Deriving the Schwarzschild metric in the weak-field regime

Question

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?

$\begingroup$ $T_{0}{}^{0} = \rho$ $\endgroup$
– Zo the Relativist
Mar 21, 2017 at 13:50 — Zo the Relativist, Mar 21, 2017 at 13:50
$\begingroup$ I have changed it, but I don't think it helps $\endgroup$
– supercoolphysicist
Mar 23, 2017 at 15:27 — supercoolphysicist, Mar 23, 2017 at 15:27

Alejandro Menaya · Accepted Answer · 2017-03-23 18:58:48Z

0

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.

answered Mar 23, 2017 at 18:58

Alejandro Menaya

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$\begingroup$ 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 $\endgroup$
– supercoolphysicist
Mar 24, 2017 at 14:56
$\begingroup$ My apologies for the misunderstood. $\endgroup$
– Alejandro Menaya
Mar 24, 2017 at 21:27

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