# General definition of an event horizon?

Horizons are in general observer-dependent. For example, in Minkowski space, an observer who experiences constant proper acceleration has a horizon.

Black hole horizons are usually defined as boundaries of regions from which no lightlike curve can reach null infinity $\mathscr{I}^+$. But how can this be interpreted in terms of an event horizon for an observer? Immortal material observers end up at timelike infinity $i^+$, not $\mathscr{I}^+$.

Is there some nice way of unifying both cases? In other words, is there a general definition of an event horizon that has both these types of horizons as special cases?

[Edited to clarify the question and remove a mistake about the dimensionality of $i^+$ versus $\mathscr{I}^+$.]

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"Horizons are in general observer-dependent." Am I missing something here? $\partial J^-(\mathscr{I}^+)$ is a topological notion; all observers should agree on which points are in the set. It's possible I'm being really confused. – Chris White Apr 25 '13 at 5:50
"Horizons are in general observer-dependent." Rather I would say that no observer will ever observe a horizon. A horizon is only defined by the behaviour at infinity and no observer can wait that long to observe it. Of course an observer can calculate where the horizon is, but that's not the same thing. – John Rennie Apr 25 '13 at 6:26
Also I'm not sure I understand why you would want to interpret a BH horizon in terms of a horizon for an observer (which I take to mean a timelike curve). What did you have in mind here? – twistor59 Apr 25 '13 at 7:07
They way I understand it, he is asking about a general definition of Horizon that has BH and Observer H as special cases. – MBN Apr 25 '13 at 10:36
@MBN: Yes, that's what I mean. I'll edit the question to try to clarify. – Ben Crowell Apr 25 '13 at 12:58

Like the horizon of distant land or sea on Earth, it's a location beyond which you'll not receive any sight of. For the Earth, it's just the bulk of the planet being in the way. This obviously observer-dependent.

For a black hole, or the cosmological boundary, it's more a matter of light can't get out, or does but is red shifted to zero frequency before it reaches the observer. (For the black hole, we assume the observer isn't falling in.)

The relation between something "inside" or "beyond" the horizon and the observer is one of following paths. The geodesic equation for null geodesics, or ray optics, must be used to study bundles of potential rays of light, and see which rays from what locations go where. It is by nature a global characteristic of the spacetime. No local knowledge of Riemann tensors or anything locally measurable can be used to define the horizon. If coordinate systems were observable instead of pure thought-stuff, then certain coordinate systems that make the math look unusually simple, such as the Schwarzschild one, may conveniently blow up at the horizon, but this theory not reality.

All immortal observers and their furniture do end up at $i^+$. In the conformal diagrams as shown in most books (Misner Thorne Wheeler, etc), $i^+$ is a point though in reality it refers to an infinite range of places.

Light emitted from mortal or immortal observers, or from anything, and assuming no funny business such as refraction though an infinite amount of matter, will end up at $\mathscr{I}^+$, no matter where, no matter how tremendously far back in the past or ahead in the future it originates. $\mathscr{I}^+$ is all places infinitely far and infinitely later, also a vast range of places but of lower dimensionality than $i^+$, despite the diagram portraying them oppositely so. It's like lat-lon projections for maps of Earth, where the North and South Poles are shown as horizontal lines.

With the screwed-up topology and dimensionality, confusion can be expected. What matters is that starting at any point inside the "event horizon" (fig 1 of the cited paper) rays of light emanating at 45 degrees and rays of massive matter (doomed observers, partners of unmatched socks, etc) heading upward, can't get out of the gray area. They can't get to $i^+$ except for light starting exactly right on the event horizon heading radially outward, but this is an edge case of measure zero.

Possibly useful text for further reading: General Relativity by Wald, chapters 11 and 12 in particular.

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Thanks for correcting my mistake about the dimensionality of $i^+$ versus $\mathscr{I}^+$. I'll edit the question to remove the mistake. But I don't think your answer addresses my question. – Ben Crowell Apr 25 '13 at 13:01