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

 Quote by PAllen Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).
You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.

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 Quote by pervect You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.
Q-reeus introduced hoops (to simplify? to detect orientation? not quite sure). So, I want to emphasize it will never work.

The simplest thing you can do with spheres to detect curvature is up for discussion. I don't think anyone here so far claims to know what the simplest construction involving spheres or parts of spheres that would detect curvature is.

[The hard part of detecting curvature from purely spatial measurements is avoiding embedding artifacts. Anything dependent on a particular foliation doesn't cut it. So, if one hypothesized that volume to surface area ration differed from 1/3, you would have to show that there does not exist any foliation in which the ratio is 1/3.]
 Recognitions: Science Advisor Staff Emeritus Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice. I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up. Another minor issue is that the Riemann of a plane only has 1 component, but the Riemann of a three-space should have 3. So the circle-packing tells us as much as we can know about the curvature of a plane, but sphere-packing doesn't tell us everything about the curvature of some particular spatial slice.

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 Quote by pervect Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice. I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up.
Kinda sorta. You bring up a good point, we've been using the word "curvature" without always being clear about what kind.

When I said that tidal gravity is the same as curvature, I meant specifically *spacetime* curvature. (I said so explicitly at least once.)

There is also, as you say, the curvature of a spatial slice. That, of course, depends on how you cut the slice, so to speak, out of spacetime. Also, as you note, the sphere packing, which measures what I've been calling the K factor, is not a complete measure even of the spatial curvature. (Also, as I've noted, the measurement you describe samples the K factor over a range of radial coordinates, or sphere areas, so it's more complicated than just measuring the K factor between two spheres that are very close together. I'm trying to stick to the "local" case, where K is effectively constant, until we get that sorted out, before bringing in variation in K.)

(Another minor point is that what you described is the *intrinsic* curvature of the spatial slice; there is also the extrinsic curvature of the slice, which is something else again.)

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 Quote by pervect Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice. I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up. Another minor issue is that the Riemann of a plane only has 1 component, but the Riemann of a three-space should have 3. So the circle-packing tells us as much as we can know about the curvature of a plane, but sphere-packing doesn't tell us everything about the curvature of some particular spatial slice.
Well I was interested in something intrinsic. I see no reason you can't construct a non-euclidean spacelike 3-surface in Minkowski flat spacetime. What would be the physical significance of that? Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric). So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.

Separately, I don't know if every 3-surface with non-vanishing Riemann tensor must deviate from the Euclidean sphere area/volume ratio. Have you determined that this is true?

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 Quote by PeterDonis Kinda sorta. You bring up a good point, we've been using the word "curvature" without always being clear about what kind. When I said that tidal gravity is the same as curvature, I meant specifically *spacetime* curvature. (I said so explicitly at least once.) There is also, as you say, the curvature of a spatial slice. That, of course, depends on how you cut the slice, so to speak, out of spacetime. Also, as you note, the sphere packing, which measures what I've been calling the K factor, is not a complete measure even of the spatial curvature. (Also, as I've noted, the measurement you describe samples the K factor over a range of radial coordinates, or sphere areas, so it's more complicated than just measuring the K factor between two spheres that are very close together. I'm trying to stick to the "local" case, where K is effectively constant, until we get that sorted out, before bringing in variation in K.) (Another minor point is that what you described is the *intrinsic* curvature of the spatial slice; there is also the extrinsic curvature of the slice, which is something else again.)
I agree with everything you wrote, though I'm still not sure which one of the various aspects of curvature is of interest. I suspect that the idea is just to overall describe curvature.

I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it. So I'm mostly interested in intrinsic curvature. I suppose that the extrinsic curvature might be of some use if you're doing ADM stuff, but it's outside the scope of my current interests.

As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor, and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition.

There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature.

The remaining part is the magneto-gravitic part, that describes frame dragging effects.

So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.

Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.

If we focus on the deviation between a reference geodesic and nearby geodesics (via the geodesic deviation equation), we can neatly categorize the various parts of the Bel decompositon as follows:

The geodesic deviation (the relative acceleration between nearby geodesics) will depend on both the spatial separation (and be proportional to it), and will also depend on the relative velocity (said velocity being measured in the fermi-normal frame of the reference geodesic).

