Where charges settle on a sphere

In 1904 J. J. Thomson, who had discovered the electron seven years earlier, proposed a model of the atom in which electrons sit inside a sphere of positive charge, and he asked how they would arrange themselves. Stripped of the physics, the question is about points on a sphere: place $N$ equal charges on the surface, each repelling every other with a force inversely proportional to the square of their distance, and find the arrangement in which the total energy

$$E = \sum_{i < j} \frac{1}{|x_i - x_j|}$$

is as small as possible. The model of the atom was abandoned within a decade. The geometric problem was not, and it is still unsolved for most $N$.

It is a natural question to ask of the Platonic solids. Why the list of perfect solids stops at five found the five shapes whose corners are as evenly spread over a sphere as symmetry allows: 4, 6, 8, 12 and 20 corners, each corner exactly like every other. If any arrangements of charges should be the most evenly spread, it should be these. Three of them are. Two are not, and the reason the cube and the dodecahedron fail says something about spreading points on a sphere that symmetry alone does not.

Where charges settle #

The minimum energy can be found by imitating the physics. Start the charges at random positions, compute the force on each from all the others, move each a little in the direction of its force, keeping it on the sphere, and repeat, accepting a move only if it lowers the energy. From many random starts the charges settle into arrangements, and the lowest one found is a candidate for the minimum.

The energies reached agree with the published minima to nine decimal places: 3.674234614 for four charges, 9.985281374 for six, 19.675287861 for eight, 49.165253058 for twelve and 150.881568334 for twenty. Four charges settle at the corners of a regular tetrahedron, six at those of an octahedron, twelve at those of an icosahedron. Eight settle into two squares of four, one above the other, with the upper square turned through 45 degrees against the lower — a square antiprism — and twenty settle into an arrangement with only a modest amount of symmetry, not a dodecahedron.

For small $N$ the answer has been proved. Minima for 4, 5, 6 and 12 charges are established theorems, the case of five, a triangular bipyramid, having been settled only in 2010 by Richard Schwartz with a computer-assisted proof. For most larger $N$ the minima are known only as the best arrangements anyone has found, and the published tables are records rather than theorems.

A landscape with one valley, then many #

For the numbers of charges in the first figure the search is easy to trust. Each of eight random starts, for 4, 6, 8, 12 and 20 charges, descends to the same arrangement with the same energy, so the energy landscape — the energy as a function of every charge’s position — has, as far as any start can tell, a single valley. That changes as the number grows. For 72 charges three random starts settle into two different arrangements with different energies, and for 110 charges two starts settle into two; each is a local minimum, an arrangement from which every small movement raises the energy, and only one of them can be the lowest.

Computations by others have found the number of local minima growing roughly exponentially with $N$, into the thousands for a few hundred charges. That is what makes the problem hard. A descent finds a valley, not the deepest valley, and the only way to gain confidence that the deepest has been found is to start from very many places and see the same answer recur. The highest point on the sphere is an eigenvalue found a problem on the sphere with the opposite character — maximising a quadratic form, where every local maximum is the global one and linear algebra finds it directly. Coulomb’s energy has no such structure, and nothing finds its minimum except search.

Five charges, and a proof that took a computer #

Five is the first number with no Platonic solid to compare against, and it shows how delicate the comparisons are. Two arrangements are natural candidates. The triangular bipyramid puts two charges at the poles and three round the equator, 120 degrees apart; the square pyramid puts one charge at a pole and four in a square below it, at whatever height is best. The bipyramid’s energy is 6.474691 and the best square pyramid’s is 6.483661, a difference of less than a seventh of one per cent, and every random start settles into the bipyramid.

Proving that the bipyramid is the true minimum, against every arrangement of five points and not just these two, took until 2010. Richard Schwartz divided the space of all five-point configurations into a very large number of small regions and showed, with exact computer arithmetic, that in every region the energy is provably above the bipyramid’s except near the bipyramid itself. It is the same kind of proof that later settled Rupert’s problem in the negative for the Noperthedron in a cube through a hole in a cube: no search can rule anything out, so a proof of optimality has to cover the whole space of possibilities at once, and with five points that space has seven dimensions after removing rotations.

