# Surface area of an event horizon and irreducible mass

1. May 15, 2010

### NanakiXIII

1. The problem statement, all variables and given/known data

Show that a black hole's event horizon will never shrink due to a Penrose process.

2. Relevant equations

$$M_{ir}^2 =\frac{1}{2}(M^2 + \sqrt{M^4-J^2})$$

3. The attempt at a solution

It is easy to show that the irreducible mass will never decrease, and I know of a result that states that the surface area of the event horizon is proportional to the square of the irreducible mass,

$$A = 16 \pi M_{ir}^2.$$

This proves what is asked, but I am stuck on deriving the relation between the surface area and the irreducible mass.

2. May 16, 2010

### NanakiXIII

Since the metric components for $\phi$ and $\theta$ are the same for Schwarzchild as for Euclidean, shouldn't the surface of the horizon still just be $4 \pi r^2$, with the radius at $r=2M$? I'm not sure where the irreducible mass comes into play.

3. May 16, 2010

### stevebd1

I'm not a homework helper so hopefully I'm not stepping out of line but the correct equation for area for a black hole is-

$$A=4 \pi \left(r_+^2+a^2)$$

where $r_+=M+\sqrt{(M^2-a^2-Q^2)}$ where $Q$ is charge (which doesn't seem relevant in this case) and $a$ is the spin parameter in metres.

Last edited: May 16, 2010
4. May 16, 2010

### NanakiXIII

I don't think you need to be a recognized homework helper to help out. I don't mind, in any case.

I'm not doing anything with charge, and I'm not sure what a spin parameter is, so I'm assuming I don't need to use that either. Then your equation reduces to what I quoted. But where does this equation come from?

5. May 16, 2010

### diazona

Of course you don't, otherwise how would anyone become a recognized homework helper in the first place?

6. May 16, 2010

### stevebd1

In your OP, you have J which represents spin or angular momentum. In geometric units J=a·M where a=j/mc (where j is angular momentum and m is mass in SI units) and M=Gm/c2. a is sometimes described as angular momentum per unit mass, a/M also produces a unitless number between 0 and 1 and provides some idea of how high the spin is, 1 being the maximum (or maximal). You should also find that $A = 16 \pi M_{ir}^2$ and $A=4 \pi \left(r_+^2+a^2)$ are equivalent.

Last edited: May 16, 2010
7. May 16, 2010

### NanakiXIII

Ah, yes, I see the equivalence now. However, I still do not understand why this expression gives the surface area of the event horizon. Apparently, the event horizon is a sphere with radius $2M_{ir}$, but why? I thought it was a sphere with radius $2M$.

8. May 16, 2010

### stevebd1

For a static black hole, the reduced circumference is equal to the coordinate radius (2M), even though the coordinate radius for a rotating black hole reduces (see the equation for r+), the reduced circumference takes on an arc like property due to frame dragging and is therefore greater than the coordinate radius. $\sqrt(r_+^2 +a^2)$ is not an exact solution for the reduced circumference but it represents it when calculating the area.

9. May 16, 2010

### NanakiXIII

Could you show me more quantitatively how this works? I'm guessing this all has something to do with the metric. I'm a little puzzled, however, since I've not actually learned what the metric for a rotating black hole is. (Kerr metric?) I also don't know what a reduced circumference is.

10. May 16, 2010

### NanakiXIII

I just had a look and if you just assume that the event horizon happens where the $rr$ component of the Kerr metric blows up, you find that the radius is $r = M + \sqrt{M^2-a^2}$. This is what Wikipedia (http://en.wikipedia.org/wiki/Kerr_metric#Important_surfaces) does. First of all, it seems ad hoc to me, I don't know why this might be a good radius to choose. Second, the radius is apparently $\sqrt{r^2 + a^2} = \sqrt{2Mr}$ and not just $r$.

11. May 16, 2010

### stevebd1

This is the same as the equation in post #3, $r_+=M+\sqrt{(M^2-a^2-Q^2)}$, just minus the charge and represents the coordinate radius for the event horizon. You could say that the coordinate distance is a radial distance where as the reduced circumference is the path that light would take, the coordinate distance being a geometric measurement, the reduced circumference being a geodesic. For a static black hole, these are equal because there is no frame dragging, but for a rotating black hole, light that was originally on a radial path, spirals into the BH and the path it takes is longer than the geometric radius from where it originated. Its this distance (reduced circumference) that is referred to when calculating surface area of the EH, centripetal acceleration of objects rotating the BH, etc. as it is (for want of a better expression) a truer distance.

