向量场的旋度与散度（三）：不同坐标系中的梯度、旋度与散度

上篇 -- 向量场的旋度与散度（二）：常用恒等式及其证明

柱坐标系的梯度、旋度与散度

梯度

根据定义，梯度就是函数值增加最快的“速度”，包括方向和大小。很容易可以想到梯度就应该是函数$f(\rho,\varphi,z)$在$\boldsymbol{\hat{\rho}},\boldsymbol{\hat{\varphi}},\boldsymbol{\hat{z}}$三个方向上偏导数的线性叠加。结合

$$\begin{cases}f_\rho=\lim_{\Delta{\rho}\rightarrow 0}\frac{f(\rho+\Delta{\rho},\varphi,z)-f(\rho,\varphi,z)}{\Delta{\rho}}=\frac{\partial{f}}{\partial{r}}\\f_\varphi=\lim_{\Delta{\varphi}\rightarrow 0}\frac{f(\rho,\varphi+\Delta{\varphi},z)-f(\rho,\varphi,z)}{\rho\Delta{\varphi}}=\frac{1}{\rho}\frac{\partial{f}}{\partial{\varphi}}\\f_z=\lim_{\Delta{z}\rightarrow 0}\frac{f(\rho,\varphi,z+\Delta{z})-f(\rho,\varphi,z)}{\Delta{z}}=\frac{\partial{f}}{\partial{z}}\end{cases}\tag{1}$$ 可以得到 $$\nabla{f}=\frac{\partial{f}}{\partial{\rho}}\boldsymbol{\hat{\rho}}+\frac{1}{\rho}\frac{\partial{f}}{\partial{\varphi}}\boldsymbol{\hat{\varphi}}+\frac{\partial{f}}{\partial{z}}\boldsymbol{\hat{z}}\tag{2}$$ 这里$\boldsymbol{\hat{\varphi}}$与$\boldsymbol{\hat{\rho}},\boldsymbol{\hat{z}}$系数不同的原因在于$\varphi$本身表示的是角度，而$\rho,z$表示的都是长度。

散度

在系列第一节，我们在直角坐标系里推到过向量场散度的表达式，当时我们的推导是直接考虑长方体的通量，进而推导出散度。这是仅限于直角坐标系的做法，更普遍的做法应该是取体积微元计算通量密度，只不过直角坐标系中长方体的微元比较好处理。
在柱坐标系中，我们很容易知道体积微元$\text{d}V=\rho\text{d}\rho\text{d}\varphi\text{d}z$，接下来就需要求向量场$\boldsymbol{F}=F_\rho\boldsymbol{\hat{\rho}}+F_\varphi\boldsymbol{\hat{\varphi}}+F_z\boldsymbol{\hat{z}}$在点$P(\rho,\varphi,z)$附近的通量。由于取得每一个面的面积很小，所以可以将向量场在每个面上看成是均匀的，于是有三个方向上的通量

