Timelike curves on event horizon

In summary, "Timelike curves on event horizon" explores the theoretical implications of paths that objects can take near the event horizon of a black hole. It discusses how these curves challenge our understanding of spacetime and causality, especially in relation to general relativity. The analysis highlights the complexities of time travel and the potential for closed timelike curves, raising questions about the nature of time and the structure of the universe in extreme gravitational fields.
  • #1
Antarres
209
102
I want to show that it is impossible to construct a timelike curve between two points on the event horizon of a black hole. This should be an obvious fact, for example, by considering any particular model of a black hole, one can go to coordinates that extend over the horizon and by observing the deformation of the light cones as one approaches the horizon, one would notice that they become degenerate, which indicates that the horizon is a null hypersurface.

However, I want to prove this without assuming from the start that the horizon is a null hypersurface, but to simply use the formal definition:

The event horizon is the boundary of a region from which it is impossible to reach null infinity by any causal path.

Ideally, this impossibility could come out of this definition directly, however I don't see yet how it is impossible to have a timelike path that stays on the horizon(by this definition then it wouldn't be able to reach null infinity since it stays on the horizon).

I have a feeling this shouldn't be hard to see, but for some reason my mind got stuck on it, so any help would be appreciated.
 
Physics news on Phys.org
  • #2
Antarres said:
I want to prove this without assuming from the start that the horizon is a null hypersurface …
The event horizon is the boundary of a region from which it is impossible to reach null infinity by any causal path.
Isn’t the boundary of a region defined by causal paths necessarily null?
 
  • Like
Likes PeroK
  • #3
Dale said:
Isn’t the boundary of a region defined by causal paths necessarily null?
Yes, it is. Wald, Chapter 8, on causal structure, goes into the details. See, for example, the corollary to Theorem 8.1.2, p. 191. Or, if those details aren't enough, the subject is treated in even more detail in Hawking & Ellis. :wink:
 
  • #4
Antarres said:
I have a feeling this shouldn't be hard to see
Intuitively it should be fairly easy to see that the statement @Dale made in post #2 is plausible (it's basically the statement that the speed of light is the fastest that any causal influence can travel), but proving it rigorously is not quite so simple. The references I gave in post #3 are based on quite a lot of mathematical work on the causal structure and properties of spacetime; it wasn't something that was easy to complete.
 
  • Like
Likes PeroK
  • #5
Yes, it definitely seems plausible, what @Dale said.

I looked into Wald, for the theorem you mentioned, but it doesn't exactly clarify what I want to prove(at least not yet). For example, I was able to prove that horizon is definitely swept by null geodesics, however not every surface swept by null geodesics is a null hypersurface. Adding a causal condition that it is an achronal surface completes the proof.

The horizon being an achronal surface basically amounts to what I was asking for. The theorem in Wald basically says that any chronological future of a point is bounded by null curves, and the corollary states that in case those curves aren't timelike they must be null. However, it doesn't look to me that I can decide solely based on that that I don't have a timelike curve on the horizon, because the definition of the horizon only applies to the exterior, which means that basically exterior of the horizon is bounded by null curves, which implies that null curves are tangent to the horizon. But this does not imply that they are also normal to the horizon. Of course every null vector is normal to itself, but it should also be normal to all the other spacelike vectors(the whole tangent space of the horizon), for it to be defined as a normal.

So it should be the case that it is impossible to find two independent null paths along the horizon, but rather that there exists only one null direction. I'll try looking at Hawking & Ellis for more details, but if someone is aware of a simple proof of this fact, that would be nice.
 
  • #6
Consulting Hawking&Ellis proved to be illuminating. Proposition 6.3.1 proves exactly what I stated in the last post from definition.

Thank you very much for the reference, I refrained from using it thinking that there are less contrived proofs but this works just fine.
 
  • #7
Antarres said:
The theorem in Wald basically says that any chronological future of a point is bounded by null curves
Or the chronological past. I believe Wald remarks that you can switch "future" and "past" in pretty much all of these theorems and they are still valid.

Antarres said:
the corollary states that in case those curves aren't timelike they must be null.
No, it says that if the curves are in the causal future (or past) but not in the chronological future (or past), then they must be null. Such curves are in the boundary of the causal (or chronological) future or past (because the interior of the causal future or past is the chronological future or past).

The event horizon is the boundary of a causal past--the causal past of future null infinity. Therefore, by the above, it must be null.
 
  • #8
Indeed, I was a bit sloppy when writing the last post about interpretation of those theorems, thanks for correcting me.

It looked at first that this proves that event horizon must be generated by null geodesics, but not further than that. However, this corollary also implies that there's no timelike curves tangent to the horizon, which makes the tangent space of the horizon consist of a basis of one null direction + 2 spacelike ones(since having no timelike curves implies having exactly one null direction). Then its orthogonal complement has a null direction, so it is a null hypersurface, like you say. The proposition in Hawking & Ellis that I mention looked to me like it gives a direct consequence of this being the case, but you're right, the same can follow from this corollary.

Thanks for the help in clarifying that.
 
  • Like
Likes PeterDonis and Dale
Back
Top