Minimize Energy for Charge Distributions

Question

I am considering [positive] charge distributions $\rho:M\rightarrow\mathbb{R}_+$ (nonnegative reals) with unit charge $\int_M\rho=1$ for convenience. Here $M$ is a nice-enough region, say a submanifold of $\mathbb{R}^n$ (or perhaps simply a metric space ?).

The electrostatic energy is (up to constant factor) $W=\int_M\rho V$, where $V(r)=\frac{\rho(r)}{r}$ is the potential at distance $r$. We can rewrite this $W=\int_M \frac{\rho(x)\rho(x')}{|x-x'|}dx\ dx'$.

What charge distributions minimize the energy for a given region?

In dimension $n=3$, for a bounded region $M\subset\mathbb{R}^3$, charges are being placed on a conductor (which by definition is an object where charges are free to move). Charges seek minimum potential $V$, and end up moving out and lieing on the boundary $\partial M$. One way to see this is via Earnshaw's theorem, which says that there is no stable equilibrium for a collection of charges acted upon only by electrostatic forces (hence the minimum is attained on a boundary, which provides a normal force). A minimum potential (for a collection of charges) corresponds to a minimum energy $W$, since general $\rho=\frac{1}{n}\sum^n_{i=1}\delta(x-x_i)$ gives $W=\frac{1}{n}\sum_iV(x_i)$.

So in this case I know that any $\rho$ will be zero on the interior of $M$... but how does $\rho$ behave on $\partial M$? I seem to have reduced the problem to compact manifolds without boundary. A brief electric-field argument shows that the boundary is actually at an equipotential $V_0$, which gives the minimum $W$.

Is it true/obvious that $\rho=\frac{1}{4\pi}$ on $M=S^2$ ? (yes, see Henry Cohn's answer below)

But for dimension $n=1,2$ the charges do NOT all run to the boundary. It is a fact (Am. J. of Phys. 61, 1993 by R. Friedberg) that the distribution on a conducting disk is not zero on the interior. And charge on a conducting needle does not all go to the ends (Am. J. of Phys. 64, 1996 by D. Griffiths). **These are the n-dimensional objects, so that the boundary of a needle is two points and the boundary of a disk is a circle.

Sorry if my questions/thoughts are too broad... I was just randomly proposing a physics problem to myself. I will clarify / be more specific if necessary.

Answer 1

The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit.

It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See http://www.ams.org/notices/200410/fea-saff.pdf for a very nice exposition and further references. For example, Figure 5 from that paper (included here thanks to Joseph O'Rourke) shows the limiting distribution for particles on a torus under an inverse $s$-th power law:
          

In this example, for $s \ge 2$ you get a uniform distribution, which is the default behavior when the energy for a continuous charge distribution diverges. For $s < 2$ the particles converge to the continuous distribution that minimizes energy. When $s<1$, this distribution is not even supported on the entire torus.

You don't see these phenomena for the sphere, because of its symmetry, but they are typical for less symmetric manifolds.

Incidentally, the behavior of 1 and 2 dimensions is not so strange. The charges do indeed end up on the boundary, if one uses a harmonic potential function (for example, a logarithmic potential in $\mathbb{R}^2$). The difficulty is that the Coulomb potential is not harmonic in $\mathbb{R}^1$ or $\mathbb{R}^2$. One way of thinking about it is that if you view the needle or disk as sitting inside $\mathbb{R}^3$, then the charges do all end up on the boundary, because the boundary in $\mathbb{R}^3$ is the entire set.

More generally, in $\mathbb{R}^n$, if you use an inverse $s$-th power law for the potential function, then all the charge will be on the boundary if $s \le n-2$ (because the potential function is superharmonic and therefore satisfies the minimum principle). When $s > n-2$, that does not happen.