A Curl-Free Field on a Punctured Plane

Introduction

A vector field can have zero curl at every point in its domain without being conservative. The qualification “in its domain” matters: a hole in the domain can prevent the conclusions that hold for curl-free fields on simply connected domains.

We will examine one field on the plane with the origin removed. Its curl vanishes everywhere it is defined, but its line integral around a counterclockwise circle is $2\pi$. This gives a concrete reason that a curl test must be used together with information about the domain.

The Field and Its Domain

Definition: The punctured plane is the set $\mathbb{R}^2\setminus\{(0,0)\}$, consisting of all points in the plane except the origin. On this domain, consider the vector field $$ \mathbf{F}(x,y)=\left\langle \frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right\rangle. $$

The denominator $x^2+y^2$ is zero exactly when $x=y=0$, so the formula does not define a vector there. At every other point the denominator is positive, and both components are defined. Write $P(x,y)=-y/(x^2+y^2)$ and $Q(x,y)=x/(x^2+y^2)$, so that $\mathbf{F}=\langle P,Q\rangle$.

For a two-dimensional field $\langle P,Q\rangle$ with continuously differentiable components, its scalar curl is defined by $\operatorname{curl}\mathbf{F}=\partial Q/\partial x-\partial P/\partial y$. We use that definition at points where the field and the required partial derivatives are defined.

The missing origin is not a minor technical detail. The domain is not the entire plane: a circle surrounding the origin cannot be filled in by points of the punctured plane without crossing the missing point. We will see that this hole allows the field to have zero curl locally while still having nonzero circulation around it.

Computing the Curl

Worked Example: Verify That the Curl Vanishes

Compute $\partial Q/\partial x$ and $\partial P/\partial y$ at an arbitrary point $(x,y)\ne(0,0)$, then compare them.

For the first component needed in the curl, use the quotient rule, differentiating with respect to $x$ and holding $y$ fixed:

\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right) = \frac{(x^2+y^2)-2x^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.

For the other component, differentiate with respect to $y$, holding $x$ fixed:

\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right) = \frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.

Both partial derivatives have the same value at every point in the punctured plane. Therefore,

\operatorname{curl}\mathbf{F}(x,y) = \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2} - \frac{y^2-x^2}{(x^2+y^2)^2} =0.

Check: The calculation applies only where $x^2+y^2\ne0$. It establishes zero curl at every point in the field’s domain; it does not assign a curl at the excluded origin.

The equality of the partial derivatives is a local calculation: it compares how the field’s components change near each point where they are defined. It does not by itself determine the integral along every closed curve in the whole domain.

Line Integrals Around Circles

For a parametrized curve $\mathbf{r}(t)=\langle x(t),y(t)\rangle$, the work integral of a vector field along the curve is $\int \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt$, with the parameter running from the initial point to the final point. The order of the parameter values records the curve’s orientation.

Worked Example: Integral Around a Counterclockwise Circle

Let a circle of radius $R>0$, centered at the origin, be traversed counterclockwise once. Use the parametrization $\mathbf{r}(t)=\langle R\cos t,R\sin t\rangle$ for $0\le t\le2\pi$. Since $R>0$, this entire circle stays in the punctured plane.

The derivative of the parametrization is

\mathbf{r}'(t)=\langle -R\sin t,R\cos t\rangle.

At the point $\mathbf{r}(t)$, the denominator of the field is $R^2\cos^2t+R^2\sin^2t=R^2$. Thus

\mathbf{F}(\mathbf{r}(t)) = \left\langle \frac{-R\sin t}{R^2}, \frac{R\cos t}{R^2} \right\rangle = \left\langle -\frac{\sin t}{R}, \frac{\cos t}{R} \right\rangle.

The dot product with the tangent vector is

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = \left(-\frac{\sin t}{R}\right)(-R\sin t) + \left(\frac{\cos t}{R}\right)(R\cos t) = \sin^2t+\cos^2t =1.

Therefore the line integral is

\int_0^{2\pi}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{2\pi}1\,dt = 2\pi.

Check: The result is independent of the positive radius $R$, because the radius factors cancel in the dot product. The calculation also exhibits the direction: the parametrization runs counterclockwise as $t$ increases from $0$ to $2\pi$.

In fact, the same computation shows that every counterclockwise circle centered at the origin, regardless of its positive radius, has integral $2\pi$. This is not a consequence of a general theorem about curl; it follows directly from the parametrization and the field’s formula.

Orientation Changes the Sign

Reversing a curve reverses the sign of its line integral. Here that can be checked directly, without relying only on a general orientation rule. Use the same circle parametrization, but let $t$ run from $0$ down to $-2\pi$. This describes one clockwise traversal, with oriented interval $[0,-2\pi]$.

Worked Example: Traverse the Circle Clockwise

For the circle parametrization already computed, the field dotted with the velocity is $1$ for every parameter value. Hence the clockwise integral is

\int_0^{-2\pi}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{-2\pi}1\,dt = -2\pi.

