Interesting question about the Thomson problem for arbitrary number of electrons

Question

This question is crossposted from here I believe this is a pretty hard question and so I decided to repost the question in the Math Overflow forum. If there is something wrong with doing this, I am sorry in advance and this question can be deleted.

Anyway, the question goes as follows:

I am looking forward to answering the question

Show that for $4$ or more points in the Thomson problem, a configuration of equidistant points in the equator line DOES NOT give a minimal configuration.

For some context and in really fast terms, the Thomson problem consists in solving the optimization problem

$$ \min_{x_1, \dots, x_N \in \mathbb R^3} \quad U(\mathbf x) = \sum_{1 \leqslant i < j \leqslant n} \frac{1}{\| x_i-x_j\|} \qquad \text{ subject to } \| x_i\|^2 = 1, \forall i \in \{1, \dots, n\}.$$

I was able to solve the $4$ points case by looking at this answered question, which seems to answer this question indirectly. In fact, it is shown that the optimal configuration must only be a tetraedron, which can't be represented by equally distant points in the equator line.

For more points, I started by thinking about how we can distribut equally distant points in the equator line. After talking to the professor who gave me this task, we came to the conclusion that "equally distant" isn't the most appropriate word for what we wish. To get a sense of what I am talking, for $4$ points, one possible distribution of the so called "equally distant" points is

After some thinking process about this, one comes to the conlcusion that for arbitrary $N$ points, this equally distributed points can be described by $$ x_{k+1} = \left( \cos\left( \frac{2\pi}{N}k \right), \sin\left( \frac{2\pi}{N}k \right), 0 \right) , \quad \text{ for } k = 0,\dots,N-1.$$

Note that every other possibility is just obtained by rotating this points. But from here I don't know how to proceed. I have tried to think about lifting one of the points while keeping the other ones fixed, but it seems to me (after some calculations) that this would make the funcion value to increase instead of decreasing.

I would apreciatte any hints or help regarding this topic.

Thanks for any help in advance.

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UPDATE.

After some computations, I was able to find the distance $\| x_i - x_j \|,$ for every pair $(i,j)$ (recall the vectors $x_{k+1}$ defined above). This yileds the objective function value with the equally distant points in the equator. The formula I derived for the distance is:

$$ \| x_i - x_j \| = \sqrt 2 \sqrt{1-\cos\left( \frac{2\pi}{N}(j-i) \right)}. $$

This leads the objective function (calculated in the points $x_{k+1}$) to be

$$ \frac{\sqrt 2}{2} \sum_{1 \leqslant i < j \leqslant N} \frac{1}{\sqrt{1-\cos\left( \frac{2\pi}{N}(j-i) \right)}}.$$

I don't know how to proceed after this thought.

FURTHER UPDATE.

Based on the latest update, I was able to "simplify" the expression a little bit more, removing the indexes $i$ and $j$. To do this, I counted exactly how many times the difference $j-i = k$ would occur for arbitrary $k \in [1,N-1]$ (turns out it is $N-k$). By doing this, the value of the objective function (calculated in the point $x_{k+1}$) becomes just

$$ U(\mathbf x) = \frac{\sqrt 2}{2} \sum_{1 \leqslant i < j \leqslant N} \frac{1}{\sqrt{1-\cos \left( \frac{2\pi}{N}(j-i) \right)}} = \frac{\sqrt 2}{2} \sum_{k=1}^{N-1} \frac{N-k}{\sqrt{1-\cos \left( \frac{2\pi}{N}k \right)}}.$$

Again, I will keep thinking how to proceed from here.

Answer 1

We only need to consider $n\ge5$.

Let us move each of the $n$ "equidistant points" on the equator slightly towards one of the two poles of the globe, so that after such a movement the points on the sphere become \begin{equation*} x_k(t):=\Big(\sqrt{1-t^2 z_k^2}\,\cos\frac{ 2\pi k}{n}, \sqrt{1-t^2 z_k^2}\,\sin\frac{2\pi k}{n},t z_k\Big), \end{equation*} where (say) $k=0,\dots,n-1$, the $z_k$'s are real numbers (to be chosen later), and $t\ge0$ is small. So, at $t=0$ we get the original $n$ "equidistant points" on the equator line.

The corresponding potential energy is \begin{equation*} u(t):=\sum_{0\le i<j\le n-1}\frac1{\|x_i(t)-x_j(t)\|}, \end{equation*} so that $u(0)$ is the initial potential energy, of $n$ "equidistant points" on the equator. Obviously, $u'(0)=0$. So, consider \begin{equation*} u''(0)=\sum_{0\le i<j\le n-1} \frac{2 z_i z_j-(z_i^2+z_j^2) \cos \dfrac{2 \pi(i-j)}{n}} {2^{3/2}\Big(1- \cos \dfrac{2 \pi(i-j)}{n}\Big)^{3/2}}. \end{equation*}

We want to choose the $z_k$'s so that $u''(0)<0$. It appears that the alternating choice $z_k:=(-1)^k$ for all $k$ will do.

However, it is somewhat more convenient to choose the $z_k$'s at random. Namely, let the $z_k$'s be independent Rademacher random variables, so that $P(z_k=\pm1)=1/2$ for each $k$. Then, after taking the expectation, the term $z_i z_j$ disappears: \begin{equation*} E u''(0)=-\frac{s_n}{\sqrt2}, \end{equation*} where \begin{equation*} s_n:=\sum_{0\le i<j\le n-1}a_{j-i}=\sum_{1\le k\le n-1}(n-k)a_k, \end{equation*} \begin{equation*} a_k:=a_{n,k}:=A(k/n),\quad A(x):=\frac{\cos 2 \pi x} {(1- \cos 2 \pi x)^{3/2}}. \end{equation*}

So, it suffices to show that $s_n>0$ for $n\ge5$.

Using the substitution $y:=\cos 2 \pi x$, we see that for all $x\in(0,1)$ \begin{equation*} A(x)\ge a_{\min}:=-\frac1{2\sqrt2} \end{equation*} Also, if an integer $k$ is such that $1\le k\le \lfloor n/8\rfloor=:k_n$, then $\cos2\pi\frac kn\ge\cos\frac\pi4=1/\sqrt2$ and hence $a_k\ge a_*:=\dfrac{1/\sqrt2} {(1- 1/\sqrt2)^{3/2}}$. So, \begin{equation*} \begin{aligned} s_n&\ge a_*\sum_{1\le k\le k_n}(n-k)+a_{\min}\sum_{1\le k\le n-1}(n-k) \\ &=\frac{(2n-1)k_n-k_n^2}{2 \sqrt{2} \left(1-\frac{1}{\sqrt{2}}\right)^{3/2}}-\frac{n^2-n}{4 \sqrt{2}} \\ &>\frac{(2n-1)(n/8-1)-(n/8-1)^2}{2 \sqrt{2} \left(1-\frac{1}{\sqrt{2}}\right)^{3/2}}-\frac{n^2-n}{4 \sqrt{2}} \\ &>-4.006 n + 0.345 n^2 >0 \end{aligned} \end{equation*} for $n\ge12$. That $s_n>0$ for $n=5,\dots,11$ is easy to check. $\quad\Box$