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Tensor determinant using box product

  1. May 27, 2016 #1

    hotvette

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    Homework Helper

    1. The problem statement, all variables and given/known data
    Using index notation only (i.e. don't expand any sums) show that:
    \begin{align*}
    &\text{(a) } \epsilon_{ijk} \det \underline{\bf{A}} = \epsilon_{mnp} A_{mi} A_{nj} A_{pk} \\
    & \text{(b) } \det \underline{\bf{A}} = \epsilon_{mnp} A_{m1} A_{n2} A_{p3}
    \end{align*}

    2. Relevant equations

    \begin{align*}
    &\text{(1) }\epsilon_{ijk} \det \underline{\bf{A}} = [\underline{\bf{A}} \underline{e}_i, \underline{\bf{A}} \underline{e}_j, \underline{\bf{A}} \underline{e}_k] \\
    & \text{(2) } \underline{\bf{A}} = A_{ij} \underline{e}_i \otimes \underline{e}_j \\
    & \text{(3) } \underline{\bf{A}} \underline{e}_j= (A_{pq} \underline{e}_p \otimes \underline{e}_q)\underline{e}_j
    = (\underline{e}_q \cdot \underline{e}_j) A_{pq} \underline{e}_p = \delta_{qj} A_{pq} \underline{e}_p = A_{pj} \underline{e}_p
    \end{align*}

    3. The attempt at a solution
    I got part (a) by using (3) for each term of the box product, substituting each of those expressions into the box product, and evaluating.

    What baffles me is (b). If I multiply [itex] \epsilon_{ijk} [/itex] times (b) and compare to (a), the net result is:
    \begin{align*}
    &\epsilon_{mnp} (\epsilon_{ijk} A_{m1} A_{n2} A_{p3}) = \epsilon_{mnp} (A_{mi} A_{nj} A_{pk}) \\
    & \epsilon_{ijk} A_{m1} A_{n2} A_{p3} = A_{mi} A_{nj} A_{pk}
    \end{align*}

    But I have no idea what to do with that (the indices look really goofy). Any tips?
     
    Last edited: May 27, 2016
  2. jcsd
  3. May 27, 2016 #2
    You can't go from the first line to the second line by dividing by ##\epsilon_{mnp}## throughout - because there is an summation involved.

    To arrive at (b) from (a), consider the value of ##\epsilon_{123}## :wink:
     
  4. May 27, 2016 #3

    hotvette

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    Homework Helper

    Ah, thanks. I was applying standard algebra techniques, which I now see is a mistake. Gosh, the answer is soooooo simple! Thanks.

    P.S. I'd fiddled with this thing for probably 2 hours. At one point I was on the right track by looking at how to eliminate [itex] \epsilon_{ijk} [/itex] from (a) to isolate the determinant. I thought about multiplying both sides by [itex] \epsilon_{ijk} [/itex] to get [itex] 6 \det(A) = xxx [/itex] but that didn't look promising. Drat!
     
    Last edited: May 27, 2016
  5. May 27, 2016 #4
    Haha, I guess the trick is to realise that the only way you can have explicit labels such as 1,2 and 3 appearing in the indices is by setting them to be so - multiplying by other stuff with arbitrary indices can never give you specific indices.
     
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