Check: Reversing the orientation changes the counterclockwise value $2\pi$ to $-2\pi$. The sign agrees with the fact that the oriented parameter now decreases from $0$ to $-2\pi$; it is not an interval described by the inequalities $0\le t\le-2\pi$.

Two Paths with the Same Endpoints

A field is conservative if it has a potential function on its domain, meaning a differentiable scalar function $f$ such that $\nabla f=\mathbf{F}$ there. The fundamental theorem for line integrals implies that a conservative field has the same line integral along any two suitable paths with the same starting and ending points. The next calculation demonstrates that this field does not have that property.

Worked Example: Compare the Upper and Lower Semicircles

Consider paths from $(1,0)$ to $(-1,0)$. Both semicircles below lie entirely in the punctured plane. Use $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, for which the earlier circle calculation gives $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=1$.

For the upper semicircle, $t$ increases from $0$ to $\pi$. Its endpoints are $\mathbf{r}(0)=(1,0)$ and $\mathbf{r}(\pi)=(-1,0)$, and its integral is

\int_0^\pi \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^\pi 1\,dt = \pi.

For the lower semicircle from the same starting point to the same ending point, $t$ runs from $0$ down to $-\pi$, on the oriented interval $[0,-\pi]$. Its endpoints are $\mathbf{r}(0)=(1,0)$ and $\mathbf{r}(-\pi)=(-1,0)$, and its integral is

\int_0^{-\pi} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{-\pi}1\,dt = -\pi.

The two paths have the same endpoints but their integrals differ by $\pi-(-\pi)=2\pi$. Check: Following the upper semicircle and then returning along the lower semicircle in reverse gives a counterclockwise full circle. The resulting closed-curve integral is $\pi-(-\pi)=2\pi$, in agreement with the circle computation.

Why Zero Curl Does Not Guarantee Conservativeness Here

On a simply connected domain, the earlier curl-free criterion says that a continuously differentiable planar field with zero curl is conservative. The hypothesis about the domain is essential. The punctured plane has a hole, and the circle around that hole is a closed curve that remains in the domain. The nonzero integral around this circle prevents the field from being conservative on the entire punctured plane.

Theorem: If a field has a potential on its domain, then its line integral around every closed piecewise-smooth curve in that domain is zero. Therefore, one closed curve with a nonzero integral is enough to show that the field has no potential on that domain.

For this field, the counterclockwise circle integral is $2\pi\ne0$. Consequently, no single-valued differentiable potential on the whole punctured plane can have this field as its gradient. This conclusion is compatible with the zero-curl calculation: the two facts describe different properties, and the hole in the domain is precisely why the usual simply connected criterion cannot be applied.

The central distinction is between local and global information. Curl is computed from derivatives near each point. A closed-curve integral can detect how the field behaves around a hole, even though every point along the curve has a neighborhood on which the curl calculation is valid.

A Common Mistake

A tempting but invalid argument is: “The curl is zero everywhere in the domain, so the field must be conservative, and every closed-curve integral must be zero.” The first calculation is correct, but the conclusion omits a necessary domain condition. For the counterclockwise unit circle, the actual calculation is

\int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}1\,dt = 2\pi,

not zero. The correction is to check the hypotheses of the curl-free criterion before using it. Here the origin is excluded, the domain is not simply connected, and the explicitly computed nonzero loop integral shows that the field is not conservative on the whole domain.

Another common source of error is to ignore orientation. The same circle has integral $2\pi$ counterclockwise and $-2\pi$ clockwise. A parametrization must have its parameter limits in the direction of travel; when the parameter decreases, write an oriented integral such as $\int_0^{-2\pi}$, rather than an impossible pair of inequalities.

Why This Example Matters

This field provides a useful test of what a curl calculation can and cannot establish. The derivatives prove that curl vanishes at every point where the field exists. They do not erase the missing origin or force the circulation around it to vanish. Integrating around a closed curve adds global information that the pointwise derivative test alone does not capture.

When applying a curl-free test, keep these questions together:

Where is the field defined, and where are its derivatives defined?
Does the theorem being used require a simply connected domain?
Can a closed curve in the domain surround a missing point?
What does a direct integral along a representative closed curve show?

For the field considered here, the answers are decisive: the origin is missing, the circle around it is allowed, and its counterclockwise integral is $2\pi$. Thus zero curl is true, but global path independence is not.

Check Your Understanding

Answer these questions using the field $\mathbf{F}(x,y)=\langle-y/(x^2+y^2),x/(x^2+y^2)\rangle$.

Why is the origin excluded from the domain, and what is the curl at points of the punctured plane?
For a circle of radius $R>0$ traversed counterclockwise once, what are the field, velocity, and dot product along the parametrization?
What is the integral around a clockwise circle, and how should its oriented parameter interval be written?
What are the integrals along the upper and lower semicircles from $(1,0)$ to $(-1,0)$?
How does the nonzero circle integral show that the field cannot have a potential on the whole punctured plane?