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More precisely, are event horizons null geodesic congruences, meaning they admit a local parametrisation $x^\mu(t,s_i)$ (with $i=1,\ldots,D-2$) such that:

For any constant $s_i$ the $x^\mu(t,s_i)$ curve is a null geodesic, meaning if $V^\mu = \frac{\partial x^\mu}{\partial t}$, then $V^2 = 0$ and $V^\mu \nabla_\mu V^\nu = 0$.

Equivalently, if I drop a light ray on any event on an event horizon, is there always an initial momentum such that it remains on the horizon for a finite time (in the past or in the future)?

In addition, is the answer sensible to the dimension of spacetime? The type of event horizon (black hole, cosmological...)?

At first I thought this was obvious by definition, but then after some thought I realised it really isn't. An event horizon is essentially defined as an envelope of past-directed null geodesics from future null infinity, so there's no obvious reason for which the horizon itself should be made out of such null geodesics. (Compare to the trivial example of a circle, which is the envelope of the family of all lines tangent to it, yet it's not a line itself...). However, I can't think of a counterexample.

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An event horizons is always generated by null geodesics. Because it is defined as the boundary of the past of future null infinity $\mathscr{I}^+$, it is a null hypersurface, which is always generated by null geodesics (i.e. define a null geodesic congruence). The only subtlety you have to worry about are caustics, which are where null generators enter the event horizon, and make the horizon look less smooth at certain points. But these do not spoil the fact that every point on the event horizon lies on a null geodesic that remains on the event horizon indefinitely to the future.

Some discussions of this can be found in Wald on/around pg. 311.

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  • $\begingroup$ And on page 194 – it seems theorem 8.1.6 is the crucial result. $\endgroup$
    – gj255
    May 8, 2018 at 22:02
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Yes, under certain conditions. A theorem due to Hawking states that in a stationary, analytic, asymptotically flat vacuum black hole spacetime, the event horizon is a Killing horizon, which in particular means it is a null hypersurface, which in particular means it is a null geodesic congruence.

EDIT: asperanz's answer below is better, in my opinion.

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  • $\begingroup$ up to the null hypersurface I'm great, but why would that imply it's a congruence of null geodesics? $\endgroup$ May 7, 2018 at 12:18
  • $\begingroup$ For example, if I define the surface $x^\mu = (\lambda,\cos\lambda, \sin\lambda,\sigma)^T$ in Minkowski space, $\lambda, \sigma$ parameters, that's a null hypersurface ($\partial_\lambda$ is null) but it contains no null geodesics. $\endgroup$ May 7, 2018 at 12:28
  • $\begingroup$ @RiccardoAntonelli That's a good counterexample, save for the fact that it's not a hypersurface in the sense of having dimension $n-1$ (and so the normal is not unique). You can find a proof that the normal to a null hypersurface generates null geodesics in Harvey Reall's notes on black holes. $\endgroup$
    – gj255
    May 7, 2018 at 12:51
  • $\begingroup$ ah so the key is the codimension 1! Thank you very much, I get it now. $\endgroup$ May 7, 2018 at 13:41

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