What is the escape velocity in the Kerr metric?

Question

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?

Possible duplicate of Escape velocity from a rotating black hole — Red Act, Mar 15 '18 at 23:57

score -2 · Answer 1 · answered Mar 15 '18 at 23:35

-2

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.

answered Mar 15 '18 at 23:35

Gareth Meredith

600

You have misread the question. It doesn't mention black holes. – ProfRob Mar 16 '18 at 00:05
I am interested in escape velocities from points outside any event horizon. – ProfRob Mar 16 '18 at 00:12

What is the escape velocity in the Kerr metric?

1 Answers1