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Vector product

  1. May 4, 2005 #1
    I looked up the formula for [tex]\nabla \times (\vec{A} \times \vec{B})[/tex]:

    [tex]\nabla \times (\vec{A} \times \vec{B}) = \vec{A}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{A}) + (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B}[/tex]

    What does a vector followed by a del mean? Mathworld says that in the context of a unit vector it's the directional derivative. It's unclear to me how this works because then you have a vector times a vector for the last two terms. Can someone please clarify how this works?
  2. jcsd
  3. May 4, 2005 #2
    Edit: IGNORE!!

    [tex]{\vec A} \cdot \nabla = \nabla \cdot {\vec A}.[/tex]

    The directional derivative is a scalar.
    Last edited: May 5, 2005
  4. May 4, 2005 #3
    Ah, ok. Dot products are commutative. Thanks.
  5. May 5, 2005 #4


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    The expression [itex](\vec A \cdot \vec \nabla )[/itex] is an operator, like [itex]\vec \nabla[/itex]. It has to work on a vector. You can see what it does by treating [itex]\vec \nabla[/itex] as a genuine vector:

    [tex]\vec A \cdot \vec \nabla = A_x \frac{\partial}{\partial x}+A_y \frac{\partial}{\partial y}+ A_z\frac{\partial}{\partial z}[/tex]

    Ofcourse, you can always check it by writing out the components :P
    Last edited: May 5, 2005
  6. May 5, 2005 #5
    oops, yeah, I'm insane! Though I should note that it can actually also work on a scalar.
    Last edited: May 5, 2005
  7. May 5, 2005 #6


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