Multipole expansion

Remarks

The series converges absolutely for $r > R$, since $\abs{P_l} \le 1$ on $[-1, 1]$. So the $l$-th integral term is at most $(R^l / r^{l+1}) \int \abs{\rho(y)} \, dy$, a geometric bound. It comes from the expansion $1/\abs{x - y} = \sum_l \abs{y}^l P_l(\cos \gamma) / r^{l+1}$ for $\abs{y} < \abs{x}$. In Cartesian form the first three terms are $$\varphi(x) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q}{r} + \frac{p \cdot x}{r^3} + \frac{1}{2} \sum_{i,j=1}^{3} Q_{ij} \frac{x_i x_j}{r^5} + \cdots \right),$$ with total charge $Q$, dipole moment $p$, and quadrupole moment $Q_{ij} = \int \rho(y) (3 y_i y_j - \abs{y}^2 \delta_{ij}) \, dy$. Example: charges $+q$ at $(0,0,d/2)$ and $-q$ at $(0,0,-d/2)$ have $Q = 0$ and $p = (0,0,qd)$, so far away $\varphi \approx q d \cos \theta / (4 \pi \varepsilon_0 r^2)$, with $\theta$ the polar angle of $x$. Non-example: inside the ball, $r < R$, the series need not converge; there one uses the interior expansion in powers $r^l$. Using the addition theorem, $P_l(\cos \gamma)$ splits into products of spherical harmonics of $x$ and of $y$, which gives the spherical multipole moments.