# 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 $(r,\theta )$ the standard polar coordinates.

### To Cartesian coordinates
#### From polar coordinates

${\begin{aligned}x&=r\cos \theta \\y&=r\sin \theta \\[5pt]{\frac {\partial (x,y)}{\partial (r,\theta )}}&={\begin{bmatrix}\cos \theta &-r\sin \theta \\\sin \theta &{\phantom {-}}r\cos \theta \end{bmatrix}}\\[5pt]{\text{Jacobian}}=\det {\frac {\partial (x,y)}{\partial (r,\theta )}}&=r\end{aligned}}$

#### From log-polar coordinates

${\begin{aligned}x&=e^{\rho }\cos \theta ,\\y&=e^{\rho }\sin \theta .\end{aligned}}$
By using complex numbers $(x,y)=x+iy'$, the transformation can be written as $x+iy=e^{\rho +i\theta }$
That is, it is given by the complex exponential function.

#### From bipolar coordinates

${\begin{aligned}x&=a{\frac {\sinh \tau }{\cosh \tau -\cos \sigma }}\\y&=a{\frac {\sin \sigma }{\cosh \tau -\cos \sigma }}\end{aligned}}$

#### From 2-center bipolar coordinates

${\begin{aligned}x&={\frac {1}{4c}}\left(r_{1}^{2}-r_{2}^{2}\right)\\[1ex]y&=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-\left(r_{1}^{2}-r_{2}^{2}+4c^{2}\right)^{2}}}\end{aligned}}$

#### From Cesàro equation

${\begin{aligned}x&=\int \cos \left[\int \kappa (s)\,ds\right]ds\\y&=\int \sin \left[\int \kappa (s)\,ds\right]ds\end{aligned}}$

### To polar coordinates
#### From Cartesian coordinates

${\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=\arctan \left|{\frac {y}{x}}\right|\end{aligned}}$ Note: solving for $\theta '$ returns the resultant angle in the first quadrant (0<θ< π/2 {\textstyle 0<\theta <{\frac {\pi }{2}}}). To find $\theta ,$ one must refer to the original Cartesian coordinate, determine the quadrant in which $\theta$ lies (for example, (3,−3) [Cartesian] lies in QIV), then use the following to solve for $\theta :$
$\theta ={\begin{cases}\theta '&{\text{for }}\theta '{\text{ in QI: }}&0<\theta '<{\frac {\pi }{2}}\\[1.2ex]\pi -\theta '&{\text{for }}\theta '{\text{ in QII: }}&{\frac {\pi }{2}}<\theta '<\pi \\[1.2ex]\pi +\theta '&{\text{for }}\theta '{\text{ in QIII: }}&\pi <\theta '<{\frac {3\pi }{2}}\\[1.2ex]2\pi -\theta '&{\text{for }}\theta '{\text{ in QIV: }}&{\frac {3\pi }{2}}<\theta '<2\pi \end{cases}}$
The value for $\theta$ must be solved for in this manner because for all values of $\theta$, $\tan \theta$ is only defined for −π/2 <θ<+π/2 {\textstyle -{\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 ${\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=2\arctan {\frac {y}{x+r}}\end{aligned}}$

#### From 2-center bipolar coordinates

${\begin{aligned}r&={\sqrt {\frac {r_{1}^{2}+r_{2}^{2}-2c^{2}}{2}}}\\\theta &=\arctan \left[{\sqrt {{\frac {8c^{2}(r_{1}^{2}+r_{2}^{2}-2c^{2})}{r_{1}^{2}-r_{2}^{2}}}-1}}\right]\end{aligned}}$
Where 2c is the distance between the poles.

