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A $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is conservative

Statement

Let $U \subseteq R^{3}$ be open and star-shaped and let $F : U \to R^{3}$ be $C^{1}$ with $curl F = 0$ on $U$ . Then a $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is exact, conservative and path-independent: there is a $C^{2}$ function $ϕ : U \to R$ with $F = \nabla ϕ$ , any two piecewise- $C^{1}$ paths in $U$ with the same endpoints give $F$ the same vector line integral, and

$\int_{γ} F \cdot d r = 0$

for every closed piecewise- $C^{1}$ path $γ$ in $U$ . Conversely, a field exact on such a set has vanishing curl, so on a star-shaped open subset of $R^{3}$ vanishing curl and exactness are equivalent.

Facts & Assumptions

Given: The star-shaped open set $U \subseteq R^{3}$ with a star centre $a \in U$ , and the $C^{1}$ field $F : U \to R^{3}$ with $curl F = 0$ on $U$ .

[F1]

A nonempty open set $U \subseteq R^{n}$ is star-shaped with respect to $a \in U$ when $a + t (x - a) \in U$ for every $x \in U$ and $0 \leq t \leq 1$ (Star-shaped open subsets of Euclidean space).

[F2]

For a continuous field $F$ on an open $U \subseteq R^{n}$ , a $C^{1}$ function $ϕ : U \to R$ is a potential when $F = \nabla ϕ$ ; $F$ is conservative when it has a potential, and path-independent when any two piecewise- $C^{1}$ paths in $U$ with the same initial and terminal points have equal vector line integrals (Piecewise-C1 path-connected domains, potential functions, conservative fields, and path independence).

[F3]

The curl of a $C^{1}$ field $F$ on an open $U \subseteq R^{3}$ is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[L1]

A $C^{1}$ field on an open subset of $R^{3}$ is closed if and only if its curl vanishes identically (A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes).

[L2]

Let $U \subseteq R^{n}$ be open and star-shaped and let $F : U \to R^{n}$ be $C^{1}$ . Then the five conditions that $F$ be closed, exact, conservative, path-independent, and give every closed piecewise- $C^{1}$ path in $U$ zero integral are equivalent (On a star-shaped open domain, closed, exact, conservative, path-independent, and zero-loop are equivalent).

Proof

technique · direct

1.1

The field $F$ is $C^{1}$ on the open set $U \subseteq R^{3}$ and its curl vanishes identically, so by the reverse direction of [L1] it is closed.

givenL1F3

1.2

By [F1] the set $U$ is nonempty, open and star-shaped with respect to its centre $a$ . With $n = 3$ these are exactly the hypotheses [L2] places on the domain, and $F$ is $C^{1}$ as [L2] requires of the field.

givenF1

2.1

By steps 1.1 and 1.2, [L2] applies and its first condition holds, so all five hold: $F$ is exact, hence there is a $C^{2}$ function $ϕ$ on $U$ with $F = \nabla ϕ$ ; $F$ is conservative, so it has a potential in the sense of [F2]; $F$ is path-independent; and every closed piecewise- $C^{1}$ path in $U$ gives $F$ integral zero.

step 1.1step 1.2L2F2

3.1

For the converse reading, suppose instead that $F$ is exact on $U$ . Then the first condition of [L2] holds by the same equivalence, so $F$ is closed, and the forward direction of [L1] makes $curl F$ vanish identically. Together with step 2.1 this gives the stated equivalence between vanishing curl and exactness on a star-shaped open subset of $R^{3}$ .

step 2.1L1L2F2∎

Remarks

The hypothesis on the domain is doing work. Star-shapedness is not a convenience: the companion examples page gives a $C^{1}$ field with vanishing curl on a connected open subset of $R^{3}$ that has no potential. What fails there is exactly [F1], since no point of the complement of a line is a star centre for it.
Why the potential is $C^{2}$ and not merely $C^{1}$ . Exactness in Exact and closed C1 vector fields asks for a $C^{2}$ potential, which is what makes all mixed second partial derivatives of $ϕ$ available and continuous; a conservative field in the sense of [F2] is only required to have a $C^{1}$ one. Step 2.1 supplies the stronger form because [L2] does.

Depends on

Used by

A curl-free C¹ field on the complement of a line that is not conservative Counterexample

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Corral, Vector Calculus, chapter 4 (LibreTexts) (standard reference, not scraped)
J. Feldman, A. Rechnitzer and E. Yeager, CLP-4 Vector Calculus (University of British Columbia), section 4.1 (standard reference, not scraped)

A C1C1 field with vanishing curl on a star-shaped open subset of R3R3 is conservative