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Levi-Civita Identity help

  1. Aug 29, 2011 #1
    Can somebody show me how

    [itex]\epsilon_{mni}a_{n}(\epsilon_{ijk}b_j c_{k})[/itex]

    Turns in to

    [itex]\epsilon_{imn}\epsilon_{ijk}a_{n}b_j c_{k}[/itex]


    Something about the first [itex]\epsilon[/itex] I'm not seeing here when the terms are moved around.
     
    Last edited: Aug 29, 2011
  2. jcsd
  3. Aug 29, 2011 #2

    vela

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    The tensor changes sign when you transpose two indices, right? A cyclic permutation is an even number of transpositions, so [itex]\epsilon_{mni} = \epsilon_{nim} = \epsilon_{imn}[/itex].
     
  4. Aug 30, 2011 #3
    When [itex]\epsilon_{ijk}[/itex] and [itex]a_{n}[/itex] change places the [itex]\epsilon_{mni}[/itex] changes to a cyclic permutation that is still positive and [itex]\epsilon_{mni} =\epsilon_{imn}=\epsilon_{nim} [/itex] but each one of these will give a different final answer.

    I don't see how [itex]\epsilon_{mni} [/itex] turns to [itex]\epsilon_{imn} [/itex] when [itex]\epsilon_{ijk} [/itex] doesn't change.

    This is from a problem proving the A X (B X C) = (A*C)B-(A*B)C identity.
     
  5. Aug 30, 2011 #4

    vela

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    You can swap [itex]a_n[/itex] and [itex]\epsilon_{ijk}[/itex] because real numbers commute. Swapping them has nothing to do with reordering the indices of [itex]\epsilon_{mni}[/itex].

    [itex]\epsilon_{mni} = \epsilon_{imn}[/itex] for all i, m, and n, so you can simply replace [itex]\epsilon_{mni}[/itex] with [itex]\epsilon_{imn}[/itex] in the summation. There's no relabeling of indices going on if that's what you think is happening.
     
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