Talk:PlanetPhysics/Laplacian in Spherical Coordinates
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[edit source]%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: Laplacian in Spherical Coordinates %%% Primary Category Code: 02.40.Dr %%% Filename: LaplacianInSphericalCoordinates.tex %%% Version: 21 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}
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The \htmladdnormallink{Laplacian}{http://planetphysics.us/encyclopedia/LaplaceOperator.html} \htmladdnormallink{operator}{http://planetphysics.us/encyclopedia/QuantumSpinNetworkFunctor2.html} in spherical coordinates is
\begin{equation} \nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin \theta \frac{\partial}{\partial \theta}\right) + \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 \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} {\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 \htmladdnormallink{unit vector}{http://planetphysics.us/encyclopedia/PureState.html} 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 \htmladdnormallink{identity}{http://planetphysics.us/encyclopedia/Cod.html} $$ cos( 90 - \theta ) = sin \theta $$
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 \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} $ \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 \htmladdnormallink{motion}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} 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 \htmladdnormallink{magnitudes}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html} 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 \htmladdnormallink{gradient}{http://planetphysics.us/encyclopedia/Gradient.html} fits into the definition of the differential for a \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} 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
$$ \nabla^2 = \nabla \cdot \nabla = \left( \frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \frac{\hat{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta}\right) \cdot \left(\frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \frac{\hat{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta}\right) $$
Let us break it up by components to make it easy to view, so carrying out only part of the \htmladdnormallink{dot product}{http://planetphysics.us/encyclopedia/DotProduct.html} our 1st term is
$$ \hat{r} \cdot \left( \frac{\partial \hat{r}}{\partial r} \frac{\partial}{\partial r} + \hat{r} \frac{\partial^2}{\partial r^2} + \frac{\partial \hat{\theta}}{\partial r} \frac{1}{r} \frac{\partial}{\partial \theta} +\hat{\theta} \frac{\partial}{\partial r} \frac{1}{r} \frac{\partial}{\partial \theta} + \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\partial \hat{\phi}}{\partial r} \frac{1}{r sin \theta} \frac{\partial}{\partial \phi} + \hat{\phi}\frac{\partial}{\partial r}\left(\frac{1}{r sin \theta}\right) \frac{\partial}{\partial \phi} + \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial r \partial \phi}\right) $$
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
$$ \hat{r} \cdot \hat{\theta} = 0 $$ $$ \hat{r} \cdot \hat{\phi} = 0 $$
This leaves only the second term
\begin{equation} \hat{r} \cdot \hat{r} \frac{\partial^2}{\partial r^2} = \frac{\partial^2}{\partial r^2} \end{equation}
Now onto the $\hat{\theta}$ term
$$ \frac{\hat{\theta}}{r} \cdot \left( \frac{\partial \hat{r}}{\partial \theta} \frac{\partial}{\partial r} + \hat{r} \frac{\partial^2}{\partial \theta \partial r} + \frac{\partial \hat{\theta}}{\partial \theta} \frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\theta} \frac{\partial}{\partial \theta} \frac{1}{r} \frac{\partial}{\partial \theta} + \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial \theta^2} + \frac{\partial \hat{\phi}}{\partial \theta} \frac{1}{r sin \theta} \frac{\partial}{\partial \phi} + \hat{\phi}\frac{\partial}{\partial \theta}\left(\frac{1}{r sin \theta}\right) \frac{\partial}{\partial \phi} + \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial \theta \partial \phi}\right) $$
The dot product gets rid of the 2nd, 3rd, 7th and 8th terms, while the 4th and 6th terms are zero when taking the derivatives
$$ \frac{\partial}{\partial \theta} \frac{1}{r} = 0 $$ $$ \frac{\partial \hat{\phi}}{\partial \theta} = 0 $$
Two terms remain, for the first term we need to calculate
$$ \frac{\partial \hat{r}}{\partial \theta} = cos \theta \, cos \phi \hat{i} + cos \theta \, sin \phi \hat{j} - sin \theta \hat{k} = \hat{\theta} $$
This yields
\begin{equation} \frac{\hat{\theta}}{r} \cdot \hat{\theta} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r} \end{equation}
The 5th term is just the dot product
\begin{equation} \frac{\hat{\theta}}{r} \cdot \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial \theta^2} = \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \end{equation}
Finally, the $\hat{\phi}$ part of the dot product is
$$ \frac{\hat{\phi}}{r sin \theta} \cdot \left( \frac{\partial \hat{r}}{\partial \phi} \frac{\partial}{\partial r} + \hat{r} \frac{\partial^2}{\partial \phi \partial r} + \frac{\partial \hat{\theta}}{\partial \phi} \frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\theta} \frac{\partial}{\partial \phi} \frac{1}{r} \frac{\partial}{\partial \theta} + \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial \phi \partial \theta} + \frac{\partial \hat{\phi}}{\partial \phi} \frac{1}{r sin \theta} \frac{\partial}{\partial \phi} + \hat{\phi}\frac{\partial}{\partial \phi}\left(\frac{1}{r sin \theta}\right) \frac{\partial}{\partial \phi} + \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial \phi^2}\right) $$
Once again the dot product makes the 2nd, 4th and 5th terms zero. The first term derivative
$$ \frac{\partial \hat{r}}{\partial \phi} = -sin \theta \, sin \phi \hat{i} + sin \theta \, cos \phi \hat{j} = sin \theta \hat{\phi} \frac{\partial}{\partial r} $$
this leads to
\begin{equation} \frac{\hat{\phi}}{r sin \theta} \cdot sin \theta \hat{\phi} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r} \end{equation}
The 3rd term derivative is
$$ \frac{\partial \hat{\theta}}{\partial \phi} = -cos \theta \, sin \phi \hat{i} + cos \theta \, cos \phi \hat{j} = cos \theta \hat{\phi} $$
this leads to
\begin{equation} \frac{\hat{\phi}}{r sin \theta} \cdot cos \theta \hat{\phi} \frac{1}{r} \frac{\partial}{\partial \theta} = \frac{cos \theta}{r^2 sin \theta} \frac{\partial}{\partial \theta} \end{equation}
The 6th term can be seen to be zero because the derivative of $\hat{\phi}$ with respect to $\phi$ is a vector perpendicular to $\hat{\phi}$ (feel free to carry out this calculation), so the dot product will be zero.
The 7th term derivative is zero
$$ \frac{\partial}{\partial \phi} \left(\frac{1}{r sin \theta}\right) = 0 $$
Then we keep the 8th term
\begin{equation} \frac{\hat{\phi}}{r sin \theta} \cdot \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial \phi^2} = \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \end{equation}
Putting the results from (3),(4),(5),(6),(7) and (8) we get
$$ \nabla_{sph}^{2} = \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{cos \theta}{r^2 sin \theta} \frac{\partial}{\partial \theta} + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$
Certainly, we could finish here, but let us combine some terms and notice the following relationships (check these for yourself)
$$ \left[\frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r}\right] = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) $$
$$ \left[\frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} + \frac{cos \theta}{r^2 sin \theta} \frac{\partial}{\partial \theta}\right] = \frac{1}{r^2 sin \theta} \frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right) $$
This gives us the equation given in (1), the Laplacian in spherical coordinates
$$ \nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$
{\bf References}
[1] Marsden, J., Tromba, A. "Vector Calculus" Fourth Edition. W.H. Freeman Company, 1996.
[2] Ellis, R., Gulick, D. "Calculus" Harcourt Brace Jovanovich, Inc., Orlando, FL, 1991.
[3] Benbrook, J. "Intermediate Electromagnetic Theory", lecture notes, University of Houston, Fall 2002.
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