I'm going to go ahead and post what I have on the spherical shell scenario. I don't have a complete analytical expression for the g_tt metric coefficient, but I have expressions for g_rr (I posted that previously but I'll post it again here) and the derivatives of everything, so it's clear enough how things work.
First of all, a correction to what I said a while back in post #147 when I posted the four components of the conservation law. When I said tangential stress was completely "uncoupled" from the other components, I was forgetting about the connection coefficient terms in the covariant derivatives. When those are included, tangential stress actually does appear in the modified TOV equation (i.e., hydrostatic equilibrium) for non-isotropic stress (i.e., radial pressure unequal to tangential stress--the tangential stress components still have to be equal by spherical symmetry). I should have remembered that because the presence of the tangential stress in the modified TOV equation is critical for keeping the shell stable (as I said in a previous thread, and as we'll see below).
Once again, we are assuming a static, spherically symmetric spacetime, and so we have five total unknown quantities: the metric coefficients g_tt and g_rr, and the three SET components T_00, T_11, and T_22 = T_33. We'll give easier names to these; in order, they are J(r), K(r), rho(r), p(r), and s(r). As mentioned previously, we have three equations relating these quantities, the three non-trivial components of the EFE (or, equivalently, two EFE components and one non-trivial conservation law, the equation for hydrostatic equilibrium, which is what we'll actually use). So we ought to be able to specify two arbitrary functions and then the rest will be determined. We'll assume in what follows that we know rho(r) and s(r) and are trying to determine the others in terms of them.
(Note: my notation above is a bit different from most of the literature; usually, g_tt and g_rr are written as exponentials, g_tt = exp(Phi) and g_rr = exp(Lambda) for example. This makes some of the equations a bit easier to calculate with for complex problems, but here we're more concerned with the general physical behavior, so I'm keeping J(r) and K(r) as we defined them in a previous thread. This means some of my formulas will look a bit different than the ones in the literature, but they're still describing the same physics.)
The first equation is the 0-0 component of the EFE: it reads:
G_{00} = \frac{1}{r^{2}} \left( 1 - \frac{1}{K} \right) - \frac{1}{r} \frac{d}{dr} \left( \frac{1}{K} \right) = 8 \pi \rho
This simplifies to:
\frac{1}{r^{2}} \frac{d}{dr} \left[ r \left( 1 - \frac{1}{K} \right) \right] = 8 \pi \rho
We define the quantity in brackets as 2m(r), where "m" is a new function of r whose physical interpretation we will see in a moment (of course the name telegraphs it, but bear with me
); we then see that
\frac{dm}{dr} = 4 \pi \rho r^{2}
This means that dm/dr is the "mass added at radius r"; so integrating dm/dr should give us the "total mass inside radius r". We then see that
K(r) = \left( 1 - \frac{2m(r)}{r} \right)^{-1}
which is what I posted before. In other words, purely from the first (0-0) component of the EFE, without looking at anything else, we see that the K factor, the metric coefficient g_rr, depends *only* on the mass inside radius r, and that depends *only* on the function rho(r). It does *not* depend on any other SET components. (Of course, if rho(r) were not one of our "known" functions, but if we instead assumed, say, that we knew p(r) and s(r), then rho(r) and hence m(r) would still depend, indirectly, on p(r) and s(r), since we would be solving for rho in terms of them. But the final result would still be that g_rr "sees" only the mass inside radius r.)
So much of the discussion in the previous thread, about whether stresses were of the "right" magnitude to make the K factor go back to 1 from outside to inside the shell, was really irrelevant. The K factor automatically goes to 1 as m(r) goes to zero, just from the above.
The next equation is the 1-1 component of the EFE, which I'll rewrite using m(r) instead of K(r):
G_{11} = - \frac{1}{r^{2}} \frac{2m}{r} + \frac{1}{r} \left( 1 - \frac{2m}{r} \right) \frac{1}{J} \frac{dJ}{dr} = 8 \pi p
which easily rearranges to
\frac{dJ}{dr} = 2 J \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)}
We'll set this aside for a moment.
