# 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

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

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 ($M \approx 10^{18}~\mathrm{kg}$) 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?

• 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
• journals.aps.org/prl/abstract/10.1103/PhysRevLett.33.558 – Count Iblis Jul 7 '16 at 20:58
• If you have a charged black body, would you call the radiation it emits "thermal"? If you heat it up to a high enough temperature, it will start emitting electrons in preference to positrons. – Peter Shor Sep 14 '16 at 21:46
• @lurscher please refer to this paper arxiv.org/abs/hep-th/0602146 – Spoilt Milk Oct 26 '16 at 17:37

As long as the black hole is large enough (and with $Q \ll M$) and the charge and mass you throw in are small, all the Hawking emission will be thermal. However, remember that Hawking emission will actually be negligible (the thermal spectra will be peaked in wavelengths of the order of the inverse temperature $T_H$ so it will almost only emmits photons and neutrinos). Thus, no discharge will be produced by Hawking emission.
There is still another mechanism that does discharge the black hole if its charge is large enough (you cannot reach extremality, where $Q = M$, this is the third law of black hole thermodynamics). This mechanism is the Schwinger pair production. The idea is that when the charge of the black hole is large enough, it makes it energetically possible to create an electron-positron pair from the vacuum OUTSIDE (but near) the horizon. The black hole would then eat the oposite charge and emit the same charge, so it would get discharged.
So, to answer your second question, even for a $10^{18}$kg BH the radius would be of 1 nanometer, still much bigger than the Compton wavelength of the electron, so it would just Hawking emit in the soft X-ray spectrum + neutrinos. The black hole would not get discharged through the Hawking emission.