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In Wald's GR, Komar integral is Eq. (11.2.9):

$$M=-\frac{1}{8\pi}\int_S\epsilon_{abcd}\nabla^c\xi^d$$

$S$ can be chosen as a 2-sphere, the boundary of a spacelike hypersurface $\Sigma$ such that the timelike Killing vector $\xi^a=(\partial_t)^a$ is normal to $\Sigma$. Here, the coordinate system is the one corresponding to the following metric

$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$

Therefore, the volume element is $\epsilon_{abcd}=r^2\sin\theta (dt)_a\wedge(dr)_b\wedge(d\theta)_c\wedge(d\phi)_d$ and therefore,

$$M=-\frac{1}{8\pi}\int_Sr^2\sin\theta (dt)_a\wedge(dr)_b\wedge(d\theta)_c\wedge(d\phi)_d\nabla^c(\partial_t)^d\\ =\frac{1}{8\pi}\int r^2\sin\theta \nabla^r(\partial_t)^td\theta d\phi$$

Here, $\nabla^r(\partial_t)^t=g^{rr}\Gamma^t_{rt}=M/r^2$. You can check Sean Carroll's GR to find the Christoffel symbols.

Then, $M=\frac{1}{8\pi}\int M\sin\theta d\theta d\phi=M/2$, a contradiction. What is wrong with my calculation?

Qmechanic
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Drake Marquis
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1 Answers1

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I lost one term, which is the one containing $\nabla^t(\partial_t)^r=g^{tt}\Gamma^r_{tt}=-M/r^2$, so this term is

$$ -\frac{1}{8\pi}\int r^2\sin\theta \nabla^t(\partial_t)^r d\theta d\phi=\frac{M}{8\pi}\int \sin\theta d\theta d\phi=M/2 $$

Add this term to previous one and correct!

Drake Marquis
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