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Vector Calc: Div Identity

  1. Aug 11, 2009 #1
    1. The problem statement, all variables and given/known data
    div(grad f x grad g)=0. I need to prove this somehow.

    2. Relevant equations

    3. The attempt at a solution

    I dont really know where to even start this at >.< any help is greatly appreciated.
  2. jcsd
  3. Aug 11, 2009 #2


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    Write grad(f) and grad(g) out in components and take their cross product. Now take div of that. Use that mixed partial derivatives are equal e.g. d^2(f)/(dxdy)=d^2(f)/(dydx). That will get the job done.
  4. Aug 12, 2009 #3
    Personally, I like using index notation for this type of problem; it's a good deal more compact and, once you get used to it, very transparent. Instead of writing out all the components longhand, write [tex] \nabla f \times \nabla g = \tilde{\varepsilon}^{ijk} f_{,i} g_{,j} \hat{\mathbf{e}}_{k} [/tex] , where [tex] \tilde{\varepsilon} [/tex] is the Levi-Civita symbol (not being careful about index placement here, since we're in flat space). Then you have
    \nabla \cdot (\nabla f \times \nabla g) &= \tilde{\varepsilon}^{ijk} (f_{,i} g_{,j})_{,k}\\
    &= \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} + \tilde{\varepsilon}^{ijk} f_{,i} g_{,j,k} \textrm{.}
    Both summands in the last equation vanish. Indeed, let [tex] P = \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} [/tex] . Then, since partials commute, we have [tex] P = \tilde{\varepsilon}^{ijk} f_{,k,i} g_{,j} [/tex] ; by a trivial relabeling of dummy indices, this gives [tex] P = \tilde{\varepsilon}^{kji} f_{,i,k} g_{,j} = -\tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} = -P [/tex] (since [tex] \tilde{\varepsilon} [/tex] is completely antisymmetric), so [tex] P = 0 [/tex]. The logic for the other summand is identical.
  5. Aug 13, 2009 #4


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    Alternatively, if you've already proven the following identities in class, you can simply combine them with [itex]\textbf{A}=\mathbf{\nabla}f[/itex] and [itex]\textbf{B}=\mathbf{\nabla}g[/itex] and the result becomes apparent.



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