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Proof using Permutation Symbols

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Proove that...
    (AxB)x(CxD)=(A.BxD)C-(A.BxC)D=(A.CxD)B-(B.CxD)A
    using Permutation Symbols

    2. Relevant equations


    3. The attempt at a solution
    I am confused about what to do after the third line from 'vela's response' (Post #2 from the reference link below).

    Reference https://www.physicsforums.com/threa...rmutation-tensor-and-kroenecker-delta.454568/
     
  2. jcsd
  3. Feb 23, 2015 #2

    stevendaryl

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    The problem is straight-forward, just tedious. You just apply the following rules:

    1. [itex](X \times Y)^c = \epsilon_{abc} X^a Y^b[/itex]
    2. [itex](X \cdot Y) = \delta_{ab} X^a Y^b[/itex]
    3. [itex]\epsilon_{abc} = \epsilon_{bca} = \epsilon_{cab} = -\epsilon_{bac} = -\epsilon{acb} = - \epsilon_{cba}[/itex]
    4. [itex]\epsilon_{abc} \epsilon_{ade} = \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd}[/itex]
    5. [itex]\epsilon_{abc} \delta_{ae} = \epsilon_{ebc}[/itex]
    To get the ball rolling, rewrite what you're being asked to prove in terms of components:

    [itex]((A \times B) \times (C \times D))^c = (A \cdot (B \times D)) C^c - (A \cdot (B \times C)) D^c = (A \cdot (C \times D)) B^c - (B \cdot (C \times D)) A^c[/itex]
     
    Last edited: Feb 23, 2015
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