[NOTES/EM-03005]-Multipole Expansion of Potential

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1 Multipole expansion of potential

The electric potential due to system of point charges \(q_\alpha\) located at positions \(\vec{r}_\alpha\) is given by \begin{equation}\label{Eq01} \phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\sum_\alpha \frac{q_\alpha}{|\vec{r}-\vec{r}_\alpha|}. \end{equation} For a continuous charge distribution density \(\rho(\vec{r})\), the potential is given by \begin{equation}\label{Eq02} \phi(\vec{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}\,')}{|\vec{r}-\vec{ r }\, ^\prime|}d^3r\,'. \end{equation} To know the potential or the electric field at large distances we need not know the distribution of charges, or the charge density everywhere. It is sufficient to know a few 'multipole moments' of the charge distribution. Let us assume that the charge distribution is confined to a finite volume \(V\). In \EqRef{EQ02}, \(\vec{r}\,^\prime \) refers to a point in volume \(V\) and \(\vec{r}\) to a field point where we are interested in finding the electric field or the potential. Then \begin{equation}\label{Eq03} |\vec{r}-\vec{r}\,^\prime| = \sqrt{r^2-2\vec{r}\cdot\vec{r}\,^\prime + \vec{r}\,^{\prime2} } = r\Big(1- 2\frac{\vec{r}\cdot\vec{r}\,^\prime}{r^2} + \frac{r\,^{\prime2}}{r^2}\Big)^{1/2}. \end{equation} and we derive an expansion of \eqRef{EQ02} for large \(r\) by expanding the r.h.s. of \EqRef{EQ03} in powers of \((r\,^\prime/r)\) as follows. \begin{eqnarray} \frac{1}{|\vec{r}-\vec{r}\,^\prime|} &=& \frac{1}{r} \Big(1- 2\frac{\vec{r}\cdot\vec{r}\,^\prime}{r^2} + \frac{r\,^{\prime2}}{r^2}\Big)^{-1/2}.\\ &=& \frac{1}{r}\Big[ 1 - \frac{1}{2} \Big(-\frac{2\vec{r}\cdot\vec{r}\,^\prime}{r^2} +\frac{r\,^{\prime 2}}{r^2}\Big) \nonumber\\
&&\qquad+ \frac{1}{2}\big(-\frac{1}{2}\big)\big(-\frac{1}{2}-1\big) \Big(- 2 \frac{\vec{r}\cdot\vec{r}\,^\prime}{r^2}+ \frac{r\,^{\prime2}}{r^2} \Big)^2 + \ldots  \Big]\\ &=&\frac{1}{r}\Big[1+ \frac{\vec{r}\cdot\vec{r}\,'}{r^2} +\frac{1}{2}\frac{1}{r^4}\Big( 3(\vec{r}\cdot\vec{r}\,')^2 - r^ 2 r\,^{\prime2}\Big) + \ldots \Big]. \end{eqnarray}

2 Quadrupole moments in 1,2,3 notation:

In the 1-2-3 notation for vectors and using Einstein summation convention we can rewrite the expressions for quadrupole moment as \begin{equation} Q_{ij} = \frac{1}{2} \int \rho(\vec{x})\big(3x_ix_j-|x|^2\delta_{ij}\big) d^3x. \end{equation} In the \(x-y-z\) notation, for a system of point charges the components of quadrupole moment are given by \begin{eqnarray} Q_{xx} &=& \sum_\alpha (3x_\alpha^2 - \vec{r}_\alpha^2) q_\alpha\\ Q_{yy} &=& \sum_\alpha (3y_\alpha^2 - \vec{r}_\alpha^2) q_\alpha\\ Q_{zz} &=& \sum_\alpha (3z_\alpha^2 - \vec{r}_\alpha^2) q_\alpha\\ Q_{xy} &=& Q_{yx} = \sum_\alpha 3 x_\alpha y_\alpha q_\alpha\\ Q_{yz} &=& Q_{zy} = \sum_\alpha 3 y_\alpha z_\alpha q_\alpha\\ Q_{zx} &=& Q_{xz} = \sum_\alpha 3 z_\alpha x_\alpha q_\alpha \end{eqnarray}

3 Large distance expansion of the potential --- The first three terms

Thus the potential has the following expansion in powers of \(r\). \begin{eqnarray} \phi(\vec{r}) &=& \frac{1}{4\pi\epsilon_0} \frac{1}{r} \int \rho(\vec{r}') d^3r'\\ && + \frac{1}{4\pi\epsilon_0} \frac{1}{r^3} \int \rho(\vec{r}^\prime) \vec{r}^\prime d^3r' \\ && + \frac{1}{4\pi\epsilon_0} \frac{1}{2} \int \frac{3(\vec{r} \cdot \vec{r}^\prime)^2 - (r^2)(r^{\prime 2})}{r^5} \rho(\vec{r}^\prime) d^3r^\prime + \ldots \end{eqnarray} If we define a unit vector \(\hat{n}\) by \(\vec{r} = r \hat{n}\) we can write the above expansion as \begin{eqnarray} \boxed{ \phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\left( \frac{q}{r} + \frac{\hat{n}\cdot\vec{p}}{r^2} + \frac{\hat{n}\cdot \overleftrightarrow{Q} \cdot{n}}{r^3} + \ldots \right)} \end{eqnarray} and in dyadic notation we write \begin{eqnarray} \vec{p} &=& \int\vec{r}\,^\prime \rho(\vec{r}\,^\prime) \,d^3r\,^\prime \\ \overleftrightarrow{Q}&=& \frac{1}{2} \int \big(3 (\overleftarrow{r}^\prime )( \overrightarrow{r}^\prime) -\overleftrightarrow{I} r^{\,\prime 2}\big)\rho(\vec{r}\,^\prime)\, d^3 r\,^\prime. \end{eqnarray} It should be noted that out of the nine components only six components of quadrupole moment are independent because the quadrupole moment tensor is symmetric (\(Q_{xy}=Q_{yx}, Q_{yz}= Q_{zy}, Q_{zx}= Q_{xz}\))

4 Properties of Dipole and Quadrupole Moment

The value of the dipole moment, in general, depends on the choice of origin. If the coordinate axes are translated by a vector \(\vec{a}\), the position of charge \(q_\alpha\) w.r.t. new coordinate axes becomes \(\vec{r}^\prime_\alpha \vec{r}_\alpha -\vec{a}\) and the expression for the dipole moment w.r.t. the new coordinate axes takes the form \begin{eqnarray} \vec{p}^\prime &=& \sum_\alpha q_\alpha \vec{r}_\alpha^\prime\\ &=& \sum_\alpha\big( q_\alpha (\vec{r}_\alpha-\vec{a}\big)\\ &=& \vec{p}- \vec{a} \sum_\alpha q_\alpha. \end{eqnarray} Thus we see that the dipole moment of a charge distribution does not change if the total charge of the system \(q= \sum_\alpha q_\alpha\) vanishes. One can similarly show that the quadrupole moment of a system of charges is independent of choice of origin if the total charge and the dipole moment of the system vanish.