PlanetPhysics/Transformation Between Cartesian Basis Vectors and Polar Basis Vectors

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

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

the corresponding transformation matrix is

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

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 ${\displaystyle ({\hat {e}}_{x},{\hat {e}}_{y})}$ as the bar coordinates, then the transformation components from a polar basis vector ${\displaystyle ({\hat {e}}_{r},{\hat {e}}_{\theta })}$ to a cartesian basis vector ${\displaystyle ({\hat {e}}_{x},{\hat {e}}_{y})}$ is calculted as

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

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

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

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

The components of cartesian basis vectors to polar basis vectors transform the same way, but now the polar coordinates have the bar

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

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

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

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

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

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

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