LaplacianInSphericalCoordinates 2005-12-30 20:39:07 2006-01-01 23:30:58 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 \begin{equation} \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} \end{equation} 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 $$ cos( 90 - \theta ) = sin \theta $$ 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 $$ x = cos \phi \, sin \theta $$ the y component follows the x component through $$ sin \phi = \frac{y}{|\hat{r}_{xy}|} $$ which yields $$ y = sin \phi \, sin \theta $$ The vector in spherical coordinates is then \begin{equation} {\bf r} = r \hat{r} = r[ sin \theta \, cos \phi \, \hat{i} + sin \theta \, sin \phi \, \hat{j} + cos \theta \, \hat{k}] \end{equation} Now describing the unit vectors of a moving particle $ \hat{r}, \hat{\theta}, \hat{\phi}$ shown in the spherical coordinates figure is a little more tricky. If a particle moves in the $\hat{r}$ direction, it only moves in or out along r so $$ \frac{\partial {\bf r}}{ \partial r} = k \hat{r} $$ For $\hat{\theta}$ and $\hat{\phi}$, think of uniform circular motion like a record, so they can be calculated from $$ \hat{\theta} = \frac{\frac{ \partial {\bf r}}{\partial \theta}}{| \frac{ \partial {\bf r}}{\partial \theta}|} $$ $$ \hat{\theta} = \frac{\frac{ \partial {\bf r}}{\partial \phi}}{| \frac{ \partial {\bf r}}{\partial \phi}|} $$ Next, take the partial derivatives of (2) and then calculate their magnitudes to get the above unit vectors. $$ \frac{ \partial {\bf r}}{\partial \theta} = r [cos \theta \, cos \phi \hat{i} + cos \theta \, sin \phi \hat{j} - sin \theta \hat{k}] $$ $$ \frac{ \partial {\bf r}}{\partial \phi} = r [-sin \theta \, sin \phi \hat{i} + sin \theta \, cos \phi \hat{j}] $$ $$ | \frac{ \partial {\bf r}}{\partial \theta}| = \sqrt{ {\bf x} \cdot {\bf x}} = \sqrt{r^2 [cos^2 \theta \, cos^2 \phi + cos^2 \theta \, sin^2 \phi + sin^2 \theta]} $$ $$ | \frac{ \partial {\bf r}}{\partial \theta}| = r \sqrt{cos^2 \theta (cos^2 \phi + sin^2 \phi) + sin^2 \theta]} $$ Using the basic trig identity $$ | \frac{ \partial {\bf r}}{\partial \theta}| = r $$ $$ | \frac{ \partial {\bf r}}{\partial \phi}| = r \sqrt{sin^2 \theta (sin^2 \phi + cos^2 \phi)]} $$ $$ | \frac{ \partial {\bf r}}{\partial \phi}| = r sin \theta $$ Then plug in these values to get $$ \hat{\phi} = -sin \phi \hat{i} + cos \phi \hat{j}$$ $$ \hat{\theta} = cos \theta \, cos \phi \hat{i} + cos \theta \, sin \phi \hat{j} - sin \theta \hat{k} $$ and from before we had $$ \hat{r} = sin \theta \, cos \phi \hat{i} + sin \theta \, sin \phi \hat{j} + cos \theta \hat {k} $$ The generalized differential for curvilinear coordinates is $$ d {\bf r} = \frac{ \partial {\bf r}}{\partial u_1} du_1 + \frac{ \partial {\bf r}}{\partial u_2} du_2 + \frac{ \partial {\bf r}}{\partial u_3} du_3 $$