Q-reeus said:
Well it would be nice to expand on that a bit. Best I could make of Einstein tensor is that it is divergenceless - so much for surmising about divergence as conceivable factor. Given the shell spherical symmetry and static state, can the operation of said tensor within shell wall be expressed entirely in terms of potential and gradients thereof, preferably in polar form? I think all but the T00 term is operative as source on rhs, yes? So there should when it's all broken down, only be potential term and derivatives at work? So the specialness of matter region re tangent contraction should be explicable just in those terms. I think.
Some clarifications:
* The entire stress-energy tensor is the "source" on the RHS of the Einstein Field Equation (EFE). The LHS is the Einstein tensor, which is defined in terms of the Ricci tensor, which is a contraction of the Riemann curvature tensor.
* For the non-vacuum region in question, a reasonable stress-energy tensor would be a "perfect fluid" with non-zero pressure:
[tex]T_{00} = \rho[/tex]
[tex]T_{11} = T_{22} = T_{33} = p[/tex]
where [itex]\rho[/itex] is the energy density and [itex]p[/itex] is the pressure. The off-diagonal components are all zero. For the spacetime to be static and spherically symmetric, both [itex]\rho[/itex] and [itex]p[/itex] must be functions of the radial coordinate r only. Also, [itex]\rho[/itex] should be positive everywhere in the non-vacuum region, but [itex]p[/itex] will not be; a negative pressure is a tension, and in order for the non-vacuum region to be stable, there will have to be tension somewhere, to keep it from falling apart (a positive pressure everywhere would be possible only if the non-vacuum region went all the way down to the center, r = 0, without any hollow interior).
* The fact that the stress-energy tensor (and therefore the Einstein tensor) is divergenceless is one way of expressing local conservation of energy; the divergence of the stress-energy tensor, over some small piece of spacetime, is just the net energy going in or out of that piece of spacetime. The divergence being zero just means energy in equals energy out; i.e., energy is conserved. So you're right that this probably doesn't help much for this particular problem.
* You can still define a "potential" in the non-vacuum region, because you can do that for any spherically symmetric geometry. And you can still view the gradient of this potential as being the "acceleration due to gravity". I *think* that the effects on the metric, including its tangential part, in the non-vacuum region can be expressed in terms of the potential, but I'm not certain; the metric coefficients may be more complicated than that in the non-vacuum region. The only metric coefficient that I'm pretty well certain can be expressed entirely in terms of the potential is [itex]g_{00}[/itex], the timelike coefficient.