Black Hole Questions.

[Willem F Esterhuyse]

We have that for an observer outside a black hole, objects takes an infinite amount of time to cross a Black Hole event horizon. Shouldn't we then see no gravity waves from merging Black Holes? The gravity waves should take an infinity of time to come out of the event horizon.

Since time stops at the event horizon of a black hole, shouldn't we see a frozen image of the star that collapsed? This instead of a flat cloud with a hole in it?

[joigus]

This is, I think, an interesting question that can be answered on a number of levels.

1) Mathematically: You solve the equations and see that GR really does predict these distorsions of space-time escaping away from the colliding BHs at speed c.

2) Empirically: LIGO experiment did confirm this prediction

3) Intuitive afterthoughts, thoughtful phrasings of what the equations might be telling us, and why the naive reasoning might fail:

Gravitational waves do not behave like EM radiation or fluxes of charged particles at all. Upon further reflection, there's actually no reason why they should. First, they correspond to a highly non-linear, highly non-static situation, in which you absolutely can't see them as "objects" running away from a static horizon. I'm not sure that, during the collision, the horizon is even properly defined, or smoothly defined in any clear way. You can look upon them, though, as distorsions of space-time that, to make things even more complicated, propagate as degrees of freedom of the Weyl tensor (the part of Riemann curvature tensor that doesn't have to vanish, even in the vacuum.)

I'm certain that radiation and particles do not propagate as degrees of freedom outside of the Einstein tensor.

I expect this answer you won't find 100% satisfactory or understandable, but I hope it is enough to convince you that you can't think of GW in terms of radiation escaping from a static horizon. They are, in a manner of speaking, distorsions of space-time itself that run away from the initial colliding distorsions. 

Put quote marks as you see fit especially in the last sentence.

I hope that's helpful.

3 hours ago, Willem F Esterhuyse said:

Since time stops at the event horizon of a black hole, shouldn't we see a frozen image of the star that collapsed? This instead of a flat cloud with a hole in it?

No, we see nothing. Light escaping from immediate vicinity of BH's horizon is extremely red-shifted, thereby undetectable. Light from inside can't get out. 

[Markus Hanke]

5 hours ago, Willem F Esterhuyse said:

Shouldn't we then see no gravity waves from merging Black Holes?

You probably mean gravitational waves - gravity waves are a phenomenon in fluid dynamics, and have no relation to black holes.

The answer to your question is threefold:

1. Gravitational radiation during BH mergers does not originate only at the event horizon, but results from the quadrupole moment of the binary system as a whole. Any kind of binary system - irrespective of what kind of objects it is comprised of - will be a source of such radiation. It is, in that sense, a global phenomenon of such spacetimes, and its source cannot be localised to any one particular point or region, including the event horizon. That being said, the geometry of the horizons reflect the geometry of all the rest of this spacetime (in very complex ways), so observing the wave forms of the radiation field allows you to extrapolate what happens at the horizon during the merger. This whole process is really a global one, and doesn’t just happen at the horizon.

2. The diverging in-fall time you are referring to applies to Schwarzschild black holes, but the spacetime in a binary system of in-spiralling BHs is not of the Schwarzschild type, not even approximately. Figuring out the precise in-fall time of a test particle from far away into one of these BHs is a highly non-trivial task, that can only be done numerically, but my guess is that it wouldn’t be infinite at all (one of the necessarily prerequisites for an infinite in-fall time is asymptotic flatness, which does not hold in this type of spacetime); it would also explicitly depend on where and when the test particle begins its free fall.

3. Even for Schwarzschild BHs, the infinite in-fall time applies only to test particles moving on time-like or null geodesics of the undisturbed background, ie it applies only to test particles whose own gravity can be neglected. This, however, is not the case for gravitational waves, which will couple to the background curvature in non-linear ways. To put it differently, a spacetime that contains gravitational radiation cannot have Schwarzschild geometry, and thus the infinite in-fall time does not necessarily follow. Even in cases where the gravitational waves are weak enough so that the background could still approximately be treated as Schwarzschild (which is not the case for a binary BH system!), the wave fronts wouldn’t propagate along the same trajectories as free-falling test particles, due to non-linear interactions with the background curvature.

5 hours ago, Willem F Esterhuyse said:

Since time stops at the event horizon of a black hole

No it doesn’t. The length of the world line of a test particle free-falling from far away and crossing the horizon is finite and well defined; spacetime at and around the event horizon is smooth and regular, so time proceeds as normal there.

The thing with this is that in curved spacetimes, there is an important difference between coordinate in-fall time and proper in-fall time. The coordinate in-fall time is what a distant observer will calculate and observe, based on his own instruments, which are not themselves located at the horizon; the numerical value of this will depend on which observer you choose. The proper in-fall time, on the other hand, is what is directly measured by a clock that is attached to the freely falling test particle itself; by definition, it equals the length of that particle’s world line through spacetime. 

For the case of a test particle freely falling into a Schwarzschild black hole, a far-away stationary observer will determine a coordinate in-fall time that diverges (goes to infinity). However, the proper in-fall time of that same test particle is finite and well defined, so the particle reaches the horizon in a finite amount of time as measured by its own clock, and continues falling through the horizon and into the singularity (also in a finite, well defined amount of time).

Because of the way that proper time is defined mathematically, all observers agree on it. On the other hand though, coordinate time is always specific to a chosen observer, and not valid anywhere else.

In curved spacetimes, time is a purely local phenomenon; a far-away observer does not share any concept of simultaneity with processes that happen at the (for him) distant horizon.

[Mordred]

Well covered Markus I was going you reply many of the details you have above however you covered everything I was going to say. Along with additional detail +1

[MigL]

To put it in layman's terms ...
Gravitational waves are the 'ringing' of space-time ( outside any possible event horizon ) from the inward spiral/collision of two or more massive objects.
Not only Black Holes, but neutron stars, white dwarfs, and even bowling balls ( but would be undetectably weak ).

As for 'infinities' and other timings related to Black Holes, your frame of reference matters, and we have different names for these timings depending on who is doing the observing.
( see Markus' explanation for more detail ).