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Divergence of the curl problem question

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data
    if a vector can be written as the curl of another vector, its divergence vanishes. Can you justify the statement: "any vector field whose divergence vanishes identically can be written as the curl of some other vector"?

    2. Relevant equations

    Prove this by construction. Let [tex]{\nabla}{\cdot}{\vec{V}}=0[/tex] and try to find any vector, U, for which [tex]{\vec{V}}={\nabla}{\cdot}{\vec{U}}[/tex]
    This amounts to showing that you have enough freedom to pick as components of \vec{U}, functions which satisfy some simple differetial equations. Simplify somewhat by trying to find a vector, [itex]{\vec{U}}[/itex], for which, say, [itex]{\vec{U_z}}=0[/itex]

    3. The attempt at a solution

    I know that for [tex]{\nabla}{\cdot}{\vec{V}}=0[/tex] we have to have for example, a vector field such as: [tex]{\vec{V}}={yz}\hat{x}+{xz}\hat{y}+{xy}\hat{z}[/tex] so that when we do [tex]{\nabla}{\cdot}{\vec{V}}=0[/tex]. I'm confused if the problem is just asking to prove the divergence of the curl is equal to 0 which I have already done a few homework problems ago or if its asking for something different here...because I'm confused by the hint that it's giving me (in relevant equations).
    Last edited by a moderator: Jan 14, 2013
  2. jcsd
  3. Jan 14, 2013 #2
    I believe there is a typo in your relevant equations. They want you to show that if
    [tex]\nabla \cdot \vec{V} = 0[/tex]
    for any particular vector field [itex]\vec{V}[/itex], then you can always find a vector field [itex]\vec{U}[/itex] such that
    [tex]\nabla \times \vec{U} = \vec{V}[/tex]
    They want you to construct a vector field U by x, y and z components that satisfies the second equation, using the equation for the divergence of V being 0 as the only given fact.
    This is actually similar to a proof you may have done earlier: in 3-dimensional space, if [itex]\vec{A}\cdot\vec{B}=0[/itex], then there is a vector [itex]\vec{C}[/itex] such that [itex]\vec{A}\times\vec{C}=\vec{B}[/itex].
    Last edited: Jan 14, 2013
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