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Math Help - Some complicated vector calculus..

  1. #1
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    Some complicated vector calculus..

    Not sure if the description fits the question I have, but here it goes:

    "If a vector function f(x,y,z) is not irrotational but the product of f and a scalar function g(x,y,z) is irrotational, show that then:

    \mathbf{f}\cdot \boldsymbol{\triangledown} \times \mathbf{f}=0"

    I've been working at this for a couple hours now... I was able to find that since:

    \boldsymbol{\triangledown} \times (g\mathbf{f})=0
    and
    \boldsymbol{\triangledown} \times (g\mathbf{f})=g(\triangledown \times \mathbf{f})+(\triangledown g)\times \mathbf{f},

    then

    g(\triangledown \times \mathbf{f})=-(\triangledown g)\times \mathbf{f}


    I don't even know if this is in the right direction. Help!
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  2. #2
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    Re: Some complicated vector calculus..

    Quote Originally Posted by limddavid View Post
    Not sure if the description fits the question I have, but here it goes:

    "If a vector function f(x,y,z) is not irrotational but the product of f and a scalar function g(x,y,z) is irrotational, show that then:

    \mathbf{f}\cdot \boldsymbol{\triangledown} \times \mathbf{f}=0"

    I've been working at this for a couple hours now... I was able to find that since:

    \boldsymbol{\triangledown} \times (g\mathbf{f})=0
    and
    \boldsymbol{\triangledown} \times (g\mathbf{f})=g(\triangledown \times \mathbf{f})+(\triangledown g)\times \mathbf{f},

    then

    g(\triangledown \times \mathbf{f})=-(\triangledown g)\times \mathbf{f}


    I don't even know if this is in the right direction. Help!
    Ya, this looks good! I will use the identity that

    \mathbf{u} \cdot \left(\mathbf{v}\times \mathbf{w}) = \left(\mathbf{u} \times \mathbf{v}\right) \cdot \mathbf{w}

    From what you have

    \triangledown \times \mathbf{f}=-\frac{\triangledown g}{g} \times \mathbf{f} }=  \mathbf{f} \times \frac{\triangledown g}{g} .

    So, \mathbf{f} \cdot \triangledown \times \mathbf{f} = \mathbf{f}\cdot \left( \mathbf{f} \times \frac{\triangledown g}{g}\right) = \left( \mathbf{f} \times \mathbf{f}\right) \cdot \frac{\triangledown g}{g}\right) = 0 , since \mathbf{f} \times \mathbf{f} = 0.
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  3. #3
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    Re: Some complicated vector calculus..

    Perfect. Now it's clear. thank you!
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