I'm trying to find the error in my logic here.

Let's say we are given the Laplace operator in polar coordinates:

$$ \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial}{\partial \theta^2} \tag{1} $$

and we're interested in transforming back to Cartesian coordinates. We could first make use of the usual coordinate transformation, namely

$$ u = r \sin(\theta), ~~v = r \cos(\theta), \tag{2}$$

to write the partial derivatives in polar coordinates as follows,

$$ \frac{\partial}{\partial r} = \sin(\theta) \frac{\partial}{\partial u} + \cos(\theta)\frac{\partial}{\partial v}, \tag{3}$$

$$ \frac{\partial}{\partial \theta} = r \cos(\theta) \frac{\partial}{\partial u} - r \sin(\theta)\frac{\partial}{\partial v}. \tag{4}$$

I think we should then rewrite all $\theta$ and $r$ in terms of $u$ and $v$, making use of $r^2 = u^2 + v^2$, so that in particular

$$ \sin(\theta) = \frac{u}{\sqrt{u^2 + v^2}}, \tag{5}$$ $$ \cos(\theta) = \frac{v}{\sqrt{u^2 + v^2}} \tag{6}.$$

With this, (3) and (4) then become

$$ \frac{\partial}{\partial r} = \frac{u}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial u} + \frac{v}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial v}, \tag{3*}$$

$$ \frac{\partial}{\partial \theta} = v \frac{\partial}{\partial u} - u \frac{\partial}{\partial v}. \tag{4*}$$

Our next task is to then compute $\frac{\partial^2}{\partial r^2}$ and $\frac{\partial^2}{\partial \theta^2}$, which we can carefully do by multiplying the right-hand sides of (3*) and (4*), keeping in mind that the operators will act on the coefficients.

However, executing this and adding the terms together yields nothing that looks remotely near the known form, namely $$\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2}.$$ So I'm wondering, what am I missing here?

In addition to this particular example, I'm interested in any general information you have about transforming partial derivatives from polar to Cartesian coordinates.

  • 4
    $\begingroup$ Wikipedia says the third term in your expression for the polar Laplace operator should have an additional factor of $\frac{1}{r^2}$. $\endgroup$ Mar 15, 2012 at 7:27
  • $\begingroup$ @Qiaochu: Indeed-- twas quite late when I posted this. I have corrected it now. $\endgroup$ Mar 15, 2012 at 17:46

2 Answers 2


Here is my proof: $$ \frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}$$ so by applying the product rule

$$ \frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}\right)= \frac{\partial^2}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2+ \frac{\partial}{\partial r}\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2}{\partial \theta^2}\left(\frac{\partial \theta}{\partial x}\right)^2+ \frac{\partial}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}$$ $$ +\frac{\partial^2 }{\partial r \partial \theta}\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}$$

You get the same relation with $y$. Now you just have to calculate the derivatives of $r,\theta$ with respect to $x,y$. You have $$ r=\sqrt{x^2+y^2},\ \theta=\arctan \frac{y}{x}$$.

What follows is a simple calculus exercise on derivatives. You just need to prove that $$\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2=1, $$ etc.(the relations you need so that when you sum $\frac{\partial^2}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$ you'll get the polar form of the laplacian).

Going on the lines you started, you shouldn't change $\sin \theta,\cos \theta$ in terms of $x,y$. Just calculate the next derivative with respect to $r,\theta$, using appropriately the formula of the partial derivative of composition of functions.

  • $\begingroup$ Beni, thanks for this. I feel like I'm missing something here, though. Is there a reason why the mixed partial derivatives do not appear in the second equation? [which I believe you meant to write as $\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x} \right) $] $\endgroup$ Mar 18, 2012 at 6:12
  • $\begingroup$ @AmeliaYzaguirre: Yes, I forgot the mixed derivatives. Sorry about that. $\endgroup$ Mar 18, 2012 at 11:22
  • $\begingroup$ @BeniBogosel Isn't the mixed derivative supposed to have a factor 2? $\endgroup$
    – Michiel
    Aug 20, 2014 at 6:51

Using a bit of differential geometry. The key is the writing $\Delta f = \text{div grad}f$: if we manage to express div and grad in a coordinate-independent manner, we can write $\Delta$ easily in any coordinate system, be it cartesian or polar.

