# A Metric induced on Kerr event horizon

1. Apr 6, 2017

### Jonsson

Hello there,

Suppose $\Delta = r^2 + 2GMr + a^2$ and $\rho^2 = r^2 + a^2 \cos ^2 \theta$. The Kerr metric is
$$ds^2 = - (1 - \frac{2GMr}{\rho^2})dt^2 - \frac{4GMar\sin^2 \theta}{\rho^2} d t d \phi + \frac{\rho^2}{\Delta} dr^2 + \rho^2 d \theta^2 + \frac{\sin^2 \theta}{\rho^2} \left[ (r^2 + a^2)^2 - a^2 \Delta \sin^2 \theta \right] d \phi^2$$

I want to determine the area of the inner horizon of the black hole. But first I need to determine the induced metric on the horizon. Suppose that the radius of the innermost horizon is $r_*$. My tutor stated without proof that the induced metric was
$$ds^2 = \rho_*^2 d \theta^2 + \frac{\sin ^2 \theta}{\rho_*^2} (r_*^2 + a^2) d \phi^2$$
How do one compute this induced metric from the Kerr metric?

Kind regards,
Marius

2. Apr 6, 2017

### martinbn

On the horizon $t$ and $r$ are constants, so $dt=0$ and $dr=0$.

3. Apr 6, 2017

### Staff: Mentor

Also $\Delta = 0$ on the horizon.

4. Apr 6, 2017

### Jonsson

That's right. Got it now. Thanks

5. Apr 6, 2017

### Staff: Mentor

This isn't quite correct. It should be $\Delta = r^2 - 2GMr + a^2$.

6. Apr 14, 2017

### stevebd1

You're probably working through things yourself but if you want a check, the equation for the surface area of both the inner and outer horizon in Kerr metric is-
$$A_{\pm}=\pm4\pi\left(r^2_{\pm}+a^2\right)$$
where
$$r_{\pm}=M\pm\sqrt{M^2-a^2}$$
where $\pm$ denotes the outer ($+$) and inner ($-$) horizon.

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