我们经常需要在一些问题中研究坐标系的关系，这里讲讲最常见的极坐标和直角坐标的雅克比矩阵的推导。以二维坐标为例，三维坐标也是同理。

直角坐标表示为 ( x , y ) (x,y) (x,y)，极坐标表示为 ( ρ , φ ) (\rho,\varphi) (ρ,φ)，它们之间有如下的关系： ρ 2 = x 2 + y 2 , φ = arctan ⁡ y x ; x = ρ cos ⁡ φ , y = ρ sin ⁡ φ \begin{aligned} \rho^2=x^2+y^2,\quad &\varphi=\arctan\frac{y}{x};\\ x=\rho\cos\varphi,\quad&y=\rho\sin\varphi \end{aligned} ρ2=x2+y2,x=ρcosφ,​φ=arctanxy​;y=ρsinφ​

向量X和向量Y的微分映射由雅克比矩阵来刻画，给定两个向量 x = ( x 1 , x 2 , ⋯   , x n ) T \mathbf{x}=(x_1,x_2,\cdots,x_n)^T x=(x1​,x2​,⋯,xn​)T， y = ( y 1 , y 2 , ⋯   , y m ) T \mathbf{y}=(y_1,y_2,\cdots,y_m)^T y=(y1​,y2​,⋯,ym​)T

{ d x 1 = ∂ x 1 ∂ y 1 d y 1 + ∂ x 1 ∂ y 2 d y 2 + ⋯ + ∂ x 1 ∂ y m d y m d x 2 = ∂ x 2 ∂ y 1 d y 1 + ∂ x 2 ∂ y 2 d y 2 + ⋯ + ∂ x 2 ∂ y m d y m ⋮ d x n = ∂ x n ∂ y 1 d y 1 + ∂ x n ∂ y 2 d y 2 + ⋯ + ∂ x n ∂ y m d y m \begin{aligned} \begin{cases} \mathrm{d}x_1=\dfrac{\partial x_1}{\partial y_1}\mathrm{d}y_1+\dfrac{\partial x_1}{\partial y_2}\mathrm{d}y_2+\cdots+\dfrac{\partial x_1}{\partial y_m}\mathrm{d}y_m\\ \mathrm{d}x_2=\dfrac{\partial x_2}{\partial y_1}\mathrm{d}y_1+\dfrac{\partial x_2}{\partial y_2}\mathrm{d}y_2+\cdots+\dfrac{\partial x_2}{\partial y_m}\mathrm{d}y_m\\ \vdots\\ \mathrm{d}x_n=\dfrac{\partial x_n}{\partial y_1}\mathrm{d}y_1+\dfrac{\partial x_n}{\partial y_2}\mathrm{d}y_2+\cdots+\dfrac{\partial x_n}{\partial y_m}\mathrm{d}y_m\\ \end{cases} \end{aligned} ⎩ ⎨ ⎧​dx1​=∂y1​∂x1​​dy1​+∂y2​∂x1​​dy2​+⋯+∂ym​∂x1​​dym​dx2​=∂y1​∂x2​​dy1​+∂y2​∂x2​​dy2​+⋯+∂ym​∂x2​​dym​⋮dxn​=∂y1​∂xn​​dy1​+∂y2​∂xn​​dy2​+⋯+∂ym​∂xn​​dym​​​

