Exploring Light's Path Between Two Colliding Black Holes

[Iamu]

I'm going to have trouble stating this question exactly in the language of GR, but I'm going to try my best.

We have a manifold with two identical black holes falling into each other from a large distance with no angular momentum, both spin and around their collective center of mass. A ray of light travels directly between them, spatially. Let's say they're far enough apart that their event horizons aren't touching, but the curvature is exactly symmetrical on either side of the path of the light ray. My best guess would be that the path of zero proper time branches into two between the two black holes; one path is pulled towards the "left" hole and one path is pulled towards the "right" hole. Hopefully, I'm right so far.

What happens to the energy and momentum that was traveling along the light ray's path before it branched? Is it divided equally? Can we set up a relativistic dual-slit experiment in a similar way?

I'm basing my assumption that the path splits on the case where the light passes directly between the black holes just as their event horizons touch, or directly between the singularities after the first contact between the event horizons. The light is bound to orbit or fall into a hole after it touches its event horizon, I figure; so which one does it fall into? Then again, I'm thinking that the proximity of the other black hole will gradually reduce the spatial extension from either singularity towards the edge of its horizon towards the other singularity, and they might never touch until the singularities did. Before I work myself in circles, can anyone please shed some light?

 

[Phrak]

My best guess would be that the path of zero proper time branches into two between the two black holes; one path is pulled towards the "left" hole and one path is pulled towards the "right" hole.

Well, no. An idealized light beam, that is zero cross-sectional area, won't branch into two beams. A real beam will be spread out by the gravitational gradients. Quantum physicists will differ.

 

[Iamu]

Well, no. An idealized light beam, that is zero cross-sectional area, won't branch into two beams. A real beam will be spread out by the gravitational gradients. Quantum physicists will differ.

So I'd assume that the singularities and their horizons touch at the same instant, as the holes fall into each other, then? I'm guessing this, because otherwise, the horizons intersect, first, and then light can potentially be behind both horizons at once.

 

[bcrowell]

The equivalence principle says that spacetime is locally Minkowski. In Minkowski space, two events determine a geodesic, in the same way that two points determine a line in Euclidean geometry. Therefore you can't have forking geodesics.

Another way to get at this is that geodesics obey a differential equation, and there are standard uniqueness theorems for solutions of differential equations.

The example in #1 is stated with perfect symmetry, so by symmetry the geodesic simply goes between the black holes, assuming the two black holes continue to be separate for long enough. But if this is a vacuum solution, then the two black holes are going to merge, and if that happens early enough, the geodesic will terminate on the newly merged singularity.

 

[Iamu]

The equivalence principle says that spacetime is locally Minkowski. In Minkowski space, two events determine a geodesic, in the same way that two points determine a line in Euclidean geometry. Therefore you can't have forking geodesics.

Another way to get at this is that geodesics obey a differential equation, and there are standard uniqueness theorems for solutions of differential equations.

This makes sense. I realize my question was a little confused. Thank you for the clear answer.

The example in #1 is stated with perfect symmetry, so by symmetry the geodesic simply goes between the black holes, assuming the two black holes continue to be separate for long enough. But if this is a vacuum solution, then the two black holes are going to merge, and if that happens early enough, the geodesic will terminate on the newly merged singularity.

So we can't have forking geodesics. But then, I'm wondering how the singularities combine. I don't think the event horizons can ever overlap before the singularities touch, because the event horizon is defined as the boundary from which light can't escape falling into the singularity. So if the geodesic can't branch, it seems to follow that the ray of light can't be behind two different event horizons at once, falling into both singularities unavoidably.

Taking a guess about it from a Newtonian point of view, the gravitational potential energy of a test particle drops off to zero as you approach the point directly between them from the direction of either singularity, and it increases on the far side of either as the two approach. This remains the case no matter how close the points get, so I'm thinking that each one's horizon shrinks towards its respective singularity on the side facing the opposite hole, and expands out farther from the singularity on the other side. So light can pass directly between the singularities until--what happens? I'm thinking that the event horizons can't overlap until the singularities touch, exactly because a geodesic can't fork. So do we have a naked singularity instead, at their point of contact?

Also, could someone please point me to some resources on black hole mergers?

 

[bcrowell]

But then, I'm wondering how the singularities combine. I don't think the event horizons can ever overlap before the singularities touch, because the event horizon is defined as the boundary from which light can't escape falling into the singularity. So if the geodesic can't branch, it seems to follow that the ray of light can't be behind two different event horizons at once, falling into both singularities unavoidably.

