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[parent] transformation between cartesian basis vectors and polar basis vectors (Example)

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

$\displaystyle \bar{T}_{i} = T_{j}\frac{\partial x^{j}}{\partial \bar{x}^{i}}$ (1)

the corresponding transformation matrix is

$\displaystyle A_{ij} = \frac{\partial x^{j}}{\partial \bar{x}^{i}}$ (2)

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 $(\hat{e}_x,\hat{e}_y)$ 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

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

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

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

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

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 components of covariant basis vectors in cartesian coordinates and polar coordinates transform between each other according to

$\displaystyle \left[ \begin{array}{c} \hat{e}_x \ \hat{e}_y \end{array} \righ... ...ht] \left[ \begin{array}{c} \hat{e}_r \ \hat{e}_{\theta} \end{array} \right] $

$\displaystyle \left[ \begin{array}{c} \hat{e}_r \ \hat{e}_{\theta} \end{array... ...ay} \right] \left[ \begin{array}{c} \hat{e}_x \ \hat{e}_y \end{array} \right]$



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Cross-references: systems, matrix, tensor, vector
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This is version 6 of transformation between cartesian basis vectors and polar basis vectors, born on 2007-07-04, modified 2007-07-06.
Object id is 255, canonical name is TransformationBetweenCartesianBasisVectorsAndPolarBasisVectors.
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Physics Classification: 02.40.Hw (Classical differential geometry)
 04.20.Cv (Fundamental problems and general formalism)
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