Multipole Expansion

Consider a bounded charge distribution that lies inside the sphere $r=a$ . It follows that $\rho=0$ in the region $r>a$ . According to the previous three equations, the electrostatic potential in the region $r>a$ takes the form

$$\phi({\bf r}) = \frac{1}{\epsilon_0}\sum_{l=0,\infty}\sum_{m=-l,+l} \frac{q_{l,m}^{\,\ast}}{2\,l+1}\,\frac{Y_{l,m}(\theta,\varphi)}{r^{\,l+1}},$$ (339)

where the

$$q_{l,m}^{\,\ast}= \int r^{\,l}\,\rho(r,\theta,\varphi)\,Y_{l,m}(\theta,\varphi)\,dV$$ (340)

are known as the multipole moments of the charge distribution $\rho({\bf r})$ . Here, the integral is over all space. Incidentally, the type of expansion specified in Equation (340) is called a multipole expansion.

The most important $q_{l,m}^{\,\ast}$ are those corresponding to $l=0$ , $l=1$ , and $l=2$ , which are known as monopole, dipole, and quadrupole moments, respectively. For each $l$ , the multipole moments $q_{l,m}^{\,\ast}$ , for $m=-l$ to $+l$ , form an $l$ th-rank tensor with $2\,l+1$ components. However, Equation (310) implies that

$$q_{l,m}^{\,\ast} = (-1)^m\,q_{l,-m}.$$ (341)

Hence, only $l+1$ of these components are independent.

For $l=0$ , there is only one monopole moment. Namely,

$$q_{0,0}^{\,\ast}=\int \rho({\bf r}')\,Y_{0,0}^{\,\ast}(\theta,\varphi)\,dV = \frac{1}{\sqrt{4\pi}}\int\rho({\bf r})\,dV = \frac{Q}{\sqrt{4\pi}},$$ (342)

where $Q$ is the net charge contained in the distribution, and use has been made of Equation (312). It follows from Equation (340) that, at sufficiently large $r$ , the charge distribution acts like a point charge $Q$ situated at the origin. That is,

$$\phi({\bf r})\simeq \phi_0({\bf r})=\frac{q_{0,0}^{\,\ast}}{\epsilon_0}\,\frac{Y_{0,0}(\theta,\varphi)}{r} = \frac{Q}{4\pi\,\epsilon_0\,r}.$$ (343)

By analogy with Equation (195), the dipole moment of the charge distribution is written

$${\bf p} = \int \rho({\bf r})\,{\bf r}\,dV.$$ (344)

The three Cartesian components of this vector are

$$p_x = \int \rho({\bf r})\,x\,dV= \int\rho({\bf r})\,r\,\sin\theta\,\cos\varphi\,dV,$$ (345)

$$p_y = \int \rho({\bf r})\,y\,dV = \int\rho({\bf r})\,r\,\sin\theta\,\sin\varphi\,dV,$$ (346)

$$p_z = \int \rho({\bf r})\,z\,dV = \int\rho({\bf r})\,r\,\cos\theta\,dV.$$ (347)

On the other hand, the spherical components of the dipole moment take the form

$$q_{1,-1}^{\,\ast} =\left(\frac{3}{8\pi}\right)^{1/2}\int \rho({\bf r})\,r\,\sin\the... ...+{\rm i}\,\varphi}\,dV = \left(\frac{3}{8\pi}\right)^{1/2}\,(p_x+{\rm i}\,p_y),$$ (348)

$$q_{1,0}^{\,\ast} = \left(\frac{3}{4\pi}\right)^{1/2}\int\rho({\bf r})\,r\,\cos\theta\,dV = \left(\frac{3}{4\pi}\right)^{1/2}p_z,$$ (349)

$$q_{1,+1}^{\,\ast} =-\left(\frac{3}{8\pi}\right)^{1/2}\int \rho({\bf r})\,r\,\sin\th... ...{\rm i}\,\varphi}\,dV = -\left(\frac{3}{8\pi}\right)^{1/2}\,(p_x-{\rm i}\,p_y),$$ (350)

where use has been made of Equations (313)-(315). It can be seen that the three spherical dipole moments are independent linear combinations of the three Cartesian moments. The potential associated with the dipole moment is

$$\phi_1({\bf r}) = \frac{1}{3\,\epsilon_0}\left(\frac{q_{1,-1}^{\,... ...}+q_{1,0}^{\,\ast}\,r\,Y_{1,0}+q_{1,+1}^{\,\ast}\,r\,Y_{1,+1}}{r^{\,3}}\right).$$ (351)

