Path Independence and the Four Theorems

What "path independent" actually means

We have a vector field $\mathbf{u} = P(x,y)\,\mathbf{i} + Q(x,y)\,\mathbf{j}$ defined on some domain $D$ in the plane. Pick two points $A$ and $B$ in $D$. The line integral $\int_A^B \mathbf{u}\cdot d\mathbf{r}$ depends, in principle, on three things: the field, the endpoints, and the curve $C$ joining them.

We say the integral is independent of path in $D$ if for every pair of endpoints $A, B$ in $D$, every piecewise smooth curve from $A$ to $B$ gives the same value. The integral cares only about $A$ and $B$, not the route.

Here's the everyday picture. You hike from your tent to the summit. The total work done against gravity depends only on the elevation gain - not the trail you picked. That's path independence for gravity. Friction, on the other hand, eats more energy on a long zig-zag than on a direct climb - friction is not path-independent, and we can feel the difference in our knees.

So when does this happen? Kaplan answers with four theorems, each adding a different angle. Let's take them one at a time. Each unlocks the next.

Theorem I - path independent means there's a potential

The first theorem is the deepest one, and it goes both directions.

Let's see why both directions work. The reverse direction first - it's quick.

Reverse ($\mathbf{u} = \nabla f \Rightarrow$ path independent): if $\mathbf{u} = \nabla f$, then along any smooth curve $\mathbf{r}(t)$,

$$\mathbf{u}\cdot \frac{d\mathbf{r}}{dt} = \nabla f \cdot \frac{d\mathbf{r}}{dt} = \frac{d}{dt}\, f\bigl(\mathbf{r}(t)\bigr)$$

by the chain rule. So

$$\int_A^B \mathbf{u}\cdot d\mathbf{r} = \int_a^b \frac{d}{dt} f(\mathbf{r}(t))\,dt = f(B) - f(A).$$

The right-hand side mentions only the endpoints. Done.

Forward (path independent $\Rightarrow$ a potential exists): here's the construction Kaplan uses, and it's the genius step. Pick a basepoint $(x_0, y_0)$ in $D$ and define

This formula is well-defined precisely because the integral is path-independent - the right-hand side doesn't care which path we use, only the endpoints. So $f$ is an honest function of $(x,y)$.

Now we have to show $\partial f/\partial x = P$ and $\partial f/\partial y = Q$. Trick: to compute $\partial f/\partial x$ at a fixed point, pick a path that ends with a horizontal segment. Walk from $(x_0,y_0)$ to some intermediate point $(x_1, y)$, then move horizontally along $y = $ const from $(x_1, y)$ to $(x, y)$. On the horizontal piece $dy = 0$, so

$$f(x, y) = (\text{stuff not depending on } x) + \int_{x_1}^{x} P(t, y)\,dt.$$

The fundamental theorem of calculus then gives $\partial f/\partial x = P(x, y)$. The same trick with a vertical segment gives $\partial f/\partial y = Q(x, y)$. So $\nabla f = \mathbf{u}$, and we have a potential.

Theorem I is the structural theorem - it says path independence is the same as having a potential. The next three theorems are about how to recognize path independence in practice.

Theorem II - closed loops give zero

Theorem I gave us a structural picture. Theorem II gives an equivalent operational test: instead of checking every pair of paths from $A$ to $B$, we just check closed loops.

The proof is a one-liner picture. Take any closed loop $C$ in $D$. Pick two points $A, B$ on it. The loop splits into two arcs from $A$ to $B$ - call them $C_1$ and $C_2'$ (the second one traversed forward). Going around the loop is the same as going $A\to B$ along $C_1$, then $B \to A$ along the reverse of $C_2'$:

$$\oint_C \mathbf{u}\cdot d\mathbf{r} \;=\; \int_{C_1} \mathbf{u}\cdot d\mathbf{r} \;-\; \int_{C_2'} \mathbf{u}\cdot d\mathbf{r}.$$

If the integral is path-independent, those two integrals are equal, and the difference is zero. Conversely, if every closed-loop integral is zero, then for any two paths $C_1, C_2$ from $A$ to $B$, gluing $C_1$ to the reverse of $C_2$ gives a closed loop with integral zero - so the two paths give the same value.

