1
$\begingroup$

This question already has an answer here:

In the non-rotating Schwarzschild metric there is a straightforward expression for the escape velocity in the radial direction, defined from the point of view of a stationary observer at that radius, which is equivalent to the Newtonian result.

My questions are: is there an equivalent expression in the Kerr metric, how does it depend on the spin parameter, in which reference frame would that be defined and how does it depend on escape direction with respect to the spin axis?

$\endgroup$

marked as duplicate by Red Act, Jon Custer, Qmechanic Mar 16 '18 at 5:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

-2
$\begingroup$

For all black holes, Kerr or not, the escape velocity is c. This is because gravity has been allowed to become strong enough, that an event horizon is formed. We actually have analogue black holes in nature, known as sonic black holes in which the velocity is not specially-associated to the speed of light c. I know that the angular momentum of a black hole may vary with size... there seems to be some evidence for that, but I think I just haven't understood your question properly.

$\endgroup$
  • $\begingroup$ You have misread the question. It doesn't mention black holes. $\endgroup$ – Rob Jeffries Mar 16 '18 at 0:05
  • $\begingroup$ I am interested in escape velocities from points outside any event horizon. $\endgroup$ – Rob Jeffries Mar 16 '18 at 0:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.