Angular Momentum of Kerr Black Hole

[Philosophaie]

Homework Statement

How do you find the Angular Momentum, J, if you are given the Angular Velocity, ω of a Kerr Black Hole.

Homework 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

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.

 

[stevebd1]

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

[tex]v_T=\omega R[/tex]

where

[tex]R=\frac{\Sigma}{\rho}\sin\theta[/tex]

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

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

[tex]\omega=\frac{2Mra}{\Sigma^2}[/tex]

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

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