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Math Help - Problem with Curl of Curl

  1. #1
    Newbie Schrodingerscat's Avatar
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    Problem with Curl of Curl

    I have been studying vector calculus, and came across what was to me an odd identity:

    \nabla\times\nabla\times\vec{F}=\nabla(\nabla\bull  et\vec{F})-\nabla^2\vec{F}

    I noticed that the two terms on the right hand side are of different character.

    The first term \nabla(\nabla\bullet\vec{F}), is the gradient of the divergence of the field, which is a vector quantity.

    The second term however, \nabla^2\vec{F} is a scalar quantity.

    So how are we supposed to do addition between a vector and a scalar?

    Something like a complex number?

    Any help with understanding this would be great.

    Thank you.
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  2. #2
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    Why do you say that \nabla ^2 \overrightarrow F is a scalar?
    By common understanding, \nabla ^2  = \nabla  \cdot \nabla  = \frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }} + \frac{{\partial ^2 }}{{\partial z^2 }} which is a scalar.
    Therefore \nabla ^2 \overrightarrow F is a vector.
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  3. #3
    Newbie Schrodingerscat's Avatar
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    Isn't the Laplacian Operator the divergence of the gradient? Divergence is a scalar quantity. That was my point.

    \nabla^2\vec{F}=\nabla\bullet\nabla\vec{F}

    Addendum:

    I stand corrected, you cannot take the Laplacian of a vector field. I am not entirely sure what the book is saying now.
    Last edited by Schrodingerscat; June 30th 2008 at 12:58 PM.
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  4. #4
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    As I said above, by common understanding \nabla ^2  = \nabla  \cdot \nabla  = \frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }} + \frac{{\partial ^2 }}{{\partial z^2 }} which is a scalar.
    Therefore it is scalar operator applied to a vector field: \left( {\nabla  \cdot \nabla } \right)\vec F


    To quote Davis & Snider’s text “to interpret (\nabla ^2) \overrightarrow F as div grad is rather strained”.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Schrodingerscat View Post
    I have been studying vector calculus, and came across what was to me an odd identity:

    \nabla\times\nabla\times\vec{F}=\nabla(\nabla\bull  et\vec{F})-\nabla^2\vec{F}

    I noticed that the two terms on the right hand side are of different character.

    The first term \nabla(\nabla\bullet\vec{F}), is the gradient of the divergence of the field, which is a vector quantity.

    The second term however, \nabla^2\vec{F} is a scalar quantity.

    So how are we supposed to do addition between a vector and a scalar?

    Something like a complex number?

    Any help with understanding this would be great.

    Thank you.
    Maybe your misunderstanding was purely notational, but for those out there viewing this, I believe this is gotten anti-climactically by the triple product

    a\times(b\times{c})=b(a\cdot{c})-c(a\cdot{b})
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  6. #6
    Forum Admin topsquark's Avatar
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    In case it helps the components of
    \nabla ^2 \overrightarrow F
    are
    \left ( \nabla ^2 \overrightarrow F \right ) _i = \nabla ^2 (F_i)

    (This is what I was taught. I believe that it is standard.)

    -Dan

    Edit: By the way, thank you Plato for a better way to indicate a vector. The \vec{ } command seems kind of "wimpy" to me.
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