Einstein Field Eqn Solution for Finite Extended Mass

Question

As is well known, the Schwarzschild metric is a vacuum solution derived by setting $T^{\mu \nu} = 0$, and assuming spherical symmetry and a static condition. Then, the final solution contains a scaling parameter that ends up being determined by the mass $M$ of the body (although there was no "body" modeled at the outset). This is the metric used to model the spacetime around a star, planet, or other isolated spherical object.

My question is, what solution is obtained (numerically or analytically) if the EFE are solved for a nonzero $T^{\mu \nu}$ with spherical symmetry? E.g. a delta function, or finite sized sphere. Can anyone elaborate on such a solution, and how it differs from the Schwarzschild?

Answer 1

You need more details. The general metric for a spherically symmetric solution is $$\mathrm{d}s^2 = - e^{2\phi(t,r)}\mathrm{d}t^2 + e^{2\psi(t,r)}\mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$ and the functions $\phi$ and $\psi$ are, in principle, arbitrary (meaning they need to be fixed by solving the EFE). This means there is a lot of freedom in the solution and it is difficult to give a complete response. In particular, the time-dependence makes the problem quite difficult.

If you also impose that the solution is stationary, then the metric simplifies to $$\mathrm{d}s^2 = - e^{2\phi(r)}\mathrm{d}t^2 + e^{2\psi(r)}\mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2,$$ which is actually much easier. The most general solutions is the anisotropic generalization of the Tolman–Oppenheimer–Volkoff solution. The TOV solution goes a step further when imposing symmetries and requests that the tangential and radial pressures are all equal, which simplifies the problem slightly more. Without this assumption, the functions $\phi$ and $\psi$ can still be freely specified (or obtained by solving the EFE).

These solutions are often used as models for relativistic stars. They can differ considerably from Schwarzschild in the interior (the exterior is always Schwarzschild due to Birkhoff's theorem). For instance, there is no need for a horizon, one can have negative masses without having singularities, etc.

The case of a homogeneous star (constant energy density) is one of the few cases that one can solve analytically. This is sometimes referred to as the Schwarzschild star (Schwarzschild worked on this solution soon after the original black hole solution) and is discussed in many books in general relativity. For example, Wald's book.

Answer 2

There is already a good answer from Nickolas Alves.

In this answer we just point out that the Birkhoff's theorem states that a spherically symmetric solution is static, and a (not necessarily thin) vacuum shell (i.e. a region with no mass/matter) corresponds to a radial branch of the Schwarzschild solution in some radial interval $r \in I:=[r_1, r_2]$.

In particular, an exterior vacuum branch $r \in I:=[r_1,\infty]$ is still described by the Schwarzschild solution even if there is a finite extended mass/matter distribution inside $r \in [0, r_1]$.