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Einstein Summation Convention, Levi-Civita, and Kronecker delta

  1. Sep 7, 2009 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Evaluate the following sums, implied according to the Einstein Summation Convention.
    [tex]\begin{array}{l}
    \delta _{ii} = \\
    \varepsilon _{12j} \delta _{j3} = \\
    \varepsilon _{12k} \delta _{1k} = \\
    \varepsilon _{1jj} = \\
    \end{array}[/tex]

    3. The attempt at a solution
    [tex]
    \begin{array}{l}
    \delta _{ii} = \delta _{11} + \delta _{12} + \delta _{13} + \delta _{21} + \delta _{22} + \delta _{23} + \delta _{31} + \delta _{32} + \delta _{33} = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 =3 \\
    \varepsilon _{12j} \delta _{j3} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{13} + \delta _{23} + \delta _{33} } \right) = \left( {0 + 0 + 1} \right)\left( {0 + 0 + 1} \right) = \left( 1 \right)\left( 1 \right) = 1 \\
    \varepsilon _{12k} \delta _{1k} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{11} + \delta _{12} + \delta _{13} } \right) = \left( {0 + 0 + 1} \right)\left( {1 + 0 + 0} \right) = \left( 1 \right)\left( 1 \right) = 1 \\
    \varepsilon _{1jj} = \left( {\varepsilon _{111} + \varepsilon _{122} + \varepsilon _{133} } \right) = 0 + 0 + 0 = 0\\
    \end{array}
    [/tex]

    Am I doing these right? Thanks!
     
  2. jcsd
  3. Sep 8, 2009 #2

    gabbagabbahey

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    Your final result is correct, but [itex]\delta_{ii}=\delta_{11}+\delta_{22}+\delta_{33}[/itex]

    Not quite, [itex]\varepsilon _{12j} \delta _{j3} =\varepsilon _{121} \delta _{13}+\varepsilon _{122} \delta _{23}+\varepsilon _{123} \delta _{33}[/itex]

    Same problem with this one.

    Good.
     
  4. Sep 8, 2009 #3

    tony873004

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    Thank you very much!
     
  5. Sep 8, 2009 #4
    [itex]\epsilon_{12j} \delta_{j1}=\epsilon_{121}=0[/itex] by defn.
     
  6. Sep 8, 2009 #5

    tony873004

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    I'm not sure if I get this then. Am I supposed to be multiplying the epsilon and delta results together? Or is the answer gabbagabbahey gave for #2 the final answer. By multiplying, I get 1 for your example. Here's my latest attempt at the original questions:
    Thanks, gabbagabbahey and latentcorpse!
    [tex]
    \begin{array}{l}
    \delta _{ii} = \delta _{11} + \delta _{22} + \delta _{33} = 1 + 1 + 1 = 3 \\
    \varepsilon _{12j} \delta _{j3} = \varepsilon _{121} \delta _{13} + \varepsilon _{122} \delta _{23} + \varepsilon _{123} \delta _{33} = 0\left( 0 \right) + 0\left( 0 \right) + 1\left( 1 \right) = 1 \\
    \varepsilon _{12k} \delta _{1k} = \varepsilon _{121} \delta _{11} + \varepsilon _{122} \delta _{12} + \varepsilon _{123} \delta _{13} = 0\left( 1 \right) + 0\left( 0 \right) + 1\left( 0 \right) = 0 \\
    \varepsilon _{1jj} = \left( {\varepsilon _{111} + \varepsilon _{122} + \varepsilon _{133} } \right) = 0 + 0 + 0 = 0 \\
    \end{array}
    [/tex]
     
  7. Sep 8, 2009 #6

    gabbagabbahey

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    For some reason (perhaps just as an example?), latentcorpse calculated [itex]\varepsilon_{12j} \delta_{j1}[/itex] instead of [itex]\varepsilon_{12j} \delta_{j3}[/itex]

    And your latest attempt looks good to me!:approve:
     
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