The event horizon is defined as the boundary from which light can't escape to infinity. You can have horizons without singularities and singularities without horizons.

 

[JesseM]

The event horizon is defined as the boundary from which light can't escape to infinity. You can have horizons without singularities and singularities without horizons.

But in the case of two black holes colliding, there won't be any horizons without singularities or singularities without horizons, the two event horizons will just combine into one event horizon (containing both singularities) at the moment after they "touch", they don't "overlap" in the manner of a Venn diagram.

 

[bcrowell]

But in the case of two black holes colliding, there won't be any horizons without singularities or singularities without horizons, the two event horizons will just combine into one event horizon (containing both singularities) at the moment after they "touch", they don't "overlap" in the manner of a Venn diagram.

Right, but I think Iamu's reasoning was based on the idea that there had to be a one-to-one correspondence between singularities and horizons. You can have two singularities with a single horizon.

 

[Iamu]

Right, but I think Iamu's reasoning was based on the idea that there had to be a one-to-one correspondence between singularities and horizons. You can have two singularities with a single horizon.

I was thinking that a test particle directly between the singularities, at any distance the singularities happen to be apart, would not accelerate relative to the whole system's center of mass.

I'm not quite getting this. If their event horizons combine before the singularities touch, then, let's say that singularities are close enough that they share an event horizon. We aim a beam of light directly between the singularities as previously. I'd imagine that the light experiences a continual boost, but I don't see why it would be bent spatially towards either singularity. Doesn't the light travel straight through between them, and potentially back out of the "horizon?" The light gains energy approaching and loses it leaving, but I think we can make the difference arbitrarily small by appropriately selecting the distances and velocity of the singularities relative to each other, energy of the light, etc.. Or does selecting these parameters this way preclude them sharing a horizon?

 

[JesseM]

I was thinking that a test particle directly between the singularities, at any distance the singularities happen to be apart, would not accelerate relative to the whole system's center of mass.

I'm not quite getting this. If their event horizons combine before the singularities touch, then, let's say that singularities are close enough that they share an event horizon. We aim a beam of light directly between the singularities as previously. I'd imagine that the light experiences a continual boost, but I don't see why it would be bent spatially towards either singularity. Doesn't the light travel straight through between them, and potentially back out of the "horizon?" The light gains energy approaching and loses it leaving, but I think we can make the difference arbitrarily small by appropriately selecting the distances and velocity of the singularities relative to each other, energy of the light, etc.. Or does selecting these parameters this way preclude them sharing a horizon?

I think once the event horizons combine, it would be inevitable that the two singularities would merge into one (though I'm not sure if this follows directly from the Penrose-Hawking singularity theorems, so perhaps someone more well-versed in GR can comment on how this could be proved/disproved). So if the two black holes are of equal mass and you use a coordinate system where they both are mirror images of one another in the collision, I'd guess that the two singularities will meet at the "center" where the light beam is.

 

[bcrowell]

I think once the event horizons combine, it would be inevitable that the two singularities would merge into one (though I'm not sure if this follows directly from the Penrose-Hawking singularity theorems, so perhaps someone more well-versed in GR can comment on how this could be proved/disproved). So if the two black holes are of equal mass and you use a coordinate system where they both are mirror images of one another in the collision, I'd guess that the two singularities will meet at the "center" where the light beam is.

IIRC the Penrose singularity theorem doesn't really specify much other than the existence of at least one singularity. The no-hair theorems would apply here, and they would tell you that if the external properties of the whole system settle down, they must settle down into the state of a Kerr black hole. Given Iamu's symmetric initial conditions, it would have to have zero angular momentum, so it would be a Schwarzschild black hole externally. I'm not sure that that strictly implies that the two singularities have to merge, but it seems very plausible, since there's no rotation -- I would expect it to happen in a time on the order of the Schwarzschild radius corresponding to either mass.

As for a photon coming in along the axis, it seems like in principle it could either hit a merged singularity on its first pass through the center of symmetry, or conceivably it could oscillate a few times before the singularities got done sluggishly falling together. The oscillating case seems unlikely to me, because I would expect the spacetime inside the horizon to have the same property that a Schwarzschild solution has: that the forward light cone lies strictly in the direction of smaller r.