Use Maxima to find Laplacian in polar coordinates

Here I'd like to share how to find the Laplacian in polar coordinates with the help of Maxima.
It also serves as a good exercise for learning how to use Maxima.

Those who have learned electromagnetism, or quantum mechanics,
would be familiar with 2-, 3-dimensional Laplacian.

A problem in physics sometimes concerns with the case with rotational invariance,
it is natural to move on to discussion on polar coordinates, rather than the Cartesian coordinates.

However, the Laplacian in polar coordinates appears a bit cumbersome,
while in Cartesian coordinates it takes such a simple form.

Its derivation is rather tedious;
this "sine" combines with that "cosine" and together with another, they cancel ... !! ... !!
spending an hour!!!

This kind of simple but lengthy work shall be done by computer.

Maxima is such a free software enabling you to do it.

The following article provides an essence and a flavor of how to use Maxima:

pianofisica.hatenablog.com

In this article, I'd like to show you how to find the Laplacian
on 2-, 3-dim. polar coordinates with the help of Maxima.

Additionally, I'll demonstrate 4-dimensional case,
and show the result for 5- and 6-dimensional cases.
(It is an advantage of using computer that an extension is easily available)

2-dimensional Laplacian
3-dimensional Laplacian
4-dimensional Laplacian
5-dimensional Laplacian
6-dimensional Laplacian

2-dimensional Laplacian

As a warm-up, let us rewrite the 2-dimensional Laplacian

$\displaystyle{\triangle^{(2)}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}}$

in polar coordinates $(r,\theta)$ . The relation among $(x,y)$ and $(r,\theta)$ is given by

$x=r\cos\theta\\ y=r\sin\theta$

or equivalently

$\displaystyle{r=\sqrt{x^2+y^2}\\ \theta=\arctan\left(\frac{y}{x}\right)}$

From the chain rule of the differential it follows that

$\displaystyle{\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array} \right)=\left(\begin{array}{c} \frac{\partial r}{\partial x}\frac{\partial }{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial }{\partial \theta} \\ \frac{\partial r}{\partial y}\frac{\partial }{\partial r}+\frac{\partial \theta}{\partial y}\frac{\partial }{\partial \theta} \end{array}\right)=R_2\left(\begin{array}{c}\frac{\partial }{\partial r}\\ \frac{\partial }{\partial \theta} \end{array}\right)}$

Here I introduced a matrix

$\displaystyle{R_2=\left(\begin{array}{cc} \frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y} \end{array}\right)}$

hence

$\displaystyle{ \left( \begin{array}{c} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{array} \right) = \left( \begin{array}{c} \left(R_2\right)_{11}\frac{\partial }{\partial r}+\left(R_2\right)_{12}\frac{\partial }{\partial \theta} & \left(R_2\right)_{21}\frac{\partial }{\partial r}+\left(R_2\right)_{22}\frac{\partial }{\partial \theta} \end{array}\right) }$

This matrix $\ R_2$ is calculated by putting the following source in Maxima

kill(all)$ assume(r>0)$
r(x,y):=sqrt(x^2+y^2)$ theta(x,y):=atan(y/x)$
r2:matrix([diff(r(x,y),x),diff(theta(x,y),x)],[diff(r(x,y),y),diff(theta(x,y),y)])$
R2:trigsimp(subst(r*sin(theta), y, subst(r*cos(theta), x, r2)));

which is found to be

$\displaystyle{R_2=\left(\begin{array}{cc} \cos\theta & -\frac{\sin \theta}{r} \\ \sin\theta & \frac{\cos\theta}{r} \end{array}\right)}$

Let $\ f=f(r,\theta)$ be an arbitrary function, and define

$\displaystyle{T_2=\left(\begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array} \right)=R_2\left(\begin{array}{c} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \end{array} \right)}$

Then, we can find the Laplacian by calculating

$\displaystyle{ \triangle^{(2)} f =\left(\begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{array} \right) \left(\begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array} \right) =\sum_{i=1}^2\left((R_2)_{i1}\frac{\partial}{\partial r}+(R_2)_{i2}\frac{\partial}{\partial \theta}\right)(T_2)_i}$

Putting in Maxima to evaluate the above mentioned object

depends(f,[r,theta])$
T2:R2.matrix([diff(f,r)],[diff(f,theta)]);
Lap2:expand(trigsimp(sum(
R2[i][1]*diff(T2[i][1],r)+R2[i][2]*diff(T2[i][1],theta), i, 1, 2)));

we obtain

$\displaystyle{\triangle^{(2)}=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}}$

3-dimensional Laplacian

The idea of finding 3-dimensional Laplacian is a straightforward extension of 2-dimensional case:

$\displaystyle{\triangle^{(3)}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}}$

The relation between the Crtesian coordinates $(x,y,z)$ and polar coordinates $(r,\theta,\varphi)$ is given by

