Surface gravity of Kerr black hole

  • #1
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##?
 

Answers and Replies

  • #2
PeterDonis
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how did they get

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

if the ##\Delta=0## on the horizon?
You have to take the derivative of your expression for ##\chi^{\mu} \chi_{\mu}## before you plug in values for quantities at the horizon.
 

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