How Does the Kronecker Delta Function as a Substitution Operator?

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bossman007
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Homework Statement



[PLAIN]http://postimage.org/image/s7m1kohst/ [/PLAIN]


Homework Equations



The Kronecker Delta = 1 ; if i=j

The Kronecker Delta = 0 ; i (not equal) j

The Attempt at a Solution



I have no idea what to do from here, or even if I did this first step right?
[PLAIN]http://postimage.org/image/t3d3u9qg7/ [/PLAIN]
 
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Still no luck on my own :(
 
bossman007 said:
Still no luck on my own :(

Okay so expanding your sum :

[tex]\sum_{i=1}^{3}a_iδ_{ij} = a_1δ_{1j} + a_2δ_{2j} + a_3δ_{3j}[/tex]

Note that the only term which survives is the term [itex]a_jδ_{jj}[/itex] where i=j, but [itex]δ_{jj} = 1[/itex] as per the Delta Kronecker. So [itex]a_jδ_{jj} = a_j[/itex]

Are you sure that's the question? I feel as if you're missing something. Some important intervals are not mentioned here.
 
Thats the entire problem...im still confused from ur response, but I feel u gave me a good nudge in the right direction. thx
 
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That's the entire problem...hmmm
 
I solved it for good, I understand it now. many thanks. The Kronecker Delta has the role of a substitution operator, basically replacing a repeated indice