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

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


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

The Killing vector that is null at the event horizon is


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

Now I got the same norm of the Killing vector


And now I should use this equation


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


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 $\kappa$?

  • 4
    $\begingroup$ You need to do this in a coordinate system that isn't singular on the horizon. $\endgroup$ Commented Sep 19, 2013 at 13:55
  • $\begingroup$ So Boyer-Lindquist coordinates are out of the question? :\ But I was under the impression that the calculation is done in B-L :\ $\endgroup$
    – dingo_d
    Commented Sep 19, 2013 at 14:58
  • $\begingroup$ I saw now in book 'Black Holes: An Introduction', that I should use ingoing Kerr coordinates :\ $\endgroup$
    – dingo_d
    Commented Sep 19, 2013 at 15:07
  • $\begingroup$ @dingo_d : I see the formula $\begin{align} \kappa^2 = -\frac{1}{2} \left ( \nabla^a \chi^b \right ) \left ( \nabla_a \chi_b \right ) \end{align}$, in this note, which is citing Wald (12.5.14), and this formula works with the standard Schwarzschild metrics, even it this metrics is singular on the horizon. $\endgroup$
    – Trimok
    Commented Sep 19, 2013 at 18:13
  • $\begingroup$ Well it says that that formula follows from the one I put, so it should work. I'm trying with ingoing Kerr coordinates but I'm getting up nowhere. I'll try with that one and see where that gets me. $\endgroup$
    – dingo_d
    Commented Sep 19, 2013 at 18:28

4 Answers 4


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.

  • $\begingroup$ You can use *emphasis word* to create italics, no need to use \textit{} here. $\endgroup$
    – Kyle Kanos
    Commented Jan 31, 2015 at 2:39
  • $\begingroup$ The way you obtain that $\kappa = \frac{f'}{2}$ for a Schwarzschild Black Hole fundamentally depends on changing the coordinate system to EF coordinates, such that it is well defined at the horizon. $\endgroup$ Commented Jun 23, 2020 at 14:12

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.


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

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.

  • 1
    $\begingroup$ I was trying to calculate $C^2$. Now as both $\chi^\mu \chi_\mu$ and $g^{rr}$ are zero at the horizon I used L'Hospital. But the final result gives $C^2<0$. Can you point out how to calculate it? $\endgroup$ Commented Oct 9, 2016 at 14:03
  • $\begingroup$ I do not think this works, you are only guaranteed that $\xi_\mu$ is proportional to $\partial_\mu r$ at the event horizon, away from it, they are actually no proportional ($\partial_\mu r$ is a spacelike vector outside the black hole, while $\xi_\mu$ is timelike). This means that taking the limit from outside the horizon to the horizon is not valid. Like @ProfShonku pointed out, you get $C^2 < 0$ which is not sensible. $\endgroup$ Commented Jun 23, 2020 at 14:23

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...

  • $\begingroup$ The derivation is straightforward with the killing vector so long as your coordinate system is non-singular. See the coordinate system given in Hawking and Ellis. $\endgroup$ Commented Dec 19, 2013 at 23:35

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