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

by Q-reeus
Tags: handle, mass, metric, shell, spherical, transition
P: 1,115
 Quote by PeterDonis 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.
See now I wrongly claimed to have 'perfectly understood' your previous #13 posting. Being unfamiliar with K and J as standard objects in GR, had thought them merely convenient and arbitrary symbols used on an ad hoc basis here. Alright I now finally get it that 'r' as SC's radial coordinate is not really equivalent to physical r, but a derived quantity based on K operating on A. Very confusing but I suppose something has to go in the interests of 'local invariance'. 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. Which gets to the next part.

 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.
An important one in my mind in untangling SC 'r' from 'actual' r, and so for tangent length. 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. Redshift is abberation free in GR so not a problem. You are now quite aware the test sphere was to be considered locally stress free. So only effect distant observer determines is underlying metric distortion. Given all that, and to the extent SM is correct, am I not entitled to expect as distant observer to see a distorted test sphere as oblate spheroid as per the Sr = J (or close to it), St = 1? This is an important part of my reality check list.
 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: $$J = 1 + 2 \phi$$ where $\phi$ 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.
Too late now to re-edit #1, but from now on will be sticking with more standard usage. Using V as label for 'the Einstein potential' -g00 rather than the redshift factor J you have used, created possible confusion with it's more familiar use; as the Newtonian potential -GM/r. Apologies for any confusion caused.
Physics
PF Gold
P: 6,166
 Quote by Q-reeus 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.
This is true, and I didn't try to calculate what the actual stresses inside the globe due only to its own gravity would be.

 Quote by Q-reeus 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.
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.
Physics
PF Gold
P: 6,166
 Quote by Q-reeus Does this not create a circular definition paradox though: r defined in terms of A, but A expressed in terms of r?
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.

 Quote by Q-reeus 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.
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.

 Quote by Q-reeus 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.
In order to account for gravitationally induced optical distortion, you first have to have a model that says what it is, so it can be compensated for. That's the part that I don't think is very simple: you have to evaluate what a null geodesic will look like as it passes through spacetime with varying curvature.

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.
P: 1,115
 Quote by PeterDonis ...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.
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. As I surmised in #1, there I asume no 'directionality' to the contribution from stress of an element of matter any different to it's mass? Very different when motion/flow is involved of course, but that does'nt concern our situation.
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P: 17,333
 Quote by Q-reeus 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?!
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.
Physics
PF Gold
P: 6,166
 Quote by Q-reeus 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.
For determining the inward "pull of gravity", you are correct; the relevant quantity is $\rho + 3 p$, and the pressure p would be negligible in the case we're discussing. However, the "pull of gravity" comes under the heading of "time components" of the curvature. We're talking about "space components", where the energy density does not come into play.
P: 1,115
 Quote by DaleSpam 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.
Something is drastically not adding up here imo. 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?
Physics
PF Gold
P: 6,166
 Quote by Q-reeus 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?
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.
P: 1,115
 Quote by PeterDonis 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.
What I meant was not just that r is defined as A = 4*pi*r2, but that from the SC's the θ and phi tangent components are themselves functions of r. So it still seems circular. One must somehow have a good handle on r in order to have that 'primitive' of A determined, surely.
 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.
Is there not a clear sense in which one is optical - light bending, while the other is inherent - the metric distortion as physical 'object'?
P: 1,115
 Quote by PeterDonis 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.
Here's where the issue seems to be hitting the fan. I entirely meant shell matter's contribution to the exterior, SM region, and said so explicitly in #25. So no confusion on my part there. Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall. There is a severe logical chasm imo. Exterior, *presumably* anisotropic SM generated by essentially shell matter exclusively. Transition to interior flat region demanding 'steep' gradients in at least one spatial metric component. Relative pressure arbitrarily small in self-gravitating case for 'small' shell. Are we trying to pull a rabbit out of a hat, folks?
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! Bed time.
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PF Gold
P: 6,166
 Quote by Q-reeus 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!
But why would the appearance of a test sphere to someone far away make any difference, when we've already established that, to an observer right next to the test sphere, it would appear spherical, *not* distorted? To me, that local observation is a much better gauge of whether space is isotropic than the observation of someone far away, particularly when the light going to the faraway observer could be distorted by the variation in gravity in between.

