0

Consider the Schwarzschild metric

$$ds^2 = -f(r)dt^2 + \frac{1}{f(r)}dr^2+r^2 d\Omega^2,$$

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?

Qmechanic
  • 201,751
jboy
  • 575

2 Answers2

0

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.

G. Smith
  • 51,534
0

  1. 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 $$ ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,\tag{1}$$ 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.

  2. 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.

Qmechanic
  • 201,751