a charge distribution with rotational symmetry - schematic.In rotational symmetry, the multipole moments of a charge distribution are described by just one component per order \(l\). This is a drastic difference to the usual \(2l+1\) independent components. Get to know this simplifications with two examples and derive it for any charge distribution with rotational symmetry!

Problem Statement

A slab superconductor with a constant magnetic field.A charge distribution shall obey rotational symmetry with respect to the \(z\) axis. Proof that its multipole moments are entirely given by its components with respect to the axis of symmetry. So, the dipole moment just depends on its \(z\) component \(p_{z}\), the quadrupole moment on \(Q_{zz}\). and so on. Let us approach this very useful conclusion in two steps:

1. Examples:

a) Show that the dipole moment is given by \(\mathbf{p}=p_{z}\mathbf{e}_{z}\).
b) Calculate that the cartesian quadrupole tensor is diagonal and that \(Q_{xx}=Q_{yy}=-\frac{1}{2}Q_{zz}\) holds.

2. Proof of the general conclusion:

a) Derive that the general solution of the electrostatic potential outside of the charge distribution is given by\[\phi\left(r,\theta\right)    =    \frac{1}{4\pi\varepsilon_{0}}\sum_{l=0}^{\infty}\frac{b_{l}P_{l}\left(\cos\theta\right)}{r^{l+1}}\]using a separation of variables in spherical coordinates. Note that the so-called Legendre Polynomials \(P_{l}\left(\cos\theta\right)\) are given by the differential equation \[\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}P_{l}\left(\cos\theta\right)+l\left(l+1\right)P_{l}\left(\cos\theta\right)    =    0\ ,\] where only a solution exists if \(l\in\mathbb{N}_{0}=\left\{ 0,\ 1,\ 2,\ \dots\right\}\).
b) Use \(P_{l}\left(\cos\theta=1\right)=1\ \forall l\) and \[\frac{1}{\sqrt{1-2\xi\cos\theta+\xi^{2}}}    \equiv    \sum_{l=0}^{\infty}P_{l}\left(\cos\theta\right)\xi^{l}\]to show that the multipole moments are given by\[b_{l}    =    \int P_{l}\left(\cos\theta^{\prime}\right)r^{\prime l}\rho\left(r^{\prime},\theta^{\prime}\right)dV^{\prime}\ .\]

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