Multipole Expansion of the Electrostatic Potential

Abstract

The potential of a localized charge distribution at large distance can be expanded as a series of multipole terms.

Appendix

2.1.1 Higher Order Terms in the Multipole Expansion of the Potential

We have already seen the multipole expansion of the potential limited to the second order. To get the general expression⁵ with all the terms of the expansion, we write the potential at a point $P(\mathbf {r})=P(x,y,z)$ from a continuous charge distribution, limited in space, described by the density $\rho ( \mathbf {r^{\prime }})=\rho (x^{\prime },y^{\prime },z^{\prime })$. The distance of the point P from the elementary volume $d\tau ^{\prime }$ in the point $\mathbf {r^{\prime }}$ is:

$$ \vert \mathbf {r} - \mathbf {r^{\prime }} \vert =\varDelta r=\sqrt{\left( x - x^{\prime } \right) ^2+\left( y - y^{\prime } \right) ^2+\left( z - z^{\prime } \right) ^2}\,. $$

We set the origin of the frame inside the volume of the charge distribution or nearby. For a distance r large compared with the dimensions of the volume, we can expand the distance $\vert \mathbf {r} - \mathbf {r^{\prime }} \vert$ as a Taylor series:

$$ {1 \over {\vert \mathbf {r} - \mathbf {r^{\prime }} \vert }}={1 \over r}+ x^{\prime }_\alpha \left[ {{\partial ~}\over {\partial ~x^{\prime }_\alpha }}\left( {1 \over { \varDelta r}}\right) \right] _{x^{\prime }_\alpha =0} + {1 \over {2!}} x^{\prime }_\alpha x^{\prime }_\beta \left[ {{\partial ^2~}\over {\partial x^{\prime }_\alpha \partial x^{\prime }_\beta }}\left( {1 \over {\varDelta r}} \right) \right] _{x^{\prime }_\alpha =x^{\prime }_\beta =0}+ \cdots $$

$$ \cdots +{1 \over {n!}} x^{\prime }_\alpha x^{\prime }_\beta x^{\prime }_\gamma \ldots \left[ {{\partial ^n~}\over {\partial x^{\prime }_\alpha \partial x^{\prime }_\beta \partial x^{\prime }_\gamma \ldots }}\left( {1 \over {\varDelta r}} \right) \right] _{x^{\prime }_\alpha =x^{\prime }_\beta = x^{\prime }_\gamma \ldots =0} $$

where we assume the sum over $\alpha$, $\beta$, $\gamma , \ldots =1, 2, 3$, with $x_1 =x,~x_2=y,~x_3=z$.

The potential due to the charge distribution is:

$$\begin{aligned} V_{0}(\mathbf {r})={1 \over {4 \pi \epsilon _0}} \int _{\tau } {{\rho (\mathbf {r^{\prime }} )} \over {\vert \mathbf {r} - \mathbf {r^{\prime }} \vert }} d\tau ^{\prime } \end{aligned}$$

(2.9)

and by substituting the expression for $1 / \vert \mathbf {r} - \mathbf {r^{\prime }} \vert$ given before, we get:

$$ V_{0}(\mathbf {r})={1 \over {4 \pi \epsilon _0}} {1 \over r} \int _\tau \rho ( \mathbf {r}~^{\prime }) d \tau ^{\prime } + {1 \over {4 \pi \epsilon _0}} \left[ {{\partial ~}\over {\partial ~x^{\prime }_\alpha }}\left( {1 \over { \varDelta r}}\right) \right] _{x^{\prime }_\alpha =0} \int _\tau x^{\prime }_\alpha \rho ( \mathbf {r}~^{\prime }) d \tau ^{\prime } $$

$$ + {1 \over {4 \pi \epsilon _0}} {1 \over {2!}} \left[ {{\partial ^2~}\over {\partial x^{\prime }_\alpha \partial x^{\prime }_\beta }}\left( {1 \over {\varDelta r}} \right) \right] _{x^{\prime }_\alpha =x^{\prime }_\beta =0} \int _\tau x^{\prime }_\alpha x^{\prime }_\beta ~\rho ( \mathbf {r}~^{\prime }) d \tau ^{\prime }+ \cdots $$

