# How to shorten the derivation of the Laplacian in polar coordinates?

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.

• Honestly, yours is the easiest method I found on the internet. – Tachyon209 May 2 '20 at 20:02

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}

• I am familiar with differential forms, but not with the Hodge star operator. Still, thanks! Your calculation looks much neater than mine, so it seems like good motivation for learning about the Hodge star operator. – pyon Oct 9 '20 at 10:01
• Now that I skimmed the Wikipedia article, it looks like an operator that gives an naturally induced between forms of complementary degrees, in a way that is induced by a metric (without which, of course, the Laplacian makes no sense). It is definitely a very neat thing! – pyon Oct 9 '20 at 10:06

Apparently, Math.SE does not allow me to “delete for real” my previous wrong answer, so I will try to provide a non-wrong answer.

Let $$\newcommand \t \theta F(x,y) = f(r,\t)$$, where $$(r,\t) = g(x,y)$$. Of course, in our case, $$g$$ is not a globally well-defined function, but who cares? Differential calculus is all about working locally. By the chain rule, $$F_{xx} = (f \circ g)_{xx} = (\nabla f \cdot g_x)_x = (\nabla f)_x \cdot g_x + \nabla f \cdot g_{xx} = g_x^T \cdot Hf \cdot g_x + \nabla f \cdot g_{xx}$$

Of course, analogously, we have $$F_{yy} = g_y^T \cdot Hf \cdot g_y + \nabla f \cdot g_{yy}$$

Adding everything, the Laplacian expands to $$F_{xx} + F_{yy} = g_x^T \cdot Hf \cdot g_y + g_y^T \cdot Hf \cdot g_y + \nabla f \cdot (g_{xx} + g_{yy})$$

Call the sum of the first two terms $$A$$ and call the last term $$B$$.

In our particular case, $$r^2 = x^2 + y^2$$ and $$\tan \t = y/x$$. Thus, $$r_x = x/r, \qquad r_y = y/r, \qquad \sec^2 \t \cdot \t_x = -y/x^2, \qquad \sec^2 \t \cdot \t_y = 1/x$$

Rearranging the last two expressions, we have $$\t_x = -y/r^2, \qquad \qquad \qquad \t_y = x/r^2$$

By the chain rule, $$A$$ reduces to $$f_{rr} (r_x^2 + r_y^2) + 2f_{r\t} (r_x \t_x + r_y \t_y) + f_{\t\t} (\t_x^2 + \t_y^2) = f_{rr} + \frac 1 {r^2} f_{\t\t}$$

Differentiating once again, we have $$r_{xx} = 1/r, \qquad r_{yy} = 1/r, \qquad \t_{xx} = 2yr_x /r^3, \qquad \t_{yy} = -2xr_y / r^3$$

Rearranging the last two expressions, we have $$\t_{xx} = 2xy / r^4, \qquad \qquad \qquad \t_{yy} = -2xy / r^4$$

By the chain rule, $$B$$ reduces to $$f_r \cdot (r_{xx} + r_{yy}) + f_\t \cdot (\t_{xx} + \t_{yy}) = \frac 1r f_r$$

Combining everything, we have

$$F_{xx} + F_{yy} = f_{rr} + \frac 1r f_r + \frac 1 {r^2} f_{\t\t}$$

• Even if we suppose that manipulating $\nabla$ like it's a vector is valid, why is $$\frac{1}{r} \nabla \cdot r \nabla = \nabla \cdot \nabla,$$ i.e. why can the $r$ and the differential operator be commuted? – Chappers Jun 26 '17 at 13:45
• @Chappers: Sorry, you are right. I updated my answer. – pyon Aug 2 '20 at 4:21
• Yes, that's much better. – Chappers Aug 2 '20 at 13:56