# Kerr-Newman black holes and infinite charge

Recall the first law of BH thermodynamics

$dM=\frac{\kappa}{8\pi} dA + \Omega dJ + \Phi dQ$

Now, let's consider the Reissner-Nordstrom solution $J=0$ such that $m>Q$ but only slightly greater. Suppose I have a small bit of charge $dQ$ which I bring in from infinity to the BH horizion.

1) Question: Is the extremal BH solution $m=Q$ possible?

I would think that if we consider the work to be done to bring $dQ$ from $\infty$ to BH horizon, this would blow up $\Phi$ and cause M to go to $\infty$ as well. Therefore, I would guess the extremal solution cannot be achieved in this scenario.

2) Now, this extremal solution DOES seem possible in the Kerr solution, but how would this affect $J$ and $\Omega$ ?

-
–  Qmechanic Apr 29 '12 at 17:03
Sorry, if you want to bring some charge, you're talking about Kerr-Nerman solutions, not just Kerr solutions, right? –  Luboš Motl Apr 29 '12 at 18:18
To be clear, yes, I meant Kerr-Newman. My mistake... –  eherrtelle59 Apr 30 '12 at 16:26
I changed the post title accordingly. Any help Lubos? –  eherrtelle59 Apr 30 '12 at 18:09

## 1 Answer

This is the "third law" of Black hole thermodynamics. I put third law in quotes, because it isn't the statement that the entropy vanishes at zero temperature, but the original statement that the specific heats become tiny at zero temperature, so that the zero temperature state is unacheivable.

The extremal limit is perfectly cold, the surface gravity diverges. This limit requires extremely adiabatic insertion of the extra charge, otherwise you put in more energy in the process than charge. The way to approach the extremal limit is to throw in charged particles, and collect neutral Hawking radiation, and the end state is a nearly zero temperature, nearly extremal state.

Penrose analyzed many such situations in the late 1960s early 1970s trying to violate cosmic censorship (Q>M is a naked singularity). His conclusion was that the third law is real. This is born out by modern formulations, where to make a perfect extremal brane you need to cool down the near-horizon field theory to exactly zero temperature, which is a usual third-law difficulty.

-
And how would this effect the Kerr-Newman solution? We have to take into account rotation as well...Perhaps I'm missing something –  eherrtelle59 Apr 30 '12 at 17:21
@eherrtelle59: rotation only marginally changes the picture--$\kappa$ depends on $M$, $Q$ and $J$ in such a way that there is always a well-defined zero temperature point, and this point is unacheivable. I have yet to see as satisfying a classical proof of the third law as I have of the zeroth through second laws of black hole dynamics, but there is no known way to add charge and angular momentum to a black hole in such a way to make $\kappa=0$, except for some very precisely fine-tuned (i.e., a set of measure zero in the parameter space) sets of parameters. –  Jerry Schirmer Apr 30 '12 at 18:27