Showing the self energies of $N$ uniformly charged disks is proportional to $N^{3/2}$

Question

How would I go about doing this?

I assume it is some integral I have to solve, but I have no idea what.

(Note:Not a physicist so please excuse incompetence with regard standard notation.)

Context is I want to estime the energy of N point particles spread over the unit sphere. This is an equation of the for $E(N)=N^2/2 - aN^{3/2}$. I know the $N^2/2$ comes from the uniform charge density on the sphere, and have been told that the $N^{3/2}$ comes from "recovering the energy of distribution of point charges, therefore subtracting the self energies of a et of N uniformly charges disks, which can be shown to be proportional to $N^{3/2}$''.

I get what it is doing, but I don't know how to show it is proportional to $N^{3/2}$. I have already managed to show the $N^2/2$ term though. Also I don't need to find a, I just need to show the power.

Answer 1

This sound likes a question on estimating the large $N$ behavior of minimal energy in Thomson's Problem. The $N^{\frac32}$ dependence you are asking really come from the $\frac{1}{\sqrt{N}}$ dependence of the radius of the uniformly charged disk.

Suppose we have placed $N$ unit point charges as uniform as possible on the unit sphere. To compute the total energy, we pick one of the charge and asks what are the contribution from the remaining $N-1$ charges.

To first approximation, we approximate the remaining $N-1$ charges distributed continuous and uniformly over the whole sphere. We know in term of energy, this is equivalent to the configuration where all the $N-1$ charges are concentrated in the center. This give us an estimate of self energy:

$$\frac12 N (N-1)$$

To improve the approximation, we need to ask where the error comes from. Between the continuous and discrete configuration, the biggest difference is in the discrete configuration, two point charge is separated by a finite distance while in the continuous case, the $N-1$ charge can get as close as it like to the test charge.

So for second approximation, we subdivide the unit sphere into $N$ disks and one charge for each disk. For the test charge, we place it in the center of its own disk. For the remaining $N-1$ charge, we assume they are uniformly distributed in their own disks.

In this configuration, the charge distribution is equivalent to have $N$ (instead of $N-1$) charges distributed continuous and uniformly over the sphere "minus" a uniformly distributed unit charged disk centered at the test charge.

Let $R$ be the radius of the unit charged disk. Since we have $N$ such disk, we could fix $R$ by demanding:

$$N ( \pi R^2 ) = 4 \pi \implies R = \frac{2}{\sqrt{N}}$$

The contribution from the uniformly distributed unit charge disk we need to subtract out becomes:

$$\frac{1}{\pi R^2} \int_0^{R} \frac{1}{r} rdrd\theta = \frac{2}{R} = \sqrt{N}$$

and to second approximation, the self energy becomes

$$\frac12 N ( N - \sqrt{N} ) = \frac12 N^2 - \frac12 N^{\frac32}$$

As we have mentioned before, the $N^{\frac32}$ dependence come from the $\frac{1}{\sqrt{N}}$ dependence of $R$. Furthermore, $a \sim \frac12$ in this approximation.

To justify whether this approximation make sense. One can grab the smallest known energy of placing $N$ point charges on sphere from the wiki page of Thomson Problem and compare it against the formula $\frac12 N^2 - a N^{\frac32}$. I have done that and it give me an estimate for $a \sim 0.55$. Not bad for such a simple approximation.

If you want to have a better estimate of $a$ for large $N$, I believe you can replace the uniformly charged disks by organizing the charges locally in a triangular lattice. However, I don't know the actual reference that carry out this computation.

Answer 2

I think the logic is as follows, from what I understand, though it's a little dodgy.

We approximate the self-energy of $N$ uniformly spaced point particles on a fixed sphere by a uniform distribution for $N$ large. Since we know $E$ goes like $Q^2$ for $Q$ the total charge here, we get your leading order contribution.

The error we made is roughly that of smearing out each of the $N$ charges over a small area, and including this self-energy in our computation. So we expect to have to take away $N\times$ the energy of each region. Since the total surface area is fixed, the area per particle goes like $1/N$ and the radius like $N^{-1/2}$.

The self-energy of a disk of charge density $\sigma$ and radius $R$ goes like $\sigma^2 R^3$. (The potential can be found as $V$ going like $\sigma R$, so $\int_0^R \sigma V r\mathrm d r$ goes like $\sigma^2 R^3$.) Since $\sigma\propto 1/R^2$ the energy goes like $1/R \propto N^{1/2}$.

Then the fact that there are $N$ such disks gives the $N^{3/2}$ answer you wanted.