The Thomson problem

The Thomson problem

Definitions and mathematical statement

For $N\ge1$, choose pairwise distinct points $x_1,\ldots,x_N$ on the unit sphere $S^2\subset{\mathbb{R}}^3$. In units with equal unit charges and Coulomb constant one, the interaction energy is \[ E_N(x_1,\ldots,x_N)=\sum_{1\le i<j\le N}\frac1{\|x_i-x_j\|_2}. \] The Thomson problem asks to determine minimizing configurations for every $N$, up to rigid motions and permutation of the charges. Distances are Euclidean chord distances, not spherical arc lengths. Coincident charges have infinite energy. Existence of a minimizer or a numerical configuration with low energy does not give its rigorous global characterization; degeneracy of minimizers is allowed, and no uniqueness claim is built into the question.

Formulation and concept references: [S1].

Short English statement

For every number of equal charges, which arrangements on the unit sphere minimize their total Coulomb repulsion?

Research attempt

Scope and method.

The inherited source [S1] concerns the Coulomb problem and the difficulty of identifying global minimizing configurations. A low numerical value is only an upper bound on the minimum. The present attempt supplies exact certificates for $N\leq4$ and $N=6$, a first- and second-variation test, and a global leading-order estimate. These do not classify the minimizing arrangements for arbitrary $N$. All distances below are chord distances, as in the selected statement.

Existence without an accidental uniqueness claim.

Extend the energy to $(S^2)^N$ by the value $+\infty$ at collisions. Each pair potential is lower semicontinuous, and so is their finite sum. The product of spheres is compact, hence the extended energy attains its minimum. There are distinct configurations with finite energy, so a minimizer contains no collision. For $N=1$ the minimum is zero and every point is equivalent under rotation.

If a finite-energy configuration has energy at most $U$, every term in its positive energy sum is at most $U$. Thus \[ |x_i-x_j|\geq U^{-1}\quad(i\ne j). \tag{391.1} \] This elementary separation bound is sufficient to put a bounded-energy search in a compact collision-free region. It is generally too weak to identify an optimal configuration, but prevents a numerical search from being justified using a compactness statement that ignores collisions.

A centroid bound sharp for the small simplexes.

Let $M=N(N-1)/2$ and $s=\sum_i x_i$. Direct expansion yields \[ \sum_{i<j}|x_i-x_j|^2=N\sum_i|x_i|^2-|s|^2=N^2-|s|^2\leq N^2. \tag{391.2} \] The function $r\mapsto r^{-1/2}$ is strictly convex and decreasing on $(0,\infty)$. Jensen's inequality and $391.2$ therefore give \[ E_N\geq M\left(\frac1M\sum_{i<j}|x_i-x_j|^2\right)^{-1/2} \geq\frac{M^{3/2}}N. \tag{391.3} \] Equality requires both $s=0$ and equal squared distances. For $N=2,3,4$, these requirements are realized by an antipodal pair, a regular equilateral triangle on a great circle, and a regular tetrahedron. Their exact energies are respectively \[ \frac12,\qquad\sqrt3,\qquad 3\sqrt{3/2}. \tag{391.4} \] This also proves the characterization in these cases: the Gram matrix has diagonal one and off-diagonal $-1/(N-1)$, and equal Gram matrices are related by an orthogonal map on their linear spans, extending to an orthogonal map of $\mathbb R^3$.

For $N>4$ the equality Gram matrix has rank $N-1>3$, so equality in $391.3$ is impossible. This is an explicit obstruction to extrapolating the regular-simplex proof: the desired equidistant point set does not fit in the ambient space. A strict inequality without a quantitative gap gives no replacement characterization.

A quadratic certificate proving the octahedron case.

