06 Multipole Expansion
From this article on, we will be using V (or \varphi) instead of \Phi (or \phi) to represent electric potential, not to be messed up with flux (or the azimuth angle in spherical coordinates).
Multipole Expansion
In an arbitrary distribution of electric charges \rho(\bm{r'}), we write down the potential:
V(\bm{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{\rcurs}\rho(\bm{r'})\d \tau' \tag{1}
To expand it with Taylor series, we may use the cosine theorem:
\rcurs^2 = r^2 + r'^2 - 2rr'\cos \alpha
where \alpha is the angle between \bm{r} and \bm{r'}. Then, we can write down \rcurs:
\rcurs = r\sqrt{1+\epsilon}
in which
\epsilon \equiv (\frac{r'}{r})^2 - 2\cos \alpha \cdot (\frac{r'}{r}) \tag{2}
Then, using Taylor expansion,
(1+\epsilon)^{-1/2} = 1 - \frac{1}{2}\epsilon + \frac{1}{2!} \cdot \frac{1}{2}\frac{3}{2}\epsilon^2 - \frac{1}{3!} \cdot \frac{1}{2}\frac{3}{2}\frac{5}{2}\epsilon^3 - ...
Using Equation 2, we get:
\begin{aligned} \frac{1}{\rcurs} &= \frac{1}{r} (1 - \frac{1}{2}(\frac{r'}{r})^2 + (\frac{r'}{r})\cos\alpha + \frac{3}{8}(\frac{r'}{r})^4 - \frac{3}{2}\cos\alpha (\frac{r'}{r})^3+4\cos^2 \alpha (\frac{r'}{r})^2 + ...) \\ &= \frac{1}{r} (1 + \frac{r'}{r}\cos\alpha + (\frac{r'}{r})^2(\frac{3\cos^2\alpha-1}{2}) + ...) \end{aligned}
We might notice that the coefficients of (r'/r)^k are the Legendre polynomials P_k(\cos\alpha) (which originates from the eigenfunction/spectrum of the Laplacian operator \nabla^2). Thus,
\frac{1}{\rcurs} = \frac{1}{r}\sum_{n=0}^\infty (\frac{r'}{r})^n P_n(\cos\alpha) \tag{3}
and
\begin{aligned} V(\bm{r}) =& \frac{1}{4\pi \epsilon_0} (\frac{1}{r}\int \rho(\bm{r'})\d \tau' + \frac{1}{r^2}\int r'\cos\alpha\rho(\bm{r'})\d\tau' \\ &+ \frac{1}{r^3}\int r'^2(\frac{3\cos^2\alpha-1}{2})\rho(\bm{r'})\d\tau' + ... \\ &+ \frac{1}{r^{n+1}}\int r'^n P_n(\cos\alpha)\rho(\bm{r'})\d\tau'+...) \end{aligned}
Thus, we succesfully expanded the original, tricky integral into terms of 1/r^n.
Special Cases
Monopole
Q = \int \rho(\bm{r'})\d \tau' is obviously the total charge.
Dipole
\begin{aligned} V_{\text{di}}(\bm{r}) &= \frac{1}{4\pi \epsilon_0} \frac{1}{r^2}\int r'\cos\alpha\rho(\bm{r'})\d\tau'\\ &= \frac{1}{4\pi \epsilon_0} \frac{\hat{\bm{r}}}{r^2} \cdot \int \bm{r'}\rho(\bm{r'})\d\tau' \end{aligned}
This works because of the distribution law of dot product. Thus, we define
\bm{p} = \int \bm{r'}\rho(\bm{r'})\d\tau' \tag{4}
as the dipole moment, so that
V_{\text{di}}(\bm{r}) = \frac{1}{4\pi \epsilon_0} \frac{\hat{\bm{r}} \cdot \bm{p}}{r^2} \tag{5}
If we zoom out far enough, or the source area is small enough, we can consider a so-called perfect dipole where it shrinks to a single point, namely the origin. We may as well set up the coordinates so that \bm{p} points up at \hat{\bm{z}} direction. Then,
V_{\text{di}}(\bm{r}) = \frac{1}{4\pi \epsilon_0} \frac{p\cos\theta}{r^2}
It would be more convenient to derive the electric field using spherical coordinates:
\begin{cases} \begin{aligned} E_r &= -\frac{\partial V}{\partial r} =\frac{1}{4\pi \epsilon_0} \frac{2p\cos\theta}{r^3}\\ E_{\theta} &= -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{1}{4\pi \epsilon_0} \frac{p\sin\theta}{r^3}\\ E_{\phi} &= -\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi} = 0 \end{aligned} \end{cases}
Quadrupole
Similarly,
\begin{aligned} r'^2(\frac{3\cos^2\alpha-1}{2}) &= \frac{3}{2}(\hat{\bm{r}}\cdot\bm{r'})^2-\frac{1}{2}\bm{r'}^2\\ &= \frac{1}{2} (3r'_ir'_j-r'^2\delta_{ij})\hat{r}_i\hat{r}_j \end{aligned}
Thus, with the tensor formalism, we define
\tilde{\bm{Q}}_{ij} = \int \rho(\bm{r'})(3r'_ir'_j-r'^2\delta_{ij})\d \tau'
as the 2D, 3x3, symmetric, 0-trace quadrupole moment tensor;
V_{\text{quad}}(\bm{r}) = \frac{1}{4\pi \epsilon_0}\frac{1}{2r^3} \tilde{\bm{Q}}_{ij}\hat{r}_i\hat{r}_j
or, we can rewrite the tensor summation part into the matrix form:
\begin{aligned} \tilde{\bm{Q}}_{ij}\hat{r}_i\hat{r}_j &= \bm{r}^{\text{T}}Q\bm{r}, \\ \bm{r}^{\text{T}} &= (\hat{r}_1, \hat{r}_2, \hat{r}_3) \end{aligned}