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Homework Help: Tensors, Levi-Civita symbol

  1. Sep 11, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    If [tex]\epsilon _{ijjk}[/tex] is the Levi-Civita symbol:
    1)Demonstrate that [tex]\sum _{i} \epsilon _{ijk} \epsilon _{ilm}=\delta _{jl} \delta _{km} -\delta _{jm} \delta _{kl}[/tex].
    2)Calculate [tex]\sum _{ij} \epsilon _{ijk} \epsilon _{ijl}[/tex].
    3)Given the matrix M, calculate [tex]\sum _{ijk} \sum _{lmn} \epsilon _{ijk} \epsilon _{lmn} M_{il} M_{jm} M_{kn}[/tex].

    2. Relevant equations

    Maybe some properties on tensors but I'm not sure.

    3. The attempt at a solution
    I'm trying to start with 1) first. I'm so new with tensors that I don't understand well what I have to do.
    I know that Levi-Civita symbol is either worth -1, 0 or 1 though I didn't understand what is an even and odd permutation of ijk. And I know Kronecker's delta which is worth either 0 or 1, depending if i=j or not.
    So starting with [tex]\sum _{i} \epsilon _{ijk} \epsilon _{ilm}[/tex], can I assume i to go from 1 to 3? And what about j and k? Fixed constants which are either 1, 2 or 3?
     
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  3. Sep 12, 2010 #2

    vela

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    (132) would be an odd permutation of (123) because you can obtain it using an odd number of transpositions (swap of a pair), e.g. swap 2 and 3. (312) on the other hand would be an even permutation of (123) because obtaining it requires an even number of transpositions, e.g. swap 1 and 3 to get (321) and then swap 1 and 2 to get (312).

    http://en.wikipedia.org/wiki/Parity_of_a_permutation
    Yes, the index i runs from 1 to 3; j, k, l, and m are fixed. There aren't too many combinations of j, k, l, and m to check since you can easily see that most are 0.
     
    Last edited: Sep 12, 2010
  4. Sep 12, 2010 #3

    fluidistic

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    Thank you very much. Now I understand and I still find that there are lots of choices (4^4 I think) even though most of them are worth 0, I must show all. Part 1) solved.
    By the way I had seen the wikipedia article but didn't understand. Your words were much clearer to me. Now I'll try part 2.


    Edit: I'm a bit confused on the notation of the sum. There are 9 cases. I take the first as [tex]k=l=1[/tex]. Is [tex]\sum _{ij} \epsilon _{ijk} \epsilon _{ijl}[/tex] worth [tex]\epsilon _{111}\epsilon _{111}+\epsilon _{211}\epsilon _{211}+\epsilon _{311}\epsilon _{311}+\epsilon _{121}\epsilon _{121}+\epsilon _{131}\epsilon _{131}+\epsilon _{231}\epsilon _{231}+\epsilon _{321}\epsilon _{321}=0[/tex]?
     
    Last edited: Sep 12, 2010
  5. Sep 12, 2010 #4

    vela

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    The only non-vanishing terms are the last two, and they're both equal to 1, so the sum is equal to 2.
     
  6. Sep 12, 2010 #5

    fluidistic

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    Indeed you're right, I made a mistake in the last term of the sum, I considered it as -1 instead of 1. My question was about if my sum was right since I don't recall having used a double subindex in sums before.
    Now I'll try to tackle part 3).
    Nevermind, I see it's only a huge arithmetic "mess". I don't see any trick but to calculate the whole sum...
    Thanks for all, and you've taught me what was an even and odd permutation. :)
     
    Last edited: Sep 12, 2010
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