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- Thread starter cgstu
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tiny-tim

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Hi cgstu! Welcome to PF!

(try using the X

In δ

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Hi cgstu! Welcome to PF!

(try using the X_{2}tag just above the Reply box )

In δ_{ij}δ_{kk}, which indices are you summing over?

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

δ

δ

δ

δ

δ

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!

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_{ij}δ_{kk} is shorthand for ∑_{k} δ_{ij}δ_{kk}.

Nooo …

the "Einstein summation convention" is that only

In this case, k is repeated (ie, there's

In other words, δ

See http://en.wikipedia.org/wiki/Einstein_summation_convention" [Broken] for details.

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thanks tiny tim,

so if I understand correctly now the answer should be three?

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, δ

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