# Multipole expansion

...a series expansion in powers of $1/r$ for the field far from a charge distribution.

### Binomial expansions

First, a little mathematical stretching...

We shall shortly have great interest in approximating things like: $$(1+\delta)^p,$$ where $\delta$ is some quantity which is small compared to $1$.

This can be found from a Taylor expansion: $$\begineq f(x) &=& f(a)+f'(a)(x-a)+f''(a)(x-a)^2/2 + f'''(a)(x-a)^3/6+...\\ &=& f(a) + \sum_{n=1}^\infty f^n\frac{(x-a)^n}{n!}.\endeq$$ Note the equal sign!

The idea is to expand $f(\delta)=(1+\delta)^p$ around $a=0$. This Mathematica command gives you the first 4 terms in a Taylor expansion of $f(\delta)$ for values of $\delta$ close to 0):

Try it:

1. According to this series, what is the lowest-order-in-$\delta$ approximation to $1/(1+\delta)$? (That is... 1+___$\delta$+....)
2. What about $1/\sqrt{1+\delta}$?
3. What about $1/\sqrt{1-\delta}$?
4. For $p=-1/2$ write out the first four terms of the expansion.

## Potentials, far from the charge

Imagine that we have some charge distribution $\rho$ that is concentrated in a small region around the origin. That is, assume beyond $r\gt a$, that $\rho(\myv r)=0$. What is the approximate field (or potential) due to this charge distribution at large distances ($r \gg a$)?

### Monopole (point charge at a distance)

We've seen that, at distances $r \gg a$ the potential of such a charge distribution looks like a point charge, that is, $V=\frac{1}{4\pi \epsilon_0}\frac{Q}{r}$, where $Q = \int \rho(\myv r') d \tau'$.

But what if the total charge sums to 0. Is there no field??

### Dipole

Two opposite charges separated by a distance $d$ form a dipole.

Clearly for the two charges pictured, the charges sum to zero. At large distances, $r \gg d$ we won't have a field that looks like a point charge. But it doesn't seem like it should just vanish either.

So lets find the 'leading order' $r$-dependence: $$V(\myv r) = \frac{1}{4\pi \epsilon_0}(q/\rr_+ -q/\rr_-) = \frac{q}{4\pi \epsilon_0}(1/\rr_+ -1/\rr_-).$$

According to the diagram, $$\rr_{+(-)} = r -(+) \frac{d}{2}\cos \theta= r(1-(+)\frac{d}{2r}\cos \theta).$$

So... $$\begineq 1/\rr_+ -1/\rr_-&=&\frac{1}{r}\left[\left(1-\frac{d}{2r}\cos \theta\right)^{-1} - \left(1+\frac{d}{2r}\cos \theta\right)^{-1}\right]\\ & \approx & \frac{1}{r}[(1+\frac{d}{2r}\cos \theta) - (1-\frac{d}{2r}\cos \theta)]\\ &=&\frac{d}{r^2}\cos \theta\endeq.$$ I've used a binomial expansion of $(1+\delta)^{-1} \approx 1-\delta$. Substituting this into the expression for the potential: $$V_\text{dipole}(\myv r) = \frac{1}{4\pi \epsilon_0} \frac{qd}{r^2}\cos \theta.$$

This dipole is characterized by a direction and a 'strength', which we can keep track of with the dipole moment:

$$\myv p \equiv q \myv d.$$

Unlike electric field lines, $\myv p$ points from negative to positive charge.

In terms of $\myv p$, the potential can be written... $$V_\text{dipole}(\myv r) = \frac{1}{4\pi \epsilon_0} \frac{\myv p \cdot \myv r}{r^3} \propto \frac{1}{r^2}.$$

Calculating the leading $r$-dependence of some other charge configurations in this way we find this trend...

### Expansion in powers of 1/r

Let's say that we have a charge distribution clustered around the origin... that is, beyond some radius $a$, $\rho(r'>a)=0$. The potential from the charge distribution is exactly: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{\rr} \rho(\myv r') d \tau',$$ and perhaps for ($r \gg a$) we can expand this in powers of $1/r$?

We'll use the law of cosines this time to express the length of $\myv\rr=\myv r-\myv r'$ exactly as:

$$\rr^2 = r^2 +r'^2 - 2\myv r \cdot \myv r' =r^2\left(1+\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right] \right).$$

Taking the square root, $\rr = r (1+[])^{1/2}$. Far from the charge distribution, the quantity $1 \gg []$, so we'll use our binomial distribution to figure: $$\begineq 1/\rr &=&\frac{1}{r}(1+[])^{-1/2}\\ &=& \frac{1}{r}\left(1-\frac{1}{2}[]+\frac{3}{8}[]^2-\frac{5}{16}[]^3 +...\right)\\ &=& \frac{1}{r}\left(1-\frac{1}{2}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right] +\frac{3}{8}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right]^2\right.\\ & & \left.-\frac{5}{16}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right]^3 +...\right)\\ &=& \frac{1}{r}\left(1 + \left(\frac{r'}{r}\right)\uv r \cdot \uv r' +\left(\frac{r'}{r}\right)^2(3(\uv r \cdot \uv r')^2-1)/2\right.\\ & &\left.+\left(\frac{r'}{r}\right)^3 (5(\uv r \cdot \uv r')^3 -3\uv r \cdot \uv r')/2 +...\right) .\endeq$$

