PlanetPhysics/Laplacian in Spherical Coordinates

The Laplacian operator in spherical coordinates is $$ \nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$

The derivation is fairly straight forward and begins with locating a vector {\mathbf 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 {\mathbf 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 $$ {\mathbf r} = r \hat{r} = r[ sin \theta \, cos \phi \, \hat{i} + sin \theta \, sin \phi \, \hat{j} + cos \theta \, \hat{k}] $$

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

$$ \frac{\partial {\mathbf r}}{ \partial r} = k \hat{r} $$

For $$\hat{\theta}$$ and $$\hat{\phi}$$, think of uniform circular motion like a record, so they can be calculated from

$$ \hat{\theta} = \frac{\frac{ \partial {\mathbf r}}{\partial \theta}}{| \frac{ \partial {\mathbf r}}{\partial \theta}|} $$

$$ \hat{\theta} = \frac{\frac{ \partial {\mathbf r}}{\partial \phi}}{| \frac{ \partial {\mathbf r}}{\partial \phi}|} $$

Next, take the partial derivatives of (2) and then calculate their magnitudes to get the above unit vectors.

$$ \frac{ \partial {\mathbf r}}{\partial \theta} = r [cos \theta \, cos \phi \hat{i} + cos \theta \, sin \phi \hat{j} - sin \theta \hat{k}] $$

$$ \frac{ \partial {\mathbf r}}{\partial \phi} = r [-sin \theta \, sin \phi \hat{i} + sin \theta \, cos \phi \hat{j}] $$

$$ | \frac{ \partial {\mathbf r}}{\partial \theta}| = \sqrt{ {\mathbf x} \cdot {\mathbf x}} = \sqrt{r^2 [cos^2 \theta \, cos^2 \phi + cos^2 \theta \, sin^2 \phi + sin^2 \theta]} $$

$$ | \frac{ \partial {\mathbf r}}{\partial \theta}| = r \sqrt{cos^2 \theta (cos^2 \phi + sin^2 \phi) + sin^2 \theta]} $$

Using the basic trig identity

$$ | \frac{ \partial {\mathbf r}}{\partial \theta}| = r $$

$$ | \frac{ \partial {\mathbf r}}{\partial \phi}| = r \sqrt{sin^2 \theta (sin^2 \phi + cos^2 \phi)]} $$

$$ | \frac{ \partial {\mathbf r}}{\partial \phi}| = r sin \theta $$

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

$$ d {\mathbf r} = \frac{ \partial {\mathbf r}}{\partial u_1} du_1 + \frac{ \partial {\mathbf r}}{\partial u_2} du_2 + \frac{ \partial {\mathbf r}}{\partial u_3} du_3 $$

For spherical coordinates we have

$$ u_1 = r $$ $$ u_2 = \theta $$ $$ u_3 = \phi $$

and from earlier we learned

$$ | \frac{ \partial {\mathbf r}}{\partial r}| \hat{r} = \frac{ \partial {\mathbf r}}{\partial r} $$

$$ | \frac{ \partial {\mathbf r}}{\partial \theta}| \hat{\theta} = \frac{ \partial {\mathbf r}}{\partial \theta} $$

$$ | \frac{ \partial {\mathbf r}}{\partial \phi}| \hat{\theta} = \frac{ \partial {\mathbf r}}{\partial \phi} $$

Plugging these values into the generalized differential yields

$$ d {\mathbf r} = dr \hat{r} + r d \theta \hat{\theta} + r sin \theta d \phi \hat{\phi} $$

The next major step is to see how the gradient fits into the definition of the differential for a function f

$$ df = \frac{ \partial f}{\partial r} dr + \frac{ \partial f}{\partial \theta} d \theta + \frac{ \partial f}{\partial \phi} d \phi $$ $$ df = \nabla f \cdot d {\mathbf r} $$

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 dot product 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

$$ \hat{r} \cdot \hat{r} \frac{\partial^2}{\partial r^2} = \frac{\partial^2}{\partial r^2} $$

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

$$ \frac{\hat{\theta}}{r} \cdot \hat{\theta} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r} $$

The 5th term is just the dot product

$$ \frac{\hat{\theta}}{r} \cdot \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial \theta^2} = \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} $$

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

$$ \frac{\hat{\phi}}{r sin \theta} \cdot sin \theta \hat{\phi} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r} $$

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

$$ \frac{\hat{\phi}}{r sin \theta} \cdot cos \theta \hat{\phi} \frac{1}{r} \frac{\partial}{\partial \theta} = \frac{cos \theta}{r^2 sin \theta} \frac{\partial}{\partial \theta} $$

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

$$ \frac{\hat{\phi}}{r sin \theta} \cdot \frac{\hat{\phi}}{r sin \theta} \frac{\partial^2}{\partial \phi^2} = \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$

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

$$ \nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $$

{\mathbf 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.