0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ $T_{0}{}^{0} = \rho$ $\endgroup$ Mar 21, 2017 at 13:50
  • $\begingroup$ I have changed it, but I don't think it helps $\endgroup$ Mar 23, 2017 at 15:27

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$ Mar 24, 2017 at 14:56
  • $\begingroup$ My apologies for the misunderstood. $\endgroup$ Mar 24, 2017 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.