my comments here: https://www.physicsforums.com/showthread.php?t=545002 or if you prefer, you or I could start a new thread regarding whether Lorentz Transformations are purely a local phenomenon.

 

[PeterDonis]

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames

Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame?

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame". So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means. So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

I'll look at what you posted in the other thread and make further comments on that if need be.

 

[PeterDonis]

I'll look at what you posted in the other thread and make further comments on that if need be.

Quick answers to the questions you posed in the other thread; I won't comment in that thread unless it seems warranted.

(1) my answer is (b)

(2) my answer could be (a), but only in the vacuous sense; in the presence of gravity there are no flat regions of spacetime, there is some curvature everywhere. (b) is the strictly correct answer, but as I said in an earlier post, for a given accuracy of measurement there will be some finite region where the deviations from flatness are not observable, and within that region a Lorentz transformation on a local coordinate patch of "flat" Minkowski coordinates will work

(3) my answer is (b), but "similar" has to be interpreted carefully; Fredrik's comments in an earlier post in that thread on the properties a coordinate chart has to have to admit these transformations are worth reading

(4) my answer is "it depends"; (a) is correct *only* if the coordinate chart meets the requirements for having a rotation transformation in the first place, per Fredrik's post; otherwise the answer is undefined because there is no well-defined assignment of coordinates to objects beyond a small local coordinate patch; (b) is never correct, if the transformation is valid at all when extended to distant objects it will change their coordinate positions

 

[Passionflower]

I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

 

[JDoolin]

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Inertial movement is movement in a straight line.

Free-fall is not inertial movement. It's accelerated movement.

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

 

[PeterDonis]

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Ah, ok. We have a terminology problem. See below.

Inertial movement is movement in a straight line.

Not in GR. In SR, yes, but that's only because SR assumes spacetime is flat. In GR, where spacetime can be curved, inertial motion has to be defined physically. An object is moving inertially if it feels no force, i.e., is weightless, i.e., is in free fall.

Free-fall is not inertial movement. It's accelerated movement.

In Newtonian terms, yes. But that's because Newtonian physics defines "force" and "acceleration" differently than GR does. In GR, an object in free fall, that is weightless, *feels* no force, and therefore there is no force on it in any coordinate-independent, invariant sense, and it is *not* accelerated. More precisely, it is not accelerated in the coordinate-independent sense of "acceleration": the covariant derivative of its 4-velocity with respect to its proper time is zero. This is sometimes called proper acceleration or 4-acceleration. And since "force" in GR is defined as the object's rest mass times its proper acceleration, a freely falling object is experiencing zero force in GR.

The object is "accelerated" with respect to the Earth, but that is coordinate acceleration, not proper acceleration; similarly, the "force" of gravity in Newtonian terms is not a "force" in GR, because it does not cause any 4-acceleration of the object. That's how I was using the terms, and how they are standardly used in GR.

You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

In GR, yes. More precisely, the object's *worldline* is a geodesic of the curved spacetime around the Earth, so it's as "straight" as any worldline can be in that spacetime. I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.

 

[Passionflower]

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

 

[Passionflower]

I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.

I disagree with that.

Whether a free falling object's path in space is straight depends solely on the chosen coordinates. In simple terms, the Sun just as much goes around the Earth as the Earth goes around the Sun it simply depends on the point of view.

 

[JDoolin]

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.



Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame".

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means.

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

In the very next instant, the body will be comoving with an entirely different GLOBAL Lorentz Frame.

So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

It's like this. If I have a function whose domain is the real numbers, and its range is the real numbers, then I would say that function is valid globally. If I have a function whose domain is 0 to 1 and range is 0 to 1, then I would say the function is valid only locally.

The Lorentz Transformations take as input and output global inertial reference frames, representing every event that ever has and ever will happen in the entire universe.

 

[JDoolin]

It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

 

[PeterDonis]

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

Because SR requires initially parallel geodesics (freely falling worldlines) to remain parallel. Otherwise the whole mechanism for setting up Lorentz frames does not work.

