Surface gravity of Kerr black hole

[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

[latex]\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[/latex]

With

[latex]\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2[/latex],

[latex]\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}[/latex]

The Killing vector that is null at the event horizon is

[latex]\chi^\mu=\partial_t+\Omega_H\partial_\phi[/latex]

where[latex] \Omega_H[/latex] is angular velocity at the horizon.

Now I got the same norm of the Killing vector

[latex]

\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}

[/latex]

And now I should use this equation

[latex]\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu[/latex]

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

[latex]\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta[/latex]

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

How do they get to the end result of [latex]\kappa[/latex]?

[xyzt]

 

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

[latex]\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[/latex]

With

[latex]\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2[/latex],

[latex]\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}[/latex]

The Killing vector that is null at the event horizon is

[latex]\chi^\mu=\partial_t+\Omega_H\partial_\phi[/latex]

where[latex] \Omega_H[/latex] is angular velocity at the horizon.

Now I got the same norm of the Killing vector

[latex]

\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}

[/latex]

And now I should use this equation

[latex]\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu[/latex]

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

[latex]\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta[/latex]

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

How do they get to the end result of [latex]\kappa[/latex]?

 

What you need is to FIRST calculate the derivative and THEN to apply the boundary conditions. You are doing the steps in reverse order.