$x=r\sin\theta\cos\varphi\\ y=r\sin\theta\sin\varphi \\ z=r\cos\theta$

$\displaystyle{r=\sqrt{x^2+y^2+z^2}\\ \theta=\arctan\left(\frac{\sqrt{x^2+y^2}}{z}\right)\\ \varphi=\arctan\left(\frac{y}{x}\right)}$

Let $\ f=f(r,\theta,\varphi)$ be an arbitrary function, and introduce

$\displaystyle{R_3=\left(\begin{array}{ccc} \frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} & \frac{\partial \varphi}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y} & \frac{\partial \varphi}{\partial y} \\ \frac{\partial r}{\partial z} & \frac{\partial \theta}{\partial z} & \frac{\partial \varphi}{\partial z} \end{array}\right) \\ T_3=\left(\begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{array} \right)=R_3\left(\begin{array}{c} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \\ \frac{\partial f}{\partial \varphi} \end{array} \right)}$

Then one can find the Laplacian via

$\displaystyle{\triangle^{(3)} f=\sum_{i=1}^3\left((R_3)_{i1}\frac{\partial}{\partial r}+(R_3)_{i2}\frac{\partial}{\partial \theta}+(R_3)_{i3}\frac{\partial}{\partial \varphi}\right)(T_3)_i}$

Put the following source in Maxima

kill(all)$ assume(r>0)$ assume(sin(theta)>0)$
r(x,y,z):=sqrt(x^2+y^2+z^2)$ theta(x,y,z):=atan(sqrt(x^2+y^2)/z)$ phi(x,y,z):=atan(y/x)$
r3:matrix(
[diff(r(x,y,z),x),diff(theta(x,y,z),x),diff(phi(x,y,z),x)],
[diff(r(x,y,z),y),diff(theta(x,y,z),y),diff(phi(x,y,z),y)],
[diff(r(x,y,z),z),diff(theta(x,y,z),z),diff(phi(x,y,z),z)])$
R3:trigsimp(
subst(r*cos(theta), z, subst(r*sin(theta)*sin(phi), y, subst(r*sin(theta)*cos(phi), x, r3))));
depends(f,[r,theta,phi])$
T3:R3.matrix([diff(f,r)],[diff(f,theta)],[diff(f,phi)])$
Lap3:expand(trigsimp(sum(
R3[i,1]*diff(T3[i][1],r)+R3[i,2]*diff(T3[i][1],theta)+R3[i,3]*diff(T3[i][1],phi), i, 1, 3)));

one finds

$\displaystyle{\triangle^{(3)}=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\cos\theta}{r^2\sin\theta}\frac{\partial}{\partial \theta} +\frac{1}{r^2\sin^2\theta}\frac{\partial}{\partial \varphi}}$

4-dimensional Laplacian

Finally the 4-dimensional Laplacian

$\displaystyle{\triangle^{(4)}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial w^2}}$

on polar coordinates $(r,\theta,\varphi,\psi)$ whose relation to the Cartesian coordinates $(x,y,z,w)$ is

$x=r\sin\theta\sin\varphi\sin\psi \\ y=r\sin\theta\sin\varphi\cos\psi\\ z=r\sin\theta\cos\varphi\\ w=r\cos\theta$

$\displaystyle{r=\sqrt{x^2+y^2+z^2+w^2}\\ \theta=\arctan\left(\frac{\sqrt{x^2+y^2+z^2}}{w}\right)\\ \varphi=\arctan\left(\frac{\sqrt{x^2+y^2}}{z}\right)\\ \psi=\arctan\left(\frac{\sqrt{x^2}}{y}\right)}$

Putting the following into Maxima

kill(all)$ assume(r>0)$ assume(sin(theta)>0)$ assume(sin(phi)>0)$ assume(sin(psi)>0)$
r(x,y,z,w):=sqrt(x^2+y^2+z^2+w^2)$ theta(x,y,z,w):=atan(sqrt(x^2+y^2+z^2)/w)$
phi(x,y,z,w):=atan(sqrt(x^2+y^2)/z)$ psi(x,y,z,w):=atan(sqrt(x^2)/y)$
r4:matrix(
[diff(r(x,y,z,w),x),diff(theta(x,y,z,w),x),diff(phi(x,y,z,w),x),diff(psi(x,y,z,w),x)],
[diff(r(x,y,z,w),y),diff(theta(x,y,z,w),y),diff(phi(x,y,z,w),y),diff(psi(x,y,z,w),y)],
[diff(r(x,y,z,w),z),diff(theta(x,y,z,w),z),diff(phi(x,y,z,w),z),diff(psi(x,y,z,w),z)],
[diff(r(x,y,z,w),w),diff(theta(x,y,z,w),w),diff(phi(x,y,z,w),w),diff(psi(x,y,z,w),w)])$
R4:trigsimp(
subst(r*cos(theta), w,  subst(r*sin(theta)*cos(phi), z, 
subst(r*sin(theta)*sin(phi)*cos(psi), y, subst(r*sin(theta)*sin(phi)*sin(psi), x, r4)))));
depends(f,[r,theta,phi,psi])$
T4:R4.matrix([diff(f,r)],[diff(f,theta)],[diff(f,phi)],[diff(f,psi)])$
Lap4:expand(trigsimp(sum(
R4[i,1]*diff(T4[i][1],r)+R4[i,2]*diff(T4[i][1],theta)+
R4[i,3]*diff(T4[i][1],phi)+R4[i,4]*diff(T4[i][1],psi), i, 1, 4)));

