Multipole Expansion of Potential and Dipole Moments
Multiple Expansion
When a distribution of charges is viewed from sufficiently far away, it appears almost as though it were a point charge. In other words, if the total charge of the charge distribution is $Q$, it would feel as if there’s a single point charge with charge $Q$ when viewed from afar. This means that the potential can be approximated as $\dfrac{1}{4\pi\epsilon_{0}} \dfrac{Q}{r}$.
But if the total charge is $0$, then the question arises as to whether it’s correct to approximate the potential as $0$. Multiple expansion is the answer to how to express the potential as an approximate formula when the total charge is $0$. Expressing the potential $V(\mathbf{r})$ as a series with respect to $\dfrac{1}{r^{n}}$ is called a multipole expansion.
The potential at position $\mathbf{r}$ is as follows.
$$ \begin{equation} V(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{1}{\cR}\rho (\mathbf{r}^{\prime})d\tau^{\prime} \label{1} \end{equation} $$
Where $\bcR = \mathbf{r} - \mathbf{r}^{\prime} (\cR = \left| \bcR \right|)$ is the separation vector.
By using the law of cosines to find the magnitude of $\cR$,
$$ \begin{align*} \cR ^2 =&\ r^2+(r^{\prime})^2-2rr^{\prime}\cos\alpha \\ =&\ r^2\left[1+\left(\dfrac{r^{\prime}}{r}\right)^2-2\dfrac{r^{\prime}}{r}\cos\alpha\right] \\ =&\ r^2\left[1+\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha\right)\right] \end{align*} $$
For convenience, let’s completely substitute the second term in the angle brackets with $\epsilon=\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right)$. Then the following holds true.
$$ \cR=r\sqrt{1+\epsilon} \implies \dfrac{1}{\cR} = \dfrac{1}{r}(1+\epsilon)^{-1/2} $$
If $\mathbf{r}$ is a place very far from the charge distribution, then $\dfrac{r^{\prime}}{r}$ becomes a very small value and $\epsilon \ll 1$ is valid.
If $|x| < 1$, then,
$$ (1 + x )^{\alpha} = 1 + \alpha x + \dfrac{\alpha (\alpha-1)}{2!}x^{2} + \dfrac{\alpha (\alpha-1)(\alpha-2)}{3!}x^{3} + \cdots $$
Therefore, $(1+\epsilon)^{-1/2}$ can be solved as a binomial series.
$$ \dfrac{1}{\cR} = \dfrac{1}{r}(1+\epsilon)^{-1/2} = \dfrac{1}{r}\left( 1- \dfrac{1}{2}\epsilon+\dfrac{3}{8}\epsilon ^2 -\dfrac{5}{16}\epsilon ^3 +\cdots \right) $$
Substituting $\epsilon$ back into its original form,
$$ \dfrac{1}{\cR}=\dfrac{1}{r}\left[ 1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) ^2 -\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) ^3 +\cdots \right] $$
Organizing according to each order of $\dfrac{r^{\prime}}{r}$ gives the following. For detailed process, see the Appendix.
$$ \dfrac{1}{\cR}=\dfrac{1}{r}\left[ 1+\left(\dfrac{r^{\prime}}{r}\right)\left( \cos\alpha \right) +\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{3\cos^2\alpha -1 }{2}\right) +\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right) +\cdots \right] $$
Here, each bracket can be expressed as the series in the form of $\sum \limits_{n=0}^{\infty} a_{n}\left( \dfrac{r^{\prime}}{r}\right)^n$. Each coefficient $a_{n}$ is as follows.
$$ \begin{align*} a_{0} =&\ 1 \\ a_{1} =&\ \cos\alpha \\ a_{2} =&\ \dfrac{3\cos^2\alpha-1}{2} \\ a_{3} =&\ \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right) \\ \vdots & \end{align*} $$
This is the same as the Legendre polynomial $P_{n}(\cos \alpha)$ for $\cos\alpha$. Therefore, summarizing,
$$ \dfrac{1}{\cR}=\dfrac{1}{r}\sum\limits_{n=0}^{\infty}\left( \dfrac{r^{\prime}}{r}\right)^n P_{n}(\cos\alpha) $$
Substituting this into the potential formula $\eqref{1}$ and bringing out $r$, which is independent of the integration, gives us the following.
