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url: /electrodynamics/lesson-10/index.md
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对于 Coulomb 特解做展开：
$$
\varphi(x)=\int\_C\frac{\rho(x')}{4\pi\varepsilon|x'-x|}\text{d}^3x'
$$
展开考虑的是 $|x'|/|x|$ 为小量. 令 $r'=|\vec{x}-\vec{x}'|$，固定 $x$ (场点) 不变，把 $x'$ 在原点附近展开，则
$$
\begin{aligned}
\frac{1}{r}&=\left.\frac{1}{r}\right|*{\vec{x}'=0}+\left\[x'\left.\frac{\partial}{\partial x'}\left(\frac{1}{r}\right)\right|*{\vec{x}'=0}+y'\left.\frac{\partial}{\partial y'}\left(\frac{1}{r}\right)\right|*{\vec{x}'=0}+z'\left.\frac{\partial}{\partial z'}\left(\frac{1}{r}\right)\right|*{\vec{z}'=0}\right]+\cdots\\\\
&\overset{r\_{\vec{x}'=0}=R}{=}\frac{1}{R}+(\vec{x}'\cdot\nabla)\frac{1}{R}+\frac{1}{2}\vec{x}'\vec{x}':\nabla\nabla\frac{1}{R}+\cdots
\end{aligned}
$$
偶极项：
$$
\varphi = \frac{1}{4\pi\varepsilon}\frac{\vec{p}\cdot\vec{x}}{R^3}
$$
四极项：
$$
\varphi=\frac{q}{4\pi\varepsilon}\frac{(\vec{x}\_0'\vec{x}\_1'+\vec{x}\_1'\vec{x}\_0')(3\vec{x}\vec{x}-R^2\bold{I})}{2R^5}
$$

> 这是双偶极子共线对顶的这种情况的领头阶，实际上也可以用四电荷方框的电荷分布来计算领头阶.

如果定义四极矩 $D = \sum\_{m,n}q\_{n}\vec{x}\_n\vec{x}\_m$，可以改写四极项.

对于连续电荷分布，有
$$
\varphi=\frac{1}{4\pi\varepsilon}\int\_{V'}\rho(\vec{x}')\left\[\frac{1}{R}-\vec{x}'\cdot\nabla\frac{1}{R}+\frac{1}{2!}\vec{x}\vec{x}:\nabla\nabla\frac{1}{R}+\cdots\right]\text{d}\tau'
$$
令：
$$
Q=\int\_{V'}\rho(\vec{x}')\text{d}\tau',,\quad\vec{p}=\int\_{V'}\rho(\vec{x}')\vec{x}'\text{d}\tau',,\quad\bold{D}=\int\_{V'}3\vec{x}'\vec{x}'\rho(\vec{x}')\text{d}\tau'
$$
上式改写为
$$
\varphi=\frac{1}{4\pi\varepsilon}\int\left\[\frac{Q}{R}-\vec{p}\cdot\nabla\frac{1}{R}+\frac{1}{6}\bold{D}:\nabla\nabla\frac{1}{R}+\cdots\right]\text{d}\tau'
$$
一个小区域内连续分布的电荷在远处激发的电场等于一系列多极子在远处激发的场的叠加.