Kronecker delta expansion

  • Thread starter cgstu
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
tiny-tim
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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:
 
  • #3
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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:

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.
 
  • #4
tiny-tim
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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:
 
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  • #5
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thanks tiny tim,
so if I understand correctly now the answer should be three?
 
  • #6
tiny-tim
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thanks tiny tim,
so if I understand correctly now the answer should be three?
No, δkk = 3, so δijδkk … ? :smile:
 

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