The third equation is the modified TOV equation for hydrostatic equilibrium. The key change from the standard TOV equation is that the pressure is not isotropic; we allow radial and tangential stress to be different. That adds an extra term to the normal TOV equation; we have
- \frac{dp}{dr} = \left( \rho + p \right) \frac{1}{2J} \frac{dJ}{dr} + \frac{2}{r} \left( p - s \right)
Substituting for 1/2J dJ/dr using the second EFE component above, we obtain
- \frac{dp}{dr} = \left( \rho + p \right) \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)} + \frac{2}{r} \left( p - s \right)
This allows us to solve for p in terms of known quantities (since we know m(r) from above and we said s(r) was known as well). However, if there is an actual analytical solution for the above, I haven't been able to find one (I don't think there is in the general case). In the special case of rho = constant, MTW give a solution of the standard TOV equation (without the last term on the RHS, i.e., assuming isotropic pressure) for a spherical star, but the equation for p is still pretty messy and I'm not sure exactly how they arrived at it. With the last term on the RHS added in, even their solution for constant rho may no longer work.
But we can still see some things just by looking at the above equation for dp/dr. First of all: in the standard case, where we have a spherical star with matter all the way into r = 0, we can have isotropic pressure because - dp/dr can be positive all the way into the center. In the shell case, however, that won't work; we must have p = 0 at both the outer and inner surfaces of the shell. That means we *need* the last term on the RHS to have a static equilibrium at all, because the (p - s) factor needs to change sign at some point within the shell in order to change the sign of dp/dr. That's why I said in that earlier thread that tangential stresses are key to keeping the shell stable; more precisely, I should have said tangential stresses that go from positive on the inner surface (so p - s can be less than zero) to negative on the outer surface (we'll see why that has to be the case in a moment) are needed to keep the shell stable if there is an interior vacuum region inside.
Why must s be negative on the outer surface? Because, if we cut a "slice" through the center of the shell, and do a force balance on it similar to what is done in the Ehlers paper, we will find that the following must hold (since there is no pressure in the interior vacuum region, so the only force that can balance is the tangential stress integrated over the shell):
\int_{a}^{b} 2 \pi r s(r) dr = 0
where r = a > 0 is the shell's inner radius and r = b > a is the shell's outer radius. This condition requires that, since s is positive on the inner surface (which we've seen it has to be to make dp/dr change sign), it must be negative on the outer surface. Physically, this makes sense because we would expect the shell's material to be compressed tangentially on the inner surface and stretched tangentially on the outer surface.
Now let's go back to the equation for dJ/dr; I'll write it with a changed sign so we are looking at what happens to J as we go *inward* through the shell, from outer to inner surface:
- \frac{dJ}{dr} = - 2 J \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)}
Again, I don't have an analytical expression for J itself from this, since I don't have one for p; but just from looking at the above we can see two things: (1) J continues to decrease as we go inward through the shell; but (2) as we approach the inner surface of the shell, - dJ/dr -> 0 smoothly (because m and p both go to zero smoothly). So there is a smooth transition from J decreasing through the shell to J being constant throughout the inner vacuum region.
(Note that dK/dr does *not* make a smooth transition at the shell boundaries, even though K itself does. As far as I can tell, this is OK: the "junction conditions" that have been mentioned before do not require that dK/dr be continuous, only that K itself is continuous. They *do* require that dJ/dr be continuous as well as J, which it is; physically, it seems to me this is because dJ/dr contributes to hydrostatic equilibrium whereas dK/dr does not, so a discontinuity in dJ/dr would case a discontinuity in the "acceleration due to gravity".)
So to sum up:
(1) The "K" factor is determined entirely by how much mass is *inside* radius r; so as you descend through the shell, K goes smoothly back to 1 from its value at the outer shell surface. Neither radial nor tangential stress has any effect on K.
(2) Hydrostatic equilibrium for a shell with an interior vacuum region requires that p = 0 at both the outer *and* inner surface, which in turn requires that tangential stress be unequal to radial stress, and that it go from negative on the shell outer surface to positive on the inner surface, and integrate to zero over the shell.
(3) The "J" factor continues to decrease smoothly through the shell, at a rate determined by both the mass and the radial pressure, becoming constant in the interior vacuum region. Tangential stress has no effect on J (except indirectly by its effect on the radial pressure profile).
