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

1. Oct 17, 2011

### Q-reeus

This is really a continuation from another thread but will start here from scratch. Consider the case of a static thin spherical mass shell - outer radius rb, inner radius ra, and (rb-ra)/ra<< 1, and with gravitational radius rs<< r(shell). According to majority opinion at least, in GR the exterior spacetime for any r >= rb is that given by SM (Schwarzschild metric) as expressed by the SC's (Schwarzschild coordinates) http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric. In the empty interior region r=< ra, a flat MM (Minkowski metric) applies. Owing probably to it's purely scalar nature, there seems little controversy as to the temporal component transition, as expressed as frequency redshift scale factor Sf = f'/f (f' the gravitationally depressed value as seen 'at infinity'). In terms of the potential V = (1-rs/r)1/2, with rs = 2GMc-2, one simply has Sf = V, which for any r=< ra has it's r parameter value 'frozen' at r = ra. There is a smooth transition from rb to ra that depends simply on V only. So far, so broadly reasonable.

What of the spatial metric components? In terms of corresponding scale factors Sr, St that operate on the radial and tangent spatial components respectively, it is readily found from the SC's that Sr = V, St = 1, everywhere in the exterior SM region. Sr = V is physically reasonable and identical to frequency redshift factor Sf here. Whether according to GR Sf and Sr diverge for some reason in the transition to the interior MM region is not clear to me, but that would be 'interesting'.
An apparently consensus view that St dives relatively steeply from St = 1 at r = rb, to some scaled fraction of V at r=< ra. [A previous attempt here at PF at the shell transition problem found something different; an invariant St = 1 for all r, but an Sr that jumped back from V to 1 in going from rb to ra. That gave flat interior spatial metric with unity scale factors Sr, St (i.e. 'at infinity' values) but a redshift factor Sf = V]. Seems though the consensus view is for an equally depressed Sr, St in the interior MM region, the exact value of Sr, St, and Sf relative to V is unclear from recent entries on that matter, where parameters A, B, were used as equivalent to Sf, (Sr, St) respectively.

I wish to focus on the question of physical justification for St diving from unity to some fraction of V. Given that an invariant St = 1 for all r >= rb (SM regime) stipulates complete independence on V or any combination of it's spatial derivatives ∂V/∂r, ∂2V/∂r2, etc. In the shell wall region, where shrinkage of St apparently occurs, there is owing to a nonzero stress-energy tensor T a different relative weighting of V and all it's derivatives to that applying for r => rb, but otherwise, only the weighting factors are different. What permits variance of St in one case, but not the other? An answer was that in the shell wall where T is nonzero, the Einstein tensor G (http://en.wikipedia.org/wiki/Einstein_tensor) operates and this is the explanation.

That seems unsatisfying, and should be justifiable at a basic, 'bare kernel' level. By that is meant identify the 'primitives' from which everything in G can be derived, and show what particular combo leads to a physically justifiable variance of St, applying nowhere but the shell wall region. So what are the 'primitives'. I would say just V and it's spatial derivatives, which owing to the spherical symmetry, are of themselves purely radial vector quantities (but obviously not all their combinations as per div, curl etc).

One caveat here is to nail down the relevant source of V - taken simply as total mass M exterior to rb. Has been pointed out that for a stable shell there must be pressure p present in addition to just rest matter density ρ, ie T = ρ + p, = T00 + T11+T22+T33. My assumption is that for a mechanically stable thin shell of normal material, ρ >>> p, and so to a very good approximation, just use ρ. If one feels the p terms should be included regardless, then I would further suggest they will act here just as a tiny addition to ρ. That is, the contribution of T11 for instance in some element of stressed matter introduces no 'directionality' per se to the potential, yes?

Can't think of any physical quantity - relevant to this case anyway - that could not be expressed as some function of the above primitives. But recall, all these primitives exist in the region exterior to rb, where St is strictly = 1!. Only remotely relevant quantity I can think of that *may* be zero in the exterior region but obviously nonzero in the wall region is the three divergence nabla2V. And that could account for a varying St? Can't imagine how. So something real, real special has to be pulled out of the hat imo. So special I consider it impossible, but open to be shown otherwise. Seems to me the anomaly is intractable in GR and a cure requires a theory where isotropic contraction of spatial and temporal components apply. Then and only then the transition issue naturally resolves. But that's my opinion. So, any GR pro willing to give this a go, let's get on with the show!

