TITLE: How to shorten the derivation of the Laplacian in polar coordinates?
QUESTION [6 upvotes]: Yesterday, I computed the formula for the Laplacian in polar coordinates.
$
\newcommand{p}[2]{\frac {\partial #1} {\partial #2}}
\newcommand{s}[2]{\p {^2 #1} {#2 ^2}}
\newcommand{m}[3]{\p {^2 #1} {#2 \partial #3}}
\newcommand{po}[0]{\p {}}
\newcommand{so}[0]{\s {}}
\newcommand{mo}[0]{\m {}}
\newcommand{t}[0]{\theta}
\newcommand{w}[1]{\left( #1 \right)}
\newcommand{ct}[0]{\cos \t}
\newcommand{st}[0]{\sin \t}
\newcommand{f}[1]{\frac #1 r}
\newcommand{g}[1]{\frac #1 {r^2}}
\newcommand{jx}[2]{\ct #1 - \f \st #2}
\newcommand{jy}[2]{\st #1 + \f \ct #2}
\newcommand{px}[2]{\jx {\po r #1} {\po \t #2}}
\newcommand{py}[2]{\jy {\po r #1} {\po \t #2}}
\newcommand{pox}[0]{\px {} {}}
\newcommand{poy}[0]{\py {} {}}
\newcommand{one}[0]{\sin^2\t + \cos^2\t}
\newcommand{zero}[0]{\st\ct - \st\ct}
$
\begin{align*}
x & = r \ct \\ y & = r \st \\ r^2 & = x^2 + y^2
\end{align*}
First, I used the chain rule to relate $\po r$ and $\po \t$ to $\po x$ and $\po y$:
\begin{align*}
\po x = \p r x \po r + \p \t x \po \t = \pox \\
\po y = \p r y \po r + \p \t y \po \t = \poy \\
\end{align*}
Applying each of $\po x$ and $\po y$ to itself:
\begin{align*}
\so x + \so y
  & = \po x {\w {\po x}} + \po y {\w {\po y}} \\
  & = \px {\w \pox} {\w \pox}  \\
  & \qquad + \py {\w \poy} {\w \poy} \\
  & = \ct \w {\ct \so r + \g \st \po \t - \f \st \mo r \t} \\
  & \qquad+ \f \st \w {\st \po r - \ct \mo \t r + \f \ct \po \t + \f \st \so \t} \\
  & \qquad + \st \w {\st \so r - \g \st \po \t + \f \ct \mo r \t} \\
  & \qquad + \f \ct \w {\ct \po r + \st \mo \t r - \f \st \po \t + \f \ct \so \t} \\
  & = \w \one \so r + \g \zero \po \t \\
  & \qquad - \f \zero \mo r \t + \f \one \po r \\
  & \qquad - \f \zero \mo \t r + \g \zero \po \t \\
  & \qquad + \g \one \so \t \\
  & = \so r + \f 1 \po r + \g 1 \so \t \\
\end{align*}
This was long and tedious. Is there some way to make the calculation shorter.

REPLY [4 votes]: If you are familiar with the language of differential forms, the Laplacian is
$$\star d \star d,$$
where $\star$ is the Hodge star operator and $d$ is exterior derivative.
In polar coordinates, the volume form is $r dr \wedge d\theta$.
The Hodge star operator sends the volume form to $1$ ($\star r dr \wedge d\theta = 1$) and acts on the $1$-form basis as follows:
$$\star dr = r d\theta,\\  \star r d\theta = -dr.$$
So for a function $f$ we have:
\begin{align}
\star d \star d f
&=\star d \star (f_r dr + \frac{f_\theta}{r} rd\theta)\\
&=\star d (f_r r d\theta-\frac{f_\theta}{r} dr)\\
&=\star (f_{rr}r dr \wedge d\theta + f_r dr \wedge d\theta - \frac{f_{\theta\theta}}{r} d\theta \wedge dr)\\
&=\star (f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2}) r dr \wedge d\theta\\
&=f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2}
\end{align}