# “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"?

• The horizon isn’t an actual object that is moving. But an actual object near the horizon would have to move in the opposite direction to stay in place, like a swimmer swimming against a current. – G. Smith Aug 4 at 19:04
• Experts don't describe the spin of a black hole in terms of a percentage of the speed of light. Not sure where the linked news stories came up with that wording. The news stories don't cite their source, but they might be referring to the paper Constraining Quasar Relativistic Reflection Regions and Spins with Microlensing. That paper uses the usual spin parameter that you mentioned (the ratio of angular momentum to the square of the mass, in natural units), which has a maximal value of $1$. It doesn't use the phrase "spinning at X% of the speed of light." – Chiral Anomaly Aug 4 at 20:09
• Spinning black holes do NOT produce a "current" that drags stuff along like tornados do. Objects falling in are given a boost in the direction of the spin, BUT objects that are launched away are given a boost against the spin direction. Objects held stationary will not be pulled tangentially. – Kevin Kostlan Aug 5 at 6:07

## 1 Answer

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.