写成矩阵的形式就是：

( d x 1 d x 2 ⋮ d x n ) = [ ∂ x 1 ∂ y 1 ∂ x 1 ∂ y 2 ⋯ ∂ x 1 ∂ y m ∂ x 2 ∂ y 1 ∂ x 2 ∂ y 2 ⋯ ∂ x 2 ∂ y m ⋮ ⋮ ⋮ ∂ x n ∂ y 1 ∂ x n ∂ y 2 ⋯ ∂ x n ∂ y m ] ( d y 1 d y 2 ⋮ d y m ) \begin{pmatrix} \mathrm{d}x_1\\ \mathrm{d}x_2\\ \vdots\\ \mathrm{d}x_n \end{pmatrix} =\begin{bmatrix} \dfrac{\partial x_1}{\partial y_1} & \dfrac{\partial x_1}{\partial y_2} & \cdots & \dfrac{\partial x_1}{\partial y_m}\\ \dfrac{\partial x_2}{\partial y_1} & \dfrac{\partial x_2}{\partial y_2} & \cdots &\dfrac{\partial x_2}{\partial y_m} \\ \vdots & \vdots & & \vdots\\ \dfrac{\partial x_n}{\partial y_1} & \dfrac{\partial x_n}{\partial y_2} & \cdots &\dfrac{\partial x_n}{\partial y_m} \end{bmatrix}\begin{pmatrix} \mathrm{d}y_1\\ \mathrm{d}y_2\\ \vdots\\ \mathrm{d}y_m \end{pmatrix} ​dx1​dx2​⋮dxn​​ ​= ​∂y1​∂x1​​∂y1​∂x2​​⋮∂y1​∂xn​​​∂y2​∂x1​​∂y2​∂x2​​⋮∂y2​∂xn​​​⋯⋯⋯​∂ym​∂x1​​∂ym​∂x2​​⋮∂ym​∂xn​​​ ​ ​dy1​dy2​⋮dym​​ ​

其中的矩阵

∂ ( x 1 , x 2 , ⋯   , x n ) ∂ ( y 1 , y 2 , ⋯   , y m ) = [ ∂ x 1 ∂ y 1 ∂ x 1 ∂ y 2 ⋯ ∂ x 1 ∂ y m ∂ x 2 ∂ y 1 ∂ x 2 ∂ y 2 ⋯ ∂ x 2 ∂ y m ⋮ ⋮ ⋮ ∂ x n ∂ y 1 ∂ x n ∂ y 2 ⋯ ∂ x n ∂ y m ] \frac{\partial(x_1,x_2,\cdots,x_n)}{\partial(y_1,y_2,\cdots,y_m)}=\begin{bmatrix} \dfrac{\partial x_1}{\partial y_1} & \dfrac{\partial x_1}{\partial y_2} & \cdots & \dfrac{\partial x_1}{\partial y_m}\\ \dfrac{\partial x_2}{\partial y_1} & \dfrac{\partial x_2}{\partial y_2} & \cdots &\dfrac{\partial x_2}{\partial y_m} \\ \vdots & \vdots & & \vdots\\ \dfrac{\partial x_n}{\partial y_1} & \dfrac{\partial x_n}{\partial y_2} & \cdots &\dfrac{\partial x_n}{\partial y_m} \end{bmatrix} ∂(y1​,y2​,⋯,ym​)∂(x1​,x2​,⋯,xn​)​= ​∂y1​∂x1​​∂y1​∂x2​​⋮∂y1​∂xn​​​∂y2​∂x1​​∂y2​∂x2​​⋮∂y2​∂xn​​​⋯⋯⋯​∂ym​∂x1​​∂ym​∂x2​​⋮∂ym​∂xn​​​ ​

就是雅克比矩阵。我们称从坐标 y \mathbf{y} y(分母)到 x \mathbf{x} x(分子)的雅克比矩阵。

这个比较简单，利用关系 x = ρ cos ⁡ φ , y = ρ sin ⁡ φ x=\rho\cos\varphi,y=\rho\sin\varphi x=ρcosφ,y=ρsinφ，

∂ x ∂ ρ = cos ⁡ φ , ∂ x ∂ φ = − ρ sin ⁡ φ ∂ y ∂ ρ = sin ⁡ φ , ∂ y ∂ φ = ρ cos ⁡ φ \begin{aligned} \dfrac{\partial x}{\partial \rho}=\cos\varphi, & \dfrac{\partial x}{\partial \varphi}=-\rho\sin\varphi\\ \dfrac{\partial y}{\partial \rho}=\sin\varphi, &\dfrac{\partial y}{\partial \varphi}=\rho\cos\varphi \end{aligned} ∂ρ∂x​=cosφ,∂ρ∂y​=sinφ,​∂φ∂x​=−ρsinφ∂φ∂y​=ρcosφ​

