- #1
yoghurt54
- 18
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Homework Statement
\nabla \times f \vec{v} = f (\nabla \times \vec{v}) + ( \nabla f) \times \vec{v}
Use with Stoke's theorem
\oint _C \vec{A} . \vec{dr} = \int \int _S (\nabla \times \vec{A}) . \vec{dS}
to show that
\oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f
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
\vec{A} = f \vec{v}
And subsituted into Stokes' theorem
\oint _C f\vec{v} . \vec{dr} = \int \int _S (\nabla \times f \vec{v}) . \vec{dS}
which gives
\oint _C f\vec{v} . \vec{dr} = \int \int _S f ((\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} ). \vec{dS}
My problem is that I got to this stage, and thought that this would work really great if
\nabla \times \vec{v} = 0.
So I got stuck on this for a while, looked at my textbook, and they said that we let
\vec{A} = f \vec{v} where \vec{v} is a CONSTANT VECTOR.
Hence, the curl of a constant vector is zero, and the RHS becomes
\int \int _S f ( \nabla f) \times \vec{v} ). \vec{dS}
which we rearrange with the triple scalar product identity to give
\int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}
So now the thing looks like this:
\oint _C f\vec{v} . \vec{dr} = \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}<br />
and cancelling \vec{v} from both sides as it's constant, gives us:
\oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f
The problem
My problem is that this restricts the kind of vector field \vec{A} 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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