V.电势多极展开与电多极矩

• 原文: https://zhuanlan.zhihu.com/p/62355838
• 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

为保持版面整洁,$★$备注写在另一篇文章中:正樹：V.电势多极展开与电多极矩-注释部分

#再次说明今后{\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}'.

#在求和时为了记述方便起见约定$123$表征$xyz$即: \vec{x}=\left( x,y,z \right)=\left( {x}_{1},{x}_{2},{x}_{3} \right)=\sum\limits_{i=1}^{3}{x}_{i}{\vec{e}_{i}

#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.

#下面记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.

✦电势多极展开:

对电势计算式\varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{r}d{V}'} 中的 $\frac{1}{r}$ 做幂级数展开有:$坐标系原点必须在电荷分布区域内★[iii]$

\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}'}

可记作: \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]

再进一步简化可记作: \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{\Rightarrow }{\mathop{D}\,:\nabla \nabla \frac{1}{R}+\cdot \cdot \cdot \right] $★[i]感觉推导不够显然或者不懂多元展开请参看注释$

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}\ \ \ \bigstar [ii]

$Q=总电荷$ $\vec{P}=电偶极矩$ $D_{ij}=电四极矩矩阵元$

(至于为何$D_{ij}$定义里有个3, 参看本文最后证明只有五个独立分量时带有❤符号的式子会用到)

总而言之上面提到的将 $\frac{1}{r}$ 展开是将三元幂级数展开公式:

f\left( {\vec{x} \right)=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

中的将[{\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}']而将 $\frac{1}{r}$ 展开的. $为何能如此代入呢?★[iii]$

#限制坐标原点取在区域{V}'}内是否是必须呢? 前面展开的函数是 f\left( r \right)=f\left( \left| \vec{x}-{\vec{x}' \right| \right)=\frac{1}{r} 而不是 \varphi \left( {\vec{x} \right)=\frac{1}{4\pi {\varepsilon }_{0}\iiint_{V}'}{\frac{\rho \left( {\vec{x}'} \right)}{r}d{V}'} 这里是有必要提一下的, 因为电势仅是 {\vec{x} 的函数, 对于确定的 {\vec{x}, 势能 \varphi \left( {\vec{x} \right) 就只是个数罢了. 而 $\frac{1}{r}$ 是 \vec{x}-{\vec{x}' 的函数, 对于确定的 {\vec{x}它还会随着 ${\vec{x}}^{\prime }$而改变. 因为是相对位置的函数, 所以原点的选取实际上是不会影响结果的, 但展开点会从 $\vec{x}-0$ 变成 \vec{x}-{\vec{x}_{0}^{\prime } 而变得麻烦些. (这里的 {\vec{x}_{0}^{\prime }是区域{V}'}内任选一点. 如果原点在区域{V}'}内的话, 方便起见肯定选原点啊.)

✦电多极矩:

前面得到的展开式: \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]

多极展开可以简记为: \varphi \left( {\vec{x} \right)={\varphi }^{\left( 0 \right)}\left( {\vec{x} \right)+{\varphi }^{\left( 1 \right)}\left( {\vec{x} \right)+{\varphi }^{\left( 2 \right)}\left( {\vec{x} \right)+\cdot \cdot \cdot 其中:

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[]A. {\varphi }^{\left( 0 \right)}=\frac{1}{4\pi {\varepsilon }_{0}\frac{Q}{R} 零阶近似,是把系统总电荷 $Q$ 全集中在原点产生的电势.

其中 Q=\iiint_{V}'}{\rho \left( {\vec{x}'} \right)d{V}'}

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[]B. {\varphi }^{\left( 1 \right)}=-\frac{1}{4\pi {\varepsilon }_{0}\vec{P}\cdot \nabla \vec{R}=\frac{1}{4\pi {\varepsilon }_{0}\frac{\vec{P}\cdot \vec{R}{R}^{3} 一阶近似,是系统电偶极矩 {\vec{P} 产生的电势.