我们可以写出雅克比矩阵 ∂ ( x , y ) ∂ ( ρ , φ ) = [ ∂ x ∂ ρ ∂ x ∂ φ ∂ y ∂ ρ ∂ y ∂ φ ] = [ cos ⁡ φ − ρ sin ⁡ φ sin ⁡ φ ρ cos ⁡ φ ] \dfrac{\partial(x,y)}{\partial(\rho,\varphi)}=\begin{bmatrix} \dfrac{\partial x}{\partial \rho} & \dfrac{\partial x}{\partial \varphi}\\ \dfrac{\partial y}{\partial \rho} &\dfrac{\partial y}{\partial \varphi} \end{bmatrix}=\begin{bmatrix} \cos\varphi &-\rho\sin\varphi\\ \sin\varphi &\rho\cos\varphi \end{bmatrix} ∂(ρ,φ)∂(x,y)​= ​∂ρ∂x​∂ρ∂y​​∂φ∂x​∂φ∂y​​ ​=[cosφsinφ​−ρsinφρcosφ​]

这里有两种方法。

利用关系 ρ 2 = x 2 + y 2 , φ = arctan ⁡ y x \rho^2=x^2+y^2,\quad \varphi=\arctan\frac{y}{x} ρ2=x2+y2,φ=arctanxy​，我们可以对上式直接应用求导

对于第一个式子： ρ = x 2 + y 2 \rho=\sqrt{x^2+y^2} ρ=x2+y2 ​

直接求导有：

∂ ρ ∂ x = 2 x 2 x 2 + y 2 = x ρ = cos ⁡ φ ∂ ρ ∂ y = 2 y 2 x 2 + y 2 = y ρ = sin ⁡ φ \frac{\partial\rho}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\rho}=\cos\varphi\\ \frac{\partial\rho}{\partial y}=\frac{2y}{2\sqrt{x^2+y^2}}=\frac{y}{\rho}=\sin\varphi ∂x∂ρ​=2x2+y2 ​2x​=ρx​=cosφ∂y∂ρ​=2x2+y2 ​2y​=ρy​=sinφ

对于第二个式子直接求导有：

∂ φ ∂ x = − y x 2 1 + y 2 x 2 = − y x 2 + y 2 = − y ρ 2 = − sin ⁡ φ ρ ∂ φ ∂ y = 1 x 1 + y 2 x 2 = x x 2 + y 2 = x ρ 2 = cos ⁡ φ ρ \frac{\partial \varphi}{\partial x}=\frac{-\dfrac{y}{x^{2}}}{1+\dfrac{y^{2}}{x^{2}}}=\frac{-y}{x^{2}+y^{2}}=\frac{-y}{\rho^2}=\frac{-\sin\varphi}{\rho}\\ \frac{\partial \varphi}{\partial y}=\frac{\dfrac{1}{x}}{1+\dfrac{y^{2}}{x^{2}}}=\frac{x}{x^{2}+y^{2}}=\frac{x}{\rho^2}=\frac{\cos\varphi}{\rho} ∂x∂φ​=1+x2y2​−x2y​​=x2+y2−y​=ρ2−y​=ρ−sinφ​∂y∂φ​=1+x2y2​x1​​=x2+y2x​=ρ2x​=ρcosφ​

当然也可以用全微分的方法来求解，我们对第一个式子全微分：

2 ρ d ρ = 2 x d x + 2 y d y 2\rho\mathrm{d}\rho=2x\mathrm{d}x+2y\mathrm{d}y 2ρdρ=2xdx+2ydy

于是得到

d ρ = x ρ d x + y ρ d y \mathrm{d}\rho=\frac{x}{\rho}\mathrm{d}x+\frac{y}{\rho}\mathrm{d}y dρ=ρx​dx+ρy​dy

于是有： ∂ ρ ∂ x = x ρ = cos ⁡ φ , ∂ y ∂ ρ = y ρ = sin ⁡ φ \dfrac{\partial \rho}{\partial x}=\frac{x}{\rho}=\cos\varphi, \dfrac{\partial y}{\partial \rho}=\frac{y}{\rho}=\sin\varphi ∂x∂ρ​=ρx​=cosφ,∂ρ∂y​=ρy​=sinφ

