3

This an attempt to improve my recent question What happens to the singularities of two black holes in the moment they merger?. What is the background talking about „singularities“ or „masses in the center?“ In Schwarzschild black holes and astrophysical black holes (Oppenheimer Snyder collapse) the mass is in the singularity which is a point in time and not part of the manifold. „Mass in a point“ means that General Relativity breaks down (keywords here are infinite curvature and geodesic incompleteness) and so the challenging question is how to avoid the singularity so that the mass is in the center somehow (planck scale) and not in a point. Such black holes are called „physical“ or „real“ black holes sometimes.

The answer could be loop quantum cosmology. There are numerous papers, .e.g this one:

We find the novel result that all strong singularities are resolved for arbitrary matter. … The effective spacetime is found to be geodesically complete for particle and null geodesics in finite time evolution. Our results add to a growing evidence for generic resolution of strong singularities using effective dynamics in loop quantum cosmology by generalizing earlier results on isotropic and Bianchi-I spacetimes.

Now what happens in the moment of the mergering? It seems quite clear that in the classical case (Schwarzschild black hole) the newly formed black hole being strongly deformed though has one singularity. Can we assume the same for real black holes whith their masses in the center? But how then do we explain that two seperately located masses are unified instantaneously? Is it more reasonable to assume the the two masses move towards to each other during the ringdown? Any clarification will be highly appreciated.

timm
  • 1,525
  • 6
  • 16
  • are you asking this question from the view of an external observer, or from the view of the singularities? – Árpád Szendrei Apr 28 '18 at 10:07
  • 1
    Shouldn't this question have an answer when just considering classical general relativity? Why do we need to drag quantum mechanics into it? Presumably, the classical black hole and the quantum gravity black hole will only differ very close to the singularity. – Peter Shor Apr 28 '18 at 11:47
  • Peter, it makes a difference if the BH contains vacuum or matter in the center, as theorists believe. – timm Apr 28 '18 at 15:11
  • Arpad, I was thinking of a diagram which shows the entire spacetime, e.g. like the "trousers", see my comment to Peter. – timm Apr 28 '18 at 15:49

1 Answers1

2

First, when talking about black holes, the word "instantaneous" doesn't make any sense. Because of general relativity, the times of events at different locations can't be compared. In some sense, everything inside the black hole happens after the outside universe ends. This isn't a very useful view, but "instantaneous" doesn't make any better sense in any of the other ways of looking at it.

The only kind of black hole we actually understand the insides of is the Schwartzchild black hole. If we take two Schwartzchild black holes and collide them head on to get a Schwartzchild black hole, then the point singularities must also collide head on; presumably they fuse to form a new point singularity (exactly how this happens will depend on quantum gravity, which we don't understand).

We have equations for the metric inside a rotationally symmetric Kerr (rotating) black hole, which has a very strange ring singularity inside it. However, the space-time inside a Kerr black hole is unstable, so real Kerr black holes should not be rotationally symmetric or contain ring singularities inside.

Now, if you take two Schwartzchild black holes and collide them slightly off-center, you get a Kerr black hole. This Kerr black hole is not going to have the rotationally symmetric (and completely unphysical) ring singularity in it; the only thing that can possibly happen is that it will contain the original two singularities of the original black holes, which are now rotating around each other. We don't understand the metric inside this type of Kerr black hole.

Next, suppose you take two of these realistic Kerr black holes and collide them so as to get rid of the angular momentum and end up with a Schwartzchild black hole. There are four singularities in the original two black holes, and I would guess they go through some complicated dynamics and end up colliding to form one singularity. Nobody knows the details of this process, though ... it's beyond our capability to compute numerically right now.

Peter Shor
  • 11,274
  • Yes, true, I was hoping that the  trousers topology would at least be helpful to understand qualitatively what‘s happening. Perhaps to obtain an answer requires perturbation theory

    For the ringdown, black hole perturbation theory can be used.

     – timm Apr 28 '18 at 14:57
  • If we could numerically model the collision of two realistic Kerr black holes to obtain a Schwartzchild one, I think we could give you a pretty decent answer to your question. However, as far as I know, this is currently beyond the scope of our numerical solvers. – Peter Shor May 04 '18 at 13:17
  • I think so too. My hope was - disregarding such details - to at least heuristically come up with an answer. – timm May 04 '18 at 16:30
  • „if we take two Schwartzchild black holes and collide them head on to get a Schwartzchild black hole, then the point singularities must also collide head on.“ Good point which highlites the problem. In this scenario the masses are at a point in time, not at a location describable by coordinates. Can masses collide which don‘t belong to the manofold? In contrast the masses are at locations in „real“ black holes before and after having collided. That‘s why I made this difference. – timm Jul 03 '18 at 08:31