# Question about Komar integral derivation in Wald

I have a question about derivation 11.2.10 in Wald (page 289). Here is a screenshot of the relevant passage:

I don't get the step $$-\frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f}\epsilon_{deab} = \frac{1}{4\pi}\int _{\Sigma}R_{ab}n^{a}\xi^{b}dV.$$ Could someone explain to me how he gets that equality? Thank you very much in advance.

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It is possible that the full expression for the LHS term would be :

$$-\frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f}\epsilon_{deab} dx^e \wedge dx^a \wedge dx^b$$

If so, and taking in account that the normal $n_d$ to the surface $\Sigma$ is simply defined by : $$n_d ~dV= -\epsilon_{deab} dx^e \wedge dx^a \wedge dx^b$$

For the explanation of the minus sign, we have to take in account that the $\Sigma$ surface is a space-like hypersurface, so the normal is a time-like vector, that is $n_d n^d = -1$ (With Wald conventions). Suppose, for one moment, that we are in a flat metric, and that the $\Sigma$ surface is the ordinary 3-spatial volume. Following the choice of the author of " $n^d$ as the unit future pointing normal to $\Sigma$", this means $n^0 = 1$. But because $n_d n^d = -1$, this means $n_0 = -1$. This is the justification of the minus sign.

So, We finally get :

$$\frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f} n_d ~dV = \frac{1}{4\pi}\int _{\Sigma}R_{ab}n^a\xi^{b} ~dV$$

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Wald uses (-,+,+,+), so you need to choose the minus sign in your second to last equation. –  Jerry Schirmer Jul 4 '13 at 18:48