$$ +{1 \over {4 \pi \epsilon _0}} {1 \over {n!}} \left[ {{\partial ^n~}\over {\partial x^{\prime }_\alpha \partial x^{\prime }_\beta \partial x^{\prime }_\gamma \ldots }}\left( {1 \over {\varDelta r}} \right) \right] _{x^{\prime }_\alpha =x^{\prime }_\beta = x^{\prime }_\gamma \ldots =0} \int _\tau x^{\prime }_\alpha x^{\prime }_\beta x^{\prime }_\gamma ~\ldots ~\rho ( \mathbf {r}~^{\prime }) d \tau ^{\prime }. $$

In this expression we can recognise the terms corresponding to the total charge, to the dipole and to the quadrupole moments that we have already seen and then the general form of the $2^n$-pole.

2.1.2 Expansion in Terms of Spherical Harmonics

The multipole expansion of the potential from a charge distribution limited in space, can be also expressed in series of spherical harmonics.⁶

If $\mathbf {r^{\prime }}$ gives the position of a point inside a sphere of radius R, and $\mathbf {r}$ that of a point outside, we can write for $1/{\vert \mathbf {r} - \mathbf {r^{\prime }} \vert }$ the expansion in terms of the spherical harmonics $Y_{lm}\left( \theta ,\varphi \right)$:

$$ {1 \over {\vert \mathbf {r} - \mathbf {r^{\prime }} \vert }}=4 \pi \sum _{l=0}^{\infty } \sum _{m=-l}^{m=l} {1 \over {2l+1}} {{(r^{\prime })^l} \over {r^{l+1}}} Y_{lm}^{*}\left( \theta ^{\prime },\varphi ^{\prime } \right) ~Y_{lm}\left( \theta ,\varphi \right) \,. $$

If the charge distribution $\rho (r^{\prime })$ is confined inside the sphere of radius R we can substitute this expansion in (2.9) and we get:

$$ V_{0}(\mathbf {r})={1 \over {4 \pi \epsilon _0}} \left\{ 4 \pi \sum _{l=0}^{\infty } \sum _{m=-l}^{m=l} {1 \over {2l+1}} {1 \over r^{l+1}} \left[ \int Y_{lm}^{*}\left( \theta ^{\prime },\varphi ^{\prime } \right) (r^{\prime })^l \rho (\mathbf {r^{\prime }})~d \tau ^{\prime } \right] Y_{lm}\left( \theta ,\varphi \right) \right\} $$

and introducing the multipole moments:

$$ q_{lm}=\int Y_{lm}^{*}\left( \theta ^{\prime },\varphi ^{\prime } \right) (r^{\prime })^l \rho (\mathbf {r^{\prime }})~d \tau ^{\prime } $$

we get the expansion:

$$ V_{0}(\mathbf {r})={1 \over {4 \pi \epsilon _0}} \left\{ 4 \pi \sum _{l=0}^{\infty } \sum _{m=-l}^{m=l} {1 \over {2l+1}} {1 \over r^{l+1}} ~q_{lm}~ Y_{lm}\left( \theta ,\varphi \right) \right\} . $$

Exercise With the formulas for the first spherical harmonics reported below, write the first three terms of the multipole expansion in spherical coordinates and compare with those expressed in cartesian coordinates.

$$ l=0 \quad \quad Y_{00}={1 \over {\sqrt{4 \pi }}} $$

$$ l=1 \quad \quad Y_{11}=-~\sqrt{{3 \over {8 \pi }}}~\sin ~\theta ~e^{i \varphi } $$

$$ Y_{10}=\sqrt{{3 \over {4 \pi }}}~\cos ~\theta $$

$$ l=2 \quad \quad Y_{22}={1 \over 4}\sqrt{{ 15 \over {2 \pi }}}~\sin ^2 \theta ~e^{2i \varphi } $$

$$ Y_{21}=-~\sqrt{{ 15 \over {8 \pi }}}~\sin \theta ~\cos \theta ~e^{i \varphi } $$

$$ Y_{20}= \sqrt{{ 5 \over {4 \pi }}}~\left( {3 \over 2} \cos ^2 \theta -{1 \over 2} \right) $$

with the relation:

$$ Y_{l-m}\left( \theta ,\varphi \right) =(-1)^m Y_{lm}^{*}\left( \theta ,\varphi \right) \,. $$

Problems

2.1

Find the dipole moment of the system of four point charges q at (a, 0, 0), q at (0, a, 0), $-q$ at $(-a,0,0)$ and $-q$ at $(0,-a,0)$.