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Then you should definitely be alarmed and troubled, because that's exactly what SR does claim. If you disagree, please provide an explicit counterexample: a geodesic path in a spacetime in which the laws of SR apply, that is not straight.

Of course it's easy to find geodesic paths in a curved spacetime that are not "straight" in the sense you're using the term, but the laws of SR don't apply in those spacetimes, precisely because they are curved. I've already given a specific example of such a law: SR requires initially parallel geodesics to remain parallel. (Or, in more ordinary language, SR requires that two objects, both in free fall and weightless, feeling no force, which are at rest relative to each other at one instant of time, must remain at rest relative to each other at all times.) In a curved spacetime, this law is violated, as my example of bodies falling towards Earth made clear.

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

Exactly. And that violates the laws of SR. Therefore, SR only applies "locally" in a curved spacetime--in a small enough region that the changes in "velocity" you speak of are not observable within the accuracy of measurement being used.

Btw, it's also worth noting that you speak of "changing velocity" without defining what that means. The 4-velocity of a freely falling body does *not* change in the coordinate-independent sense I gave in my last post: its covariant derivative with respect to the body's proper time is zero. So if you think its velocity is changing, what is it changing relative to? Any such definition of velocity "changing" will be a coordinate-dependent definition. The GR definition is not; it's a genuine physical observable (whether or not the body feels acceleration).

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

And how are the coordinates in this supposed global Lorentz frame to be defined? Can you specify a way to do it that is both consistent with all the laws of SR, *and* works in a curved spacetime, where gravity is present? If so, please elucidate.

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

If by "locations" you mean "coordinates", then yes, mathematically, this is what the LT does. That's the easy part. You have completely glossed over the hard part, *assigning* those coordinates in the first place in a way that is consistent with all the laws of SR. If you think you can do that in a curved spacetime, again, please elucidate.

 

[pervect]

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

Well, there's never (or at least not usually) a straight line in our actual universe, because space-time is in general not flat. So geodesics are as close as we come.

Geodesics are of course described by the geodesic equation.

<br /> \frac{d^2x^\lambda }{dt^2} + \Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{dt}\frac{dx^\nu }{dt} = 0<br />

Futhermore, if we multiply through by m, and replace the right hand side by a force, this is about as close as GR comes to Newton's equations of motion.

Specifically, if we have a test particle moving under the action of an external non-gravitational force, (for instance, an electric field), we can write in GR

<br /> m\,\frac{d^2x^\lambda }{ds^2} + m\,\Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{ds}\frac{dx^\nu }{ds} = F <br />

So, it shouldn't be too surprising that the Christoffel symbols act pretty much like forces. In particular, we can identify some of them as being equal to what we used called the "force" of gravity in Newtonian theory.

And in GR, we can replace solving F=ma with solving the geodesic equations (well, there are occasions where this doesn't work, and we have to worry about the Paperpatrou equations, but this is rare)

BUT

As I remarked earlier, the components of the Christoffel symbols transform in a complex way. From wiki http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=455650120 they transform like:

<br /> \overline{\Gamma^k_{ij}} =<br /> \frac{\partial x^p}{\partial y^i}\,<br /> \frac{\partial x^q}{\partial y^j}\,<br /> \Gamma^r_{pq}\,<br /> \frac{\partial y^k}{\partial x^r}<br /> + <br /> \frac{\partial y^k}{\partial x^m}\, <br /> \frac{\partial^2 x^m}{\partial y^i \partial y^j} <br />

which is not the tensor transformation law. So it's convenient to think of Christoffel symbols as "forces" in anyone particular coordinate system that you want to work in, but it's a mistake to think they'll transform in the same manner as the forces in flat space-time that you may be thinking of them as being analogous to.

 

[JDoolin]

A couple of notes that I wanted to throw out before I go to work:

But let's look at the other extreme now, from the very edge of the distribution.

(1) I wanted to point out that this distribution is NOT an equipartition of rapidity, but an equipartition of v/(c sqrt(1-1/v/c^2)

I think if I redo it with equipartition of rapidity, we'll have distribution that looks the same (density-wise) in the center after Lorentz Transformation.

