# Angular Momentum of Kerr Black Hole

1. Feb 26, 2014

### Philosophaie

1. The problem statement, all variables and given/known data
How do you find the Angular Momentum, J, if you are given the Angular Velocity, ω of a Kerr Black Hole.

2. Relevant equations
J = I*ω
$I = mass*r^2$
Event Horizon:
$r_+ = M + (M^2 − (J/mass/c)^2)^{1/2}$
Static Limit and the Ergosphere:
$r_0 = M + (M^2 − (J/mass/c)^2*cos^2θ)^{1/2}$
where
$M=G*MassOfBlackHole/c^2$
c=Speed Of Light

3. The attempt at a solution
Unsure where the Moment of Inertial mass radius is.
The Black Hole is not visible to the naked eye.
Is it the mass and radius of the observer or is it the mass and radius of the Black Hole itself that the Event Horizon, Static Limit and all the orbiting Stars, planets,etc. are based on.

Last edited: Feb 26, 2014
2. Mar 2, 2014

### stevebd1

I know when calculating tangential velocity in Kerr metric, r is the reduced circumference (R)-

$$v_T=\omega R$$

where

$$R=\frac{\Sigma}{\rho}\sin\theta$$

and $\Sigma=\sqrt((r^2+a^2)^2-a^2\Delta \sin^2\theta)$, $\Delta= r^{2}+a^{2}-2Mr$, $\rho=\sqrt(r^2+a^2 \cos^2\theta)$ and $a=J/M$ so I'm guessing you would use the reduced circumference when considering inertia.

Though in order to calculate the reduced circumference you need $a$ which means you need to know J. If the only info you have is $\omega$ and M, then the equation for the frame dragging rate is-

$$\omega=\frac{2Mra}{\Sigma^2}$$

where $\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta$, replace $a$ with J/M and rearrange relative to J, which might be quadratic.

Source-
'Compact Objects in Astrophysics' by Max Camenzind page 379

Last edited: Mar 2, 2014