LaplacianInSphericalCoordinates 2005-12-30 20:39:07 2006-01-02 19:45:29 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 $$ d {\bf r} = \frac{ \partial {\bf r}}{\partial u_1} du_1 + \frac{ \partial {\bf r}}{\partial u_2} du_2 + \frac{ \partial {\bf r}}{\partial u_3} du_3 $$ For spherical coordinates we have $$ u_1 = r $$ $$ u_2 = \theta $$ $$ u_3 = \phi $$ and from earlier we learned $$ | \frac{ \partial {\bf r}}{\partial r}| \hat{r} = \frac{ \partial {\bf r}}{\partial r} $$ $$ | \frac{ \partial {\bf r}}{\partial \theta}| \hat{\theta} = \frac{ \partial {\bf r}}{\partial \theta} $$ $$ | \frac{ \partial {\bf r}}{\partial \phi}| \hat{\theta} = \frac{ \partial {\bf r}}{\partial \phi} $$ Plugging these values into the generalized differential yields $$ d {\bf r} = dr \hat{r} + r d \theta \hat{\theta} + r sin \theta d \phi \hat{\phi} $$ The next major step is to see how the gradient fits into the definition of the differential for a function f $$ df = \frac{ \partial f}{\partial r} dr + \frac{ \partial f}{\partial \theta} d \theta + \frac{ \partial f}{\partial \phi} d \phi $$ $$ df = \nabla f \cdot d {\bf r} $$ So we see that the following must be equal and we need to solve for the gradient's components. $$ \frac{ \partial f}{\partial r} dr = \nabla_r f dr $$ $$ \frac{ \partial f}{\partial \theta} d \theta = \nabla_{\theta} f d \theta $$ $$ \frac{ \partial f}{\partial \phi} d \phi = \nabla_{\phi} f d \phi $$ Therfore our scale factors are $$ \nabla_r = 1 $$ $$ \nabla_{\theta} = \frac{1}{r} $$ $$ \nabla_{\phi} = \frac{1}{r sin \theta} $$ which finally gives us the gradient in spherical coordinates $$ \nabla_{sph} = \frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \frac{\hat{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta} $$ The last tedious calculation is then the Laplacian, which is our goal $$ \nabla^2 = \nabla \cdot \nabla = ( \frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \frac{\hat{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta}) \cdot (\frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \frac{\hat{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta}) $$ Let us break it up by components to make it easy to view, so carrying out only part of the dot product our 1st term is $$ \hat{r} \cdot ( \frac{\partial \hat{r}}{\partial r} \frac{\partial}{\partial r} + \hat{r} \frac{\partial^2}{\partial r^2} + \frac{\partial \hat{\theta}}{\partial r} \frac{1}{r} \frac{\partial}{\partial \theta} +\hat{\theta} \frac{\partial}{\partial r} \frac{1}{r} \frac{\partial}{\partial \theta} + \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\partial \hat{\phi}}{\partial r} \frac{1}{r sin \theta} \frac{\partial}{\partial \phi} + \hat{\phi}\frac{\partial}{\partial r}(\frac{1}{r sin \theta}) \frac{\partial}{\partial \phi} + \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial r \partial \phi} $$ While this looks a little scary, all but the 2nd term is zero. The first term is zero because there is no r in $\hat{r}$ $$ \hat{r} = sin \theta \, cos \phi \hat{i} + sin \theta \, sin \phi \hat{j} + cos \theta \hat {k} $$ so taking the derivative with respect to r yeilds zero. Similarly, the 3rd and 6th term are zero when taking the partial derivatives. The 4th,5th,7th and 8th terms are zero because the dot product of two orthogonal vectors ($90^o$) is zero so $$ \hat{r} \cdot \hat{\theta} = 0 $$ $$ \hat{r} \cdot \hat{\phi} = 0 $$