in collaboration with Vladimir A. Yudin

How $N$ equal charges will arrange on a sphere trying to minimize the potential energy of the system?

Thomson considered this problem in an attempt to explain the planetary atom model. He performed experiments on finding the best arrangements for small numbers of charges.

Since the advent of computers many numerical experiments have been carried out. However only at the end of the XX century some special cases were rigorously solved.

Let’s consider forces that influence the charges. We fix all the charges except for one and consider the latter.

An interaction of two charges is given by Coulomb's law. The interaction force is inversely proportional to the square of the distance between the charges. This means the closer the charges to each other the greater is the force.

The resultant force is calculated by perpendicular and the tangent. The former push the charge out of the sphere’s surface. And as all charges are confined to the sphere’s surface
this component has no effect on the charge movement. The tangent component defines the direction and the velocity of movement.

When the tangent component vanishes (so only perpendicular component remains) the charge stop moving.

So for the system of unfixed charges to be static the forces acting upon each charge must de perpendicular to the sphere’s surface.

Let’s consider $N$ equal charges confined to a sphere. And let’s find their distribution on the sphere.

We consider the cases when the problem was solved rigorously.

$N=2$ Two charges will be located at the diametrically opposite points.

$N=3$ Consider a plane passing through three charges. This plane and the sphere intersect in a circle. So our charges will be on the circle, hence they should be on the great circle and hence they should be
locatedin the vertices of a regular triangle.

$N=4$ Four charges will be located in the veêtices of a regular tetrahedron.

It is easy to see that in cases of $N=2,3,4$ we have equal pairwise distances. And so one can use well-known inequalities of arithmetic mean, geometric mean and harmonic mean to prove the minimality of the
potential energy. These inequalities give sharp lower estimates for the potential energy of the system because they turn to equalities whenever we have extremal configurations.

$N=6$ Six charges extremal configuration is that charges are located at the vertices of a regular octahedron (you can easily imagine it as points of intersection of a sphere
and coordinate axes).

$N=12$ Twelve charges will be located at the vertices of an icosahedron. (An icosahedron is a regular polyhedron having 12 polyhedron vertices, 30 polyhedron edges, and 20
triangle faces).

Surprisingly but Thomson problem has been rigorously solved only for 2,3,4,6,12 charges on a sphere.

As we have just seen solution of the Thomson problem for $N = 4, 8, 12$ are tetrahedron, octahedron and icosahedron respectively. And what about other two regular polyhedrons?

For $N=8$ the problem is still open. But one can easyly prove that the polyhedron vertices of the inscribed cube is not the solution. If we antiprism is better. Is this
configuration the best? It is not proven so far.

$N=20$ In case of 20 charges as well as for 8 charges one can produce disposition of less potential energy then that of dodecahedron.

Consider unsolved cases. For 5 charges optimality of this disposition have been unsuccessful.

By the case of 5 charges we shall illustrate the equilibrium shape notion. As we know Thomson problem is to find configuration with minimum potential energy. It appears, some other stabilizing configurations
exist. They said to be equilibrium configurations. But they are unstable equilibrium configurations that is if you remove one of the charges a little the charges will scatter.

For example if all the charges are originally located on the great circle they will always remain there. Since all forces of interaction lie in the equatorial plane. The charges will be at the vertices of a regular pentagon.

Consider another equilibrium configuration. Let’s place four charges at the square’s vertices and the fifth one at the perpendicular to the square plane. The first four charges remain at rest and the fifth one
which will be shifting in vertical line until a minimum energy configuration is achieved. So we will get a quadrangular pyramid.

The number of equilibrium configurations rises swiftly with the increasing of the number of charges. This make difficulties for the computational modeling research even to cutting edge computers.

Thomson problem have important applications in spaces of other dimensions.

In two-dimensional space(i.e. plane) the "spheres" are circles. So we have $N$ equal charges on the circle. The minimum energy configuration is the vertices of a regular
N-sided polygon.

In spaces of higher dimensions the problem is rigorously solved only occasionally.

For example, for 120 charges in 4-dimensional space the solution is a regular 120-faced polyhedron.

A curious result was received in 24-dimensional space. Minimum energy configuration for 196569 charges is provided by celebrated Leech lattice.

#### Exact solutions of particular cases of the problem in higher dimensions and with different potentials:

A. V. Kolushov, V.A. Yudin Extremal dispositions of points on a unit sphere // Analysis Mathematica. 1997. V. 23, N 1.

Н. Н. Андреев Расположение точек на сфере с минимальной энергией // Труды Математического института им. В.А. Стеклова РАН. 1997. Т. 219. С. 27-31.

Н.Н. Андреев Минимальный дизайн 11-го порядка на трехмерной сфере // Математические заметки. 2000. Т. 67, N 4. С. 489-497.