Multipole Expansion: Electrostatics

Question

Why in multipole expansion (or the terms therein) goes as mono-, di-, quadru-, octu-, or more specifically why are they in powers of 2? Why can't we have hexapole for instance?

Answer 1

1. When one solves Laplace's equation $$0~=~\nabla^2\Phi~=~\frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial \Phi}{\partial r} - \frac{1}{r^2}L^2\Phi $$in spherical coordinates in the bulk (away from a central region with sources), it separates in an angular problem on the $2$-sphere $S^2$ and a radial problem.

2. Since the rotation group $SO(3)$ acts on the $2$-sphere $S^2$, the angular solutions are representations of the Lie group $SO(3)$, namely linear combinations of the spherical harmonics. All finite-dimensional irreducible representations $V_{\ell}$ of the Lie group $SO(3)$ are characterized by an integer spin $\ell\in\mathbb{N}_0$, which are related to the $2^{\ell}$-pole term. Here the Casimir $L^2$ has eigenvalue $\ell(\ell+1)$. The irrep $V_{\ell}\equiv\underline{\bf 2\ell\!+\!1}$ has dimension $2\ell\!+\!1$. E.g. $\ell=0$ is a monopole, $\ell=1$ is a dipole, $\ell=2$ is a quadrupole, and so forth.

3. The main point is that potentials for any number of charges can be classified according to above scheme. The corresponding radial $2^{\ell}$-pole solution falls off as $ r^{-(\ell+1)}$. See also related Phys.SE posts here and here.

Answer 2

Think about what it takes to construct each multipole with point charges that all have the same magnitude.

1. A monopole takes one point charge.
2. The three basic dipoles take two - one positive, one negative, oriented in a combination of the three orthogonal directions.
3. The five basic quadrupoles are all constructed from various combinations of two dipoles (four monopoles).
4. The seven basic octupoles are all constructed from two quadruoples ($8$ monopoles).
5. etc.

In other words, in order to construct a multipole of a given order you need to arrange the charges in such a way as to cancel all of the lower order multipoles. Thus the easiest way to construct a multipole of order $n$ is to take a multipole of order $n-1$, copy it, offset the copy along an axis of (a)symmetry, then flip all of the charges of the copy. That is, put a $2^{n-1}$-pole next to a $2^{n-1}$-pole in such a way that the $2^{n-1}$-pole of the combination is zero. That requires $2^{n-1} + 2^{n-1}=2^n$ monopoles of equal charge magnitude.

For an explanation of the number of basic types of each multipole, see the second item in @Qmechanic's answer.