# An approach to the Thomson problem for $N$ electrons embedded on the surface of a sphere

I asked this question on the mathematics board but it received no attention and it received no attention so I am lead to believe that I posted on the wrong board.

The Thomson problem is concerned with minimizing the electrostatic potential of classical electrons on a sphere which is equivalent to maximing the distance between points on a sphere.

By defining the position of a point on a unit sphere centered at the origin with the pair of angles $(\alpha_n, \beta_n)$ with $(0\leq \alpha_n\leq \pi, \ 0\leq \beta_n< 2\pi)$ such that:

$$x_n= \cos(\alpha_n),\\ y_n=\sin(\alpha_n)\sin(\beta_n),\\ z_n=\sin(\alpha_n)$$

I can define the distance between two points $k$ and $l$ as $$D(k,l) = \sqrt {(\cos(\alpha_k) - \cos(\alpha_l))^2 + (\sin(\alpha_k) - \sin(\alpha_l))^2 + (\cos(\alpha_k) \sin(\beta_k) - \cos(\alpha_l) \sin(\beta_l))^2} .$$

By taking the sum of all these distances $T(n)=\sum_{k=0}^{n} \sum_{l=0}^{k}D(k,l)$ we obtain a function of $2n$ variables.

Finding the maximum of this function through differentiation and by substituting appropriate values for the pairs $(\alpha_n,\beta_n)$ where neccesary one should be able to find a solution/class of solutions for any $n$.

I just wanted to know if there are any flaws in this approach as Wikipedia states that the Thomson problem has no known solution.

• It's a multi-dimensional problem that involves $$2N-3$$ variables. By fixing the first charge at the north pole and the second charge at Prime meridian with varying latitude, every additional charge add $$2$$ degree of freedom into the system.

• There're plenty of local minima or equilibrium solutions for large $$N$$. For example, $$N=8$$ there's one configuration as charges locating at vertices of a cube, however it's unstable equilibrium. The known but unproven lowest energy state is staggered conformation of two squares that gives $$D_{4d}$$ symmetry.
See more on T. Erber and G. M. Hockney, "Complex Systems: Equilibrium Configurations of $$N$$ Equal Charges on a Sphere ($$2\le N\leq 112$$)", Advances in Chemical Physics, Volume $$98$$, pp. $$495–594$$, $$1997$$.

• The accepted lowest energy configurations only up to $$5$$-electron case had been proven.

• The topology of sphere is special: It's bounded but with no boundaries. Everywhere on the surface is isotropic, unless you mark different positions with charges. Every charges can be assigned as north pole $$(\theta_1,\phi_1)=(0,0)$$. So for one configuration has $$C_1$$ symmetry, there're $$2N$$ ways of writing the coordinates $$(\theta_i,\phi_i)$$. The second charge may be chosen as the closest to the first so that $$(\theta_2,\phi_2)=(\theta_2,0)$$ with $$\theta_2$$ is lowest among other $$\theta_{i\ge 3}$$. The factor of $$2$$ is due to enantiomer nature of $$C_1$$ so we need mirror version for those $$N$$ combinations.

• The actual computer program is coded by means of calculating the repulsion to see how small perturbation of subsequent motion changes the energy. The points are initially generated randomly on the sphere.

For $$N=3$$ and assume you don't know it's equilateral,

\begin{align*} \mathbf{r}_{1} &= (0,0,1) \\ \mathbf{r}_{2} &= (\sin \theta, 0, \cos \theta) \\ \mathbf{r}_{3} &= (\sin \alpha \cos \beta,\sin \alpha \sin \beta,\cos \alpha) \\ E &= \sum_{i

• $$\theta = \pi$$, $$\left \{ \begin{array}{ccccl} 0 &=& \dfrac{\partial E}{\partial \theta} &=& -\dfrac{1}{4} \tan \dfrac{\alpha}{2} \sec \dfrac{\alpha}{2} \cos \beta \\ 0 &=& \dfrac{\partial E}{\partial \alpha} &=& \dfrac{1}{4} \left( \sec \dfrac{\alpha}{2} \tan \dfrac{\alpha}{2}- \csc \dfrac{\alpha}{2} \cot \dfrac{\alpha}{2} \right) \\ \end{array} \right.$$ No valid $$\alpha$$ in this case

• $$\beta = \pi$$, $$\left \{ \begin{array}{ccccl} 0 &=& \dfrac{\partial E}{\partial \theta} &=& -\dfrac{1}{4} \left( \cot \dfrac{\theta}{2} \csc \dfrac{\theta}{2}+ \cot \dfrac{\alpha+\theta}{2} \csc \dfrac{\alpha+\theta}{2} \right) \\ 0 &=& \dfrac{\partial E}{\partial \alpha} &=& -\dfrac{1}{4} \left( \cot \dfrac{\alpha}{2} \csc \dfrac{\alpha}{2}+ \cot \dfrac{\alpha+\theta}{2} \csc \dfrac{\alpha+\theta}{2} \right) \end{array} \right.$$ $$\alpha=\theta=\dfrac{2\pi}{3}$$ is the solution in this case

• The cases for $$\alpha=\pi$$ and $$\beta=0$$ are omitted here.

• Most symbolic works are done on Mathematica.

Can you feel how your method is not practical for large $$N$$?

• You should be able to do something like, rationally parametrizing the trig functions and feeding the result into Groebners algorithm. I have been working on this approach and it should work but ... the order of the polynomials grows very fast. Unless that is defeated I don't consider it practical in general (unless you are "mainlining" supercomputers. And it would take some development to render it into something humans could tolerate to validate – rrogers Feb 2 '19 at 21:50