V.电势多极展开与电多极矩-注释部分

• 原文: https://zhuanlan.zhihu.com/p/62371285
• 发布日期: 2019-04-24
• 分类: 电动力学

文章收录于:电动力学前情回顾

这是文章的注释部分,正文请转到:正樹：V.电势多极展开与电多极矩

\bigstar [i]\ \ 将\varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{r}d{V}'} 中的 $\frac{1}{r}$ 展开是如何完成的呢?

#如果感觉下面内容有些恶意不妨先转到:正樹：多元函数幂级数展开的n种常用等价形式

先回忆一下多元函数的幂级数展开:\begin{align} & f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{s}{\Delta {x}_{i}{\left. \frac{\partial f\left( {\vec{\xi } \right)}{\partial {\xi }_{i} \right|}_{\vec{\xi }=\vec{a}+}\sum\limits_{i,j=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}{\left. \frac{\partial }^{2}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j} \right|}_{\vec{\xi }=\vec{a} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i,j,k=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}{\left. \frac{\partial }^{3}f\left( {\vec{\xi } \right)}{\partial {\xi }_{i}\partial {\xi }_{j}\partial {\xi }_{k} \right|}_{\vec{\xi }=\vec{a}+\cdot \cdot \cdot } \\ \end{align}

自然可以写成:f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\sum\limits_{i=1}^{s}{\Delta {x}_{i}\frac{\partial f\left( {\vec{a} \right)}{\partial {a}_{i}+}\sum\limits_{i,j=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}{2!}\frac{\partial }^{2}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}+\sum\limits_{i,j,k=1}^{s}{\frac{\Delta {x}_{i}\Delta {x}_{j}\Delta {x}_{k}{3!}\frac{\partial }^{3}f\left( {\vec{a} \right)}{\partial {a}_{i}\partial {a}_{j}\partial {a}_{k}+\cdot \cdot \cdot }即 f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\cdot \cdot \cdot +\Delta {x}_{s}\frac{\partial }{\partial {a}_{s} \right)}^{n}f\left( {\vec{a} \right)}

特别地,当 $s=3$ 时: [即三元函数时]

f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta {x}_{1}\frac{\partial }{\partial {a}_{1}+\Delta {x}_{2}\frac{\partial }{\partial {a}_{2}+\Delta {x}_{3}\frac{\partial }{\partial {a}_{3} \right)}^{n}f\left( {\vec{a} \right)}=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{n}f\left( {\vec{a} \right)}

一般也就展开到二阶: f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\Delta \vec{x}\cdot {\nabla }_{\vec{a}f\left( {\vec{a} \right)+\frac{1}{2!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{2}f\left( {\vec{a} \right)+\cdot \cdot \cdot

接下来将[{\vec{x} 类比于 \vec{x}-{\vec{x}' 或 $r$] , [展开点 {\vec{a}类比于{\vec{x} 或 $R$] , [$\Delta \vec{x}$ 类比于-{\vec{x}']

再利用上述结论: \begin{align} & \frac{1}{r}=\frac{1}{\left| \vec{x}-{\vec{x}' \right|}=f\left( \vec{x}-{\vec{x}' \right)=f\left( {\vec{x} \right)+\left( -{\vec{x}' \right)\cdot \nabla f\left( {\vec{x} \right)+\frac{1}{2}{\left( -\vec{x}\cdot \nabla \right)}^{2}f\left( {\vec{x} \right)\cdot \cdot \cdot \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}{\left( {x}'}_{1}\frac{\partial }{\partial {x}_{1}+{x}'}_{2}\frac{\partial }{\partial {x}_{2}+{x}'}_{3}\frac{\partial }{\partial {x}_{3} \right)}^{\text{2}\frac{\text{1}{R}+\cdot \cdot \cdot \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}\sum\limits_{i,j=1}^{3}{x}'}_{i}{x}'}_{j}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+}\cdot \cdot \cdot \\ \end{align}

#记R=\left| {\vec{x} \right|且做多极展开时满足$r\approx R=x\gg x^{\prime }$即电荷区域很集中且场点很远.

#{\vec{x}'={x}'{\vec{e}_{x}+{y}'{\vec{e}_{y}+{z}'{\vec{e}_{z}表源点, 而{\vec{x}表场点,且记\vec{r}=\vec{x}-{\vec{x}'.

#此外$\nabla$与{\nabla }'}也是完全不一样的东西,分别是对场坐标与源坐标求偏导.