2.2

Write the potential for the system of three point charges: two charges $+q$ in the points (0, 0, a) and $(0,0,-a)$, and a charge $-2q$ in the origin of the frame. Find the approximate form of this potential at distance much larger than a. Compare the result with the potential from the main term in the multipole expansion.

2.3

Two segments cross each other at the origin of the frame and their ends are at the points $(\pm a, 0, 0)$ and $(0,\pm a,0)$. They have a uniform linear charge distribution of opposite sign. Write the quadrupole term for the potential at a distance $r \gg a$.

2.4

Calculate the quadrupole term of the expansion for the potential from two concentric coplanar rings charged with q and $-q$ and with radii a and b.

2.5

Write the interaction energy of two electric dipoles $\mathbf {p}_1$ and $\mathbf {p}_2$ with their centers at distance r.

2.6

Using the result of the previous problem write the force between the electric dipoles $\mathbf {p}_1$ and $\mathbf {p}_2$ at distance r. Then consider the force when the dipoles are coplanar oriented normal to their distance and they are parallel or antiparallel. Determine also the force when the dipoles are on the same line and oriented in the same or in the opposite direction.

2.7

Two coplanar electric dipoles have their centers a fixed distance r apart. Say $\theta$ and $\theta ^{\prime }$ the angles the dipoles make with the line joining their centers and show that if $\theta$ is fixed, they are at equilibrium when

$$ \tan \theta ^{\prime } = -{1 \over 2} \tan \theta \,. $$

Solutions

2.1

The dipole moment of the system has components:

$$p_x=\sum _1^4 q_i x_i =2qa~~~~~p_y=\sum _1^4 q_i y_i =2qa~~~~~p_z=\sum _1^4 q_i z_i =0\,.$$

The dipole moment is $\mathbf {p}=(2qa,2qa,0)$ with module $p=2q\sqrt{2} a$. This is the moment of an elementary dipole with opposite charges 2q located in the centers of the positive and the negative charges which are distant $\sqrt{2} a$.

2.2

In spherical coordinates the potential depends only on the distance r from the origin and on the angle $\theta$. Adding the potentials from the three charges we have:

$$ V(r,\theta )= {1 \over {4 \pi \epsilon _0}} \left[ -{{2q}\over r}+{q \over {[r^2-2ra\cos \theta +a^2]^{1\over 2}}} +{q \over {[r^2+2ra\cos \theta +a^2]^{1\over 2}}}\right] $$

and expanding in power series as in (2.4) we find:

$$\begin{aligned} V(r,\theta )= {1 \over {4 \pi \epsilon _0}}{{qa^2}\over r^3}\left( 3\cos ^2 \theta -1\right) \,. \end{aligned}$$

(2.10)

In the multipole expansion for the system of the three charges the first non null term is the quadrupole moment. It is easy to find the components: $Q_{xz}=Q_{yz}=Q_{xy}=0$, $Q_{xx}=Q_{yy}=-qa^2$ and $Q_{zz}=2qa^2$ so that the potential is:

$$ V(x,y,z)= {1 \over {4 \pi \epsilon _0}}{1 \over r^5}\left( Q_{xx}x^2+Q_{yy}y^2+Q_{zz}z^2\right) $$

$$ ={1 \over {4 \pi \epsilon _0}}{{qa^2}\over r^5}\left( 2 z^2 -x^2 -y^2\right) $$

that is the formula (2.10) written in cartesian coordinates.

