# Surface gravity of Kerr black hole

1. Dec 19, 2014

### Ganesh Ujwal

I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand.

Firstly, the metric is given by

$\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2$

With

$\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2,$
$\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}$

The Killing vector that is null at the event horizon is

$\chi^\mu=\partial_t+\Omega_H\partial_\phi$

where $\Omega_H$ is angular velocity at the horizon.

Now I got the same norm of the Killing vector

$\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}$

And now I should use this equation

$\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu$

And I need to look at the horizon. Now, on the horizon $\omega=\Omega_H$ so my first term in the norm is zero, but, on the horizon $\Delta=0$ too, so how are they deriving that side, and how did they get

$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta$

if the $\Delta=0$ on the horizon? Since $\rho$ and $\Sigma$ both depend on $r$, and even if I evaluate them at $r_+=M+\sqrt{M^2-a^2}$ they don't cancel each other.

How do they get to the end result of $\kappa$?

2. Dec 19, 2014

### Staff: Mentor

You have to take the derivative of your expression for $\chi^{\mu} \chi_{\mu}$ before you plug in values for quantities at the horizon.