5.3 Conservative Fields

A vector field assigns an arrow $\mathbf{F}(x, y) = (P, Q)$ to every point — a wind map, a force map. The natural integral of a field is work: along a curve $C$ traced by $\mathbf{r}(t)$ ,

$\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\, dt,$

which totals the field's push along your direction of travel. The deepest question you can ask about a field is whether it is somebody's gradient: $\mathbf{F} = \nabla f$ for a potential $f$ . Such conservative fields inherit a fundamental theorem — work equals potential difference, $f(\text{end}) - f(\text{start})$ — so work is path-independent, and around any closed loop it is exactly zero. One loop is therefore a courtroom: the figure puts two fields on trial.

A flat carpet of arrows seen from almost straight overhead, crossed by a pink circle of radius 1.5 centered at the origin. A slider k switches the carpet between two fields. With k at 0 the arrows all point radially outward, brightening from purple to yellow away from the center, so along the pink circle they cross it at right angles everywhere; with k at 1 the arrows instead run along the circle rather than across it. Comparing how each field meets the loop is the scene's whole game: crossing everywhere means zero circulation, running along it means circulation that never cancels. — An overhead view of a flat carpet of arrows crossed by a pink circle of radius $1.5$ centered at the origin. Slider $k$ chooses the field. At $k = 0$ it is $\mathbf{F} = (2x, 2y)$ , pointing radially outward everywhere, its arrows brightening from purple to yellow with distance from the center. At $k = 1$ it is $\mathbf{F} = (-y, x)$ , whose arrows run *along* the circle instead of across it.

Explore

Follow the pink circle with your eye. The radial field's arrows cross it at a right angle at every single point — the field has no component along the walk anywhere. Predict the circulation around the loop before computing it.
Move slider $k$ to $1$ to swap in the rotational field. Now the arrows run along the circle. At $(1.5, 0)$ the field is $(0, 1.5)$ , pointing toward $+y$ , so it turns counterclockwise in the $xy$ -plane: at every point of a counterclockwise walk, the field is at your back. Can contributions like that cancel? Predict this circulation's sign.
Both fields have constant strength on the circle — the arrows along it share one color. Compute the two strengths at radius $1.5$ by hand. Then compare the circle's arrows at $k = 0$ and $k = 1$ : they wear the same color. The figure scales color and arrow length within whichever field is on screen, so the picture cannot compare the two fields' strengths — only your computation can.
The radial field is a gradient: find, by inspection, a function $f$ with $\nabla f = (2x, 2y)$ , and describe its graph. Walking any closed loop on that surface, why must you end at the height you started?
Try the same for $(-y, x)$ : what would $f_x = -y$ and $f_y = x$ demand of the mixed second derivatives $f_{yx}$ and $f_{xy}$ ? Clairaut's theorem says those must be equal for any twice-differentiable $f$ — conclude what the example below confirms by direct integration.

\left[ 2 \cdot x,\; 2 \cdot y \right]

The gradient of the radial potential ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Step 4 asks you to find a potential by inspection. This is the check: the gradient of $f = x^2 + y^2$ is exactly the radial field the figure draws, so that field is conservative and the loop integral must vanish.

\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}

The derivative matrix of the rotational field ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Step 5's argument, computed. If $(-y, x)$ were $\nabla f$ , this matrix would be $f$ 's Hessian — and Clairaut's theorem forces a Hessian to be symmetric. The computation returns the rows $(0, -1)$ and $(1, 0)$ . The off-diagonal entries are $-1$ and $1$ . They differ, so no such $f$ exists and the rotational field cannot be conservative, no potential-hunting required.

The verdict, computed both ways

For a conservative field the fundamental theorem for line integrals does all the work, and the geometric tell is the one in step 1: a gradient field crosses its own potential's level curves squarely (Chapter 4), so against a loop that happens to be a level curve it musters no push at all. The rotational field fails a cheaper test: comparing cross-partials, $\partial Q/\partial x - \partial P/\partial y = 1 - (-1) = 2 \neq 0$ , while every gradient field must score zero there. But the most honest verdict is a line integral, computed below — and its value is worth keeping: the same number returns in the next section wearing a theorem.

Example 1 Two fields, one loop, two verdicts

Compute the circulation of $\mathbf{F} = (2x, 2y)$ and of $\mathbf{F} = (-y, x)$ counterclockwise around the circle $\mathbf{r}(t) = (1.5\cos t,\ 1.5\sin t)$ , $t \in [0, 2\pi]$ .

Show solution

The velocity is $\mathbf{r}'(t) = (-1.5\sin t,\ 1.5\cos t)$ . For the radial field, $\mathbf{F}(\mathbf{r}(t)) = (3\cos t,\ 3\sin t)$ , so

$\mathbf{F}\cdot\mathbf{r}' = -4.5\sin t\cos t + 4.5\sin t\cos t = 0$

at every $t$ : the integrand is identically zero and the circulation is $0$ — perpendicularity, in numbers. Consistently, $\mathbf{F} = \nabla f$ for $f = x^2 + y^2$ (check: $f_x = 2x$ , $f_y = 2y$ ), and a closed loop on the bowl $f$ ends at its starting height. For the rotational field, $\mathbf{F}(\mathbf{r}(t)) = (-1.5\sin t,\ 1.5\cos t)$ , which is exactly $\mathbf{r}'(t)$ : the field rides the walk perfectly, and

$\mathbf{F}\cdot\mathbf{r}' = 2.25\sin^2 t + 2.25\cos^2 t = 2.25,$

a constant. The circulation is $2.25 \cdot 2\pi = 4.5\pi \neq 0$ — so $(-y, x)$ is not conservative, no potential exists, and the cross-partials test's nonzero score of $2$ was telling the truth. Note $2.25$ is the squared radius: this field's circulation grows with the loop, a hint the next section explains.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.