I'm trying to follow Wald's derivation of the Komar mass on page 288. We start with an asymptotically flat static space-time with killing vector ##\xi^{a}##. Let ##\Sigma## be a hypersurface orthogonal to ##\xi^{a}## and let ##S\subseteq \Sigma## be a topological 2-sphere. The quantity ##F = \int _{S}V^{-1}\xi^{a}N^{b}\nabla_{a}\xi_{b}dA## can be seen as the total force that must be exerted by an observer at infinity to keep stationary a unit surface mass density distributed over ##S##; here ##N^{a}## is the outward normal to ##S## lying in ##\Sigma## (so that ##N^{a}## is orthogonal to ##\xi^{a}##), ##V## is the redshift factor, and ##dA = \epsilon_{ab}## is the natural volume element on ##S## induced by the space-time metric.(adsbygoogle = window.adsbygoogle || []).push({});

Define the normal bi-vector to ##S## by ##N^{ab} = 2V^{-1}\xi^{[a}N^{b]}## then it is easy to see, by virtue of killing's equation ##\nabla_{a}\xi_{b} = \nabla_{[a}\xi_{b]}##, that ##\int _{S}V^{-1}\xi^{a}N^{b}\nabla_{a}\xi_{b}dA = \frac{1}{2}\int _{S}N^{ab}\nabla_{a}\xi_{b}dA ##. Wald then chooses the orientation of the natural volume element ##\epsilon_{abcd}## on space-time associated with the space-time metric so that ##\epsilon_{abcd} = -6N_{[ab}\epsilon_{cd]}## and then writes ##\frac{1}{2}\int _{S}N^{ab}\nabla_{a}\xi_{b}dA = -\frac{1}{2}\int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d}##. This final equality is what is confusing me; he doesn't really showwhythis equality actually holds.

In an effort to show the equality myself, I had to go through some less than elegant calculations. First note that ##-6N_{[ab}\epsilon_{cd}] = \frac{1}{4}\epsilon_{abcd}\epsilon^{ijkl}N_{ij}\epsilon_{kl}## (see the formulas in appendix B of Wald). Then, ##\epsilon_{abcd}\nabla^{c}\xi^{d} = \frac{1}{4}\epsilon_{abcd}\epsilon^{ijkl}N_{ij}\epsilon_{kl}\nabla^{c} \xi^{d}## so multiplying both sides by ##\epsilon^{ab}## we have ##\epsilon_{abcd}\epsilon^{ab}\nabla^{c}\xi^{d} = \frac{1}{4}\epsilon_{abcd}\epsilon^{ijkl}N_{ij}\epsilon_{kl}\epsilon^{ab}\nabla^{c}\xi^{d} = \frac{1}{2}\epsilon_{abcd}\epsilon^{ijab}N_{ij}\nabla^{c}\xi^{d} = -2\delta^{[i}_{c}\delta^{j]}_{d}N_{ij}\nabla^{c}\xi^{d} = -2N^{cd}\nabla_{c}\xi_{d}##. Multiplying both sides by ##\epsilon_{ij}## we find that ##\epsilon_{abcd}\epsilon^{ab}\epsilon_{ij}\nabla^{c}\xi^{d} = 2\epsilon_{ijcd}\nabla^{c}\xi^{d} = -2N^{cd}\nabla_{c}\xi_{d}\epsilon_{ij}\Rightarrow N^{cd}\nabla_{c}\xi_{d}\epsilon_{ab} = -\epsilon_{abcd}\nabla^{c}\xi^{d}## finally giving us ##F = \frac{1}{2}\int _{S}N^{ab}\nabla_{a}\xi_{b}dA = \frac{1}{2}\int _{S}N^{cd}\nabla_{c}\xi_{d}\epsilon_{ab} = -\frac{1}{2}\int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d}##.

As you can see, this seems like more calculations than would be needed to show an equality that Wald simply writes down and doesn't explicitly show himself so I'm wondering frantically if I am going the long route in showing it through calculations i.e. is there something that I'm missing completely that makes the aforementioned equality trivial with no need for the calculations I did above; is this why Wald simply writes down the equality and doesn't show it? Does anyone know if there is something that makes the above equality immediate without all this work? Thank you very much in advance.

EDIT: Also, does anyone know why given the natural volume element ##\epsilon_{abcd}## associated with the space-time metric, we can always choose its orientation so that ##\epsilon_{abcd} = -6N_{[ab}\epsilon_{cd]}## i.e. where does this equality even come from? Wald claims we can do this (as stated above) but I can't immediately see why.

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# Quick question, Komar integral derivation in Wald

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