Minimize Energy for Charge Distributions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:21:26Z http://mathoverflow.net/feeds/question/80731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80731/minimize-energy-for-charge-distributions Minimize Energy for Charge Distributions Chris Gerig 2011-11-12T04:17:45Z 2011-11-13T15:56:12Z <p>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 ?).</p> <p>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 $W=\int_M \frac{\rho(x)\rho(x')}{|x-x'|}dx\ dx'$.</p> <p><strong>What charge distributions minimize the energy for a given region?</strong></p> <p>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 $W=\frac{1}{n}\sum_iV(x_i)$.</p> <p>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 $W$.</p> <p>Is it true/obvious that $\rho=\frac{1}{4\pi}$ on $M=S^2$ ? (yes, see Henry Cohn's answer below)</p> <p>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 <em>disk</em> is not zero on the interior. And charge on a conducting <em>needle</em> 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.</p> <p>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.</p> http://mathoverflow.net/questions/80731/minimize-energy-for-charge-distributions/80732#80732 Answer by Henry Cohn for Minimize Energy for Charge Distributions Henry Cohn 2011-11-12T04:54:49Z 2011-11-13T15:56:12Z <p>The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit. </p> <p>It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See <a href="http://www.ams.org/notices/200410/fea-saff.pdf" rel="nofollow">http://www.ams.org/notices/200410/fea-saff.pdf</a> 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: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://cs.smith.edu/~orourke/MathOverflow/Fig5TorusNearOpt.jpg" alt="Torus Fig 5"></p> <p>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 $s &lt; 2$ the particles converge to the continuous distribution that minimizes energy. When $s&lt;1$, this distribution is not even supported on the entire torus.</p> <p>You don't see these phenomena for the sphere, because of its symmetry, but they are typical for less symmetric manifolds.</p> <p>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.</p> <p>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.</p>