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List of canonical coordinate transformations

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This is a list of canonical coordinate transformations.

Contents

[edit] 2-Dimensional

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

[edit] 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

[edit] 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 θ, one must refer to the original Cartesian coordinate, determine the quadrant in which θ lies (ex (3,-3) [Cartesian] lies in QIV), then use the following to solve for θ:

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 θ must be solved for in this manner because for all values of θ, arctanθ is only defined for -\frac{\pi}{2}<\theta<+\frac{\pi}{2}

Note that one can also use

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

[edit] 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}

[edit] To Cartesian coordinates from two-center bipolar coordinates[1]

Main article: 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}

[edit] 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.

[edit] 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

[edit] 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

[edit] 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

[edit] 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 θ has a range of 360° the same considerations as in polar (2 dimensional) coordinates apply whenever an arctangent of it is taken. φ 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, φ 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 φ 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.

[edit] To Cartesian coordinates

[edit] From spherical coordinates

Main article: spherical coordinates
{x}=\rho \, \sin\theta \, \cos\phi \quad
{y}=\rho \, \sin\theta \, \sin\phi \quad
{z}=\rho \, \cos\theta \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 \;

[edit] From cylindrical coordinates

Main article: 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 \;

[edit] To Spherical coordinates

[edit] 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{-(x^2+y^2)}{\rho^2\sqrt{x^2+y^2}}\\ \frac{-y}{x^2+y^2} & \frac{x}{x^2+y^2} & 0\\ \end{pmatrix}

[edit] From cylindrical coordinates

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

[edit] To cylindrical coordinates

[edit] From Cartesian coordinates

r=\sqrt{x^2 + y^2}
\theta=\arctan\frac{y}{x} + \pi u_0(-x) \, \operatorname{sgn} y
h=z \quad
\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}{\sqrt{x^2+y^2}}&\frac{x}{\sqrt{x^2+y^2}}&0\\ 0&0&1 \end{pmatrix}

[edit] 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

[edit] 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'-z'y'')^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)}

[edit] References

  1. ^ Weisstein, Eric W.. "Bipolar Coordinates." Treasure Troves. 26 May 1999. Sociology and Anthropology China. 14 Feb 2007 [1]
Retrieved from "http://en.wikipedia.org/wiki/List_of_canonical_coordinate_transformations"
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