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

Actually, since I posted the assessment in #8, I owe you a mea culpa, Q-reeus; after thinking this over and thinking about DaleSpam's comment, I think my assessment in #8 was incorrect, because I was comparing apples and oranges. The Earth is not a hollow sphere made of steel. The case you just gave in the above quote is a much better test case, and as we'll see, it does *not* support the claim that pressure is negligible.

I did say in #8 that the key factor is not whether pressure is small compared to energy density, but whether the ratio of pressure to energy density is small compared to the "correction factor" in the metric coefficients. So let's estimate those numbers for a "toy globe". Let's suppose our "toy globe" is a spherical shell made of steel; its total radius is 1 meter, and the shell is 0.1 meter thick (with a hollow interior). We'll assume that the stress components inside the shell are of the same order as atmospheric pressure, or about 10^5 pascals. (They may actually be somewhat larger, but we'll use the lower bound since that will make the pressure/energy density ratio as small as possible.) The mass density of steel is about 8000 kg/m^3.

Energy density of the shell = 8000 kg/m^3 x c^2 = 7 x 10^19 J/m^3

Ratio of pressure to energy density = about 10^-15

Volume of the shell = 4/3 pi x (1^3 - 0.9^3) = about 1 m^3 (how convenient!).

Mass of the shell = 8000 kg

Correction to metric coefficient at shell surface = 2GM / c^2 r
= (2 x 6.67 x 10^-11 x 8000) / (9 x 10^16 x 1) = about 10^-23

So I was wrong in post #8; for a typical object made of ordinary materials, the pressure is *not* negligible; the pressure/energy density ratio is many orders of magnitude *larger* than the correction to the metric coefficients for the object. So there is plenty of "room" for the spatial parts of the stress-energy tensor inside the object to "correct" the metric to flat from the outside to the inside of the shell. I should note, though, that even the above computation is not really "correct", in that I haven't actually tried to compute any components of the field equation; I've just done a quick order of magnitude estimate of the same quantities I estimated in post #8, to show that the relationship I talked about there doesn't hold for an ordinary object.

There is a definite meaning to "radial distance", yes: in principle, we could line up our tiny measuring objects from some given sphere with area A, all the way to the center of the whole scenario, and count them. The radial distance measured this way, if we start from a sphere in one of the regions where K > 1 (EV or NV), will be larger than the area A divided by 4 pi. However, the exact relationship between the two will be complicated, because it has to take into account how K varies from the sphere with area A all the way to the center of the scenario. It's simpler to state things in terms of the differential area and volume as I did because those quantities are "local", at least in terms of radial movement (and since we're assuming spherical symmetry and time independence, everything can vary only as a function of radial movement), so K can be assumed constant for any given pair of spheres with areas A and A + dA.

But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates.

Originally Posted by Q-reeus:
"Gets back I suppose to that other thread where I asked for what a distant observer will see through a telescope - a distorted or undistorted test sphere. A sensibly and physically real Sr implies oblate spheroid will be observed."
This is a more complicated question as well because it requires you to evaluate the path of the light rays from the object to the distant observer, and the "scaling factor" and time dilation factor will change through the intervening spacetime. It may well be that you are correct that the anisotropy I described (more volume between spheres A and A + dA than Euclidean geometry would lead you to expect) will be seen by a distant observer as a test sphere appearing distorted, but it's not as straightforward a question as you seem to think it is.

But my whole point is that K and J are the "primitives". K and J are coordinate-independent; they can be directly measured in terms of local observations (K is how much the volume between two spheres with area A and A + dA exceeds the Euclidean value, and J is the observed gravitational redshift/blueshift factor at a given sphere with area A). If by "V" you mean what you have been calling the "potential", the coordinate-independent definition of that would be made in terms of J, the "gravitational redshift" factor, via the usual definition:

[tex]J = 1 + 2 \phi[/tex]

where [itex]\phi[/itex] is the potential in units where c = 1, and with the usual convention that the potential is zero at "infinity" and negative in a bound system such as this one. But this potential is not what is directly observed; that's J.

Still struggling to reply effectively to your two very helpful earlier posts, but will deal with this one now. I can't see any mea culpa. First, had in mind the relevant stresses in the globe are those owing only to gravitational self-interaction (i.e. - it's floating out there in space). Vastly smaller than and nothing to do with any atmosperic pressure.

But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10^-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10^-23. So rho contributes a fraction (1-10^-15), while p's contribute a fraction 10^-15.

Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r?

