Does path independence of a line integral prove existence of a scalar potential?

[Mathmos6]

Homework Statement

Hi there - I'm wondering about how you can actually show the existence of a scalar potential for an irrotational vector field E - if $\nabla \times E = 0$ everywhere, then how does one show there exists a scalar potential $\phi(x)$ such that $E=- \nabla \phi$?

The Attempt at a Solution

By Stokes' theorem we can see that $\int_C E dx = \int_S \nabla \times E dS = 0$ everywhere so our integral is path independent, but does path independence necessarily prove the existence of a scalar potential?

Thanks a lot, Mathmos6

 

[elect_eng]

There is a well-known vector identity:

$$\bigtriangledown \times \bigtriangledown\phi=0$$

 

[Dick]

Sure it does. Now pick a point A and define the potential at any point X as the path integral of E from A to X. It's well defined because of path independence.

 

[Mathmos6]

Ah, fair enough - thanks Dick!