6.3 Conservative Vector Fields

Big idea

The work integral of the previous section depends, in general, on the whole path. Two routes from one point to another can give two different answers, and there is no shortcut around computing each one. This section identifies the fields for which that dependence disappears. For them the integral is decided by the endpoints alone, and evaluating it costs one subtraction.

The fields with that property are exactly the gradient fields, and the reason is a single-variable theorem in disguise. Along a path, the composite $f(\mathbf{r}(t))$ is a function of one variable, and the chain rule says its derivative is $\nabla f(\mathbf{r}(t))\cdot \mathbf{r}'(t)$, which is precisely the integrand of the work integral for $\mathbf{F} = \nabla f$. So the work integral is the integral of a derivative, and the fundamental theorem of calculus finishes the job: the answer is the change in $f$ between the ends.

This is the first appearance of the pattern that organizes the rest of the chapter. An integral of a derivative over a region equals something evaluated on the boundary of that region. Here the region is a curve, its derivative operator is the gradient, and its boundary is the pair of endpoints, where a plus sign sits on the terminal point and a minus sign on the initial one. Green’s theorem, Stokes’ theorem and the divergence theorem all repeat this sentence with the region raised one dimension at a time.

The practical questions are then: given a field, how do you tell whether it is a gradient field, and if it is, how do you find the function? The test is a comparison of mixed partial derivatives, and the construction is an integration followed by a correction. The test carries a hypothesis about the shape of the domain, and that hypothesis is not decoration.

Decoder

If a vector field is the gradient of a scalar function on a region, then its line integral along any piecewise smooth curve inside that region equals the value of the function at the terminal point minus its value at the initial point.

Read the three clauses in order. “Is the gradient of a scalar function” is the hypothesis, and it is a strong one: most fields fail it. “Any piecewise smooth curve inside that region” is the payoff, since the shape of the curve has vanished from the conclusion. “Terminal minus initial” is the arithmetic, and it also explains the sign rule from the previous section: swapping the endpoints negates the difference.

Two consequences follow immediately and are used constantly. Integrals of a conservative field are path independent, because any two curves with the same endpoints produce the same difference. And the integral around any closed curve is zero, because the terminal point is the initial point and the difference is $f(A) - f(A)$. On an open connected region these two statements and conservativity are all equivalent, so any one of them can serve as the definition.

The classic mistake is to run the cross-partials test, see that it passes, and declare the field conservative without looking at the domain. The test detects a local obstruction only. If the domain has a hole in it, a field can pass the test everywhere and still have nonzero circulation around a loop enclosing the hole.

Definitions and results

Conservative field and potential. A vector field $\mathbf{F}$ on a region $D$ is conservative if there is a scalar function $f$ on $D$ with $\mathbf{F} = \nabla f$. Such an $f$ is a potential function for $\mathbf{F}$, and it is unique up to an additive constant on a connected region.

Fundamental theorem for line integrals. If $f$ has continuous partial derivatives on an open region containing a piecewise smooth curve $C$ running from $A$ to $B$, then $$ \int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A) $$

Path independence. $\int_C \mathbf{F}\cdot d\mathbf{r}$ is path independent on $D$ when its value depends only on the endpoints of $C$. On an open connected region, $\mathbf{F}$ is conservative if and only if it is path independent, if and only if $\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$ for every closed curve $C$ in $D$.

Simply connected regions. An open connected region is simply connected when every closed curve in it can be shrunk to a point without leaving the region. A disk and the whole plane qualify. A disk with its center removed, or an annulus, does not.

Cross-partials test in the plane. If $\mathbf{F} = \langle P, Q\rangle$ is conservative on an open region, then $\partial P/\partial y = \partial Q/\partial x$ there, because both equal the mixed second partial of a potential. Conversely, if $P$ and $Q$ have continuous partials on a simply connected open region and $\partial P/\partial y = \partial Q/\partial x$, then $\mathbf{F}$ is conservative on that region.

Cross-partials test in space. For $\mathbf{F} = \langle P, Q, R\rangle$ the conditions are $$ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \qquad \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}, \qquad \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} $$ with the same simply connected hypothesis for the converse.

Reconstructing a potential. Antidifferentiate $P$ with respect to $x$, allowing an unknown function of the remaining variables as the constant. Differentiate the result with respect to $y$, set it equal to $Q$, and solve for that unknown. Repeat with $z$ in space. If the equations are inconsistent, the field was not conservative.

