# List of common coordinate transformations

This is a list of some of the most commonly used coordinate transformations.

## 2-Dimensional

Let (x, y) be the standard Cartesian coordinates, and r and θ the standard polar coordinates.

### To Cartesian coordinates from polar coordinates

$x=r\,\cos\theta \quad$
$y=r\,\sin\theta \quad$
$\frac{\partial(x, y)}{\partial(r, \theta)} = \begin{pmatrix} \cos\theta & -r\,\sin\theta \\ \sin\theta & r\,\cos\theta \end{pmatrix}$
$\det{\frac{\partial(x, y)}{\partial(r, \theta)}} = r$

### To polar coordinates from Cartesian coordinates

$r=\sqrt{x^2 + y^2}$
$\theta^\prime = \arctan\left|\frac{y}{x}\right|$

Note: solving for $\theta^\prime$ returns the resultant angle in the first quadrant ($0<\theta<\frac{\pi}{2}$). To find $\theta$, one must refer to the original Cartesian coordinate, determine the quadrant in which $\theta$ lies (ex (3,-3) [Cartesian] lies in QIV), then use the following to solve for $\theta$:

For $\theta^\prime$ in QI:
$\theta = \theta^\prime$
For $\theta^\prime$ in QII:
$\theta= \pi - \theta^\prime$
For $\theta^\prime$ in QIII:
$\theta = \pi + \theta^\prime$
For $\theta^\prime$ in QIV:
$\theta = 2\pi - \theta^\prime$

The value for $\theta$ must be solved for in this manner because for all values of $\theta$, $\tan\theta$ is only defined for $-\frac{\pi}{2}<\theta<+\frac{\pi}{2}$, and is periodic (with period $\pi$). This means that the inverse function will only give values in the domain of the function, but restricted to a single period. Hence, the range of the inverse function is only half a full circle.

Note that one can also use

$r=\sqrt{x^2 + y^2}$
$\theta^\prime = 2 \arctan \frac{y}{x+r}$

### To Cartesian coordinates from log-polar coordinates

Main article: Log-polar coordinates
$\begin{cases}x = e^\rho\cos\theta, \\ y = e^\rho\sin\theta.\end{cases}$

By using complex numbers $(x,y)=x+iy'$, the transformation can be written as

$x + iy = e^{\rho+i\theta} \,$

i.e. it is given by the complex exponential function.

### To log-polar coordinates from Cartesian coordinates

$\begin{cases} \rho = \log\sqrt{ x^2 + y^2}, \\ \theta = \arctan \frac{y}{x}. \end{cases}$

### To Cartesian coordinates from bipolar coordinates

Main article: bipolar coordinates
$x = a \ \frac{\sinh \tau}{\cosh \tau - \cos \sigma}$
$y = a \ \frac{\sin \sigma}{\cosh \tau - \cos \sigma}$

### To Cartesian coordinates from two-center bipolar coordinates

$x = \frac{r_1^2-r_2^2}{4c}$
$y = \pm \frac{1}{4c}\sqrt{16c^2r_1^2-(r_1^2-r_2^2+4c^2)^2}$[1]

### To polar coordinates from two-center bipolar coordinates

$r = \sqrt{\frac{r_1^2+r_2^2-2c^2}{2}}$
$\theta = \arctan \left[ \sqrt{\frac{8c^2(r_1^2+r_2^2-2c^2)}{r_1^2-r_2^2}-1}\right]$

Where 2c is the distance between the poles.

### To Cartesian coordinates from Cesàro equation

Main article: Cesàro equation
$x = \int \cos \left[\int \kappa(s) \,ds\right] ds$
$y = \int \sin \left[\int \kappa(s) \,ds\right] ds$

### Arc length and curvature from Cartesian coordinates

$\kappa = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}$

$s = \int_a^t \sqrt { x'^2 + y'^2 }\, dt$

### Arc length and curvature from polar coordinates

$\kappa=\frac{r^2+2r'^2-rr''}{(r^2+r'^2)^{3/2}}$

$s = \int_a^\phi \sqrt { 1 + y'^2 }\, d\phi$

## 3-Dimensional

Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with θ the angle measured away from the +Z axis. As φ has a range of 360° the same considerations as in polar (2 dimensional) coordinates apply whenever an arctangent of it is taken. θ has a range of 180°, running from 0° to 180°, and does not pose any problem when calculated from an arccosine, but beware for an arctangent. If, in the alternative definition, θ is chosen to run from −90° to +90°, in opposite direction of the earlier definition, it can be found uniquely from an arcsine, but beware of an arccotangent. In this case in all formulas below all arguments in θ should have sine and cosine exchanged, and as derivative also a plus and minus exchanged.

All divisions by zero result in special cases of being directions along one of the main axes and are in practice most easily solved by observation.

