Different forms of Stokes' theorem

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Homework Help Overview

The discussion revolves around different forms of Stokes' theorem, specifically applying it to a vector field defined as ##\vec V=\vec a \phi##. Participants are analyzing the left-hand side and right-hand side of the theorem and exploring the implications of vector calculus operations on the components involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to equate the left-hand side and right-hand side of Stokes' theorem by evaluating the curl of the vector field and questioning the assumptions about the terms involved, particularly regarding the behavior of constant vectors and gradients.

Discussion Status

Some participants have provided insights and confirmations regarding the mathematical expressions, while others are still questioning the validity of their reasoning and the correctness of their proofs. There is an ongoing exploration of the implications of the vector operations involved.

Contextual Notes

Participants are navigating potential uncertainties in their proofs and are seeking validation of their mathematical reasoning. There is a reference to external resources that may influence their understanding of vector calculus concepts.

MatinSAR
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Homework Statement
Find the different forms using ##\vec V=\vec a \phi## and ##\vec V=\vec a \times \vec P## for constant ##\vec a##.
Relevant Equations
Stokes' theorem.
What am I trying to do for ##\vec V=\vec a \phi## :
##R.H.S= \oint \vec V \cdot d \vec \lambda=\oint \vec a \phi \cdot d \vec \lambda=\vec a \cdot \oint \phi d \vec \lambda ##

##L.H.S= \iint_S \vec \nabla \times \vec V \cdot \vec d \sigma=\iint_S \vec \nabla \times (\vec a \phi) \cdot \vec d \sigma=\iint_S (\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a) \cdot \vec d \sigma= ?##
I think ##\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a## should be 0. why is this wrong?

##\phi \vec \nabla \times \vec a## is 0 because ##\vec a## is a constant vector.
##(\vec \nabla \phi) \times \vec a## is 0 because ##\vec \nabla## acts on both ##\phi## and ##\vec a## so it should be zero.

Edit:
Now I think ##(\vec \nabla \phi) \times \vec a## is not 0 because ##\vec \nabla## acts only on ##\phi## so we can rewrite it as ##- \vec a \times (\vec \nabla \phi).##
 
Last edited:
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MatinSAR said:
Now I think ##(\vec \nabla \phi) \times \vec a## is not 0 because ##\vec \nabla## acts only on ##\phi## so we can rewrite it as ##\vec a \times (\vec \nabla \phi).##
Right.
 
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haruspex said:
Right.
Thanks for the reply @haruspex .
##L.H.S= \iint_S \vec \nabla \times \vec V \cdot d \vec \sigma=\iint_S \vec \nabla \times (\vec a \phi) \cdot d \vec \sigma=\iint_S (\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a) \cdot d \vec \sigma=##
##- \iint_S \vec a \times (\vec \nabla \phi) \cdot d \vec \sigma=- \iint_S \vec a \cdot (\vec \nabla \phi) \times d \vec \sigma=\vec a \cdot \iint_S d \vec \sigma \times (\vec \nabla \phi) ##
L.H.S = R.H.S
##\vec a \cdot \iint_S d \vec \sigma \times (\vec \nabla \phi) = \vec a \cdot \oint \phi d \vec \lambda ##
##\iint_S d \vec \sigma \times (\vec \nabla \phi) = \oint \phi d \vec \lambda##

I hope I won't have problem with other part. (##\vec V=\vec a \times \vec P##)
 
I've managed to prove 2nd part using what I've learnt here : https://www.physicsforums.com/threads/vector-operators-grad-div-and-curl.1057533/

But I'm not sure if my proof is mathematically true or it is nonsense. Picture of my work:
2023_12_23 3_15 PM Office Lens.jpg


I would be grateful if someone could point out the problem with my proof.
 

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