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A Tensor Problem: A skew-symmetric tensor and another tensor

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    If [itex]A_{ij}[/itex] is a skew-symmetric tensor, and [itex]B_{ij}[/itex] is a second-order tensor, evaluate the expression

    [tex](B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}[/tex]

    and express the final answer in its simplest form.

    2. Relevant equations

    For a skew-symmetric tensor, [itex]A_{ik}=-A_{ki}[/itex]


    3. The attempt at a solution

    I'm stuck and unsure what's the first step. I notice that the expression in the bracket looks similar to what happens when two Levi-Civita symbols come together to form an expression of two pairs of the Kronecker delta. Other than that I'm quite lost. Can I get a tip please?
     
  2. jcsd
  3. Oct 3, 2012 #2
    You are probably summing over all the repeated indices, right? Remember then that the dummy indices are arbitrary, and you can for example swap k and i if you feel like it. Using this, maybe you can write the expression into a form where you take BijBkl as a common factor, multiplying some expression containing the tensor A.
     
  4. Oct 3, 2012 #3
    Here's what I have so far:

    [tex]
    (B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}\\
    =B_{ij} B_{1l}A_{i1}+B_{ij}B_{2l}A_{i2}+B_{il} B_{1l}A_{i1}+B_{il}B_{2j}A_{i2}\\
    =B_{1j} B_{1l}A_{11}+B_{2j} B_{1l}A_{21}+B_{1j}B_{2l}A_{12}+B_{2j}B_{2l}A_{22}+B_{1l} B_{1l}A_{11}+B_{2l} B_{1l}A_{21}+B_{1l}B_{2j}A_{12}+B_{2l}B_{2j}A_{22}\\
    =2B_{1j} B_{1l}A_{11}+2B_{2l}B_{2j}A_{22}[/tex]

    So the final answer that I can give is... [tex]2\sum_i B_{ij} B_{il} A_{ii}[/tex] or in the Einstein summation, [tex]2B_{mj} B_{ml} A_{nn}[/tex] with nn no sum.

    If this is correct, is there any other way to write this without the no sum?
     
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