Last edited: Oct 17, 2011
2. Oct 17, 2011

### atyy

3. Oct 17, 2011

### Staff: Mentor

Don't know if you saw this in the other thread:
The tangential pressure components of the stress energy tensor. There is no tangential pressure in the vacuum region, there is in the matter region.

Here is a arxiv paper you may like. It uses an analytical model for the shells, so it is not the usual "step function" you would normally consider, but it describes things like the radial and transverse pressures:
http://arxiv.org/abs/0911.4822

4. Oct 17, 2011

### Q-reeus

Thanks for the links but need some expert translation here. I get that extrinsic curvature jumps across a thin boundary proportional to the boundary 'surface' stress-energy density (link 2 - the others are highly technical). Now specifically how does the extrinsic curvature relate to tangent metric factor St for shell case? There should surely be a relatively simple function in ρ and r; St(rho,r) = f(rho,r) which can/must be decomposable into just V and derivatives thereof. Maybe it's there in those links, but if so someone has to point to just where. If not, I'm still inclined to think maths divorced from physics - "It's all about imposing enough boundary conditions, old chap".

5. Oct 17, 2011

### Q-reeus

Sorry but I missed it! While the link is interesting, I get that the models are about highly compressible matter shells where pressure is mostly not an appreciable source of T itself and the complexity derives from the highly non-uniform matter density profiles. Maybe some cases there imply 'extreme' pressures, but either way, it's just an addition to rho surely given static conditions. At any rate for my shell scenario, as mentioned there, a typical self-supporting shell of say metal will always have rho many orders of magnitude in excess of any p contributions (anti-matter bomb vs firecracker). So it still gets back to explaining the case of just a shell of uniform matter density as only source of T and G - stress is not worth stressing over.

6. Oct 17, 2011

### Staff: Mentor

I don't think you can make this assumption in general; I think it will only hold if the potential at the inner surface of the shell is close to 0 (where 0 is the potential at infinity)--equivalently, the radius of the shell (inner or outer) is much greater than 2M, where M is the total mass of the shell (as measured by an observer far away from it). If the shell radius is not very large compared to M (meaning the potential within the shell is significantly less than 0, or equivalently the t-t and r-r metric coefficients are significantly different from 1), then the state of the matter in the shell will be much more like white dwarf or neutron star material than like ordinary metal, and the pressure in the material will have to be comparable to its energy density or the shell will not be stable. (Also, as I mentioned in the previous thread where this topic was discussed, the pressure will have to be negative--a tension--in some portion of the shell or it will fall in on itself.)

7. Oct 17, 2011

### Q-reeus

Yes that's the condition I was talking about in specifying in #1 that rs/r << 1. The aim is to avoid 'strong gravity' complications and as much as possible all the fierce math that goes with that. A bare minimum scenario that draws out the relevant factors.
I felt like commenting on that over there, but here's the chance. This seems strange. I can only see positive biaxial tangent stress. So you are saying there is need of a radial tensile component?

8. Oct 17, 2011

### Staff: Mentor

Ok, understood. Basically, we're assuming that the spherical shell can hold itself together and be made of "normal" materials for which the energy density is much larger than any stress component. The corrections to the metric coefficients will all be small, but as long as those corrections are still large compared to the ratio of pressure to energy density, which seems reasonable, we should be fine.

(Quick back of the envelope calculation to verify that it's reasonable: metric coefficient correction for the Earth is of order 10^-6, the ratio of r to M. For a typical material like steel, pressure components are of order 10^9 in SI units--gigapascals--and energy density is of order 10^20 in SI units--one kilogram per cubic meter is 9 x 10^16, water is 1000 kg/m^3, and steel is several times the density of water. So the ratio of pressure to energy density in a typical structural material is of order 10^-11, five orders of magnitude smaller than the metric correction for r a million times M. Looks like we're OK.)

No, I'm saying there has to be a tangential hoop stress, which will be a tension, not a pressure, so it will be a negative quantity in the stress-energy tensor. See here:

http://en.wikipedia.org/wiki/Cylinder_stresses

This talks about cylinders, not spheres, but the general idea is the same for spheres, including the sign of the stress. The Wiki page isn't very careful about signs, but the convention for the stress-energy tensor is clear: pressure (compressive stress) is positive and tension (tensile stress--stretching) is negative.

