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Homework Help: Indicial notation proof

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Use the identity
    [tex]
    \epsilon_{ijk} \epsilon_{klm} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}
    [/tex]
    as a shortcut to obtain the following results:
    a) [tex]\epsilon_{ilm} \epsilon_{jlm}= 2\delta_{ij}[/tex]
    b) [tex]\epsilon_{ijk} \epsilon_{ijk} = 6[/tex]
    2. Relevant equations


    3. The attempt at a solution
    I tried to solve that by solving the determinant [tex]\epsilon_{ilm} \epsilon_{jlm}= [\delta][/tex] but the result just became zero. or even, I tried to change i=k and j=l in the first equation but the result was zero as well. I don't know what should I do, I'm really stuck.
     
  2. jcsd
  3. Sep 9, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member


    This is incorrect, on the LHS only the indices [itex]i[/itex], [itex]j[/itex], [itex]l[/itex], and[itex]m[/itex] are free indices ([itex]k[/itex] is being summed over, so it is called a dummy index), so only those indices should appear on the RHS side of the identity. Instead, you should have:

    [tex]\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}
    [/tex]

    Start by using the fact that [itex]\epsilon_{ijk}[/itex] is invariant under cyclic transposition of indices so that [itex]\epsilon_{jlm}=\epsilon{mjl}[/itex] and hence, [itex]\epsilon_{ilm} \epsilon_{jlm}=\epsilon_{ilm} \epsilon_{mjl}[/itex], which is in the same form as your identity with [itex]j \to l[/itex], [itex]k\to m[/itex], [itex]l\to j[/itex] and [itex]m\to l[/itex]... What does that give you (show your calculatiuons)?
     
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