According to Carroll (6.3 Killing horizons):

In general, if a Killing vector field $\chi^\mu$is null along some hypersurface $\Sigma$, we say that $\Sigma$ is a Killing horizon of $\chi^\mu$.

where a Killing vector field is a vector field such that the metric is conserved i.e. $\chi^\mu=(\partial_{\sigma_*})^\mu\implies \partial_{\sigma_*}g_{\mu\nu}=0$.

He then goes on to say, under the necessary assumptions, that 'Every event horizon of $\Sigma$ in a stationary, asymptotically flat spacetime is a killing horizon for some Killing vector field $\chi^\mu$.'

This last fact is not clear to me and I generally don't understand what Killing horizons have to do with event horizons. This makes it hard for me to understand why Killing horizons are even useful.

This question is related but doesn't answer why these two types of horizons are related.


1 Answer 1


The following is just my quick and dirty understanding after skimming Hawking and Ellis pp. 323-340, and looking at this talk, and this SE question.

We want to talk about the notion that gravitational collapse is complete, and a black hole has reached an equilibrium.

To a distant observer, this is defined by whether the spacetime is stationary, i.e., has a timelike Killing vector "near infinity."

To an observer who intends to poke and prod the black hole from nearby, we want some other definition. However, this is hard, because if we poke the black hole with a stick, we lose our stick. We would like to say that somehow this observer doesn't see the horizon "changing over time," but the horizon is a null surface, so this doesn't quite make sense. You can't visit the horizon once and then visit it again later to see if it's changed. The way we get around this is by the definition of a Killing horizon. A black hole in equilibrium is one whose event horizon is also a Killing horizon. (Not all Killing horizons are event horizons. A light cone in Minkowski space is a Killing horizon.)

The Hawking rigidity theorem says that if the spacetime is a vacuum spacetime and is analytic and regular, then if a distant observer says it's in equilibrium (stationarity), then a nearby observer does as well (horizon is Killing). This is a non-obvious theorem, because due to time dilation, the distant observer may have a hard time judging whether infalling matter has really fallen through the horizon "now."

Another way of stating the result of the theorem is that we actually have two independent Killing vectors at the horizon: the null one and another one that constitutes the extension of the timelike one that exists at infinity.

It's clearly necessary that we assume a vacuum spacetime, or at least have some kind of energy condition. For if not, then we could write down a metric for something like a Schwarzschild metric that oscillates in size, and there would be some matter field that would allow this, which we could find from the Einstein field equations.

There seems to be a lot of work going into strengthening the theorem by weakening the assumptions about analyticity and so on. An early version by Carter, before Hawking's, required axial symmetry. Hawking's proof looks long and high-tech to me, and it seems to involve some case splitting between spinning and static black holes, the former having an ergosphere which seems to complicate things. People seem to suspect that analyticity isn't essential (or maybe can be replaced by something more physically motivated?). Analyticity makes the proofs easier because apparently a theorem by Nomizu says that if there's analyticity, we can successfully extend the timelike Killing vector from infinity to the entire manifold (although it will not necessarily be timelike everywhere, e.g., inside the horizon of a Schwarzschild black hole).

  • $\begingroup$ This nicely explains why Killing horizons are useful but my confusion lies in the fact that I don't understand why Killing horizons and event horizons are related. Why does having a null Killing vector have any physical meaning? What does having a null Killing vector do that makes it an event horizon? (in some cases) $\endgroup$ Commented Jan 17, 2021 at 23:31
  • $\begingroup$ A light cone in Minkowski space is a Killing horizon Only in (1+1)D. But it is a conformal Killing horizon in higher dimensions. $\endgroup$
    – A.V.S.
    Commented Jan 18, 2021 at 13:58

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