Prerequisite 正交曲线坐标系中的矢量算符
球坐标系中标量函数 $u(r, \theta, \phi)$ 和矢量函数 $ \boldsymbol{\mathbf{v}} (r, \theta, \phi)$ 的梯度，散度，旋度和拉普拉斯算符的公式如下．其中 $r$ 是极径，$\theta $ 是极角，$\phi $ 是方位角．
梯度算符
\begin{equation}
\boldsymbol\nabla u = \frac{\partial u}{\partial r} \hat{\boldsymbol{\mathbf{r}}} + \frac{1}{r} \frac{\partial u}{\partial \theta} \hat{\boldsymbol{\mathbf{\theta}}} + \frac{1}{r\sin \theta } \frac{\partial u}{\partial \phi} \hat{\boldsymbol{\mathbf{\phi}}}
\end{equation}
散度算符
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{v}} = \frac{1}{r^2} \frac{\partial}{\partial{r}} (r^2 v_r) + \frac{1}{r\sin \theta} \frac{\partial}{\partial{\theta}} (\sin\theta v_\theta) + \frac{1}{r\sin \theta} \frac{\partial v_\phi}{\partial \phi}
\end{equation}
旋度算符
\begin{equation} \begin{aligned}
\boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{v}} = & \frac{1}{r\sin \theta} \left[ \frac{\partial}{\partial{\theta}} (\sin \theta v_\phi) - \frac{\partial}{\partial{\phi}} \right] \hat{\boldsymbol{\mathbf{r}}} + \frac1r \left[\frac{1}{\sin \theta} \frac{\partial v_r}{\partial \phi} - \frac{\partial}{\partial{r}} (r v_\phi) \right] \hat{\boldsymbol{\mathbf{\theta}}} \\
&+ \frac1r \left[ \frac{\partial}{\partial{r}} (r v_\theta) - \frac{\partial v_r}{\partial \theta} \right] \hat{\boldsymbol{\mathbf{\phi}}}
\end{aligned} \end{equation}
拉普拉斯算符
\begin{equation}
\boldsymbol{\nabla}^2 u = \boldsymbol{\nabla}\boldsymbol{\cdot} ( \boldsymbol\nabla u) = \frac{1}{r^2} \frac{\partial}{\partial{r}} \left(r^2 \frac{\partial u}{\partial r} \right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial{\theta}} \left(\sin \theta \frac{\partial u}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^{2}{u}}{\partial{\phi}^{2}}
\end{equation}
拉普拉斯算符的分解
为了书写方便本书中定义两个算符 $ \boldsymbol{\nabla}^2 _r$ 和 $ \boldsymbol{\nabla}^2 _\Omega$ 满足
\begin{equation}
\boldsymbol{\nabla}^2 = \boldsymbol{\nabla}^2 _r + \frac{ \boldsymbol{\nabla}^2 _\Omega}{r^2}
\end{equation}
其中 $ \boldsymbol{\nabla}^2 _r$ 只对 $r$ 求偏导，$ \boldsymbol{\nabla}^2 _\Omega$ 只对 $\theta,\phi$ 求偏导．$ \boldsymbol{\nabla}^2 _\Omega$ 还可以进一步分解为
\begin{equation}
\boldsymbol{\nabla}^2 _\Omega = \boldsymbol\nabla _\Omega \boldsymbol\cdot \boldsymbol\nabla _\Omega
\end{equation}
其中（详见 “球坐标系中的角动量算符
”）
\begin{equation}
\boldsymbol\nabla _\Omega = - \left(\sin\phi \frac{\partial}{\partial{\theta}} + \cot\theta\cos\phi \frac{\partial}{\partial{\phi}} \right) \hat{\boldsymbol{\mathbf{x}}}
+ \left(\cos\phi \frac{\partial}{\partial{\theta}} - \cot\theta \sin\phi \frac{\partial}{\partial{\phi}} \right) \hat{\boldsymbol{\mathbf{y}}}
+ \frac{\partial}{\partial{\phi}} \hat{\boldsymbol{\mathbf{z}}}
\end{equation}
推导
球坐标系中，位置矢量可以表示为
\begin{equation}
\boldsymbol{\mathbf{r}} = r \hat{\boldsymbol{\mathbf{r}}} = r\sin\theta\cos\phi\, \hat{\boldsymbol{\mathbf{x}}} + r\sin\theta\sin\phi\, \hat{\boldsymbol{\mathbf{y}}} + r\cos\theta \hat{\boldsymbol{\mathbf{z}}}
\end{equation}
同样，球坐标系的三个单位矢量由三个坐标增加的方向确定
\begin{equation}
\left\{\begin{aligned}
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial r} &= \sin\theta\cos\phi\, \hat{\boldsymbol{\mathbf{x}}} + \sin\theta\sin\phi\, \hat{\boldsymbol{\mathbf{y}}} + \cos\theta\, \hat{\boldsymbol{\mathbf{z}}} \\
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial \theta} &= r\cos\theta\cos\phi\, \hat{\boldsymbol{\mathbf{x}}} + r\cos\theta\sin\phi\, \hat{\boldsymbol{\mathbf{y}}} - r\sin\theta\, \hat{\boldsymbol{\mathbf{z}}} \\
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial \phi} &= -r\sin\theta\sin\phi\, \hat{\boldsymbol{\mathbf{x}}} + r\sin\theta\cos\phi\, \hat{\boldsymbol{\mathbf{y}}}
\end{aligned}\right. \end{equation}
归一化得三个单位矢量为
\begin{equation}
\begin{cases}
\hat{\boldsymbol{\mathbf{r}}} = \sin\theta\cos\phi\, \hat{\boldsymbol{\mathbf{x}}} + \sin\theta\sin\phi\, \hat{\boldsymbol{\mathbf{y}}} + \cos\theta\, \hat{\boldsymbol{\mathbf{z}}} \\
\hat{\boldsymbol{\mathbf{\theta}}} = \cos\theta\cos\phi\, \hat{\boldsymbol{\mathbf{x}}} + \cos\theta\sin\phi\, \hat{\boldsymbol{\mathbf{y}}} - \sin\theta\, \hat{\boldsymbol{\mathbf{z}}} \\
\hat{\boldsymbol{\mathbf{\phi}}} = -\sin\phi\, \hat{\boldsymbol{\mathbf{x}}} + \cos\phi\, \hat{\boldsymbol{\mathbf{y}}}
\end{cases}
\end{equation}
不难验证这三个单位矢量两两间内积为零，即两两垂直，所以球坐标系也属于正交曲线坐标系．对比
eq. 9 与
eq. 10 ，有
\begin{equation}
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial r} = \hat{\boldsymbol{\mathbf{r}}} \qquad
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial \theta} = r \hat{\boldsymbol{\mathbf{\theta}}} \qquad
\frac{\partial \boldsymbol{\mathbf{r}} }{\partial \phi} = r\sin\theta\, \hat{\boldsymbol{\mathbf{\phi}}}
\end{equation}
所以位置矢量的微分可以表示为
\begin{equation}
\,\mathrm{d}{ \boldsymbol{\mathbf{r}} } = \hat{\boldsymbol{\mathbf{r}}} \,\mathrm{d}{r} + r \hat{\boldsymbol{\mathbf{\theta}}} \,\mathrm{d}{\theta} + r\sin\theta \hat{\boldsymbol{\mathbf{\phi}}} \,\mathrm{d}{\phi}
\end{equation}
代入
eq. 1 到
eq. 6 即可完成推导．