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Divergence of a cross product help

  1. Mar 13, 2013 #1
    I need to prove the identity: [itex]\nabla[/itex]([itex]\vec{A} \times \vec{B}[/itex])=[itex]\vec{B} \bullet[/itex]([itex]\nabla \times \vec{A}[/itex]) - [itex]\vec{A} \bullet[/itex]( [itex]\nabla \times \vec{B}[/itex])

    I need to prove for an arbitrary coordinate system, meaning I have scaling factors.

    The proof should be quite straight forward if you use the levi chevita symbol but so far this is what i have:

    [itex]\nabla[/itex]([itex]\vec{A} \times \vec{B}[/itex])=[itex]\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}[/itex]( [itex]\frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}[/itex]) = [itex]\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}[/itex]( [itex]\frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}[/itex]) [itex]\ast \hat{e_{i}} \bullet \hat{e_{i}}[/itex]

    But I can't see how to take it into the desired format
    I think I should squeeze in a kroncker delta somehow, but I'm not sure.
     
  2. jcsd
  3. Mar 13, 2013 #2

    vela

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    Do you know how to do the proof without the scaling factors?
     
  4. Mar 13, 2013 #3
    Yes, I just tried and it works fine even without the addition of unit vectors and the delta.
    My problem is that I cannot take the scaling factors in or out of the derivative.
    Perhaps I missed somthing?
     
  5. Mar 13, 2013 #4

    SteamKing

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    Would these scaling factors perhaps be constants? Or are they functions of one or more of the coordinates?
     
  6. Mar 13, 2013 #5
    Of course they are functions, consider for instance spherical coordinates
     
  7. Mar 13, 2013 #6

    vela

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    I'd try simplifying ##\frac{h_1 h_2 h_3}{h_i}## to ##h_j h_k##. Then I think you just have to apply the product rule and grind it out.
     
  8. Mar 13, 2013 #7
    This is my idea too but how can take the scaling factor out of the derivative?
     
  9. Mar 13, 2013 #8

    lurflurf

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    I don't know why you want to fool with h or moving it. Just use the chain rule to hold each vector fixed in turn.

    $$\dfrac{\mathrm{d}}{\mathrm{dt}}u=\dfrac{\mathrm{dx}}{\mathrm{dt}}\dfrac{\partial}{\partial x}u+\dfrac{\mathrm{dy}}{\mathrm{dt}} \dfrac{\partial} {\partial y}u$$

    in your example

    $$\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}) =\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} (B_{k})_\mathrm{fixed}) +\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} (A_{j})_\mathrm{fixed} B_{k}) $$
     
  10. Mar 13, 2013 #9
    Ok but this doesn't prove the identity, if you look at the definition of the curl in curvilinear coordinates
     
  11. Mar 13, 2013 #10

    lurflurf

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    Once variables are fixed they can move through derivatives and the identity is easily proven

    $$\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}) =(B_{k})_\mathrm{fixed} \frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} ) +(A_{j})_\mathrm{fixed} \frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} B_{k}) $$
     
  12. Mar 13, 2013 #11

    vela

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    Hint:
    $$\nabla\cdot(\vec{A} \times \vec{B})
    = \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial x_i} [h_j h_k \epsilon_{ijk} A_{j} B_{k}]
    = \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial x_i} [\epsilon_{ijk} (h_j A_j) (h_k B_k)]$$
     
  13. Mar 13, 2013 #12
    thanks a lot
     
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