# Proof that black hole's irreducible mass never decreases

1. May 17, 2010

### NanakiXIII

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

In a Penrose process, a mass $\delta M < 0$ and an angular momentum $\delta J < 0$ are added to a black hole. Given that

$$\delta J \leq \frac{\delta M}{\Omega}$$

where $\Omega$ is the angular velocity of the horizon, show that the irreducible mass never decreases, i.e. $\delta M_{irr} > 0$.

2. Relevant equations

$$M_{irr}^2 = \frac{1}{2} \left[ M^2 + \sqrt{M^4 - J^2} \right]$$

3. The attempt at a solution

The function for the irreducible mass is weakly decreasing for decreasing $M$ and strongly increasing for decreasing $J$, so I imagine that if $J$ decreases enough compared to the decrease in $M$, you indeed get that the irreducible mass can only rise. That is, I can imagine that the given inequality might lead to the requested conclusion. I have not been able to prove that this particular inequality works. For one, I'm not sure how to relate the angular velocity to anything else, which I think might be crucial. Perhaps I should somehow relate it to the angular momentum, but how?

EDIT: I've found an expression saying

$$\Omega = -\frac{g_{t\phi}}{g_{\phi\phi}}.$$

I don't understand this, can anyone explain why this is true?

Last edited: May 17, 2010
2. May 18, 2010

### stevebd1

I can tell you this is the frame dragging rate as viewed from infinity which increases as r reduces.

$$g_{\phi\phi}}\ =\frac{\Sigma^2}{\rho^2}\sin^2\theta\ =\frac{(r^2+a^2)^2-a^2\Delta \sin^2\theta}{\rho^2}\ \sin^2\theta$$

$$g_{t\phi}=-\frac{2Mra\ \sin^2\theta}{\rho^2}$$

where $\Delta= r^{2}+a^{2}-2Mr$ and $\rho^2=r^2+a^2 \cos^2\theta$

This can reduce to-

$$-\frac{g_{t\phi}}{g_{\phi\phi}}\ =\ \omega\ =\ \frac{2Mra}{\Sigma^2}$$

Multiply by c for SI units (rad/s). $\Omega$ is normally used to represent the frame dragging (as observed from infinity) at the event horizon, normally denoted with a + or - to represent the outer or inner horizon, $\omega$ is normally used to represent the frame dragging rate else where.

Last edited: May 18, 2010
3. May 18, 2010

### NanakiXIII

Yes, I had discovered this is how it is regarded, but I do not understand why that expression is valid. Apparently the Kerr metric can also be regarded as a co-rotating frame that is rotating at an angular velocity of

$$\Omega = -\frac{g_{t \phi}}{g_{\phi \phi}}$$

but I do not see why this is true.

4. May 18, 2010

### stevebd1

It has something to do with setting the angular momentum equation for a particle in Kerr space time to zero and finding out that there is still angular motion. Below is a very reduced angular momentum equation for a particle in space time around a BH with a/M=1 (i.e. maximal), it is also only relative to the equatorial plane-

$$\frac{L}{m}=R^2\frac{d \phi}{d \tau}-\frac{2M^2}{r}\frac{dt}{d \tau}$$

where

$$R^2=r^2+M^2+\frac{2M^3}{r}$$

where R is the reduced circumference for a maximal Kerr BH at the equator.

If we set L/m to zero, we get-

$$\frac{d \phi}{dt}=\frac{2M^2}{rR^2}$$

which gives results equivalent to $\omega$ in post #2 when applied to a Kerr BH with a/M=1 at the equator.

source-
http://www.eftaylor.com/pub/SpinNEW.pdf" [Broken] page F-14

Last edited by a moderator: May 4, 2017
5. May 18, 2010

### NanakiXIII

That seems fair enough. I'm wondering, however, whether the expression with the quotient of metric components might not be something more general, i.e. valid for any metric, not just the Kerr metric.

6. May 18, 2010

### stevebd1

They are general, the equations reduce to the Schwarzschild solution when a=0, though I would say $g_{t\phi}$ is fairly specific to Kerr and Kerr-Newman metric.

Last edited: May 18, 2010
7. May 18, 2010

### NanakiXIII

Well, $g_{t\phi}$ is just 0 for a static black hole, so that would lead to an angular velocity of 0, which makes sense. But that's not proof that it's a general relation.

8. May 19, 2010

### stevebd1

I don't know if this answers your question and you may already be aware of this but the metric tensor is a 4 x 4 matrix with $t, r, \theta, \phi$ running left to right, top to bottom, hence $g_{tt}, g_{rr}$, etc. correspond to points in the matrix. Though the annotations might change, this is pretty much applied to most metrics. As there is a component in Kerr metric that relates to time and change in longitude, a $g_{t \phi}$ component is introduced which has two points in the matrix, hence Kerr metric can be expressed-

$$ds^2 = g_{tt} dt^2 + g_{rr}dr^2 + g_{\phi \phi} d\phi^2 + g_{\theta\theta} d\theta^2 + 2 g_{t \phi} dt d\phi$$

where $g_{t \phi}$ is counted twice (see link below).

Source-
http://www.astro.ku.dk/~milvang/RelViz/000_node12.html" [Broken]

Last edited by a moderator: May 4, 2017
9. May 19, 2010

### NanakiXIII

Yes, I'm aware of what a metric tensor is (not really a matrix, actually ;)), that's not what confused me. I simply did not understand where the expression for $\Omega$ comes from and was wondering whether it held for any metric. I've tracked the expression down in literature and I know now to some extent where it comes from, though I'm waiting for my teacher to provide some more details. If you care to know, I can post what I know here.