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Permutation symbol - indicial notation

  1. Sep 21, 2008 #1
    1. The problem statement and attempt at solution

    Hey, I'm still trying to get my head around indicial notation. I'm finding it quite hard..


    I think this is somewhat right, but I don't know if the answer is clear enough..

    Any hints/comments are greatly appretiated!

    Thank you
    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2


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    It would help if we knew what [itex]\theta[/itex] was! The angle between [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex]?

    I think you need to add two things:
    1) How you got from [itex](a_i\hat{e}_i\times b_j\hat{e}_j)^2[/itex] to [itex]a_i^2b_j^2(\epsilon_{ijk})^2\hat{e}_k[/itex]

    2) Why [itex]\epsilon_{ijk}[/itex] is equal to [itex]sin(\theta)[/itex]
  4. Sep 21, 2008 #3
    Something important to note is that when you square an expression with indices, you can't re-use any of the "dummy" indices (i.e. the ones you sum over). You have to write out the second factor using different symbols for the dummy indices. You can't just raise everything to the power 2.
    e.g. If you take the square of
    [tex]v_i = \epsilon_{ijk}a_j b_k[/tex]
    you get
    [tex]v_i^2 = (\epsilon_{ijk}a_j b_k)(\epsilon_{imn}a_m b_n)[/tex]
    You CANNOT write
    [tex]v_i^2 = (\epsilon_{ijk})^2(a_j)^2( b_k)^2[/tex]
    Write the summations out in full for a small example and you'll see that this is true.
    Last edited: Sep 21, 2008
  5. Sep 22, 2008 #4
    you write that:
    v_i = \epsilon_{ijk}a_j b_k


    [tex] (\epsilon_{ijk}a_j b_k) = (\epsilon_{ijk}a_i b_j)[/tex] ?

    As for me not changing the indices when I took the square, that's exactly what I did :redface:

    Thank you!
  6. Sep 22, 2008 #5
    [tex] (\epsilon_{ijk}a_j b_k) = (\epsilon_{ijk}a_i b_j)[/tex] ?
    No. On the left hand side you are implying a summation over the repeated indices j and k. On the right hand side you are implying summation over i and j. This is not the same thing. Furthermore, in
    [tex] v_i = \epsilon_{ijk}a_j b_k [/tex],
    since i is not repeated in the left hand side, you are implying that i is fixed.
    Do a google on the Einstein summation convention for more information.
  7. Sep 23, 2008 #6
    v_i^2 = (\epsilon_{ijk}a_j b_k)(\epsilon_{imn}a_m b_n)

    [tex] \epsilon_{ijk}\epsilon_{imn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km} [/tex]

    [tex] (\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km})(a_ja_mb_kb_n) [/tex]

    [tex] a_ja_jb_kb_k - a_ja_kb_kb_j = a_j^2b_k^2 - a_ja_kb_kb_j [/tex]

    I'm so incredibly stuck :uhh:
  8. Sep 23, 2008 #7


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    [tex]= (\delta_{jm}a_j)(\delta_{kn}b_n)a_mb_n- (\delta_{jn}a_j)(\delta_{km}a_m)b_kb_n[/tex]
    [tex]a_ma_m b_nb_n- a_na_kb_kb_n= a^2b^2- (a\cdot b)^2[/tex]

    [itex]a_na_n[/itex] is a sum. It is equal to [itex]|a|^2[/itex] not [itex]a_n^2[/itex].
  9. Sep 24, 2008 #8
    Alright, I get that but am still not able to solve the problem
    Think I'll call it quits for now and do a bit more reading.
    Looks like my head isn't ready for this yet :p

    Thanks a lot though!
  10. Sep 24, 2008 #9


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    I notice that you still haven't told us what [itex]\theta[/itex] is. If it is, in fact, the angle between [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex], and you are using a and b to denote the lengths of those two vectors, then you probably want to use the very fundamental property that [itex]\vec{a}\times\vec{b}= ab sin(\theta)[/itex].
  11. Sep 25, 2008 #10
    Oh sorry, I totally forgot. Yes, theta is the angle between a and b.

    v = (a x b) = [tex] (a_je_j x b_ke_k) = absin(\theta) [/tex]

    [tex] v_i = (\epsilon_{ijk}a_jb_k)=absin(\theta) [/tex]

    [tex] v_i^2 = (\epsilon_{ijk})^2a_j^2b_k^2=a^2b^2sin^2(\theta) [/tex]

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