05. Coordinate Systems and Jacobians

Jacobian Matrix

Suppose $x=x(u,v)$ and $y=y(u,v)$
We have a transformation o coordinates

$$(x,y)\to (u,v)$$

By Chain Rule then

$$\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial f}{\partial x}& \frac{\partial f}{\partial y}\end{array}\right)\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ & \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)$$

where the Jacobian Matrix is

$$\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ & \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)$$

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Jacobian

$$\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$$

Note that the Jacobian is equivalent to an area scale factor

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5.1 Plane Polar Coordinates

Plane Polar Coordinates

For $P=(x,y)\in {\mathbb{R}}^2$ and $(x,y)\ne (0,0)$ then

Let $r$ be the distance from $(0,0)$ and
Let $\theta$ be the anti-clockwise angle that $\theta$ that $\overrightarrow{OP}$ makes with the $x-$axis

Hence

$$x=r\cos \theta ,y=r\sin \theta$$

Where $r\in [0,\infty )$ ad $\theta \in [0,2\pi )$

Jacobian

Jacobian Matrix

$$\left(\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)$$

Jacobian

$$\frac{\partial (x,y)}{\partial (r,\theta )}=r$$

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Property of Jacobians

Let $r,s$ be functions of variables $u,v$ which are functions of $x,y$
Then

$$\frac{\partial (r,s)}{\partial (x,y)}=\frac{\partial (r,s)}{\partial (u,v)}\cdot \frac{\partial (u,v)}{\partial (x,y)}$$

Proof

$$\begin{gathered} \begin{array}{rl}\frac{\partial (r,s)}{\partial (x,y)}& =\left| \\ \begin{array}{cc}\frac{\partial r}{\partial x}& \frac{\partial r}{\partial y}\\ \frac{\partial s}{\partial x}& \frac{\partial s}{\partial y}\end{array}\right|\\ & \\ & =\left|\begin{array}{cc}\frac{\partial r}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial r}{\partial v}\frac{\partial v}{\partial x}& \frac{\partial r}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial r}{\partial v}\frac{\partial v}{\partial y}\\ & \\ \frac{\partial s}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial s}{\partial v}\frac{\partial v}{\partial x}& \frac{\partial s}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial s}{\partial v}\frac{\partial v}{\partial y}\end{array}\right|\\ & =\left|\left(\begin{array}{cc}\frac{\partial r}{\partial u}& \frac{\partial r}{\partial v}\\ \frac{\partial s}{\partial u}& \frac{\partial s}{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)\right|\\ & =\left|\begin{array}{cc}\frac{\partial r}{\partial u}& \frac{\partial r}{\partial v}\\ \frac{\partial s}{\partial u}& \frac{\partial s}{\partial v}\end{array}\right|\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|\\ & =\frac{\partial (r,s)}{\partial (u,v)}\cdot \frac{\partial (u,v)}{\partial (x,y)}\end{array} \end{gathered}$$

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"Inverse"-like result of Jacobians corollary

$$\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}=1$$

Proof

Take $r=x$ and $s=y$ in the proof above

Note that the stronger result holds as

$$\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

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5.2 Parabolic Coordinates

Parabolic Coordinates

For coordinates $u,v$ given in terms of $x,y$

$$x=\frac{1}{2}(u^2-v^2)y=uv$$

Jacobian

$$\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ & \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|=\left|\begin{array}{cc}u& -v\\ v& u\end{array}\right|=u^2+v^2$$

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5.3 Cylindrical Polar Coordinates

Cylindrical Polar Coordinates

Extend plane polar coordinates to three dimensions by $z$-coordinates
So that

$$x=r\cos \theta ,y=r\sin \theta ,z=z$$

Inverse Transformation

$$r=\sqrt{x^2+y^2},\tan \theta =\frac{y}{x},z=z$$

Jacobian

Jacobian Matrix

$$\left(\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial z}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial z}\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial z}\end{array}\right)=\left(\begin{array}{ccc}\cos \theta & -r\sin \theta & 0\\ \sin \theta & r\cos \theta & 0\\ 0& 0& 1\end{array}\right)$$

Hence the Jacobian is

$$\frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r{\cos }^2\theta +r{\sin }^2\theta =r$$

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5.4 Spherical Polar Coordinates

Spherical Polar Coordinates

Let $(x,y,z)$ be the Cartesian Coordinates for a general point $P\in {\mathbb{R}}^3$

For $P\ne (0,0,0)$
Let $r$ be the distance between $P$ and $O$
Let $\theta$ be the angle from the z-axis to the position vector $\overrightarrow{OP}$
Let $\varphi$ be the angle when you convert $(x,y)$ to coordinates $(\rho \cos \varphi ,\psi \sin \varphi )$

Hence in terms of the spherical coordinates $(r,\theta ,\varphi )$ we have

$$x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi ,z=r\cos \theta$$

where $r\ge 0$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$

Inverse Transformation

$$r=\sqrt{x^2+y^2+z^2},\tan \theta =\frac{\sqrt{x^2+y^2}}{z},\tan \varphi =\frac{y}{x}$$

Jacobian

Jacobian Matrix

$$\left(\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial \varphi }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial \varphi }\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial \varphi }\end{array}\right)=\left(\begin{array}{ccc}\sin \theta \cos \varphi & r\cos \theta \cos \varphi & -r\sin \theta \sin \varphi \\ \sin \theta \sin \varphi & r\cos \theta \sin \varphi & r\sin \theta \cos \varphi \\ \cos \theta & -r\sin \theta & 0\end{array}\right)$$

Jacobian

$$\frac{\partial (x,y,z)}{\partial (r,\theta ,\varphi )}=r^2\sin \theta$$

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