The Komar mass, a conserved quantity associated with an asymptotically time-like Killing vector, for stationary spacetime is $$M_K=\frac{-1}{8\pi}\int\int\nabla^{\mu}\xi^{\nu}_{(t)}dS_{\mu\nu},$$ using the definition of surfcae element $dS_{\mu\nu}=-2n_{[\mu}k_{\nu]}\sqrt{\sigma}d^2\theta$, where $n_{\mu}$ and $k_{\nu}$ are TL and SL unit normal vector to closed hypersurface 2-boundary. $$M_K=\frac{1}{8\pi}\int\int\nabla^{\mu}\xi^{\nu}_{(t)}(n_{\mu}k_{\nu}-n_{\nu}k_{\mu})\sqrt{\sigma}d^2\theta.$$ Here, $n_{\mu}=-\frac{\delta^{t}_{\mu}}{|g^{tt}|^{1/2}}$ and $k_{\nu}=\frac{\delta^{r}_{\nu}}{|g^{rr}|^{1/2}}$. And using the definition of Killing vector $\nabla^{\mu}\xi^{\nu}_{(t)}+\nabla^{\nu}\xi^{\mu}_{(t)}=0$. $$\Rightarrow M_{K}=\frac{1}{4\pi}\int\int\nabla^{\mu}\xi^{\nu}_{(t)}n_{\mu}k_{\nu}\sqrt{\sigma}d^2\theta$$ $$=\frac{-1}{4\pi}\int\int\frac{1}{|g^{tt}g^{rr}|^{1/2}}\nabla^{t}\xi^{r}_{(t)}\sqrt{\sigma}d^2\theta$$ $$M_K=\frac{-1}{4\pi}\int\int\frac{1}{|g^{tt}g^{rr}|^{1/2}}(g^{tt}\Gamma^{r}_{tt}+g^{t\phi}\Gamma^{r}_{\phi t})\sqrt{\sigma}d^2\theta.$$ I got the correct answer for Kerr-Newman black hole but with an overall negative sign. Please correct me, where I am getting wrong?

Secondly, if we do not use the definition of Killing vectors ($\nabla^{\mu}\xi^{\nu}_{(t)}+\nabla^{\nu}\xi^{\mu}_{(t)}=0$), then we get $$M_K=\frac{1}{8\pi}\int\int\frac{1}{|g^{tt}g^{rr}|^{1/2}}(-\nabla^{t}\xi^{r}_{(t)}+\nabla^{r}\xi^{t}_{(t)})\sqrt{\sigma}d^2\theta.$$ And using the definition of covariant derivatives, we get $$M_K=\frac{-1}{8\pi}\int\int\frac{1}{|g^{tt}g^{rr}|^{1/2}}(g^{tt}\Gamma^{r}_{tt}+g^{t\phi}\Gamma^{r}_{\phi t}-g^{rr}\Gamma^{t}_{rt})\sqrt{\sigma}d^2\theta.$$ Which vanishes for Kerr-Newman black hole. Please correct me, what I am missing.

  • $\begingroup$ What metric signature are you using for the Kerr-Newman metric? (That could potentially explain the overall minus sign. $\endgroup$ – mmeent Dec 5 '19 at 10:19
  • $\begingroup$ I am working with (-,+,+,+) metric sign convention. $\endgroup$ – undergrad Dec 5 '19 at 13:53
  • $\begingroup$ In that case, I believe the formula for the Komar mass should not include the overall minus sign in your first formula. (See e.g. Caroll eq. 6.38) $\endgroup$ – mmeent Dec 5 '19 at 15:09

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