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Contracted Epsilon Identity

  1. Sep 23, 2015 #1
    Hi,

    I am confused about how I arrive at the contracted epsilon identity. [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

    1. The problem statement, all variables and given/known data


    Show that [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    From the relation between the Levi-civita symbol and the Kronecker delta, I compute [itex]\epsilon_{ijk} \epsilon_{imn}[/itex] by finding the determinant of the following matrix.

    [itex]\epsilon_{ijk} \epsilon_{imn} = det \left[ \begin{array}{cccc} \delta_{ii} & \delta_{im} & \delta_{in} \\ \delta_{ji} & \delta_{jm} & \delta_{jn} \\ \delta_{ki} & \delta_{km} & \delta_{kn} \end{array} \right][/itex] which yields

    [itex]\epsilon_{ijk} \epsilon_{imn} = \delta_{ii} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) - \delta_{im} (\delta_{ji} \delta_{kn} - \delta_{jn} \delta_{ki}) + \delta_{in} (\delta_{ji} \delta_{km} - \delta_{jm} \delta_{ki})[/itex]


    I am confused about how to progress.

    Thanks for any help you can give.
     
  2. jcsd
  3. Sep 23, 2015 #2

    Geofleur

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    For starters, what does ## \delta_{ii} ## equal? Note that you need to sum over repeated indices.
     
  4. Sep 23, 2015 #3
    [tex]\delta_{ii} = 3[/tex]

    This seems to be the only repeated indice.
     
  5. Sep 23, 2015 #4

    RUber

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    Note that your first term in the expansion: ##\delta_{ii} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})## looks a lot like the final result you are looking for.
    Then, the challenge should be to show that ##- \delta_{im} (\delta_{ji} \delta_{kn} - \delta_{jn} \delta_{ki}) + \delta_{in} (\delta_{ji} \delta_{km} - \delta_{jm} \delta_{ki}) = 0## in all cases. To do this, think about what must be true for any term to not be zero, and show that it implies another opposite term must also not be zero.
     
  6. Sep 23, 2015 #5

    RUber

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    You seem to be using a different definition of the Kronecker delta? Usually the only possible outcomes are 0 or 1.
     
  7. Sep 23, 2015 #6
    I thought the idea was that [itex]\delta_{ii}[/itex] implied the summation of [itex]\delta_{11}, \delta_{22}, and \delta_{33}[/itex], which are each respectively equal to 1.
     
  8. Sep 23, 2015 #7

    Geofleur

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    That's right, ## \delta_{ii} = \delta_{11} + \delta_{22} + \delta_{33} = 3 ##. Now eliminate the ## \delta ##'s in front of the other two terms and see what happens.
     
  9. Sep 23, 2015 #8

    RUber

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    I see. I was thinking one term at a time, rather than the sum over the terms.
    In that case, I get the same result as Geofleur.
     
  10. Sep 23, 2015 #9
    I am unsure of how to evaluate these deltas.

    [itex]\delta_{im} = 0[/itex] unless [itex]i = m[/itex] so do they both just vanish?
     
  11. Sep 23, 2015 #10

    Geofleur

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    They do vanish unless ## i = m ##. In that case, the ##\delta##'s in front become ones, and the ## i ##'s inside the parenthesized terms become ## m ##'s.
     
  12. Sep 23, 2015 #11

    Geofleur

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    Note that the ##i##'s are repeated and thus being summed over.
     
  13. Sep 23, 2015 #12
    ahh, I see.

    so I have that [itex]\epsilon_{ijk} \epsilon_{imn} = 3 \delta_{jm} \delta_{kn} - 3 \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} + \delta_{jn} \delta_{km} + \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/itex]

    Thank you very much for your help!
     
  14. Sep 23, 2015 #13

    Ray Vickson

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    The factor ##\epsilon_{ijk}## vanishes unless ##ijk## is a permutation of ##123##, so for any pair ##j \neq k## the required ##i## is uniquely determined. Then, for that ##i##, ##\epsilon_{imn}## vanishes unless ##mn## is a permutation of ##jk##. Thus, for a nonzero term on the left, we need either ##j = m## and ##k = n## (in which case the left-hand-side is ##(\pm1 )^2 = +1##, or ##j = n## and ##k = m## (in which case the left-hand-side is ##(-1)(+1) = -1##). That is, the nonzero values of the left-hand-side are the same as the nonzero values of ##\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}##. The same is true of the zero values, so the two sides must be equal.
     
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