Proof of transformational symmetry

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SUMMARY

The discussion focuses on proving that a symmetric second-order tensor, denoted as D, retains its symmetry under transformation into any coordinate system. The transformation law for second-order tensors is given by D'_{pq} = a_{pr}a_{qs}D_{rs}. The proof demonstrates that D'_{pq} = D'_{qp} holds true, confirming the symmetry of D' after applying the transformation. The key step involves recognizing that the transposition of the tensor affects the indices differently than the transformation coefficients.

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Homework Statement


Using indical notation, prove that D retains it's symmetry when transformed into any other coordinate system, i.e. D'_{pq} = D'_{qp} (where D is a symmetric 2nd order tensor)


Homework Equations


D'_{pq} = a_{pr}a_{qs}D_{rs} (law of transformation for 2nd order tensors)


The Attempt at a Solution


D_{pq} = D_{qp} (as D is symmetric)

D'_{pq} = a_{pr}a_{qs}D_{rs}
D'^{T}_{pq}=(a_{pr}a_{qs}D_{rs})^{T}
D'_{qp} = a_{qs}a_{pr}D_{sr} (can someone please explain why when you transpose this, the a's swaps position but the D swaps indices?)
D&#039;_{pq} =a_{pr}a_{qs}D_{rs} (swapping p<=>q, s<=>r)

We can see that D&#039;_{pq} is in the same form of D&#039;_{qp}, thus D&#039;_{pq} = D&#039;_{qp}
 
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