# Can a super-extremal charged black hole be made out of electrons only?

In a previous Question it was argued that it would be impossible to add enough charge to a black hole to make it pass the extremal black hole limit since adding charge would increase the mass of the black hole due to the electrostatic field energy (and thus mass) that would be added as the charge is added.

Note that an electron cannot be a black hole but if an electron were a black hole it would be a super-extremal black hole per this wikipedia article: Basically the Schwarzchild radius for the electron's mass is $r_s = 1.35 \times 10^{-57} m$ whereas the charge radius of the electron is $r_q = 9.15 \times 10^{-37} m$. So since $$\frac{r_q}{r_s} \approx 10^{21} \gt 1$$ an electron, if it was a black hole, would be a super-extremal black hole by a large margin. In words, the electrons mass is completely negligible compared to its charge.

An uncharged black hole can be constructed out of matter at any given mass density by simply constructing a big enough sphere of that matter. This is true because a sphere of radius $R$ with a constant (low?) density will have a mass $M$ that is $\propto R^3$ whereas the Schwartzchild radius is $\propto M \propto R^3$. So as $R$ increases the radius of the sphere will eventual be less than the Schwartzchild radius.

So, can we make a super-extremal charged black hole by using a very large sphere of radius $R$ that is made out of electrons uniformly distributed with (low) charge density $\rho$?

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I was hoping the answer was yes, but as I did the calculations I convinced myself it would not work, so I am documenting my calculations by asking and answering this question. I hope this might be interesting to other people and I hope you will check my calculations. If my answer is wrong, please correct me! – FrankH Oct 23 '11 at 8:13
It should be added that there is no reason to suppose that the relation Q=M for extreme black holes is correct for black holes the size of electrons. Electrons are quantum black holes in string theory--- they are strings--- and the lightest black hole excitation in a charged string theory always is on the other side of the extremality bound. Whether you call it a black hole is a matter of convention, but you can't use classical GR to describe something that small. – Ron Maimon Oct 23 '11 at 17:44
Yes, @ron, you are right. The electron cannot be a black hole for lots of reasons. The electrons Schwartzchild radius is many orders of magnitude smaller than the Planck length so classical GR will not apply. I will edit the question to clarify that. Thanks. Any opinion on my answer would be appreciated. – FrankH Oct 23 '11 at 18:17

The answer is no. As stated, the electrons mass will be ignored. If we assume the mass of the sphere is just due to the electrostatic field energy of the uniformly charged sphere, can we create a black hole and can the charge to mass ratio $\frac{Q}{M}$ be super-extremal ($\frac{Q}{M} \gt 1$ in appropriate units).
Now the charge of the sphere is $$Q \propto \rho R^3$$ the electric field strength $E(r)$ as a function of the radius $r$ is $E(r) \propto \rho r$ and the mass $M$ due to the electrostatic field energy is $$M \propto \int_0^R E^2(r) d^3r \propto \rho^2 R^5$$ Therefore the ratio of charge to electrostatic field energy-mass is $$\frac{Q}{M} \propto \rho^{-1} R^{-2}$$
Consider the case of constant $\rho$ with $R$ increasing: Since $M \propto R^5$ increases rapidly, eventually a black hole will form. However the ratio $\frac{Q}{M}$ will decrease as $R$ increases, so it will not be an extremal black hole.
Now consider the case of constant $R$ with $\rho$ increasing: Since $M \propto \rho^2$ a black hole will eventually form. but again the ratio $\frac{Q}{M}$ will decrease as $\rho$ increases, so it will not be an extremal black hole.