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Help! Proof of an identity?

  1. Oct 20, 2005 #1
    I tried to prove

    div(F x G) = G.(curlF) - F.curl(G)

    and ended up getting the right hand side equaling twice the left hand side, with no idea where i'd gone wrong :(

    can someone show me how to prove it correctly?

    and, if you have time.. see if you can pick where i went wrong?

    this is how I attempted to do it:

    (I'm going to stop with the bolding for vectors because it's too annoying)

    F x G = (f2g3 - f3g2)i + (f3g1 - f1g3)j + (f1g2 - f2g1)k

    div(F x G) = d/dx(f2g3 - f3g2) + d/dy(f3g1 - f1g3) + d/dz(f1g2 - f2g1)

    curlF = (df3/dy - df2/dz)i + (df1/dz - df3/dx)j + (df2/dx - df1/dy)k

    G.(curlF) = d/dx(f2g3 - f3g2) + d/dy(f3g1 - g3f1) + d/dz(f1g2 - f2g1)

    i can see i'm already in trouble here... this is the left hand side already...

    then I used the same method for F.(curlG) which gets me the negative of G.(curlF) , so that when I take it from G.(curlF) I get twice the left hand side...

    Any help will be greatly appreciated
  2. jcsd
  3. Oct 20, 2005 #2


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    Your expression for G.(curlF) is just wrong. Correct expression x derivative term is g3(df2/dx)-g2(df3/dx). When you work out the rest you will see the original formula is correct.
  4. Oct 20, 2005 #3
    ...still stuck :(

    I understand what you corrected me on, but once i've worked out G.(curlF) - F.(curlG) I can't see what you could do to it to make it the left hand side of the equation.

    G.(curlF) - F.(curlG) = g1(df3/dy - df2/dz) + g2(df1/dz - df3/dx) + g3(df2/dx - df1/dy) - f1(dg3/dy - dg2/dz) - f2(dg1/dz - dg3/dx) - f3(dg2/dx - dg1/dy)


    then where do I go from there?
  5. Oct 20, 2005 #4


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    Let's take the d/dx term. You have
    -g2df3/dx+g3df2/dx+f2dg3/dx-f3dg2/dx, which is
    d(f2g3)/dx-d(f3g2)dx, which is what you want.
  6. Oct 20, 2005 #5
    ah... thanks a bunch...


    g3df2/dx + f2dg3/dx = d(f2g3)dx

    Is that just something I should've known? or are there some intermediate steps there?
  7. Oct 21, 2005 #6

    matt grime

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    it's the product rule. traditionally learning that d(uv)/dx = vdu/dx+udv/dx comes well before you get onto div and curl.
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