Replacing the $M$ coefficient in Schwarzschild metric with a distribution $M(r)$

Question

Consider the Schwarzschild metric
$$ds^2 = -f(r)dt^2 + frac{1}{f(r)}dr^2+r^2 dOmega^2,$$
where $f(r)=1-frac{2M}{r}$. I take it that $M$, although not really the mass of a black hole, is the coefficient that is closely associated with the mass. I'm curious as to what will happen if in the metric I directly replace $M$ with a distribution $M(r)$ and what happens to things like null and timelike geodesics, its horizons, etc.
My question is: Is it a valid starting point of inquiry to replace $M$ with a distribution $M(r)$? Or are there any physical rules that will make this problem ill-posed?

G. Smith · Answer

What will happen if in the metric I directly replace $M$ with a distribution $M(r)$?

It will no longer be a vacuum solution of Einstein’s field equations.

Qmechanic · Answer

Let us for simplicity work in units where the speed of light $c=1$ is equal to one, and assume that there is no cosmological constant $Lambda=0$. A spherically symmetric vacuum solution to the EFE of the form
$$ ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 dOmega^2,tag{1}$$
and such that it asymtotically becomes Minkowski space
$$ -g_{tt}(r!=!infty)~=~ 1~=~g_{rr}(r!=!infty), tag{2}$$
is then uniquely given by
$$ -g_{tt}(r)~=~ 1-frac{R_S}{r} ~=~frac{1}{g_{rr}(r)},tag{3} $$
where $R_S$ is a length parameter, i.e. a constant, cf. Birkhoff's theorem and this Phys.SE post.

In particular, OP's function $M(r)$ can therefore not depend on $r$. For the interpretation of $M$ as the mass of the black hole, see this Phys.SE post.

Replacing the $M$ coefficient in Schwarzschild metric with a distribution $M(r)$

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