Question

(a) Show that the quadrupole term in the multipole expansion can be written as

${\mathit{V}}_{\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{d}}\mathbf{\left(}\overrightarrow{\mathbf{r}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{\sum }_{\mathbf{i}\mathbf{,}\mathbf{j}\mathbf{-}\mathbf{1}}^{\mathbf{3}}\widehat{{\mathbf{r}}_{\mathbf{i}}}\widehat{{\mathbf{r}}_{\mathbf{j}}}{\mathit{Q}}_{\mathbf{i}\mathbf{j}}$ ............(1)

(in the notation of Eq. 1.31) where

${\mathit{Q}}_{\mathbf{ij}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\mathbf{[}\mathbf{3}\mathbf{r}{\mathbf{\text{'}}}_{\mathbf{j}}\mathbf{r}{\mathbf{\text{'}}}_{\mathbf{j}}\mathbf{-}{\mathbf{(}\mathbf{r}\mathbf{\text{'}}\mathbf{)}}^{\mathbf{2}}{\mathbf{\delta }}_{\mathbf{ij}}\mathbf{]}\mathit{\rho }\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}\mathbf{\text{'}}}\mathbf{\right)}\mathit{d}\mathit{\tau }\mathbf{\text{'}}$ ..........(2)

Here

${\mathit{\delta }}_{\mathbf{i}\mathbf{j}}\mathbf{=}\{\begin{array}{l}1\\ 0\end{array}\begin{array}{cccc}\mathbf{i}\mathbf{f}& \mathbf{i}& \mathbf{=}& \mathbf{j}\\ \mathbf{i}\mathbf{f}& \mathbf{i}& \mathbf{\ne }& \mathbf{j}\end{array}$ ..........(3)

is the Kronecker Delta and $Q_{ij}$is the quadrupole moment of the charge distribution. Notice the hierarchy

${\mathit{V}}_{\mathbf{mon}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{Q}}{\mathbf{r}}\mathbf{}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{dip}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{}}\frac{\widehat{{\mathbf{r}}_{\mathbf{j}}{\mathbf{p}}_{\mathbf{j}}}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{quad}}\mathbf{\left(}\hat{\mathbf{r}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{\mathbf{r}{\mathbf{3}}^{}}\mathbf{\sum }_{\mathbf{i}\mathbf{}\mathbf{j}\mathbf{-}\mathbf{1}}^{\mathbf{3}}\hat{{\mathbf{r}}_{\mathbf{i}}\widehat{{\mathbf{r}}_{\mathbf{j}}}}\mathit{Q}\mathit{i}\mathit{j}\mathbf{}\mathbf{}\mathbf{}\mathbf{;}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

The monopole moment (Q) is a scalar, the dipole moment $\left(\stackrel{\rightharpoonup }{p}\right)$ is a vector, the quadrupole moment $\left(Q_{ij}\right)$ is a second rank tensor, and so on.

(b) Find all nine components$Q_{ij}$ of for the configuration given in Fig. 3.30 (assume the square has side and lies in the x-y plane, centered at the origin).

(c) Show that the quadrupole moment is independent of origin if the monopole and

dipole moments both vanish. (This works all the way up the hierarchy-the

lowest nonzero multipole moment is always independent of origin.)

(d) How would you define the octopole moment? Express the octopole term in the multipole expansion in terms of the octopole moment.

Short answer

Answer

(a) We have proved that the quadrupole term in the multipole expansion can be written in terms of the quadrupole moment as

$V_{quad}\left(\hat{r}\right)=\frac{1}{8\pi {\epsilon }_0}\frac{1}{r^3}\int \rho \left(\stackrel{\rightharpoonup }{r\text{'}}\right)d\tau \text{'}\left(3{\cos }^2\theta \text{'}-1\right)r{\text{'}}^2$.

(b) For the charge configuration shown in Fig. 3.30, we have found out the nine components of the quadrupole moment. We have obtained$Q_{xy}=Q_{yx}=\frac{3q{a}^2}{2}$. The rest of the terms are zero.

(c) We have proved that the quadrupole moment is independent of the choice of origin if the monopole and dipole moments vanish.

(d) We have derived an expression for the octopole moment as

Step-by-step solution

Given data

The quadrupole moment can be written as

.

The Kronecker Delta function can be defined as