# Electron black hole?

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Imagine you build a sort of "Dyson sphere", except there's nothing inside it yet, and instead of solar panels, it's a dense array of powerful linear particle accelerators, which are pointed inward and supplied with humongous amounts of electrons.

Now suppose the accelerators start shooting electrons into the center, slowly at first, then faster and faster, in such huge quantities and at such high energies that the electrons get compressed and form a black hole.

My question is, what effect would the high charge-to-mass ratio of the electrons and their electrostatic pressure have on the Hawking radiation coming from the black hole, on any other properties, or I guess whether it's even possible to create the black hole in the first place?

Edited by Lorentz Jr
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There are valid reasons why 'charged' Black Holes would not be plausable.
They would tend to attract opposite charges  much more strongly than gravitationally.

Similarily, once you start to accumulate charge, it becomes more and more difficult to add additional charge, and gravity is certainly not strong enough to overcome the repulsion.

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10 minutes ago, Lorentz Jr said:

whether it's even possible to create the black hole in the first place?

You should be able to calculate the electrostatic potential energy of an assembly of electrons that’s dense enough to be a black hole.

Nuclei aren’t black holes, and the EPE there is measured in MeV for just a few protons. For 10 electrons you have a Schwarzschild radius of about 10^-56 m. Since the EPE is 1/r, you have to get the additional ~40 orders of magnitude of size reduction, which raises that EPE by the same factor.

So it’s going to be at least 10^45 eV

I’d put that in the “not possible” category

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Just now, MigL said:

They would tend to attract opposite charges

That's why I like this problem! I don't think the charge can even be visible outside the black hole, since it acts through light.

Just now, MigL said:

gravity is certainly not strong enough to overcome the repulsion.

That's why the accelerators have to force the electrons together by shooting them at each other at high energies.

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1 minute ago, Lorentz Jr said:

That's why I like this problem! I don't think the charge can even be visible outside the black hole, since it acts through light.

Charge, along with mass/energy and angular momentum, is a classically conserved quantity in Black Hole formation.

There are solutions to the EFE for a charged non-rotating Black Hole.
See here     Reissner–Nordström metric - Wikipedia

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27 minutes ago, Lorentz Jr said:

That's why I like this problem! I don't think the charge can even be visible outside the black hole, since it acts through light.

$e^-+e^-\overbrace{\rightarrow }^{photon} e^-+e^-$ There is the Moeller scattering of two electrons fired at generating a scattering event of two electons. You will subsequently generate an off shell mediator photon. So you will be generating light, however assuming that you can generate sufficient mass density to have the mass below the Schwartzchild radius the photons would not escape. See Swansont's reply above

However electrons will repel each other so simply firing them into a specific region will not be sufficient. I don't think a magnetic trap will suffice either as the EM field interactions will cause further scatterings and particle generation/ annihilation events.

However lets assume you can generate a BH by some method of containment. The Hawking radiation will be fastest the smaller the BH. So the initial BH will want to decay extremely quickly however Hawking radiation only applies when the Blackbody temperature of the BH is higher than the Blackbody of the surrounding region... Not sure how that would work as your firing beams into the confined space.

Edited by Mordred
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Is it correct to say, that for an outside observer the electrons (as anything else) will never disappear behind the BH horizon, and their charges will never disappear behind it, too?

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1 hour ago, MigL said:

Reissner–Nordström metric - Wikipedia

Perfect. From the black-hole section, the limit is $2r_q = r_s$.

$4 r_q^2 = r_s^2$

$Q^2 G / \pi \epsilon_0 = 4 G^2 M^2$

$Q^2/M^2 = 4 G \pi \epsilon_0$

$Q/M = 2 \sqrt{\pi G \epsilon_0} = 2 \sqrt{\pi (6.67*10^{-11} N m^2/kg^2) (8.85*10^{-12} C^2/N m^2)}$

$Q/M = 8.6*10^{-11} C/kg$

For the electron: $\displaystyle{q/m = \frac{1.6*10^{-19} C}{9.1*10^{-31} kg} = 1.76*10^{-11} C/kg}$

So it looks like electrons are well within the limit. Which is strange, because what has a higher charge/mass ratio than electrons?

Does the electrostatic energy somehow decrease the effective mass? You'd think it would make it harder to create the BH one way or another.

1 hour ago, swansont said:

You should be able to calculate the electrostatic potential energy

What effect does that have?

Is it correct to say, that for an outside observer the electrons (as anything else) will never disappear behind the BH horizon, and their charges will never disappear behind it, too?

