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Einstein notation and the permutation symbol

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data
    This is my first exposure to Einstein notation and I'm not sure if I'm understanding it entirely. Also I added this class after my instructor had already lectured about the topic and largely had to teach myself, so I ask for your patience in advance...

    The question is:
    Evaluate the following expression: εijkaiaj


    2. Relevant equations
    a ^ b = ai ei ^ bj ej = aibj (ei ^ ei) = aibj εijk ek

    Where I'm following his notation that ^ represents the cross product of the two vectors


    3. The attempt at a solution
    Now, just going off what I have seen so far in the handout he has posted, I believe the answer to be

    εijkaiaj = (a ^ a)k or, εijkaiaj is the kth component of a ^ b and because the expression is a vector crossed with itself it is equal to zero

    But what does it mean to be the kth component of a cross product? Honestly I'm working backward from a similar to an example he has in the handout and making the assumption that the reason the ek component is absent from the expression is because it is the kth component of the cross product, but from what I have to reference I cannot say with any degree of certainty if that is true and it makes me uncomfortable. Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 14, 2014 #2

    pasmith

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    Homework Helper

    For a vector, [itex]a_k[/itex] is the component corresponding to [itex]\mathbf{e}_k[/itex]. Thus [itex]\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3\mathbf{e}_3[/itex]. If you work out the cross product of [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex] you'll find that the [itex]\mathbf{e}_1[/itex] component is [itex]a_2b_3 - a_3b_2 = \epsilon_{ij1}a_i b_j[/itex] and similarly for the other components. Thus [itex]\epsilon_{ijk}a_i b_j[/itex] is the [itex]\mathbf{e}_k[/itex] component of [itex]\mathbf{a} \times \mathbf{b}[/itex].

    You can get that [itex]\epsilon_{ijk}a_ia_j = 0[/itex] more easily by observing that swapping the dummy indices [itex]i[/itex] and [itex]j[/itex] changes the sign of [itex]\epsilon_{ijk}[/itex] but doesn't change the sign of [itex]a_ia_j[/itex]; thus [itex]\epsilon_{ijk}a_ia_j = \epsilon_{jik}a_ja_i = -\epsilon_{ijk}a_ia_j = 0[/itex]. The same argument shows that if [itex]T_{ij}[/itex] is any symmetric tensor then [itex]\epsilon_{ijk}T_{ij} = 0[/itex].
     
  4. Sep 14, 2014 #3
    pasmith,
    Thank you for clarifying, that definitely helps! But it does bring up another question for me though; again, I'm in the early stages of learning about this notation, and know that switching the indices changes εijk to εjik = -εijk, but why does it equal zero?
     
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