Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On many websites focused on physics, (say ) they like to represent differential operators in different coordinates. I.e. going from the standard basis to polar coordinates they would write: $$\frac{\partial }{\partial x} = \frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta}.$$ Here is my understanding and I would like some validation or corroboration: If our function $f$ is assumed to be independent of coordinates (and so it should be in a real life application such as in physics), then the derivatives of $f$ in different basis relate to each other. We know that $x=r\cos\theta$ and $y= r\sin\theta$ and so in an abuse of notation we may write $$f(x,y) = f(r\cos\theta,r\sin\theta) :=g(r,\theta)$$ and rename the $g$ to $f$ in an abuse of notation because we are identifying them as the same output (but with a different basis representing their domains). If the maps between the coordinates are smooth enough (and in this case away from 0), we may use the chain rule to compute $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$$ By plugging in for $r_x$ and $\theta_x$ and "erasing" the $f$ from both sides, we obtain the "change of variables" for the differential operator. Now because we know by definition of applying the operator, $$(\frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta})f = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x},$$ does this serve as sufficient justification for this notation? Further why may we them use such methods algebraically such $$\frac{\partial^2}{\partial x^2}=(\frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta})(\frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta})$$ and expanding keeping in mind left and right multiplication (composition!) may not be commutative?

share|cite|improve this question
up vote 1 down vote accepted

Your basic idea here is correct. Here's a way to go about it more formally.

Let $p = xe_1 + y e_2$ be a point, and let $p' = k(p)$ be a general, potentially nonlinear transformation map. For example, $p' = re_1 + \theta e_2$. This picture is somewhat different from what you'd expect for a coordinate system transformation; it's more like we're actively deforming the cartesian grid so that $r$ lies along one axis and $\theta$ along the other, but the mathematics is essentially the same.

Now then, let $F(p) = F'(p')$ be some scalar field. Note that $F(p) = (F' \circ k)(p)$ by definition. Let's look at some derivatives. For some vector $a$, we have

$$(a \cdot \nabla) F = (a \cdot \nabla)(F' \circ k) = [a \cdot \nabla k] \cdot \nabla' F'$$

The quantity $a \cdot \nabla k$ defines a linear operator called the Jacobian, which we can denote as $\underline k_p(a)$ and with transpose $\overline k_p(a)$. In particular, note that this implies

$$a \cdot \nabla F = a \cdot \overline k_p(\nabla') F'$$

Or more generally,

$$\nabla = \overline k_p(\nabla')$$

Now then, in polar coordinates, $\underline k_p(a)$ is given by

$$\begin{align*} \underline k_p(e_1) &= \frac{\partial p'}{\partial x} = e_1 \frac{\partial r}{\partial x} + e_2 \frac{\partial \theta}{\partial x} \\\underline k_p(e_2) &= \frac{\partial p'}{\partial y} = e_1 \frac{\partial r}{\partial y} + e_2 \frac{\partial \theta}{\partial y} \end{align*}$$

We can find $\partial/\partial x = e_1 \cdot \nabla = \underline k_p(e_1) \cdot \nabla'$. Note that $\nabla' = e_1 \partial/\partial r + e_2 \partial/\partial \theta$, so we get

$$\begin{align*} e_1 \cdot \nabla F &= \underline k_p(e_1) \cdot \nabla' F' \\&= \left( e_1 \frac{\partial r}{\partial x} + e_2 \frac{\partial \theta}{\partial x} \right) \cdot \left(e_1 \frac{\partial F'}{\partial r} + e_2 \frac{\partial F'}{\partial \theta} \right) \\ &= \frac{\partial r}{\partial x} \frac{\partial F'}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial F' }{\partial \theta} \end{align*}$$

As required.

Now, what happens when you want to do a second derivative? Well, the $\partial r/\partial x$ factor will naturally be a function of $x,y$ and not $r,\theta$, so you will have to invert $k$ to express the whole function $\partial F'/\partial r$ purely in terms of $r, \theta$. Once you do that (and do the same with the $\partial \theta/\partial x$ factor), you can take a second derivative directly.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.