# 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

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

#### From log-polar coordinates

{\displaystyle {\begin{aligned}x&=e^{\rho }\cos \theta ,\\y&=e^{\rho }\sin \theta .\end{aligned}}}

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

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

I.e., it is given by the complex exponential function.

#### From bipolar coordinates

{\displaystyle {\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

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

#### From Cesàro equation

{\displaystyle {\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

{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta ^{\prime }&=\arctan \left|{\frac {y}{x}}\right|\end{aligned}}}

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

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

The value for ${\displaystyle \theta }$  must be solved for in this manner because for all values of ${\displaystyle \theta }$ , ${\displaystyle \tan \theta }$  is only defined for ${\displaystyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}$ , and is periodic (with period ${\displaystyle \pi }$ ). 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

{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta ^{\prime }&=2\arctan {\frac {y}{x+r}}\end{aligned}}}

#### From 2-center bipolar coordinates

{\displaystyle {\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

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

### Arc-length and curvature

#### In Cartesian coordinates

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

#### In polar coordinates

{\displaystyle {\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 [1], 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

{\displaystyle {\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:

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

#### From cylindrical coordinates

{\displaystyle {\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:

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

### To spherical coordinates

#### From Cartesian coordinates

{\displaystyle {\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:

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

#### From cylindrical coordinates

{\displaystyle {\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

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

#### From spherical coordinates

{\displaystyle {\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

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