2.3

For the given charge distribution it is easy to see that $Q_{xz}=Q_{yz}=Q_{xy}=0$ and by simple calculations we find:

$$ Q_{xx}={1 \over 3} q a^2~~~~~~~~~Q_{yy}=-{1 \over 3} q a^2~~~~~~~~~Q_{zz}=0 $$

so that the quadrupole potential is:

$$ V(x,y,z)={1 \over {4 \pi \epsilon _0}} {{q a^2} \over 3} ~ {{(x^2 - y^2)}\over {(x^2+y^2+z^2)^{ 5 \over 2}}}\,. $$

2.4

We consider the two rings on the plane $z=0$ with their centers in the origin. The total charge of the system is zero and, for the symmetry of the charge distribution with respect to the origin, also the dipole moment is null. The quadrupole term is the first non null term. It is evident that $Q_{xz}=Q_{yz}=0$ and by simple integrals we can get:

$$ Q_{xx}=Q_{yy}={q \over 4}(a^2 - b^2)~~~~~~~~Q_{zz}={q \over 2}(b^2 - a^2)~~~~~~~~Q_{xy}=0 $$

so that the first term of the potential expansion is:

$$ V(x,y,z)={1 \over {4 \pi \epsilon _0}} {{q (a^2 - b^2)} \over 4} ~ {{(x^2 + y^2-2z^2)}\over {(x^2+y^2+z^2)^{ 5 \over 2}}}.$$

2.5

The interaction energy of the two dipoles is equal to the potential energy of $\mathbf {p}_2$ in the field of $\mathbf {p}_1$. Saying $\mathbf {r}$ the vector from the center of $\mathbf {p}_1$ to that of $\mathbf {p}_2$, we can write:

$$ U_{21}=-\mathbf {p}_2 \cdot \mathbf {E}_1 (\mathbf {r})=\mathbf {p}_2 \cdot \mathbf {\nabla }V_1= \mathbf {p}_2 \cdot \mathbf {\nabla }\left( {1 \over {4 \pi \epsilon _0}} {{\mathbf {p}_1 \cdot \mathbf {r}} \over {r^3}}\right) ={1 \over {4 \pi \epsilon _0}} \mathbf {p}_2 \cdot \left[ {{\mathbf {\nabla }(\mathbf {p}_1 \cdot \mathbf {r})} \over {r^3}}+ (\mathbf {p}_1 \cdot \mathbf {r}) \mathbf {\nabla }{1 \over r^3}\right] $$

and since:

$$ \mathbf {\nabla }(\mathbf {p}_1 \cdot \mathbf {r})= \mathbf {p}_1~~~~~~~~\mathbf {\nabla }{1 \over r^3}=-3{\mathbf {r} \over r^5}~~~~~ (\text {note that}~~\mathbf {\nabla }{1 \over r^n}=-n{\mathbf {r} \over r^{n+2}}) $$

we get:

$$\begin{aligned} U_{21}=U_{12}={1 \over {4 \pi \epsilon _0}} \left[ {{\mathbf {p}_1 \cdot \mathbf {p}_2 } \over r^3}-3{{(\mathbf {p}_1 \cdot \mathbf {r})(\mathbf {p}_2 \cdot \mathbf {r})} \over r^5}\right] \end{aligned}$$

(2.11)

that is symmetric in $\mathbf {p}_1$ and $\mathbf {p}_2$.

2.6

From the solution of the previous problem and from the formula (2.2) for the force on a dipole in an electric field we get:

$$ \mathbf {F}_2=-\mathbf {\nabla }U_{21}=\mathbf {\nabla }(\mathbf {p}_2 \cdot \mathbf {E}_1(\mathbf {r})) $$

$$ =-{1 \over {4 \pi \epsilon _0}}\left[ ( \mathbf {p}_1 \cdot \mathbf {p}_2 )\mathbf {\nabla }{1 \over r^3} -3 (\mathbf {p}_1 \cdot \mathbf {r})(\mathbf {p}_2 \cdot \mathbf {r}) \mathbf {\nabla }{1 \over r^5} -3 {{\mathbf {\nabla }[ (\mathbf {p}_1 \cdot \mathbf {r})](\mathbf {p}_2 \cdot \mathbf {r})]} \over r^5} \right] $$

