# Event horizon - physical or chart-dependent?

1. Jan 27, 2009

### DrGreg

In another thread it was said:

(my bold)

In Rindler coordinates a "Rindler horizon" forms at a fixed distance below the Rindler observer (accelerating uniformly "vertically upwards" and Born-rigidly in flat spacetime, in the absence of gravity). This horizon has many of the properties of a Schwarzchild horizon around a black hole: a free-falling object takes infinite coordinate time to drop to it, nothing (not even light) passes vertically upward through the horizon, the coordinate speed of light at the horizon is zero, and so on. Yet, it can't be claimed (can it?) that the Rindler horizon "is an objective physical truth that all coordinate systems agree on" because in Minkowski coordinates nothing unusual happens at all.

Now I guess this may depend on precisely how you define what an "event horizon" actually is. Maybe the Rindler horizon doesn't count as an event horizon under some definitions.

So, is an event horizon an artefact of a coordinate chart?

2. Jan 27, 2009

### George Jones

Staff Emeritus
Consider the region of spacetime from which it is possible for a photon to escape to (future null) infinity. Now consider the complement of this region, i.e.,the region of spacetime from which it is not possible for a photon to escape to infinity. This latter region is the black hole region.

The boundary between the two regions is the event horizon. This is the definition of "event horizon" that relativists use.

The Rindler horizon is not an event horizon since from any event in Minkowski spacetime, it's possible for a photon to escape to infinity. The Rindler horizon is an example of a Killing horizon. The Schwarzschild horizon is both an event horizon and a Killing forizon.

3. Jan 28, 2009

### DrGreg

Thanks, George.

I've seen the technical definition of a Killing vector, but never really got an intuitive grasp of what it is. Looking at Wikipedia's inadequate article on Killing horizon, it seems that essentially it's the fact that the vector (1,0,0,0) (i.e. $\partial / \partial t$) is null on the horizon that characterises it as a "Killing horizon". Or to put it another way, $g_{00}$ vanishes on the horizon.

The definition of event horizon (referred to as "absolute horizon" on Wikipedia) raises the question of what exactly "escaping to infinity" means. I haven't thought about this too deeply, but my initial thought is that a Rindler observer would think that photons below the Rindler horizon don't escape to infinity, even though the Minkowski observer thinks they do. As all coordinate systems are equally good, why is the Minkowski observer correct and the Rindler observer wrong? In any case, what is infinite in one coordinate system could be finite in another. (E.g. think of the stereographic projection of the infinite complex plane onto a finite sphere.)

How do you escape to infinity in a coordinate-independent way?

4. Jan 28, 2009

### robphy

5. Jan 29, 2009

### DrGreg

I don't see how that answers my question. Did you read my whole post or just the last sentence? The preceding sentences set the context.

(If we ignore the left half of the diagram, the rest roughly represents Rindler coords and shows light to the left of the horizon apparently failing to escape to infinity, even though if we redrew the diagram in Minkowski coordinates the light would escape to infinity. The two observers appear to disagree on where infinity is; how do we know one of them is wrong?)

6. Jan 29, 2009

### xantox

A black hole event horizon is observer-independent since it depends only on the asymptotic structure of spacetime, not on the specific worldline of a particular observer – like it would be the case for a Rindler horizon. The event horizon is the boundary of all events, in an asymptotically flat spacetime, which do not lie in the causal past of future null infinity. Note that future null infinity is an asymptotic region of spacetime which all observers' worldlines not trapped in a black hole must intersect at some point.