If $f$ is a scalar field on $\mathbb{R}^2$, how do I interpret the quantity $x\frac{\partial f}{\partial y} - y\frac{\partial f}{\partial x}$? Suppose $f$ is a scalar field on $\mathbb{R}^2$ and can be written as $\phi(x^2 + y^2)$ for some differentiable function $\phi:\mathbb{R} \to \mathbb{R}$.
I am trying to show that $$x\frac{\partial f}{\partial y} - y\frac{\partial f}{\partial x} = 0.$$
It’s easy to show this directly computing the partial derivatives of $f$: the left side becomes $2xy - 2xy$, but I don’t have any intuition for what the quantity above means. Is it related to the Curl of a vector field?
 A: The left hand side is the generator of rotation of $f$ around the $z$ axis.
Define $R_\theta f : \mathbb{R}^2 \to \mathbb{R}$ by
$$R_\theta f(x,y) = f(x\cos\theta-y\sin\theta, y\cos\theta+x\sin\theta).$$
This corresponds to a rotation of the field $f$ by an angle $\theta$ in the $xy$-plane (in 3 dimensions: around the $z$ axis).
Then,
$$
\frac{d}{d\theta}R_\theta f(x,y) = (-x\sin\theta-y\cos\theta)\frac{\partial f}{\partial x} + (-y\sin\theta+x\cos\theta)\frac{\partial f}{\partial y}
$$
so
$$
\left. \frac{d}{d\theta}R_\theta f(x,y) \right|_{\theta=0}
= -y\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial y}.
$$
In quantum physics, the operator $-y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}$ (modulo some factors) corresponds to the $z$ component of orbital angular momentum.
A: The equation $x\,\partial_yf - y\,\partial_x f = 0$ can be written as $$\nabla f(x,y) \cdot (-y,x) = 0,$$ where $\cdot$ is the dot product. The left hand side is the directional derivative in the direction orthogonal to the position vector $(x,y)$, and it's vanishing. This tells you that $f$ changes only in the direction of position vector. This implies that $f$ is constant on circles. We can actually see this more directly from $f(x,y) = \phi(x^2+y^2)$ since $f$ depends only on the length of $(x,y)$, so $f$ gives the same value for all the points on the same circle.
Given a non-zero $(x,y)$, we can also express the starting equation as: there is $\lambda\in\mathbb R$ such that $$\nabla f(x,y) = \lambda\, (x,y).$$
This follows from the fact that both $\nabla f(x,y)$ and $(x,y)$ are orthogonal to $(-y,x)$, and $\mathbb R^2$ is $2$-dimensional. This $\lambda$ is easy to determine from $f(x,y) = \phi(x^2+y^2)$. Let $r = \|(x,y)\|$ be the length of $(x,y)$. Then $\lambda = \phi'(r^2)$. Furthermore, the length of $\nabla f(x,y)$ depends only on the length of $(x,y)$: $$\| \nabla f(x,y) \| = r|\phi'(r^2)|,$$ Thus, the vector field $\nabla f$ is rotationally symmetric.
Hopefully, all of this gives you some sense of what kind of field $f$ is.
A: This is the curl (or the 2D equivalent) of the vector function $(-xf, -yf)$.  Don't know if that answers the "interpretation" you need or not.
A: There's a nice idea from differential geometry, which is to think of partial derivaties $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ as vector fields.
How so? Think of directional derivatives from calculus:
$$\left.\frac{\partial f}{\partial x}\right|_p=\lim_{t\to 0}\frac{f(p+te_1)-f(p)}{t}$$
$$\left.\frac{\partial f}{\partial y}\right|_p=\lim_{t\to 0}\frac{f(p+te_2)-f(p)}{t}$$
The derivations $\left.\frac{\partial}{\partial x}\right|_p,\left.\frac{\partial}{\partial y}\right|_p$ at $p$ are naturally associated with $e_1,e_2$ (canonical basis of $\Bbb{R}^2$).
Since each point $p$ gives you a vector, $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ give you vector fields.
So the idea is: if $X$ is a vector field, $X|_p$ is a vector and $X|_p(f)$ tells you how $f$ changes at $p$ in the direction $X|_p$.
Another example of a vector field is $V:=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$. If $p:=a_1e_1+a_2e_2$, notice that $V|_p$ is just $p$ itself:
$$V|_p=a_1\left.\frac{\partial}{\partial x}\right|_p+a_2\left.\frac{\partial}{\partial y}\right|_p= a_1e_1+a_2e_2=p$$
In your case the vector field in question is $W:=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}$. Notice that $W$ is perpendicular to $V$. Indeed,
$$\langle V|_p,W|_p\rangle=\langle a_1e_1+a_2e_2,-a_2e_1+a_1e_2\rangle=0.$$
($-a_2e_1+a_1e_2$ is the rotation of $a_1e_1+a_2e_2$ in $90º$ anticlockwise)
Now we understand $-y\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y}=0$ as $W|_p(f)=0$. This means that $f$ does not change when you move perpendicularly to your position vector. In other words, $f$ does not change when you keep the same distance from the origin as you move. In other words, $f$ does not change when you move in a circle. By the way: these are different ways of saying that $f$ is a function of $x^2+y^2$.
In physics, this is the idea of a conservative force: if you want to know the potential difference gained in a trajectory, you only need to know the starting point $A$ and the final point $B$, no matter what happened between $A$ and $B$.
(if you think about gravity, you get $f(x,y)=\frac{GMm}{d^2}=\frac{GMm}{x^2+y^2}$)
