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Is the curl of a div. free vector field perpendicular to the field?

  1. Aug 31, 2013 #1
    Hi PF-members.

    My intuition tells me that: Given a divergence free vector field [itex] \mathbf{F} [/itex], then the curl of the field will be perpendicular to field.
    But I'm having a hard time proving this to my self.

    I'know that : [itex] \nabla\cdot\mathbf{F} = 0 \hspace{3mm} \Rightarrow \hspace{3mm} \exists\mathbf{A}: \mathbf{F} = \nabla\times\mathbf{A} [/itex]

    Therefore : [itex] \mathbf{F}\cdot(\nabla\times\mathbf{F}) = 0 \hspace{3mm} \Rightarrow \hspace{3mm} [\nabla\times\mathbf{A}]\cdot[\nabla\times(\nabla\times\mathbf{A})] = 0 [/itex]

    But I can't prove that this actually equals zero... Please help!!
  2. jcsd
  3. Aug 31, 2013 #2
    Off the top of my head (as in, there are probably less artificial examples):

    Let [itex]\mathbf{F} = (x-z,y-z,-2z)[/itex]. As [itex]\nabla \cdot \mathbf{F} = 1+1-2=0[/itex], the field is divergence-free. However, [itex]\nabla \times \mathbf{F} = (1,-1,0)[/itex], so, clearly, [itex]\mathbf{F}\cdot (\nabla \times \mathbf{F})\not{\equiv}0[/itex] and by contradiction, your intuition seems to be wrong.
    Last edited: Aug 31, 2013
  4. Aug 31, 2013 #3
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