Connection between Laplacian and volume element in spherical and cylindrical polar coordinates

Question

I know that the volume element in

• spherical polar coordinates is: $\bbox[5px,border:3px solid green]{r^2} \sin\theta \mathrm{d}\phi\mathrm{d}\theta\mathrm{d}r$
• in cylindrical polar coordinates: $\bbox[5px,border:3px solid red]{r}\mathrm{d}\theta\mathrm{d}r\mathrm{d}z$

I also know that the component of the Laplacian involving the $r$ derivative is [source]:

• in spherical polar coordinates: $$\frac{1}{\bbox[5px,border:3px solid green]{r^{2}}} \frac{\partial}{\partial r}\left(\bbox[5px,border:3px solid green]{r^{2}} \frac{\partial f}{\partial r}\right)$$
• in cylindrical polar coordinates: $$\frac{1}{ \bbox[5px,border:3px solid red]{r}} \frac{\partial}{\partial r}\left( \bbox[5px,border:3px solid red]{r} \frac{\partial f}{\partial r}\right)$$

Notice that whatever the function of $r$ I had in the volume element, that same function appears in the Laplacian (ie $r^2$ and $r$).

---

I hardly think it is a coincidence. What is the deeper connection between the Laplacian and the volume element?

My guess is that it has to do with the metric of SPC and CPC. I have done a somewhat introductory undergraduate physics course in General Relativity, so if there is an answer witout using higher math level than that, it'd be appreciated.

Answer 1

This connection arises from the divergence theorem, also known as Gauss's theorem https://en.wikipedia.org/wiki/Divergence_theorem $$\int d^3\vec r \,\bigl(\vec\nabla,\vec f(\vec r)\bigr)=\oint\bigl(\vec f(\vec r),d\vec S\bigr)$$
It is useful to draw a small cube in spherical coordinates ($d V=r^2\sin\theta d r d\theta d\phi$) and evaluate the flow of $\vec f(\vec r)$ through the surface of this element of volume. According to Gauss theorem for a small element of volume $$dV\,(\vec\nabla, \vec f(\vec r))=dV div(\vec f)\approx (f_{r+dr}dS_{r+dr}-f_{r}dS_{r})+(f_{\phi+d\phi}dS_{\phi+d\phi}-f_{\phi}dS_{\phi})+(f_{\theta+d\theta}dS_{\theta+d\theta}-f_{\theta}dS_{\theta})$$ where $\vec f(\vec r)=f_{r}\vec e_r+f_{\phi}\vec e_\phi+f_{\theta}\vec e_\theta$ and $dS_\alpha$ is the element of surface perpendicular to the corresponding vector $\vec e_{\alpha}$.

$$f_{r+dr}dS_{r+dr}-f_{r}dS_{r}\approx f_{r+dr}\cdot(r+dr)^2\sin\theta d\theta d\phi-f_{r}\cdot(r)^2\sin\theta d\theta d\phi$$ $$f_{\phi+d\phi}dS_{\phi+d\phi}-f_{\phi}dS_{\phi}\approx f_{\phi+d\phi}\cdot rdr d\theta-f_{\phi}\cdot rdr d\theta$$ $$(f_{\theta+d\theta}dS_{\theta+d\theta}-f_{\theta}dS_{\theta})\approx f_{\theta+d\theta}r\sin(\theta+d\theta)drd\phi-f_{\theta}r\sin(\theta)drd\phi$$

Dividing all by $dV$ we get the approximate expression for divergence of the field $$div (\vec f)\approx\frac{1}{r^2\sin\theta d r d\theta d\phi}\Bigl(\frac{\partial}{\partial r}\bigl(r^2f_{r}\bigr)\sin\theta d r d\theta d\phi\,+\,\frac{\partial}{\partial \phi}\bigl(f_{\phi}\bigr)r d r d\theta d\phi+\frac{\partial}{\partial \theta}\bigl(\sin\theta f_{\theta}\bigr)rd r d\theta d\phi\Bigr)$$ $$=\frac{1}{r^2}\frac{\partial}{\partial r}\bigl(r^2f_{r}\bigr)+\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\bigl(f_{\phi}\bigr)+\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}\bigl(\sin\theta f_{\theta}\bigr)$$ Taking $\vec f(\vec r)=\vec\nabla g(\vec r)=\vec e_r\frac{\partial}{\partial r}g+\vec e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}g+\vec e_\theta\frac{1}{r}\frac{\partial}{\partial\theta}g \,$ we get the expression for Laplacian in spherical coordinates.

Answer 2

What is the deeper connection between the Laplacian and the volume element?

Good catch. You are basically seeing the results of the Voss Weyl Formula for divergence. It is a generalized expression for divergence which is independent of coordinates. In a way you can think of it as an ice cream which can change flavor to one's taste (maybe practical purpose too). Here is the formula in 3 dimensions:

$$ \text{div} F = \frac{1}{\sqrt{Z}} \sum_{i=1}^3\frac{\partial }{\partial Z^i} (\sqrt{Z} F^i)$$

Where $\sqrt{Z}$ is the volume element of that coordinate system, $Z^i$ represents the coordinates of the coordinate system $F^i$ is the component of the vector

Combine this expression with the expression for general gradient and it's pretty clear what the connection is:

$$ \text{div} \text{grad} F = \frac{1}{\sqrt{Z} } \sum_{i=1}^3\frac{\partial}{\partial Z^i}(\sqrt{Z} \sum_{j=1}^3 Z^{ij} \frac{\partial F}{ \partial Z^j})$$

$Z^{ij}$ is something known as the contravariant metric tensor,

Reference: Details of all the above can be found Page-110 of Pavel Grinfield's Tensor Calculus Book