I know that for any conservative vector field the following is true
$$\oint_C \nabla f \ \mathrm \cdot \ d\vec r = 0$$
However, my question is: if the integral on a closed loop of a vector field evaluates to $0$, then must it be conservative?
Or algebraically put:
Suppose that $\vec F$ is a vector field (we don't know if it's conservative) and $C:\vec r(t)$ is a closed path.
Are the following statements equivalent?
$$\vec F=\nabla f\quad \Longleftrightarrow\quad \oint_C \vec F \ \mathrm \cdot \ d\vec r = 0$$
or should it be $$\vec F=\nabla f\quad \implies\quad \oint_C \vec F \ \mathrm \cdot \ d\vec r = 0$$ instead?
I would like to add the following comment to my question:
For any vector field not of the form $\vec F = \nabla f$, is it the case that $$\oint_C \vec F \ \mathrm \cdot \ d\vec r \neq 0\quad \forall C$$ Or it possible that a vector field of that same form has $$\oint_C \vec F \ \mathrm \cdot \ d\vec r = 0$$ over some $C$.
Best Answer
The statements are indeed equivalent as long as the domain $U \subset \Bbb R^n$ of $\vec F$ is path connected (but not necessarily simply connected!). As the proof here explains: if $\vec F: U \to \Bbb R^n$ satisfies $\oint_C \vec F\cdot d \vec r = 0$ for all closed loops in the domain, then an associated potential function $f:U \to \Bbb R$ can be constructed as follows.
Select a fixed element $\vec a \in U$. For every $\vec x \in U$, there exists a path $\gamma[\vec a, \vec x]$ connecting $\vec a$ and $\vec x$. Define $$ f(\vec x) = \int_{\gamma[\vec a, \vec x]} \vec F(\vec r) \cdot d \vec r. $$ Because $\vec F$ is conservative (i.e. because of the integral condition), this function is well defined (i.e. we get the same result no matter which path we choose). With this $f$, we indeed find that $\nabla f = \vec F$.