How Is the Komar Mass of a Schwarzschild Geometry Calculated?

[mongolianbeef]

Homework Statement

The Komar mass of a Schwarzschild geometry can be written as $\frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \nabla_{\alpha} \xi^{\beta}dA$, where $n^{\alpha}$ and $\sigma_{\beta}$ are timelike and spacelike normal vectors respectively. How does one actually go about evaluating this integral?

Homework Equations

The Attempt at a Solution

I've simplified it down to $\frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \Gamma^{\beta}_{\alpha t} dA$ but I have no idea how to continue from there. Wouldn't $n^{\alpha}\sigma_{\beta}$ be zero since they are orthogonal? Also, what is dA? I know it includes factors from the metric but which ones?

 

[clamtrox]

Also, what is dA? I know it includes factors from the metric but which ones?

It's the surface element of the 2-surface you defined, $dA = \sqrt{g^{(2)}} d^2x$ where $g^{(2)}$ is the determinant of induced metric on this surface.

How does one actually go about evaluating this integral?

So in Schwarzschild, I can choose my vectors n and σ in the obvious way, normalizing them to n²=-1, σ²=+1,
$$n = (-\sqrt{1-\frac{2M}{r}}, 0, 0, 0)$$
$$\sigma = (0, \sqrt{1-\frac{2M}{r}}^{-1}, 0, 0)$$
Then the 2-surface is just a sphere, with $\sqrt{-g^{(2)}} = r^2 \sin(\theta)$, $dA = r^2 d\Omega$, so I get
$$I = \frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \Gamma^{\beta}_{\alpha t} dA = \frac{1}{4\pi}\int_{S}n^{t} g_{rr} \sigma^{r} \Gamma^{r}_{t t} dA = \frac{1}{4\pi}\int_{S} -\sqrt{1-\frac{2M}{r}}(1-\frac{2M}{r})^{-1}(1-\frac{2M}{r})^{-1/2} \frac{M}{r^2} (1-\frac{2M}{r}) dA$$
$$= -\frac{1}{4\pi}\int_{S} \frac{M}{r^2} r^2 d\Omega = -\frac{M}{4\pi}\int_S d\Omega = -M$$
So maybe there's a sign wrong somewhere, or something, but otherwise it goes like that :)