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How does GR handle metric transition for a spherical mass shell? 
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#127
Oct2711, 12:00 PM

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I wonder if the following would be true (it seem intuitively plausible based on Synge's results combined with Peter Donis's findings; I wouldn't rely on this without calculating it, though):
Take a ruler with marks r1 and r2 on it. In a region of curvature, use it to lay out two concentric spherical surfaces of a chosen solid angle. The relation of surface areas to r1, r2, and solid angle will be strictly Euclidean. But what about the volume between them measured with Peter's little marble idea? My guess is that it will not match the Euclidean prediction. 


#128
Oct2711, 01:02 PM

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[Hardening my stance on this. Comments in #123 apply: "While your concentric circles around north pole analogy in #99 talked in terms of perimetertoradius ratio, one could equally talk in terms of an enclosed surface areatoperimeter ratio of a dished annulus (numerically different, but having in common dependence on surface curvature). Once you see it the latter way, the hoop thing springs out as a more evident manifestation that local phenomena will exist, which is why I used it." There is a marble count excess for dished annulus sitting on 2sphere (2D analogue of spherical shells experiencing 3curvature). Same general effect must apply to a hoop, and likewise 3D container] 


#129
Oct2711, 01:20 PM

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This is why I keep saying we should stick to the local case first; it avoids introducing confounding factors like tidal gravity that are *not* the same as the K factor. If you keep muddling these things together, you will keep on being confused. Once again: the K factor does *not* cause any stress on objects. Therefore, the K factor *cannot* change the physical size of a container; that would require causing stress on the container's walls. This is a basic point of the physics involved, and if it's not clear, we need to stick to the local case until it is. (If you insist on something about the nonvacuum case, inside a solid object, the stressenergy tensor is nonzero, so yes, there are additional forces "pulling" on a small object. But those forces also are *not* the K factor; they are related to it, in the sense that they are also functions of the radial coordinate r, but they are *not* the same. We can try to work that out after we get the "local", vacuum case clear.) 


#130
Oct2711, 01:23 PM

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#131
Oct2711, 03:10 PM

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Where did this start? Oh yes  your argument in #126 that varying K, just over the radius of a hoop, throws out the ability to accurately determine local curvature via marble count: "Yes, but as I pointed out, to calculate the count, you need to know the K factor for a whole range of "r" values, from r = 0 out to the "r" of the hoop, which is its circumference divided by 2 pi. This brings in additional complications which are not present if you consider the area between two nearby circles of circumference C and C + dC." Are we yet again misunderstanding each other's words? I took the above to mean, since it was not specified any more clearly, that 'varying r' was from the center of the hoop to it's periphery. And you meant something different? What exactly? If not, you are saying gradient of K locally matters re count  just as I thought. If not, how should one take it to mean? By the time that bit was written, you were quite aware of what I meant by hoop. "Let's take your analogy of north pole  or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that noneuclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed  we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg  with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?" Notice  one hoop, sampling a varying surface curvature. As it does so, the marble packing density alters  gaps will open  and to maintain packing density, one every now and then 'pops an extra one in'. Hope this part at least is perfectly bedded down. Sheesh. 


#132
Oct2711, 04:49 PM

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Peter is correct that if there is curvature of spacetime in a region, then there is tidal gravity. There may also be noneuclidean effects on space alone, these being much harder to detect. 


#133
Oct2711, 04:58 PM

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#134
Oct2711, 04:59 PM

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I should note that by "tidal gravity" I include *any* effect that causes initially parallel geodesics to converge or diverge. That's exactly what the Riemann curvature tensor captures. 


#135
Oct2711, 04:59 PM

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#136
Oct2711, 05:09 PM

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However, the last part of that paragraph is *not* about the local issue: 


#137
Oct2711, 05:21 PM

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#138
Oct2711, 05:31 PM

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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.] 


#139
Oct2711, 06:15 PM

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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 spacetime 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 threespace should have 3. So the circlepacking tells us as much as we can know about the curvature of a plane, but spherepacking doesn't tell us everything about the curvature of some particular spatial slice. 


#140
Oct2711, 07:06 PM

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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.) 


#141
Oct2711, 08:03 PM

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


#142
Oct2711, 08:27 PM

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I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of spacetime 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 spacetime 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 electrogravitic 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 topogravitic part. This describes the purely spatial part of the curvature. The remaining part is the magnetogravitic 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 spacetime, 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 ferminormal 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 electrogravitic 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. 


#143
Oct2711, 08:49 PM

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#144
Oct2711, 09:03 PM

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I recognize now that my terminology may be nonstandard. 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. 


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