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Trouble understanding dual tensors

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Using 26.40, show that a pseudovector p and antisymmetric second rank tensor (in three dimensions) A are related by: $$ {A}_{ij} = {\epsilon}_{ijk}{p}_{k} $$

    2. Relevant equations

    26.40: $$ {p}_{i} = \frac{1}{2}{\epsilon}_{ijk}{A}_{jk} $$

    3. The attempt at a solution

    This isn't a homework question (though it may tread rather close to one). The book I'm working out of is Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence. I'm having trouble understanding their demonstration of the association of any arbitrary three dimensional second rank antisymmetric tensor with a pseudovector.

    The step they use that I'm having trouble with is (the initial contraction and substitution of the kronecker delta identity wasn't a problem):

    $$ {\epsilon}_{ijk}{p}_{k} = \frac{1}{2}({\delta}_{il}{\delta}_{jm} - {\delta}_{im}{\delta}_{jl}){A}_{lm} $$

    And then:

    $$ {\epsilon}_{ijk}{p}_{k} = \frac{1}{2}({A}_{ij}-{A}_{ji}) = {A}_{ij} $$

    The very last equality I understand, since Aij are the components of an antisymmetric tensor. I don't get exactly what they did with the delta functions though. In attempting to recreate their work, I get to:

    $$ {\epsilon}_{ijk}{p}_{k} = \frac{1}{2}(1-{\delta}_{ij}{\delta}_{ji}){A}_{ij} $$

    Which reduces to zero on both sides if i = j, and if i != j:

    $$ {\epsilon}_{ijk}{p}_{k} = \frac{1}{2}{A}_{ij} $$

    The step they appear to be taking is:

    $$ {\delta}_{ij}{\delta}_{ji}{A}_{ij} = {A}_{ji} $$

    And I can't seem to work out that equality.
     
  2. jcsd
  3. Nov 22, 2015 #2

    Samy_A

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    Science Advisor
    Homework Helper

    Assuming that the right-hand side expression is summed over l and m, the only ##\delta## terms that survive are those where ##i=l, j=m## (first product of ##\delta##'s) and ##i=m, j=l## (second product of ##\delta##'s).
    So you get: $$ {\epsilon}_{ijk}{p}_{k} = \frac{1}{2}({\delta}_{ii}{\delta}_{jj}{A}_{ij} - {\delta}_{ii}{\delta}_{jj}{A}_{ji})= \frac{1}{2}({A}_{ij}-{A}_{ji})$$
     
    Last edited: Nov 22, 2015
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