Conservative vector field

Definition

Let $U \subseteq \mathbb{R}^n$ be open, and let $F \colon U \to \mathbb{R}^n$ be a continuous vector field. The field $F$ is conservative on $U$ if there is a function $\phi \colon U \to \mathbb{R}$ of class C^1 with $F = \nabla \phi$ on $U$, that is, $F_i = \partial \phi / \partial x_i$ for $i = 1, \dots, n$.

Remarks

A function $\phi$ with $\nabla \phi = F$ is a scalar potential of $F$. Physics texts often use $-\phi$ instead. The field $F(x, y) = (2 x y, x^2)$ is conservative on $\mathbb{R}^2$, with $\phi(x, y) = x^2 y$. The field $F(x, y) = (-y, x)$ is not conservative on $\mathbb{R}^2$. Its line integral around the unit circle is $2\pi \ne 0$, which contradicts path independence. If $F$ is conservative and of class $C^1$, then its curl is zero. The converse holds on convex sets but fails in general.

Explanations

L1 human

A conservative field is the slope field of some landscape $\phi$. Going around a loop in a landscape brings you back to the same height. So the total work of the field around the loop is zero.