Why it is the mass instead of the mass distribution used in Schwarzschild metric?

Question

Schwarzschild metric is given by (Carroll, chapter 5) $$ ds^2=-\left(1-\frac{2GM}{r}\right)\mathrm{d}t^2+\left(1-\frac{2GM}{r}\right)^{-1}\mathrm{d}r^2+r^2d\Omega^2 $$ where $M$ is the total mass of a gravitating body. According to the text book, this can be derived from the Einstein equation in vacuum $$ R_{\mu\nu}=0 $$ using the spherically symmetric condition.

My question is why there is the $M$, instead of a term which looks like $\int\rho(r)\ \mathrm{d}V$ or $\int T_{\mu\nu}(r)\ \mathrm{d}V'$ shows up in Schwarzschild metric? Isn't it the inverse square relation ($F\propto r^{-2}$) that makes it possible in Newtonian gravitation to treat spherically symmetric mass distribution as a mass point? So how does this come true here in GR? Thank you in advance!

Answer 1

Birkhoff's theorem tells us that:

any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. This means that the exterior solution (i.e. the spacetime outside of a spherical, nonrotating, gravitating body) must be given by the Schwarzschild metric.

So that means that the variation of density with radial distance is irrelevant. As long as we are outside the body all we need to know is the total mass.

If we are inside the body then the variation of density with distance does matter, and in that case the geometry is given by the Schwarzschild interior metric not the Schwarzschild metric.

Answer 2

My question is why there is the M, instead of a term which looks like ∫ρ(r) dV or ∫Tμν(r) dV′ shows up in Schwarzschild metric?

The Schwarzschild metric describes only the part of spacetime without matter and must continuously connect to metric of the interior part with matter.

Denoting energy density of matter $\varepsilon$ and $\kappa$ Einstein's gravitation constant the corresponding solution for metric component $g^{-1}_{rr}$ reads

\begin{equation} \label{e-2lambda} g^{-1}_{rr} =\left\{\begin{array}{rcl} 1-{\displaystyle \frac{\kappa}{r}\int_{0}^{r}\varepsilon(\tilde{r})~\tilde{r}^2~d\tilde{r}}, & ~~~~\mbox{for} & 0\leq r \leq R \\ \\1-{\displaystyle \frac{\kappa}{r}\int_{0}^{R}\varepsilon(\tilde{r})~\tilde{r}^2~d\tilde{r}} & ~~~~ \mbox{for} & R\lt r \lt \infty. \end{array}\right.\tag{1} \end{equation}

For matter like for example perfect fluid energy density is proportional to the rest mass density. It yields $\varepsilon=\rho~c^2$. In this case the constant in vacuum part of solution $(r>R) $ can be expressed by the total rest mass $M$: \begin{equation} \frac{8 \pi G}{c^4} \int_{0}^{R} c^2~\rho(\tilde{r})~\tilde{r}^2~ d\tilde{r}= \frac{2 G}{c^2} \int_{0}^{V_{R}} \rho(\tilde{r})~ dV = \frac{2 G M}{c^2}\equiv r_{S} \tag{2} \end{equation} where are $dV= 4 \pi~\tilde{r}^2~ d\tilde{r}$ and $ V_{R}=4/3 \pi~R^3$