其中 \vec{P}=\iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}

注意垂直符号标在了中间R上而不是r-上,这点还蛮重要.因为如果标在了r-上的话,当考察点M在电荷连线中垂线上的话明明r+=r-差值l*cosθ却不为零.这显然不符合逻辑,不甚严谨.上图是电偶极子在远处( $R\gg l$ 所以两个夹角近似相等为 $\theta$ )产生电势的示意图,电偶极子被定义为等量异号的两个点电荷中由负电荷指向正电荷的位矢与电荷量的乘积,即: $\vec{P}=Q\vec{l}$ . 而电偶极矩也是类似的定义,即负电荷指向正电荷的位矢改为负电中心指向正电中心的矢量.所谓电中心的概念类似于力学中重心的概念,无非就是把质量改成电荷量.

\vec{P}\cdot \vec{R}=PR\cos \theta \Rightarrow \cos \theta =\frac{\vec{P}\cdot \vec{R}{QlR} ---- [i] {\varphi }_{M}=\frac{1}{4\pi {\varepsilon }_{0}\left( \frac{Q}{r}_{+}-\frac{Q}{r}_{-} \right)=\frac{Q}{4\pi {\varepsilon }_{0}\left( \frac{r}_{-}-{r}_{+}{r}_{+}{r}_{-} \right)\approx \frac{Q}{4\pi {\varepsilon }_{0}\left( \frac{l\cos \theta }{R}^{2} \right)\overset{代入}\left[ i \right]}{\mathop{=}\,\frac{1}{4\pi {\varepsilon }_{0}\frac{\vec{P}\cdot \vec{R}{R}^{3}

前面提到的电偶极矩似乎有两个定义: \vec{P}=\iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'} 与 $\vec{P}=Q\vec{l}$

其实二者是等价的, 在电中性的体系中:

正电中心的坐标是: ${\vec{X}}_{+}=\frac{{\iiint }_V^{\prime }}{{\vec{x}}^{\prime }{\rho }_{+}d{V}^{\prime }}{\iiint }_V^{\prime }{\rho }_{+}d{V}^{\prime }=\frac{{\iiint }_V^{\prime }}{{\vec{x}}^{\prime }{\rho }_{+}d{V}^{\prime }}Q$

负电中心的坐标是: ${\vec{X}}_{-}=\frac{{\iiint }_V^{\prime }}{{\vec{x}}^{\prime }{\rho }_{-}d{V}^{\prime }}{\iiint }_V^{\prime }{\rho }_{-}d{V}^{\prime }=\frac{{\iiint }_V^{\prime }}{{\vec{x}}^{\prime }{\rho }_{-}d{V}^{\prime }}-Q$

其中 ${\rho }_{+},{\rho }_{-}$分别为正负电荷密度(${\rho }_{+}\ge 0,{\rho }_{-}\le 0$),在正电荷分布的位置 ${\rho }_{-}=0$, 在负电荷分布的位置 ${\rho }_{+}=0$, 显然电荷密度 $\rho \left({\vec{x}}^{\prime }\right)={\rho }_{+}\left({\vec{x}}^{\prime }\right)+{\rho }_{-}\left({\vec{x}}^{\prime }\right)$. 而$Q$是体系带的等量正负总电荷的绝对值. \begin{align} & \vec{P}=Q\vec{l}=Q\left( {\vec{X}_{+}-{\vec{X}_{-} \right)=Q{\vec{X}_{+}-Q{\vec{X}_{-} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\iiint_{V}'}{\vec{x}'{\rho }_{+}d{V}'+\iiint_{V}'}{\vec{x}'{\rho }_{-}d{V}'=\iiint_{V}'}{\vec{x}'\rho }d{V}' \\ \end{align}

从上面的推导可以看出:①虽然涉及到了位置矢量的累加但却结果却不依赖于坐标原点,因为本质上是计算带等量电荷的正负电荷中心, 然后电偶极矩它们的相对位矢的函数, 而相对位置是不依赖于坐标系选取的. ②电荷分布 $\rho \left({\vec{x}}^{\prime }\right)$关于原点对称的体系电偶极矩为 0.

电偶极矩计算式推广到 \vec{P}=\iiint_{V}'}{\vec{x}'\rho \left( {\vec{x}'} \right)d{V}'}之后也可以计算非电中性的体系了,不过这时候的极矩跟原点的选择有关了, 因为总有配不上对的电荷.

