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Basic tensor equation

  1. Sep 19, 2015 #1
    I'm new to working with tensors, and feel a bit uneasy about the nomenclature. I picture words like antisymmetry in terms of average random matrices where no symmetry can be found at all. However, if I understand it correctly, antisymmetry is a type of symmetry, but where signs are inverted. So with this nomenclature Reagan, in that famous joke, is right to label anticommunists as a type of communists. But, anyway that's just linguistic mishmash.

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

    [itex]A_{ij} = k \epsilon_{ijk} a_k[/itex]

    k is a constant. [itex]\vec{a}[/itex] is a vector.

    What must k be to fulfill [itex]A_{ij} A_{ij} = |\vec{a}|^2[/itex] ?

    3. The attempt at a solution

    The dual use of k also confuses me somewhat but from what I understand the only constant is the non-index k.

    Is it correct to assume that [itex]\vec{a}[/itex] is taken to exist in three dimensions, from the use of the 3-index levi cevita?

    [itex]A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijk} a_k = k^2 \epsilon_{ijk} \epsilon_{ijk} a_k a_k = 6 k^2 a_k a_k[/itex]

    So if k is [itex]\pm \frac{1}{\sqrt{6}}[/itex] it appears to fulfill the requirements, but this isn't the correct answer.
  2. jcsd
  3. Sep 19, 2015 #2


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    Gold Member

    Hey, the first equality on the last equation you wrote violates the Einstein summation convention.
  4. Sep 19, 2015 #3

    Would this be a correction?

    [itex]A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijn} a_n [/itex]

    Meaning I need to find [itex]\epsilon_{ijk} \epsilon_{ijn}[/itex]? It seems like [itex]n \neq k[/itex] terms amount to zero, because it implies that there's a duplicate index number in one of the levi-civitas, leaving only ones involving [itex]a_k^2[/itex]'s.

    With an interpretation of k like this:
    [itex]A_{ij} = k \sum_k \epsilon_{ijk} a_k[/itex]

    I see I have double ij on both sides though, which may violate the summation convention still.

    I know there's an identity that relates [itex]\epsilon_{ijk} \epsilon_{lmn}[/itex] to Kroneckers. And I can show that both sides are equal, but I don't understand how one would arrive at that equality by "brute force", i.e. what's the natural thought behind its derivation? I see that the positive contributions come from the permutations that return 1*1 (but not -1*-1 mysteriously enough - do they get redundant? edit: yes they seem to contribute a double) and that the negatives come from 1*(-1).

    [itex]\epsilon_{ijk} \epsilon_{lmn}= \delta_{il} \delta_{jm} \delta_{kn} + \delta_{im} \delta_{jn} \delta_{kl} + \delta_{in} \delta_{jl} \delta_{km} - \delta_{im} \delta_{jl} \delta_{kn} - \delta_{il} \delta_{jn} \delta_{km}- \delta_{in} \delta_{jm} \delta_{kl}[/itex]

    In the books they never explain how they construct the above in the first place. It's a kind of a jump. I mean it's possible to memorize it by using mnemonics (rotation of indices etc), but I would like to understand how to derive it from a blank piece of paper, or do I have to accept that the "wheel has been invented" on this matter?

    One attempt I made at that was to list all levi civitas that return 1, i.e. : {(1,2,3), {2,3,1}, {3,1,2}), then to pair them with (i,j,k) and (l,m,n). So, with i=1, j=2, k=3 I looked up what a selection of l, m and n would have to be, for example 1=m, 2=n, 3=l. Another way would be to make some kind of truth table to make the levi civita - kronecker translation.
    Last edited: Sep 19, 2015
  5. Sep 19, 2015 #4

    Ray Vickson

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    Let's say ##A_{ij} = c \epsilon_{ijk} a_k##. Now ##\epsilon \neq 0## requires that ##i,j,k## all be different. Your quantity
    [tex] A_{ij} A_{ij} = c^2 \sum_{i,j: i \neq j} \sum_{k,n} \epsilon_{ijk} \epsilon_{ijn} a_k a_n [/tex]
    Nonzero terms in the sum need ##k \neq i,j## and ##n \neq i,j##, hence ##k = n##. That is,
    [tex] A_{ij}A_{ij} = c^2 \sum_{i,j: i \neq j} \sum_{k: k \neq i,j} (\epsilon_{ijk})^2 a_k^2, [/tex]
    and ##\epsilon^2 = 1## for the non-zero terms. For each pair ##(i,j)## there is only one ##k##, and as we vary the pair ##(i,j)## we sweep over all ##k = 1,2,3##. Furthermore, there will be an equal number of ##k=1##, ##k=2## and ##k=3## terms. Thus, we have ##A_{ij}A_{ij} = K c^2 \sum_{k=1}^3 a_k^2##, with a constant ##K## that is not too difficult to figure out.
  6. Sep 19, 2015 #5
    Thanks, it's a bit tricky, but I think I get this part now. It's probably a good idea to write it out in sums like you do, at least at the moment. I get K=2, which will lead the the correct result.
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