Answer to Question #174404 in Electricity and Magnetism for Hafimu

Question

I'm a second year physics undergraduate, and currently learning using Vanderlinde's Classical Electromagnetics Theory. One of the books question I found is:

2-12 Show that the quadrupole term of the multipole expansion of the potential can be written

$V_{3}=\frac{1}{8\pi\varepsilon_{0}}\sum q_i\vec{r}^{\left(i\right)}\cdot \vec{\nabla }\left[\vec{r}^{\left(i\right)}\cdot \vec{\nabla }\left(\frac{1}{r}\right)\right]$

How should I solve this? Where do I start?

Answer 1

We can start by defining multipole expansion. It is a mathematical series representing a function that depends on angles. it is usually two angles on a sphere. These series are useful because they can often be truncated, meaning that only the first few terms need to be retained for a good approximation to the original function.

Proof: The scalar potential $V(\vec{r})$ at a point $\vec{r}$ due to a system of changes is given by

$V(\vec{r})=\frac{1}{4 \pi \epsilon_o}\int_{v'} \frac{\rho \vec{r' dV'}}{ |\vec{r}-\vec{r'}|}$

We assume that the point is at a large distance from the charge distribution. If ${V'}$ varies over the charge distribution, then $r>>r'$

Since, $V(\vec{r})=\frac{1}{4 \pi \epsilon_o}\int_{v'} \frac{\rho \vec{r' dV'}}{ |\vec{r}-\vec{r'}|}$

Then ${ |\vec{r}-\vec{r}'|}=[r^2-2\vec{r}\vec{r}'+r^2]^{0.5}=[1-\frac{2\hat{v}\vec{r}'}{r}+(\frac{r'}{r})^2]^{0.5}$

Where $\hat{v}\equiv\frac{\vec{r}'}{r}$

Then using the fact that r is much larger than r', we can write

$\frac{1}{{ |\vec{r}-\vec{r}'|}}=\frac{1}{r}\frac{1}{[1-\frac{2\hat{v}\vec{r}'}{r}+(\frac{r'}{r})^2]^{0.5}}$

and using the binomial expansion

$\frac{1}{[1-\frac{2\hat{v}\vec{r}'}{r}+(\frac{r'}{r})^2]^{0.5}}=1+\frac{\hat{r}-\vec{r}}{r}+ \frac{1}{2r^2}(3(\hat{r} .\vec{r}-r'^2)+0(\frac{\vec{r}'}{r})^3$

We neglect the third spherical term. Hence, the above potential can be written

$V(\vec{r})=\frac{1}{4 \pi\epsilon_o r}\int_{v'} \rho (\hat{r}) [1+\frac{\hat{r}-\vec{r}}{r}+ \frac{1}{2r^2}(3(\hat{r} .\vec{r}-r'^2)+0(\frac{\vec{r}'}{r})^3] dV'$

We write it as $V(v)=V_{mon}(v)+V_{dip}(v)+V_{quad}(v)+...$

$V_{mon}(\vec{r})=\frac{1}{4 \pi\epsilon_o r}\int_{v'} \rho (\hat{r}) dV'$

$V_{dip}(\vec{r})=\frac{1}{4 \pi\epsilon_o r^2}\int_{v'} \rho (\hat{r}) [(\hat{r} -r')] dV'$

$V_{quad}(\vec{r})=\frac{1}{8 \pi\epsilon_o r^3}\int_{v'} \rho (\hat{r})[(3(\hat{r}-r'^2)-r'^2] dV'$

Now writing $\rho(r')$ in terms of discrete charge distribition

$\rho (r')=\frac{Total \space change}{Total \space volume}=\frac{\sum dq_i}{\sum dv_i}$

and using the properties of integration of divergence

$\int_V \vec{\bigtriangledown}.\frac{\vec{r}}{r^3}dV=\int_s \frac{\vec{r}}{r^3}ds=\frac{1}{r^2}\int_s ds=4 \pi$

$\vec{\bigtriangledown}.\frac{\vec{r}}{r^3}=\vec{\bigtriangledown}. {\triangledown}\frac{\vec{-1}}{r}=- {\triangledown}^2\frac{\vec{1}}{r}=4 \pi \delta (r) \implies \delta( \vec{r})= \frac{-1}{4 \pi}{r}{\triangledown}^2 \frac{1}{r}$

We can rewrite the quadrupole term as

$V_3= \frac{1}{8 \pi\epsilon_o}\sum q_i \vec{r}^{(i)}\vec{\bigtriangledown}[\vec{r}^{(i)}\vec{\bigtriangledown} \frac{1}{r}]$