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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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  • It would seem? charged BH's emit charged particles 'The emission of charged particles by BH is physically equivalent to the spontaneous emission by an excited state' https://www.researchgate.net/publication/51989625_Black_Hole_Evaporation_in_a_Noncommutative_Charged_Vaidya_Model also https://physics.stackexchange.com/q/46870/ seems to answer this? – Mr Anderson Dec 30 '19 at 02:52
  • @MrAnderson there isn't a real answer there, but one says "Thus, no discharge will be produced by Hawking emission." So does the BH evaporate its charge (in any other way)? – Árpád Szendrei Dec 30 '19 at 03:18
  • Szendrei, I'm sure you know more than I do, but how BH charge is evaporated/shed seems to be discussed in detail also here https://physics.stackexchange.com/q/490524/ – Mr Anderson Dec 30 '19 at 03:28
  • @MrAnderson yes thank you but those aren't really answering it either. "For an observer far from a black hole the Hawking radiation appears thermal and random." And the other says the BH would lose mass faster then charge. I am asking whether it really loses charge if it just emits photons. – Árpád Szendrei Dec 30 '19 at 03:36
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    Yes, if a black hole has positive charge it will produce more positively charged particles than negatively charged ones, causing its net charge to decrease. – knzhou Dec 30 '19 at 08:01
  • @knzhou thank you can you please tell me are you saying that if a BH has a certain net charge, then the particles emitted will have a higher probability of being that charge, so the net charge of the BH will decrease? – Árpád Szendrei Dec 30 '19 at 16:21
  • Yes, that’s right. – knzhou Dec 30 '19 at 19:30

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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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  • Thank you so much, so basically the probability for (the charge of) the emitted particles equals, so the net charge of the BH remains. – Árpád Szendrei Dec 30 '19 at 16:20
  • Yes, for the rare case where a pair is produced there is no preference which charge leaves. It will depend on the drection of the pair and the probability equal for the two charges. – anna v Dec 30 '19 at 18:05