Quick question, Komar integral derivation in Wald

[WannabeNewton]

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.

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 show why this 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.

 

[VeryConfusedP]

Wannabe:

Obviously I can't help you with your problem (though I wish I could), but who is Wald?

 

[WannabeNewton]

...but who is Wald?

Hi VC! Wald is the author of a graduate textbook in General Relativity titled "General Relativity" - Robert M. Wald

 

[DimReg]

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.

I don't think I know as much as you do about this, but I'll offer my thoughts in the hopes that it helps.

It seems to me that the freedom to make this choice is in the definition of N^a. N^a seems to be an orientation of the surface, so I would guess Wald is just following a procedure to make an orientation of the whole manifold out of an orientation for a hypersurface within the manifold. This doesn't seem always possible to me, for example a moebius strip isn't orientable. But intuitively, I would guess you can choose orientable 1d strips of the moebius strip, which might be a counter example to whether or not you can always make such a choice.

There might be some conditions on the manifold from casual structure that make such a choice possible.

 

[WannabeNewton]

Hi Dim, thanks for responding. Thankfully, I figured out what he was saying. Note that the unit normal to $\Sigma$ is given by $n^{a} = V^{-1}\xi^{a}$ and so, based on expression B.2.24 in Wald (page 434), we have that $\frac{1}{4}\epsilon_{abcd} = V^{-1}\xi_{[a}\epsilon_{bcd]}$ where $\epsilon_{bcd}$ is the induced volume element on $\Sigma$. Applying the result again we have that $\epsilon_{abcd} = 12V^{-1}\xi_{[a} N_{[b} \epsilon_{cd]]}$. The wedge product is associative so $\xi\wedge (N\wedge \epsilon_{\Sigma}) = (\xi\wedge N)\wedge \epsilon_{\Sigma}$ thus we have that $\xi_{[a} N_{[b} \epsilon_{cd]]} = \frac{1}{2}N_{[ab}\epsilon_{cd]}$ giving us $\epsilon_{abcd} = 6N_{[ab}\epsilon_{cd]}$. From here, I guess since the natural volume element $\epsilon_{abcd}$ is only unique up to a sign (i.e. up to orientation), he just flips the orientation of $\epsilon_{abcd}$ to get $\epsilon_{abcd} = -6N_{[ab}\epsilon_{cd]}$.

 

[Infrared]

Is $M = -\frac{1}{8\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d}$ an invariant of the space-time in the sense that it is independent of the choice of $S$? It is not obvious to me why this is the case, if it is so. More generally, does any global characterization of any kind of mass of a stationary space-time necessarily have to be independent of the integration region?

 

[WannabeNewton]

It is certainly not obvious from inspection that the Komar mass is independent of the choice of topological 2-sphere in the vacuum region of the asymptotically flat stationary space-time but it can be shown using Stokes' theorem (note the emphasis on the fact that the independence of the choice of topological 2-sphere is only true in general in the vacuum region).

To see this, first note the following: \epsilon^{abef}\nabla_{e}\{\epsilon_{abcd}\nabla^{c}\xi^{d}\} = \nabla_{e}\{\epsilon^{abef}\epsilon_{abcd}\nabla^{c}\xi^{d}\} = -4\nabla_{e}\{\delta^{[e}_{c}\delta^{f]}_{d}\nabla^{c}\xi^{d}\} = -4\nabla_{e}\nabla^{e}\xi^{f} = 4R^{f}{}{}_{d}\xi^{d} = 0 where I have used the fact that $\nabla^{c}\xi^{d} = \nabla^{[c}\xi^{d]}$ and that $R_{ab} = 0$ in vacuum space-time (this is from Einstein's equations). Hence \epsilon_{ijkf}\epsilon^{abef}\nabla_{e}\{\epsilon_{abcd}\nabla^{c} \xi^{d} \} = 0 = -6\delta^{[a}_{i}\delta^{b}_{j}\delta^{e]}_{k}\nabla_{e}\{\epsilon_{abcd}\nabla^{c}\xi^{d}\} = -6\nabla_{[k}\{\epsilon_{ij]cd}\nabla^{c}\xi^{d}\} so we have that $\nabla_{[e}\{\epsilon_{ab]cd}\nabla^{c}\xi^{d}\} = 0$ identically in the vacuum region of the asymptotically flat stationary space-time.

Now applying Stokes' theorem in the vacuum region to a 3 dimensional volume $\Sigma$ bounded by two topological 2-spheres $S$ and $S'$ we see that \int _{\Sigma}\nabla_{[e}\{\epsilon_{ab]cd}\nabla^{c}\xi^{d}\} = 0 = \int _{S'}\epsilon_{abcd}\nabla^{c}\xi^{d} - \int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d} i.e. the Komar mass is independent of the choice of topological 2-sphere.

As far as your last question goes, no that is not a necessary requirement. The Bondi mass for example will depend on the integration region (it will codify the energy flux due to outgoing gravitational radiation so it will change if we take time translations of the integration region).