The deviation turns out to be quadratic with respect to the velocity. The terms independent of velocity, presnet at zero velocity, will give rise to the electro-gravitic part of the tensor, and are described by the tidal forces.

The parts that are proportional to velocity describe the magnetic part.

The parts that are proportional to velocity squared are due to the spatial curvature (i.e. the topogravitic part of the tensor). They're rather analogous to the v^2/R type forces that an object moving in a circular path of radius R with velocity v experiences.

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 Quote by PAllen Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric).
You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head.

 Quote by PAllen So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.
Provided it isn't one of the special cases.

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 Quote by pervect I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it.
Not the extrinsic 3-curvature of a spatial slice taken out of a 4-d spacetime. But I agree extrinsic curvature isn't very interesting compared to intrinsic; certainly not for the topic of this thread.

 Quote by pervect As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor,
Agreed.

 Quote by pervect and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition. There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature. The remaining part is the magneto-gravitic part, that describes frame dragging effects. So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.
Well, the term "tidal gravity" may be a bit ambiguous. I posted earlier the definition I was using: *anything* that causes geodesic deviation. All parts of the Riemann tensor describe this in some form; you describe how later in your post.

I recognize now that my terminology may be non-standard. I first saw it in Kip Thorne's Black Holes and Time Warps, and it made sense to me. But my arguments in this thread are essentially unchanged if "tidal gravity" is taken to describe just the electrogravitic part of the full curvature tensor, as you say. It's still true that what I am calling the K factor is not the same as tidal gravity, and is not equivalent to it, nor is it equivalent to curvature in full.

 Quote by pervect Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.
I think this is true, yes.

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 Quote by PeterDonis You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head. Provided it isn't one of the special cases.
Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.

Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.

 Quote by PeterDonis You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat.
Yes an observer traveling radially at escape velocity observes his space as flat in that the distance between two r-coordinate values is exactly the difference. Other observers, who are also on a radial path or stationary, observe this distance differently. One can obtain this distance by applying the Lorentz factor (local speed wrt a free falling at escape velocity observer) before integration between these two r-coordinates.

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 Quote by PAllen Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.
Yes, I agree.

 Quote by PAllen Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.
Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true.

To be more precise in stating what this means: inside the horizon, the vector $\frac{\partial}{\partial T}$ corresponding to Painleve time T *and* the vector $\frac{\partial}{\partial r}$ corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector $\frac{\partial}{\partial T}$, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to $\frac{\partial}{\partial T}$ *outside* the horizon either, because of the dTdr term in the Painleve metric.)

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 Quote by PeterDonis Yes, I agree. Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true. To be more precise in stating what this means: inside the horizon, the vector $\frac{\partial}{\partial T}$ corresponding to Painleve time T *and* the vector $\frac{\partial}{\partial r}$ corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector $\frac{\partial}{\partial T}$, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to $\frac{\partial}{\partial T}$ *outside* the horizon either, because of the dTdr term in the Painleve metric.)
Fascinating. After a little thought I agree. So I guess the SC geometry everywhere has too many symmetries to preclude construction of flat spatial slices.

 Quote by PAllen Originally Posted by Q-reeus: "Unless you can prove that the marble filled hoop will *not* experience changed packing density (restraint = fixed marble count) in heading towards pointy end, you have to face the fact that locally observed effects are present. And one possible *interpretation* by a local flat-land observer, who can't discern curvature directly, is varying hoop size, or alternately, shrinking marbles. Stresses can't explain it, but effects normally put down to changing container size and/or marble size are there. All you have to do to end that argument, is what I asked above - can the hoop packing density/number be independent of surface curvature?" Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).
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.

 Quote by pervect You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.
And in #139:
 "Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice. I think thats what was wanted, though I haven't been following in detail and the thread is too long to try and catch up."
Full marks from me for carefully reading what I have been arguing - yes got the drift admirably.
 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?

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