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

 

[PeterDonis]

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

The cone's surface (barring the singularity at the apex) is still "flat" in the sense that it has zero intrinsic curvature. It has nonzero *extrinsic* curvature--the way in which it is embedded in the surrounding 3-D space is curved--but that's not what we're talking about in this discussion. If you calculated the connection coefficients in the formulas that pervect was writing, working with the Cartesian coordinates you assigned to the cone when it was laid out flat and you drew a straight line on it, they were zero when the cone was laid out flat and they are still zero when the cone is wrapped back up into a cone. So the cone is still flat in the intrinsic sense. (Technically, you could construct coordinates on the cone where the connection coefficients were nonzero, but the point is that you don't *have* to--there will always be *some* coordinate chart where they are all zero, whether the cone is laid out flat or wrapped up. The same is *not* true for a sphere--see below.)

Now try the same thought experiment with a sphere. You can't do it. There is no way to "lay a sphere out flat" and draw straight lines on the flat version, and then wrap it all back up into a sphere again. It's impossible--any such operation will distort the surface and invalidate its geometric invariants. That's one way of expressing the fact that a sphere has nonzero *intrinsic* curvature. The connection coefficients in pervect's formulas are nonzero on a sphere (i.e., there is *no* coordinate chart on the sphere that makes them all zero). And that's the kind of curvature we're talking about when we say that gravity is curvature of spacetime.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

No, you are not even seeing that. Light is bent by gravity. This has been measured for light passing close by the Sun: take two stars that are a little further apart in the sky than the apparent diameter of the Sun, when the Sun is in a different part of the sky--say the are separated by an angle a. When the Sun is in between them in the sky, they are separated by a *larger* angle, a + da, because the Sun bends their light towards itself. And what's true for the Sun is true for any large gravitating body.

The spacetime we are living in has intrinsic curvature because of gravity; it is a spacetime analogue of something like a sphere (actually more like a saddle, but the same argument I gave for a sphere would apply to a saddle too). It is *not* the spacetime analogue of something like a cone or a cylinder that can be laid out flat and wrapped up again without changing its intrinsic geometry. And in the presence of intrinsic curvature, all your intuitions about how things work in flat Euclidean space, or flat Minkowskian spacetime, are simply wrong on any large scale; they only approximately work over very small patches. You can only treat the Earth's surface as flat over a small area, and you can only treat spacetime as flat over a small region.

 

[JDoolin]

I have made several further posts in the other thread.

https://www.physicsforums.com/showthread.php?t=545002

I think I have explained myself better there.

I'm trying to find the most concise description of our disagreement: You seem to take the attitude that infinite straight lines do not exist. And I take the attitude that infinite straight lines do exist, or at least are definable. In my thinking, regardless of the curvature of paths caused by gravity, it is always possible to imagine what would happen if matter wasn't there. In your thinking, regardless of how we try to imagine what would happen if matter wasn't there, there would always be more matter, screwing things up. Now I won't argue with that, but the question is not whether we can really map the straight lines with perfect accuracy. The question is whether those straight lines exist at all. I think they do. You think they don't.

I explain why I think they do in the other thread.

 

[PeterDonis]

I'll comment in the other thread.

 

[Dale]

I will do that. It will take a few days.

This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

 

[George Jones]

This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

 

[Q-reeus]

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

Try #17 where the specs were set out.

 

[George Jones]

I am looking for a new post, written in succinct scientific style (without a lot commentary), of the form

"For a matter distribution given by ..., calculate ..."

 

[JDoolin]

Here's an idea for a non-stationary matter distribution:

Assume you have 5000 particles, originating from an event (t=0,x=0) and each is assigned a random x and y rapidity between -3 and 3, traveling away from each other in the xy-plane. Assume that these particles move with constant velocity until time t=1.

At time t=1, the particles spontaneously develop mass. Given one of the particles, calculate the force on this particle resulting from the other 4999 particles.

(The proper-time/coordinate-time of the spontaneous development of mass, and the delay before that mass is detected, would need to be more fully described to answer the question.)