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: $$(1+x)^p,$$ where $x$ is some quantity which is small compared to $1$.
This can be found from a Taylor expansion (or Power Series 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 power series is always correct. But it is most useful if $a$ is a number which is close to the values of $x$ that we're interested in. In this case, we're interested in values of $x$ which are small compared to 1, that is, $x$ is close to 0.
So we'll expand
$f(x)=(1+x)^p$ around the value $a=0$. The Taylor series for $f(x)=(1+x)^p$ will look a lot simpler...
$$\begineq f(x) &= f(0)+f'(0)(x-0)+f''(0)(x-0)^2/2 + f'''(0)(x-0)^3/6+...+f^n()(x-0)^n/n!+...\\
&= f(0) + f'(0)x+f''(0)x^2/2 + f'''(0)(x-0)^3/6+...f^nx^n/n!\\
&= f(0)+px+(p)(p-1)x^2/x+(p)(p-1)(p-2)x^3/3!+...
.\endeq$$
This Sagemath command gives you the first 3 terms in a Taylor expansion of $f(x)=(1+x)^p$, for values of $x$ close to 0):
- According to this series, what is the lowest-order-in-$x$ approximation to $1/(1+x)$? That is... $$(1+x)^{-1}=1+{\color{purple}\text{__?__}}\,x+\text{[terms of order x^2 and higher]}.$$
- What about $1/\sqrt{1+x}$?
- What about $1/\sqrt{1-x}$?
- 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... $$V_\text{dipole}(\myv r) = \frac{1}{4\pi \epsilon_0} \frac{\myv p \cdot \myv r}{r^3} =\frac{1}{4\pi \epsilon_0} \frac{qdr\cos\theta}{r^3} = \frac{1}{4\pi \epsilon_0} \frac{qd}{r^2}\cos\theta. $$
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. By that, we mean beyond some radius $a$, $\rho(r'>a)=0$. The potential due to this charge distribution is exactly: $$V(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{\rr} \rho(\myv r') d \tau'.$$ But 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 in square brackets is much smaller than 1: $1 \gg []$. Now, going beyond the binomial expansion, the full Taylor series starts out like this: $$\begineq 1/\rr &=\frac{1}{r}(1+[])^{-1/2}\\ &= \frac{1}{r}\left(1-\frac{1}{2}[]+\frac{3}{8}[]^2-\frac{5}{16}[]^3 +...\right)\\ &= \frac{1}{r}\left(1-\frac{1}{2}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right] +\frac{3}{8}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right]^2\right.\\ &\ \ \ \ \ \ \ \left.-\frac{5}{16}\left[ \frac{r'^2}{r^2} - \frac{2\myv r \cdot \myv r'}{r^2} \right]^3 +...\right)\\ &= \frac{1}{r}\left(1 {\color{red}-\frac12\frac{r'^2}{r^2} } {\color{blue}+ \frac12\frac{2\myv r \cdot \myv r'}{r^2} } +\frac38 \left(\frac{r'^2}{r^2}\right)^2 {\color{purple} -\frac38 2\left(\frac{2\myv r \cdot \myv r'}{r^2}\right) \frac{r'^2}{r^2} } {\color{red}+\frac38 \left( \frac{2\myv r\cdot\myv r'}{r^2}\right)^2 }\\ \right.\\ & \ \ \ \ \ \ \left. -\frac{5}{16}\left(\frac{r'^2}{r^2}\right)^3 +\frac{5}{16}3\left(\frac{r'^2}{r^2}\right)^2\left( \frac{2\myv r\cdot\myv r'}{r^2}\right) -\frac{5}{16}3\left(\frac{r'^2}{r^2}\right)\left( \frac{2\myv r\cdot\myv r'}{r^2}\right)^2 {\color{purple}+\frac{5}{16}\left( \frac{2\myv r\cdot\myv r'}{r^2}\right)^3 }+... \right) &= X\\ &= \frac{1}{r}\left(1 {\color{blue}+ \left(\frac{r'}{r}\right)\uv r \cdot \uv r' } {\color{red}+\left(\frac{r'}{r}\right)^2\frac12(3(\uv r \cdot \uv r')^2-1)}\right.\\ &\ \ \ \ \ \ \ \left.{\color{purple}+\left(\frac{r'}{r}\right)^3 (5(\uv r \cdot \uv r')^3 -3\uv r \cdot \uv r')/2 } +...\right) .\endeq$$
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 3.27
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 ($r\gg d$) 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
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