9. Oct 17, 2011

### Staff: Mentor

Since I contributed to the "apparent consensus" in the other thread, I would like to post here after thinking this over some more. One key thing that I don't think I captured correctly in the other thread is that statements like "St dives relatively steeply from St = 1 at r = rb, to some scaled fraction of V at r=< ra" are (at least I believe they are) coordinate-dependent. So I think it is helpful to start by first describing as much as we can independently of any coordinates, entirely in terms of "physical" quantities that are coordinate-invariant. I will do that here for the description I gave in the other thread of how the "packing" of objects between two spheres close together varies as we descend through the three "regions" of this scenario: the exterior vacuum region (EV), the non-vacuum "shell" region (NV), and the interior vacuum region (IV).

Start somewhere in the EV, well above the outer surface of the shell, but still far enough from "infinity" that the metric differs significantly from the Minkowski metric. Take two concentric 2-spheres, both centered exactly on the "shell" (but much larger), one with physical area A, the other with slightly larger physical area A + dA. By "physical area" I mean the area measured by covering the 2-sphere with very small identical objects and counting the objects (and by "identical objects" I mean that if you bring any two of them to the same location they will be exactly the same shape and size).

Pack the volume between these two spheres with the same identical little objects (which we take to be exactly spherical so they have the same dimension in all directions). Euclidean geometry would lead us to predict that the volume dV between the two spheres that we will find from this packing is given by:

$$dV = \frac{1}{4} \sqrt{\frac{A}{\pi}} dA$$

However, when we actually do the packing, we will find the physical volume dV between the two spheres is *larger* than this, by some factor K; and K will get *larger* as we descend in region EV, getting closer to the outer surface of region NV. Finally, at the outer surface of region NV, K will have reached some value K_o > 1.

As we continue to descend through NV, we find the opposite effect now taking place; the factor K begins to get smaller, and when we reach the inner surface of NV, we find that it is now 1, the same value it has at "infinity". In other words, K_i = 1.

Once we are inside region IV, the K factor does not change; spacetime is flat. So since K is 1 at the inner surface of NV, it is 1 throughout IV. If we take any pair of spheres centered in IV, with areas A and A + dA, the volume between them, as shown by packing with identical little objects, will be exactly as given by the above formula.

I should also include what happens to the "rate of time flow" as we descend through the three regions. (This "rate of time flow" would be observable as gravitational redshift/blueshift, so substitute that in wherever I say "time flow" if that makes it easier to see what I mean.) In region EV, time flow, relative to its rate "at infinity", is slowed by the same factor K that volume between two spheres is increased from its Euclidean value. So at the outer surface of NV, time flow, relative to infinity, is multiplied by a factor J_o = 1 / K_o. However, as we descend through region NV, time flow continues to slow, so when we reach the inner surface of NV, time flow relative to infinity is multiplied by a factor J_i < J_o. Therefore, we can see that J_i is *not* equal to 1 / K_i; the relationship between J and K that held in region EV no longer holds in region NV. And since the rate of time flow is the same throughout NV, the relationship between J and K that held in EV doesn't hold in IV either.

I won't comment at this point on how any of the above relates to the descriptions in terms of coordinate systems; I'll save that for another post. But I think the above summarizes the physical observations that would apply in the scenario.

10. Oct 17, 2011

### Staff: Mentor

I think this is incorrect. The stress should be exactly the right magnitude to cause the change in the metric you are interested in. If it is a thin massive shell it will be very high and if it is a thick low-density shell then it will be lower. The density will clearly be larger, but that doesn't mean that the lateral stresses are negligible, particularly since they affect different components of the stress-energy tensor and it is precisely these different components that are causing your concern.

I feel like you are deliberately ignoring a clear physical source of the effect you are interested in for no reason other than your a-priori assumption that it cannot possibly be large enough.

11. Oct 17, 2011

### Q-reeus

Check!
Still not getting that (a side issue in our context, but still). Substitute air-filled balloon for the shell, and internal gas pressure (outward acting) and there one must have tangential hoop tension for stability. But with only inwardly acting self-gravity acting on the shell, hoop stresses must be positive to avoid collapse, surely? The link you gave points out that in the thick wall case, shear stresses owing to gradient of hoop stress as function of radius can matter, but not important for the thin wall case. I just dug this link up: http://arxiv.org/abs/gr-qc/0505040 p11-12 may be of use.

12. Oct 17, 2011

### Q-reeus

No. I think the assessment in #8 settles it nicely. In the #1 scenario, pressure is piddling - the matter is with the effect of matter. If you want, I can pin it down to the shell = a standard 'toy globe' sitting on a typical professor of GR's shelf. Work out the relative contributions there!