$$\text{d}\Phi_\rho=F_\rho(\rho+\text{d}\rho,\varphi,z)(\rho+\text{d}\rho)\text{d}\varphi\text{d}z-F_\rho(\rho,\varphi,z)\rho\text{d}\varphi\text{d}z\tag{3.1}$$ $$\text{d}\Phi_\varphi=F_\varphi(\rho,\varphi+\text{d}\varphi,z)\text{d}\rho\text{d}z-F_\varphi(\rho,\varphi,z)\text{d}\rho\text{d}z\tag{3.2}$$ $$\text{d}\Phi_z=F_z(\rho,\varphi,z+\text{d}z)\rho\text{d}\rho\text{d}\varphi-F_z(\rho,\varphi,z)\rho\text{d}\rho\text{d}\varphi\tag{3.3}$$ 三个式子同时比上体积微元$\text{d}V$得到 $$\frac{\text{d}\Phi_\rho}{\text{d}V}=\frac{F_\rho(\rho+\text{d}\rho,\varphi,z)(\rho+\text{d}\rho)\text{d}\varphi\text{d}z-F_\rho(\rho,\varphi,z)\rho\text{d}\varphi\text{d}z}{\rho\text{d}\rho\text{d}\varphi\text{d}z}\\=\frac{1}{\rho}\frac{(\rho F_\rho)\vert_{\rho}^{\rho+\text{d}\rho}}{\text{d}\rho}\\=\frac{1}{\rho}\frac{\partial(\rho F_\rho)}{\partial{\rho}}\tag{4.1}$$ $$\frac{\text{d}\Phi_\varphi}{\text{d}V}=\frac{F_\varphi(\rho,\varphi+\text{d}\varphi,z)\text{d}\rho\text{d}z-F_\varphi(\rho,\varphi,z)\text{d}\rho\text{d}z}{\rho\text{d}\rho\text{d}\varphi\text{d}z}\\=\frac{1}{\rho}\frac{F_\varphi\vert_{\varphi}^{\varphi+\text{d}\varphi}}{\text{d}\varphi}\\=\frac{1}{\rho}\frac{\partial{F_\varphi}}{\partial{\varphi}}\tag{4.2}$$ $$\frac{\text{d}\Phi_z}{\text{d}V}=\frac{F_z(\rho,\varphi,z+\text{d}z)\rho\text{d}\rho\text{d}\varphi-F_z(\rho,\varphi,z)\rho\text{d}\rho\text{d}\varphi}{\rho\text{d}\rho\text{d}\varphi\text{d}z}\\=\frac{F_z\vert_{z}^{z+\text{d}z}}{\text{d}z}\\=\frac{\partial{F_z}}{\partial{z}}\tag{4.3}$$ 综合三式得到 $$\nabla\cdot\boldsymbol{F}=\frac{1}{\rho}\frac{\partial(\rho F_\rho)}{\partial{\rho}}+\frac{1}{\rho}\frac{\partial{F_\varphi}}{\partial{\varphi}}+\frac{\partial{F_z}}{\partial{z}}\tag{4}$$

旋度

旋度的处理思路同样需要取微元，不同之处在于需要对$\boldsymbol{\hat{\rho}},\boldsymbol{\hat{\varphi}},\boldsymbol{\hat{z}}$三个方向分别处理。先考虑绕$\boldsymbol{\hat{\rho}}$的旋度

$$(\nabla\times\boldsymbol{F})_\rho=\frac{-F_z(\rho,\varphi,z)\text{d}z-F_\varphi(\rho,\varphi,z+\text{d}z)\rho\text{d}\varphi}{\rho\text{d}\varphi\text{d}z}\\+\frac{F_z(\rho,\varphi+\text{d}\varphi,z)\text{d}z+F_\varphi(\rho,\varphi,z)\rho\text{d}\varphi}{\rho\text{d}\varphi\text{d}z}\\=\frac{1}{\rho}\frac{F_z\vert_{\varphi}^{\varphi+\text{d}\varphi}}{\text{d}\varphi}+\frac{F_\varphi\vert_{z+\text{d}z}^{z}}{\text{d}z}\\=\frac{1}{\rho}\frac{\partial{F_z}}{\partial{\varphi}}-\frac{\partial{F_\varphi}}{\partial{z}}\tag{5.1}$$ 在另外两个方向上的旋度也不难得到，这里就不展开说明了 $$(\nabla\times\boldsymbol{F})_\varphi=\frac{\partial{F_\rho}}{\partial{z}}-\frac{\partial{F_z}}{\partial{\rho}}\tag{5.2}$$ $$(\nabla\times\boldsymbol{F})_z=\frac{\partial{F_\varphi}}{\partial{\rho}}-\frac{1}{\rho}\frac{\partial{F_\rho}}{\partial{\varphi}}\tag{5.3}$$ 三式相加并适当变形可以得到 $$\nabla\times\boldsymbol{F}=(\frac{1}{\rho}\frac{\partial{F_z}}{\partial{\varphi}}-\frac{\partial{F_\varphi}}{\partial{z}})\boldsymbol{\hat{\rho}}+(\frac{\partial{F_\rho}}{\partial{z}}-\frac{\partial{F_z}}{\partial{\rho}})\boldsymbol{\hat{\varphi}}+(\frac{\partial{F_\varphi}}{\partial{\rho}}-\frac{1}{\rho}\frac{\partial{F_\rho}}{\partial{\varphi}})\boldsymbol{\hat{z}}\\=\frac{1}{\rho}[(\frac{\partial{F_z}}{\partial{\varphi}}-\frac{\partial(\rho F_\varphi)}{\partial{z}})\boldsymbol{\hat{\rho}}+(\frac{\partial{F_\rho}}{\partial{z}}-\frac{\partial{F_z}}{\partial{\rho}})\rho\boldsymbol{\hat{\varphi}}+(\frac{\partial(\rho F_\varphi)}{\partial{\rho}}-\frac{\partial{F_\rho}}{\partial{\varphi}})\boldsymbol{\hat{z}}]\\=\frac{1}{\rho}\left\vert\begin{matrix}\boldsymbol{\hat{\rho}}&\rho\boldsymbol{\hat{\varphi}}&\boldsymbol{\hat{z}}\\\frac{\partial}{\partial{\rho}}&\frac{\partial}{\partial{\varphi}}&\frac{\partial}{\partial{z}}\\F_\rho&\rho F_\varphi&F_z\end{matrix}\right\vert\tag{5}$$

