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Laplacian in Spherical Coordinates (Definition)

The Laplacian operator in spherical coordinates is

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

The derivation is fairly straight forward and begins with locating a vector r in spherical coordinates as shown in the figure.

Figure: Spherical Coordinates
\includegraphics[scale=.698]{SphericalCoordinates.eps}

The z component of the unit vector in direction of r is given from the simple right triangle

$\displaystyle cos \theta = \frac{z}{\vert \hat{r} \vert} $

Since a unit vector has a length of 1, the z component is

$\displaystyle z = cos \theta $

To get the x component, we need to get the $\hat{r}$ projected onto the xy-plane

$\displaystyle cos( 90 - \theta) = \frac{ \vert\hat{r}_{xy}\vert}{ \vert\hat{r}\vert } $

using the trig identity

$\displaystyle cos( 90 - \theta ) = sin \theta $

the projected unit vector is

$\displaystyle \vert\hat{r}_{xy}\vert = sin \theta $

Finally, the x component is reached through the right triangle

$\displaystyle cos \phi = \frac{x}{\vert\hat{r}_{xy}\vert }$

giving

$\displaystyle x = cos \phi \, sin \theta $

the y component follows the x component through

$\displaystyle sin \phi = \frac{y}{\vert\hat{r}_{xy}\vert} $

which yields

$\displaystyle y = sin \phi \, sin \theta $

The vector in spherical coordinates is then

$\displaystyle {\bf r} = r \hat{r} = r[ sin \theta \, cos \phi \, \hat{i} + sin \theta \, sin \phi \, \hat{j} + cos \theta \, \hat{k}]$ (2)

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

$\displaystyle \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

$\displaystyle \hat{\theta} = \frac{\frac{ \partial {\bf r}}{\partial \theta}}{\vert \frac{ \partial {\bf r}}{\partial \theta}\vert} $

$\displaystyle \hat{\theta} = \frac{\frac{ \partial {\bf r}}{\partial \phi}}{\vert \frac{ \partial {\bf r}}{\partial \phi}\vert} $

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

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

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

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \theta}\vert = \sqrt{ {\b... ...{r^2 [cos^2 \theta \, cos^2 \phi + cos^2 \theta \, sin^2 \phi + sin^2 \theta]} $

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \theta}\vert = r \sqrt{cos^2 \theta (cos^2 \phi + sin^2 \phi) + sin^2 \theta]} $

Using the basic trig identity

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \theta}\vert = r $

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \phi}\vert = r \sqrt{sin^2 \theta (sin^2 \phi + cos^2 \phi)]} $

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \phi}\vert = r sin \theta $

Then plug in these values to get

$\displaystyle \hat{\phi} = -sin \phi \hat{i} + cos \phi \hat{j}$

$\displaystyle \hat{\theta} = cos \theta \, cos \phi \hat{i} + cos \theta \, sin \phi \hat{j} - sin \theta \hat{k} $

and from before we had

$\displaystyle \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

$\displaystyle d {\bf r} = \frac{ \partial {\bf r}}{\partial u_1} du_1 + \frac{ ... ...tial {\bf r}}{\partial u_2} du_2 + \frac{ \partial {\bf r}}{\partial u_3} du_3 $

For spherical coordinates we have

$\displaystyle u_1 = r $

$\displaystyle u_2 = \theta $

$\displaystyle u_3 = \phi $

and from earlier we learned

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial r}\vert \hat{r} = \frac{ \partial {\bf r}}{\partial r} $

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \theta}\vert \hat{\theta} = \frac{ \partial {\bf r}}{\partial \theta} $

$\displaystyle \vert \frac{ \partial {\bf r}}{\partial \phi}\vert \hat{\theta} = \frac{ \partial {\bf r}}{\partial \phi} $

Plugging these values into the generalized differential yields

$\displaystyle 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

$\displaystyle df = \frac{ \partial f}{\partial r} dr + \frac{ \partial f}{\partial \theta} d \theta + \frac{ \partial f}{\partial \phi} d \phi $

$\displaystyle 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.

