Conservative Vector Fields

A conservative field is simply a field that's secretly a gradient: there's some function $f$ with $\vec F = \nabla f$. The name comes from physics, where these are exactly the forces that conserve energy: whatever work they take from you on the way out, they give back on the way home. The test $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ looks abstract, but it's just two partial derivatives compared: how the horizontal push changes as you move up, and how the vertical push changes as you move right. If they match, there's no twist, and the field is a gradient.

From the last lesson, gradient fields are wonderful: work is just $f(\text{end}) - f(\text{start})$. So what can we do when someone hands us a field? We need a quick test, and a way to find $f$.

Why the test works. If $\vec F = \nabla f$, then $P = f_x$ and $Q = f_y$, so $P_y = f_{xy}$ and $Q_x = f_{yx}$. By Clairaut's theorem from partial derivatives, mixed partials are equal, so $P_y = Q_x$. Every gradient field passes. The converse also holds, as long as the region has no holes; Green's theorem, the next lesson, shows why.

What the test is really measuring. The difference $Q_x - P_y$ measures how much the field twists around each point. Picture a tiny paddle wheel dropped in the flow: if the push on its top differs from the push on its bottom, it spins. A gradient field never spins a paddle wheel, so going around a loop gains nothing. There's an animation further down that adds a twist to a gradient field and shows the loop work appear.

Finding the potential:

1. Integrate $P$ with respect to $x$: $f = \int P\,dx + g(y)$. The "constant" of integration can depend on $y$, since anything with only $y$ in it disappears when you take $\partial/\partial x$.
2. Differentiate that in $y$ and set it equal to $Q$ to find $g'(y)$.
3. Integrate to get $g(y)$.

A common mistake. It's tempting to integrate $P$ in $x$, integrate $Q$ in $y$, and add the two results. That double-counts any term containing both variables: for $\langle 2xy, x^2\rangle$ you'd get $2x^2y$ instead of $x^2y$. Find $g(y)$ the careful way, and check your $f$ by taking its gradient.

In 3D, the test is $\operatorname{curl}\vec F = \vec 0$, three cross-partial checks at once; it's coming up in the next unit.

Physicists check whether a force is conservative before talking about potential energy; non-conservative forces like friction turn mechanical energy into heat.