Proof of the 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!