Module 43: Conservative fields and Green’s theorem

A. Prerequisites and observable goals#

B. The central idea: some line integrals depend only on endpoints#

A conservative vector field is a gradient, $\vv F=\grad\phi$, for a scalar potential $\phi$ on its domain. Along a path, the chain rule makes $\vv F\cdot\vv r'$ a derivative of $\phi(\vv r(t))$, so the work becomes a difference of potential values. Other fields retain path information; Green’s theorem can turn a closed planar path integral into an area integral describing interior rotation or outflow.

C. Potential, domain, and boundary theorems#

For a differentiable potential and a piecewise smooth path from $A$ to $B$,

On an open connected domain with sufficient smoothness, being conservative is equivalent to path independence and to zero integral around every closed path. For a $C^1$ planar field $\vv F=(P,Q)$, a necessary condition is $P_y=Q_x$. On a simply connected open domain it is also sufficient. A domain with a hole requires additional care: equality of cross partials alone is not enough. [O3Conservative]

For a positively oriented (counterclockwise) piecewise smooth simple closed boundary $C$ enclosing a suitable planar region $D$, and $P,Q\in C^1$ on an open neighborhood of $D$, Green’s theorem states

Its outward flux form is

For a region with holes, orient the outer boundary counterclockwise and each inner boundary clockwise, keeping the region on the left as you traverse. The theorem applies when the field is smooth on the region and a neighborhood, not necessarily inside excluded holes. [O3Green]

D. Why local contributions cancel inside a region#

Partition a region into small cells. Circulation around adjacent cells cancels on shared edges because those edges are traversed in opposite directions. Only the outer boundary remains. The small-cell circulation per area approaches $Q_x-P_y$. In flux form, outward flow across one cell’s shared side is inward flow to its neighbor; again the internal contributions cancel. This cancellation explains the orientation convention and foreshadows the three-dimensional boundary theorems.

For a potential, the ordinary chain rule gives $\dd\phi(\vv r(t))/\dd t=\grad\phi\cdot\vv r'$. Integrating proves the endpoint theorem directly. A zero derivative test on a punctured domain cannot create a single-valued potential if a closed loop has nonzero work around the hole.

E. Choose between a potential and a region integral#

For any path, first check whether finding a potential is easy. Integrate one component and include an unknown function of the other coordinates, then differentiate to match the remaining components. For a closed planar path, consider Green when the enclosed region is simpler than the boundary parameterization. Verify the field has no singularities on the required region. For holes, include every boundary component with its correct direction rather than filling a forbidden hole silently.

F. Worked examples#

Building a potential. For $\vv F=(2xy,x^2+2y)$, integrate $\phi_x=2xy$ in $x$ to get $\phi=x^2y+g(y)$. Matching $\phi_y=x^2+g'(y)=x^2+2y$ gives $g=y^2+C$. Thus work from $(0,0)$ to $(1,2)$ is $(1^2\cdot2+4)-0=6$, independent of path within $\R^2$.

A hole counterexample. On $\R^2\setminus\{0\}$, the field $\vv F=(-y/(x^2+y^2),x/(x^2+y^2))$ satisfies $P_y=Q_x$ but has circulation $2\pi$ around the counterclockwise unit circle. It is not globally conservative there. Green cannot be applied to the full enclosed disk because the field is undefined at its center.

Area from a boundary. Choosing $P=0,Q=x$ makes $Q_x-P_y=1$, so for positive orientation, $A(D)=\oint_C x\dd y$. The symmetric form is $A=\tfrac12\oint_C(x\dd y-y\dd x)$. Reversing orientation changes the signed result, not the geometric area.

Computing bridge: Potential energy and the domain of the model#

A force derived from a potential can give path-independent work on the stated domain. In T05, the central force excludes the origin, and the simulation must reject a singular state rather than divide by zero. Checking a derivative identity at ordinary points does not remove a hole in a field’s domain. The practical connection is to record model restrictions alongside invariants such as energy, and not use a conservation argument where its assumptions fail.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 43#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Retrieve M36.02 and M42.05. Explain why curl-free behavior on a domain with a hole is not by itself a global potential. Advance when you can choose between potential evaluation, direct parameterization, and Green, while stating which region must be free of singularities.