Area of the event horizon of a rotating black hole

Question

The Kerr metric for a black hole of mass $M$ and angular momentum $J = aM$ is

$$ds^{2} = - \frac{\Delta(r)}{\rho^{2}}(dt-a\sin^{2}\theta d\phi)^{2} + \frac{\rho^{2}}{\Delta(r)}dr^{2} + \rho^{2} d\theta^{2} + \frac{1}{\rho^{2}}\sin^{2}\theta (adt - (r^{2}+a^{2}) d\phi)^{2},$$

where $\Delta(r) = r^{2} + a^{2} - 2Mr$, $\rho^{2} = r^{2} + a^{2} \cos^{2}\theta$ and $- M < a < M$.

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The event horizon is at $r_{+} = M + \sqrt{M^{2} - a^{2}}$. This is obtained by solving the equation $\Delta(r) = 0$.

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How do you use this to compute the area of the horizon?

My idea is to simplify the metric to obtain

$$ds^{2} = - \left( \frac{r^{2} + a^{2} - 2Mr - a^{2} \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dt^{2} + \left( \frac{r^{2} + a^{2} \cos^{2}\theta}{r^{2} + a^{2} - 2Mr} \right) dr^{2} - \left( \frac{4aMr \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dtd\phi + \left( r^{2} + a^{2} \cos^{2}\theta \right) d\theta^{2} + \sin^{2}\theta \left( \frac{(a^{2} + r^{2})^{2} - a^{2} \sin^{2}\theta (a^{2}-2Mr+r^{2}) }{r^{2} + a^{2} \cos^{2}\theta} \right) d\phi^{2}.$$

Then, I think that the area of the horizon is given by

$$A = \int d\theta\ d\phi\ g_{\phi\phi}g_{\theta\theta}|_{r=r_{+}}.$$

Am I wrong?

Answer 1

If you write the element surface as:
$d\sigma$ = $dl_{\theta}$ $dl_{\phi}$
you should have:
$dl_{\theta}$ = $\sqrt{g_{\theta\theta}}$ $d\theta$
$dl_{\phi}$ = $\sqrt{g_{\phi\phi}}$ $d\phi$
Hence the horizon area should be:
A = $\int d\theta\ d\phi\ \sqrt{g_{\theta\theta}} \sqrt{g_{\phi\phi}} |_{r=r_{+}}$
Note: $ds^2$ is a squared distance in spacetime

Answer 2

The correct formula is actually [cf. Carroll equation (6.107)]

$$ A = \int \sqrt{|\gamma|} \, d\theta \, d\phi $$

where $|\gamma|$ is the determinant of the induced metric.