2
$\begingroup$

I thought that polar coordinates ($r, \phi$) can be viewed as a special case of cylindrical coordinates ($\rho, \phi, z$) with $z=0$, or as spherical coordinates ($r, \theta, \phi$) with $\theta=\frac{\tau}{4}$.

(Note: $\tau \equiv 2\pi$)

The Laplacians for these coordinate systems are:

$$\nabla^2 f_{\text{polar}} = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho \frac{\partial f}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2}$$ $$\nabla^2 f_{\text{cyl}} = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho \frac{\partial f}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2} $$ $$\nabla^2 f_{\text{sph}} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial f}{\partial r}\right)+\frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f}{\partial \theta}\right) +\frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial \phi^2} $$

More specifically, say that we have a function $f:\mathbb{R}^3\to\mathbb{R}$, which can be expressed in spherical or cylindrical, ie. $f_{\text{cyl}}(\rho,\phi,z)=f_{\text{sph}}(r,\theta,\phi)$. Assume that in the XY-plane, $\frac{\partial f_{\text{cyl}}}{\partial z}=0$, which implies $\frac{\partial f_{\text{sph}}}{\partial \theta}=0$. In the XY-plane, these functions then obey the differential equations

$$\nabla^2 f_{\text{cyl}} = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho \frac{\partial f_{\text{cyl}}}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2 f_{\text{cyl}}}{\partial \phi^2}$$

and

$$\nabla^2 f_{\text{sph}} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial f_{\text{sph}}}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 f_{\text{sph}}}{\partial \phi^2} $$

But if they follow different differential equations, how can $f_{\text{cyl}}=f_{\text{sph}}$?

$\endgroup$

1 Answer 1

1
$\begingroup$

OK, I think I found my mistake.

Even though $$\left.\frac{\partial f_{\text{sph}}}{\partial \theta}\right|_{\theta=\frac{\tau}{4}} = 0$$ we can still have $$\left.\frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f_{\text{sph}}}{\partial \theta}\right) \right|_{\theta=\frac{\tau}{4}}\neq0\,\,.$$

For example, try $f_{\text{sph}}=\sin(\theta)$. Therefore, the Laplacian in spherical can not be simplified as in the question.

For the same reason, I also should have stated that $\left.\frac{\partial^2 f_{\text{cyl}}}{\partial z^2}\right|_{z=0}=0$ if I wanted to simplify the Laplacian for cylindrical coordinates.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .