Xavius said:
My question is this, could a black hole ever rotate with such high speed that it flattens out to the point that some of the material of the black hole travels outside the event horizon? I assume not but don't know how to do the math to check.
From what I could gather, as gravity increases relative to the metric the closer you get to the black hole, the centripetal acceleration decreases relative to the metric. For instance, the Newtonian equation for overall gravitational acceleration taking into account centripetal acceleration is-
a_t=a_g-a_c
where a
t is total gravitational acceleration, a
g is gravitational acceleration and a
c is the centripetal acceleration of the orbiting object. This can be rewritten as-
a_t=\frac{Gm}{r^2}-\frac{v_t^2}{r}
where G is the gravitational constant, m is the mass of the central object, r is the radius between the two objects and v
t is the tangential velocity of the orbiting object.
The relativistic version of this (i.e. for a static black hole) is-
a_t=\frac{Gm}{r^2}\frac{1}{\sqrt{1-2M/r}}-\frac{v_t^2}{r}\sqrt{1-2M/r}
Where M=Gm/c
2 This ties with the statement that once inside the event horizon, all world lines lead to the singularity. Centripetal acceleration becomes zero at the EH and negative once inside the EH. (Note: If you set a
t as zero, v
t as c and solve for r, you will get the photon sphere radius).
This gets a little more complex for rotating black holes as frame dragging is taken into account but the principle still applies. It also explains why objects that approach the ergosphere (which is spinning at c relative to infinity) are still pulled into the BH. The whole idea of centripetal acceleration reducing in extreme gravity is demonstrated to some extent in the following paper-
Geometric transport along circular orbits in stationary axisymmetric spacetimes
http://arxiv.org/abs/gr-qc/0407004
