PlanetPhysics/Laplacian in Spherical Coordinates

The [[../LaplaceOperator/|Laplacian]] [[../QuantumSpinNetworkFunctor2/|operator]] in spherical coordinates is

$\underset{sph}{\overset{2}{\mathrm{\nabla }}}=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^{2}si{n}^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$

The derivation is fairly straight forward and begins with locating a [[../Vectors/|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 [[../PureState/|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 [[../Cod/|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

$cos\varphi =\frac{x}{|{\hat{r}}_{xy}|}$

giving

$x=cos\varphi sin\theta$

the y component follows the x component through

$sin\varphi =\frac{y}{|{\hat{r}}_{xy}|}$

which yields

$y=sin\varphi sin\theta$

The vector in spherical coordinates is then

$𝐫=r\hat{r}=r[sin\theta cos\varphi \hat{i}+sin\theta sin\varphi \hat{j}+cos\theta \hat{k}]$

Now describing the unit vectors of a moving [[../Particle/|particle]] $\hat{r},\hat{\theta },\hat{\varphi }$ 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

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

For $\hat{\theta }$ and $\hat{\varphi }$, think of uniform circular [[../CosmologicalConstant2/|motion]] like a record, so they can be calculated from

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

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

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

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

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

$|\frac{\partial 𝐫}{\partial \theta }|=\sqrt{𝐱\cdot 𝐱}=\sqrt{r^2[co{s}^2\theta co{s}^2\varphi +co{s}^2\theta si{n}^2\varphi +si{n}^2\theta ]}$

$|\frac{\partial 𝐫}{\partial \theta }|=r\sqrt{co{s}^2\theta (co{s}^2\varphi +si{n}^2\varphi )+si{n}^2\theta ]}$

Using the basic trig identity

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

$|\frac{\partial 𝐫}{\partial \varphi }|=r\sqrt{si{n}^2\theta (si{n}^2\varphi +co{s}^2\varphi )]}$

$|\frac{\partial 𝐫}{\partial \varphi }|=rsin\theta$

Then plug in these values to get

$\hat{\varphi }=-sin\varphi \hat{i}+cos\varphi \hat{j}$ $\hat{\theta }=cos\theta cos\varphi \hat{i}+cos\theta sin\varphi \hat{j}-sin\theta \hat{k}$

and from before we had

$\hat{r}=sin\theta cos\varphi \hat{i}+sin\theta sin\varphi \hat{j}+cos\theta \hat{k}$

The generalized differential for curvilinear coordinates is

$d𝐫=\frac{\partial 𝐫}{\partial u_1}d{u}_1+\frac{\partial 𝐫}{\partial u_2}d{u}_2+\frac{\partial 𝐫}{\partial u_3}d{u}_3$

For spherical coordinates we have

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

and from earlier we learned

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

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

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

Plugging these values into the generalized differential yields

$d𝐫=dr\hat{r}+rd\theta \hat{\theta }+rsin\theta d\varphi \hat{\varphi }$

The next major step is to see how the [[../Gradient/|gradient]] fits into the definition of the differential for a [[../Bijective/|function]] f

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

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

$\frac{\partial f}{\partial r}dr={\mathrm{\nabla }}_{r}fdr$ $\frac{\partial f}{\partial \theta }d\theta ={\mathrm{\nabla }}_{\theta }fd\theta$ $\frac{\partial f}{\partial \varphi }d\varphi ={\mathrm{\nabla }}_{\varphi }fd\varphi$

Therfore our scale factors are

${\mathrm{\nabla }}_r=1$ ${\mathrm{\nabla }}_{\theta }=\frac{1}{r}$ ${\mathrm{\nabla }}_{\varphi }=\frac{1}{rsin\theta }$

which finally gives us the gradient in spherical coordinates

${\mathrm{\nabla }}_{sph}=\frac{\partial }{\partial r}\hat{r}+\frac{\partial }{\partial \theta }\frac{\hat{\theta }}{r}+\frac{\partial }{\partial \varphi }\frac{\hat{\varphi }}{rsin\theta }$

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

${\mathrm{\nabla }}^2=\mathrm{\nabla }\cdot \mathrm{\nabla }=(\frac{\partial }{\partial r}\hat{r}+\frac{\partial }{\partial \theta }\frac{\hat{\theta }}{r}+\frac{\partial }{\partial \varphi }\frac{\hat{\varphi }}{rsin\theta })\cdot (\frac{\partial }{\partial r}\hat{r}+\frac{\partial }{\partial \theta }\frac{\hat{\theta }}{r}+\frac{\partial }{\partial \varphi }\frac{\hat{\varphi }}{rsin\theta })$

Let us break it up by components to make it easy to view, so carrying out only part of the [[../DotProduct/|dot product]] our 1st term is

