Proof of the existence of a scalar potential

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Homework Help Overview

The discussion revolves around proving the existence of a scalar potential for an irrotational vector field E, specifically when it is given that the curl of E is zero everywhere.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how path independence, derived from Stokes' theorem, relates to the existence of a scalar potential. Some participants suggest that defining the potential through a path integral from a reference point is a valid approach.

Discussion Status

Participants are exploring the relationship between path independence and the existence of a scalar potential. Some guidance has been offered regarding the definition of potential using path integrals, but there is no explicit consensus on the proof itself.

Contextual Notes

The discussion is constrained by the need to show the existence of a scalar potential without assuming prior knowledge of certain vector identities or theorems.

Mathmos6
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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 [itex]\nabla \times E = 0[/itex] everywhere, then how does one show there exists a scalar potential [itex]\phi(x)[/itex] such that [itex]E=- \nabla \phi[/itex]?

The Attempt at a Solution


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

Thanks a lot, Mathmos6
 
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There is a well-known vector identity:

[tex]\bigtriangledown \times \bigtriangledown\phi=0[/tex]
 
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.
 
Ah, fair enough - thanks Dick!
 

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