- List of canonical coordinate transformations
This is a list of canonical coordinate transformations.
2-Dimensional
Let (x, y) be the standard
Cartesian coordinates , and r and θ the standardpolar coordinates .To Cartesian coordinates from polar coordinates
:x=r,cos heta quad:y=r,sin heta quad:frac{partial(x, y)}{partial(r, heta)} =egin{pmatrix}cos heta & -r,sin heta \sin heta & r,cos hetaend{pmatrix}
:det{frac{partial(x, y)}{partial(r, heta) =r
To polar coordinates from Cartesian coordinates
:r=sqrt{x^2 + y^2}:heta^prime = arctanleft|frac{y}{x} ight|Note: solving for heta^prime returns the resultant angle in the first quadrant (0< heta
). To find heta, one must refer to the original Cartesian coordinate, determine the quadrant in which heta lies (ex (3,-3) [Cartesian] lies in QIV), then use the following to solve for heta: :For heta^prime in QI::heta = heta^prime
:For heta^prime in QII::heta= pi - heta^prime
:For heta^prime in QIII::heta = pi + heta^prime
:For heta^prime in QIV::heta = 2pi - heta^prime
The value for heta must be solved for in this manner because for all values of heta, arctan heta is only defined for frac{pi}{2}< heta<+frac{pi}{2}
Note that one can also use:r=sqrt{x^2 + y^2}:heta = 2 arctan frac{y}{x+r}
To Cartesian coordinates from bipolar coordinates
:x = a frac{sinh au}{cosh au - cos sigma}
:y = a frac{sin sigma}{cosh au - cos sigma}
To Cartesian coordinates from two-center bipolar coordinates [Weisstein, Eric W.. "Bipolar Coordinates." "Treasure Troves". 26 May 1999. Sociology and Anthropology China. 14 Feb 2007 [http://bbs.sachina.pku.edu.cn/Stat/Math_World/math/b/b233.htm] ]
: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}
To polar coordinates from two-center bipolar coordinates
:r = sqrt{frac{r_1^2+r_2^2-2c^2}{2
:heta = arctan left [ sqrt{frac{8c^2(r_1^2+r_2^2-2c^2)}{r_1^2-r_2^2}-1} ight]
Where 2"c" is the distance between the poles.
To Cartesian coordinates from Cesàro equation
:x = int cos left [int kappa(s) ,ds ight] ds
:y = int sin left [int kappa(s) ,ds ight] ds
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
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 }, dphi
3-Dimensional
Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with φ 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.
To Cartesian coordinates
From spherical coordinates
:x}= ho , sin heta , cosphi quad :y}= ho , sin heta , sinphi quad :z}= ho , cos heta quad
:frac{partial(x, y, z)}{partial( ho, heta, phi)} =egin{pmatrix}sin hetacosphi& hocos hetacosphi & - hosin hetasinphi \sin hetasinphi & hocos hetasinphi & hosin hetacosphi \cos heta & - hosin heta & 0end{pmatrix}So for the volume element::dx;dy;dz=det{frac{partial(x, y, z)}{partial( ho, heta, phi) d ho;d heta;dphi = ho^2 sin heta ; d ho ; d heta ; dphi ;
From cylindrical coordinates
:x}={r} ,cos heta:y}={r} , sin heta:z}={h} ,
:frac{partial(x, y, z)}{partial(r, heta, h)} =egin{pmatrix}cos heta & -rsin heta & 0 \sin heta & rcos heta & 0 \ 0 & 0 & 1end{pmatrix}So for the volume element::dx;dy;dz=det{frac{partial(x, y, z)}{partial(r, heta, h) dr;d heta;dh ={r}; dr ; d heta ; dh ;
To Spherical coordinates
From Cartesian coordinates
:ho}=sqrt{x^2 + y^2 + z^2}
:heta}=arctan left( frac{sqrt{x^2 + y^2{z} ight)=arccos left( {frac{z}{sqrt{x^2 + y^2 + z^2} ight)
:phi}=arctan left( {frac{y}{x ight)= arccos left( frac{x}{sqrt{x^2+y^2 ight) = arcsin left( frac{y}{sqrt{x^2+y^2 ight)
:frac{partial( ho, heta, phi)}{partial(x, y, z)} =egin{pmatrix} frac{x}{ ho} & frac{y}{ ho} & frac{z}{ ho} \frac{xz}{ ho^2sqrt{x^2+y^2 & frac{yz}{ ho^2sqrt{x^2+y^2 & frac{-(x^2+y^2)}{ ho^2sqrt{x^2+y^2\frac{-y}{x^2+y^2} & frac{x}{x^2+y^2} & 0\end{pmatrix}
From cylindrical coordinates
:ho}=sqrt{r^2+h^2}:heta}= heta quad:phi}=arctanfrac{r}{h}
:frac{partial( ho, heta, phi)}{partial(r, heta, h)} =egin{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( ho, heta, phi)}{partial(r, heta, h)} = frac{1}{sqrt{r^2+h^2
To cylindrical coordinates
From Cartesian coordinates
:r=sqrt{x^2 + y^2}:heta=arctanfrac{y}{x} + pi u_0(-x) , operatorname{sgn} y :h=z quad
:frac{partial(r, heta, h)}{partial(x, y, z)} =egin{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&1end{pmatrix}
From spherical coordinates
:r = ho sin phi ,:heta = heta ,:h = ho cos phi ,
:frac{partial(r, heta, h)}{partial( ho, heta, phi)} =egin{pmatrix}sinphi & 0 & hocosphi \0 & 1 & 0 \cosphi & 0 & - hosinphiend{pmatrix}
:detfrac{partial(r, heta, h)}{partial( ho, heta, phi)} = - ho
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
:au=frac{z(x'y"-y'x")+z"(xy'-x'y)+z'(x"y-x"'y")}{(x'^2+y'^2+z'^2)(x"^2+y"^2+z"^2)}
References
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