OK, I have to show the following:
$$ \nabla \times \left( \frac{1}{r^2} \hat r \right) = 0$$
This should be pretty easy, but I wanted to be sure I was doing this correctly.
I set up the matrix:
$$
\begin{bmatrix}
\hat r & \hat \theta & \hat \phi \\
\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\
\frac{1}{r^2} & 0 & 0 \\
\end{bmatrix}
=\left(\frac{\partial}{\partial \theta}(0)-\frac{\partial}{\partial \phi}(0)\right)\hat r-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \phi}(\frac{1}{r^2})\right)\hat \theta-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \theta}(\frac{1}{r^2})\right)\hat \phi$$
which leaves me with 0 because $\frac{\partial}{\partial \theta}(\frac{1}{r^2})$ and $\frac{\partial}{\partial \phi}(\frac{1}{r^2})$ are both zero.
This is correct, yes? I know this is ridiculously simple a problem but I want to make sure I did not forget everything I learned last semester. (Also, I was curious if there is a more rigorous proof, tho this is for a phys and not a math class).
Edit: BTW this is in spherical (I think — the assignment uses $\hat r$ so I am going with that).
Best Answer
Any vector field that can be expressed in the form $f(r)\mathbf{\hat r}$ must necessarily have zero curl (where the function is smooth, at least).
This can be seen by noting that, if you have a scalar field $g(r)$, and you take its gradient, you get $g'(r)\mathbf{\hat r}$. As such, with $f(r)=g'(r)$, you get the vector field. Now, you have $$\nabla \times f(r)\mathbf{\hat r} = \nabla \times \nabla g(r) = 0$$
In particular, for $f(r)=\frac1{r^2}$, you have $g(r)=-\frac1r$.
Note that this only applies if the function is independent of $\theta$ and $\phi$.