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.

Friction is path-dependent - walking $2$m or $100$m matters because friction does work along every step. The formal definition above is just the math saying "no friction-like behaviour: only endpoints count."

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

It turns out path independence is equivalent to having a potential - a single scalar function that explains the whole field. 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

We want a faster check than "try every pair of paths." It turns out we don't need pairs - closed loops are enough. 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: $P_y = Q_x$

If a potential exists, the components $P$ and $Q$ can't be independent - they're tied together by the equality of mixed partials. Here's the cheap necessary check. 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

Now we want the converse. Most of the time the necessary condition is also sufficient - but not always, and the not always is what trips students up. 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.

Before we trust Theorem IV: does the necessary condition ever hold without the converse? It turns out yes, and the canonical example is so clean it's worth seeing.

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).