View source for PlanetPhysics/Transformation Between Cartesian Basis Vectors and Polar Basis Vectors

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

<blockquote><math>
\bar{T}_{i} = T_{j}\frac{\partial x^{j}}{\partial \bar{x}^{i}}
</math></blockquote>

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

<blockquote><math>
A_{ij} = \frac{\partial x^{j}}{\partial \bar{x}^{i}}
</math></blockquote>

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

<math> r = \sqrt{ x^2 + y^2 } </math>
<math> \theta = tan^{-1}\left( \frac{y}{x} \right) </math>

and for polar to cartesian

<math> x = r \cos \theta </math>
<math> y = r \sin \theta </math>

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

<math> A_{11} = \frac{\partial {x}^{1}}{\partial \bar{x}^{1}} = \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}</math>

<math> A_{12} = \frac{\partial {x}^{2}}{\partial \bar{x}^{1}} = \frac{\partial \theta}{\partial x} = -\frac{y}{x^2 + y^2}</math>

<math> A_{21} = \frac{\partial {x}^{1}}{\partial \bar{x}^{2}} = \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}</math>

<math> A_{22} = \frac{\partial {x}^{2}}{\partial \bar{x}^{2}} = \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2}</math>

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

<math> B_{11} = \frac{\partial {x}^{1}}{\partial \bar{x}^{1}} = \frac{\partial x}{\partial r} = \cos \theta</math>

<math> B_{12} = \frac{\partial {x}^{2}}{\partial \bar{x}^{1}} = \frac{\partial y}{\partial r} = \sin \theta</math>

<math> B_{21} = \frac{\partial {x}^{1}}{\partial \bar{x}^{2}} = \frac{\partial x}{\partial \theta} = -r \sin \theta</math>

<math> B_{22} = \frac{\partial {x}^{2}}{\partial \bar{x}^{2}} = \frac{\partial y}{\partial \theta} = r \cos \theta</math>

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

<math> \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] </math>

<math> \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]</math>

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