Q-reeus said:
What would really impress is knowing what BT enforces about the specific behavour of SET terms for say the shell of #1. Knowing that would clear up much, but it seems beyond the reach of anyone.
BT doesn't say anything specific about the SET terms in the "interior" region; the whole point is that as long as there is an exterior *vacuum* region, and as long as the spacetime is spherically symmetric, the exterior vacuum *must* be Schwarzschild. The whole reason the theorem is so general and powerful is that it makes *no* assumptions whatsoever about the interior region, other than spherical symmetry.
Part of the problem here may be that you have not considered just how restrictive the assumption of *exact* spherical symmetry is. It is really that assumption, all by itself, that is enforcing restrictions on the SET terms. Think about what has to be constrained to ensure exact spherical symmetry: all motions must be radial, and radial motions cannot vary *at all* with angular coordinates. Basically, the whole problem is reduced to two dimensions from four; t and r are the only coordinates of interest, and energy density, radial momentum density, radial pressure, and tangential pressure are the only other variables of interest. That is a huge reduction in complexity from the general problem, and a huge restriction on possible solutions.
Also, if you look at the conservation laws for the SET, you see something else: tangential stress is completely uncoupled from the other variables. Here's the generic conservation equation again:
[tex]\nabla_{b} T^{ab} = 0[/tex]
I.e., the covariant divergence of the SET is zero. But this is really four equations, one for each coordinate (t, r, theta, phi) (the range of the index a in the above; the index b is summed over all four coordinates). So the above expands to:
[tex]\nabla_{0} T^{00} + \nabla_{1} T^{01} + \nabla_{2} T^{02} + \nabla_{3} T^{03} = 0[/tex]
[tex]\nabla_{0} T^{10} + \nabla_{1} T^{11} + \nabla_{2} T^{12} + \nabla_{3} T^{13} = 0[/tex]
[tex]\nabla_{0} T^{20} + \nabla_{1} T^{21} + \nabla_{2} T^{22} + \nabla_{3} T^{23} = 0[/tex]
[tex]\nabla_{0} T^{30} + \nabla_{1} T^{31} + \nabla_{2} T^{32} + \nabla_{3} T^{33} = 0[/tex]
Spherical symmetry forces many of these components to be zero; what we are left with is the following, making the substitutions [itex]T^{00} = \rho[/itex] (energy density), [itex]T^{01} =\mu[/itex] (radial momentum density), [itex]T^{11} = p[/itex] (radial pressure/stress), [itex]T^{22} = T^{33} = t[/itex] (tangential pressure/stress):
[tex]\nabla_{t} \rho + \nabla_{r} \mu = 0[/tex]
[tex]\nabla_{t} \mu + \nabla_{r} p = 0[/tex]
[tex]\nabla_{\theta} t = 0[/tex]
[tex]\nabla_{\phi} t = 0[/tex]
As you can see, there are *no* equations relating t to any other variables. (The last two equations simply confirm our prescription that there are no tangential variations in stress.) What this means is that *no* changes in any other SET components can be driven by changes in t. But again, it is exact spherical symmetry that drives that constraint (the fact that no tangential momentum flow can exist--since by the above equations, you can see that tangential momentum flow is what would be required to "exchange" tangential stress with energy density, as radial momentum flow allows "exchange" between radial pressure and energy density by the first two equations).