# Deriving the Schwarzschild metric in the weak-field regime

I am trying to derive the weak-field Schwarzschild metric, but starting from the same form as Schwarzschild:

$ds^2=-(1+2\Phi(r))dt^2+(1-2\Psi(r))dr^2 +r^2 d\Omega^2$

which has $R=-2\partial_r^2 \Phi(r)$ and $R_{00}=\partial_r^2\Phi(r)$. When I use $T^0_0=\rho$, then I get to the equation

$T^0_0=R^0_0 -\frac{1}{2} \delta^0_0 R = -\partial_r^2\Phi(r)+\partial_r^2 \Phi(r)=0$

i.e. $\rho=0$. Is it impossible to derive taking a non-vacuum solution, or have I made a mistake somewhere?

• $T_{0}{}^{0} = \rho$ – Jerry Schirmer Mar 21 '17 at 13:50
• I have changed it, but I don't think it helps – supercoolphysicist Mar 23 '17 at 15:27

## 1 Answer

Schwartzschild spacetime is a vacum solution of Einstein equation, whatever you get won't be Schwarzschild's solution.

Furthermore, with this tool you can calculate the Einstein tensor. For the metric you have proposed:

$$G_{00} = \frac{2(r\partial_r\Psi(r)-2\Psi^2(r)+\Psi(r))}{r^2(2\Psi(r)-1)^2}\propto T_{00}$$

which is not the $G_{00}$ you got. Probably some mistake during the calculation lead to that.

• To be precise, I meant the weak field solution for a spherical object, e.g. here physicspages.com/2015/01/11/… In the weak field limit, this should reduce to Schwarzschild – supercoolphysicist Mar 24 '17 at 14:56
• My apologies for the misunderstood. – Alejandro Menaya Mar 24 '17 at 21:27