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Levi-civita and symmetric tensor

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric.


    2. Relevant equations



    3. The attempt at a solution
    The first bit I think is just like the proof that a symmetric tensor multiplied by an antisymmetric tensor is always equal to zero.
    [tex]\epsilon_{ijk} = - \epsilon_{jik}[/tex] As the levi-civita expression is antisymmetric and this isn't a permutation of ijk.
    [tex]\epsilon_{ijk}a_{ij} = - \epsilon_{jik}a_{ij}[/tex]
    [tex]\epsilon_{ijk}a_{ij} = - \epsilon_{jik}a_{ji}[/tex] As [tex]a_{ji}[/tex] is symmetric.
    [tex]\epsilon_{ijk}a_{ij} = - \epsilon_{ijk}a_{ij}[/tex] Swapping dummy indicies.
    [tex]2\epsilon_{ijk}a_{ij} = 0[/tex]
    [tex]\epsilon_{ijk}a_{ij} = 0[/tex]

    I think that proves the "if [tex]a_{ij}[/tex] is symmetric" part, but not the "and only if" part. There's a section in my text that claims a tensor is symmetric in similar based on "contracting with [tex]epsilon_{kqp}[/tex]" and using the relationship between the Levi-civita equation and the Kronecker delta:
    [tex]\epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}[/tex]
    But I didn't understand what it meant by that.

    So, here goes:
    [tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = \epsilon_{kqp}(0) = 0[/tex]

    [tex](\epsilon_{kqp}\epsilon_{ijk})a_{ij} = (\delta_{qi}\delta_{pj} - \delta_{qj}\delta_{pi})a_{ij}[/tex]

    [tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = (\delta_{qi}\delta_{pj})a_{ij} - (\delta_{qj}\delta_{pi})a_{ij}[/tex]

    [tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = a_{qp} - a_{pq}[/tex]
    Which is zero only where a is symmetric, right?

    If someone could please tell me if I'm on the right track or if I've done something completely wrong, that'd be really handy. I've been stuck on this one for hours and only just thought of those last two lines while I was writing out the equations on here, but I really don't know if it's right.
     
    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 3, 2009 #2
    All tensors can be separated into a symmetric and antisymmetric part.

    Tij = bSij + aAij , sometimes written Tij = bS(ij) + aA[ij].

    Can you prove this?

    Operate the Levi-Civita tensor on each part.
     
  4. Oct 4, 2009 #3
    I can't prove that. It's stated without proof in my course readings and I haven't managed to figure it out. I don't know how to operate the Levi-Civita tensor either (yes, I'm in trouble, I know).
    So is this a way to answer the whole question? And what I had done isn't right at all?
     
  5. Oct 4, 2009 #4
    OK, you don't have to do the proof, but can just use it. That's plenty good enough.

    I think you should try the direct approch to get a feel for how tensor multiplication works by substituting 1, 2, and 3 into the Levi-Civita tensor.

    Remember, [itex]\epsilon_{ijk}[/itex] = 1 for even permutations of (ijk), [itex]\epsilon_{ijk}[/itex] = -1 for odd permutations of of (ijk), and [itex]\epsilon_{ijk}[/itex] = 0 for every other combination.

    All together, there are only 6 non-zero elements in [itex]\epsilon_{ijk}[/itex]. You need to write them out for [itex]\epsilon_{ijk}S_{ij}[/itex] and [itex]\epsilon_{ijk}A_{ijk}[/itex]. I'm using S to represent a symmetric tensor and A for an antisymmetric tensor.

    [tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + 4\ more\ tems = 0[/tex]
     
  6. Oct 5, 2009 #5
    Ummmmerrrrrr:
    [tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + \epsilon_{231}S_{23} + \epsilon_{321}S_{32} + \epsilon_{213}S_{21}+ \epsilon_{132}S_{13}[/tex]

    [tex]\epsilon_{ijk}S_{ij} = (1)S_{12} + (1)S_{31} + (1)S_{23} + (-1)S_{32} + (-1)S_{21} + (-1)S_{13}[/tex]

    [tex]\epsilon_{ijk}S_{ij} = S_{12} + S_{31} + S_{23} - S_{23} - S_{12} - S_{31} = 0[/tex] because [tex]S_{ij}=S_{ji}[/tex] ?

    Following the same logic for the antisymmetric side:
    [tex]\epsilon_{ijk}A_{ij} = \epsilon_{123}A_{12} + \epsilon_{312}A_{31} + \epsilon_{231}A_{23} + \epsilon_{321}A_{32} + \epsilon_{213}A_{21}+ \epsilon_{132}A_{13}[/tex]

    [tex]\epsilon_{ijk}A_{ij} = (1)A_{12} + (1)A_{31} + (1)A_{23} + (-1)A_{32} + (-1)A_{21} + (-1)A_{13}[/tex]

    [tex]\epsilon_{ijk}A_{ij} = A_{12} + A_{31} + A_{23} + A_{23} + A_{12} + A_{31} = 2A_{12} + 2A_{31} + 2A_{23}[/tex] because [tex]A_{ij} = -A_{ji}[/tex]

    So, from here I can say that [tex]\epsilon_{ijk}T{ij} = 0[/tex] whenever [tex]T_{ij}[/tex] is a symmetric tensor, and if [tex]\epsilon_{ijk}T_{ij} = 0[/tex], [tex]T_{ij}[/tex] must be symmetric, as if [tex]T_{ij}[/tex] had an antisymmetric component then [tex]\epsilon_{ijk}T_{ij}[/tex] would not equal zero. (So zero antisymmetric component in [tex]T_{ij} = bS_{ij} + aA_{ij}[/tex] .)
    Maybe? That looks (to me) like it's complete. I think. :redface:
     
  7. Oct 5, 2009 #6

    Hurkyl

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    eijkSij should have three values, one for each possible value of k....

    (P.S. your book doesn't use upper indices?!?!)
     
  8. Oct 5, 2009 #7
    Well done, Meggle. Especially since you had to encode it all in mathtext :q.

    What Hurkyl has pointed out is that εijkSij is a vector. After εijk is contracted with Sij there is still one index left over.

    We could better write the equation in question as εijkSij = 0k just to remember this.

    You're equations for εijkSij would then look like this.

    [tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + \epsilon_{231}S_{23} + \epsilon_{321}S_{32} + \epsilon_{213}S_{21}+ \epsilon_{132}S_{13}[/tex]

    [tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}(S_{12} - S_{21}) + \epsilon_{312}(S_{31} - S_{13}) + \epsilon_{231} (S_{23} - S_{32})[/tex]

    [tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}0_{12} + \epsilon_{312}(0_{31}) + \epsilon_{231} (0_{23})[/tex]

    [tex]\epsilon_{ijk}S_{ij} = 0_3 + 0_2 + 0_1[/tex]

    [tex]\epsilon_{ijk}S_{ij} = 0_k[/tex]
     
    Last edited: Oct 5, 2009
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