and then:

$$ \mathbf {F}_2= {1 \over {4 \pi \epsilon _0}}\left[ 3 {{( \mathbf {p}_1 \cdot \mathbf {p}_2 ) \mathbf {r} + \mathbf {p}_1 (\mathbf {p}_2 \cdot \mathbf {r})+(\mathbf {p}_1 \cdot \mathbf {r}) \mathbf {p}_2 } \over r^5} -15 {{(\mathbf {p}_1 \cdot \mathbf {r})(\mathbf {p}_2 \cdot \mathbf {r})\mathbf {r}} \over r^7} \right] \,. $$

This formula is symmetric in $\mathbf {p}_1$ and $\mathbf {p}_2$ but $\mathbf {r}$ is directed from $\mathbf {p}_1$ to $\mathbf {p}_2$ so we get $\mathbf {F}_1$ changing $\mathbf {r}$ to $-\mathbf {r}$ and we have $\mathbf {F}_1= -\mathbf {F}_2$ as expected.

For two coplanar parallel dipoles normal to the line joining their centres:

$$ \text {with same direction} (\uparrow ~\uparrow ): \mathbf {F}_2 = {3 \over {4 \pi \epsilon _0}}{{p_1 p_2} \over r^4} \hat{\mathbf {r}}~~~~~\text {a repulsive force} $$

$$ \text {with opposite direction} (\uparrow ~\downarrow ): ~~~\mathbf {F}_2 = - {3 \over {4 \pi \epsilon _0}}{{p_1 p_2} \over r^4} \hat{\mathbf {r}}~~~~~\text {an attractive force} $$

for the dipoles on the same line

$$ \text {with same direction} (\rightarrow ~\rightarrow ): \mathbf {F}_2 = - {6 \over {4 \pi \epsilon _0}}{{p_1 p_2} \over r^4} \hat{\mathbf {r}}~~~~~\text {an attractive force} $$

$$ \text {with opposite direction} (\rightarrow ~\leftarrow ): \mathbf {F}_2 = {6 \over {4 \pi \epsilon _0}}{{p_1 p_2} \over r^4} \hat{\mathbf {r}}~~~~~\text {a repulsive force.} $$

2.7

For the two coplanar dipoles shown in Fig. 2.6 the interaction energy (2.11) becomes:

$$ U_{int}={ 1 \over {4 \pi \epsilon _0}} {{p_1 p_2} \over r^3}\Bigl [ \cos (\theta ^{\prime } - \theta ) -3 \cos \theta \cos \theta ^{\prime } \Bigr ]\,. $$

At fixed $\theta$ we find the minimum of this energy solving the equation:

$$ {{\partial U} \over {\partial \theta ^{\prime }}}={ 1 \over {4 \pi \epsilon _0}} {{p_1 p_2} \over r^3} \Bigl [ - \sin (\theta ^{\prime } - \theta ) + 3 \cos \theta \sin \theta ^{\prime } \Bigr ]=0\,. $$

The solution is:

$$ \tan \theta ^{\prime } = - {1 \over 2} \tan \theta $$

$$ \text {with the condition}~~~~ {{\partial ^2 U} \over {\partial {\theta ^{\prime }}^2}}={ 1 \over {4 \pi \epsilon _0}} {{p_1 p_2} \over r^3} \Bigl [ 2 \cos \theta \cos \theta ^{\prime } - \sin \theta \sin \theta ^{\prime } \Bigr ] > 0\,. $$

For $\theta = \pi /2$ the minimum is at $\theta ^{\prime } =- \pi /2$, for $\theta = 0$ at $\theta ^{\prime } =0$ and for $\theta = \pi$ at $\theta ^{\prime } =- \pi$.

Fig. 2.6

Coplanar dipoles $\mathbf {p}_1$ and $\mathbf {p}_2$ with their centers at fixed distance r and orientations $\theta$ and $\theta ^{\prime }$ with respect to r