球坐标系的梯度、旋度与散度

梯度

求球坐标系的梯度和求柱坐标系的思路是类似的，我们只要注意到$\boldsymbol{\hat{r}},\boldsymbol{\hat{\theta}},\boldsymbol{\hat{\phi}}$三个方向上不同的长度表达式就可以得到

$$\nabla{f}=\frac{\partial{f}}{\partial{r}}\boldsymbol{\hat{r}}+\frac{1}{r}\frac{\partial{f}}{\partial{\theta}}\boldsymbol{\hat{\theta}}+\frac{1}{r\sin{\theta}}\frac{\partial{f}}{\partial{\phi}}\boldsymbol{\hat{\phi}}\tag{6}$$

散度

不难得到体积微元$\text{d}V=r^2\sin{\theta}\text{d}r\text{d}\theta\text{d}\phi$，通量分别为

$$\text{d}\Phi_r=F_r(r+\text{d}r,\theta,\phi)(r+\text{d}r)^2\sin{\theta}\text{d}\theta\text{d}\phi-F_r(r,\theta,\phi)r^2\sin{\theta}\text{d}\theta\text{d}\phi\tag{7.1}$$ $$\text{d}\Phi_\theta=F_\theta(r,\theta+\text{d}\theta,\phi)r\sin(\theta+\text{d}\theta)\text{d}r\text{d}\phi-F_\theta(r,\theta,\phi)r\sin{\theta}\text{d}r\text{d}\phi\tag{7.2}$$ $$\text{d}\Phi_\phi=F_\phi(r,\theta,\phi+\text{d}\phi)r\text{d}r\text{d}\theta-F_\phi(r,\theta,\phi)r\text{d}r\text{d}\theta\tag{7.3}$$ 三个式子分别得到通量密度后相加得散度 $$\nabla\cdot\boldsymbol{F}=\frac{1}{r^2}\frac{\partial(r^2F_r)}{\partial{r}}+\frac{1}{r\sin{\theta}}\frac{\partial(\sin{\theta}F_\theta)}{\partial{\theta}}+\frac{1}{r\sin{\theta}}\frac{\partial{F_\phi}}{\partial{\phi}}\tag{7}$$

