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Homework Help: Vector calculus: divergence of a cross product

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to prove the identity div (a x b) = b dot (curl a) - a dot (curl b)

    3. The attempt at a solution

    I've done the proof about 10 times now, and everytime I get the left hand of the identity equal to this:
    (all the d's are partial derivatives)
    d(a3b1)/dx - d(a2b1)/dx + d(a3b1)/dy - d(a1b3)/dy + d(a1b2)/dz - d(a2b1)/dz
    where vector a = a1i + a2j + a3k and vector b = b1i + b2j + b3k
    When I do the right hand side I get exactly the same thing above but doubled. So in affect I'm deriving 1 = 2. I'm sure there is an easy identity to manipulate the cross and dot products, but the brute force method should work and it's not, and I'm am completely lost as to where.
  2. jcsd
  3. Jan 26, 2010 #2


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    Hi elimenohpee! :smile:

    (have a curly d: ∂ and try using the X2 tag just above the Reply box :wink:)
    Why? :confused:

    You should. for example, be getting b1(∂a3/∂y - ∂a2/∂z) on the left, which is b1(curl a)1 :smile:
  4. Aug 3, 2010 #3
    I'm having the exact same problem, here's mine re-done with some tex:

    Prove [tex]\nabla \cdot \left( A \times B \right) = B \cdot \left( \nabla \times A \right) - \left( \nabla \times B \right) [/tex]
    Where A, B, C are vectors

    I started by working with the LHS, by finding the cross product then finding the divergence.

    [tex] \nabla \cdot \Left( \left( A_y B_z - A_z B_y \right) i + \left( A_z B_x - A_x B_z \right) j + \left( A_x B_y - A_y B_x \right) \Right) k [/tex]

    Then taking the partial derivatives [tex]\frac{\partial}{\partial x}[/tex] onto i (and y onto j, z onto k) I keep ending up with zero! Either my working of the cross product is wrong, or my partial derivatives are.
    Last edited: Aug 3, 2010
  5. Aug 3, 2010 #4


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    Your cross product is fine, so you're messing up the differentiation. The first term in the divergence will be

    [tex]\partial_x (A_yB_z-A_zB_y) = (\partial_x A_y) B_z + A_y (\partial_x B_z) - (\partial_x A_z)B_y - A_z(\partial_x B_y)[/tex]

    Is that what you got?
  6. Aug 3, 2010 #5
    I guess I don't know how to do partial deriv. properly, but I can see how you got that. So the second term would be
    [tex]\partial_y (A_zB_x-A_xB_z) = (\partial_y A_z) B_x + A_z (\partial_y B_x) - (\partial_y A_x)B_z - A_x(\partial_y B_z)[/tex]

    But from there do the partial derivatives just drop? So does the above 2nd term become
    [tex]B_x + A_z - B_z - A_x[/tex]

    I'm confused between
    [tex]\partial_x x = 1 [/tex]
    [tex]\partial_x y = ? [/tex]
    [tex]\partial_x xy = x \partial_x y + y \partial_x x = y [/tex]
    Last edited: Aug 3, 2010
  7. Aug 3, 2010 #6


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    No, the components of the vectors are functions of x, y, and z. You can't really do anything more with them. Now it's a matter of rearranging the terms until it looks like the other side of the equation. For instance, you can combine two of the terms you have so far like this:

    [tex]-A_z(\partial_x B_y)+A_z(\partial_y B_x) = -\A_z(\partial_x B_y - \partial_y B_x) = -A_z (\nabla \times B)_z[/tex]

    which is part of [itex]A\cdot(\nabla \times B)[/itex].
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