The reduced circumference is derived from Kerr metric. An exact solution (relative to the azimuth) is-

$$\sqrt{g_{\phi\phi}}\ =\ \varpi=\frac{\Sigma}{\rho}sin\theta$$

where

$$\] \Sigma^2=(r^2+a^2)^2-a^2\Delta sin^2\theta\\ \\ \Delta= r^{2}+a^{2}-2Mr\\ \\ \rho^2=r^2+a^2 cos^2\theta \[$$

symbols you'll recognise if you look at Kerr metric (which I recommend). You'll find at the equator, regardless of spin, the reduced circumference for the outer event horizon (r+) equals 2M. Why simply $(r_+^2 +a^2)$ is used to represent this in the surface area calculation isn't crystal clear but according to a number of sources, it appears to be the norm and is equivalent to equation $A = 16 \pi M_{ir}^2$. Regarding your OP, you might be worth crunching some numbers to see how M compares to Mir for a black hole with spin.

Last edited: May 16, 2010
12. May 16, 2010

### NanakiXIII

I'm afraid this is altogether making very little sense to me. The reduced circumference, it's a radial measure, I understand from your description, then why is it called a circumference? And why is it given by $g_{\phi\phi}$?

13. May 16, 2010

### stevebd1

$g_{\phi\phi}$ is part of the Kerr metric language. Reduced circumference is derived (reduced) from measuring the circumference of a stationary ring in the equatorial plane concentric to the black hole and dividing it by $2 \pi$. This gets a little more complex for a rotating black hole. I highly recommend 'Exploring Black Holes' by Taylor and Wheeler as it covers both static and rotating black holes.

Another equation for surface area for a rotating black hole is-

$$A=4\pi\left(2M^2+2M\sqrt{M^2-a^2}\right)$$

which is also equivalent to the other two equations mentioned above.

source-
http://edoc.ub.uni-muenchen.de/6024/1/Deeg_Dorothea.pdf" [Broken] page 12

Last edited by a moderator: May 4, 2017
14. May 16, 2010

### NanakiXIII

Ah, now I have an idea of what this reduced circumference means. I suppose it might make sense to use it. It seems odd, however, since the reduced circumference depends on $\theta$, that we're still talking about a sphere. Shouldn't we be integrating instead of just saying $A = 4 \pi r^2$?

15. May 16, 2010

### NanakiXIII

On something of a side note, I also tried calculating the surface area via

$$A = \int_0^{2\pi} \int_0^{\pi} \sqrt{g_{\theta\theta} g_{\phi\phi}} d\theta d\phi$$.

I threw this into Mathematica and what came out was very ugly, but there is one interesting bit. The result basically (Mathematica also requires a set of elaborate conditions which I don't understand) looks like

$$2\pi(r^2 + a^2) + G(r,a,M)$$

where $G[/tex] stands for Garbage and consists of very large and imaginary (i.e. non-real) expressions. I won't pretend to understand what this all means, but it seems interesting that if you throw away the garbage, you get the right expression except for a factor 2. 16. May 17, 2010 ### stevebd1 $$A = \int_S dS = \int_0^{2\pi} \int_0^{\pi} \sqrt{g_{\theta\theta} g_{\phi\phi}}\ d\theta d\phi$$ where [itex]g_{\theta\theta}=\rho^2$ and $g_{\phi\phi}$ is as post #11

$$A= 2 \pi \int_0^{\pi} \sin\theta\sqrt{\left(r^2+a^2\right)^2-a^2\Delta\sin^2\theta}\ d\theta$$

which produces results identical to $A=4 \pi \left(r_+^2+a^2)$ when $r=r_+$.

Source-
http://liu.diva-portal.org/smash/record.jsf?pid=diva2:19868" page 20

Last edited by a moderator: Apr 25, 2017
17. May 17, 2010

### NanakiXIII

I finally got it! The integral you posted, stevebd1, is what I was trying to solve. When you try to do that, it comes out very ugly. However, if you first substitute $r = r_+$, the integrand becomes trivial (this actually amazed me, I hadn't foreseen this at all) and you find a neat expression which, after a lot of fiddling, finally comes out to be

$$A = 16 \pi M_{ir}^2$$!

Thank you for your support in getting to grips with this. Though in the end it came down mostly to some quirky mathematics, I appreciate our discussion as I have become more familiar with several concepts and gained a few insights.