"Black hole spins at $X$% of the speed of light", what does that mean? I've seen a few news stories recently (example, example) about some black holes spinning at X% of the speed of light. What does that mean? What exactly is moving at that speed, and with respect to what?
The answers I could find on physics stack exchange and wikipedia talk instead about the spin parameter, the angular momentum as a percent of the maximum (GM^2/c), above which the Kerr solution would have a naked singularity.
So what's the physical meaning of statements like "black hole spins at X% of the speed of light" or "black hole spins Y times a second"?
 A: The velocity such articles talk about is the angular displacement (${\rm{d}} \phi$) times the local circumference divided by $2 \pi$ (the so called radius of gyration $ \bar{\omega} = \sqrt{ | g_{ \phi \phi} | } $ ) divided by the elapsed coordinate time of an observer at infinity (${\rm{d}}t$), so
$$v = \sqrt{ | g_{ \phi \phi} | }  \cdot \frac{{\rm{d}} \phi}{{\rm{d}} t} = \bar{\omega} \cdot \omega$$
If you set the spin parameter $a=1$, $\theta=\pi/2$ and $r=r_{+}=1+\sqrt{1-a^2}$, you get
$$ g_{ \phi \phi} = \frac{a^2 \left(a^2+(r-2) r\right) \sin ^4 \theta -\left(a^2+r^2\right)^2 \sin ^2 \theta }{a^2 \cos ^2 \theta +r^2} = -4$$
and for ${\rm{d}}\phi/{\rm{d}}t$ we choose the frame dragging angular velocity
$$ \omega = | \frac{g_{t \phi}}{g_{\phi \phi}} | =\frac{2 a r}{\left(a^2+r^2\right)^2-a^2 \sin ^2 \theta  \left(a^2+r^2-2 r\right)} = \frac{1}{2}$$
because due to the gravitational time dilation in the frame of the far away observer everything at the horizon is frozen in place and corotating with the dragged space, so the result is
$$ v = \bar{\omega} \cdot \omega = \sqrt{|-4|} \cdot \frac{1}{2} = 1$$
and since we used natural units of $\rm G=M=c=1$, the velocity with maximum spin would be $\rm c$.
If you talk about the local velocity a testparticle would need to travel against the direction of the black hole's rotation in order to stay stationary with respect to the fixed stars you already get $\hat{v}= \rm c$ at the outer ergosphere, but that $\hat{v}$ is the travelled local distance with respect to a ZAMO divided by the testparticle's proper time $\tau$, which would be infinite velocity at the horizon.
This is why inside the ergosphere even light which locally travels retrogradely still movel progradely with respect to the distant coordinate bookkepper and the fixed stars.
In this video the velocity of the horizon is calculated by a different approach (unfortunately it is in german, but the math on the blackboard is international anyway).
The coordinates used in this posting and the linked articles and video are Boyer Lindquist coordinates.