The metric and the inverse metric are $$ g_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&1 \end{array} \right)=g_{ab}^{-1} $$ in cartesian coordinates and $$ g'_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&r^2 \end{array} \right),\ \ \ g'^{-1}_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&r^{-2} \end{array} \right) $$ in polar coordinates. The volume form is $\omega=dx\wedge dy=r\, dr\wedge d\varphi$ and the differential of a function is $$ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy,\ \ \ dF= \frac{\partial F}{\partial r}dr + \frac{\partial F}{\partial \varphi}d\varphi. $$ Now, div is given by $ d[\omega(E)]=\text{div}(E) \omega, $ for any vector field $E=E^x \partial_x + E^y\partial_y$ or $B=B^r \partial_r + B^\varphi\partial_\varphi$; so $$ d[\omega(E)] = d[E^x dy - E^y dx]=\left(\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}\right) dx\wedge dy\implies\text{div}(E)=\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}. $$ Or $$ d[\omega(B)] = d[B^r r\,d\varphi - B^\varphi r\,dr] =\left(\frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}\right)r\, dr\wedge d\varphi\implies\text{div}(B)= \frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}. $$ And grad$f=g^{-1}df$, so $$ \text{grad}f = \frac{\partial f}{\partial x}\partial_x + \frac{\partial f}{\partial y}\partial_y \ \text{ and } \ \text{grad}F = \frac{\partial F}{\partial r}\partial_r + \frac{1}{r^2}\frac{\partial F}{\partial \varphi}\partial_\varphi. $$ Finally, $\Delta=$div grad: $$ \Delta f = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2} \ \text{ and } \ \Delta F = \frac{1}{r}\left( r \frac{\partial F}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 F}{\partial \varphi^2}. $$

  • $\begingroup$ Could you explain how one obtains the metric in polar coordinates (I always get $dx^2+dy^2=2xdx+2ydy=dr^2$, what am I doing wrong)? Then there is an explicit expression of the Laplacian in local coordinates which only uses the metric tensor, right? Thanks in advance. $\endgroup$ Sep 12, 2018 at 0:10
  • 1
    $\begingroup$ @LukeMathwalker Of course: $dx=dr \cos\varphi - r \sin\varphi\, d\varphi$, $dy=dr \sin\varphi+r\cos\varphi\, d\varphi$; this gives you the matrix $M$ with $M_{11}=\cos\varphi$, $M_{12}=-r\sin\varphi$, $M_{21}=\sin\varphi$ and $M_{22}=r\cos\varphi$. Then, $g'=M^T g M$. Concerning your calculation, you should not interpret $dx^2$ as $d(x\cdot x)$ but rather as $dx\otimes dx$. $\endgroup$
    – Brightsun
    Sep 12, 2018 at 17:16
  • 1
    $\begingroup$ @LukeMathwalker I think you are referring to $\sqrt{\mathrm{det}(g)} \Delta F = \partial_\mu (\sqrt{\mathrm{det}(g)} g^{\mu\nu} \partial_\nu F)$, where I assumed the Einstein summation convention on the indices $\mu$, $\nu$. One can show that this formula coincides with the one given above whenever the volume form is $dx^1\wedge \cdots \wedge dx^n$ in those coordinates $x^1$, $\ldots$, $x^n$ in which the metric tensor reads diag$(1,1,\ldots,1)$. The above approach has the advantage of also deriving the expressions for the divergence and the gradient. $\endgroup$
    – Brightsun
    Sep 12, 2018 at 17:21
  • $\begingroup$ Ah okay thank you! This helped me solving a question I asked here as well math.stackexchange.com/questions/2910656/…. Sadly I had to write the $w_2,w_3$ occuring there as $\cos(t),\sin(t)$ in order to transform it like you described. I would be glad if you could take a short look on that. (This is the last question, I promise :)). $\endgroup$ Sep 13, 2018 at 22:43
  • 1
    $\begingroup$ @LukeMathwalker B. Schutz, Geometrical methods of mathematical physics. $\endgroup$
    – Brightsun
    Sep 14, 2018 at 19:28

You must log in to answer this question.