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Math Help - vector integeral calculus

  1. #1
    Junior Member piglet's Avatar
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    vector integeral calculus

    am i right in saying the following

     \phi \bigtriangledown  = \bigtriangledown  \phi where  \phi is some scalar field?

    Im equating  \bigtriangledown\times (\phi A) = (\phi \bigtriangledown ) \times A + (\bigtriangledown  \phi) \times A

    and wondering if this is the same as

     A \times (\phi A) = 2 ((\bigtriangledown \phi) \times A)
    Last edited by piglet; April 12th 2010 at 07:27 AM. Reason: fixed latex. Piglet- please check for correctness.
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  2. #2
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    Erm, isn't A\times(\phi A) = \phi(A\times A) = 0?
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  3. #3
    Junior Member piglet's Avatar
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    To be honest im really just looking for the difference between
    \phi.\bigtriangledown  and  \bigtriangledown.\phi

    if there is any difference. I thought they would equate to the same thing?

    I fixed my original post as well as i had a small but significant mistake in the formula
    Last edited by piglet; April 12th 2010 at 07:29 AM. Reason: mistake in original post
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  4. #4
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    No, they are not the same thing. They are not even the same "kind" of thing.

    \nabla \phi is the vector function \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{j}.

    \phi \nabla is not a function at all but rather a "vector operator". If f is a real valued function then \phi \nabla f= \phi(\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}).

    If \vec{F} is a vector valued function, say \vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} then \phi\nabla\cdot \vec{F}= \phi(\frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z})

    We could also write \phi\nabla\times\vec{F}= \phi((\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z})\vec{i}- (\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z})\vec{j}+ (\frac{\partial g}{\partial x}- \frac{\partial x}{\partial y})\vec{k})
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  5. #5
    Junior Member piglet's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    No, they are not the same thing. They are not even the same "kind" of thing.

    \nabla \phi is the vector function \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{j}.

    \phi \nabla is not a function at all but rather a "vector operator". If f is a real valued function then \phi \nabla f= \phi(\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}).

    If \vec{F} is a vector valued function, say \vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} then \phi\nabla\cdot \vec{F}= \phi(\frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z})

    We could also write \phi\nabla\times\vec{F}= \phi((\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z})\vec{i}- (\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z})\vec{j}+ (\frac{\partial g}{\partial x}- \frac{\partial x}{\partial y})\vec{k})
    Thanks a great deal for your help. I really appreciate the time you take to answer the questions so thoroughly.

    Piglet
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