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Thread: Stokes' Theorem

  1. #1
    Super Member craig's Avatar
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    Stokes' Theorem

    Show, using Stokes' Theorm, that

    $\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0$

    Firstly Stokes gives us:

    $\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}$

    Expanding $\displaystyle \nabla \times [\phi (\nabla \phi)]$ gives:

    $\displaystyle \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi$

    The second part of this is zero for any vector, so this leaves the integral as:

    $\displaystyle \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}$

    Anyone any ideas?

    Cheers in advance!
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  2. #2
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    Quote Originally Posted by craig View Post
    Show, using Stokes' Theorm, that

    $\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0$

    Firstly Stokes gives us:

    $\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}$

    Expanding $\displaystyle \nabla \times [\phi (\nabla \phi)]$ gives:

    $\displaystyle \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi$

    The second part of this is zero for any vector, so this leaves the integral as:

    $\displaystyle \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}$

    Anyone any ideas?

    Cheers in advance!
    Remember that $\displaystyle \phi(x,y,z)$ is a scalar function!

    $\displaystyle \displaystyle \nabla \phi =\frac{\partial \phi}{\partial x}\mathbf{i}+\frac{\partial \phi}{\partial y}\mathbf{j}+\frac{\partial \phi}{\partial z}\mathbf{k}$

    $\displaystyle \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)$

    Expand out what is inside the parenthesis and you will get zero (As long as $\displaystyle \phi$ is twice continuously differentiable)

    Remember 2nd continuous partial derivatives commute Clairaut's theorem

    Symmetry of second derivatives - Wikipedia, the free encyclopedia
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Remember that $\displaystyle \phi(x,y,z)$ is a scalar function!

    $\displaystyle \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)$
    Ahh of course! I was thinking of it as a vector function.

    If we take $\displaystyle \displaystyle \left(\nabla \times \nabla \phi \right)$, we get,

    $\displaystyle \displaystyle \left(\frac{\partial}{\partial y} \frac{\partial \phi}{\partial z} - \frac{\partial}{\partial z} \frac{\partial \phi}{\partial y},... \right) = \left(0,0,0 \right)$

    Hence the integral is equal to zero!

    Many thanks for that
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