Why a Curl-Free Vector Field Is Not Always Conservative

A vector field is conservative when it is the gradient of a scalar function, and that single property has practical consequences: the field is independent of path, its circulation around every closed curve is zero, and line integrals in it reduce to a difference of potential values. The useful question is how to decide, from the components alone, whether a given field has the property.

What the property actually requires

For $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ to be conservative there must be one function $f$ with $\dfrac{\partial f}{\partial x}=P$ and $\dfrac{\partial f}{\partial y}=Q$ on the whole domain. Mixing the two equations gives a necessary condition, because

$\frac{\partial P}{\partial y}=\frac{\partial^{2}f}{\partial y\,\partial x}=\frac{\partial^{2}f}{\partial x\,\partial y}=\frac{\partial Q}{\partial x}.$

So a conservative field always has equal cross partials. The question is when the converse holds.

The test, and the part of it people skip

The converse is a theorem with a hypothesis about the domain: if $P$ and $Q$ have continuous first partial derivatives on an open, simply connected region and $P_{y}=Q_{x}$ throughout that region, then the field is conservative there. "Simply connected" means the region has no hole: any closed loop inside it can be shrunk to a point without leaving the region. A disk or the whole plane is simply connected; a plane with a point removed is not.

That hypothesis is not decoration. It is the difference between a test that works and a test that gives the wrong answer.

A counterexample that makes the point

Consider the field defined on the punctured plane, every point except the origin:

$\mathbf{F}(x,y)=\frac{-y\,\mathbf{i}+x\,\mathbf{j}}{x^{2}+y^{2}},\qquad (x,y)\neq(0,0).$

Writing $r^{2}=x^{2}+y^{2}$, the cross partials are

$\frac{\partial P}{\partial y}=-\frac{1}{r^{2}}+\frac{2y^{2}}{r^{4}},\qquad \frac{\partial Q}{\partial x}=\frac{1}{r^{2}}-\frac{2x^{2}}{r^{4}},$

and their difference is $\dfrac{2}{r^{2}}-\dfrac{2(x^{2}+y^{2})}{r^{4}}=0$. The cross partials are equal at every point where the field is defined, so anyone who applies the test without reading its hypothesis concludes that the field is conservative.

The circulation disagrees. Traversing the unit circle counterclockwise once, $\mathbf{F}$ at that radius is $-\sin t\,\mathbf{i}+\cos t\,\mathbf{j}$ and the velocity is $-\sin t\,\mathbf{i}+\cos t\,\mathbf{j}$, so the integrand is $\sin^{2}t+\cos^{2}t=1$ and the circulation is $\int_{0}^{2\pi}dt=2\pi$. A field whose circulation around a closed curve is nonzero is not conservative. The test failed because the domain is the plane with a hole, and the missing hypothesis — not the arithmetic — is what went wrong.

Intuitively, the would-be potential is the polar angle, which increases by $2\pi$ after one lap and therefore cannot be a single-valued function on the punctured plane.

Getting the verdict right in practice

• Name the domain before applying the test. Say "on $\mathbb{R}^{2}$" or "on the disk $x^{2}+y^{2}<4$", not just "the field is conservative". A verdict without a domain is incomplete, because the same component formula can be conservative on one region and not on another.
• Check whether any component is undefined somewhere. A denominator that vanishes removes points from the domain, and a punctured domain is the usual sign that the hole hypothesis matters.
• Remember the theorem is an equivalence on nice domains. On an open, simply connected region, equal cross partials and conservativity are the same statement, so a single failing point is enough to reject the field.
• If the field is conservative, the potential is determined only up to a constant, because adding a constant to $f$ does not change its gradient.

Why the distinction matters downstream

Path independence, the fundamental theorem for line integrals, and the vanishing of circulation all depend on conservativity rather than on equal cross partials. A field like the counterexample above is path dependent in a way that is easy to miss: two different routes between the same two points can differ by a multiple of $2\pi$, so a single numerical answer for "the" line integral does not exist.