The kerr metric describes the frame-dragged space just outside a spinning uncharged black hole. I have read in popular science articles that frame dragging is like a stick spinning in treacle causing treacle around the stick to move with the rotation.But if another stick span nearby in the opposite direction treacle would bulge upwards where the two areas of opposite spin met. Does space get distorted like this for two spinning black holes or does something stop it from doing so? Is the Kerr metric conserved?
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4$\begingroup$ The moment you added a second black hole, the Kerr metric became completely invalid. The Einstein field equations are highly nonlinear; the sum of two solutions is not, generally, also a solution. $\endgroup$– controlgroupCommented Oct 24 at 17:29
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$\begingroup$ @controlgroup. Surely this depends on how close the black holes get? I am asking how much distortion of Kerr metric would be caused? $\endgroup$– user441992Commented Oct 24 at 17:31
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2$\begingroup$ It doesn’t make sense to talk about metrics being “destroyed” or “conserved”. The overall metric simply evolves. If the spins were opposite but equal in magnitude I would expect it to evolve to a Schwarzschild metric. $\endgroup$– GhosterCommented Oct 24 at 17:31
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$\begingroup$ @BenWyvis The Kerr metric describes one rotating black hole, not two. You can't just add spacetimes together except in the weak-field limit (which Kerr objects definitely do not satisfy). $\endgroup$– controlgroupCommented Oct 24 at 17:32
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$\begingroup$ It doesn’t make sense to say “where the frames meet”. There aren’t two distinguishable “frames”. The Schwarzschild metric would apply at $t=\infty$, assuming I’m right that it settles into a Schwarzschild metric. Calculating this would take a supercomputer. $\endgroup$– GhosterCommented Oct 24 at 17:41
1 Answer
While the Kerr metric is a solution for a single black hole, it is not difficult to see intuitively what happens in this scenario.
Frame dragging is the effect of the time dilation. Time of things close to the horizon is nearly frozen, as we observe from outside, so there is practically no relative motion there. Simply speaking, what is near the horizon, stays near the horizon and follows the horizon’s movement whether it rotates, flies forward with the black hole, or expands.
If two rotating black holes are momentarily at a close distance, then the the time dilation created by each adds up in a non-linear way. The horizon is where the time dilation is maximal. So when the time dilation increases near the horizon due to the close presence of another black hole, the horizon expands toward the other black hole as a narrow trunk. This happens with both black holes, so the gap between their horizons quickly narrows.
This is a runaway process. The closer the horizons become, the stronger they are pulled together. Once the trunks extend toward each other, there is nothing to stop them. And once they touch, it’s already one bigger black hole still yet of a weird shape, but no force in the universe can break its two parts apart.
So we will assume for the purposes of this question that the black holes are still far enough apart and are not merging yet. The horizons just have a little hump each in the area of the gap, but are not extending yet as trunks. Matter rotating with each horizon will continue doing so. When it gets to the gap, the time dilation induced by the other black hole will slow down this rotation slightly. We would see it as the matter going over the hump.
As the black holes become closer, these humps will expand into trunks. The matter rotating with the horizon will go all the way up and then down the trunk thus substantially slowing down its movement in the direction around the black hole. This slowing down is due to the time dilation induced by the other black hole that distorts the shape of the horizon.
Hence the answer is that the frame dragging around each black hole remains, albeit in a very distorted shape. The only exception is a single point exactly in the middle between the black holes. A particle located exactly there will be in an unstable equilibrium like a pencil standing on its tip and ready to fall in one direction or another. However momentarily there this particle would not experience a rotational frame dragging due to the symmetry of the setup.
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$\begingroup$ Here is a mathematically correct video from numerical gravity illustrating how the horizons of two black holes extend toward each other and merge: m.youtube.com/watch?v=Y1M-AbWIlVQ $\endgroup$ Commented Oct 26 at 5:43
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$\begingroup$ In the video the orbit precesses because of spin.How does spin cause precession in this case? $\endgroup$– user441992Commented Oct 26 at 9:25
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$\begingroup$ @BenWyvis I think this would be an interesting topic to post a separate question. This way you would get more insight from others in answers and comments. $\endgroup$ Commented Oct 29 at 4:11
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$\begingroup$ Here is a video of a head-on collision of two spinning black holes: m.youtube.com/watch?v=4nM6kf2OAFw $\endgroup$ Commented Nov 12 at 17:11