贡献者: addis
与直角坐标系不同的是,极坐标系中的 $ \hat{\boldsymbol{\mathbf{r}}} $ 与 $ \hat{\boldsymbol{\mathbf{\theta}}} $ 都是坐标的函数,即 $ \hat{\boldsymbol{\mathbf{r}}} = \hat{\boldsymbol{\mathbf{r}}} (r,\theta)$,$ \hat{\boldsymbol{\mathbf{\theta}}} = \hat{\boldsymbol{\mathbf{\theta}}} (r,\theta)$,它们对坐标的偏导如下:
\begin{equation}
\left\{\begin{aligned}
\frac{\partial \hat{\boldsymbol{\mathbf{r}}} }{\partial r} &= 0\\
\frac{\partial \hat{\boldsymbol{\mathbf{r}}} }{\partial \theta} &= \hat{\boldsymbol{\mathbf{\theta}}} \end{aligned}\right. ~~~~,
\qquad
\left\{\begin{aligned}
\frac{\partial \hat{\boldsymbol{\mathbf{\theta}}} }{\partial r} &= 0\\
\frac{\partial \hat{\boldsymbol{\mathbf{\theta}}} }{\partial \theta} &= - \hat{\boldsymbol{\mathbf{r}}} \end{aligned}\right. ~~~~.
\end{equation}
这是容易理解的,若一个单位矢量绕着它的起点逆时针转动,那么它的终点的速度的方向必然是它本身逆时针旋转 90 度的方向,而大小等于矢量模长乘以角速度
。
1. 证明
如果令极轴方向的单位矢量为 $ \hat{\boldsymbol{\mathbf{x}}} $,令其逆时针旋转 $\pi/2$ 的矢量为 $ \hat{\boldsymbol{\mathbf{y}}} $,则
\begin{equation}
\hat{\boldsymbol{\mathbf{r}}} = \cos \theta \, \hat{\boldsymbol{\mathbf{x}}} + \sin \theta \, \hat{\boldsymbol{\mathbf{y}}} ~,
\end{equation}
\begin{equation}
\hat{\boldsymbol{\mathbf{\theta}}} = \cos\left(\theta +\pi/2\right) \, \hat{\boldsymbol{\mathbf{x}}} + \sin\left(\theta +\pi/2\right) \, \hat{\boldsymbol{\mathbf{y}}}
= -\sin \theta \, \hat{\boldsymbol{\mathbf{x}}} + \cos \theta \, \hat{\boldsymbol{\mathbf{y}}} ~,
\end{equation}
所以
\begin{equation}
\left\{\begin{aligned}
\frac{\partial \hat{\boldsymbol{\mathbf{r}}} }{\partial r} &= 0\\
\frac{\partial \hat{\boldsymbol{\mathbf{r}}} }{\partial \theta} &= - \sin \theta \, \hat{\boldsymbol{\mathbf{x}}} + \cos \theta \hat{\boldsymbol{\mathbf{y}}} = \, \hat{\boldsymbol{\mathbf{\theta}}} \end{aligned}\right. ~~~~,
\qquad
\left\{\begin{aligned}
\frac{\partial \hat{\boldsymbol{\mathbf{\theta}}} }{\partial r} &= 0\\
\frac{\partial \hat{\boldsymbol{\mathbf{\theta}}} }{\partial \theta} &= - \cos \theta \, \hat{\boldsymbol{\mathbf{x}}} - \sin \theta \, \hat{\boldsymbol{\mathbf{y}}} = - \hat{\boldsymbol{\mathbf{r}}}
\end{aligned}\right. ~~~~.\end{equation}
事实上,由于 $ \hat{\boldsymbol{\mathbf{r}}} $ 与 $ \hat{\boldsymbol{\mathbf{\theta}}} $ 都只是 $\theta$ 的函数,也可以把偏导符号改成导数符号
\begin{equation}
\frac{\mathrm{d}{ \hat{\boldsymbol{\mathbf{\theta}}} }}{\mathrm{d}{\theta}} = - \hat{\boldsymbol{\mathbf{r}}} ~,
\qquad
\frac{\mathrm{d}{ \hat{\boldsymbol{\mathbf{r}}} }}{\mathrm{d}{\theta}} = \hat{\boldsymbol{\mathbf{\theta}}} ~.
\end{equation}