13. Oct 17, 2011

### Q-reeus

Sure but is that any more than an alternate way of saying Sr varies as Sr = V (in my notation the -ve sign for V is absent), whereas St is an invariant 1, for all r=< rb? I'm failing to see the need to work only in relative areas and volumes. There must surely be some definite, unambiguos meaning to r here. 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.
We have reached the flat MM interior, fine.
Understand perfectly what you are saying here, but for the moment I have to treat it as just assertion as to relative evolution of K and J. I will be convinced when it can be shown operation of the 'primitives' V and and any combination of derivatives of V, lead to this explicitly. Must be going!:zzz:

14. Oct 17, 2011

### Staff: Mentor

Figures 2 and 3 of the paper I cited above show less than 1 order of magnitude difference. In any case, the fact that the t-t component of the stress energy tensor is large is completely irrelevant if you are interested in the r-r, theta-theta, or phi-phi components of the curvature.

Last edited: Oct 17, 2011
15. Oct 17, 2011

### Staff: Mentor

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.

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:

$$J = 1 + 2 \phi$$

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.

16. Oct 17, 2011

### Staff: Mentor

Ok, now for the "future post" to talk some about coordinates. First, the easy stuff: the spacetime is static, spherically symmetric, and asymptotically flat, so we can easily define angular coordinates $\theta$, $\phi$ in the usual manner, and a time coordinate $t$ that corresponds to the asymptotic Minkowski time coordinate "at infinity". The metric will then be independent of $t$, and the only dependence on the angular coordinates will be the $r^{2} sin^{2} \theta d\phi^{2}$ term in the tangential part, which won't be an issue in this problem (and in fact I'll be writing the tangential part as $d\Omega^{2}$ to make clear that there's nothing to see there). So basically the only coordinate where there is anything interesting to figure out is the radial coordinate.

The most familiar radial coordinate for these types of scenarios is the Schwarzschild radial coordinate, which I'll label $r$; it is defined by

$$A = 4 \pi r^{2}$$

for a 2-sphere with physical area A. In terms of this radial coordinate, the metric must take the form:

$$ds^{2} = - J(r) dt^{2} + K(r) dr^{2} + r^{2} d\Omega^{2}$$

where J and K are the gravitational redshift and "packing excess" coefficients I defined in my previous post, which must be functions of r only because of what we said above about the other coordinates. Two things are immediately evident:

(1) For this definition of the radial coordinate, the coefficient in front of the tangential part of the metric *must* be $r^{2}$, and nothing else, because we *defined* r that way (so the area of the sphere at r is $4 \pi r^{2}$). (Btw, this observation also shows why K in the metric above *has* to be the "packing excess" coefficient I defined in my previous post.)

(2) In region IV (interior vacuum), K = 1, so the spatial part of the metric with this radial coordinate is exactly in the Minkowski form. The only difference, in terms of this radial coordinate, between IV and the asymptotically flat region "at infinity" is the coefficient J, the gravitational redshift or "time dilation" factor.

However, there is another way to define the radial coordinate, the "isotropic" way; the idea here is that we want the relationship between the radial coordinate and actual physical distance to be the same in all directions (as it obviously is not for the Schwarzschild r coordinate). We'll call this isotropic radial coordinate $R$, and the metric in terms of it looks like this:

$$ds^{2} = - J(R) dt^{2} + L(R) \left( dR^{2} + R^{2} d\Omega^{2} \right)$$

I've put a different coefficient, L(R), in front of the spatial part because now, as you can see, it multiplies *all* of the spatial metric, instead of just the radial part. So the area of a sphere at R is

$$A = 4 \pi R^{2} L$$

What does the function L(R) look like? We can stipulate without loss of generality that it must go to 1 at infinity, so that R and r become the same there. We also expect L to get larger as we go deeper into region EV, closer to the outer surface of region NV. But what about in regions NV and IV? I'm still working on that, so I'll follow up with another post.

17. Oct 18, 2011

### Staff: Mentor

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.

18. Oct 18, 2011

### 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. 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?! I have a feeling there is some thought of mixing up assumed elastic, material distortions with the underlying, vastly smaller metric ones. The relative contribution of p to the latter is always insignificant for 'steel globes', no? Am I missing something here? More later.

19. Oct 18, 2011

### Q-reeus

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.

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.
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.

20. Oct 18, 2011

### Staff: Mentor

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.

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.