PlanetPhysics/Transformation Between Cartesian Basis Vectors and Polar Basis Vectors

From the definition of a covariant [[../Vectors/|vector]] (covariant [[../Tensor/|tensor]] of rank 1)

${\overline{T}}_i=T_j\frac{\partial x^j}{\partial {\overline{x}}^i}$

the corresponding transformation [[../Matrix/|matrix]] is

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

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

$r=\sqrt{x^2+y^2}$ $\theta =ta{n}^{-1}\left(\frac{y}{x}\right)$

and for polar to cartesian

$x=r\cos \theta$ $y=r\sin \theta$

So if we designate $({\hat{e}}_x,{\hat{e}}_y)$ as the bar coordinates, then the transformation components from a polar basis vector $({\hat{e}}_r,{\hat{e}}_{\theta })$ to a cartesian basis vector $({\hat{e}}_x,{\hat{e}}_y)$ is calculted as

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

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

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

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

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

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

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

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

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

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

$\left[\begin{array}{c}{\hat{e}}_x\\ {\hat{e}}_y\end{array}\right]=\left[\begin{array}{cc}\frac{x}{\sqrt{x^2+y^2}}& -\frac{y}{x^2+y^2}\\ \frac{y}{\sqrt{x^2+y^2}}& \frac{x}{x^2+y^2}\end{array}\right]\left[\begin{array}{c}{\hat{e}}_r\\ {\hat{e}}_{\theta }\end{array}\right]$

$\left[\begin{array}{c}{\hat{e}}_r\\ {\hat{e}}_{\theta }\end{array}\right]=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -r\sin \theta & r\cos \theta \end{array}\right]\left[\begin{array}{c}{\hat{e}}_x\\ {\hat{e}}_y\end{array}\right]$