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Proposition: the event horizon and the apparent horizon of a black hole always coincide.

As a reminder: the event horizon is defined as the boundary of the closure of the causal past of future null infinity, i.e. the boundary of the closure of the set of points which can reach infinity by a timelike path (so the boundary of the black hole). The apparent horizon is defined as the boundary of the set of trapped surfaces, where a trapped surface is defined as a surface for which all outgoing null rays converge.

Question: The above proposition is known to be false. Can someone give a counterexample? As always for counterexamples, the simpler the better!

I know the following:

  • For a counterexample we necessarily need to consider a non-stationary black hole, as Hawking proved that the proposition is true for stationary ones.
  • The apparent horizon is always inside the black hole. But doesn't this imply that there are points inside the black hole from which light rays can diverge?
  • Apparently, one counterexample is given by stellar collapse, though I fail to understand this counterexample, i.e. I don't recognize the points which are outside the apparent horizon and inside the black hole.
  • Unlike the event horizon, the apparent horizon is observer dependent. Wikipedia states: "For example, it is possible to slice the Schwarzschild geometry in such a way that there is no apparent horizon, ever, despite the fact that there is certainly an event horizon." Clearly they don't coincide then, but doesn't this contradict the fact that they do coincide for stationary spacetimes?

So basically, I know some differences between the two concepts, but I cannot think of a concrete example for which we can identify the apparent horizon and the event horizon, and see that they are different.

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  • $\begingroup$ You can look at Hawking-Ellis Proposition 9.2.8, the comments after it, and the figure on the next page(figure 59). I don't know if that figure satisfies you as an example. $\endgroup$ – MBN Jan 16 '14 at 9:52
  • $\begingroup$ I checked that figure before. I understand it provides a counterexample, though I don't understand why the apparent horizon is at the specific place where it's depicted there. $\endgroup$ – ScroogeMcDuck Jan 16 '14 at 21:09
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One simple counterexample is the observer moving in the Minkowski spacetime with constant acceleration. This observer perceives an apparent horizon - Rindler horizon, while there is no event horizon at all (and therefore no black hole). The existence of this apparent horizon leads to the Unruh effect - black-body radiation measured by such observer.

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    $\begingroup$ Rindler horizon is not an apparent horizon in the sense of the OP: It is not the boundary of the set of trapped surfaces, where a trapped surface is defined as a surface for which all outgoing null rays converge. $\endgroup$ – Valter Moretti Jan 16 '14 at 15:12
  • $\begingroup$ It's constant acceleration from the observers standpoint. $\endgroup$ – Kevin Kostlan Jan 16 '14 at 15:54
  • $\begingroup$ @V.Moretti: Agreed. Can this possibly lead to another type of counterexamples: there could be an event horizon but no trapped surfaces since null rays rays do not converge? $\endgroup$ – user23660 Jan 16 '14 at 16:01
  • $\begingroup$ @V.Moretti: Another thought: we possibly could provide convergence to null rays normal to Rindler horizon by placing the accelerating observer inside a contracting cosmological solution. $\endgroup$ – user23660 Jan 16 '14 at 16:42
  • $\begingroup$ I think there is no way to use Rindler horizon: Rindler horizon is in Minkowski spacetime and therein null rays are null segments, it is impossible to have trapped surfaces. Maybe you write Rindler horizon but you actually mean Killing horizon (?). In general, however, the existence of trapped surfaces does not depend on the observer. I am not an expert on these counterexamples however my feeling is that apparent horizons and event horizons are different, though related, notions so it should be possible to find out a counterexample in the literature. $\endgroup$ – Valter Moretti Jan 16 '14 at 16:49
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The Vidya metric is a counterexample. Any non-stationary horizon is as well.

$$ds^2 = -f dv^2 + 2 dv dr + r^2 d\Omega^2$$

Sections 5.1.7 and 5.1.8, starting p. 171 of Poisson's "Relativist's Toolkit".

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Consider a collapsing shell of dust in what is otherwise flat spacetime. Outside is Schwarzschild spacetime, inside it is Minkowski spacetime. There is a moment in time when the shell reaches its Schwarzschild radius $r = 2M$. The event horizon forms earlier, at the centre, and expands outward at the speed of light to reach $r = 2M$ at the same time the shell does. After that, the event horizon remains at $r = 2M$ forever.

The apparent horizon forms when the shell crosses $r = 2M$, and splits with one apparent horizon remaining at $r = 2M$, and another apparent horizon following the shell down to $r \rightarrow 0$. So the latter does not correspond with an event horizon. Figure from Visser (2014)

This example is taken from an excellent pedagogical paper by Matt Visser, "On the observability of horizons" (2014), see his Figure 1.

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