7
$\begingroup$

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

Wald p289

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.

$\endgroup$

1 Answer 1

8
$\begingroup$

[EDIT](Precisions for the sign in the definition of the normal)

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 :

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

$\endgroup$
2
  • 2
    $\begingroup$ Wald uses (-,+,+,+), so you need to choose the minus sign in your second to last equation. $\endgroup$ Jul 4, 2013 at 18:48
  • $\begingroup$ @WannabeNewton : I have made an edit in the answer to add some precision about the choice of the sign in the definition of the normal $\endgroup$
    – Trimok
    Jul 5, 2013 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.