Getting stuff out of a black hole using a second one

[michelcolman]

Matter and even information in a black hole (beyond the event horizon) is supposedly lost forever because nothing can get out.

What if a second black hole were to pass close by? With such a trajectory that it's able to get away again, but close enough to partially undo the first hole's gravity? Wouldn't that allow stuff to get out again? (by flattening out the spacetime in between the two holes)

For example, suppose you are just beyond the event horizon of a black hole, in a spaceship with enormously powerful engines, you are struggling in vain to try to get out but can't even hold your position, then a second hole passes by at high speed and its gravity is just enough to help you get to the middle in between the two holes. Then, as the holes separate again, you just stay in the middle until the holes are far enough away to allow you to escape. Free at last!

You probably don't even need a second black hole, any sufficiently massive object (neutron star, maybe even a planet if you are a really short distance beyond the event horizon) could be enough to push the event horizon in a little bit and allow you to escape.

Wouldn't that work? And wouldn't that, theoretically, allow all supposedly lost information to be retrieved, which means, even if this possibility is only theoretical and highly impractical, that the black hole doesn't have such an enormous entropy after all?

 

[George Jones]

No, this doesn't happen.

As two black holes approach each other, their event horizons don't disappear. If they get close enough, their separate event horizons will merge into one event horizon, i.e., the two black holes merge into one black hole, but nothing inside either horizon ever finds itself outside.

 

[michelcolman]

But suppose the holes get so close together that the event horizons intersect, but neither hole is actually inside the other's event horizon. Wouldn't it then be possible to go from one hole towards the other?

For example, suppose you are just a few inches beyond the horizon of the first hole, the second hole passes by just outside the event horizon of the first hole, on a curved trajectory that goes back into outer space again. Their event horizons intersect, but most of the mass-energy of each hole is still outside of the other's event horizon. Meanwhile, while the two holes are close together, you are actually closer to the second hole than to the first one. So I would imagine the second hole would drag you away with it. Right?

Now adjust the situation a little bit so that, when the holes are at their closest together, you happen to be right in the middle between them. None of the two holes has more right to "claim" you, so you should be able to stay in the middle as the holes separate again.

If intersecting event horizons are a problem, I think you can even get there without the event horizons intersecting. Imagine you are in between two black holes. Both holes have the same mass, and you are exactly in the middle. Imagine the distance between the two event horizons is very small, a few meters, and you are in that area. You get the same amount of attraction from both holes, so you are not being pulled towards one or the other. Your local space-time is flat as seen by an outside observer. Now, supposedly, if you move just a few meters to one side, you would suddenly find yourself inside the event horizon of one of the holes in an extremely curved bit of space-time. How can that be possible? Surely the curvature will increase as you get closer to one hole, but not from nothing to "even light can't get away" in a few meters?

I would imagine the event horizons of the individual holes to be pushed in because of the opposing curvature from the other hole. You would be inside the combined event horizon, but no longer in the event horizon of the individual hole. Then, as the holes separate again, the combined event horizon will disappear.

I can't see any reason why this shouldn't work?

 

[George Jones]

Once the event horizons intersect, there is only one doughnut-shaped event horizon, i.e.,the two black hole have merged into a single black hole. Everything inside the two separate black holes is now inside the single, merged black hole and cannot get out. Also, the merged black hole cannot split into two black holes. Once the black holes have merged, they cannot separate again.

 

[A.T.]

Imagine you are in between two black holes. Both holes have the same mass, and you are exactly in the middle. Imagine the distance between the two event horizons is very small, a few meters, and you are in that area. You get the same amount of attraction from both holes, so you are not being pulled towards one or the other. Your local space-time is flat as seen by an outside observer.

Zero total attraction, doesn't mean flat space-time. The space-time in the center of a mass sphere is not flat, according to the interior Schwarzschild solution. Gravitational potential just has a local minimum there. In your case it has a local maximum along the line connecting the black holes, and a minimum perpendicular to that line. At such extremal points the attraction is zero because it is connected to the first derivate of the potential. The curvature is however connected to the second derivate and doesn't have to be zero.

 

[michelcolman]

Once the event horizons intersect, there is only one doughnut-shaped event horizon, i.e.,the two black hole have merged into a single black hole. Everything inside the two separate black holes is now inside the single, merged black hole and cannot get out. Also, the merged black hole cannot split into two black holes. Once the black holes have merged, they cannot separate again.

OK, I guess that's another big difference between Newtonian and relativistic gravity then. The escape velocity from a set of two bodies is higher than the escape velocity between the two bodies (since each half's own mass does not count when calculating the attraction on that half) and I thought the same would be true for the black hole: the event horizon would only apply to small parts, not if the two halves were separating.

Thinking about this a bit more, I can see where I went wrong. Time itself stops at the event horizon, so there's really no way in which the two situations could be considered similar in any way.

Thanks!

Michel

 

[michelcolman]

Zero total attraction, doesn't mean flat space-time. The space-time in the center of a mass sphere is not flat, according to the interior Schwarzschild solution. Gravitational potential just has a local minimum there. In your case it has a local maximum along the line connecting the black holes, and a minimum perpendicular to that line. At such extremal points the attraction is zero because it is connected to the first derivate of the potential. The curvature is however connected to the second derivate and doesn't have to be zero.

Let me get this straight, because I'm getting confused...

If two black holes approach each other (but not so close as to merge, event horizons not even nearly touching), what happens to the space in between?

Imagine you are just outside the event horizon of a black hole, struggling to get away, and a second hole passes by in the vicinity (but not really close by).

On the one hand, the attraction from the second hole should reduce the force pulling you towards the first one (through tidal effects). But on the other hand, the second hole's gravity could increase curvature to the point where time stands still, effectively moving the event horizon outwards to include you?

(So I get a different conclusion depending on whether I define the event horizon as the place where you can just get away from the hole, or the place where time stands still due to extreme curvature)