The closeness of the two energies is also why the answer depends on the force. For forces that fall off much more steeply with distance than Coulomb’s, the balance tips and the square pyramid wins; Schwartz located the crossover in a later paper. Coulomb’s law happens to be on the bipyramid’s side. That the winner depends on the force at all is a reminder that “evenly spread” is not one notion but a family of them, one for each way of measuring how much two nearby points mind each other’s company; for some of them the answer is a symmetric solid, and for others it is not, even with only five points.

Three Platonic solids are minima and two are not #

The corners of each Platonic solid, treated as charges, have an energy that can be computed exactly and compared with the minimum.

The tetrahedron, octahedron and icosahedron are exactly the minima. The cube holds 0.0655 more energy than the best arrangement of eight charges, and the dodecahedron 0.9171 more than the best of twenty. The three that succeed have triangular faces; the two that fail have square and pentagonal ones. That is the pattern in a sentence: charges on a sphere settle with each charge surrounded by a ring of neighbours, and a ring of neighbours around every charge, covering the sphere, makes triangles. A square face leaves its two diagonals as long, wasted gaps where two charges could be brought closer to the opposite pair; a pentagonal face leaves even more.

Symmetry and minimal energy are different properties. Every corner of a cube is exactly like every other, and the cube is the most symmetric arrangement of eight points that exists; the antiprism has less symmetry and less energy. The physics does not care about symmetry for its own sake. It cares about distances, and the antiprism’s distances are better: its shortest distance between two charges is a little longer than the cube’s edge, so the nearest pairs, which contribute most to the energy, repel each other less, and the pairs further apart are spread more evenly between near and far. Symmetry is a consequence the minimum sometimes has, not a cause.

Turning half a cube lowers the energy #

The difference between the cube and the antiprism can be watched continuously. Take eight charges as two squares of four, one above the other, turn the upper square through some angle against the lower, and for each angle choose the separation of the squares that gives the least energy.

With no turn, the best separation makes exactly a cube, energy 19.74077. As the upper square turns, the energy falls, steadily and without a bump, to 19.67529 at 45 degrees, the square antiprism, which is the minimum for eight charges. Turning by half a corner places each upper charge over a gap in the lower square instead of directly over a charge, and the two squares can then come closer together — their best separation falls from 0.577 to 0.560 of the radius — while every charge keeps its distance from the charges nearest it. The cube’s faces are squares; the antiprism’s sides are triangles, eight of them around the middle, and the energy has found the triangles.

A film of charge, and the cost of making it grainy #

For many charges, the energy is dominated by a simple term. If the charge were spread smoothly over the sphere as a continuous film, its energy would be $N^2/2$ for a total charge of $N$, since a uniform film’s potential is the same everywhere. Point charges do slightly better, because a charge does not repel itself, and the saving grows with $N$.

Divided by $N^{3/2}$, the saving settles near a constant: $-0.541$ at four charges, $-0.551$ at forty, heading towards about $-0.5523$, the value fitted to computed minima for thousands of charges. The $N^{3/2}$ has a physical reading. Each charge is surrounded by a patch of sphere of area about $4\pi/N$, and its own share of the film’s self-energy, which the point charge avoids, is of order its charge squared over the patch’s radius, $1/\sqrt{4\pi/N}$; summed over $N$ charges that gives a saving of order $N \cdot \sqrt N$. The constant in front depends on how well the patches can be shaped, and its value, about $0.5523$, is known only numerically. Whether it has a closed form, perhaps related to the energy of a perfect triangular lattice in the plane, is not settled.

Twelve defects that no arrangement can avoid #

As $N$ grows, the charges organise themselves into an almost perfect triangular mesh, each charge with six neighbours, like atoms in a flat crystal. But a sphere is not flat, and a perfect six-neighbour mesh cannot cover it.

