事情是这样的，有 Jacobian 雅各比矩阵算转换坐标系的，但没人算单位矢量转换矩阵，其实只需要一点点微积分的只是就可以推导，比代数法好算很多，几乎是直写的。有一点点技巧在里面，道理并不复杂。

假设某函数从 $f : ℝ_n → ℝ_m$， 从 $x ∈ ℝ_n$ 映射到 向量 $f(x) ∈ ℝ_m$， 其雅可比矩阵是一 $m×n$ 的矩阵，也就是从 $ℝ_n 到 ℝ_m$ 的线性映射，表现了一个多变数向量函数的最佳线性逼近。

$\mathbf{J}=\left[\begin{array}{lll}\frac{\partial \mathbf{f}}{\partial x_{1}} & \cdots & \frac{\partial \mathbf{f}}{\partial x_{n}}\end{array}\right]=\left[\begin{array}{ccc}\frac{\partial f_{1}}{\partial x_{1}} & \cdots & \frac{\partial f_{1}}{\partial x_{n}} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{m}}{\partial x_{1}} & \cdots & \frac{\partial f_{m}}{\partial x_{n}}\end{array}\right]$ , 矩阵的分量表示成：$\mathbf{J}_{i j}=\frac{\partial f_{i}}{\partial x_{j}}$.

1. 圆柱坐标系与直角坐标系间坐标分量的矩阵关系

圆柱坐标系到直角坐标系的转化，结果由 $\mathbf{F}: \mathbb{R}^{+} \times[0, \pi) \times[0,2 \pi) \rightarrow \mathbb{R}^{3}$ 函数给出, 其分量为 :

$$ \begin{aligned} &x= \rho \cos \phi \\ &y= \rho \sin \phi \\ &z= z \end{aligned} $$

$$ \mathbf{J}_{\mathbf{F}}(\rho, \phi, z)=\left[\begin{array}{lll} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial z} \end{array}\right]=\left[\begin{array}{ccc} \cos \phi & -\rho\sin \phi & 0 \\ \sin \phi & \rho \cos \phi & 0 \\ 0 & 0 & 1 \end{array}\right] $$

结合梯度公式：$\nabla f=\frac{\partial f}{\partial \rho} \vec{e}_{\rho}+\frac{1}{r} \frac{\partial f}{\partial \phi} \vec{e}_{\phi}+\frac{\partial f}{\partial r} \vec{e}_{z}$

$$ \left[\begin{array}{c}\vec{e}_{x} \\ \vec{e}_{y} \\ \vec{e}_{z}\end{array}\right]= \left[\begin{array}{l}\frac{\partial \vec{e}}{\partial x} \\ \frac{\partial \vec{e}}{\partial y} \\ \frac{\partial \vec{e}}{\partial z}\end{array}\right] = \left[\begin{array}{lll} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial z} \end{array}\right] \left[\begin{array}{l}\frac{\partial \vec{e}}{\partial \rho} \\ \frac{\partial \vec{e}}{\partial \phi} \\ \frac{\partial \vec{e}}{\partial z}\end{array}\right] =\mathbf{J}_{\mathbf{F}}(\rho, \phi, z)\left[\begin{array}{c}\vec{e}_{\rho} \\ \rho\vec{e}_{\phi} \\ \vec{e}_{z}\end{array}\right] \\ =\left[\begin{array}{ccc} \cos \phi & -\rho\sin \phi & 0 \\ \sin \phi & \rho \cos \phi & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c}\vec{e}_{\rho} \\ \rho\vec{e}_{\phi} \\ \vec{e}_{z}\end{array}\right] \\=\left[\begin{array}{ccc}\cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{c}\vec{e}_{\rho} \\ \vec{e}_{\phi} \\ \vec{e}_{z}\end{array}\right] $$

2. 球坐标系与直角坐标系间坐标分量的矩阵关系

球坐标系到直角坐标系的转化，结果由 $\mathbf{F}: \mathbb{R}^{+} \times[0, \pi) \times[0,2 \pi) \rightarrow \mathbb{R}^{3}$ 函数给出, 其分量为 :

$$ \begin{aligned} &x=r \sin \theta \cos \phi \\ &y=r \sin \theta \sin \phi \\ &z=r \cos \theta \end{aligned} $$

此坐标变换的雅可比矩阵是

$$ \mathbf{J}_{\mathbf{F}}(r, \theta, \phi)=\left[\begin{array}{lll} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{array}\right]=\left[\begin{array}{ccc} \sin \theta \cos \phi & r \cos \theta \cos \phi & -r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & -r \sin \theta & 0 \end{array}\right] $$

结合梯度公式：$\nabla f=\frac{\partial f}{\partial r} \vec{e}_{r}+\frac{1}{r} \frac{\partial f}{\partial \theta} \vec{e}_{\theta}+\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \vec{e}_{\phi}$

$$ \left[\begin{array}{c}\vec{e}_{x} \\ \vec{e}_{y} \\ \vec{e}_{z}\end{array}\right]= \left[\begin{array}{l}\frac{\partial \vec{e}}{\partial x} \\ \frac{\partial \vec{e}}{\partial y} \\ \frac{\partial \vec{e}}{\partial z}\end{array}\right] = \left[\begin{array}{lll} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{array}\right] \left[\begin{array}{l}\frac{\partial \vec{e}}{\partial r} \\ \frac{\partial \vec{e}}{\partial \theta} \\ \frac{\partial \vec{e}}{\partial \phi}\end{array}\right] =\mathbf{J}_{\mathbf{F}}(r, \theta, \phi)\left[\begin{array}{c}\vec{e}_{\rho} \\ r\vec{e}_{\theta} \\ rsin\theta \vec{e}_{\phi}\end{array}\right] \\ =\left[\begin{array}{ccc} \sin \theta \cos \phi & r \cos \theta \cos \phi & -r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & -r \sin \theta & 0 \end{array}\right]\left[\begin{array}{c}\vec{e}_{\rho} \\ r\vec{e}_{\theta} \\ rsin\theta \vec{e}_{\phi}\end{array}\right] \\=\left[\begin{array}{ccc}\sin \theta \cos \phi & \cos \theta \cos \phi & -\sin \phi \\ \sin \theta \sin \phi & \cos \theta \sin \phi & \cos \phi \\ \cos \theta & -\sin \theta & 0\end{array}\right]\left[\begin{array}{c}\vec{e}_{r} \\ \vec{e}_{\theta} \\ \vec{e}_{\phi}\end{array}\right] $$