At the other edge beyond Newton with General Relativity? #2 The Schwarzschild solution is derived by using the static, spherically symmetric solution
$$ds^2=-B(r)dt^2 + A(r) dr^2 +\text{angular terms},$$
assuming a concentrated mass, therefore gaining $A=1/B$ from $R_{\mu\nu} = 0$ outside the mass
and (more importantly) from assuming $\lim A(r) = 1$ for asymptotical flatness.
My question is, would it be possible to derive a stronger gravitational law than Newton from the assumption that $\lim A(r) = 0$? Does this work or is it somehow in conflict with the original field equations? Has it been calculated out anywhere in a paper before?
 A: By Birkhoff's Theorem, the only spherically symmetric solution of the vacuum Einstein Field Equations must be the Schwarzschild solution. More precisely,

Any C^0, piecewise C^1, solution of the Einstein's empty space equations which is spherically symmetric in an open set $V$, is locally equivalent to part of the maximally extended Schwarzschild solution in $V$.

I took the formal statement from Hawking & Ellis, Appendix B (they prove for a $C^2$ metric, but mention and provide references to the stronger result).
In essence, if you want to impose the conditions

*

*Spacetime is spherically symmetric

*The stress-energy tensor vanishes on the solution of interest (for example, for $r \to + \infty$)

you automatically imposed that the metric must be the Schwarzschild metric, unless you are considering a theory of gravity different from General Relativity. Hence, you can't impose $\lim_{r \to + \infty} A(r) = 0$, that won't be a solution to Einstein's Field Equations. In other words, $\lim_{r \to + \infty} A(r) = 1$ is not an assumption, but rather a consequence.
Hence, one can't obtain a stronger Newtonian limit through the prescription you have given. It would need to contradict the Einstein Field Equations.
