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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?

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

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  • $\begingroup$ 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? $\endgroup$
    – Souradeep
    Jun 4 '19 at 15:42
  • $\begingroup$ @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. $\endgroup$
    – Sam Colley
    Jun 4 '19 at 15:55
  • $\begingroup$ So, that means that xi is just the vector (1,0,0,Omega_H) in the Kerr basis $\endgroup$
    – Souradeep
    Jun 4 '19 at 16:01
  • $\begingroup$ @Souradeep Yes :) $\endgroup$
    – Sam Colley
    Jun 4 '19 at 16:06
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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 $$\xi_\chi=g_{\chi v}\xi^v+g_{\chi\chi}\xi^\chi =\frac{(r^2+a^2)^2-\Delta a^2\sin^2\theta}{\rho^2}\sin^2\theta\Omega_H+\left(-\frac{2Mra}{\rho^2}\sin^2\theta\right)=0 ,$$ $$\xi_v=g_{v v}\xi^v+g_{v\chi}\xi^\chi =0 .$$

Two additional comments.

(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 $$\delta E_H \propto \delta E - \Omega_H \delta L.$$

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