LaplacianInSphericalCoordinates 2005-12-30 20:39:07 2005-12-31 00:03:14 Definition % 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 The Laplacian operator in spherical coordinates is $$\nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r} + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} ( sin \theta \frac{\partial}{\partial \theta}) + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$ The derivation is fairly straight forward and begins with locating a vector {\bf r} in spherical coordinates as shown in the figure. \begin{figure} \includegraphics[scale=.698]{SphericalCoordinates.eps} \caption{Spherical Coordinates} \end{figure} The z component of the unit vector in direction of {\bf r} is given from the simple right triangle $$ cos \theta = \frac{z}{| \hat{r} |} $$ Since a unit vector has a length of 1, the z component is $$ z = cos \theta $$ To get the x component, we need to get the $\hat{r}$ projected onto the xy-plane $$ cos( 90 - \theta) = \frac{ |\hat{r}_{xy}|}{ |\hat{r}| } $$ using the trig identity $$ cos( 90 - \theta ) = sin \theta $$ so the projected unit vector is $$ |\hat{r}_{xy}| = sin \theta $$ Finally, the x component is reached through the right triangle $$ cos \phi = \frac{x}{|\hat{r}_{xy}| }$$ giving $$ x = cos \phi \, sin \theta $$