# Light-like normal vectors

Can someone please show me how to mathematically establish that the normal vector to the event horizon of a Kerr Black Hole is light-like?

• Have you tried making some computations yourself? What is it in particular that you do not understand about the computation?
– Void
Jun 4 '19 at 15:09
• I am clueless. I don't know how I can find the normal vector. I don't need to do computations, I need to prove it analytically. Jun 4 '19 at 15:10
• Would you the be able to write down the normal to the horizon and compute its norm in, say, the Schwarzschild space-time?
– Void
Jun 4 '19 at 15:12
• No. I don't know that either Jun 4 '19 at 15:13
• I just need the general procedure of finding the normal vector from a metric Jun 4 '19 at 15:13

In Kerr coordinates $$(v,r,\theta,\phi)$$ define,

$$k^a=\Big(\frac{\partial}{\partial v}\Big)^a\qquad m^a=\Big(\frac{\partial}{\partial \phi}\Big)^a$$

Proposition: $$\xi^a=k^a+\Omega_Hm^a$$ is normal to $$r=r_+$$, where $$\Omega_H=\frac{a}{r_+^2+a^2}$$.

Proof (Sketch): If you determine $$\xi_{\mu}$$ in Kerr coordinates you can show that $$\xi_{\mu}dx^{\mu}|_{r=r_+}\propto dr$$ and therefore it is normal to the surface $$\{r=r_+\}$$ since $$dr(X)=0$$ for all $$X$$ in the tangent bundle to this surface. Then you just need to calculate $$\xi^{\mu}\xi_{\mu}$$ which you should find is vanishing.

• I'm sorry, I don't understand the quantities k^a and m^a. You are taking the derivative with respect to v and phi respectively. What does the a index represent? Jun 4 '19 at 15:42
• @Souradeep Ah I'm sorry, I've written them in `abstract index notation'. This is used to denote quantities that hold in any frame, i.e. here there is always a normal to the horizon in any frame. So here $\partial_v$ can be just thought of as a label for the vector that has components $(1,0,0,0)^T$ in the Kerr coordinate basis and similarly, $\partial_{\phi}$ has components $(0,0,0,1)^T$ in the Kerr coordinate basis. Jun 4 '19 at 15:55
• So, that means that xi is just the vector (1,0,0,Omega_H) in the Kerr basis Jun 4 '19 at 16:01
• @Souradeep Yes :) Jun 4 '19 at 16:06

Let me try to write down explicitly Sam Colley's proof, so that someone on the learning curve (like me^^) may get the idea more straightforwardly.

First, it is convenient to write down the Killing vector in Kerr coordinates $$(v,r,\theta,\chi)$$ as a vector (contravariant vector). To be specific, the components of $$\xi^a=k^a+\Omega_H m^a=(\partial_v)^a+\Omega_H (\partial_\chi)^a$$ read $$\xi^\mu = (1,0,0,\Omega_H),$$ with 1-form bases $$\partial_v$$ and $$\partial_\chi$$, where $$\Omega_H=\frac{a}{r_+^2+a^2}=\frac{2Mar}{(r^2+a^2)^2}$$ is the angular velocity on the horizon due to frame-drag. In other words, a particle free-falls from infinity with vanishing angular momentum $$L=0$$ will gain this amount of angular momentum as it touches the horizon.

On the other hand, the normal vector of the horizon can be obtained by the hypersurface $$f(v,r,\theta,\chi)=r-r_+$$. In this case, it is more convenient to express this normal vector as a 1-form (covariant vector) with components $$n_\mu=(0,1,0,0).$$

Then, the question reduces to show explicitly (1) $$g_{\mu\nu}\xi^\mu \propto n_\nu$$ and (2) $$\xi^\mu n_\mu =0$$.

Now, (2) is almost trivial, given the components of the vectors. For (1), one makes use of the metric tensor and takes into account that the calculations are carried out on the horizon $$r=r_+$$. To be more specific, by making use of

$$g_{\chi\chi}=\frac{(r^2+a^2)^2-\Delta a^2\sin^2\theta}{\rho^2}\sin^2\theta ,$$ $$g_{\chi v}=-\frac{2Mra}{\rho^2}\sin^2\theta ,$$ $$g_{vv}=-\left(1-\frac{2Mr}{\rho^2}\right) ,$$ where $$\Delta\equiv r^2+a^2-2Mr$$ and $$\rho^2=\Sigma\equiv r^2+a^2\cos^2\theta$$. One finds $$\xi_\chi=g_{\chi v}\xi^v+g_{\chi\chi}\xi^\chi =\frac{(r^2+a^2)^2-\Delta a^2\sin^2\theta}{\rho^2}\sin^2\theta\Omega_H+\left(-\frac{2Mra}{\rho^2}\sin^2\theta\right)=0 ,$$ $$\xi_v=g_{v v}\xi^v+g_{v\chi}\xi^\chi =0 .$$

(a) The above explicit calculations are only valid for Kerr coordinates but not for Boyer-Lindquist coordinates. Some reference seems to mix the formulae in these two coordinates and might be a source of confusion (compare, for instance, Eq.(4.5), Eq.(4.32), and Eq.(C.10) of arXiv:1501.06570). In fact, in Boyer-Lindquist coordinates, one can readily demonstrate that the above normal vector is indeed null, by noticing $$g^{\mu\nu}n_\mu n_\nu=g^{rr}=\frac{\Delta}{\rho^2}$$ and $$\Delta=0$$ on the horizon (see Eq.(2,8) of arXiv:1410.2130).
(b) A straightforward physical implication is the following. Consider a particle free-falls from infinity. It has the energy and angular momentum $$\delta E$$ and $$\delta L$$, both measured by an infinite observer. As it falls through the horizon, its energy is measured by an observer who is marginally hanging onto the horizon by an angular velocity $$\Omega_H$$, namely, whose world-line is marginally time-like (mostly null in practice). The latter, on the other hand, finds that the energy of the particle is $$\delta E_H$$. By the above result, one can show that $$\delta E_H \propto \delta E - \Omega_H \delta L.$$