Black hole electrons - why don't they evaporate?

Question

https://en.wikipedia.org/wiki/Black_hole_electron
The black hole electron theory states that a black hole with the same mass, spin, and charge of an electron would have identical properties; technically, there's evidence all electrons are or are not black holes. But a black hole the mass of an electron should evaporate very quickly; wouldn't the simple fact that electrons are persistent disprove this theory?

Answer 1

Charge is conserved. What particle could a black hole with one unit charge decay into?

While Hawking radiation remains an unobserved theoretical phenomenon many physicists think is plausible for certain reasons, it is likely more well behaved far away from the quantum gravity domain or when the entire hole is close to a quantum system. Even then the decay of charged black holes can be non-intuitively slow. And if there is no lightest charged particle to decay into, then the hole would be stuck there.

Answer 2

The term "black hole electron" is a misnomer. If you find the stationary solution of the Einstein-Maxwell equations that has the same mass, charge, and spin as observed for an electron, you would get a Kerr-Newman solution. But not all Kerr-Newman metrics describe a black hole. The Kerr-Newman solution only has an event horizon (and therefore describes a black hole) if

$$ \frac{J^2/M^2 + Q^2}{M^2} \leq 1$$ (with the spin angular momentum $J$, charge $Q$ and $M$ in natural units).

However for an electron we would have

$$ \frac{J^2/M^2 + Q^2}{M^2} \sim 10^{89},$$ which is very much not smaller than one, and therefore would not have an event horizon and would not be a black hole. (Such solutions are referred to as being over-extremal, as mentioned in the comments).

Not having an event horizon also implies there not being any Hawking radiation or Hawking temperature (the other answer calculating the Schwarzschild Hawking temperature for such an object is being a bit silly).

Nonetheless, such an object would still be subject to the Schwinger effect and superradiance, which would at first glance lead to it shedding electric charge and angular momentum. However, the electron already has the lowest charge and spin of any free particle, and therefore there are no particles available to carry away its charge or angular momentum (other than electrons).

Of course, the real problem is that electrons have been experimentally confirmed to behave like quantum objects (if not, atoms would not be stable, etc.). Consequently, they cannot be described by a classical solution to the Einstein-Maxwell equations.

More to the point, Carter's observation – that a Kerr-Newman solution with mass, charge and spin of an electron would have the correct magnetic dipole moment for an electron and a finite self-energy – means that one may suspect that such features carry over to a theory of quantum gravity. In particular, including gravity may cure some shortcomings of quantum electrodynamics such as having to put in the electron magnetic dipole moment by hand, or needing an infinite renormalization of the bare electron mass, to end up with finite effective mass.

Answer 3

Well, the reasons why "Black hole electron" is just a speculative "What IF ?" scenario is this.

Extremal charged black hole Hawking radiation temperature would be, $$ \tag 1 T_E = \frac{M_p^2}{2\pi} \frac{\sqrt{M^2-M_e^2}}{\left(M+\sqrt{M^2-M_e^2}\right)^2}, $$

where $M_e$,- extremal mass limit which for electron-charge black hole would be,

$$ \tag 2 M_e = \frac{1}{\sqrt{4\pi}} ~g~e~M_p = \text{const}$$

where $M_p$,- Planck mass and $g$,- charge gauge coupling constant.

Mass extremal limit parameter is as small as you can get for the charged black hole. If you want to decrease charged black hole mass down to electron mass, then as per (2) you need to increase it's charge up to $e\times M_p/m_e\approx 10^{22}e\approx3800~C.$

If you do not do this and only simply will "imagine" that a charged black hole can be as small as you wish,- i.e. when asymptotic mass $M^2 \lt M_e^2$, then as per (1) you will get a imaginary temperature (i.e. $T_E \in \mathbb C~$), which in physics does not make any sense and does not represent any physical system temperature, since absolute lowest temperature possible can be $0K \in \mathbb R$ (ignoring quantum fluctuations, which makes this limit slightly greater than zero.)

Hence, such "electron-black holes" can't exist, since they would have either too less mass or too less charge.