PlanetPhysics/Transformation Between Cartesian Coordinates and Polar Coordinates

From the definition of a contravariant vector (contravariant tensor of rank 1)

${\displaystyle {\bar {T}}^{i}=T^{j}{\frac {\partial {\bar {x}}^{i}}{\partial x^{j}}}}$

we get the transformation matrix from the partial derivatives

${\displaystyle A_{ij}={\frac {\partial {\bar {x}}^{i}}{\partial x^{j}}}}$

In order to calculate the transformation matrix, we need the equations relating the two coordinates systems. For cartesian to polar, we have

${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$

${\displaystyle \theta =tan^{-1}\left({\frac {y}{x}}\right)}$

and for polar to cartesian

${\displaystyle x=r\cos \theta }$

${\displaystyle y=r\sin \theta }$

So if we designate ${\displaystyle (x,y)}$ as the bar coordinates, then the transformation components from polar coordinates ${\displaystyle (r,\theta )}$ to cartesian coordinates ${\displaystyle (x,y)}$ is calculated as

${\displaystyle A_{11}={\frac {\partial {\bar {x}}^{1}}{\partial x^{1}}}={\frac {\partial x}{\partial r}}=\cos \theta }$

${\displaystyle A_{12}={\frac {\partial {\bar {x}}^{1}}{\partial x^{2}}}={\frac {\partial x}{\partial \theta }}=-r\sin \theta }$

${\displaystyle A_{21}={\frac {\partial {\bar {x}}^{2}}{\partial x^{1}}}={\frac {\partial y}{\partial r}}=\sin \theta }$

${\displaystyle A_{22}={\frac {\partial {\bar {x}}^{2}}{\partial x^{2}}}={\frac {\partial y}{\partial \theta }}=r\cos \theta }$

The components from cartesian coordinates to polar coordinates transform the same way, but now the polar coordinates have the bar

${\displaystyle B_{11}={\frac {\partial {\bar {x}}^{1}}{\partial {x}^{1}}}={\frac {\partial r}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}}$

${\displaystyle B_{12}={\frac {\partial {\bar {x}}^{1}}{\partial {x}^{2}}}={\frac {\partial r}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}}$

${\displaystyle B_{21}={\frac {\partial {\bar {x}}^{2}}{\partial {x}^{1}}}={\frac {\partial \theta }{\partial x}}=-{\frac {y}{x^{2}+y^{2}}}}$

${\displaystyle B_{22}={\frac {\partial {\bar {x}}^{2}}{\partial {x}^{2}}}={\frac {\partial \theta }{\partial y}}={\frac {x}{x^{2}+y^{2}}}}$

In summary, the {\mathbf components of contravariant vectors} in cartesian coordinates and polar coordinates transform between each other according to

${\displaystyle \left[{\begin{matrix}x\\y\end{matrix}}\right]=\left[{\begin{matrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{matrix}}\right]\left[{\begin{matrix}r\\\theta \end{matrix}}\right]}$

${\displaystyle \left[{\begin{matrix}r\\\theta \end{matrix}}\right]=\left[{\begin{matrix}{\frac {x}{\sqrt {x^{2}+y^{2}}}}&{\frac {y}{\sqrt {x^{2}+y^{2}}}}\\-{\frac {y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}\end{matrix}}\right]\left[{\begin{matrix}x\\y\end{matrix}}\right]}$