However, from Equations (313)-(315),

$$r\,Y_{1,-1} =\left(\frac{3}{8\pi}\right)^{1/2}\,(x-{\rm i}\,y),$$ (352)

$$r\,Y_{1,0} = \left(\frac{3}{4\pi}\right)^{1/2}z,$$ (353)

$$r\,Y_{1,+1} = -\left(\frac{3}{8\pi}\right)^{1/2}\,(x+{\rm i}\,y).$$ (354)

Hence,

$$\phi_1({\bf r})= \frac{1}{4\pi\,\epsilon_0}\,\frac{p_x\,x+p_y\,y+p_z\,z}{r^{\,3}}= \frac{1}{4\pi\,\epsilon_0}\,\frac{{\bf p}\cdot{\bf r}}{r^{\,3}},$$ (355)

in accordance with Equation (200). Note, finally, that if the net charge, $Q$ , contained in the distributions is non-zero then it is always possible to choose the origin of the coordinate system in such a manner that ${\bf p}={\bf0}$ .

The Cartesian components of the quadrupole tensor are defined

$$Q_{ij} = \int \rho({\bf r})\,(3\,x_i\,x_j-r^{\,2}\,\delta_{ij})\,dV,$$ (356)

for $i$ , $j=1$ , $2$ , $3$ . Here, $x_1=x$ , $x_2=y$ , and $x_3=z$ . Incidentally, because the quadrupole tensor is symmetric (i.e., $Q_{ji}=Q_{ij}$ ) and traceless (i.e., $Q_{11}+Q_{22}+Q_{33}=0$ ), it only possesses five independent Cartesian components. The five spherical components of the quadrupole tensor take the form

$$q_{2,-2}^{\,\ast} =\left(\frac{5}{96\pi}\right)^{1/2}\,(Q_{11}+2\,{\rm i}\,Q_{12}-Q_{22}),$$ (357)

$$q_{2,-1}^{\,\ast} =\left(\frac{5}{24\pi}\right)^{1/2}\,(Q_{13}+{\rm i}\,Q_{23}),$$ (358)

$$q_{2,0}^{\,\ast} =\left(\frac{5}{16\pi}\right)^{1/2}\,Q_{33},$$ (359)

$$q_{2,+1}^{\,\ast} =-\left(\frac{5}{24\pi}\right)^{1/2}\,(Q_{13}-{\rm i}\,Q_{23}),$$ (360)

$$q_{2,+2}^{\,\ast} =\left(\frac{5}{96\pi}\right)^{1/2}\,(Q_{11}-2\,{\rm i}\,Q_{12}-Q_{22}).$$ (361)

Moreover, the potential associated with the quadrupole tensor is

$$\phi_2({\bf r})= \frac{1}{5\,\epsilon_0}\sum_{m=-2,+2} \frac{q_{2... ...3}}=\frac{1}{8\pi\,\epsilon_0}\sum_{i,j=1,3}\frac{Q_{ij}\,\,x_i\,x_j}{r^{\,5}}.$$ (362)

It follows, from the previous analysis, that the first three terms in the multipole expansion, (340), can be written

$$\phi({\bf r})\simeq \phi_0({\bf r})+\phi_1({\bf r})+\phi_2({\bf r... ..._0\,r^{\,3}}+ \sum_{i,j=1,3}\frac{Q_{ij}\,x_i\,x_j}{8\pi\,\epsilon_0\,r^{\,5}}.$$ (363)

Moreover, at sufficiently large $r$ , these are always the dominant terms in the expansion.