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 $C$.

  • $\begingroup$ Some terminology: if $\int_C \vec F \cdot d\vec r = 0$ over every loop, then $\vec F$ is called a "closed" vector field (or "differential form"). If $\vec F = \nabla f$ for some scalar field $f$, then $\vec F$ is called an "exact" form. $\endgroup$ Jan 29, 2021 at 20:53
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    $\begingroup$ The statement is true if the integral is zero for all closed loop. $\endgroup$
    – DiegoMath
    Jan 29, 2021 at 20:53
  • $\begingroup$ @user My first comment is actually incorrect: a "closed" form in this context is one form which $\nabla \times \vec F = 0$ (i.e. an "irrotational" field). It is indeed true that if $\oint_C \vec F \cdot d\vec r = 0$ for all loops within the domain of $\vec F$, then $\vec F$ is the gradient of some scalar function. So yes: the two statements are equivalent $\endgroup$ Jan 29, 2021 at 21:09
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    $\begingroup$ @user256872: Of course the line integral around some particular curve $C$ can be zero even if the vector field is not conservative (just take a conservative vector field and change it, away from the curve $C$, so that it's not conservative anymore – that doesn't change the integral over $C$). $\endgroup$ Jan 29, 2021 at 21:28
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    $\begingroup$ @user256872 Yes, at least we can't have that for all loops $C$. Interestingly, the condition $\nabla \times \vec F = 0$ ensures that $\oint_C \vec F = 0$ only when the domain of $\vec F$ is simply connected. For more on that, see this article. For the example listed, $\oint_C \vec F \cdot d\vec r$ is zero whenever $C$ does not "wrap around" the $z$-axis. $\endgroup$ Jan 29, 2021 at 21:28

1 Answer 1


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 $\nabla f = \vec F$.


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