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Proof for div (u cross v)

  1. Sep 26, 2009 #1
    Please show me the way the proof this!

    div (u cross v) = v dot grad (u)- u dot grad(v) where v and u is a vector.

    The product rule doesn't seem working
     
  2. jcsd
  3. Sep 26, 2009 #2

    Hootenanny

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    The "product rule" in vector calculus is different to the product rule for scalar functions. The "vector product rule" depends on type of product you are dealing with (e.g. cross, scalar) and the type of operator that you are dealing with (e.g. divergence, curl). in fact, the relationship that you wish to prove is the product rule for the divergence of a cross product.

    The most straightforward way to prove your relation is a by direct computation. You should know that

    [tex]\boldsymbol{u}\times\boldsymbol{v} = \varepsilon_{ijk}u_jb_k[/tex]

    And that

    [tex]\text{div}\left(\boldsymbol{A}\right) = \frac{\partial A_i}{\partial x_i}[/tex]

    So all you need to do, is put the two together and compute. (Be careful with your summation indices).
     
  4. Sep 26, 2009 #3

    lurflurf

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    that should be

    [tex]\mathbf{\nabla\cdot (u\times v)=v\cdot \nabla\times u-u\cdot \nabla\times v}[/tex]

    div (u cross v) = v dot curl (u)- u dot curl(v)
    use partial differentiation
    [tex]\mathbf{\nabla\cdot (u\times v)=\nabla_u\cdot (u\times v)+\nabla_v \cdot (u\times v)}[/tex]
    div (u cross v)=div_u (u cross v)+div_v (u cross v)
    recall
    a dot b cross c=-b dot a cross c==c dot a cross b
    thus
    [tex]\mathbf{\nabla_u\cdot (u\times v)=v\cdot \nabla\times u}[/tex]
    [tex]\mathbf{\nabla_v\cdot (u\times v)=-u\cdot \nabla\times v}[/tex]
    div_u (u cross v)= v dot curl u
    div_v (u cross v)=-u dot curl v

    therefore

    div (u cross v)=v dot curl (u)- u dot curl(v)

    Introducing an arbitrary coordinate system and a double sum is not very straight forward.
    Calculus students hate epsilons and deltas. :)

    vector calculus favorite
    [tex]\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]
     
    Last edited: Sep 26, 2009
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