Perhaps I introduced confusion when I used the word "anisotropy" to refer to the fact that there is more distance between two neighboring 2-spheres in the Schwarzschild exterior region than Euclidean geometry would lead one to expect based on the areas of the spheres. The only thing that that shows to be non-isotropic is the "Euclidean-ness" of the space. I've already described in detail how that "anisotropy" goes away as you descend through the non-vacuum "shell" region; if you want a "physical" explanation of how that works, I would say it's because the non-Euclideanness of the space is due to the mass of the shell being below you, as you descend through the shell, less and less of its mass is below you. "Below" here just means "at a smaller radius", or, if I were to be careful about describing everything in terms of direct observables, "below" means "lying on a 2-sphere with a smaller area than the one you are on".

It's also worth noting, I think, that even the "non-Euclideanness" of the space, as I described it, is observer-dependent; it assumes that the non-Euclideanness is being judged using 2-spheres which are at rest relative to the shell. Observers who are freely falling through the exterior vacuum region will *not* see this anisotropy in the space that is "at rest" relative to them; in other words, if a freely falling observer were to measure the areas of two adjacent 2-spheres that were falling with him, and measure the distance between the two 2-spheres, he would find the relationship between those measurements to be exactly Euclidean.
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P: 17,333
 Quote by Q-reeus 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?
The exterior SM is a vacuum solution, so there isn't any matter contribution in the exterior SM.

 Quote by Q-reeus 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?
That transition requires non-zero spatial components of the stress-energy tensor, i.e. stress and pressure. The large contribution of the energy density is in the wrong place to matter for the spatial components of the curvature.
 Sci Advisor PF Gold P: 1,848 The diagram below of a "Flamm's paraboloid" may aid understanding the curvature of space (but not spacetime) outside the event horizon of a black hole. The surface represents a 2D cross-section through 4D spacetime involving the $r$ and $\phi$ coordinates only. In this diagram $r$ is the radius in the horizontal direction. "Ruler" distances in space (e.g. PeterDonis's method of counting small identical objects packed together) are represented by distances along the curved surface. (The vertical direction has no physical meaning at all.) Allen McCloud, Wikimedia Commons, CC-BY-SA-3.0The diagram goes all the way to the event horizon, where the surface becomes vertical at the bottom of the "trumpet". So, for the example being discussed in this thread, you need to slice off the bottom part of the surface. 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.) The mathematics of Flamm's paraboloid is discussed on Wikipedia at Schwarzschild metric: Flamm's paraboloid. P.S. Flamm's paraboloid does not represent the gravitational potential. The potential has a somewhat similar shape but it's a completely different formula.
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PF Gold
P: 6,166
 Quote by DrGreg 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?
PF Gold
P: 1,848
 Quote by PeterDonis 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?
PF Gold
P: 5,060
 Quote by DrGreg 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?
This is leading to a thought I've been having for a while, but didn't want to distract this thread. Simply, why bother with a shell at all? In this case, you have a perfectly continuous but non-smooth manifold joining the SC geometry to Minkowski geometry inside a chosen r value. The metric can be made continuous, a single global chart can be used, you just have metric derivatives undefined on one surface. This is perfectly analogous to taking a ball and cutting it in half, and treating the resulting boundary as a 2-manifold. No one would consider that a weird or inadmissable surface from a geometric point of view. As for physics, the shell-less geometry would lead to failure of EFE on the junction surface (only). It may simplify things if Q-reeus could state his fundamental issue for this simpler case - I still don't get his confusion at all. This shell-less manifold wouldn't be strictly physical, but all physics off the shell, and almost all physics going through the shell (e.g. geodesics) would be perfectly well defined.
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P: 17,333
 Quote by Q-reeus Agree with essentially all the rest above, but not the probable implication that stress in the shell can account for anything remotely significant re transition through shell wall.
Why not? Your argument regarding the relative size of the pressure and energy density is not relevant. So what logical reason could you have to disagree?

 Quote by Q-reeus There is a severe logical chasm imo.
Agreed. The reference that I posted derived the pressure terms directly from the Einstein field equations. So there is no logical gap there. But you, without touching the math, claim to know that the effect is too small. A logical chasm indeed.
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