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Kronecker delta expansion

  1. Mar 4, 2010 #1
    If given δ_ijδ_kk what would the expansion of that be? I thought it was nine but have been told that is incorrect. I know that i=j =1 else zero so I thought that the δ_kk would equal 3 times 3 from the expansion of δ_ij but that isn't the answer
  2. jcsd
  3. Mar 4, 2010 #2


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    Welcome to PF!

    Hi cgstu! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    In δijδkk, which indices are you summing over? :wink:
  4. Mar 4, 2010 #3
    Re: Welcome to PF!

    I guess thats what I am not sure of. I know that if i=j then the delta function =1 else delta =0 so my thinking was

    δ11δ11 + δ12δ11 +
    δ13δ11 + δ21δ11 +
    δ22δk11 + δ23δ11 + ...... where only when the indices matched is the entire function = 1

    δ11δ11, δ22δ11,
    δ33δ11.... etc

    This would give me a total of 9. However, this is incorrect and I do not understand why.
  5. Mar 5, 2010 #4


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    Hi cgstu! :smile:

    Nooo …

    the "Einstein summation convention" is that only repeated indices are summed over.

    In this case, k is repeated (ie, there's two of them!), so you sum over k, but i and j are not repeated, so you don't sum over them, and they'll still be in the final result.

    In other words, δijδkk is shorthand for ∑k δijδkk. :smile:

    See http://en.wikipedia.org/wiki/Einstein_summation_convention" [Broken] for details. :wink:
    Last edited by a moderator: May 4, 2017
  6. Mar 5, 2010 #5
    thanks tiny tim,
    so if I understand correctly now the answer should be three?
  7. Mar 6, 2010 #6


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    No, δkk = 3, so δijδkk … ? :smile:
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