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The dual nature of Nabla ?

  1. Feb 8, 2012 #1
    I didn't get the concept of dual or hybrid nature of nabla? I-e vector differential operator .. Is it means that nabla can produce a vector from scalar field (gradient) and scalar from vector field(divergence) ? What's the concept of Nabla's Dual nature ? Please explain..
     
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  3. Feb 8, 2012 #2

    tiny-tim

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    hi abrowaqas! :smile:

    ∇ behaves just like an ordinary vector …

    the vector itself is ∇f (where f is a scalar)​

    and you can dot-product it, or cross-product it, with a vector A …

    ∇.A (divA) or ∇xA (curlA) :wink:
     
  4. Feb 9, 2012 #3

    HallsofIvy

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    If you multiply a vector by a scalar, you get a vector. If you multiply a vector by a vector (dot product) you get a scalar. If you multiply a vector by a vector (cross product), you get vector.

    If that vector is "nabla" those three types of "multiplication" become
    [tex]\nabla f(x,y,z)= grad f(x,y,z)[/tex]
    [tex]\nabla\cdot\vec{f}(x,y,z)= div \vec{f}(x, y, z)[/tex]
    [tex]\nabla\times\vec{f}(x,y,z)= curl \vec{f}(x, y, z)[/tex]
     
  5. Feb 11, 2012 #4
  6. Feb 11, 2012 #5

    tiny-tim

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    oh, that's completely different from what i thought you were asking about! :rolleyes:

    that book (by bernard maxum) is saying that ∇ is both a vector and a derivative …

    in general (not cartesian) coordinates, the simple "dot" and "cross" procedure doesn't work …

    that's what it means by the "dual nature" of ∇ :smile:
     
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