旋度

三个方向上的旋度分别为

$$(\nabla\times\boldsymbol{F})_r=\frac{1}{r}\frac{\partial{F_\phi}}{\partial{\theta}}-\frac{1}{r\sin{\theta}}\frac{\partial{F_\theta}}{\partial{\phi}}\tag{8.1}$$ $$(\nabla\times\boldsymbol{F})_\theta=\frac{1}{r\sin{\theta}}\frac{\partial{F_r}}{\partial{\phi}}-\frac{\partial{F_\phi}}{\partial{r}}\tag{8.2}$$ $$(\nabla\times\boldsymbol{F})_\phi=\frac{\partial{F_\theta}}{\partial{r}}-\frac{1}{r}\frac{\partial{F_r}}{\partial{\theta}}\tag{8.3}$$ 因此 $$\nabla\times\boldsymbol{F}=(\frac{1}{r}\frac{\partial{F_\phi}}{\partial{\theta}}-\frac{1}{r\sin{\theta}}\frac{\partial{F_\theta}}{\partial{\phi}})\boldsymbol{\hat{r}}+(\frac{1}{r\sin{\theta}}\frac{\partial{F_r}}{\partial{\phi}}-\frac{\partial{F_\phi}}{\partial{r}})\boldsymbol{\hat{\theta}}+(\frac{\partial{F_\theta}}{\partial{r}}-\frac{1}{r}\frac{\partial{F_r}}{\partial{\theta}})\boldsymbol{\hat{\phi}}\\=\frac{1}{r^2\sin{\theta}}[(\frac{\partial(r\sin{\theta}F_\phi)}{\partial{\theta}}-\frac{\partial(rF_\theta)}{\partial{\phi}})\boldsymbol{\hat{r}}+(\frac{\partial{F_r}}{\partial{\phi}}-\frac{\partial(r\sin{\theta}F_\phi)}{\partial{r}})r\boldsymbol{\hat{\theta}}+(\frac{\partial(rF_\theta)}{\partial{r}}-\frac{\partial{F_r}}{\partial{\theta}})r\sin{\theta}\boldsymbol{\hat{\phi}}]\\=\frac{1}{r^2\sin{\theta}}\left\vert\begin{matrix}\boldsymbol{\hat{r}}&r\boldsymbol{\hat{\theta}}&r\sin{\theta}\boldsymbol{\hat{\phi}}\\\frac{\partial}{\partial{r}}&\frac{\partial}{\partial{\theta}}&\frac{\partial}{\partial{\phi}}\\F_r&rF_\theta&r\sin{\theta}F_\phi\end{matrix}\right\vert\tag{8}$$

正交曲线坐标系的梯度、旋度与散度

柱坐标系和球坐标系同属于曲线坐标系，更重要的是它们还都是正交曲线坐标系(orthogonal curvilinear coordinate system)。
正交曲线坐标系是什么样的坐标系？让我们先考虑三维空间中一个普通的曲线坐标系，设$u,v,w$分别是坐标系三个坐标。当我们固定$v,w$的取值而让$u$取遍定义域上所有取值时，可以得到一条曲线，这就是坐标曲线；当我们固定$u$的取值而让$v,w$取遍定义域上所有取值时，可以得到一条曲面，这就是坐标曲面。对于三个坐标中任意两个坐标我们都可以进行类似的操作，因此通过空间中有定义的任意一点，我们都可以构造出三条坐标曲线和三个坐标曲面。
过点$P(u_0,v_0,w_0)$，构造出三个坐标分别对应的坐标曲线$l_1,l_2,l_3$。在过$P$的$l_1$的切线上，我们取沿$u$坐标增大的方向的单位向量$\boldsymbol{\hat{u}}$作为$P$点处的$u$坐标的单位向量。同样的方法可以得到$\boldsymbol{\hat{v}}$和$\boldsymbol{\hat{w}}$。如果对于任意的点$P$都有$\boldsymbol{\hat{u}},\boldsymbol{\hat{v}},\boldsymbol{\hat{w}}$两两正交，那么以$u,v,w$为坐标的曲线坐标系就是一个正交曲线坐标系。不难发现无论是柱坐标系的$\boldsymbol{\hat{\rho}},\boldsymbol{\hat{\varphi}},\boldsymbol{\hat{z}}$单位向量，还是球坐标系的$\boldsymbol{\hat{r}},\boldsymbol{\hat{\theta}},\boldsymbol{\hat{\phi}}$单位向量都满足两两正交，因此柱坐标系和球坐标系都是正交曲线坐标系。

