- #26

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Originally posted by lethe

you want a complete proof? stokes theorem says [tex]\mbox{$\int_Cd\omega = \int_{\partial C} \omega$}[/tex]. so if [tex]\mbox{$\quad C$}[/tex]

is a circle, then [tex]\mbox{$\partial C$}[/tex] is zero, regardless of whether [tex]\mbox{$\omega$}[/tex] vanishes or not.

"if C is a circle, then [tex]\mbox{$\partial C$}[/tex] is zero" - you say? We can use Stoke's Theorem on an open surface with a closed curve as a boundary. We cannot use Stoke's Theorem on a closed surface without boundary. Likewise, we cannot apply Stoke's Theorem to a closed curve with no boundary. A boundary on an open curve would just be its end points. Then Stoke's Theorem in that case would simply reduce to the fundamental theorem of calculus. But sorry, you cannot apply Stoke's Theorem when there is no boundary. So your argument does not seem to say anything about the integral or the integrand.

Neither does this seem to say anything about the time derivative of a conserved integral. I said conserved integral, not conservative field. A conservative field means you get back what you put in when you end where you started. But I don't see how that has anything to do with a time dervative of an integral that is a constant which is not necessarily zero.

like another counter example? the electric field in the vicinity of a point charge is nonzero. this is a conservative field, so any loop integral is zero.