This is the version that matches our hook. In the gradient field, both red and blue paths gave the same number; equivalently, the closed loop "red forward, then blue backward" integrates to zero. In the vortex field, the two integrals differed - so the loop did not integrate to zero.

Theorem III - a necessary condition: $P_y = Q_x$

So far we have two equivalent ways to spot path independence: find a potential, or check that all loops give zero. Both are global - we'd need to find $f$, or somehow integrate around every loop. We want a local test we can run pointwise. Theorem III gives one.

The proof is one line if we're already holding Theorem I. Path independence gives us $f$ with $P = f_x$ and $Q = f_y$. Then

$$\frac{\partial P}{\partial y} = \frac{\partial^2 f}{\partial y\,\partial x}, \qquad \frac{\partial Q}{\partial x} = \frac{\partial^2 f}{\partial x\,\partial y}.$$

For a $C^2$ function, mixed partials are equal (Clairaut's theorem). So $P_y = Q_x$.

In the language of curl: writing the 2D scalar curl as $Q_x - P_y$, Theorem III says path-independent fields are curl-free. That's the local test we wanted - just compute one derivative and compare.

How to use it. Theorem III is a one-way street: it's a necessary condition. If you ever find a point where $P_y \neq Q_x$, the field is not path-independent - rule it out and move on. But the converse - "if $P_y = Q_x$ everywhere, then path-independent" - is not always true. That gap is the whole point of Theorem IV.

Theorem IV - the converse, with one more hypothesis

We want to turn Theorem III around: from $P_y = Q_x$ conclude path independence. Theorem IV does this, but it has to assume something extra about $D$.

Recall that simply connected means: $D$ is open, connected, and has no holes - more precisely, every simple closed curve in $D$ encloses a region that also lies entirely in $D$. The disk, the half-plane, all of $\mathbb{R}^2$, a square, a sector - all simply connected. The annulus, or the plane with the origin removed, are not.

Proof sketch. Take any simple closed curve $C$ in $D$. Because $D$ is simply connected, the region $R$ enclosed by $C$ also lies in $D$. So $P, Q$ and their partials are continuous on $R$, and Green's theorem applies:

$$\oint_C P\,dx + Q\,dy \;=\; \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \;=\; \iint_R 0\,dA \;=\; 0.$$

Every simple closed loop integrates to zero. By Theorem II, the integral is path-independent.

The simply-connected hypothesis pulls all the weight: it's exactly what guarantees $R \subseteq D$, which is exactly what makes Green's theorem applicable. Drop it, and the proof breaks - and so, as we'll now see, does the conclusion.

The four theorems on one page

Here's all four, lined up. Read across each row: hypothesis, conclusion, what role it plays.

And here's the practical workflow for "is this field conservative?":

1. Quick local test: compute the 2D scalar curl $Q_x - P_y$. If it's nonzero anywhere in $D$, the field is not path-independent. Stop.
2. If $Q_x = P_y$ everywhere on $D$, look at the shape of $D$. Simply connected? Theorem IV gives you path independence for free. Find $f$ by the standard recipe (path along axes, or guess-and-check).
3. If $D$ has holes (e.g. the punctured plane), the field is conservative locally on every simply-connected sub-region, but globally you need to test one closed loop around each hole. Nonzero on any of them ⇒ not path-independent on $D$.
4. If you have a potential $f$, evaluate any line integral as $f(B) - f(A)$. No parametrization, no curve - just two function values.

The vortex from card 5 is a beautiful one-line example of step 3: curl-free yet $\oint = 2\pi$ around the origin. The recipe in step 2 - how to actually find $f$ - lives on its own page, Finding Potentials (forthcoming).