one finds

$\displaystyle{\triangle^{(4)}={{\partial ^2}\over{\partial r^2}}+{{3}\over{r}}{{\partial }\over{\partial r}}+{{1}\over{r^2}}{{\partial^2}\over{\partial \theta^2}}+{{2\cos \theta}\over{r^2\sin \theta}}{{\partial }\over{\partial \theta}}\\\qquad\qquad\qquad+{{1}\over{r^2\sin ^2\theta}}{{\partial^2}\over{\partial \varphi^2}}+{{\cos \varphi}\over{r^2\sin ^2\theta\sin \varphi}}{{\partial}\over{\partial \varphi}}+{{1}\over{r^2\sin ^2\theta\sin ^2\varphi}}{{\partial^2}\over{\partial \psi^2}} }$

From these results, you may guess the result for 5-, 6-, ... dimension, and check it!

The results shall be ...

5-dimensional Laplacian

The polar coordinates in 5-dimension is

$x=r\sin\theta\sin\varphi\sin\psi\sin\chi\\y=r\sin\theta\sin\varphi\sin\psi\cos\chi\\z=r\sin\theta\sin\varphi\cos\psi\\w=r\sin\theta\cos\varphi\\s=r\cos\theta$

The 5-dimensional Laplacian in polar coordinate takes the form

$\displaystyle{\triangle^{(5)}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial w^2}+\frac{\partial^2}{\partial s^2}\\\quad\ \ \ \, ={{\partial^2}\over{\partial r^2}}+{{4}\over{r}}{{\partial}\over{\partial r}}+{{1}\over{r^2}}{{\partial^2}\over{\partial\theta^2}}+{{3\cos \theta}\over{r^2\sin \theta}}{{\partial}\over{\partial\theta}}+{{1}\over{r^2\sin ^2\theta}}{{\partial^2}\over{\partial\varphi^2}}+{{2\cos \varphi}\over{r^2\sin^2\theta\sin\varphi}}{{\partial}\over{\partial\varphi}}\\\qquad\quad+{{1}\over{r^2\sin^2\theta\sin^2\varphi}}{{\partial^2}\over{\partial\psi^2}}+{{\cos\psi}\over{r^2\sin^2\theta\sin^2\varphi\sin\psi}}{{\partial}\over{\partial\psi}}+{{1}\over{r^2\sin ^2\theta\sin ^2\varphi\sin ^2\psi}}{{\partial^2}\over{\partial\chi^2}}}$

and for 6-dimension

6-dimensional Laplacian

Let the polar coordinate be

$x=r\sin\theta\sin\varphi\sin\psi\sin\chi\sin\eta\\y=r\sin\theta\sin\varphi\sin\psi\sin\chi\cos\eta\\z=r\sin\theta\sin\varphi\sin\psi\cos\chi\\w=r\sin\theta\sin\varphi\cos\psi\\s=r\sin\theta\cos\varphi\\t=r\cos\theta$

and then the result is

$\displaystyle{\triangle^{(6)}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial w^2}+\frac{\partial^2}{\partial s^2}+\frac{\partial^2}{\partial t^2}\\\quad\ \ \ \, ={{\partial^2}\over{\partial r^2}}+{{5}\over{r}}{{\partial}\over{\partial r}}+{{1}\over{r^2}}{{\partial^2}\over{\partial\theta^2}}+{{4\cos \theta}\over{r^2\sin \theta}}{{\partial}\over{\partial\theta}}+{{1}\over{r^2\sin ^2\theta}}{{\partial^2}\over{\partial\varphi^2}}+{{3\cos \varphi}\over{r^2\sin^2\theta\sin\varphi}}{{\partial}\over{\partial\varphi}}\\\qquad\quad+{{1}\over{r^2\sin^2\theta\sin^2\varphi}}{{\partial^2}\over{\partial\psi^2}}+{{2\cos\psi}\over{r^2\sin^2\theta\sin^2\varphi\sin\psi}}{{\partial}\over{\partial\psi}}\\\qquad\quad+{{1}\over{r^2\sin ^2\theta\sin ^2\varphi\sin ^2\psi}}{{\partial^2}\over{\partial\chi^2}}+{{\cos\chi}\over{r^2\sin ^2\theta\sin ^2\varphi\sin ^2\psi\sin\chi}}{{\partial}\over{\partial\chi}}\\\qquad\quad+{{1}\over{r^2\sin ^2\theta\sin ^2\varphi\sin ^2\psi\sin ^2\chi}}{{\partial^2}\over{\partial\eta^2}}}$

For general $n$ -dimension, ... would it be transparent anymore?

Keywords：Maxima, Vector calculus, Laplacian