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Do the event horizon of Kerr black Hole have escape velocity is equal to the speed of light same Schwarzschild black hole?

Nuttz Multi
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Yes, the escape velocity at the event horizon of any black hole is the speed of light. That is one way to define where the event horizon is.

However, “escape velocity” is an overly Newtonian way to think about what is going on, which is: within the horizon, there are no lightlike or timelike geodesics that cross to the outside.

G. Smith
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  • Do you have a book or a paper about calculating the escape velocity of the kerr black hole? – Nuttz Multi Mar 12 '19 at 04:44
  • @NuttzMulti escape velocity is a concept from Newtonian physics that doesn't generalize very well to general relativity. – Zo the Relativist Mar 12 '19 at 05:14
  • See section 9 (especially p. 26) of https://arxiv.org/pdf/0706.0622.pdf for a calculation of the location of the Kerr event horizon. It does not involve a calculation of “escape velocity”. It considers photon orbits. – G. Smith Mar 12 '19 at 05:16
  • https://physics.stackexchange.com/questions/337906/escape-velocity-from-a-rotating-black-hole discusses Kerr escape velocity. – G. Smith Mar 12 '19 at 05:26
  • https://en.wikipedia.org/wiki/File:Orbit_around_a_rotating_Kerr_black_hole.gif claims to derive the Kerr escape velocity. I don’t know whether it is correct. – G. Smith Mar 12 '19 at 06:00