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 \(\theta \in (-\pi; \pi]\):
$$
\begin{aligned}
r &= \sqrt{x^2 + y^2} \
\theta &=
\begin{cases}
\arctan \left( \frac{y}{x} \right) \qquad \qquad \text{if } x \gt 0 \
\arctan \left( \frac{y}{x} \right) + \pi \qquad \text{ if } x \lt 0, y \ge 0 \
\arctan \left( \frac{y}{x} \right) - \pi \qquad \text{ if } x \lt 0, y \lt 0 \
\frac{\pi}{2} \qquad \qquad \qquad \qquad \text{if } x = 0, y \gt 0 \
-\frac{\pi}{2} \qquad \qquad \qquad \, \, \, \, \, \, \text{ if } x = 0, y \lt 0 \
0 \qquad \qquad \qquad \qquad \, \text{ if } x = y = 0
\end{cases}
\end{aligned}
$$
- Using the convention \(r \ge 0\) and \(\theta \in [[Polar Coordinate System|0; 2\pi)\):
$$
\begin{aligned}
r &= \sqrt{x^2 + y^2} \
\theta &=
\begin{cases}
\arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \qquad \text{ if } y > 0 \
2\pi - \arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \text{if } y < 0 \
0 \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \ge 0 \
\pi \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \lt 0
\end{cases}
\end{aligned}
$$
If \(\mathbf{p} \in \mathbb{R}^2\) has [polar coordinates]] \((r, \theta)\), then its Cartesian coordinates \((x,y)\) are
$$
\begin{aligned}
x &= r \cos \theta \
y &= r \sin \theta
\end{aligned}
$$
Proof
TODO
Theorem: Cartesian \(\leftrightarrow\) Spherical
If \(\mathbf{p} \in \mathbb{R}^3\) has Cartesian coordinates \((x,y,z)\), then its spherical coordinates \((r, \theta, \phi)\) are:
- Using the convention \(r \ge 0\), \(\theta \in [0; \pi]\) and \(\phi \in (-\pi; \pi]\):
$$
\begin{aligned}
r &= \sqrt{x^2 + y^2 + z^2} \
\theta &=
\begin{cases}
0 \qquad \qquad \qquad \hphantom{,,,,,} \text{if } x = y = z = 0 \
\arccos \frac{z}{\sqrt{x^2 + y^2 + z^2}} \hphantom{,,,,} \text{otherwise }
\end{cases} \
\phi &=
\begin{cases}
\arctan \left( \frac{y}{x} \right) \qquad \qquad \text{if } x \gt 0 \
\arctan \left( \frac{y}{x} \right) + \pi \qquad \text{ if } x \lt 0, y \ge 0 \
\arctan \left( \frac{y}{x} \right) - \pi \qquad \text{ if } x \lt 0, y \lt 0 \
\frac{\pi}{2} \qquad \qquad \qquad \qquad \text{if } x = 0, y \gt 0 \
-\frac{\pi}{2} \qquad \qquad \qquad \, \, \, \, \, \, \text{ if } x = 0, y \lt 0 \
0 \qquad \qquad \qquad \qquad \, \text{ if } x = y = 0
\end{cases}
\end{aligned}
$$
- Using the convention \(r \ge 0\), \(\theta \in [0; \pi]\) and \(\phi \in [[Spherical Coordinate System|0; 2\pi)\):
$$
\begin{aligned}
r &= \sqrt{x^2 + y^2 + z^2} \
\theta &=
\begin{cases}
0 \qquad \qquad \qquad \hphantom{,,,,,} \text{if } x = y = z = 0 \
\arccos \frac{z}{\sqrt{x^2 + y^2 + z^2}} \hphantom{,,,,} \text{otherwise }
\end{cases} \
\phi &=
\begin{cases}
\arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \qquad \text{ if } y > 0 \
2\pi - \arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \text{if } y < 0 \
0 \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \ge 0 \
\pi \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \lt 0
\end{cases}
\end{aligned}
$$
If \(\mathbf{p} \in \mathbb{R}^3\) has [spherical coordinates]] \((r, \theta, \phi)\), then its Cartesian coordinates \((x,y,z)\) are
$$
\begin{aligned}
x &= r \sin \theta \cos \phi \
y &= r \sin \theta \sin \phi \
z &= r \cos \theta
\end{aligned}
$$
Proof
TODO
Theorem: Cartesian \(\leftrightarrow\) Cylindrical
If \(\mathbf{p} \in \mathbb{R}^3\) has Cartesian coordinates \((x,y,z)\), then its Cylindrical coordinates \((\rho, \phi, z)\) are:
- Using the convention \(\rho \ge 0\) and \(\phi \in (-\pi; \pi]\):
$$
\begin{aligned}
\rho &= \sqrt{x^2 + y^2} \
\phi &=
\begin{cases}
\arctan \left( \frac{y}{x} \right) \qquad \qquad \text{if } x \gt 0 \
\arctan \left( \frac{y}{x} \right) + \pi \qquad \text{ if } x \lt 0, y \ge 0 \
\arctan \left( \frac{y}{x} \right) - \pi \qquad \text{ if } x \lt 0, y \lt 0 \
\frac{\pi}{2} \qquad \qquad \qquad \qquad \text{if } x = 0, y \gt 0 \
-\frac{\pi}{2} \qquad \qquad \qquad \, \, \, \, \, \, \text{ if } x = 0, y \lt 0 \
0 \qquad \qquad \qquad \qquad \, \text{ if } x = y = 0
\end{cases} \
z &= z
\end{aligned}
$$
- Using the convention \(\rho \ge 0\) and \(\phi \in [[Cylindrical Coordinate System|0; 2\pi)\):
$$
\begin{aligned}
\rho &= \sqrt{x^2 + y^2} \
\phi &=
\begin{cases}
\arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \qquad \text{ if } y > 0 \
2\pi - \arccos \frac{x}{\sqrt{x^2 + y^2}} \qquad \text{if } y < 0 \
0 \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \ge 0 \
\pi \qquad \qquad \qquad \qquad \hphantom{,,,,} \text{if } y = 0, x \lt 0
\end{cases} \
z &= z
\end{aligned}
$$
If \(\mathbf{p} \in \mathbb{R}^3\) has [Cylindrical coordinates]] \((\rho, \phi, z)\), then its Cartesian coordinates \((x,y,z)\) are
$$
\begin{aligned}
x &= \rho \cos \phi \
y &= \rho \sin \phi \
z &= z
\end{aligned}
$$
Proof
TODO