# 多極展開

## 電勢的多極展開式

$\Phi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_{\mathbb{V'}} \frac{\rho(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\ \mathrm{d}^3\mathbf{r}'$

### 笛卡兒多極展開

$f(\mathbf{r}')=f(\mathbf{O})+\mathbf{r}'\cdot\nabla' f(\mathbf{O})+\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 r'_{\alpha}r'_{\beta} \frac{\partial^2 f(\mathbf{O})}{\partial r'_{\alpha}\partial r'_{\beta}}+\dots$

$\frac{\partial f(\mathbf{r}')}{\partial r'_{\alpha}}=\frac{(r_{\alpha} - r'_{\alpha})}{|\mathbf{r} - \mathbf{r}'|^3}$
$\frac{\partial^2 f(\mathbf{r}')}{\partial r'_{\alpha}\partial r'_{\beta}}= \frac{3(r_{\alpha} - r'_{\alpha})(r_{\beta} - r'_{\beta})}{|\mathbf{r} - \mathbf{r}'|^5} - \frac{\delta_{\alpha\beta}}{|\mathbf{r} - \mathbf{r}'|^3}$

\begin{align} \frac{1}{|\mathbf{r} - \mathbf{r}'|} & =\frac{1}{r}+\frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} +\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1} ^3 \left(\frac{3r_{\alpha}r_{\beta}}{r^5} - \frac{\delta_{\alpha\beta}}{r^3}\right)r'_{\alpha}r'_{\beta} +\dots \\ & =\frac{1}{r}+\frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} +\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1} ^3 \left(\frac{3r_{\alpha}r_{\beta}r'_{\alpha}r'_{\beta} - r^2r'_{\alpha}r'_{\beta}\delta_{\alpha\beta}}{r^5}\right) +\dots \\ & =\frac{1}{r}+\frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} +\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1} ^3 \left(\frac{3r_{\alpha}r_{\beta}r'_{\alpha}r'_{\beta} - r_{\alpha}r_{\beta}r^{\prime 2}\delta_{\alpha\beta}}{r^5}\right) +\dots \\ \end{align}

$\Phi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_{\mathbb{V'}} \left[\frac{1}{r}+\frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} +\frac{1}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3 \frac{r_\alpha r_\beta (3 r'_\alpha r'_\beta - r^{\prime 2}\delta_{\alpha\beta})}{r^5}+\dots\right] \rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$

$q\stackrel{def}{=}\ \int_{\mathbb{V'}}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$
$\mathbf{p}\stackrel{def}{=}\ \int_{\mathbb{V'}}\mathbf{r}'\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$
$Q_{\alpha\beta}\stackrel{def}{=}\ \int_{\mathbb{V'}}(3 r'_\alpha r'_\beta - r^{\prime 2}\delta_{\alpha\beta})\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$

$\Phi(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}+\frac{\mathbf{p}\cdot\mathbf{r}}{r^3} +\frac{1}{2r^5}\sum_{\alpha=1}^3\sum_{\beta=1}^3 Q_{\alpha\beta}r_\alpha r_\beta +\dots\right)$

### 球多極展開

$\frac{1}{|\mathbf{r} - \mathbf{r}'|}=\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}\frac{4\pi}{2\ell+1}\ \frac{r^{\prime\ell}}{r^{\ell+1}}Y^*_{\ell m}(\theta',\phi')Y_{\ell m}(\theta,\phi),\qquad r'

$\Phi(\mathbf{r})=\frac{1}{\epsilon_0}\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell} \frac{Y_{\ell m}(\theta,\phi)}{((2\ell+1)r^{\ell+1}} \int_{\mathbb{V'}}Y^*_{\ell m}(\theta',\phi') r^{\prime\ell}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$

$q_{\ell m}\stackrel{def}{=}\ \int_{\mathbb{V'}}Y^*_{\ell m}(\theta',\phi') r^{\prime\ell}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$

$\Phi(\mathbf{r})=\frac{1}{\epsilon_0}\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell} \frac{q_{\ell m} Y_{\ell m}(\theta,\phi)}{(2\ell+1)r^{\ell+1}}$

\begin{align} q_{00} & =\frac{1}{\sqrt{4\pi}}\int_{\mathbb{V'}}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & =\frac{1}{\sqrt{4\pi}}\ q \\ q_{11} & = -\sqrt{\frac{3}{8\pi}}\int_{\mathbb{V'}} r' \sin{\theta'}\ e^{-i\phi'}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & = -\sqrt{\frac{3}{8\pi}}\ (p_x - ip_y) \\ q_{10} & =\sqrt{\frac{3}{4\pi}}\int_{\mathbb{V'}} r' \cos{\theta}\ \rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & = -\sqrt{\frac{3}{4\pi}}\ p_z \\ q_{22} & =\sqrt{\frac{15}{32\pi}}\int_{\mathbb{V'}} r^{\prime 2} \sin^2{\theta'}\ e^{-2i\phi'}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & =\sqrt{\frac{15}{288\pi}}\ (Q_{11}-2iQ_{12}-Q_{22}) \\ q_{21} & = - \sqrt{\frac{15}{8\pi}}\int_{\mathbb{V'}} r^{\prime 2} \sin{\theta'}\cos{\theta'}\ e^{-i\phi'}\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & = - \sqrt{\frac{15}{72\pi}}\ (Q_{13}-iQ_{33}) \\ q_{20} & =\sqrt{\frac{5}{16\pi}}\int_{\mathbb{V'}} r^{\prime 2}(\cos^2{\theta'}-1)\rho(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & & =\sqrt{\frac{5}{16\pi}}\ Q_{33} \end{align}

