# Platonic solids and charged particles

It is known that there are five Platonic solids: If, lets say, there are 4 particles with the same electricity charge and whose movement is constrained to be on a sphere, resulting forces will eventually move particles to form a tetrahedron.

I have two questions here:

1) If number of particles are 8, 6, 20, and 12, would resulting solid (in similar experiment) be hexahedron, octahedron, dodecahedron, and icosahedron, respectively?

2) What would happen if number of particles is not 4, 8, 6, 20, or 12? Would an equilibrium be reached ever? Would the equilibrium shape be unique or not? Are those solids (resulting from such experiments) widely known? What would be their other properties?

I attempted some simulations, and researche the internet but didn't reach any sound conclusion.

• Possibly related to math.stackexchange.com/questions/66365/… Apr 30 '14 at 13:22
• Electrostatic energy should be proportional to the sum of the inverse distance of all the pairs of points. So to turn this into a mathematical problem, you are trying to minimize $$\sum_{i=1}^{n-1}\sum_{j=i+1}^n\lVert p_i-p_j\rVert^{-1}$$ with the sphere constraint $\lVert p_i\rVert=1$. This makes the problem a mathematical one, while we leave it to physics.stackexchange.com to decide whether electrons actually reach and maintain that energy minimum.
– MvG
Apr 30 '14 at 13:38
• See this web page for an investigation of this problem. Apr 30 '14 at 13:59
• Cross reference to Physics SE: Shape of electric charges on sphere in equilibrium state
– MvG
Apr 30 '14 at 16:23

This is known as the Thomson problem. For $4$, $6$, and $12$ particles, the minimum energy configuration is the corresponding Platonic solid, but not for $8$ or $20$ particles. In particular, for $8$ particles a square antiprism has lower energy than a cube. Minimum energy configurations of course exist for other $n$, but finding them is a difficult numerical problem; as with most packing problems, there appears to be no general pattern for large $n$.
• No, in general if you start from an arbitrary distribution of particles and perform, say, gradient descent, you will probably only land on a local minimum, not the global minimum. This is why the problem is hard for large $n$ if you care about global minima. // Starting from a cube, actually, you may not end up anywhere because the cube is a critical point with zero gradient -- sort of like a stationary bicycle that's perfectly balanced vertically. But if you perturb the particles slightly away from the cube randomly, you should end up at the antiprism.