PlanetPhysics/Laplacian in Spherical Coordinates

The Laplacian operator in spherical coordinates is

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

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

To get the x component, we need to get the projected onto the xy-plane

using the trig identity

the projected unit vector is

Finally, the x component is reached through the right triangle

giving

the y component follows the x component through

which yields

The vector in spherical coordinates is then

Now describing the unit vectors of a moving particle shown in the spherical coordinates figure is a little more tricky. If a particle moves in the direction, it only moves in or out along r so

For and , think of uniform circular motion like a record, so they can be calculated from

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

Using the basic trig identity

Then plug in these values to get

and from before we had

The generalized differential for curvilinear coordinates is

For spherical coordinates we have

and from earlier we learned

Plugging these values into the generalized differential yields

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

So we see that the following must be equal and we need to solve for the gradient's components.

Therfore our scale factors are

which finally gives us the gradient in spherical coordinates

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

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

While this looks a little scary, all but the 2nd term is zero. The first term is zero because there is no r in

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 () is zero so

This leaves only the second term

Now onto the term

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

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

This yields

The 5th term is just the dot product

Finally, the part of the dot product is

Once again the dot product makes the 2nd, 4th and 5th terms zero. The first term derivative

this leads to

The 3rd term derivative is

this leads to

The 6th term can be seen to be zero because the derivative of with respect to is a vector perpendicular to (feel free to carry out this calculation), so the dot product will be zero.

The 7th term derivative is zero

Then we keep the 8th term

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

Certainly, we could finish here, but let us combine some terms and notice the following relationships (check these for yourself)

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

{\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.