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I will describe the question first in 2D, but my interest is in $\mathbb{R}^3$. An electron $x$ will shoot from the origin along an initial vector $v$. You know the speed $|v|$ but not the direction. Your task is to arrange a minimum number of point-charge electrons at fixed locations on an origin-centered unit circle so that, no matter in which direction $v$ is pointing, $x$ cannot escape the disk. All the charges are equal, and repel each other via inverse-square Coulomb force.

Below there are four charges on the circle. The left electron escapes, the right, which is shot along the same direction but with a lesser speed, does not. (Of course aiming $v$ along a diagonal is the best escape strategy in this configuration.)

    ElectronTracks

Now here is my question. Consider the same problem in $\mathbb{R}^3$. Given $|v|$, arrange fixed electrons on a sphere to cage-in the electron $x$ starting from the origin with velocity $v$ in any direction. It seems natural to think that an optimal arrangement is a type of disk-packing on a sphere, for example, the solutions to the Thompson problem or the Tammes problem. (Here I am mining an earlier MO question.)

          DiskPacking
          (Image from Paul Sutcliffe.)

Q. Is an optimal electron cage for a given speed $|v|$ a configuration that minimizes electrostatic potential? Or maximizes the minimum distance between any pair of electrons, i.e., an optimal disk packing on the sphere?

Answered by Robert Israel: No such cage is possible in $\mathbb{R}^3$!
($\mathbb{R}^2$ is rather different and not a reliable guide to $\mathbb{R}^3$.)

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    $\begingroup$ I think it should rather maximize the minimal potential on the sphere (which is the weakest spot in the net you can try to reach with given kinetic energy). Technically speaking, it shouldn't be even on the sphere itself but slightly inside, but all those configurations for the large number of electrons are essentially the same and not known exactly anyway. $\endgroup$
    – fedja
    Commented Sep 30, 2016 at 13:37
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    $\begingroup$ Just as an aside, I thought I'd mention that this problem reminds me a little of the polywell: en.wikipedia.org/wiki/Polywell $\endgroup$
    – Dan Piponi
    Commented Sep 30, 2016 at 19:20

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Electrostatic potential is a harmonic function on any region without charges. It has no local minimum, in fact the value at the origin is the average of the potential over a sphere centred at the origin within your "cage". Therefore it is impossible to do what you want: there will always be a path the electron can take to escape to infinity.

EDIT: Just for fun, I tried it numerically using $60$ unit charges at the vertices of a truncated icosahedron, with $100$ randomly chosen initial velocities with speed $0.01$. By time $t=40$, all but four had managed to escape the cage. An animation of the trajectories is here. The last one escaped by $t=70$.

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  • $\begingroup$ There will always be a path, Yes, but (I think) not one with the electron given a sufficiently small initial velocity (coasting from then onward, subject to the forces upon it). $\endgroup$
    – Joseph O'Rourke
    Commented Sep 30, 2016 at 16:56
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    $\begingroup$ I would think it's more likely that the union in phase space of all trajectories starting from the origin with initial speed $\epsilon$ would be dense in (if not equal to) a connected component of the surface of constant energy. If so, there will be trajectories that escape the sphere. $\endgroup$
    – Robert Israel
    Commented Sep 30, 2016 at 17:42
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    $\begingroup$ I would appreciate an intuitive explanation. Imagine $10^9$ charges on the sphere. Wouldn't they tend to confine the origin electron $x$ to the vicinity of the origin, and increase in repulsion as $x$ wanders toward the sphere? $\endgroup$
    – Joseph O'Rourke
    Commented Sep 30, 2016 at 23:51
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    $\begingroup$ In the continuum limit, the potential inside a uniformly charged sphere is constant. The repulsion from the charges on one side exactly cancels the repulsion from the other side. This goes back to Newton(for gravity rather than electrostatics). See here $\endgroup$
    – Robert Israel
    Commented Oct 2, 2016 at 5:58
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    $\begingroup$ Near the origin, the potential is approximately a constant plus a linear combination of solid spherical harmonics corresponding to the lowest nonzero moments of the charge distribution. As you approach the sphere, there is repulsion from the charges there but a net attraction to the "holes" between the charges. $\endgroup$
    – Robert Israel
    Commented Oct 2, 2016 at 19:46
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Here is Robert Israel's animation.

RIsrael

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    $\begingroup$ It looks like escape happens only through hexagonal faces. Is this always the case? And are the outer trajectories always perpendicular to a hexagonal face? Gerhard "Normal Is Easier To Spell" Paseman, 2016.09.30. $\endgroup$
    – Gerhard Paseman
    Commented Oct 1, 2016 at 1:08
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    $\begingroup$ You can't come out very close to one of the vertices because of energy. Other than that, I think most of the sphere should be accessible, although there may be focussing effects that favour the centres of the hexagonal faces, which are the points of lowest potential energy on the sphere. $\endgroup$
    – Robert Israel
    Commented Oct 2, 2016 at 20:09
  • $\begingroup$ @GerhardPaseman I posted an answer showing the equipotential surface at the origin, which, I think, confirms your conjecture that escape is possible through only the hexagonal faces. $\endgroup$
    – Don Hatch
    Commented Jul 25 at 23:11
  • $\begingroup$ @RobertIsrael see my previous comment (sorry, clumsy here-- I don't know how to cc multiple people at once, or from an answer) $\endgroup$
    – Don Hatch
    Commented Jul 25 at 23:13
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Following up on Robert Israel's animation and Gerhard Paseman's conjecture that escape happens only through hexagonal faces of the truncated icosahedron...

If I did it right, this is a rendering of the equipotential surface whose potential energy is the same as that at the origin. I used this formula from wikipedia for guidance (that is, the potential energy is proportional to the sum of inverses of the distances from each of the 60 fixed electrons).

Observe that the outside is accessible from the origin through the hexagonal faces, but not through the pentagonal faces. I think what this implies is an initially-slow-moving (low kinetic energy) electron at the origin can escape through a hexagonal face, but never through a pentagonal face.equipotential surface at the origin

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