A Conservative Field Flows from a Potential
In lesson 5 you saw one strange fact: in some fields the work is the same no matter how you route the path. These are conservative fields. A conservative field hides a secret: an invisible height landscape, a potential φ, lies underneath, and the field is the gradient of that landscape (F = ∇φ). So the work is just the difference in start and end height, and the path does not matter. The tell that reveals this is the curl. A conservative field has zero curl everywhere, and once around any closed loop the total is exactly zero. By contrast, if the curl is not zero, like in a swirl, no such potential exists at all. This lesson shows the same fact in four faces: zero curl, a potential exists, path independence, closed loop zero. The four are really one body.
The fastest test for a conservative field is the curl. Toggle between the two fields. Drag the probe over the conservative field F = ∇φ, and wherever you put it the curl (∇×F)z reads zero. The point turns green for a pass. A zero curl means there is no twist forcing things to spin, which is the sign that a potential lies underneath. Now switch to the swirl [−y, x]. Drag it anywhere and the curl locks at a constant 2, so the point turns orange for a fail. A single point already decides it: if even one spot has a nonzero curl, the field is not conservative.
When the curl is zero you can recover the invisible potential φ in full. The arrows are the proof. Every arrow of the conservative field F = ∇φ points toward the brighter, higher side, the uphill of φ. Drag the gold point. The height φ at that spot shows up as a number. This value is the height you build step by step by starting at the origin and integrating along F. Since any route of integration gives the same value (path independence), φ is fixed to one number wherever you put the point. So even given only the field F, you can rebuild the landscape φ underneath it, point by point. That is why "F = ∇φ here" follows the point everywhere.
Now prove path independence directly with two routes. Two paths from A to B are laid out: a straight segment that goes direct and a detour arc that loops high. Hit the toggle to light up one route at a time. The work along each route shows up as a number. Because the field is conservative, the two values are identical. What is more, both match φ(B) − φ(A) exactly. One difference of endpoint heights settles the whole answer. In lesson 5's swirl the two routes would have split apart. Here, no matter how you bend the path, if start and end are the same the answer is the same. That is the heart of a conservative field.
Why do the two values always agree? Because of the fundamental theorem of line integrals, the gradient theorem. Keep endpoint A fixed and drag endpoint B. A path from A to B is drawn, and the line integral along it, ∫C F·dr, and the height difference φ(B) − φ(A) appear side by side. Wherever you drag B, the two stay equal. In symbols, ∫C ∇φ·dr = φ(B) − φ(A). All the little steps you pile up along the path fold up neatly into just the difference of start and end heights. A general integral makes you follow every point of the path, but in a conservative field knowing the two endpoints alone finishes the job.
Finally, nail down that not every field has a potential. The swirl F = [−y, x] has a curl of 2 everywhere. Use the slider to grow a loop centered at the origin. The gold arrows around the rim all turn the same way. The ∮F·dr you sum once around shows up as a number, and it is never zero for any radius. It is exactly 2πr², so the larger the loop the larger it grows. But if this field had a potential φ, the gradient theorem would force a closed loop, where start equals end, to give ∮ = 0. The single fact that ∮ is not zero proves this swirl has no potential at all. It is the exact opposite of a conservative field.