List of common coordinate transformations

Let $(x,\; y)$ be the standard Cartesian coordinates, and $(r,\; \backslash theta)$ the standard polar coordinates.

$$\backslash begin\{align\}$$

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{align}

{{Main|log-polar coordinates}}

$$\backslash begin\{align\}$$

x &= e^\rho\cos\theta, \\

y &= e^\rho\sin\theta.

\end{align}

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

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

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

{{Main|bipolar coordinates}}

$$\backslash begin\{align\}$$

x &= a \frac{\sinh \tau}{\cosh \tau - \cos \sigma} \\

y &= a \frac{\sin \sigma}{\cosh \tau - \cos \sigma}

\end{align}

{{Main|two-center bipolar coordinates}}

$$\backslash begin\{align\}$$

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{align}

{{Main|Cesàro equation}}

$$\backslash begin\{align\}$$

x &= \int \cos \left[\int \kappa(s) \,ds\right] ds \\

y &= \int \sin \left[\int \kappa(s) \,ds\right] ds

\end{align}

$$\backslash begin\{align\}$$

r &= \sqrt{x^2 + y^2} \\

\theta' &= \arctan\left|\frac{y}{x}\right|

\end{align}

Note: solving for $\backslash theta\text{'}$ returns the resultant angle in the first quadrant (). To find $\backslash theta,$ one must refer to the original Cartesian coordinate, determine the quadrant in which $\backslash theta$ lies (for example, (3,−3) [Cartesian] lies in QIV), then use the following to solve for $\backslash theta:$

$$\backslash theta\; =\; \backslash begin\{cases\}$$

\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 $\backslash theta$ must be solved for in this manner because for all values of $\backslash theta$, $\backslash tan\backslash theta$ is only defined for , and is periodic (with period $\backslash pi$). 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

$$\backslash begin\{align\}$$

r &= \sqrt{x^2 + y^2} \\

\theta' &= 2 \arctan \frac{y}{x + r}

\end{align}

$$\backslash begin\{align\}$$

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{align}

Where 2c is the distance between the poles.

$$\backslash begin\{align\}$$

\rho &= \log\sqrt{x^2 + y^2}, \\

\theta &= \arctan \frac{y}{x}.

\end{align}

$$\backslash begin\{align\}$$

\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{align}

$$\backslash begin\{align\}$$

\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{align}

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.

{{Main|spherical coordinates}}

$$\backslash begin\{align\}$$

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{align}

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

{{Main|cylindrical coordinates}}

$$\backslash begin\{align\}$$

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{align}

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

{{Main|spherical coordinates}}

$$\backslash begin\{align\}$$

\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{align}

See also the article on atan2 for how to elegantly handle some edge cases.

So for the element:

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

{{Main|cylindrical coordinates}}

$$\backslash begin\{align\}$$

\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{align}

$$\backslash begin\{align\}$$

r &= \sqrt{x^2 + y^2} \\

\theta &= \arctan{\left(\frac{y}{x}\right)} \\

z &= z \quad

\end{align}

\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}

$$\backslash begin\{align\}$$

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{align}

$$\backslash begin\{align\}$$

s &= \int_0^t \sqrt{{x'}^2 + {y'}^2 + {z'}^2}\, dt \\[3pt]

\kappa &= \frac{\sqrt{\left(zy'-yz'\right)^2 + \left(xz' - zx'\right)^2 + \left(yx' - xy'\right)^2}}{\left({x'}^2 + {y'}^2 + {z'}^2\right)^\frac{3}{2}} \\[3pt]

\tau &= \frac{x\left(y'z - yz'\right) + y\left(xz' - x'z\right) + z'\left(x'y - xy'\right)}{{\left(x'y - xy'\right)}^2 + {\left(xz'- x'z\right)}^2 + {\left(y'z - y''z'\right)}^2}

\end{align}

• Geographic coordinate conversion
• Transformation matrix

• {{cite book |last=Arfken |first=George |title=Mathematical Methods for Physicists |year=2013 |url=https://books.google.com/books?id=qLFo_Z-PoGIC|publisher=Academic Press |isbn=978-0123846549 |author-link=George B. Arfken}}

{{Reflist}}

{{DEFAULTSORT:Canonical Coordinate Transformations}}

Coordinate transformations

Category:Coordinate systems

Category:Hamiltonian mechanics