Justification behind changing coordinates of a differential operator

Question

On many websites focused on physics, (say http://skisickness.com/2009/11/20/ ) 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?

Answer 1

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.