对第二个式子变换一下：

tan ⁡ φ = y x \tan\varphi=\frac{y}{x} tanφ=xy​

然后我们再求全微分：

1 cos ⁡ 2 φ d φ = − y x 2 d x + 1 x d y \frac{1}{\cos^2\varphi}\mathrm{d}\varphi=-\frac{y}{x^2}\mathrm{d}x+\frac{1}{x}\mathrm{d}y cos2φ1​dφ=−x2y​dx+x1​dy

于是得到

d φ = − y cos ⁡ 2 φ x 2 d x + cos ⁡ 2 φ x d y = − y ρ 2 d x + x ρ 2 d y = − sin ⁡ φ ρ d x + cos ⁡ φ ρ d y \mathrm{d}\varphi=-\frac{y\cos^2\varphi}{x^2}\mathrm{d}x+\frac{\cos^2\varphi}{x}\mathrm{d}y=-\frac{y}{\rho^2}\mathrm{d}x+\frac{x}{\rho^2}\mathrm{d}y=-\frac{\sin\varphi}{\rho}\mathrm{d}x+\frac{\cos\varphi}{\rho}\mathrm{d}y dφ=−x2ycos2φ​dx+xcos2φ​dy=−ρ2y​dx+ρ2x​dy=−ρsinφ​dx+ρcosφ​dy

于是有： ∂ φ ∂ x = − sin ⁡ φ ρ , ∂ φ ∂ y = cos ⁡ φ ρ \frac{\partial \varphi}{\partial x}=\frac{-\sin\varphi}{\rho}, \frac{\partial \varphi}{\partial y}=\frac{\cos\varphi}{\rho} ∂x∂φ​=ρ−sinφ​,∂y∂φ​=ρcosφ​

∂ ( ρ , φ ) ∂ ( x , y ) = [ ∂ ρ ∂ x ∂ ρ ∂ y ∂ φ ∂ x ∂ φ ∂ y ] = [ cos ⁡ φ sin ⁡ φ − sin ⁡ φ ρ cos ⁡ φ ρ ] \dfrac{\partial(\rho,\varphi)}{\partial(x,y)}=\begin{bmatrix} \dfrac{\partial \rho}{\partial x} & \dfrac{\partial \rho}{\partial y}\\ \dfrac{\partial \varphi}{\partial x}&\dfrac{\partial \varphi}{\partial y} \end{bmatrix}=\begin{bmatrix} \cos\varphi &\sin\varphi\\ \dfrac{-\sin\varphi}{\rho}&\dfrac{\cos\varphi}{\rho} \end{bmatrix} ∂(x,y)∂(ρ,φ)​= ​∂x∂ρ​∂x∂φ​​∂y∂ρ​∂y∂φ​​ ​= ​cosφρ−sinφ​​sinφρcosφ​​ ​

这里刚好是一个二阶方阵，所以可以直接对3中的雅克比矩阵求逆：

∂ ( ρ , φ ) ∂ ( x , y ) = ( ∂ ( x , y ) ∂ ( ρ , φ ) ) − 1 = [ cos ⁡ φ − ρ sin ⁡ φ sin ⁡ φ ρ cos ⁡ φ ] − 1 = [ cos ⁡ φ sin ⁡ φ − sin ⁡ φ ρ cos ⁡ φ ρ ] \dfrac{\partial(\rho,\varphi)}{\partial(x,y)}=\left(\dfrac{\partial(x,y)}{\partial(\rho,\varphi)}\right)^{-1}=\begin{bmatrix} \cos\varphi &-\rho\sin\varphi\\ \sin\varphi &\rho\cos\varphi \end{bmatrix}^{-1}{}=\begin{bmatrix} \cos\varphi &\sin\varphi\\ \dfrac{-\sin\varphi}{\rho}&\dfrac{\cos\varphi}{\rho} \end{bmatrix} ∂(x,y)∂(ρ,φ)​=(∂(ρ,φ)∂(x,y)​)−1=[cosφsinφ​−ρsinφρcosφ​]−1= ​cosφρ−sinφ​​sinφρcosφ​​ ​