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Divergent free question

  1. Nov 24, 2013 #1


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    Gold Member

    hey pf!

    so if i have a vector field [itex] \vec{V}[/itex] and i know [itex] \nabla \cdot \vec{V}=0[/itex] would i be able to express [itex] \vec{V}[/itex] in the following manner: [itex] \vec{V}= \nabla \times \vec{f}[/itex] for some [itex] \vec{f}[/itex]since we know this automatically satisfies the divergent free requirement?

    if not, what must be assumed in order to claim that such an [itex] \vec{f}[/itex] exists?

    thanks for your time!

  2. jcsd
  3. Nov 25, 2013 #2


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    Homework Helper

    Yes. You have some freedom in choosing [itex]\vec f[/itex] since [itex]\nabla \times (\vec f + \nabla \phi) = \nabla \times \vec f = \vec V[/itex] for any scalar field [itex]\phi[/itex].
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