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Homework Help: Vector identity proof using index notation

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Using index notation to prove

    [tex]\vec{\nabla}\times\left(\vec{A}\times\vec{B}\right) = \left(\vec{B}\bullet\vec{\nabla}\right)\vec{A} - \left(\vec{A}\bullet\vec{\nabla}\right)\vec{B} + \vec{A}\left(\vec{\nabla}\bullet\vec{B}\right) - \vec{B}\left(\vec{\nabla}\bullet\vec{A}\right)[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I tried converting the right side first...
    [tex]B_{a}A_{k,a}-A_{a}B{k,a}+A_{a}\epsilon_{abc}B_{b,c}-[/tex][tex]B_{a}\epsilon_{abc} A_{b,c}[/tex]

    but i don't really know if this is right... i'm very very new to the index notation system. i don't really understand the rules beyond the basics. will someone please give me some clue as to where to go from here?
  2. jcsd
  3. Sep 15, 2010 #2
    Systems of index notation vary, so you may need to give a reference or a description of your system. Using the index notation I know, the del operator is written [tex]\partial_a[/tex], so the right side becomes [tex]B^b \partial_b A^a - A^b \partial_b B^a + A^a \partial_b B^b - B^a \partial_b A^b[/tex], and the left side becomes [tex]\mathbf{\nabla}\times \epsilon_{abc} A^b B^c = \epsilon^{ade} \partial_d \epsilon_{ebc} A^b B^c[/tex]. From there, try to figure out how to contract the epsilons on the left against each other. You definitely shouldn't have any epsilons on the right side, because all you have is dot products, they're not antisymmetric.
  4. Sep 16, 2010 #3
    Thanks for your response. I'm not sure what system the professor is using. He doesn't use superscripts at all and uses commas for derivatives. He didn't give a name for it other than "index notation" though.

    is it okay to just change [tex]\epsilon_{ade}\partial_{d}\epsilon_{ebc}A_{b}B_{c}[/tex] to [tex]\epsilon_{ade}\epsilon_{ebc}\partial_{d}A_{b}B_{c}[/tex] or does this mess with the meaning?
    Last edited: Sep 16, 2010
  5. Sep 16, 2010 #4


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    That's fine. It's what makes index notation so useful in deriving identities like this.
  6. Sep 16, 2010 #5
    okay, so I've got

    [tex]\epsilon_{ade}\epsilon_{ebc}\partial_{d}A_{b}B_{c} =
    \left(\delta_{ab}\delta_{dc}-\delta_{ac}\delta_{db}\right)\partial_{d}A_{b}B_{c} =
    \partial_{d}A_{a}B_{d}-\partial_{d}A_{d}B_{a} =

    but I can't figure out how to get that to have four terms, like on the right... It seems like I'm missing something here. Do I just keep adding onto terms in the second part with A, B, & del in different orders until I get something that looks right?
  7. Sep 16, 2010 #6


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    You have to be a bit more careful. Remember the derivative acts on the product of Aa and Bc, so you need to use the product rule before you can simplify the indices using the Kronecker deltas.
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