# A Recognize horizon coordinate-independently?

1. Jul 16, 2016

### strangerep

This is probably something I should know already, but... (sigh).

OK, so we can't just look at a metric (e.g., black hole in Schwarzschild coords) and deduce something physically weird at $r = 2M$, because that's just an artifact of the coordinate system -- the curvature tensor remains finite on that surface, and there are other coord systems in which the metric looks nicer. Also, there are spacetimes (e.g., Rindler) which have horizons even though they're globally flat.

We have the weirdness that an ingoing radial geodesic can cross a black hole horizon and continue down to the singularity, but an initially-outgoing radial geodesic (started from inside the horizon) can't get out, and quickly turns around to fall inward again.

Is there a way to easily recognize such behaviour (i.e., a 1-way horizon) using coordinate-independent techniques, without having to solve and analyze the full equations of motion? E.g., some tensor condition involving Killing vectors? Degeneracy of 1 or more degrees of freedom? Other?

2. Jul 16, 2016

### Orodruin

Staff Emeritus
Did you try an alternative set of coordinates for the black hole such as Kruskal-Szekeres? In those coordinates, the event horizon is simply the region from which you can never escape the singularity. Also, locally close to the horizon, the Schwarzschild black hole behaves just as Rindler coordinates on Minkowski space (it must! equivalence principle!).

In order to separate the two you must use global properties of the space-time. For example, in the Schwarzschild case, there exists a global timelike Killing field only outside of the horizon. When you extend to KS coordinates, there is no longer such a Killing field, but when you go from Rindler to Minkowski coordinates, there is.

Careful here. There is no such thing as "radial" inside the black hole, the $r$ direction is a time direction. Going to larger values of $r$ would be like going back in time. You could of course wonder about the time direction on a geodesic inside the black hole and ask whether it goes to larger or smaller $r$ as time increases. If you look at KS coordinates, it should become clear that the geodesics from the region outside of the black hole can only go in the direction of the singularity if you keep the time ordering.

3. Jul 16, 2016

### Markus Hanke

I would say the curvature invariants formed from the Riemann tensor do the trick here, for example the Kretschmann scalar. If at least one of those invariants diverges, you are dealing with a curvature singularity, otherwise it is just an issue with your coordinate system.
I am not sure though if a diverging Kretschmann scalar is both a necessary and sufficient condition for the presence of a curvature singularity.

4. Jul 16, 2016

### Staff: Mentor

That will find (real, non-coordinate) singularities, but that's not what strangerep is looking for if I'm understanding him properly. He's looking for a way of finding horizons and there's no reason why there has to be a singularity at a horizon.

5. Jul 16, 2016

### Staff: Mentor

Y'know, I'm sure that we had a thread on this a while back, but damned if I can find it.....

I think the property that we're looking for is something along the lines of:
Consider a timelike worldline W that extends to infinity. Is there a region of spacetime such that no timelike worldline can be extended forward from any event in the region to intersect W, but there does exist at least one event in W such that a timelike worldline through that event will enter the region? If so, we would say that the region is separated from W by a horizon.

6. Jul 16, 2016

### Staff: Mentor

The Rindler horizon is not a global feature of the spacetime. It is only a horizon for a particular family of accelerating observers. And different families of such observers have different Rindler horizons. So I don't think it's useful to lump Rindler horizons in with event horizons for this discussion.

This is also not true of Rindler horizons--"outgoing" radial geodesics inside the horizon do not "turn around and fall inward again". They keep on going out. They just can't catch up to the Rindler horizon because it is also an outgoing radial geodesic that started outward before they did.

In Schwarzschild spacetime, by contrast, an "outgoing" radial geodesic inside the horizon never moves outward at all; it is moving inward (decreasing $r$ coordinate) from the instant it is created. So it never "turns around".

