Curl of a gradient is zero

Statement

Let $U \subseteq \mathbb{R}^3$ be open, and let $\phi \colon U \to \mathbb{R}$ be of class $C^2$. Then $$\nabla \times (\nabla \phi) = 0 \quad \text{on } U.$$

Remarks

Proof: the first component of the curl of $\nabla \phi$ is $\partial_y \partial_z \phi - \partial_z \partial_y \phi$, and this is zero by Schwarz's theorem. The other components are similar. So a conservative $C^1$ vector field is irrotational. This gives a quick test: if $\operatorname{curl} F \ne 0$ at some point, then $F$ has no scalar potential. In $\mathbb{R}^n$, the same argument shows $\partial_i F_j = \partial_j F_i$ for all $i, j$ when $F = \nabla \phi$ with $\phi$ of class $C^2$. The converse needs a condition on $U$. It holds on convex sets, but not on every open set.

Explanations

L1 human

A gradient field has no swirl. Going around a small loop in a landscape, you return to the same height. So the field $\nabla \phi$ has no circulation, and its curl is zero.