于是乎:\varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\rho \left( {\vec{x}'} \right)\left[ \frac{1}{R}-{\vec{x}'\cdot \nabla \frac{1}{R}+\frac{1}{2}\sum\limits_{i,j=1}^{3}{x}'}_{i}{x}'}_{j}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+\cdot \cdot \cdot \right]d{V}'} \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}\iiint_{V}'}{\rho \left( {\vec{x}'} \right)d{V}'}-\nabla \frac{1}{R}\cdot \iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}+\frac{1}{6}\sum\limits_{i,j=1}^{3}{\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'}+\cdot \cdot \cdot \right] \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}Q-\vec{P}\cdot \nabla \frac{1}{R}+\frac{1}{6}\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{\text{1}{R}+\cdot \cdot \cdot \right] \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\left[ \frac{1}{R}Q-\vec{P}\cdot \nabla \frac{1}{R}+\frac{1}{6}\overset{\rightrightarrows }{\mathop{D}\,:\nabla \nabla \frac{1}{R}+\cdot \cdot \cdot \right]

其中 \left\{ \begin{align} & Q=\iiint_{V}'}{\rho \left( {\vec{x}'} \right)d{V}'} \\ & \vec{P}=\iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'} \\ & {D}_{ij}=\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'} \\ & \overset{\Rightarrow }{\mathop{D}\,=\iiint_{V}'}{3{\vec{x}'{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'} \\ \end{align} \right. 且有 \overset{\Rightarrow }{\mathop{D}\,:\nabla \nabla \frac{1}{R}=\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{\text{1}{R}

$★[ii]$ 电四极矩是个张量,这里自然会涉及到并矢,双点乘等概念.

https://zhuanlan.zhihu.com/p/62612523

\bigstar [iii]\ \ f\left( \vec{a}+\Delta \vec{x} \right)=f\left( {\vec{a} \right)+\Delta \vec{x}\cdot {\nabla }_{\vec{a}f\left( {\vec{a} \right)+\frac{1}{2!}{\left( \Delta \vec{x}\cdot {\nabla }_{\vec{a} \right)}^{2}f\left( {\vec{a} \right)+\cdot \cdot \cdot

将[ {\vec{x} 类比于 \vec{x}-{\vec{x}' 或 $r$] , [展开点 {\vec{a} 类比于 {\vec{x} 或 $R$] , [$\Delta \vec{x}$ 类比于-{\vec{x}'] 大致情况如图:

相对于 $r或R$ 这个尺寸,电荷分布区域 {V}'} 看起来与一个点无异. 值得注意的是,原点必须取在区域内这个展开才能够在低阶项就有足够高的近似程度,因为零阶近似就是将电荷集中在原点的电势. 而区域 {V}'}相对而言是如此的小以至于原点的选择在区域 {V}'}内是等价的.

所以当你要展开函数 f\left( {\vec{r} \right)=f\left( \vec{x}-{\vec{x}' \right)=\frac{1}{\left| \vec{x}-{\vec{x}' \right|}=\frac{1}{r} 时,自然要认识到确定了整体特征的展开点就是 $\vec{r}=\vec{x}$ 这个点,也就是 {x}'}=0 的点,实际上还要认识到在计算 \varphi \left( {\vec{x} \right) 的积分 \frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{\left| \vec{x}-{\vec{x}' \right|}d{V}'} 里面 {\vec{x} 是个常数,积分变量是 ${\vec{x}}^{\prime }$.自然展开点的函数值就是 f\left( {\vec{x} \right)=\frac{1}{x}=\frac{1}{R} .所以这个修正项/微扰项$\Delta \vec{x}$ 自然是由-{\vec{x}'来充当的.

$★[iv]$ 若电荷分布具有球对称性[即\rho \left( -{\vec{x}' \right)=-\rho \left( {\vec{x}'} \right)]则有:

\iiint_{V}'}{x}'}^{2}\rho \left( {\vec{x}'} \right)d{V}'}=\iiint_{V}'}{y}'}^{2}\rho \left( {\vec{x}'} \right)d{V}'}=\iiint_{V}'}{z}'}^{2}\rho \left( {\vec{x}'} \right)d{V}'}=\frac{1}{3}\iiint_{V}'}{r}'}^{2}\rho \left( {\vec{x}'} \right)d{V}'}=0

其中 {r}'}^{2}={x}'}^{2}+{y}'}^{2}+{z}'}^{2} 所以就有 $D_{11}=D_{22}=D_{33}=0$

而 $D_{12}=D_{13}=D_{23}=0$ 更是显然, 所以电四极矩为0.