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??


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): $$\rho(\myv r') = q\delta(\myv r'-\myv d /2) - q\delta(\myv r'+\myv d /2).$$

Example: Problem 3.29

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:

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

Image credits

Emilio Labrador