Kronecker Delta Expansion: What is the Result of δijδkk?

[cgstu]

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

 

[tiny-tim]

Welcome to PF!

Hi cgstu! Welcome to PF!

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

In δ_ijδ_kk, which indices are you summing over?

 

[cgstu]

Hi cgstu! Welcome to PF!

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

In δ_ijδ_kk, which indices are you summing over?

I guess that's 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 +
δ₂₂δ_k11 + 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.

 

[tiny-tim]

Hi cgstu!

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, δ_ijδ_kk is shorthand for ∑_k δ_ijδ_kk.

See http://en.wikipedia.org/wiki/Einstein_summation_convention" for details. ​

 

[cgstu]

thanks tiny tim,
so if I understand correctly now the answer should be three?

 

[tiny-tim]

thanks tiny tim,
so if I understand correctly now the answer should be three?

No, δ_kk = 3, so δ_ijδ_kk … ?