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Proof using Levi-Civita symbol

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]\sum_{j,k}[/tex] [tex]\epsilon_{ijk}[/tex] [tex]\epsilon_{ljk}[/tex] = 2[tex]\delta_{il}[/tex]



    2. Relevant equations
    [tex]\epsilon_{ijk}[/tex] [tex]\epsilon_{ljk} = [tex]\delta_{il}[/tex]([tex]\delta_{jj}[/tex][tex]\delta_{kk}[/tex] - [tex]\delta_{jk}[/tex][tex]\delta_{kj}[/tex]) + [tex]\delta_{ij}[/tex]([tex]\delta_{jk}[/tex][tex]\delta_{kl}[/tex] - [tex]\delta_{jl}[/tex][tex]\delta_{kk}[/tex]) + [tex]\delta_{ik}[/tex]([tex]\delta_{jl}[/tex][tex]\delta_{kk}[/tex] - [tex]\delta_{jj}[/tex][tex]\delta_{kl}[/tex])


    3. The attempt at a solution

    Okay, in cases where subscripts of the Kronecker delta are equal, then [tex]\delta_{jj}[/tex] = 1.

    If the subscripts are not equal, then [tex]\delta_{il}[/tex] = 0.

    So plugging those into the parenthesis of the above equation gives me:

    [tex]\delta_{il}[/tex]([tex]\delta_{jj}[/tex][tex]\delta_{kk}[/tex]) ?

    If that is the case, then how could the two inside the parenthesis equal 2? I know I must be missing something.
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2

    learningphysics

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    In your formula, replace the [tex]\delta_{jj}[/tex], [tex]\delta_{kk}[/tex] etc... where the variables are the same... with 1.

    Also, [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex]
     
  4. Sep 11, 2007 #3
    If [tex]\delta_{ij}\delta_{jk} = \delta_{ik}[/tex] does that mean that [tex]\delta_{lk}\delta_{kj} = \delta_{lj}[/tex] and so on?
     
  5. Sep 11, 2007 #4

    learningphysics

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    Yes, exactly.
     
  6. Sep 11, 2007 #5
    Okay, I think one more question will help me get it.

    [tex]\delta_{jk}[/tex][tex]\delta_{kj}[/tex] = ?
     
  7. Sep 11, 2007 #6

    learningphysics

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    = [tex]\delta_{jj} = 1[/tex]
     
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