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[]C.{\varphi }^{\left( 2 \right)}=\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{2!}\frac{1}{3}\overset{\Rightarrow }{\mathop{D}\,:\nabla \nabla \frac{1}{R}=\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{6}\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R}

二阶近似, 是电四极矩 \overset{\Rightarrow }{\mathop{D}\, 产生的电势,其中 \left\{ \begin{align} & {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.

通过定义很容易证明系统电荷分布如果具有球对称性的话总电四极矩为0 $★[iv]$

电四极矩有九个分量: \overset{\Rightarrow }{\mathop{D}\,\text{=}\left[ \begin{matrix} {D}_{\text{11} & {D}_{12} & {D}_{13} \\ {D}_{21} & {D}_{22} & {D}_{23} \\ {D}_{31} & {D}_{32} & {D}_{33} \\ \end{matrix} \right]显然有 \left\{ \begin{align} & {D}_{\text{12}\text{=}{D}_{\text{21} \\ & {D}_{\text{13}\text{=}{D}_{\text{31} \\ & {D}_{\text{23}\text{=}{D}_{\text{32} \\ \end{align} \right.即对称阵.

各个分量产生的势能分别为: {\varphi }_{D}_{ij}=\frac{1}{4\pi {\varepsilon }_{0}\frac{D}_{ij}{6}\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R}

下面给出六个四极矩仅含有一个分量的体系, 这样可以对这九个分量有一个直观感受:

下面选取 {D}_{\text{33}体系给出电势计算公式 {\varphi }_{M}={\varphi }_{D}_{33}=\frac{1}{4\pi {\varepsilon }_{0}\frac{D}_{33}{6}\frac{\partial }^{2}{\partial {z}^{2}\frac{1}{R} 的推导过程:

如图分别在z=-b,-a,a,b处放置电荷+Q,-Q,-Q,+Q.这个体系的总电荷为0, 不难看出总电偶极矩也为0. 有两个方向相反的电偶极矩 {\vec{P}_{1},{\vec{P}_{2} 且都只有 $z$ 分量, 即 $\vec{P}=\left(0,0,P\right)$.

它们在 $M\left(M_x,M_y,M_z\right)$ 点产生的总电势为: \begin{align} & {\varphi }_{M}=-\frac{1}{4\pi {\varepsilon }_{0}{\vec{P}_{1}\cdot \nabla \frac{1}{r}_{1}-\frac{1}{4\pi {\varepsilon }_{0}{\vec{P}_{2}\cdot \nabla \frac{1}{r}_{2} \\ & \ \ \ \ \ =-\frac{1}{4\pi {\varepsilon }_{0}\left( {P}_{1}\frac{\partial }{\partial z}\frac{1}{r}_{1}+{P}_{2}\frac{\partial }{\partial z}\frac{1}{r}_{2} \right) \\ & \ \ \ \ \ =-\frac{1}{4\pi {\varepsilon }_{0}\left( Q\left( b-a \right)\frac{\partial }{\partial z}\frac{1}{r}_{1}-Q\left( b-a \right)\frac{\partial }{\partial z}\frac{1}{r}_{2} \right) \\ & \ \ \ \ \ \ \ \text{=}-\frac{1}{4\pi {\varepsilon }_{0}Q\left( b-a \right)\left( \frac{\partial }{\partial z}\frac{r}_{2}-{r}_{1}{R}^{2} \right) \\ & \ \ \ \ \ =-\frac{1}{4\pi {\varepsilon }_{0}Q\left( b-a \right)\left( \frac{\partial }{\partial z}\frac{\left( a+b \right)\cos \theta }{R}^{2} \right) \\ & \ \ \ \ \ =\frac{1}{4\pi {\varepsilon }_{0}Q\left( b-a \right)\left( a+b \right)\left( -\frac{\partial }{\partial z}\frac{M}_{z}{R}^{3} \right)=\frac{1}{4\pi {\varepsilon }_{0}Q\left( b-a \right)\left( a+b \right)\frac{\partial }^{2}{\partial {z}^{2}\frac{1}{R} \\ \end{align}

最后两个等号是分别带入了这两个结果: \left\{ \begin{align} & {M}_{z}=\vec{R}\cdot {\vec{e}_{z}=R\cos \theta \Rightarrow cos\theta =\frac{M}_{z}{R} \\ & \frac{\partial }{\partial z}\frac{1}{R}=-\frac{1}{R}^{2}\frac{\partial R}{\partial z}=-\frac{1}{R}^{2}\frac{M}_{z}{R}=-\frac{M}_{z}{R}^{3} \\ \end{align} \right.