The potential is $$\begineq V(\myv r) &=& \frac{1}{4\pi \epsilon_0} \int \left(\frac{1}{r} + \frac{\myv r\cdot \myv r'}{r^3}+\frac{1}{2}\left[\frac{3(\myv r \cdot \myv r')^2}{r^5}-\frac{r'^2}{r^3}\right]+...\right)\rho(\myv r') d \tau'\\ &=& \frac{1}{4\pi \epsilon_0} \frac{1}{r} \int \rho(\myv r') d \tau' + \frac{1}{4\pi \epsilon_0} \frac{\myv r}{r^3} \cdot \int \myv r ' \rho(\myv r') d \tau'+... \endeq$$

The first integral is apparently the total charge of the charge distribution. This is formally called the monopole moment:

$$Q_\text{monopole} \equiv \int \rho(\myv r') d \tau'.$$

The second integral is formally called the dipole moment of the charge distribution:

$$\myv p_\text{dipole} \equiv \int \myv r' \rho(\myv r') d \tau'.$$

This gives the same sort of dipole moment as the two point charges discussed earlier with the charge distribution (using dirac delta functions): $$\rho(\myv r') = q\delta(\myv r'-\myv d /2) - q\delta(\myv r'+\myv d /2).$$

Monopole moment?
Dipole moment?

## Let's get real...

We've got enough info to calculate the 3-d E-field from the expression for the dipole potential: $$V_\text{dipole} = \frac{1}{4\pi \epsilon_0} \frac{\myv r \cdot \myv p}{r^3}=\frac{1}{4\pi \epsilon_0} \frac{p \cos \theta}{r^2}.$$ where the last piece is assuming that the dipole $\myv p$ is oriented in the $\uv{z}$-direction.

$$\myv E = - \myv \grad V,$$ and using the spherical polar coordinate form of the gradient:

$$E_r = -\frac{\del V}{\del r} = \frac{2p\cos \theta}{4\pi \epsilon_0 r^3},$$ $$E_\theta = -\frac{1}{r}\frac{\del V}{\del \theta} = \frac{p \sin \theta}{\pi \epsilon_0 r^3},$$ $$E_(\phi) = -\frac{1}{r \sin \theta} \frac{\del V}{\del \phi} = 0.$$

Sketch this in the $z$-$y$-plane...

Which looks like this:

We know that this is not what the dipole field from a real dipole looks like. It ought to be more like...

...But that's OK. Remember that the multipole expansion is only supposed to be a good approximation far away from the charge distribution.

### Charge distribution leaves its origin

What happens to the dipole moment if the charge distribution is moved away from the origin?

Consider two charge distributions: $\rho_1(\myv r)$ and $\rho_2(\myv r) = \rho_1(\myv r - \myv a)$. The dipole moment of the first one is: $$\myv p_1 = \int \myv r \rho_1(\myv r) d \tau.$$

The dipole moment of the second distribution is... $$\begineq \myv p_2 &=& \int \myv r \rho_2(\myv r) d \tau=\int \myv r \rho_1(\myv r-\myv a) d \tau=\int (\myv r' + \myv a) \rho_1(\myv r') d \tau\\ &=&\int \myv r' \rho_1(\myv r') d \tau + \myv a \int \rho_1(\myv r') d \tau\\ &=&\myv p_1 + \myv a Q\endeq$$ (In that 3rd step, we've pulled a change of variables: $\myv r-\myv a=\myv r'$ where $d \tau = d \tau'$.

So, in general $\myv p_2 \neq \myv p_1$, unless the monopole term is zero.

This pattern holds for higher order monopoles: The $n$-pole term is independent of position if all the previous -pole terms have vanished.

### Multipole expansion in Legendre polynomials

In the development above, we could have written the dot product instead as: $$\myv r \cdot \myv r' = r r'\cos \theta'.$$ Putting this in to our equation for the potential in powers of $1/r$ above: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \left[\frac{1}{r} + \frac{r'}{r^2}(\cos \theta')+\frac{r'^2}{r^3}(\frac{1}{2}(3\cos^2 \theta'-1))+...\right]\rho(\myv r') d \tau'.$$

Some of these terms sure look like the Legendre polynomials. And with more patience to calculate higher terms, we find they are. So:

$$V(\myv r) = \frac{1}{4\pi \epsilon_0} \sum_{l=0}^\infty \frac{1}{r^{l+1}} \int (r')^l P_l(\cos \theta') \rho(\myv r') d \tau'.$$