Write $t=x_i\cdot x_j$ and $f(t)=(2-2t)^{-1/2}$ for $-1\leq t<1$. Consider the polynomial \[ P(t)=a+bt+ct^2,\qquad a=\frac1{\sqrt2},\quad b=\frac1{2\sqrt2},\quad c=\frac12-\frac1{2\sqrt2}>0. \tag{391.5} \] It has $P(-1)=f(-1)$, $P(0)=f(0)$ and $P'(0)=f'(0)$. In fact \[ P(t)\leq f(t)\quad(-1\leq t<1), \quad\hbox{with equality only at }t=-1,0. \tag{391.6} \] Here is a direct remainder proof. For a fixed $t$ distinct from the two interpolation nodes, subtract a suitable multiple $A(s+1)s^2$ from $f(s)-P(s)$ to make it vanish also at $s=t$. There are four zeros counting the double zero at zero, so repeated Rolle's theorem gives $6A=f'''(\xi)>0$ for an intermediate $\xi$. Therefore $f(t)-P(t)=A(t+1)t^2>0$. At the nodes equality is already built in. This proof works also for $-1<t<0$; the sign of the remainder is not guessed from a graph.

Set $Q=\sum_i x_ix_i^{\mathsf T}$. Then $\operatorname{tr}Q=N$, and \[ \sum_{i<j}t_{ij}=\tfrac12(|s|^2-N),\qquad \sum_{i<j}t_{ij}^2=\tfrac12(\operatorname{tr}(Q^2)-N). \] Since the three eigenvalues of $Q$ are nonnegative, Cauchy--Schwarz gives $\operatorname{tr}(Q^2)\geq N^2/3$. Combining these identities with $b,c>0$ proves the general lower bound \[ E_N\geq aM-\frac{bN}{2}+ \frac c2\left(\frac{N^2}{3}-N\right). \tag{391.7} \] At $N=6$ the right side is $6\sqrt2+3/2$. The six vertices $\{\pm e_1,\pm e_2,\pm e_3\}$ attain it: there are three antipodal pairs of distance two and twelve remaining pairs of distance $\sqrt2$. Thus the octahedron is a global minimizer with this exact energy.

The equality conditions characterize it. Equality in $391.6$ forces every inner product between distinct vertices to be zero or minus one. Distinct unoriented lines represented by vertices must therefore be mutually orthogonal, so there are at most three such lines. Each line contains at most two unit points. Six points force exactly three orthogonal lines with both signs present. This is the octahedron up to rigid motion and relabeling. The argument uses positivity of the quadratic coefficients to certify a global bound; it is stronger than checking a Hessian at the candidate.

What stationarity and the Hessian actually test.

For a collision-free critical configuration, tangential projection of the ordinary gradient must vanish: \[ (I-x_ix_i^{\mathsf T}) \sum_{j\ne i}\frac{x_i-x_j}{|x_i-x_j|^3}=0 \quad(1\leq i\leq N). \tag{391.8} \] The minus sign in the gradient is immaterial after setting the projection to zero. Symmetry often proves $391.8$, but it does not compare energies of different critical configurations.

Let $v_i\perp x_i$ and move each point on its great-circle geodesic with initial velocity $v_i$. At zero time its acceleration is $-|v_i|^2x_i$. With $a_{ij}=x_i-x_j$, $b_{ij}=v_i-v_j$ and $d_{ij}=|a_{ij}|$, differentiation gives the second variation \[ \sum_{i<j}\left[ \frac{3(a_{ij}\cdot b_{ij})^2}{d_{ij}^5} -\frac{|b_{ij}|^2-(1-x_i\cdot x_j)(|v_i|^2+|v_j|^2)}{d_{ij}^3} \right]. \tag{391.9} \] To verify the curvature correction, differentiate $|x_i(t)-x_j(t)|^2$ twice: its second derivative is $2|b_{ij}|^2-2(1-x_i\cdot x_j)(|v_i|^2+|v_j|^2)$. Applying the chain rule to the inverse square root gives $391.9$. At a local minimum this quadratic form is nonnegative. Global rotations produce zero directions, which must not be interpreted as instability. Conversely, a negative direction disproves local minimality. Nonnegativity alone proves neither uniqueness nor global minimality, and a floating-point nearly zero eigenvalue is not an exact certificate of either sign.

A uniform upper bound from random configurations.