上面得到的结果是: {\varphi }_{M}=\frac{1}{4\pi {\varepsilon }_{0}Q\left( {b}^{2}-{a}^{2} \right)\frac{\partial }^{2}{\partial {z}^{2}\frac{1}{R}

电四极矩分量产生电势计算公式为: {\varphi }_{M}={\varphi }_{D}_{33}=\frac{1}{4\pi {\varepsilon }_{0}\frac{D}_{33}{6}\frac{\partial }^{2}{\partial {z}^{2}\frac{1}{R} \begin{align} & {D}_{33}=\iiint_{V}'}{3{x'}_{3}^{2}\rho \left( {\vec{x}'} \right)}d{V}' \\ & \ \ \ \ \ \ =3{\left( -b \right)}^{2}Q+3{\left( -a \right)}^{2}\left( -Q \right)+3{a}^{2}\left( -Q \right)+3{b}^{2}Q \\ & \ \ \ \ \ \ =6\left( {b}^{2}-{a}^{2} \right)Q \\ \end{align} 大概就是这种感觉.

电四极矩有九个分量且是个对称矩阵, 但它独立的分量只有5个而不是6个.

简证: R\ne 0\Rightarrow {\nabla }^{2}\frac{1}{R}=0=\sum\limits_{i,j=1}^{3}{\delta }_{ij}\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{1}{R}=0\Rightarrow \sum\limits_{i,j=1}^{3}{\delta }_{ij}{r}'}^{2}\rho \left( {\vec{x}'} \right)\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{1}{R}=0

\begin{align} & {\varphi }^{\left( 2 \right)}=\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{6}\sum\limits_{i,j=1}^{3}{D}_{ij}\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R} \\ & \ \ \ \ \ =\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{6}\sum\limits_{i,j=1}^{3}{\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R} \\ & \ \ \ \ \ =\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{6}\left[ \sum\limits_{i,j=1}^{3}{\iiint_{V}'}{3{x}'}_{i}{x}'}_{j}\rho \left( {\vec{x}'} \right)d{V}'\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R}-\iiint_{V}'}{\left( \sum\limits_{i,j=1}^{3}{\delta }_{ij}{r}'}^{2}\rho \left( {\vec{x}'} \right)\frac{\partial }^{2}{\partial {x}_{i}\partial {x}_{j}\frac{1}{R} \right)}d{V}' \right] \\ & \ \ \ \ \ =\frac{1}{4\pi {\varepsilon }_{0}\frac{1}{6}\sum\limits_{i,j=1}^{3}{\left[ \iiint_{V}'}{\left( 3{x}'}_{i}{x}'}_{j}-{\delta }_{ij}{r}'}^{2} \right)\rho \left( {\vec{x}'} \right)d{V}'} \right]\frac{\partial }^{2}{\partial {x}_{j}\partial {x}_{i}\frac{1}{R} \\ \end{align}

所以电四极矩又可以被定义为: {D}_{ij}=\iiint_{V}'}{\left( 3{x}'}_{i}{x}'}_{j}-{\delta }_{ij}{r}'}^{2} \right)\rho \left( {\vec{x}'} \right)d{V}'}

其中 {r}'}^{2}={x'}_{1}^{2}+{x'}_{2}^{2}+{x'}_{3}^{2} 这样显然有 $D_{11}+D_{22}+D_{33}=0$ ❤

多了一个限制条件所以在六个独立分量的基础上又要少一个, 即只有五个独立分量.

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[]D. ${\phi }^{\left(\text{3}\right)}\text{=}$ 电八极矩

其实这个我也没处理过(笑)、一般是用不到的, 展开到电四极矩就足够精确了.

(实际上我感觉更可能是因为展开到电四极矩就足够麻烦了...)