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I Kronecker Delta summation (easy)

  1. Aug 27, 2016 #1

    joshmccraney

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    Hi PF!

    As outlined in my book ##\delta_{ij} \delta_{jk} = \delta_{ik}## but don't we sum over repeated indices (and the ##j## is repeated)? Can someone explain why we do not sum in this situation?

    Thanks!
     
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  3. Aug 27, 2016 #2

    blue_leaf77

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    Yes it is summed over ##j##. Written explicitly, the sum looks like
    $$
    \delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
    $$
    You see that if ##i\neq k##, all terms above will vanish. If ##i=k## there will be only one term surviving, therefore that sum can be written compactly as ##\delta_{ik}##.
     
  4. Aug 27, 2016 #3

    joshmccraney

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    Thanks for responding blue_leaf77! Let's assume ##i,j,k## vary from ##1,2,3##. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

    Did I miss something?
     
  5. Aug 27, 2016 #4

    PeroK

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    Why would ##i## and ##k## be summed?

    And, if you sum them on the right hand side you would get equality.
     
    Last edited: Aug 27, 2016
  6. Aug 27, 2016 #5

    joshmccraney

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    That's what blue_leaf wrote.
    Since ##i## and ##k## can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.
     
  7. Aug 27, 2016 #6

    PeroK

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    blue leaf summed only ##j##, which is a repeated index.
     
  8. Aug 27, 2016 #7

    joshmccraney

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    Ohhhhh I see now! Shoot, yea I was doing it all wrong! Thank you both!!!
     
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