Consider some number of charged particles on a closed interval, each with possibly different amounts of charge. Assuming some kind of 'friction' to dampen their motion, they will eventually find a stationary equilibrium, where the repelling force from all the other charges is balanced (except in the case of the particles on the end points, which are also acted on by the 'walls'). It seems intuitively clear that this equilibrium depends only on the final ordering of the particles, but how does one show this? What is it about the interaction potential $\frac{q_aq_b}{d}$ that makes this this solution unique?

I actually care about a slightly more complicated situation, which the previous example was meant to be a simplification of. Consider some number of charged particles on the unit circle in $\mathbb{R}^2$, each with possibly different amounts of charge. However, the force between them behaves a little weird:

  • The repelling force between two particles is not constrained to the unit circle, it cuts across the circle along a chord and tries to push the particle along the straight-line connecting the two points. However, because the particles are constrained to the circle, the effective force is the projection of this vector to the tangent space to the circle.
  • The repelling force is proportional to the inverse of the distance between two particles (along the chord), not the inverse squared. Therefore, the potential is of the form $q_aq_b\ln(d)$.

It is then not hard to show the force exerted on particle $b$ by particle $a$ is proportional to $\cot\left(\frac{\theta_b-\theta_a}{2}\right)$, and the potential is proportional to $\ln\left(\sin\left(\frac{\theta_b-\theta_a}{2}\right)\right)$.

Again, it seems intuitively clear that, for a fixed cyclic ordering of the particles, there should be a unique equilibrium solution with all forces balanced (modulo rotation of the entire picture). However, how would one show this?

It seems like there should be well known tricks for how to do this, but I don't know much about this kind of stuff.


1 Answer 1


Monotonicity is too weak a thing for uniqueness, but strict convexity is enough. If $U(X)$ is your potential on the configuration $X$ and you know that $U(tX+(1-t)Y)<tU(X)+(1-t)U(Y)$ for $0<t<1$ and $X\ne Y$, then there exists a unique stationary point, which is the global minimum. Note that $1/x$ is strictly convex on $x>0$ and so is -$\log \sin x$ on $(0,\pi)$, so this trick takes care of both your questions.

  • $\begingroup$ There might be a little bit more annoyance after this, because the potential U is a sum of a bunch of functions $F(x_a,x_b)$ (where $F$ is the pairwise interaction potential). These functions on configuration space aren't strictly convex, but they are convex, so the function $U$ is merely convex. After modding out by rotation of the entire system, the resulting function SHOULD be strictly convex, but it needs to be checked. Still, I think this is the answer I was looking for. $\endgroup$ Dec 28, 2009 at 21:12
  • $\begingroup$ The check is rather trivial: fix one charge (this mods out the rotations and makes sure that the difference in argument is always between $0$ and $2\pi$, so the half-difference is on the interval $(0,\pi)$ of convexity). If some two pairwise distances in 2 configurations differ, they already give you a strict inequality. Otherwise, all pairwise distances are the same, which can happen only if configurations are identical. $\endgroup$
    – fedja
    Dec 28, 2009 at 22:39

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