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Tensor Algebra

  1. Sep 27, 2014 #1
    Can some one write for me the Symmetric part of a third order tensor (as a tensor form)

    Thanks .
  2. jcsd
  3. Sep 27, 2014 #2


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    What do you mean by "symmetric" for a third order tensor? A second order tensor would be represented (in a given coordinate system) by a 3 by 3 matrix (or 4 by 4 if you are counting time) and it would be symmetric if and only if [itex]A_{ij}= A_{ji}[/itex]. But a third order tensor would be represented by a "3 by 3 by 3" array, [itex]A_{ijk}[/itex]. And then we can have several different kinds of "symmetry":
    [itex]A_{ijk}= A_{jik}[/itex], or [itex]A_{ijk}= A_{ikj}[/itex], or [itex]A_{ijk}= A_{kji}[/itex]. You could even have a kind of symmetry by "rotating" the indices: [itex]A_{ijk}= A_{kij}= A_{jki}[/itex].
  4. Sep 28, 2014 #3
    Ok i will explain .
    [tex]T_{[abc]} = \frac{1}{6} \big( T_{abc} -T_{acb} + T_{bca} -T_{bac} + T_{cab} -T_{cba} \big) [/tex]
    this is the anti-symmetric part of a third order tensor, i want to write me the symmetric part of a third order tensor
  5. Sep 28, 2014 #4
    [itex]T_{[abc]} = \frac{1}{6} \big( T_{abc} -T_{acb} + T_{bca} -T_{bac} + T_{cab} -T_{cba} \big) [/itex]
  6. Oct 2, 2014 #5
    The symmetric part is like the antisymmetric part, but with all signs +. Its symmetry should be easy to verify.
  7. Oct 2, 2014 #6
    Thanks you very much . Lperrich.
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