Can there be clouds of free electrons in space?

Question

Electrons are repelled from each other by the electromagnetic force, which is stronger than gravity. However, if there is a sufficiently high amount of free electrons in some region of space, could they be bound by gravity but still separated by the repellent electromagnetic force?

I have also read that the added energy from the electrostatic potentials increases gravity, up to the point where you can form a black hole in principle. But does it mean that increasing the amount of electrons in a point will increase the electrostatic potential? Could electrons be bound before turning into a black hole?

Answer 1

Adding to David_h's answer to account for general relativistic effects: in summary you can't actually even create a black hole with any arbitrarily-large number of electrons, because a black hole composed entirely of electrons would have a charge about $1.76\times10^{11}$ times greater than its mass and it would be a naked singularity according to the Reissner-Nordstrom metric.

To demonstrate: in the RN metric, the event horizon has a radius of

$$r=\frac{1}{2}\left(r_s\pm\sqrt{r_s^2-4r_Q^2}\right),$$

where $r_s$ is the classic Schwarzschild radius proportional to mass (constant of proportionality $\frac{2G}{c^2}\approx1.485\times10^{-27}$) and $r_Q$ is the corrective factor associated with the electric field, proportional to the electric charge (constant of proportionality $\sqrt{\frac{G}{4\pi \epsilon_0c^{4}}}\approx8.617\times10^{-18}$ - much larger). It's $\pm$ because there are actually two horizons for a charged black hole - an outer one, corresponding to that for a Schwarzschild black hole, and an inner one within which there is a region of pseudo-normal space - try it with the RN metric for yourself and see. Since the charge of the electron is much larger than its mass (by about 11 orders of magnitude) and the constant of proportionality for the corrective factor is also much larger than that of the Schwarzschild radius, it's easy to show that a black hole composed entirely of electrons would have an imaginary event horizon radius (which corresponds to having no event horizon, i.e. a naked singularity that cannot, under the known laws of physics, exist).

Quoth the Wikipedia article on the RN metric: "These concentric event horizons become degenerate for $2r_Q=r_s$, which corresponds to an extremal black hole. Black holes with $2r_Q > r_s$ cannot exist in nature because if the charge is greater than the mass there can be no physical event horizon (the term under the square root becomes negative)."

In other words, the repulsion of an electron cloud will always overcome its own gravity, even in the extreme case of a black hole-like object. You can also discount things like an arbitrary number of electrons just orbiting each other because again, the electromagnetic force dominates the gravitational force for electrons (and actually protons and most other known particles too) at all ranges. Only objects that are roughly electrically-neutral, like asteroids or planets or stars or galaxies, can exist in clouds.

Answer 2

I can't answer for general relativistic effects, but I would expect that you would need extraordinary initial densities of electrons for those to come into play.

In the Newtonian limit, the situation is quite clear: no, electrons alone could not be bound by gravity. As the Coulomb force and the Newtonian gravity both scale with $1/r^2$, it is clear that the repulsive coulomb force between two electrons is larger than the attractive force for any distance. Therefore, the net force on any electron in this cloud is pushing it away from the cloud, and therefore you can't have a bound system like that.

Footnote: as an exception, an electron in the center of a symmetric distribution would not experience a net force at all, but all other electrons would be repelled by it, so same result.

Answer 3

An obvious corollary to this question is "how massive would a charged particle have to be for the gravitational attraction to overcome the electrostatic repulsion", which is very easy to solve. At the crossover point as you increase mass, the forces are equal:

$$|F_\mathrm{Gravity}| = |F_\mathrm{Coulomb}| \rightarrow G\frac{m_1m_2}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$

For identical particles, $m_1=m_2=m$ and $q_1=q_2=q$, which gives $$\frac{m}{q} = \frac{1}{\sqrt{4\pi\epsilon_0G}} \approx 10^{10}\ \mathrm{kg/C}$$ For $q$ equal to the elementary charge, this gives about $10^{-9}\ \mathrm{kg}$, which is $10^{21}$ times larger than the electron mass.

Answer 4

You might want to have a look at Earnshaw's theorem. According to its Wikipedia article, "a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges." Nonetheless, if we throw gravity in the mix, we would have to redo our calculations, as there would now be an additional force acting on the system's particles. I'm assuming the configuration you describe would still not be possible as electrons are much less susceptible to gravitational interactions than they are to electrostatic ones, so you would need a huge density. Nonetheless, there exists a possibility that, due to the shielding of outermost electrons (this is, electrons from a layer exerting a force inwards on the electrons of lower layers), alongside the newly added gravitational force, such a configuration were possible