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Orbital mechanics says something falling from infinity hits at the escape velocity of the object it strikes.

The escape velocity at the event horizon is lightspeed, it must be higher for the singularity inside.

Thus we have a superluminal impact velocity. Huh? Does special relativity manage to avoid this situation?

Qmechanic
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Loren Pechtel
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    You might be referring to the "pseudo-derivation" of the Schwarzschild radius of a black hole. You can treat it like a classical mechanics problem of a light-speed escape velocity, but the actual physics is more complicated than this. It is just a coincidence that using the "pseudo-derivation" still gives Schwarzschild radius – BioPhysicist Feb 11 '19 at 18:46
  • From an outside observer, it appears they never hit due to time dilation. But then how do we get GRB and other observations? – John Alexiou Feb 11 '19 at 19:55
  • Newtonian orbital mechanics says that. Special relativity avoids the situation by not handling gravitation at all. General relativity could handle it (and does, numerically), but it is very very complicated and there are no equations for merging black holes. – m4r35n357 Feb 11 '19 at 20:38

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The OP could be asking either how fast the singularities hit or how fast the event horizons hit.

Singularities

The singularities are spacelike, so it's not well defined to ask about their velocity. The velocity of A relative to B is only something we expect to be well defined if A and B are timelike or lightlike world lines (or surfaces with similar properties). A singularity is basically the end of the world for observers inside the horizon, so asking how fast one singularity is moving relative to another is like asking how fast the world is moving relative to someone else's end of the world. It doesn't make sense. The end of the world is a time, not an object.

Event horizons

GR doesn't have global frames of reference or a way to measure the velocity of an object relative to a distant observer. See How do frames of reference work in general relativity, and are they described by coordinate systems? . So there is not any meaningful sense in which we can talk about the velocity of an event horizon as seen by some distant observer, only by an observer near the horizon. So this means that if we were hoping to get an answer about the speed at which the event horizons collide as they merge, it would have to be the speed as measured by an observer right near the point of merging.

An event horizon is a lightlike surface, so in general all observers near a horizon say that a horizon is moving at the speed of light. So the observer probably sees both horizons as moving at $c$ as they collide. This doesn't mean that one horizon is moving at $2c$ relative to the other horizon, because we don't have frames of reference moving at $c$, i.e., there can't be an observer at rest relative to one of the horizons.

  • While you can't have an observer sitting on the event horizon you could have one in orbit about one of the holes orbiting perpendicular to the axis of the collision. I do agree that c + c doesn't give 2c but even at the event horizon we should see c--yet Einstein says we can't. – Loren Pechtel Feb 12 '19 at 05:22
  • @LorenPechtel: While you can't have an observer sitting on the event horizon you could have one in orbit about one of the holes To be in orbit, you have to be at a radius greater than or equal to 3 times the Schwarzschild radius $r$ (the innermost stable circular orbit, or ISCO). So you're at a significant distance from the horizon (significant compared to $r$, which sets the scale of curvature), and therefor you don't have any meaningful way of talking about the horizon's velocity relative to you. This is the idea of the linked Q&A. –  Feb 12 '19 at 19:59