# Path independence: conservative field vs non-conservative field

Why is it so that if we have a smooth, conservative vector field, we can choose any two points within its domain and join them with any type of smooth primitive curve, we will get the same result no matter what form of the curve do we choose. For example, we can choose a straight line, parabola, circle, etc... all of them will yield the same result. Now, if the field is not conservative the result might be different for each curve that we chose. Can someone explain why this happens? In other words if the field is conservative it is path independent, why it is so?

• What's your definition of conservative? – Muphrid Feb 8 '15 at 5:05
• It has a potential function: $F = \nabla \varphi$ – user_777 Feb 8 '15 at 5:16

## 2 Answers

Apply Stokes' theorem. Over a surface $M$ which has a bounding curve $\partial M$, a vector field $F$ that is a gradient $\nabla \varphi$ obeys

$$\int_{\partial M} F \cdot d\ell = \int_M (\nabla \times F) \cdot dA = \int_M (\nabla \times \nabla \varphi) \cdot dA$$

But for any gradient, $\nabla \times \nabla \varphi = 0$. Hence this integral is zero.

If $F$ is a gradient everywhere, then we can consider any closed curve $\partial M$--or given two points, any combination of paths between those points--and still get zero, allowing us to conclude that the line integral is path independent.

One definition of conservative vector field is that the integral over any closed contour (in the relevant domain) vanishes (meaning, equals zero).

A path integral's value switches sign if you switch the orientation of the path. Now, if you take two paths from one point to another and switch one of their orientations, you get a loop. Of course you can also argue the other way, that if path integrals only depend on endpoints, then the vector field integrated over any loop vanishes and so it is conservative.

Another definition is that $F=\nabla U$ for some scalar function $U$. The gradient theorem states that if a vector field is a gradient, then path integrals depend only on endpoints. Conversely, if path integrals depend only on endpoints then we can fix $a$ in the domain and define $U$ using path integrals as $U(x)=\int_a^x F(r)\cdot dr$ and verify that $\nabla U=F$ (this is done at Wikipedia).