Diagnosing Nonconservative Fields with Loops

Why Closed Loops Are Useful

A vector field is conservative only if its line integrals are consistent with a potential: the integral between two fixed points cannot depend on the route taken. A closed curve returns to its starting point, so a conservative field must have zero line integral around it. This makes loops a practical diagnostic: a single nonzero circulation disproves conservativeness.

Curl provides complementary evidence. For a planar field, nonzero curl at a point rules out a potential on any open region containing that point. But curl alone does not detect every obstruction: a field can have zero curl throughout its domain and still have nonzero circulation around a loop. The calculations below show how to use both tests without treating either one as more conclusive than it is.

Circulation Around a Closed Curve

Let \(\mathbf{F}=(P,Q)\) be a planar vector field, and let a piecewise-smooth curve \(C\) be parameterized by \(\mathbf{r}(t)=(x(t),y(t))\), for \(a\le t\le b\). Its line integral is

\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \left[ P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t) \right]\,dt.

If \(C\) is closed, then \(\mathbf{r}(a)=\mathbf{r}(b)\). The line integral around \(C\), with the chosen direction of travel, is called its circulation. Reversing the direction changes the sign of the circulation.

Definition: The circulation of \(\mathbf{F}\) around a piecewise-smooth closed curve \(C\) is \(\displaystyle\oint_C\mathbf{F}\cdot d\mathbf{r}\). In the plane, this is \(\displaystyle\oint_C P\,dx+Q\,dy\). The curve must lie in the domain of the field.

For a continuously differentiable planar field, the scalar curl is \(Q_x-P_y\). In three dimensions, curl is the vector \(\nabla\times\mathbf{F}\). In either setting, if \(\mathbf{F}\) is the gradient of a potential on an open domain, its curl is zero there. For the planar case, this follows from equality of the mixed partial derivatives of a twice continuously differentiable potential.

Why Conservative Fields Have Zero Circulation

Suppose \(\mathbf{F}=\nabla\phi\) on a domain containing a piecewise-smooth closed curve \(C\). Along each smooth segment, the chain rule gives \(\nabla\phi(\mathbf{r}(t))\cdot\mathbf{r}'(t)=\frac{d}{dt}\phi(\mathbf{r}(t))\). Integrating each segment and adding them makes the intermediate endpoint values cancel. The remaining endpoint values are equal because the curve is closed.

\oint_C \nabla\phi\cdot d\mathbf{r} = \phi(\mathbf{r}(b))-\phi(\mathbf{r}(a)) = 0.

This gives a useful one-way test: if even one valid closed curve has nonzero circulation, the field cannot be conservative on a domain containing that curve. The converse is not established by one zero result. A field must have zero circulation around every closed curve in the domain for that condition to support path independence.

Diagnostic principle: Nonzero circulation on one closed curve proves nonconservativeness. Zero circulation on one curve proves only that the integral happened to vanish on that curve.

Worked Examples: Direct Loop Tests

Worked Example: A Circle with Nonzero Curl

Consider \(\mathbf{F}(x,y)=(-y,x)\) and the unit circle traversed counterclockwise. Use \(\mathbf{r}(t)=(\cos t,\sin t)\), for \(0\le t\le2\pi\). Then \(\mathbf{r}'(t)=(-\sin t,\cos t)\), and the field along the curve is \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t)\). Therefore,

\begin{aligned} \oint_C\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}(-\sin t,\cos t)\cdot(-\sin t,\cos t)\,dt\\ &=\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt\\ &=\int_0^{2\pi}1\,dt\\ &=2\pi. \end{aligned}

For comparison, \(P=-y\) and \(Q=x\), so \(Q_x=1\) and \(P_y=-1\). The curl is \(Q_x-P_y=1-(-1)=2\), which is nonzero everywhere. The direct loop calculation gives nonzero circulation, so the field is not conservative on any open domain containing this circle and the points where the field is defined.

Check: The parameterization returns to its starting point since \(\mathbf{r}(0)=(1,0)=\mathbf{r}(2\pi)\). The integrand is identically \(1\), whose integral over an interval of length \(2\pi\) is \(2\pi\), as calculated.

Worked Example: A Curl-Free Field Detected by a Loop

On the punctured plane \(x^2+y^2>0\), consider

\mathbf{F}(x,y)= \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right).

First check the curl. Writing \(r^2=x^2+y^2\), differentiation gives

Q_x=\frac{y^2-x^2}{(x^2+y^2)^2}, \qquad P_y=\frac{y^2-x^2}{(x^2+y^2)^2}, \qquad Q_x-P_y=0.