### To log-polar coordinates from Cartesian coordinates

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

### Arc-length and curvature
#### In Cartesian coordinates

${\begin{aligned}\kappa &={\frac {x'y''-y'x''}{\left({x'}^{2}+{y'}^{2}\right)^{\frac {3}{2}}}}\\s&=\int _{a}^{t}{\sqrt {{x'}^{2}+{y'}^{2}}}\,dt\end{aligned}}$

#### In polar coordinates

${\begin{aligned}\kappa &={\frac {r^{2}+2{r'}^{2}-rr''}{(r^{2}+{r'}^{2})^{\frac {3}{2}}}}\\s&=\int _{a}^{\varphi }{\sqrt {r^{2}+{r'}^{2}}}\,d\varphi \end{aligned}}$

## 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 , see conventions in spherical coordinates). 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

${\begin{aligned}x&=\rho \,\sin \theta \cos \varphi \\y&=\rho \,\sin \theta \sin \varphi \\z&=\rho \,\cos \theta \\{\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}&={\begin{pmatrix}\sin \theta \cos \varphi &\rho \cos \theta \cos \varphi &-\rho \sin \theta \sin \varphi \\\sin \theta \sin \varphi &\rho \cos \theta \sin \varphi &\rho \sin \theta \cos \varphi \\\cos \theta &-\rho \sin \theta &0\end{pmatrix}}\end{aligned}}$
So for the volume element: $dx\,dy\,dz=\det {\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}\,d\rho \,d\theta \,d\varphi =\rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi$

#### From cylindrical coordinates

${\begin{aligned}x&=r\,\cos \theta \\y&=r\,\sin \theta \\z&=z\,\\{\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}&={\begin{pmatrix}\cos \theta &-r\sin \theta &0\\\sin \theta &r\cos \theta &0\\0&0&1\end{pmatrix}}\end{aligned}}$
So for the volume element: $dV=dx\,dy\,dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}\,dr\,d\theta \,dz=r\,dr\,d\theta \,dz$

### To spherical coordinates
#### From Cartesian coordinates

${\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arctan \left({\frac {\sqrt {x^{2}+y^{2}}}{z}}\right)=\arccos \left({\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\right)\\\varphi &=\arctan \left({\frac {y}{x}}\right)=\arccos \left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)=\arcsin \left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)\\{\frac {\partial \left(\rho ,\theta ,\varphi \right)}{\partial \left(x,y,z\right)}}&={\begin{pmatrix}{\frac {x}{\rho }}&{\frac {y}{\rho }}&{\frac {z}{\rho }}\\{\frac {xz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&{\frac {yz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&-{\frac {\sqrt {x^{2}+y^{2}}}{\rho ^{2}}}\\{\frac {-y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}&0\\\end{pmatrix}}\end{aligned}}$
See also the article on atan2 for how to elegantly handle some edge cases.
So for the element: $d\rho \,d\theta \,d\varphi =\det {\frac {\partial (\rho ,\theta ,\varphi )}{\partial (x,y,z)}}\,dx\,dy\,dz={\frac {1}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}\,dx\,dy\,dz$

#### From cylindrical coordinates

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

### To cylindrical coordinates
#### From Cartesian coordinates

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

#### From spherical coordinates

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

### Arc-length, curvature and torsion from Cartesian coordinates

${\begin{aligned}s&=\int _{0}^{t}{\sqrt {{x'}^{2}+{y'}^{2}+{z'}^{2}}}\,dt\\[3pt]\kappa &={\frac {\sqrt {\left(z''y'-y''z'\right)^{2}+\left(x''z'-z''x'\right)^{2}+\left(y''x'-x''y'\right)^{2}}}{\left({x'}^{2}+{y'}^{2}+{z'}^{2}\right)^{\frac {3}{2}}}}\\[3pt]\tau &={\frac {x'''\left(y'z''-y''z'\right)+y'''\left(x''z'-x'z''\right)+z'''\left(x'y''-x''y'\right)}{{\left(x'y''-x''y'\right)}^{2}+{\left(x''z'-x'z''\right)}^{2}+{\left(y'z''-y''z'\right)}^{2}}}\end{aligned}}$