TransformationBetweenCartesianBasisVectorsAndPolarBasisVectors 2007-07-04 22:44:21 2007-07-04 22:59:20 Example tensor % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here From the definition of a covariant vector \begin{equation} \bar{T}_{i} = T_{j}\frac{\partial x^{j}}{\partial \bar{x}^{i}} \end{equation} the corresponding transformation matrix is \begin{equation} A_{ij} = \frac{\partial x^{j}}{\partial \bar{x}^{i}} \end{equation} In order to calculate the transformation matrix, we need the equations relating the two coordinates systems. For cartesian to polar, we have $$ r = \sqrt{ x^2 + y^2 } $$ $$ \theta = tan^{-1}\left( \frac{y}{x} \right) $$ 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 $$ A_{11} = \frac{\partial {x}^{1}}{\partial \bar{x}^{1}} = \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}$$ $$ A_{12} = \frac{\partial {x}^{2}}{\partial \bar{x}^{1}} = \frac{\partial \theta}{\partial x} = -\frac{y}{x^2 + y^2}$$ $$ A_{21} = \frac{\partial {x}^{1}}{\partial \bar{x}^{2}} = \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}$$ $$ A_{22} = \frac{\partial {x}^{2}}{\partial \bar{x}^{2}} = \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2}$$ The components of cartesian basis vectors to polar basis vectors transform the same way, but now the polar coordinates have the bar $$ B_{11} = \frac{\partial \bar{x}^{1}}{\partial x^{1}} = \frac{\partial x}{\partial r} = \cos \theta$$ $$ B_{12} = \frac{\partial \bar{x}^{2}}{\partial x^{1}} = \frac{\partial y}{\partial r} = \sin \theta$$ $$ B_{21} = \frac{\partial \bar{x}^{1}}{\partial x^{2}} = \frac{\partial x}{\partial \theta} = -r \sin \theta$$ $$ B_{22} = \frac{\partial \bar{x}^{2}}{\partial x^{2}} = \frac{\partial y}{\partial \theta} = r \cos \theta$$ In summary, the {\bf components of covariant basis vectors} in cartesian coordinates and polar coordinates transform between each other according to $$ \left[ \begin{array}{c} \hat{e}_x \\ \hat{e}_y \end{array} \right] = \left[ \begin{array}{cc} \frac{x}{\sqrt{x^2 + y^2}} & -\frac{y}{x^2 + y^2} \\ \frac{y}{\sqrt{x^2 + y^2}} & \frac{x}{x^2 + y^2} \end{array} \right] \left[ \begin{array}{c} \hat{e}_r \\ \hat{e}_{\theta} \end{array} \right] $$ $$ \left[ \begin{array}{c} \hat{e}_r \\ \hat{e}_{\theta} \end{array} \right] \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ -r \sin \theta & r \cos \theta \end{array} \right] \left[ \begin{array}{c} \hat{e}_x \\ \hat{e}_y \end{array} \right]$$