# Is the Hawking radiation of a charged black hole thermal?

Suppose you have a Schwarzschild black hole of mass $M$ and angular parameter $a = 0$ (no rotation).

Question: is it possible to throw a charge $Q$ at a faster rate than it will be re-radiated? Will the radiation profile be still thermal?

If it is thermal, it would mean that even big, cold black holes would emit a lot of energy in the form of electromagnetic radiation just to be able to get rid of the extra charge? A thermal spectra that starts at $511\: \mathrm{KeV}$ (the energy of the lowest charged particles and has very little emitted power at lower energies) would be a very weird thing to call 'thermal'.

There is a differential expression for the increase in temperature as a small charge (compared to the black hole mass) is added to the black hole that one obtains by taking the formula 11.2.17 in this page (Modern Relativity, 2005, David Waite) and deriving against $e$ and $M$ and taking

$$\delta T = ( \frac{\partial T}{\partial e} + \frac{c^2}{G} \frac{\partial T}{\partial M} ) \delta e$$

so this gives a profile for the variation of black hole temperature with charge.

Question: Is it correct to conclude that you can estimate the overall Hawking radiation of a relatively small black hole ($M \approx 10^{18} Kg$) by adding electric charge at a faster rate than it will be re-emitted by the thermal spectra?

Or is the spectra totally non-thermal, and the radiation will favour throwing away charged particles, while being electromagnetically cold?

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You probably have to solve the charged Dirac field in that metric. You should find more elections being emitted then positrons. – Prathyush Dec 14 '12 at 21:39
The real question is: how does Hawking radiation thwart our attempts at making a super-extramal hole by charging it up and allowing it to radiate, relying on the 511keV cutoff to prevent it losing charge as it loses mass? – Kevin Kostlan Jan 30 '14 at 14:18