Surface gravity of Kerr black hole

Question

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 $\kappa$?

Answer 1

The calculation can be done in this coordinate system just fine, even though it doesn't extend across the horizon. Surface gravities are very commonly computed in coordinate systems which go bad at the horizon. For example, the surface gravity of Schwarzschild

$ds^2 = -f dt^2 + f^{-1} dr^2 + r^2 d\Omega_2^2, \qquad f= 1- \frac{r_+}{r},$

is easily found to be $\kappa = \frac{f'}{2} = \frac{1}{2r_+}$.

I think your problem is that you are evaluating quantities at the horizon before taking derivatives. It's important to first take derivatives, and then to evaluate at the horizon.

Answer 2

You are right that the $(\Omega_H-\omega)^2$ term doesn't contribute. This is because it is the square of something that vanishes at the horizon: when you take the derivative, there remains a vanishing factor. As for the other term, since $\Delta$ vanishes at the horizon, this term vanishes except when the derivative hits $\Delta$. This yields the last formula you wrote.

Answer 3

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

Now use the horizon condition you will get $$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2 }{\Sigma}\partial_\nu \Delta $$

Since $\chi^\mu$ is null on horizon and a null vector is normal to itself so $\chi^\mu$ must be proportional to the normal of the horizon. A constant r surface has a normal $\partial_\mu{r}$. So $$\chi_{\mu}=C\partial_\mu{r}$$ Now our job is to find C. It can be found easily

$$g^{\mu\nu}\chi_{\mu}\chi_{nu}=C^2g^{\mu\nu}\partial_\mu{r}\partial_\nu{r}\\ =C^2g^{rr}$$

So $$C^2=\frac{\chi^{\mu}\chi_{\mu}}{g^{rr}}$$

So after the algebra is done take the horizon limit and you will find C. The rest are just few lines algebra.

Answer 4

Ok, every book I looked has this solved by looking at four velocity and four acceleration of a free particle at the horizon, so that must be it :\ Altho I'm sure there's a way to do it via Killing vector $\chi^\mu=\partial_t+\Omega_H\partial_r$.

So I'll just go through this derivation with acceleration...