要想得到正交曲线坐标系下的梯度、旋度和散度，就必须先得出坐标系中的长度元、面积元和体积元。面积元和体积元都可以通过长度元相乘求得，因此我们现在重点关注长度元。
考虑以$q_1,q_2,q_3$为坐标的正交曲线坐标系，对应的坐标曲线为$l_1,l_2,l_3$，那么三个方向的长度元就可以记为$\text{d}l_1,\text{d}l_2,\text{d}l_3$。我们不知道$\text{d}l_1$在正交曲线坐标中的表达式，但我们知道在直角坐标系里有

$$\text{d}l_1=\sqrt{(\text{d}x)^2+(\text{d}y)^2+(\text{d}z)^2}\tag{9}$$ 在$l_1$上，$q_2,q_3$的取值都为定值，因此$x,y,z$的全微分都只有和$q_1$有关的项，即 $$\begin{cases}\text{d}x=\frac{\partial{x}}{\partial{q_1}}\text{d}q_1\\\text{d}y=\frac{\partial{y}}{\partial{q_1}}\text{d}q_1\\\text{d}z=\frac{\partial{z}}{\partial{q_1}}\text{d}q_1\end{cases}\tag{10}$$ 代回$(9)$即可得 $$\text{d}l_1=\sqrt{(\frac{\partial{x}}{\partial{q_1}})^2+(\frac{\partial{y}}{\partial{q_1}})^2+(\frac{\partial{z}}{\partial{q_1}})^2}\text{d}q_1\tag{9'}$$ 对三个方向都进行这样的操作后，我们可以把根式提取出来 $$H_i=\sqrt{(\frac{\partial{x}}{\partial{q_i}})^2+(\frac{\partial{y}}{\partial{q_i}})^2+(\frac{\partial{z}}{\partial{q_i}})^2}\space(i=1,2,3)\tag{11}$$ 提取出来的$H_i$就被称为拉梅系数。
现在要表达长度元就很方便了，直接有 $$\begin{cases}\text{d}l_1=H_1\text{d}q_1\\\text{d}l_2=H_2\text{d}q_2\\\text{d}l_3=H_3\text{d}q_3\end{cases}\tag{12}$$ 面积元和体积元也不难得到 $$\begin{cases}\text{d}S_{1}=\text{d}l_2\text{d}l_3=H_2H_3\text{d}q_2\text{d}q_3\\\text{d}S_{2}=\text{d}l_3\text{d}l_1=H_3H_1\text{d}q_3\text{d}q_1\\\text{d}S_{3}=\text{d}l_1\text{d}l_2=H_1H_2\text{d}q_1\text{d}q_2\end{cases}\tag{13}$$ $$\text{d}V=\text{d}l_1\text{d}l_2\text{d}l_3=H_1H_2H_3\text{d}q_1\text{d}q_2\text{d}q_3\tag{14}$$
对于梯度、旋度和散度，有了前面柱坐标系和球坐标系的推导，我们不难总结出 $$\nabla{f}=\frac{1}{H_1}\frac{\partial{f}}{\partial{q_1}}\boldsymbol{\hat{q_1}}+\frac{1}{H_2}\frac{\partial{f}}{\partial{q_2}}\boldsymbol{\hat{q_2}}+\frac{1}{H_3}\frac{\partial{f}}{\partial{q_3}}\boldsymbol{\hat{q_3}}\tag{15}$$ $$\nabla\cdot\boldsymbol{F}=\frac{1}{H_1H_2H_3}(\frac{\partial(F_1H_2H_3)}{\partial{q_1}}+\frac{\partial(F_2H_3H_1)}{\partial{q_2}}+\frac{\partial(F_3H_1H_2)}{\partial{q_3}})\tag{16}$$ $$\nabla\times\boldsymbol{F}=\frac{1}{H_1H_2H_3}\left\vert\begin{matrix}H_1\boldsymbol{\hat{q_1}}&H_2\boldsymbol{\hat{q_2}}&H_3\boldsymbol{\hat{q_3}}\\\frac{\partial}{\partial{q_1}}&\frac{\partial}{\partial{q_2}}&\frac{\partial}{\partial{q_3}}\\H_1F_1&H_2F_2&H_3F_3\end{matrix}\right\vert\tag{17}$$
读者可以自行证明，附上柱坐标系和球坐标系的拉梅系数 $$H_\rho=1,H_\varphi=\rho,H_z=1\tag{18.1}$$ $$H_r=1,H_\theta=r,H_\phi=r\sin{\theta}\tag{18.2}$$