## 電能的多極展開式

$U=\int_{\mathbb{V}} \rho(\mathbf{r})\Phi(\mathbf{r})\ \mathrm{d}^3\mathbf{r}$

\begin{align} \Phi(\mathbf{r}) & =\Phi(\mathbf{O})+\mathbf{r}\cdot\nabla\Phi(\mathbf{O})+\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 r_{\alpha}r_{\beta} \frac{\partial^2 \Phi(\mathbf{O})}{\partial r_{\alpha}\partial r_{\beta}}+\dots \\ & =\Phi(\mathbf{O})-\mathbf{r}\cdot\mathbf{E}(\mathbf{O})-\frac{1}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 r_{\alpha}r_{\beta} \frac{\partial E_{\beta}(\mathbf{O})}{\partial r_{\alpha}}+\dots \\ \end{align}

$\Phi(\mathbf{r})=\Phi(\mathbf{O})-\mathbf{r}\cdot\mathbf{E}(\mathbf{O})-\frac{1}{6}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 (3r_{\alpha}r_{\beta}-r^2\delta_{\alpha\beta}) \frac{\partial E_{\beta}(\mathbf{O})}{\partial r_{\alpha}}+\dots$

$U=q\Phi(\mathbf{O})-\mathbf{p}\cdot\mathbf{E}(\mathbf{O})-\frac{1}{6}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 Q_{\alpha\beta} \frac{\partial E_{\beta}(\mathbf{O})}{\partial r_{\alpha}}+\dots$

## 磁向量勢的多極展開式

$\mathbf{A}(\mathbf{r})\ \stackrel{def}{=}\ \frac{\mu_0}{4\pi}\int_{\mathbb{V}'} \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\, d^3\mathbf{r}'$

$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_{\mathbb{V'}} \left[\frac{1}{r}+\frac{\mathbf{r}\cdot\mathbf{r}'}{r^3} +\frac{1}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3 \frac{r_\alpha r_\beta (3 r'_\alpha r'_\beta - r^{\prime 2}\delta_{\alpha\beta})}{r^5}+\dots\right] \mathbf{J}(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}'$

\begin{align}\int_{\mathbb{V}'}J_{\alpha}(\mathbf{r}')\, d^3\mathbf{r}' & =\int_{\mathbb{V}'}[\mathbf{J}(\mathbf{r}')\cdot\nabla']r'_{\alpha}\, d^3\mathbf{r}' =\int_{\mathbb{V}'}\nabla'\cdot[r'_{\alpha}\mathbf{J}(\mathbf{r}')]-r'_{\alpha}\nabla'\cdot\mathbf{J}(\mathbf{r}')\, d^3\mathbf{r}' \\ & =\int_{\mathbb{V}'}\nabla'\cdot[r'_{\alpha}\mathbf{J}(\mathbf{r}')]\, d^3\mathbf{r}' \\ \end{align}

$\int_{\mathbb{V}'}J_{\alpha}(\mathbf{r}')\, d^3\mathbf{r}' =\int_{\mathbb{S}'}r'_{\alpha}\mathbf{J}(\mathbf{r}')\cdot\, d\mathbf{S}'=0$

\begin{align} \int_{\mathbb{V}'}\nabla'\cdot[r'_{\alpha}r'_{\beta}J(\mathbf{r}')]\, d^3\mathbf{r}' & =\int_{\mathbb{V}'}r'_{\beta}[J(\mathbf{r}')\cdot\nabla']r'_{\alpha} +r'_{\alpha}[J(\mathbf{r}')\cdot\nabla']r'_{\beta} +r'_{\alpha}r'_{\beta}\nabla'\cdot J(\mathbf{r}')\, d^3\mathbf{r}' \\ & =\int_{\mathbb{V}'}r'_{\beta}J_{\alpha}(\mathbf{r}')+r'_{\alpha}J_{\beta}(\mathbf{r}')\, d^3\mathbf{r}' \\ & = 0 \\ \end{align}

\begin{align} \mathbf{r}\cdot\int_{\mathbb{V'}}\mathbf{r}'J_{\alpha}(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' & = \frac{1}{2}\sum_{\beta=1}^3 r_{\beta}\int_{\mathbb{V'}}r'_{\beta}J_{\alpha}(\mathbf{r}')-r'_{\alpha}J_{\beta}(\mathbf{r}')\ \mathrm{d}^3\mathbf{r}' \\ & = -\ \frac{1}{2}\left[\mathbf{r}\times\int_{\mathbb{V'}} \mathbf{r}'\times\mathbf{J}(\mathbf{r})\ \mathrm{d}^3\mathbf{r}'\right]_{\alpha} \\ \end{align}

$\mathbf{m}\stackrel{def}{=}\ \frac{1}{2}\int_{\mathbb{V}'}\mathbf{r}'\times \mathbf{J}(\mathbf{r}')\, d^3\mathbf{r}'$

$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\frac{\mathbf{m}\times\mathbf{r}}{r^3}$

## 參考文獻

1. ^ 1.0 1.1 1.2 Jackson, John David, Classical Electrodynamic 3rd., USA: John Wiley & Sons, Inc., pp. 111, 145–151, 1999, ISBN 978-0-471-30932-1
2. ^ Ross D. Adamson. The Fast Multipole Method. January 21, 1999 [December 10, 2010].