$$ V(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\sum \limits_{n=0}^{\infty} \dfrac{1}{r^{n+1}} \int (r^{\prime})^nP_{n}(\cos\alpha) \rho (\mathbf{r}^{\prime}) d\tau^{\prime} $$
Expanding this series again results in
$$ \begin{align*} V(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}} \bigg[ &\dfrac{1}{r} \int r^{\prime}\cos\alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime} \\ &+ \dfrac{1}{r^2}\int r^{\prime}\cos\alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime} + \dfrac{1}{r^3}\int(r^{\prime})^2\dfrac{3\cos^2\alpha -1 }{2}\rho (\mathbf{r}^{\prime})d\tau^{\prime} + \cdots \bigg] \end{align*} $$
The first term is the potential created by a monopole, the second term is by a dipole, and the third term is by a quadrupole. The $n$th term is related to the $2^{n-1}$-polar term.
Dipole Moment and Dipole Term
Since the multipole expansion is a series with respect to the reciprocal of $r$, usually, when $r$ is large, the monopole term is the largest. mono refers to monopole.
$$ V_{\text{mono}}(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{r} $$
If the total charge of the gathered charges is $0$, then the monopole term is $0$. Otherwise, $+$ and $-$ can pair up to make $0$, but since there can only be one monopole term, it isn’t possible. Hence, if the dipole term isn’t $0$, then the dipole term is the largest. dip refers to dipole.
$$ V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{1}{r^2}\int r^{\prime} \cos \alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime} $$
Here, since $\hat{\mathbf{r}}\cdot\mathbf{r}^{\prime}=r^{\prime}\cos\alpha$,
$$ V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{1}{r^2}\hat{\mathbf{r}}\cdot\int \mathbf{r}^{\prime} \rho (\mathbf{r}^{\prime})d\tau^{\prime} $$
This integral value is independent of $\mathbf{r}$ and is specifically named the dipole moment of the charge distribution, denoted as $\mathbf{p}$.
$$ \mathbf{p}=\int\mathbf{r}^{\prime}\rho (\mathbf{r}^{\prime})d\tau^{\prime} $$
The dipole moment allows for a simple representation of the dipole’s potential.
$$ V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{\mathbf{p}\cdot\hat{\mathbf{r}} } {r^2} $$
Appendix
$$ 1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^{2} \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right)^2 -\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right)^3 +\cdots $$
Expanding the square and cubic terms from the above equation yields
$$ 1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^2 \left[ \left( \dfrac{r^{\prime}}{r}\right)^2-\dfrac{4r^{\prime}\cos\alpha}{r}+4\cos^2\alpha \right] $$
$$ -\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left[ \left( \dfrac{r^{\prime}}{r}\right)^3 -3\left(\dfrac{r^{\prime}}{r}\right)^22\cos\alpha + 3\left( \dfrac{r^{\prime}}{r}\right)4\cos^2\alpha-8\cos^3\alpha \right] +\cdots $$
Now, organizing according to the order of $\dfrac{r^{\prime}}{r}$ gives
$$ 1+\left(\dfrac{r^{\prime}}{r}\right)\cos\alpha +\left(\dfrac{r^{\prime}}{r}\right)^2\left(-\dfrac{1}{2} +\dfrac{3}{2}\cos^2\alpha \right)+\left( \dfrac{r^{\prime}}{r} \right)^3 \left( -\dfrac{3}{2}\cos\alpha +\dfrac{5}{2}\cos^3\alpha \right) + \cdots $$
Upon organizing this gives
$$ 1+\left(\dfrac{r^{\prime}}{r}\right)\left( \cos\alpha \right) +\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{3\cos^2\alpha -1 }{2}\right) +\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right) +\cdots $$