Suppose we have a sphere of radius $r$ and mass m and a negatively charged test particle at distance d from its center, $d\gg r$. If the sphere is electrically neutral, the particle will fall toward the sphere because of gravity. As we deposit electrons on the surface of the sphere, the Coulomb force will overcome gravity and the test particle will start to accelerate away. Now suppose we keep adding even more electrons to the sphere. If we have n electrons, the distribution of their pairwise distances has a mean proportional to $r$, and there are $n(n-1)/2$ such pairs, so the binding energy is about $n^2/r$. If this term is included in the total mass-energy of the sphere, the gravitational force on the test particle would seem to be increasing quadratically with $n$, and therefore eventually overcomes the linearly-increasing Coulomb force. The particle slows down, turns around, and starts falling again. This seems absurd; what is wrong with this analysis?
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1$\begingroup$ Can you clarify why it sounds absurd? $\endgroup$– Mark EichenlaubMar 2, 2011 at 5:34
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$\begingroup$ Just because electrons are supposed to repel each other and gravity is supposed to be a weak force. I wonder if there is another argument that gives a different answer? $\endgroup$– Dan BrumleveMar 2, 2011 at 6:12
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3$\begingroup$ I think the other comments are right: this result is correct and is not actually paradoxical. But I agree it's counterintuitive, and I think it's a nice thought experiment. $\endgroup$– Ted BunnMar 2, 2011 at 14:27
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2$\begingroup$ How do you keep accumulating electrons on the sphere anyway? $\endgroup$– RaskolnikovMar 8, 2011 at 23:42
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1$\begingroup$ Apologies if I'm just stating the obvious, but: does it seem less counterintuitive to you once you consider that the requisite $n$ for which gravity dominates goes to infinity as $G_N \to 0$? $\endgroup$– Matt ReeceMar 9, 2011 at 1:27
3 Answers
The statement that the gravitational attraction will eventually dominate the coulomb repulsion as n increases is correct. You probably think the restmass of the electrons will invoke gravitational attraction, but that part is neglible for high electrondensities on the sphere. The gravitational attraction by (the curvature of spacetime caused by) bindingenergy is far greater for high densities.
I don't know where the breakpoint is, but suppose I look way past that limit, i.e. an incredible dense sphere(or shell) of electrons. If I make it dense enough, that could very well become a black hole. Note that this would be a strange black hole, since the 'mass' of such a black hole consists almost entirely of the bindingenergy, and not the restmasses of the electrons. From this point of view it might be more easily imaginable that the gravitional attraction will dominate eventually.
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4$\begingroup$ Black holes of a given mass have a maximum charge, which I believe may be much less than you'd get if they were composed solely of electrons. Nobody has taken this into account so far. $\endgroup$ Mar 10, 2011 at 16:00
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$\begingroup$ how does the maximum charge grows with mass? what happens if i add more charge than allowed? does the mass grows at least accordingly? doesn't that fix the minimum ratio of electrical charge versus gravitational mass for physical systems? $\endgroup$– lurscherMar 10, 2011 at 17:19
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$\begingroup$ @Peter Shor: there may be various upperlimits on the maximum charge of a black hole. I take one into account(below). @lurscher: yes to both. If charge is added, the mass grows at least accordingly, and yes, that fixes a maximum chargedensity. I suspect a blackhole of pure massless charged particles would exactly have that density. If you'd add more charge, you'd essentially add extra binding energy and as a result the radius and mass of the black hole would increase. So @Peter again: therefore there is maximum charge given the mass of a black hole. It could be even lower by other reasons, IDK. $\endgroup$– JBSnorroMar 10, 2011 at 18:01
The passive gravitational mass of the electron is, by experiment, less than 9% of the theoretical expected value, as you can see in my answer on this question.
The result was so unexpected that no one believed in the correcteness of the result.
I believe that the active gravitational mass of the electron will also be of the same order (or zero).
One nice theory could collapse because of one single experiment.
I am very much impressed: Almost no question do not include the Black Hole in the response. May be the case that they atract votes up. On this site there are already 2000 references to this term.
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$\begingroup$ That is interesting, however this thought experiment works just as well with protons, because it is the gravitational mass of the electric potential energy that is causing the dominant effect, not the gravitational rest mass. $\endgroup$ Mar 13, 2011 at 20:51
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$\begingroup$ @Helder: the result is so unexpected, that many continue to not believe the correctness of the result -1. $\endgroup$ Oct 8, 2011 at 21:03
As a brainstorming answer, lets calculate the binding energy in another way:
suppose we have N electron in the sphere, the electrostatic energy to bring a new electron into the sphere is $Ne/R$. If we add a new electron, only the net electrostatic potential $Ne/R$.
the net potential on far-away electrons is
$N( e - m_{e} G )/R$
what is actually quadratic in $N$ is the energy required to bind together N electrons in the sphere, it is actually
$e/R + 2e/R + ... Ne/R = N(N-1)e/2R$
This also contributes to gravitational weight, but there will be a maximum capacity where the electrons will escape the sphere
the capacitance of a sphere is given by $4 \pi \epsilon R$. So in this case the binding energy is bounded by the capacitance of the conductor used for your sphere
