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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?

Qmechanic
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tms
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    Why do you think equal positive and negative charges would be produced? – Dale Jun 05 '19 at 17:02
  • 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 – tms Jun 05 '19 at 17:22
  • 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. – Dale Jun 05 '19 at 17:47
  • similar question here https://physics.stackexchange.com/q/168891/ – anna v Jun 05 '19 at 17:58
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    It's best to think of the electromagnetic field of a charged black hole as existing outside the event horizon. See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html 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. – PM 2Ring Jun 06 '19 at 05:03

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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.

https://arxiv.org/pdf/gr-qc/9712019

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.

Aylon Pinto
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  • 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. – TimRias Mar 02 '20 at 10:38
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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.

anna v
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  • 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 – tms Jun 05 '19 at 17:33
  • 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 https://physics.stackexchange.com/questions/451618/why-does-hawking-radiation-outside-the-schwarzschild-radius-decrease-a-black-hol – anna v Jun 05 '19 at 17:53
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    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. – amateurAstro Jun 05 '19 at 17:54
  • 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. – anna v Jun 05 '19 at 17:58
  • @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. – safesphere Jun 06 '19 at 07:19
  • @safesphere according to the experimental fact of the great attraction of the black holes to each other and the merging, it is obvious that to outside observers there is a macroscopic effective newtonian gravitational attraction to black holes. Once gravity is quantized and the singularity becomes fuzzy one can talk of the charges residing there. Anyway, the charge of a charged black hole should be modeled someplace in a black hole model for outside observers, which we are. – anna v Jun 06 '19 at 10:23
  • I agree with @ameteurAstro. I don’t think that the 1/r objection is relevant. The horizon will not be far away if the Hawking radiation is hot enough to produce electron positron pairs, and even if it were the electrostatic potential should still slightly change the relative probability of escape for the two particles, I think. – Dale Jun 06 '19 at 11:00
  • @dale even if this were true, at some slight probability level ( dont forget it is quantum mechanics here) charge would still be conserved, just the black hole would be losing a bit of charge, being neutralized by the preference to positive assimilation. – anna v Jun 06 '19 at 12:41
  • @anna v yes, that is the point. As the black hole evaporates it must have a greater probability to release a like-charge particle in order for charge to be conserved once evaporation completes and there is no charge remaining – Dale Jun 06 '19 at 12:47
  • @annav The black hole solutions are vacuum solutions. They are valid only outside the horizon, because they are static, but the metric inside is not static and therefore is not a valid solution. The mass and charge are spherically symmetric parameters located below the radius of interest. Since the solutions are valid outside the horizon and are invalid inside, the mass and charge therefore are located asymptotically close to the horizon from outside. – safesphere Jun 06 '19 at 17:35
  • @annav Quantum gravity removing the singularity is a myth. The singularity is where geodesics end (or begin, e.g. in the Big Bang). Will quantum gravity remove the Big Bang? Of course not. We already know what happened, we need quantum gravity only to understand how exactly this happened. The same is with the black hole singularity (assuming it is even there). We already know what happens there - the geodesics of existence end. We need quantum gravity only to understand how exactly this happens at the microscopic level when the shrinking space is squeezed out of existence. – safesphere Jun 06 '19 at 17:52
  • @safesphere can you give a link for this claim "the mass and charge therefore are located asymptotically close to the horizon from outside. ". ? – anna v Jun 06 '19 at 18:54
  • @Dale I would expect evaporation to stop when there is no horizon any more because of the loss of mass? then the star/planet/whatever would carry the charge? – anna v Jun 06 '19 at 18:56
  • @annav Here is a competent critique explaining that the mass cannot be in the singularity, because ithe singularity is in the future and it is not possible for a mass located in the future to source gravity in the present. The mass (ADM, Komar or Abbott-Deser) belongs to the entire curved spacetime, because black holes are vacuum solutions with no massive objects: https://elblogdelcomandos.blogspot.com/2019/03/donde-se-me-habra-perdido-la-masa_28.html - I meant by my comment that the infalling objects were located at the horizon while their mass-energy becomes spread over the curved spacetime. – safesphere Jun 07 '19 at 05:00
  • @annav Consider a weightless ideal mirror box filled with EM radiation. We know such a box has inertia and therefore mass proportional to the frequency. In any frame, we see its frequency redshifted to zero at the horizon. The baryonic matter is not different either, because its mass is the energy of (virtual) massless gluons. So the mass-energy of infalling objects decreases to zero at the horizon by transferring to the mass-energy of the gravitational field. The total mass-energy remains the same, but the black hole sucks it completely out of the infalling objects suspended at the horizon. – safesphere Jun 07 '19 at 05:20
  • @safesphere Fine, this formulation I can understand, thanks It means that the charges will also be following the infalling energy/momentum in any reasonable model. – anna v Jun 07 '19 at 05:22