- #1

davi2686

- 33

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It is a nonsense use the generalized Stokes' theorem in right side of Faraday's Law?

we know this is true...[itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Rightarrow \int_{\Sigma}\vec{\nabla}\times\vec{E}\cdot d\vec{A}=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}[/itex]

i was thinking if we consider [itex] \Sigma[/itex] a smooth manifold (i really don't know if i can do that)... can i do this?

[itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Leftrightarrow \int_{\partial \Sigma}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} \Rightarrow \int_{\Sigma}\underline{d}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} [/itex]

where [itex]\int_{\partial \Sigma}E[/itex] is the integration in boundary of [itex] \Sigma[/itex] manifold and [itex] \underline{d}[/itex] the exterior derivative

i'm afraid i have done a mess with the concepts, but the doubt is in my head and it don't let me sleep...

we know this is true...[itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Rightarrow \int_{\Sigma}\vec{\nabla}\times\vec{E}\cdot d\vec{A}=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}[/itex]

i was thinking if we consider [itex] \Sigma[/itex] a smooth manifold (i really don't know if i can do that)... can i do this?

[itex]\displaystyle \oint_{\partial \Sigma}\vec{E}\cdot dl=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A}\Leftrightarrow \int_{\partial \Sigma}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} \Rightarrow \int_{\Sigma}\underline{d}E=-\int_{\Sigma} \frac{d\vec{B}}{dt}\cdot d\vec{A} [/itex]

where [itex]\int_{\partial \Sigma}E[/itex] is the integration in boundary of [itex] \Sigma[/itex] manifold and [itex] \underline{d}[/itex] the exterior derivative

i'm afraid i have done a mess with the concepts, but the doubt is in my head and it don't let me sleep...

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