$\hat{r}\cdot (\frac{\partial \hat{r}}{\partial r}\frac{\partial }{\partial r}+\hat{r}\frac{{\partial }^2}{\partial r^2}+\frac{\partial \hat{\theta }}{\partial r}\frac{1}{r}\frac{\partial }{\partial \theta }+\hat{\theta }\frac{\partial }{\partial r}\frac{1}{r}\frac{\partial }{\partial \theta }+\frac{\hat{\theta }}{r}\frac{{\partial }^2}{\partial r\partial \theta }+\frac{\partial \hat{\varphi }}{\partial r}\frac{1}{rsin\theta }\frac{\partial }{\partial \varphi }+\hat{\varphi }\frac{\partial }{\partial r}\left(\frac{1}{rsin\theta }\right)\frac{\partial }{\partial \varphi }+\frac{\hat{\varphi }}{rsin\theta }\frac{{\partial }^2}{\partial r\partial \varphi })$

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

$\hat{r}=sin\theta cos\varphi \hat{i}+sin\theta sin\varphi \hat{j}+cos\theta \hat{k}$

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

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

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

$\frac{\hat{\theta }}{r}\cdot (\frac{\partial \hat{r}}{\partial \theta }\frac{\partial }{\partial r}+\hat{r}\frac{{\partial }^2}{\partial \theta \partial r}+\frac{\partial \hat{\theta }}{\partial \theta }\frac{1}{r}\frac{\partial }{\partial \theta }+\hat{\theta }\frac{\partial }{\partial \theta }\frac{1}{r}\frac{\partial }{\partial \theta }+\frac{\hat{\theta }}{r}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{\partial \hat{\varphi }}{\partial \theta }\frac{1}{rsin\theta }\frac{\partial }{\partial \varphi }+\hat{\varphi }\frac{\partial }{\partial \theta }\left(\frac{1}{rsin\theta }\right)\frac{\partial }{\partial \varphi }+\frac{\hat{\varphi }}{rsin\theta }\frac{{\partial }^2}{\partial \theta \partial \varphi })$

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

$\frac{\partial }{\partial \theta }\frac{1}{r}=0$ $\frac{\partial \hat{\varphi }}{\partial \theta }=0$

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

$\frac{\partial \hat{r}}{\partial \theta }=cos\theta cos\varphi \hat{i}+cos\theta sin\varphi \hat{j}-sin\theta \hat{k}=\hat{\theta }$

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 $\hat{\varphi }$ part of the dot product is

$\frac{\hat{\varphi }}{rsin\theta }\cdot (\frac{\partial \hat{r}}{\partial \varphi }\frac{\partial }{\partial r}+\hat{r}\frac{{\partial }^2}{\partial \varphi \partial r}+\frac{\partial \hat{\theta }}{\partial \varphi }\frac{1}{r}\frac{\partial }{\partial \theta }+\hat{\theta }\frac{\partial }{\partial \varphi }\frac{1}{r}\frac{\partial }{\partial \theta }+\frac{\hat{\theta }}{r}\frac{{\partial }^2}{\partial \varphi \partial \theta }+\frac{\partial \hat{\varphi }}{\partial \varphi }\frac{1}{rsin\theta }\frac{\partial }{\partial \varphi }+\hat{\varphi }\frac{\partial }{\partial \varphi }\left(\frac{1}{rsin\theta }\right)\frac{\partial }{\partial \varphi }+\frac{\hat{\varphi }}{rsin\theta }\frac{{\partial }^2}{\partial {\varphi }^2})$

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

$\frac{\partial \hat{r}}{\partial \varphi }=-sin\theta sin\varphi \hat{i}+sin\theta cos\varphi \hat{j}=sin\theta \hat{\varphi }\frac{\partial }{\partial r}$

this leads to

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

The 3rd term derivative is

$\frac{\partial \hat{\theta }}{\partial \varphi }=-cos\theta sin\varphi \hat{i}+cos\theta cos\varphi \hat{j}=cos\theta \hat{\varphi }$

this leads to

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

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

The 7th term derivative is zero

$\frac{\partial }{\partial \varphi }\left(\frac{1}{rsin\theta }\right)=0$

Then we keep the 8th term

$\frac{\hat{\varphi }}{rsin\theta }\cdot \frac{\hat{\varphi }}{rsin\theta }\frac{{\partial }^2}{\partial {\varphi }^2}=\frac{1}{r^{2}si{n}^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$

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

$\underset{sph}{\overset{2}{\mathrm{\nabla }}}=\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{cos\theta }{r^{2}sin\theta }\frac{\partial }{\partial \theta }+\frac{1}{r^{2}si{n}^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$

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

$[\frac{{\partial }^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}]=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)$

$[\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{cos\theta }{r^{2}sin\theta }\frac{\partial }{\partial \theta }]=\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)$

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

$\underset{sph}{\overset{2}{\mathrm{\nabla }}}=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^{2}si{n}^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$

{\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.