# Area of event horizon and irreversible mass of Kerr black hole

1. Dec 31, 2012

### ck99

Hi everyone, and happy new year if you happen to be reading this tomorrow. Rather than partying, I am writing up 100+ pages of astrophysics lecture notes, which I think will take infinite time as I keep getting stuck on every other line.

My current problem is with the equation for the surface area of a Kerr black hole, I'm not sure if I am failing to understand this, or have made some transcription errors in my notes. (Possibly both; my lecturer writes really fast and this material is a real stretch for me!)

We have defined the radius of the event horizon as

r+ = rg + √(rg2 - a2)

where rg is GM/c2

Then we say that surface area of event horizon A is given by

A = 4∏(r+2 + a2) (no explanation given of where this equation comes from)

A = 8∏rg(rg + √(rg2 + a2)

I can't follow the algebra in that step, is it correct?

Then we expand that, just replacing rg with GM/c2 to get

A = 8∏(GM/c2)[(GM/c2) + √{(GM/c2)2 - (J/Mc)2}

Now we write area as

A = 4∏(2GM1/c2)

where M1 is defined as the "irreversible mass"

M1 = M√{(1/2)(1+√{1-(a/rg)2}}

I can't make the algebra work to equate this expression for area in terms of irreversible mass into our original expression for area, or to get the intermediate steps to work properly. I have tried looking online for some more help, but most authors wite about "irreducible mass" and I'm not sure if that's the same thing. Unfortunately I'm not clever enough to compare alternative expressions for these equations (that I have found in textbooks) to my lecture notes, and say definitively if I have written it down correctly.

Any help on this would be much appreciated!

2. Dec 31, 2012

### Staff: Mentor

To get rid of many constants, I'll use rg=1 and divide all other length units by rg and area units by rg^2.

Simplified equations:
(1) $r_+=1+\sqrt{1-a^2}$

(2) $A = 4\pi(r_+^2 + a^2)$ (probably the result of an integration over the surface)

This should be equal to
(3) $A = 8\pi(1 + \sqrt{1 + a^2)}$

Use (1) in (2):
(3b) $A = 4\pi((1+\sqrt{1-a^2})^2 + a^2) = 4\pi (1+2\sqrt{1-a^2}+(1-a^2)+a^2) = 8\pi (1+\sqrt{1-a^2})$

Looks like a typo in (3).

Expansion is not necessary here, use units with G=c=1 and keep the assumption M=1.

(4) $A=4\pi(2M_1) = 8 \pi M_1$ - this cannot be right, A goes with M^2 and not M. I think there is a square missing at your bracket:
(4b) $A=4\pi(2M_1)^2 = 16 \pi M_1^2$
(5) $M_1=\sqrt{\frac{1}{2}(1+\sqrt{1-a^2})}$

Using (5) in (4) directly gives (3b).