In every arrangement, adding up six minus the number of neighbours over all the charges gives exactly 12. In these four, every one of the twelve shortfalls is a charge with five neighbours, and all the rest have six. The rule is Euler’s formula for the sphere: a triangulation with $V$ vertices, $E$ edges and $F$ faces has $V - E + F = 2$, each face has three edges and each edge two faces, and combining these gives $\sum (6 - \deg) = 12$ exactly. Twelve pentagons, whatever the hexagons derived the same twelve from the other side, for polyhedra built of pentagons and hexagons like a football, and the shape of a cell around a random point found its flat-space twin: random cells in the plane average exactly six sides.

The same twelve can be counted in angles. A charge with six neighbours in a nearly equilateral triangulation is surrounded by six angles of about 60 degrees, a full 360, and the surface around it is nearly flat. A charge with five neighbours has only 300 degrees around it, a shortfall of 60, which is where the surface bends. Seven hundred and twenty degrees of gap found that the shortfalls of angle at the corners of any convex polyhedron add up to exactly 720 degrees, and twelve shortfalls of 60 degrees each are exactly that. The defects are where the triangulation stores the sphere’s curvature, and the sphere has exactly 720 degrees of it to store.

The icosahedron is the arrangement in which the twelve defects are all there is — twelve charges, each with five neighbours. For larger $N$ the defects spread apart, as far from one another as the sphere allows, and sit roughly at the corners of an icosahedron. For several hundred charges and more, computations by others have found that the twelve isolated defects become unstable and are replaced by short chains in which five-neighbour and seven-neighbour charges alternate, each chain still contributing a net shortfall of one; the total stays twelve, because Euler’s formula does not allow anything else.

Charges, viruses and footballs #

The same arithmetic governs much more than charges. The protein coats of many viruses are built from identical units packed on a closed shell, and the shells that result are icosahedral, with twelve five-fold positions and many six-fold ones, for the reason the charges are: identical units packing as tightly as possible on a sphere make a triangular mesh, and Euler’s formula then demands twelve exceptions. Carbon atoms in the fullerene $C_{60}$ make twelve pentagons and twenty hexagons for the same reason. In each case the twelve are forced by topology; where they sit, and whether they are isolated or spread into chains, is decided by energy.

The Thomson problem also connects to the most even placement of points on a sphere for other purposes: numerical integration, the design of antennas and the placement of satellites. Equal area on a sphere met the related question of dividing the sphere into equal pieces. The discrete version of spreading points apart is the design of codes, where sixteen spheres that fill a cube found sixteen seven-bit words so evenly spread that balls around them fill the whole space of words exactly — a case where the most symmetric arrangement and the best one coincide, as they do for the tetrahedron, octahedron and icosahedron here. Every one of these problems has the same structure, a smooth global quantity to minimise and a topological constraint the minimiser cannot escape, and the twelve defects appear in all of them.

What the pictures cannot show #

The minima here are the lowest energies found by a descent from a handful of random starts. For small $N$ they agree with the published values, and for 4, 5, 6 and 12 charges those values are proved; for 8 and 20 they are the accepted best, verified by many independent searches and not by proof. For larger $N$ a descent from a few starts can stop in a local minimum above the true one, and the larger arrangements drawn are settled states, not certified minima. The twelve-defect count holds for every one of them regardless, because it is a fact about triangulations, not about energy.

The constant $-0.5523$ is a fitted value from the literature, not computed here; the figure shows the scaled saving approaching it for forty charges, which is far too few to estimate it independently. The pictures also cannot show why the defects become chains for large $N$, since the largest arrangement drawn has only 110 charges, well below where chains appear.

Still open: the minimum for most numbers of charges #

For which $N$ is the minimum-energy arrangement of $N$ charges on a sphere known with proof? Only for a few: 2, 3, 4, 5, 6 and 12. For eight charges, where every computation finds the square antiprism, there is no proof that nothing does better, and the same holds for every $N$ beyond six except twelve. The difficulty is the landscape: the number of local minima grows exponentially with $N$, a search that finds the best of them has no way to certify it, and the proofs that exist for small $N$ use special structure that does not generalise.

Steve Smale included a closely related question, finding near-optimal points on the sphere efficiently for a logarithmic version of the energy, in his list of problems for the twenty-first century. A problem a physicist posed about electrons in 1904 has outlived the atom it was meant to explain, and it remains a standing example of how hard it is to prove that an arrangement found by search is the best one possible.