It is a nonsense use the generalized Stokes' theorem in right side of Faraday's Law?(adsbygoogle = window.adsbygoogle || []).push({});

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 dont 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 dont let me sleep...

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Can I use generalized Stokes' theorem in Faraday's law?

Loading...

Similar Threads - generalized Stokes' theorem | Date |
---|---|

I About extrinsic curvature | Jul 16, 2017 |

A Diffeomorphisms & the Lie derivative | Jan 23, 2017 |

A Two cones connected at their vertices do not form a manifold | Jan 10, 2017 |

General Conditions for Stokes' Theorem | Oct 15, 2013 |

Who discovered the general case of Stokes' Theorem? | Mar 30, 2006 |

**Physics Forums - The Fusion of Science and Community**