Now test the unit circle counterclockwise, again using \(\mathbf{r}(t)=(\cos t,\sin t)\) for \(0\le t\le2\pi\). On this circle, \(x^2+y^2=1\), so \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t)\), while \(\mathbf{r}'(t)=(-\sin t,\cos t)\). Hence

\begin{aligned} \oint_C\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi} \left[(-\sin t)(-\sin t)+(\cos t)(\cos t)\right]\,dt\\ &=\int_0^{2\pi}1\,dt\\ &=2\pi. \end{aligned}

The loop lies entirely in the field’s domain, and its circulation is nonzero. Thus the field is not conservative on the punctured plane, despite having zero curl there. The missing origin matters: the curl test is a local condition, while this loop detects a global obstruction in the domain.

Check: The denominator on the unit circle is \(1\), so the field is defined at every point of the loop. The dot product is \(1\), giving \(2\pi\) over the full parameter interval.

Worked Example: A Rectangle with Nonzero Circulation

Consider \(\mathbf{F}(x,y)=(0,x)\) and the unit square traversed counterclockwise, starting at \((0,0)\). Compute the integral on each of its four edges.

Bottom edge: \(y=0\), \(x\) goes from \(0\) to \(1\), and \(dy=0\). Since \(P=0\), the contribution \(\int P\,dx+Q\,dy\) is \(0\).
Right edge: \(x=1\), \(y\) goes from \(0\) to \(1\). Here \(Q=x=1\), so the contribution is \(\int_0^1 1\,dy=1\).
Top edge: \(y=1\), \(x\) goes from \(1\) to \(0\), and \(dy=0\). The contribution is \(0\) because \(P=0\).
Left edge: \(x=0\), \(y\) goes from \(1\) to \(0\). Here \(Q=x=0\), so the contribution is \(\int_1^0 0\,dy=0\).

Adding the four contributions gives

\oint_C\mathbf{F}\cdot d\mathbf{r}=0+1+0+0=1.

Here \(Q_x=1\) and \(P_y=0\), so the curl is \(1\) throughout the plane. The nonzero circulation directly proves that the field is not conservative.

Check: The four edge contributions sum to \(1\), and the curve returns to \((0,0)\). Each edge was traversed in the counterclockwise direction, including the top edge from right to left.

Choosing a Useful Loop

A useful diagnostic curve must be closed, piecewise smooth, and contained in the field’s domain. Choose a curve that is easy to parameterize or divide into simple segments. Circles and rectangles are often convenient. If the field has a singularity, avoid curves that pass through it; a loop around the excluded point can still be informative if every point on the loop is in the domain.

Use the calculation that is appropriate to the question. Direct parameterization gives the circulation along the chosen curve. Curl tells whether a potential is ruled out locally: for a continuously differentiable field, nonzero curl at a point contradicts the necessary condition for a potential on an open neighborhood of that point. A zero curl value, by itself, does not prove that a potential exists on the entire domain. In particular, do not use a curl-based region formula to replace a direct loop calculation here; that relationship requires a separate theorem.

Three conclusions to keep separate:

One nonzero loop integral is enough to rule out conservativeness on a domain containing the loop.
One zero loop integral does not establish conservativeness.
Zero curl is a necessary local condition for a potential, but it may not settle the question on a domain with a hole.

Common Mistake: Treating One Zero Loop as Proof

Consider \(\mathbf{F}(x,y)=(0,x^2)\) on the plane and the rectangle with corners \((-1,0)\), \((1,0)\), \((1,1)\), and \((-1,1)\), traversed counterclockwise. The bottom and top edges contribute \(0\) because \(P=0\) and \(dy=0\). On the right edge, \(x=1\), so \(Q=1\) and the contribution is \(\int_0^1 1\,dy=1\). On the left edge, \(x=-1\), so \(Q=1\), but the direction is from \(y=1\) to \(y=0\); its contribution is \(\int_1^0 1\,dy=-1\). Thus the total circulation is \(0+1+0-1=0\).

It would be incorrect to conclude that the field is conservative. Its curl is \(Q_x-P_y=2x\), which is nonzero, for example, at \((1,0)\). Therefore no potential exists on an open region containing that point. The zero circulation on this particular rectangle occurred because the contributions from its two vertical edges canceled. A different loop, or the curl test, can expose what this loop misses.

Check Your Understanding

Use the definitions and diagnostic principles from this tutorial to answer the following.

What does a nonzero circulation around one closed curve prove about a field on a domain containing that curve?
Why does zero circulation around one closed curve fail to prove that a field is conservative?
For \(\mathbf{F}=(P,Q)\), what is the planar curl, and what does a nonzero value at a point imply about a potential near that point?
Why can a field with zero curl throughout its domain still have nonzero circulation around a loop?
When computing circulation around a rectangle, how does reversing the direction on an edge affect its contribution?