### To Cartesian coordinates

#### From spherical coordinates

Main article: spherical coordinates
${x}=\rho \, \sin\theta \, \cos\phi \quad$
${y}=\rho \, \sin\theta \, \sin\phi \quad$
${z}=\rho \, \cos\phi\quad$
$\frac{\partial(x, y, z)}{\partial(\rho, \theta, \phi)} = \begin{pmatrix} \sin\theta\cos\phi & \rho\cos\theta\cos\phi & -\rho\sin\theta\sin\phi \\ \sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \\ \cos\theta & -\rho\sin\theta & 0 \end{pmatrix}$

So for the volume element:

$dx\;dy\;dz=\det{\frac{\partial(x, y, z)}{\partial(\rho, \theta, \phi)}} d\rho\;d\theta\;d\phi = \rho^2 \sin\theta \; d\rho \; d\theta \; d\phi \;$

#### From cylindrical coordinates

${x}={r} \,\cos\theta$
${y}={r} \, \sin\theta$
${z}={h} \,$
$\frac{\partial(x, y, z)}{\partial(r, \theta, h)} = \begin{pmatrix} \cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

So for the volume element:

$dx\;dy\;dz=\det{\frac{\partial(x, y, z)}{\partial(r, \theta, h)}} dr\;d\theta\;dh = {r}\; dr \; d\theta \; dh \;$

### To Spherical coordinates

#### From Cartesian coordinates

${\rho}=\sqrt{x^2 + y^2 + z^2}$
${\phi}=\arctan \left( {\frac{y}{x}} \right)= \arccos \left( \frac{x}{\sqrt{x^2+y^2}}\right) = \arcsin \left( \frac{y}{\sqrt{x^2+y^2}}\right)$
${\theta}=\arctan \left( \frac{\sqrt{x^2 + y^2}}{z} \right)=\arccos \left( {\frac{z}{\sqrt{x^2 + y^2 + z^2}}} \right)$
$\frac{\partial(\rho, \theta, \phi)}{\partial(x, y, z)} = \begin{pmatrix} \frac{x}{\rho} & \frac{y}{\rho} & \frac{z}{\rho} \\ \frac{xz}{\rho^2\sqrt{x^2+y^2}} & \frac{yz}{\rho^2\sqrt{x^2+y^2}} & -\frac{\sqrt{x^2+y^2}}{\rho^2}\\ \frac{-y}{x^2+y^2} & \frac{x}{x^2+y^2} & 0\\ \end{pmatrix}$

So for the volume element:

$d\rho\ d\theta\ d\phi=\det\frac{\partial(\rho,\theta,\phi)}{\partial(x,y,z)}dx\ dy\ dz=\frac{1}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}dx\ dy\ dz$

#### From cylindrical coordinates

${\rho}=\sqrt{r^2+h^2}$
${\phi}=\phi \quad$
${\theta}=\arctan\frac{r}{h}$
$\frac{\partial(\rho, \theta, \phi)}{\partial(r, \phi, h)} = \begin{pmatrix} \frac{r}{\sqrt{r^2+h^2}} & 0 & \frac{h}{\sqrt{r^2+h^2}} \\ \frac{-r}{r^2+h^2} & 0 & \frac{h}{r^2+h^2} \\ 0 & 1 & 0 \end{pmatrix}$
$\det \frac{\partial(\rho, \theta, \phi)}{\partial(r, \phi, h)} = \frac{1}{\sqrt{r^2+h^2}}$

### To Cylindrical Coordinates

#### From Cartesian Coordinates

$r=\sqrt{x^2 + y^2}$
$\theta = \begin{cases} 0 & \mbox{if } x = 0 \mbox{ and } y = 0\\ \arcsin(\frac{y}{r}) & \mbox{if } x \geq 0 \\ -\arcsin(\frac{y}{r}) + \pi & \mbox{if } x < 0\\ \end{cases}$
$h=z \quad$

Note that many computer systems may offer a more concise function for computing $\theta$, such as atan2(y,x) in the C language.

$\frac{\partial(r, \theta, h)}{\partial(x, y, z)} = \begin{pmatrix} \frac{x}{\sqrt{x^2+y^2}}&\frac{y}{\sqrt{x^2+y^2}}&0\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0\\ 0&0&1 \end{pmatrix}$

#### From Spherical Coordinates

$r = \rho \sin \phi \,$
$\theta = \theta \,$
$h = \rho \cos \phi \,$
$\frac{\partial(r, \theta, h)}{\partial(\rho, \theta, \phi)} = \begin{pmatrix} \sin\phi & 0 & \rho\cos\phi \\ 0 & 1 & 0 \\ \cos\phi & 0 & -\rho\sin\phi \end{pmatrix}$
$\det\frac{\partial(r, \theta, h)}{\partial(\rho, \theta, \phi)} = - \rho$

### Arc length, curvature and torsion from cartesian coordinates

$s = \int_0^t \sqrt { x'^2 + y'^2 + z'^2 }\, dt$
$\kappa=\frac{\sqrt{(z''y'-y''z')^2+(x''z'-z''x')^2+(y''x'-x''y')^2}}{(x'^2+y'^2+z'^2)^{3/2}}$
$\tau=\frac{z'''(x'y''-y'x'')+z''(x'''y'-x'y''')+z'(x''y'''-x'''y'')}{(x'^2+y'^2+z'^2)(x''^2+y''^2+z''^2)}$

## References

1. ^ Weisstein, Eric W.. "Bipolar Coordinates." Treasure Troves. 26 May 1999. Sociology and Anthropology China. 14 February 2007 [1]