Area and volume have to be based on linear measure somehow, so is it not still down to picking a linear length measure as 'the' proper yardstick? How else to climb out of this hole? My hunch is tangent spatial measure is implicitly that yardstick.

Gravitationally induced optical distortion can in principle be fully accounted for, either directly with corrective optics, or by computer processing of image data, just as e.g atmospheric distortion is compensated for in modern astronomy.

...No, this is not correct. The various components of the stress-energy tensor match up with different components of the curvature, which in turn match up with different components of the metric. So the ratio of pressure to energy density may not be relevant to the effect on the metric. That's why I said in my last post that the calculation I gave there wasn't really "correct"; the correct thing to do is to look at the Einstein Field Equation and figure out how each component of the stress-energy tensor affects the curvature, and through that the metric. That's also why DaleSpam has been saying that the pressure may be relevant even though it's much smaller than the energy density. The paper he referenced gives an example where the effects of pressure are certainly not negligible.

But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10^-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10^-23. So rho contributes a fraction (1-10^-15), while p's contribute a fraction 10^-15. Isn't that still the only important consideration on this? How can something 10^-15 times smaller than the other be overwhelmingly dominant?!

Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients.

Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.

Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution? And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?

No, because I didn't express A in terms of r; I used A to *define* r. A is the "primitive", the actual observable quantity, and I defined it in terms of covering the 2-sphere with little identical objects and counting them...

...In the sense that to cover the 2-sphere with little identical objects and count them, you have to place the objects tangentially, yes, I suppose it is. But as I said before, that's because we are assuming spherical symmetry, so the area of the 2-sphere seems like a good benchmark, or "primitive", on which to base other things.

Also, since a test sphere in this spacetime, to an observer next to it, will *not* appear distorted--it will be spherical--if it does appear distorted to an observer much further away from the black hole, that distortion would have to be "gravitationally induced optical distortion", would it not? If so, you wouldn't want to "compensate" for that, since it would eliminate the effect you are interested in.

I think you're confusing the metric and curvature in the exterior vacuum region with the metric and curvature in the non-vacuum "shell" region. Everything DaleSpam and I have been saying applies only to the metric and curvature in the non-vacuum region. In the exterior vacuum region the stress-energy tensor is zero; there is no pressure or energy density, so there is no "matter contribution" at all. The curvature of spacetime in the exterior region is determined by the overall mass of the shell, via the form of the Schwarzschild metric, but that metric is a solution of the *vacuum* Einstein Field Equation (stress-energy tensor equal to zero). The exterior metric does not depend on any details of the shell's internal structure; two shells with identical overall mass, but drastically different energy density (say one is much thicker than the other, and correspondingly much less dense), would lead to the same metric in the exterior vacuum region. But within the shell, the metric would be quite different for those two cases: the spatial metric would have to change to flat over a much shorter distance for the thin shell, so the curvature components would have to change must faster as you descended through the shell.

Now this would all go away if in fact hte SM is *actually* isotropic spatially, so are SC's fooling us? This is why I want that test sphere result so bad!

Do you agree that the overwhelmingly larger matter contribution is what will determine/generate the exterior SM? And that spatial anisotropy is inherently present there - owing to that matter contribution?

And that there *must* be significant change to at least one spatial metric component upon transition to the flat MM interior? Well how does all this square up with what you have been saying?

The interior of the shell could then be represented by a horizontal flat circular disk almost capping the bottom of the remaining trumpet, but there is then a gap to be bridged between the disk and the trumpet to represent the shell. I've no idea whether is possible to construct a curved surface to bridge that gap which would correctly represent the geometry within the shell. (I suspect it might not.)

I was thinking of the "bridge" surface as being basically a section of a torus, with the "bottom" part merging into the horizontal circular disk representing the interior, and the "top" part having just the right slope to merge into the exterior paraboloid at the appropriate circular "cut" from it. If "torus" is interpreted generally enough (basically to allow an elliptical "cross section" as well as a circular one), would that be a plausible candidate?

It certainly sounds plausible -- that's pretty much what I had in mind myself. But on the topic of embedding an arbitrary curved manifold into a higher-dimensional Euclidean space, I am a bit out of my depth. I think, in general, there's no guarantee it can be done with just one extra dimension, i.e. you usually need more. It may be just good luck that Flamm's paraboloid works in just 3 dimensions for the exterior Schwarzschild solution.

And even if you can produce a 2D curved surface in 3D space to represent the entire "shell" spacetime, I'm not sure if the "junctions" (where there's a discontinuity in the energy-momentum-stress tensor from zero to non-zero) would need to be smooth -- maybe there would a sharp "crease" in the surface at these points?