# Why is a line integral of a conservative vector field independent of path?

I am looking for intuition behind why the ability of a vector field $\vec{F}(x,y)$ to be written in the form $\vec{F}(x,y)=\nabla f$ where $f$ is simply a function of $x,y,$ and $z$ implies that a line integral on that conservative vector field is independent of path.

Furthermore, what does a conservative vector field look like visually? What properties do conservative vector fields have in the $xy$ plane that non-conservative vector fields don't?

As a window to complex analysis, does there exist a relationship between the fact that line integrals of conservative vector fields in $\mathbb{R}^n$ on closed curves is $0$ and Cauchy's Integral Theorem?

• The integral of $\nabla f\cdot dx$ is nothing but the integral of $df$, so the reason why the integral does not depend on the path is clear. Regarding the question the shape of conservative fields, I can tell you the shape they haven't. Conservative fields have vanishing rotational. That means they can't have vortices. – Dog_69 Jun 15 '18 at 9:49

## 1 Answer

Imagine you're walking on a mountainside $z = \varphi(x,y)$ and trace the path on a contour plot of the mountain. Your velocity vector at any point on that path can be split into two orthogonal components. One points parallel to the level curve; this direction (locally) isn't taking you any higher or lower. The other component points orthogonal to the level curve; this direction (locally) only takes you higher or lower.

Conservative vector fields are entirely orthogonal to the level curves of some function. There is some mountain they are only taking you up or down on. (I'm not 100% sure if the converse is true: that if your vector field is orthogonal to the level curves of some function it's conservative). There is a corresponding opposite kind, too: solenoidal vector fields are entirely parallel to the level curves of some function.

For example, $\mathbf{F}(x,y)=\langle x, y\rangle$ is a conservative vector field – the gradient of $\varphi(x,y) = \frac{1}{2}(x^2 + y^2)$. And a corresponding solenoidal vector field is $\mathbf{G}(x,y) = \langle-y,x\rangle$.

Plot of $\mathbf{F}(x,y)=\langle x, y\rangle$

Plot of $\mathbf{G}(x,y) = \langle-y,x\rangle$

As for relation to the Cauchy integral theorem: there is a proof involving Green's theorem and the Cauchy-Riemann equations, which I believe is sketched on the Wikipedia article for the Cauchy integral theorem. Essentially, the proof sneakily uses, without telling you, a differential operator called the exterior derivative. Differential forms (for our purposes, vector fields) that are in the image of the exterior derivative are "conservative" or "exact", and forms that are in the kernel of the exterior derivative are "closed". Every exact form is closed, but whether every closed form is exact depends on what set you're looking at. And it's a condition related to exactness that guarantees the existence of an "antiderivative".

If you're interested in more (this is a deep question!) check out Gamelin's book on complex analysis. Other topics worth reading about: differential forms, the generalized Stokes theorem, de Rham cohomology.