Conservation of energy. If $\mathbf{F} = -\nabla U$ is the force on a particle, the work integral equals $U(A) - U(B)$, and Newton’s second law then makes kinetic plus potential energy constant along any trajectory. Hence the name.

Worked examples

Testing, building a potential, and using it

Let $\mathbf{F}(x,y) = \langle 2xy + 3, \; x^2 - 4y\rangle$. Here $P = 2xy+3$ and $Q = x^2 - 4y$, so $P_y = 2x$ and $Q_x = 2x$. They agree, and the domain is the whole plane, which is simply connected, so $\mathbf{F}$ is conservative.

Build the potential. Antidifferentiating $P$ in $x$ gives $f(x,y) = x^2 y + 3x + g(y)$. Differentiating in $y$ gives $f_y = x^2 + g'(y)$, and this must equal $Q = x^2 - 4y$, so $g'(y) = -4y$ and $g(y) = -2y^2$. Therefore $$ f(x,y) = x^2 y + 3x - 2y^2 $$ Check by differentiating: $f_x = 2xy + 3$ and $f_y = x^2 - 4y$, which is $\mathbf{F}$.

Now any work integral is a subtraction. Along any curve from $(0,0)$ to $(2,1)$, $$ \int_C \mathbf{F}\cdot d\mathbf{r} = f(2,1) - f(0,0) = (4 + 6 - 2) - 0 = 8 $$

A field that fails the test

Let $\mathbf{F}(x,y) = \langle y^2, \; x\rangle$. Then $P_y = 2y$ and $Q_x = 1$, which disagree except on the line $y = 1/2$, so no potential exists on any open region.

Watch the failure directly. Go from $(0,0)$ to $(1,1)$ along the diagonal $\mathbf{r}(t) = \langle t,t\rangle$: the integrand is $\langle t^2, t\rangle\cdot\langle 1,1\rangle = t^2 + t$, and the integral is $\frac13 + \frac12 = \frac56$. Now go along the two legs instead. On the bottom leg $y = 0$, so $\mathbf{F} = \langle 0, x\rangle$ while $d\mathbf{r} = \langle dx, 0\rangle$, and the contribution is $0$. On the right leg $x = 1$ with $d\mathbf{r} = \langle 0, dy\rangle$, so the integrand is $Q = 1$ and the contribution is $1$. Two paths, same endpoints, values $5/6$ and $1$.

A potential in three variables

Let $\mathbf{F} = \langle yz, \; xz, \; xy\rangle$. The three cross-partial conditions hold: $P_y = z = Q_x$, $P_z = y = R_x$, and $Q_z = x = R_y$. Antidifferentiating $P$ in $x$ gives $f = xyz + g(y,z)$; matching $f_y = xz + g_y$ against $Q = xz$ forces $g_y = 0$, and matching $f_z$ against $R$ forces $g_z = 0$, so $g$ is constant and $f = xyz$.

The work along any curve from $(1,1,1)$ to $(2,3,4)$ is then $f(2,3,4) - f(1,1,1) = 24 - 1 = 23$, with no parametrization computed at all.

Passing the test without being conservative

On the plane with the origin removed, let $$ \mathbf{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \; \frac{x}{x^2+y^2} \right\rangle $$ Both partials come out to the same expression: $$ \frac{\partial P}{\partial y} = \frac{-(x^2+y^2) + 2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, \qquad \frac{\partial Q}{\partial x} = \frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} $$ Yet on the unit circle $\mathbf{r}(t) = \langle\cos t, \sin t\rangle$ the denominator is $1$, so $\mathbf{F} = \langle -\sin t, \cos t\rangle = \mathbf{r}'(t)$, and the circulation is $\int_0^{2\pi} 1\,dt = 2\pi$, not zero.

There is no contradiction. The punctured plane is not simply connected: the unit circle cannot be shrunk to a point without crossing the missing origin, so the converse half of the test does not apply. On any smaller region that avoids the origin and has no hole, such as the right half plane $x>0$, the field is conservative there, with potential $\arctan(y/x)$. The angle is a perfectly good potential locally, and fails globally only because it cannot be defined continuously all the way around.

Practice

First the recognition and construction work: apply the cross-partials test, decide whether the domain supports the converse, and produce a potential function when one exists.

Then the evaluation. These use a potential to compute work along a specified path, including closed loops where the answer is forced to be zero.

Quiz

Five items on testing for a conservative field, constructing potentials, and evaluating line integrals from endpoint values.