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Related to stokes theorem

  • Thread starter yoghurt54
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Homework Statement



[tex] \nabla \times f \vec{v} = f (\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} [/tex]

Use with Stoke's theorem

[tex] \oint _C \vec{A} . \vec{dr} = \int \int _S (\nabla \times \vec{A}) . \vec{dS} [/tex]

to show that

[tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

Homework Equations




I think the only that you really need to know is the the scalar triple product.

The Attempt at a Solution



Ok, I allowed

[tex] \vec{A} = f \vec{v} [/tex]

And subsituted into Stokes' theorem

[tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S (\nabla \times f \vec{v}) . \vec{dS} [/tex]

which gives

[tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S f ((\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

My problem is that I got to this stage, and thought that this would work really great if

[tex] \nabla \times \vec{v} = 0 [/tex].

So I got stuck on this for a while, looked at my textbook, and they said that we let

[tex] \vec{A} = f \vec{v} [/tex] where [tex] \vec{v} [/tex] is a CONSTANT VECTOR.

Hence, the curl of a constant vector is zero, and the RHS becomes

[tex] \int \int _S f ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

which we rearrange with the triple scalar product identity to give

[tex] \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v} [/tex]

So now the thing looks like this:

[tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}
[/tex]

and cancelling [tex] \vec{v} [/tex] from both sides as it's constant, gives us:

[tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

The problem

My problem is that this restricts the kind of vector field [tex] \vec{A} [/tex] can be, as it has to be a product of some function times a constant vector.

Does this make the derived identity non-universal, or is this a necessary step in order to arrive at the identity?
 
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Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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My problem is that this restricts the kind of vector field [tex] \vec{A} [/tex] can be, as it has to be a product of some function times a constant vector.

Does this make the derived identity non-universal, or is this a necessary step in order to arrive at the identity?
The vector identity you proved involves only [itex]f[/itex]. So, who cares what you restricted [itex]\textbf{A}[/itex] to be? The identity is valid for any scalar function [itex]f[/itex], and that's all you were asked to prove.
 
  • #3
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Yeah you're right. I don't know why I got hung up on this, I guess I was reading too much into it.
 

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