# Gauss's law in differential forms

by loom91
Tags: differential, forms, gauss
 P: 339 also almost immediatley after the proof using divergence theorem there is a proof by the dirac delta function in griffith i u think this is illogical(though it isn't) u can refer that proof
P: 400
 Quote by Hurkyl There's a reason I said it's important to explicitly state the statement in which you're interested. Some discussions only consider smooth vector fields, and the divergence operator you learned in your elementary calculus classes. That the integrands are equal is a easy consequence of continuity. Other contexts consider more general classes of vector fields, and even generalizations of that notion! They also use generalizations of the divergence operator. Without knowing what type of objects you are using for your fields, and what definition of divergence you are using, it is essentially impossible to provide a proof that would satisfy you!
Well, a general electric field is not required to be differentiable, or even continuous. The integral form of Gauss's law does not impose this requirement. But since the divergence operator is by definition a derivative, it seems inadequate to deal with discontinuous fields. Claiming Dirac deltas to be the derivatives of functions that can not be differentiated seems to be a dubious step. What is this generalised divergence operator you speak of? I only know of the definition using space derivatives. Thanks.

Molu

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