PlanetPhysics/Transformation Between Cartesian Coordinates and Polar Coordinates

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

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

we get the transformation [[../Matrix/|matrix]] from the partial derivatives

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

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 $(x,y)$ as the bar coordinates, then the transformation components from polar coordinates $(r,\theta )$ to cartesian coordinates $(x,y)$ is calculated as

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

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

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

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

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

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

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

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

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

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

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right]\left[\begin{array}{c}r\\ \theta \end{array}\right]$

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