1
$\begingroup$

Protons have a positive charge and hold negatively charged electrons in atoms. Thanks, quantum mechanics rules electrons don't fall to the nucleus. If we substitute the atomic nucleus with a micro black hole about radius 10^-15 meters whether electron will fall to the event horizon or will keep on any high above the event horizon by Heisenberg-uncertainty-principle? How will electrons behave according to general relativity and quantum mechanics if the black hole radius/mass increases or decreases?

$\endgroup$
2
  • 3
    $\begingroup$ The electron in a hydrogen atom has a lot of wave function at the nucleus. In the case of a black hole, well, bye bye electron. $\endgroup$
    – Jon Custer
    Sep 2, 2021 at 18:54
  • 2
    $\begingroup$ BTW, a 1e-15 m Schwarzschild BH (which has a mass of 6.73326E11 kg) has a Hawking radiation temperature ~1.82223E11 K. See vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator That's not exactly a friendly environment for a bound electron. ;) $\endgroup$
    – PM 2Ring
    Sep 2, 2021 at 19:04

1 Answer 1

3
$\begingroup$

A normal micro-black hole has a high Hawking temperature and will evaporate quickly. However, if it is extremal (that is, rotating fast enough or charged enough) the Hawking temperature is zero. Hence it has been proposed by F.C. Adams that Planck mass black extremal holes could remain stable, and since they cannot change charge without accreting particles (that would have a hard time tunnelling into their tiny event horizons) they could form "atoms" with captured electrons (or positrons) with a lifetime of $10^{49}$ years (they could also form "atoms" with orbiting protons, with lifetimes of $10^{37}$). No such atoms have been observed, so far.

The behaviour of electrons around black holes would be described by solving the Schrödinger or Dirac equation in the Schwarzschild (or Kerr-Newman) metric. My impression of the results is that for small holes they behave much like in a hydrogen atom, but with slight shifts of the energy levels.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.