Coordinate Transformations

Theorem: Cartesian $\leftrightarrow$ Polar

If $\mathbf{p}\in {\mathbb{R}}^2$ has Cartesian coordinates $(x,y)$, then its polar coordinates $(r,\theta )$ are:

• Using the convention $r\ge 0$ and $\theta \in (-\pi ;\pi ]$:

$$\begin{array}{rl}r& =\sqrt{x^2+y^2}\\ \theta & =\{\begin{array}{l}\arctan \left(\frac{y}{x}\right)\text{if }x>0\\ \arctan \left(\frac{y}{x}\right)+\pi \text{ if }x<0,y\ge 0\\ \arctan \left(\frac{y}{x}\right)-\pi \text{ if }x<0,y<0\\ \frac{\pi }{2}\text{if }x=0,y>0\\ -\frac{\pi }{2}\text{ if }x=0,y<0\\ 0\text{ if }x=y=0\end{array}\end{array}$$

• Using the convention $r\ge 0$ and $\theta \in [0;2\pi )$:

$$\begin{array}{rl}r& =\sqrt{x^2+y^2}\\ \theta & =\{\begin{array}{l}\arccos \frac{x}{\sqrt{x^2+y^2}}\text{ if }y>0\\ 2\pi -\arccos \frac{x}{\sqrt{x^2+y^2}}\text{if }y<0\\ 0\text{if }y=0,x\ge 0\\ \pi \text{if }y=0,x<0\end{array}\end{array}$$

If $\mathbf{p}\in {\mathbb{R}}^2$ has polar coordinates $(r,\theta )$, then its Cartesian coordinates $(x,y)$ are

$$\begin{array}{rl}x& =r\cos \theta \\ y& =r\sin \theta \end{array}$$

PROOF

TODO

Theorem: Cartesian $\leftrightarrow$ Spherical

If $\mathbf{p}\in {\mathbb{R}}^3$ has Cartesian coordinates $(x,y,z)$, then its spherical coordinates $(r,\theta ,\varphi )$ are:

• Using the convention $r\ge 0$, $\theta \in [0;\pi ]$ and $\varphi \in (-\pi ;\pi ]$:

$$\begin{array}{rl}r& =\sqrt{x^2+y^2+z^2}\\ \theta & =\{\begin{array}{l}0\text{if }x=y=z=0\\ \arccos \frac{z}{\sqrt{x^2+y^2+z^2}}\text{otherwise }\end{array}\\ \varphi & =\{\begin{array}{l}\arctan \left(\frac{y}{x}\right)\text{if }x>0\\ \arctan \left(\frac{y}{x}\right)+\pi \text{ if }x<0,y\ge 0\\ \arctan \left(\frac{y}{x}\right)-\pi \text{ if }x<0,y<0\\ \frac{\pi }{2}\text{if }x=0,y>0\\ -\frac{\pi }{2}\text{ if }x=0,y<0\\ 0\text{ if }x=y=0\end{array}\end{array}$$

• Using the convention $r\ge 0$, $\theta \in [0;\pi ]$ and $\varphi \in [0;2\pi )$:

$$\begin{array}{rl}r& =\sqrt{x^2+y^2+z^2}\\ \theta & =\{\begin{array}{l}0\text{if }x=y=z=0\\ \arccos \frac{z}{\sqrt{x^2+y^2+z^2}}\text{otherwise }\end{array}\\ \varphi & =\{\begin{array}{l}\arccos \frac{x}{\sqrt{x^2+y^2}}\text{ if }y>0\\ 2\pi -\arccos \frac{x}{\sqrt{x^2+y^2}}\text{if }y<0\\ 0\text{if }y=0,x\ge 0\\ \pi \text{if }y=0,x<0\end{array}\end{array}$$

If $\mathbf{p}\in {\mathbb{R}}^3$ has spherical coordinates $(r,\theta ,\varphi )$, then its Cartesian coordinates $(x,y,z)$ are

$$\begin{array}{rl}x& =r\sin \theta \cos \varphi \\ y& =r\sin \theta \sin \varphi \\ z& =r\cos \theta \end{array}$$

PROOF

TODO

Theorem: Cartesian $\leftrightarrow$ Cylindrical

If $\mathbf{p}\in {\mathbb{R}}^3$ has Cartesian coordinates $(x,y,z)$, then its Cylindrical coordinates $(\rho ,\varphi ,z)$ are:

• Using the convention $\rho \ge 0$ and $\varphi \in (-\pi ;\pi ]$:

$$\begin{array}{rl}\rho & =\sqrt{x^2+y^2}\\ \varphi & =\{\begin{array}{l}\arctan \left(\frac{y}{x}\right)\text{if }x>0\\ \arctan \left(\frac{y}{x}\right)+\pi \text{ if }x<0,y\ge 0\\ \arctan \left(\frac{y}{x}\right)-\pi \text{ if }x<0,y<0\\ \frac{\pi }{2}\text{if }x=0,y>0\\ -\frac{\pi }{2}\text{ if }x=0,y<0\\ 0\text{ if }x=y=0\end{array}\\ z& =z\end{array}$$

• Using the convention $\rho \ge 0$ and $\varphi \in [0;2\pi )$:

$$\begin{array}{rl}\rho & =\sqrt{x^2+y^2}\\ \varphi & =\{\begin{array}{l}\arccos \frac{x}{\sqrt{x^2+y^2}}\text{ if }y>0\\ 2\pi -\arccos \frac{x}{\sqrt{x^2+y^2}}\text{if }y<0\\ 0\text{if }y=0,x\ge 0\\ \pi \text{if }y=0,x<0\end{array}\\ z& =z\end{array}$$

If $\mathbf{p}\in {\mathbb{R}}^3$ has Cylindrical coordinates $(\rho ,\varphi ,z)$, then its Cartesian coordinates $(x,y,z)$ are

$$\begin{array}{rl}x& =\rho \cos \varphi \\ y& =\rho \sin \varphi \\ z& =z\end{array}$$

PROOF

TODO