# Killing vectors in General Relativity?

I'm looking to derive the surface area of the event horizon of a Schwarzschild black hole. I was just wondering if it were possible for someone to explain to me this:

$$\sqrt{g_{\theta\theta}g_{\phi\phi}}$$

I don't know what it is, or how to compute it. I don't even know what it is called? Could anyone shed some light on this.

I give the potential answerers the full right and privilege to patronise me.

I do, however, have one hunch that it is some sort of killing vector, or closely related to the killing vector...?

## 1 Answer

In general, the square root of the determinant of the metric will give the volume element. That is, $dV = \sqrt{|\det g|} dx_1 \cdots dx_n$. Here you have the metric restricted to surfaces of constant $t$ and $r$, so you will get the area element on these. Since the metric is diagonal, the determinant is just $g_{\theta\theta}g_{\phi\phi}$.

• I seem to feel that the expression in my question evaluates to $r^2\sin\theta$ in spherical polar coordinates... – DarthPlagueis Oct 22 '15 at 22:28
• Do you know of any freely available resources which could describe how to compute these metrics in more detail? – DarthPlagueis Oct 22 '15 at 22:38