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Proof of transformational symmetry

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Using indical notation, prove that D retains it's symmetry when transformed into any other coordinate system, i.e. [tex]D'_{pq} = D'_{qp}[/tex] (where D is a symmetric 2nd order tensor)


    2. Relevant equations
    [tex]D'_{pq} = a_{pr}a_{qs}D_{rs}[/tex] (law of transformation for 2nd order tensors)


    3. The attempt at a solution
    [tex]D_{pq} = D_{qp}[/tex] (as D is symmetric)

    [tex]D'_{pq} = a_{pr}a_{qs}D_{rs}[/tex]
    [tex]D'^{T}_{pq}=(a_{pr}a_{qs}D_{rs})^{T}[/tex]
    [tex]D'_{qp} = a_{qs}a_{pr}D_{sr}[/tex] (can someone please explain why when you transpose this, the a's swaps position but the D swaps indices?)
    [tex]D'_{pq} =a_{pr}a_{qs}D_{rs}[/tex] (swapping p<=>q, s<=>r)

    We can see that [tex]D'_{pq}[/tex] is in the same form of [tex]D'_{qp}[/tex], thus [tex]D'_{pq} = D'_{qp}[/tex]
     
    Last edited: Mar 16, 2010
  2. jcsd
  3. Mar 17, 2010 #2
    Anyone?
     
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