# 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 $\Omega$ ?

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– 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

@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