Let's suppose we put 5 electrons on a perfectly conducting (no resistance at all) sphere.

There's no equilibrium configuration with 5 (though there is with 2, 3, 4 or 6). So will they keep moving on the sphere for ever ?

  • $\begingroup$ If the sphere has "no resistance at all", it's pretty obvious they'll never stop moving regardless of the existence of equilibria, so why mention that? Further, it's a little unclear what sort of idealized model you want to use. By "electrons", do you mean classical point particles with charge, mass, and no other properties, or actual electrons, or something else? If you want to know about five real electrons, I imagine the thermal velocities will mostly drown out their electrostatic interactions and each one will be all over the place at room temperature. $\endgroup$ – Mark Eichenlaub Jan 18 '13 at 2:12
  • $\begingroup$ I imagine a classical point particle. Yes "no resistance at all". Because of the forces they exert on each other, they will move and accelerate and radiate energy. And if they're giving away energy, they'll have to stop. But how can they stop since there's no equilibrium configuration ? $\endgroup$ – Anarchasis Jan 19 '13 at 12:46

The answer seems to be that they end up in a non-symmetrical unstable equilibrium with three electrons equispaced on a great circle that we can call the "equator" and the other two at the poles.

This is what is observed in a pentacoordinate compound which is one with a central atom and five other groups (called "ligands") attached to the central atom. All the ligands are assumed to be identical.

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  • $\begingroup$ This was definitely my intuition, but I'm curious if there are any good visualizations of "potential energy as a function of configuration" that can demonstrate this visually. Though at best it's a 7-dimensional configuration space so that might be a challenge... $\endgroup$ – user10851 Jan 18 '13 at 4:53
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    $\begingroup$ Is this really an unstable equilibrium? If they end up in it, it seems like it should be stable... $\endgroup$ – Nathaniel Jan 18 '13 at 6:11
  • $\begingroup$ @Nathaniel Yes, it's a stable equilibrium because it has minimum energy. The N=5 case is one of the few instances of the Thomson Problem where an exact solution is known. $\endgroup$ – mmc Jan 20 '13 at 23:34
  • $\begingroup$ @mmc it looks like that link answers the OP's question - you could post a brief summary as an answer... $\endgroup$ – Nathaniel Jan 21 '13 at 1:02

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