However, this property--of "outgoing" radial geodesics moving inward, not outward--actually identifies, not the event horizon, but what is called an "apparent horizon" or "trapped surface". This kind of surface can be defined locally: it is a surface for which the expansion scalar of the congruence of outgoing null normals is negative. (That is the technical way of saying "radially outgoing light rays from the surface move inward, not outward".) In an idealized black hole spacetime where the hole is static and has always existed, the apparent horizon coincides with the event horizon, but this will not be true in more realistic cases where the hole gains mass as things fall into it.

The key issue is that the event horizon is not locally defined. It is only defined globally--it is the boundary of the region of spacetime that cannot send light signals to future null infinity. So in order to know where the event horizon is, you have to know the entire future of the spacetime.

In the idealized case where the black hole is static, I believe it is proved in Hawking & Ellis that the event horizon must be a Killing horizon, i.e., it must be the surface on which a Killing vector field of the spacetime that is timelike at infinity becomes null. But this obviously only works if there is such a KVF to begin with, and in realistic black hole spacetimes, where things are falling into the hole, there won't be such a KVF. So I don't think this will work as a shortcut for spotting an event horizon in more general spacetimes.

7. Jul 16, 2016

### strangerep

Well, before I posted my question, I spent most of the day reviewing various material. But,... from your answer,... I need to study and think about KS coords more thoroughly. IIUC, the "maximally extended spacetime" contains regions/features that might be unphysical (e.g., white holes). So I'm wondering whether KS coords replace one unphysical puzzle with another.

But isn't this just a coordinate artifact? E.g., an infalling observer, equipped with a clock and a ruler, surely does not start measuring time with his ruler at a certain point. (?)

IIUC, the infalling observer is not aware of crossing the horizon, but rather continues to see a black disc ahead (very briefly, before he is ripped apart by tidal forces). If this observer is carrying a gun, he could presumably fire bullets in what he perceives to be the "upward" direction (away from the black disc). But, if this is done inside the horizon, what happens regarding relative separation of bullet(s) and observer? I would have thought the separation between bullet and observer (as measured by the observer with his ruler) would initially increase, though (the elementary particles of) both would soon end up at the central singularity.

8. Jul 16, 2016

### Staff: Mentor

Yes and no. It is true that the coordinate $r$ is timelike inside the horizon only in Schwarzschild coordinates, so that particular property is coordinate-dependent. But it is also true that all timelike curves inside the horizon have decreasing $r$, and that property is independent of coordinates. (More precisely, this property is true in the interior of the black hole, which is usually labeled as Region II in the maximally extended spacetime, e.g., in Kruskal coordinates.)

It increases, of course. But both the bullet and the observer still have decreasing $r$ coordinates. The observer's $r$ coordinate is decreasing faster; that's why the bullet's separation from him increases.

9. Jul 17, 2016

### Markus Hanke

Ah - I misunderstood the OP. Sorry.

10. Jul 17, 2016

### strangerep

After a little more reading, I realized this is essentially the content of the Hawking rigidity theorem.

But, apparently, it is not always the case that a Killing horizon is also an event horizon: for a Kerr-Newman spacetime, the ergosphere coincides with a Killing horizon, but the ergosphere is outside the event horizon.

11. Jul 17, 2016

### Ravi Mohan

If I understand correctly, you are looking for an event horizon of a spacetime solution. According to what I have read, the safest way to get one is by computing the congruence of null geodesics. But that is what you don't want. Now there are other definitions of horizons which do use coordinate independent techniques, for instance isolated horizons, and they correspond to the event horizons in very special situations. You might find useful information in http://arxiv.org/pdf/gr-qc/0508107v2.pdf and the references in it.

12. Jul 17, 2016

### Staff: Mentor

Yes, this is a case where the two types of horizon don't match up. The event horizon of a Kerr-Newman hole is also a Killing horizon, but it's the horizon of a different KVF from the one that has the static limit (boundary of the ergosphere) as its Killing horizon.

13. Jul 17, 2016

### strangerep

After a quick skim, that certainly looks like a paper worthy of careful study. Thank you.