Hey I'm a noob to physics however after reading about black holes and Hawking radiation. If a black hole made entirely out of negatively charged particles (say this black hole is made entirely of electrons) Then when it evaporates equal parts positive and negative particles (positrons and electrons for this example) are created. Doesn't this violate the law of conservation of charge?

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    $\begingroup$ Why do you think equal positive and negative charges would be produced? $\endgroup$ – Dale Jun 5 '19 at 17:02
  • $\begingroup$ When charged virtual particles are created at the event horizon would it not be equally likely that the position falls in as that the electron falls in, meaning that on average 50% of the time positrons are made and 50% of the time electrons are made $\endgroup$ – tms Jun 5 '19 at 17:22
  • $\begingroup$ I am not sure, but I suspect not. I suspect that the negatively charged antiparticle would be electrostatically repelled from the black hole and the positive antiparticle would be electrostatically attracted back into the black hole. $\endgroup$ – Dale Jun 5 '19 at 17:47
  • $\begingroup$ similar question here physics.stackexchange.com/q/168891 $\endgroup$ – anna v Jun 5 '19 at 17:58
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    $\begingroup$ It's best to think of the electromagnetic field of a charged black hole as existing outside the event horizon. See math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… Also, Hawking radiation is almost entirely photons, until the BH is tiny. It takes a lot of energy to create an electron or positron, and the Hawking radiation of a stellar mass black hole is rather feeble. $\endgroup$ – PM 2Ring Jun 6 '19 at 5:03

There is an upper bound for the charge of a BH compared to its mass. It is given by $GM^2=Q^2$ in natural units. Where $G$ is newton's constant, $M$ the mass of the black hole and $Q$ the charge. If the charge exceeds this, it creates a naked singularity, which is forbidden by nature. This all can be found in Sean Carroll's notes on GR on pages 202-206.


Given this I think your thought experiment won't work since a BH made of only electrons will violate this but you can do the math yourself.

  • $\begingroup$ Not per se. Compressing a cloud of electrons into a small enough volume to form a black hole requires a lot energy. It turns out this energy is always sufficient to ensure the resulting black hole is sub-extremal. $\endgroup$ – mmeent Mar 2 '20 at 10:38

Hawking radiation comes from the horizon, not from the black hole itself, and the probability of + and - charge generated is equal as far as the model goes.

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).

As far as our observations and measurements go, conservation of charge is a strict law.

  • $\begingroup$ since the probability of + and - particles appearing is equal and the mass for those particles comes from the black hole, as it evaporates the mass inside the black hole decreases and equal +/- particles are made. ex. If a 1kg black hole made only out of electrons evaporates to 50% its original mass then 25% its mass-energy would be stored in positrons outside the black hole, 25% in electrons outside the black hole and the other 50% still inside the black hole, thus violating the law of conservation of charge due to the fact that the net charge of the entire system has changed $\endgroup$ – tms Jun 5 '19 at 17:33
  • $\begingroup$ The total charge entering below the horion from hawking radiation would be zero, because an equal number of positrons and electrons would be eaten up by the black hole, their pair leaving the region. Anyway net charge is counted in closed systems. A particle eating horizon is not closed to the rest of the universe. pairs are created, a plus is eaten up, a minus leaves, and vice verso. The black hole loses energy equivalent to the mass depletion, not charge. see physics.stackexchange.com/questions/451618/… $\endgroup$ – anna v Jun 5 '19 at 17:53
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    $\begingroup$ Assuming creation of an electron-positron pair near the charged black hole, it is much more likely that the positron would be attracted into the negatively charged black hole, thus reducing its charge, and the electron would escape. $\endgroup$ – amateurAstro Jun 5 '19 at 17:54
  • $\begingroup$ Not really, the horizon is kilometers away from the singularity where the black hole charge resides, and electric repulsion attraction goes as 1/r. The pair production effect depends on energy and virtual diagrams. $\endgroup$ – anna v Jun 5 '19 at 17:58
  • $\begingroup$ @annav Nothing "resides" in a singularity. A singularity is not a "place" and not an "object". The attraction does not go as $1/r$, because $r$ inside the horizon is time. While you are inside, there is no horizon, because it is in the past and no longer exists, and there is no singularity, because it is in the future and does not "exist" yet. So a singularity cannot attract anything, whether gravitationally or electromagnetically, because, again, there is no singularity in any spatial direction around you, so there is nowhere to be attracted to. $\endgroup$ – safesphere Jun 6 '19 at 7:19

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