$\displaystyle \frac{ \partial f}{\partial r} dr = \nabla_r f dr $

$\displaystyle \frac{ \partial f}{\partial \theta} d \theta = \nabla_{\theta} f d \theta $

$\displaystyle \frac{ \partial f}{\partial \phi} d \phi = \nabla_{\phi} f d \phi $

Therfore our scale factors are

$\displaystyle \nabla_r = 1 $

$\displaystyle \nabla_{\theta} = \frac{1}{r} $

$\displaystyle \nabla_{\phi} = \frac{1}{r sin \theta} $

which finally gives us the gradient in spherical coordinates

$\displaystyle \nabla_{sph} = \frac{\partial}{\partial r} \hat{r} + \frac{\parti... ...at{\theta}}{r} + \frac{\partial}{\partial \phi} \frac{\hat{\phi}}{rsin \theta} $

The last tedious calculation is then the Laplacian, which is our goal

$\displaystyle \nabla^2 = \nabla \cdot \nabla = \left( \frac{\partial}{\partial ... ...ta}}{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

$\displaystyle \hat{r} \cdot \left( \frac{\partial \hat{r}}{\partial r} \frac{\p... ...c{\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 $\hat{r}$

$\displaystyle \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

$\displaystyle \hat{r} \cdot \hat{\theta} = 0 $

$\displaystyle \hat{r} \cdot \hat{\phi} = 0 $

This leaves only the second term

$\displaystyle \hat{r} \cdot \hat{r} \frac{\partial^2}{\partial r^2} = \frac{\partial^2}{\partial r^2}$ (3)

Now onto the $\hat{\theta}$ term

$\displaystyle \frac{\hat{\theta}}{r} \cdot \left( \frac{\partial \hat{r}}{\part... ...t{\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

$\displaystyle \frac{\partial}{\partial \theta} \frac{1}{r} = 0 $

$\displaystyle \frac{\partial \hat{\phi}}{\partial \theta} = 0 $

Two terms remain, for the first term we need to calculate

$\displaystyle \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

$\displaystyle \frac{\hat{\theta}}{r} \cdot \hat{\theta} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r}$ (4)

The 5th term is just the dot product

$\displaystyle \frac{\hat{\theta}}{r} \cdot \frac{\hat{\theta}}{r} \frac{\partial^2}{\partial \theta^2} = \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}$ (5)

Finally, the $\hat{\phi}$ part of the dot product is

$\displaystyle \frac{\hat{\phi}}{r sin \theta} \cdot \left( \frac{\partial \hat{... ...i} + \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

$\displaystyle \frac{\partial \hat{r}}{\partial \phi} = -sin \theta \, sin \phi ... ...\theta \, cos \phi \hat{j} = sin \theta \hat{\phi} \frac{\partial}{\partial r} $

this leads to

$\displaystyle \frac{\hat{\phi}}{r sin \theta} \cdot sin \theta \hat{\phi} \frac{\partial}{\partial r} = \frac{1}{r} \frac{\partial}{\partial r}$ (6)

The 3rd term derivative is

$\displaystyle \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

$\displaystyle \frac{\hat{\phi}}{r sin \theta} \cdot cos \theta \hat{\phi} \frac... ...al \theta} = \frac{cos \theta}{r^2 sin \theta} \frac{\partial}{\partial \theta}$ (7)

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

$\displaystyle \frac{\partial}{\partial \phi} \left(\frac{1}{r sin \theta}\right) = 0 $

Then we keep the 8th term

$\displaystyle \frac{\hat{\phi}}{r sin \theta} \cdot \frac{\hat{\phi}}{r sin \th... ...partial \phi^2} = \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2}$ (8)

Putting the results from (3),(4),(5),(6),(7) and (8) we get

$\displaystyle \nabla_{sph}^{2} = \frac{\partial^2}{\partial r^2} + \frac{1}{r} ... ...artial \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)

$\displaystyle \left[\frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partia... ...1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) $

$\displaystyle \left[\frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} + \frac{... ...tial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right) $

This gives us the equation given in (1), the Laplacian in spherical coordinates

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

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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See Also: Laplacian in Cylindrical Coordinates, Laplacian in Cartesian Coordinates, Laplacian


Cross-references: dot product, function, gradient, magnitudes, motion, identity, unit vector, vector, operator, Laplacian
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This is version 21 of Laplacian in Spherical Coordinates, born on 2005-12-30, modified 2006-01-05.
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