Fix one point $x$ on the sphere and let $Y$ be uniform on $S^2$. The inner product $t=x\cdot Y$ is uniform on $[-1,1]$. Therefore \[ \mathbb E\frac1{|x-Y|} =\frac12\int_{-1}^1(2-2t)^{-1/2}\,dt=1. \] For $N$ independent uniform points, collisions have probability zero and $\mathbb E E_N=M$. Hence some configuration has energy at most $M$, and the minimum satisfies $E_N^{\min}\leq N(N-1)/2$. Expectation is used only for existence; it gives no explicit minimizing configuration.

A matching leading-order lower bound by smearing charges.

We give the positivity detail needed in an electrostatic lower bound. For finite signed compactly supported measures of finite Newton energy, the kernel $1/|x-y|$ is positive definite. One verification uses \[ \frac1r=\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-tr^2}\,dt. \] Each Gaussian kernel is positive definite, by its positive Fourier transform, so its quadratic energy is nonnegative. Truncating the $t$ integral and then passing to the limit proves the assertion; finite absolute energy justifies the limit for the signed measures used here. Equivalently one can first regularize them smoothly.

Let $\sigma_{x,\varepsilon}$ be probability surface measure on the radius-$\varepsilon$ sphere centered at $x$, and put $\nu=\sum_i\sigma_{x_i,\varepsilon}$. A direct angular integral gives the shell potential \[ \int\frac{d\sigma_{x,\varepsilon}(y)}{|z-y|} =\frac1{\max\{|z-x|,\varepsilon\}}. \tag{391.10} \] The self-energy of each shell is $1/\varepsilon$. Its potential is at most the point-charge potential. Averaging that inequality over a second shell and using $391.10$ again shows that the mutual energy of two shells is at most $1/|x_i-x_j|$, even if the shells overlap. Consequently \[ I(\nu):=\iint\frac{d\nu(x)d\nu(y)}{|x-y|} \leq 2E_N+N/\varepsilon. \tag{391.11} \] All of $\nu$ is supported in the radius-$R$ ball with $R=1+\varepsilon$. Let $\sigma_R$ be uniform probability measure on its boundary sphere. Its potential equals $1/R$ throughout that ball. Positivity of the energy of $\nu-N\sigma_R$ therefore yields \[ 0\leq I(\nu-N\sigma_R)=I(\nu)-N^2/R. \] Together with $391.11$, this gives \[ E_N\geq\frac{N^2}{2(1+\varepsilon)}-\frac N{2\varepsilon}. \tag{391.12} \] Choose $\varepsilon=N^{-1/2}$ and use $(1+\varepsilon)^{-1}\geq1-\varepsilon$. We obtain the entirely uniform bounds \[ \frac{N^2}{2}-N^{3/2}\leq E_N^{\min}\leq\frac{N(N-1)}2. \tag{391.13} \] In particular $E_N^{\min}/N^2\to1/2$. These bounds leave a large configuration-sensitive next-order term undetermined. Leading asymptotics alone cannot discriminate between competing finite arrangements whose energies differ far below order $N^2$.

Diagnostics, certificates and the first gap.

The accompanying checks compare floating-point energies for the small candidates with the derived formulas, numerically check the polynomial contacts and sampled minorant values, and compare the second-variation formula with finite differences away from collisions. The analytic inequality $391.6$, rather than a sampled graph, certifies the global lower bound for $N=6$.

One possible route for another $N$ is a minorizing polynomial with nonnegative harmonic coefficients and contacts at all inner products of a proposed configuration. Such a polynomial would need a verified inequality on the entire interval $[-1,1)$ and equality in every resulting moment bound. There is no reason that an arbitrary candidate admits a certificate of this particular form. Failure to find one does not establish that the candidate is wrong, and local numerical optimization does not replace one.

The results above settle $N=1,2,3,4,6$ within this attempt and give uniform leading asymptotics. The first gap toward the catalogue question is a rigorous global comparison for general configurations at every remaining $N$. No such classification follows from the centroid, Hessian or smearing bounds. The all-$N$ problem remains unresolved by this attempt.

Sources

• S1 Altschuler et al., The Thomson Problem, Physical Review Letters author manuscript page $1997$. https://web.cels.anl.gov/~zippy/publications/thomson/thomsonPRL.html. Formulation source.