List of common coordinate transformations explained

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 x &= r\cos\theta \\ y &= r\sin\theta \\[5pt] \frac &= \begin \cos\theta & -r\sin\theta \\ \sin\theta & \phantomr\cos\theta \end \\[5pt] \text = \det &= r\end

From log-polar coordinates

\begin x &= e^\rho\cos\theta, \\ y &= e^\rho\sin\theta.\end

By using complex numbers

(x,y)=x+iy'

, the transformation can be written as x + iy = e^

That is, it is given by the complex exponential function.

From bipolar coordinates

See main article: bipolar coordinates. \begin x &= a \frac \\ y &= a \frac\end

From 2-center bipolar coordinates

See main article: two-center bipolar coordinates. \begin x &= \frac\left(r_1^2 - r_2^2\right) \\[1ex] y &= \pm \frac\sqrt\end

From Cesàro equation

See main article: Cesàro equation. \begin x &= \int \cos \left[\int \kappa(s) \,ds\right] ds \\ y &= \int \sin \left[\int \kappa(s) \,ds\right] ds\end

To polar coordinates

From Cartesian coordinates

\begin r &= \sqrt \\ \theta' &= \arctan\left|\frac\right|\endNote: solving for

\theta'

returns the resultant angle in the first quadrant (0 < \theta < \frac). 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\theta' & \text \theta' \text & 0 < \theta' < \frac \\[1.2ex]\pi - \theta' & \text \theta' \text &\frac < \theta' < \pi \\[1.2ex]\pi + \theta' & \text \theta' \text& \pi < \theta' < \frac \\[1.2ex]2\pi - \theta' & \text \theta' \text &\frac < \theta' < 2\pi\end

The value for

\theta

must be solved for in this manner because for all values of

\theta

,

\tan\theta

is only defined for -\frac<\theta<+\frac, 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 r &= \sqrt \\ \theta' &= 2 \arctan \frac\end

From 2-center bipolar coordinates

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

Where 2c is the distance between the poles.

To log-polar coordinates from Cartesian coordinates

\begin \rho &= \log\sqrt, \\ \theta &= \arctan \frac.\end

Arc-length and curvature

In Cartesian coordinates

\begin \kappa &= \frac \\ s &= \int_a^t \sqrt\, dt\end

In polar coordinates

\begin \kappa &= \frac \\ s &= \int_a^\varphi \sqrt\, d\varphi\end

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 https://commons.wikimedia.org/wiki/File:3D_Spherical.svg, 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

See main article: spherical coordinates. \begin x &= \rho \, \sin\theta\cos\varphi\\ y &= \rho \, \sin\theta\sin\varphi\\ z &= \rho \, \cos\theta \\ \frac &= \begin \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\end

So for the volume element: dx\,dy\,dz = \det\,d\rho\,d\theta\,d\varphi = \rho^2 \sin\theta \,d\rho \,d\theta \,d\varphi

From cylindrical coordinates

See main article: cylindrical coordinates. \begin x &= r \, \cos\theta\\ y &= r \, \sin\theta \\ z &= z \, \\ \frac &= \begin \cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end\end

So for the volume element: dV = dx\,dy\,dz = \det\,dr\,d\theta\,dz = r \,dr \,d\theta \,dz

To spherical coordinates

See main article: spherical coordinates.

From Cartesian coordinates

\begin \rho &= \sqrt \\ \theta &= \arctan \left(\frac \right)=\arccos \left(\right) \\ \varphi &= \arctan \left(\right) = \arccos \left(\frac\right) = \arcsin \left(\frac\right) \\ \frac &= \begin \frac & \frac & \frac \\ \frac & \frac & -\frac \\ \frac & \frac & 0 \\ \end\end

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\,dx\,dy\,dz=\frac\,dx\,dy\,dz

From cylindrical coordinates

See main article: cylindrical coordinates. \begin \rho &= \sqrt \\ \theta &= \arctan\frac \\ \varphi &= \varphi \\ \frac &= \begin \frac & \frac & 0 \\ \frac & \frac & 0 \\ 0 & 0 & 1 \\ \end \\ \det \frac &= \frac\end

To cylindrical coordinates

From Cartesian coordinates

\begin r &= \sqrt \\ \theta &= \arctan \\ z &= z \quad\end

\frac = \begin \frac & \frac & 0 \\ \frac & \frac & 0 \\ 0 & 0 & 1 \end

From spherical coordinates

\begin r &= \rho \sin \varphi \\ h &= \rho \cos \varphi \\ \theta &= \theta \\ \frac &= \begin \sin\varphi & \rho\cos\varphi & 0 \\ \cos\varphi & -\rho\sin\varphi & 0 \\ 0 & 0 & 1 \\ \end \\ \det\frac &= -\rho\end

Arc-length, curvature and torsion from Cartesian coordinates

\begin s &= \int_0^t \sqrt\, dt \\[3pt] \kappa &= \frac \\[3pt] \tau &= \frac\end

See also

References