Right, because it would be really weird if the charge just disappeared. 🤔

Edited by Lorentz Jr
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2 hours ago, Lorentz Jr said:

What effect does that have?

You want to shoot electrons together. They need enough energy to get close enough to form the black hole, and the potential energy of the configuration is the minimum amount of energy it will take.

(The gravitational attraction is small, and I’m ignoring the bremsstrahlung losses as the electrons get close, which would likely be significant. And ignoring a whole host of technical obstacles.)

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1 hour ago, swansont said:

You want to shoot electrons together. They need enough energy to get close enough to form the black hole, and the potential energy of the configuration is the minimum amount of energy it will take.

Right, but how would that affect creation of a BH and its behavior, e.g. the rate of Hawking radiation? It almost seems like the extra energy would make creation of the BH easier by adding to the mass, even though that seems counterintuitive. And it looks like it's possible with electrons as long as there's not too much rotation.

And here's another one for ya: How can charge-neutral Hawking radiation be emitted by a BH full of electrons? What would be left over, with the same amount of charge but less mass/energy? 🤔

Edited by Lorentz Jr
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31 minutes ago, Lorentz Jr said:

Right, but how would that affect creation of a BH and its behavior, e.g. the rate of Hawking radiation? It almost seems like the extra energy would make creation of the BH easier by adding to the mass, even though that seems counterintuitive. And it looks like it's possible with electrons as long as there's not too much rotation.

You aren’t going to make a BH if the electrons don’t get close enough to each other, i.e. within 2x the Schwarzschild radius. Otherwise you just have a very brief moment where there’s a small electron cloud that will fly apart from the repulsion.

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36 minutes ago, swansont said:

You aren’t going to make a BH if the electrons don’t get close enough to each other

Yes, I just now figured out what you were saying. Sorry about being so slow.

Anyway, what really interest me are the relativistic and quantum-mechanical aspects of the problem. Is the charge really invisible to the outside world? And how can it evaporate through charge-neutral radiation? By "affecting the creation", I meant whether it would change the minimum density requirement.

Edited by Lorentz Jr
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1 minute ago, Lorentz Jr said:

Is the charge really invisible to the outside world?

Why would it be, if it

will never disappear behind the BH horizon

to which you agreed.

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Why would it be

Because I'm just an ignorant beginner and I could easily be wrong! 😲

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1 minute ago, Lorentz Jr said:

Because I'm just an ignorant beginner and I could easily be wrong! 😲

Ahh, me too.

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9 hours ago, Lorentz Jr said:

That's why I like this problem!

You like problems that cannot be definitively answered (i.e., verified by experiments)..

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1 hour ago, Sensei said:

You like problems that cannot be definitively answered

Come on, Sensei. You know about corner cases. They're what testing is all about! 😎

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The charge is visible to the outside world, just as mass/energy and angular momentum are. These 'quantities' are stored in the configuration and size of the Event Horizon, and are conserved in the classical model of a BH that arises from the solutions provided by GR and the EFEs.

Hawking radiation is a Quantum Mechanical result of the temperature of a BH ( all bodies at a temperature emit black body radiation ), and the BH's temperature is a result of its entropy.
See Bekenstein-Hawking entropy, which is a statistical thermodynamic treatment of micro-states, again encoded on the surface of the Event Horizon   ( Black hole thermodynamics - Wikipedia ).
Essentially, all a BH presents to the outside universe is made available only by the Event Horizon.

I have found, that unless you are a 'Stephen Hawking.', it is best not to mix classical and quantum models

Edited by MigL
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19 hours ago, Lorentz Jr said:

Is the charge really invisible to the outside world?
And how can it evaporate through charge-neutral radiation?

Okay! So it looks like the consensus is as follows:

• I like @Genady's suggestion that the charge doesn't disappear because it never reaches the event horizon. That means it doesn't move inside the black hole, so there's no need to update the electromagnetic field outside of it.
• Even though electrostatic pressure obviously resists compression of the electrons, it's temporarily compensated for by external pressure from more incoming electrons. So the electrostatic energy of the electrons probably decreases the density required to form a black hole by increasing their total mass/energy.
• Electrostatic energy may also be the answer to charge-neutral Hawking radiation. As long as radiation reduces the electron density, it can get its energy from the resulting loss of potential energy.

Thanks to @MigL for the link to info about charged black holes, and to everyone else for commenting.

Edited by Lorentz Jr

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