Penrose explained that a trapped surface is a region of spacetime where the null expansion $\theta$ is always negative. A black hole is therefore a trapped surface that cannot communicate with null infinity.

On the other hand, Hawking's area theorem says that the expansion $\theta$ is strictly non-negative on the horizon.

How can the expansion be positive on the horizon and negative immediately inside? Doesn't it have to be continuous?

  • $\begingroup$ Non-negative and positive are not the same thing. $\endgroup$
    – G. Smith
    Jul 8, 2020 at 20:04
  • $\begingroup$ That's not the answer. A generic black hole will have positive expansion on the horizon so the question remains. $\endgroup$ Jul 8, 2020 at 20:05
  • $\begingroup$ Then that information belongs in the question, not in a comment, so that the third paragraph is not a non sequitur. $\endgroup$
    – G. Smith
    Jul 8, 2020 at 20:09
  • $\begingroup$ I think you forgot to make the distinction between a trapped surface which is defined with both families of null geodesics having $0$ expansion, and the event horizon. We can prove that one is contained in the other but they need not be strictly the same if I remember correctly. Chapter 4 of damtp.cam.ac.uk/user/hsr1000/teaching.html notes on black holes explains this really well $\endgroup$ Jul 8, 2020 at 20:12
  • $\begingroup$ are you sure that the expansion of the horizon that you're using in the statement of the area theorem is the outward null expansion? Because an expanding blackhole horizon is a spacelike surface that therefore does not have a null tangent vector. $\endgroup$ Jul 8, 2020 at 21:47

1 Answer 1


Trapped surface is a quasilocal notion — a closed two-surface which has the property that the expansions in each of the two forward-in-time pointing, normal-to-the-surface, null directions are everywhere negative. One does not need to know the metric outside of immediate vicinity of a surface to determine whether it is trapped.

Event horizon, on the other hand is a global notion, to determine precisely where it is located at any given moment we would still need to now all the future evolution of our spacetime.

So, there is no contradiction. A surface could be inside event horizon and yet have a positive expansion for its null geodesics if, for example, all its outgoing null geodesics intersect in the future with infalling matter and then be trapped in a larger trapped surface of a later time slice. Consequently, for growing event horizon the expansion of null geodesic congruence just inside of it would remain positive.

Concrete example: The Vaidya spacetime. Let us consider the spherically symmetric ingoing Vaidya spacetime representing gravitational collapse of null dust matter. In Eddington-Finkelstein-like coordinates $(t,r)$ and for a simple function $m(v)$ with a finite interval of constant rate accretion, the spacetime structure could be represented by the following spacetime diagram:

image from SageManifold gallery notebook

This image and the details of calculation are available from this Sage Manifold notebook.

Since angular variables are suppressed in this figure each point represents a 2-sphere. Yellow lines are infalling matter. Green lines are null geodesics (infalling geodesics have constant advanced time $v=t+r$). Thick black line is the event horizon and red line is the trapping horizon. Trapping horizon here is the boundary of trapped region: any point of the diagram to the left of red line corresponds to a trapped surface. We see that after the last bit of matter crossed the event horizon, trapping and event horizon coincide, but before that, there are surfaces outside trapping horizon (that would thus have positive expansion of outgoing null geodesics), yet inside the event horizon (meaning outgoing null geodesics would still get trapped eventually).

A black hole is therefore a trapped surface that cannot communicate with null infinity.

This would be wrong, since this statement mixes quasilocal and global aspects. If there is no communication with asymptotic region, then trapped surface is redundant. Generally, trapped surfaces are used as a substitute for global boundaries in numerical relativity since the full evolution of spacetime is not yet known and event horizon is thus cannot be determined (yet).

For a discussion on differences between various types of boundaries of black hole spacetimes see the following paper:

  • $\begingroup$ I see. So as long as matter is still falling in, the event horizon will always be outside the trapped surface, and it's the event horizon that always grows, since this is the boundary of the region that cannot communicate with infinity. Thank you very much for this beautiful answer. $\endgroup$ Jul 13, 2020 at 9:25
  • $\begingroup$ Interesting, though, that the trapping horizon also looks like it's growing in your figure. Maybe this is only an illusion? Or maybe it grows in time but the null expansion is somehow still negative there? $\endgroup$ Jul 13, 2020 at 9:29
  • $\begingroup$ Trapping horizon is generallly not a null hypersurface, it does not correspond to a null geodesic and for spherically symmetric case it is a hypersurface where radial null geodesics that were outgoing initially turn back $dr/d\lambda=0$ (expansion is precisely zero). So yes, it is growing in this coordinates while the matter is crossing its Schwarzschild sphere. But note that the growth region is spacelike and this part of trapping horizon could for example be made into $t=\rm const$ slice fragment. $\endgroup$
    – A.V.S.
    Jul 13, 2020 at 10:57

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