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There are basically two types of charged black holes, Reissner–Nordström, and Kerr–Newman.

Next, Hawking radiation itself is mostly electromagnetic. The numerous popular articles which say that Hawking radiation is particle/anti-particle pairs in which one of each pair gets swallowed by the black hole are misleading. Hawking radiation is almost entirely electromagnetic radiation---that is, photons. You can still use the language of particle/anti-particle, but it is a bit misleading.

If Hawking radiation is true why isn't the entire universe glowing?

This answer claims that Hawking radiation is just photons. Photons are EM neutral.

Now basically there are different interpretations of Hawking radiation, but what they mainly say is that the evaporation itself does not come from the black hole itself (nothing can escape the black hole), but from near outside the horizon.

Hawking radiation comes from the horizon, not from the black hole itself If you are worrying if your hypothetical black hole completely evaporates, i.e. no more horizon, it means that the electrons will just disperse due to the electrostatic repulsion, (once the body stops being a black hole).

Violation of the Law of Conservation of Charge in black hole?

Are electrons just incompletely evaporated black holes?

Now these answers claim that BHs evaporate in a way that their leftovers are basically charged particles, that is, the charge that the BH did not evaporate stays during evaporation, until the BH's mass is not enough to create a event horizon anymore, and the BH ceases to exist anymore, and the leftover electrons just disperse.

Based on these it might be possible that the BH evaporates just photons, EM neutral particles, thus no EM charge is really evaporated from the BH, and the BH keeps its entire charge during the whole process of evaporation. At that point, as the BH's mass reduces so that it cannot create a event horizon anymore, the leftover electrons just disperse.

Question:

  1. Does a BH evaporate its EM charge?
Árpád Szendrei
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The main point is that Hawking radiation depletes the energy of the black hole so after a very long time it no longer forms an Horizon. In the particle-antiparticle illustration of Hawing radiation, it is obvious that an off mass shell photon or graviton is involved, the energy given by the energy of the black hole . Since the particle that escapes is equally probable to be a + or a - charge, there is no change in the charge carried by the black hole.

When the energy is taken away by a photon, it is obvious that the charge of the black hole remains the same. The remnant after it can no longer form a horizon will be a charged body in space, losing electrons due to the repulsion between them, as far as I can tell.If it is a positive excess charge it will be harder, due to the mass of the protons.

One should not forget that Hawking radiation is a mixture of classical General relativity, and quantization, so a definite answer of the diagrams involved will come once gravity is rigorously quantized.

Edit for completeness:

The illustrations of Hawking radiation being pair production at the horizon where one of the pair leaves and the other is eaten by the black hole, is good for an intuitive image, the energy provided by a virtual graviton from the black hole. But

Hawking radiation is black-body radiation that is predicted to be released by black holes, due to quantum effects near the black hole event horizon. It is named after the theoretical physicist Stephen Hawking

The article covers the various theories (and paradoxes) with Hawking radiation. Black body radiation in this case is photons from the horizon coming from interactions with the gravitational field of the black hole, taking away energy , and thus diminishing the mass of the black hole.

anna v
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