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Einstein summation convention proof

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Using the Einstein summation convention, prove:

    A[tex]\bullet[/tex]B[tex]\times[/tex]C = C[tex]\bullet[/tex]A[tex]\times[/tex]B


    2. Relevant equations



    3. The attempt at a solution
    I tried to follow an example from my notes, but I don't entirely understand it. Would it be possible to find out if what I've done (below) is correct, or where I went wrong?

    (B[tex]\times[/tex]C)[tex]_{i}[/tex] = [tex]\epsilon[/tex][tex]_{ijk}[/tex]B[tex]_{j}[/tex]C[tex]_{k}[/tex]

    (A[tex]\bullet[/tex]B[tex]\times[/tex]C)[tex]_{i}[/tex] = [tex]\epsilon[/tex][tex]_{ijk}[/tex]A[tex]_{i}[/tex]B[tex]_{j}[/tex]C[tex]_{k}[/tex]

    = [tex]\epsilon[/tex][tex]_{kij}[/tex]A[tex]_{i}[/tex]B[tex]_{j}[/tex]C[tex]_{k}[/tex]

    =C[tex]\bullet[/tex]A[tex]\times[/tex]B

    Thanks!
    PS. I'm not very familiar with Latex and couldn't get the symbols to line up properly.. sorry! They should all be subscripts.
     
  2. jcsd
  3. Sep 28, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think thats about right. The thing is just to show that e_{ijk}=e_{jki}, right?
     
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