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Mixed symmetry property and degrees of freedom

  1. Jul 26, 2012 #1
    How can I calculate degrees of freedom of a rank (o,3) tensor, Aabc, that is mixed symmetry and antisymmetric in the first 2 indices? By mixed symmetry I mean this:
  2. jcsd
  3. Jul 28, 2012 #2


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    You have 3*3*3=27 options for [itex]A_{ijk}[/itex].

    Now by antisymmetry terms like A_{abc}=-A_{bac};
    A_{cab}=-A_{acb}; A_{bca}=-A_{cba};

    So only 27-3=24 terms are independent, now after the mixed symmetry we are left with:
    24-1=23 (cause one term depends on the other two).

    So we are left with 5 dof.

    Hope I helped somehow.
    Obviously that terms with the same first two indices are zero, and we have 3*2=6 such terms, so we are left with 23-6=17 dof.
    After that we have terms like:
    A_{abb}=-A_{bab}; A_{acc}=-A_{cac} ; A_{cbb}=-A_{bcb}; A_{caa}=-A_{aca} ; A_{bcc}=-A_{cbc} ; A_{baa}=-A_{aba}
    Which means in the end:
    17-6=11 dof.

    I hope I counted right this time. :-D
    Last edited: Jul 28, 2012
  4. Jul 28, 2012 #3


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    BTW this raises a nice programming task of how to compute some arbitrary tensor of rank [itex]0\choose n[/itex] with the above constraints.

    Pitty I am not that good programmer.
  5. Aug 7, 2012 #4
    For a rank (0,3) tensor, Aabc, without any constraint, degrees of freedom are 216, a,b,c = 0, ..., 6.

    If this tensor is antisymmetric in the first 2 indices, degrees of freedom dicrease to 90.

    If it is mixed symmetry, the number of constraint equations are:

    [itex]\frac{n(n-1)(n-2)}{3!}[/itex], a,b,c=0, ..., n.

    For our example n = 6 so, the number of constraint equations are 20 therefore degrees of freedom are 70.
    Last edited: Aug 7, 2012
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