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A Kronecker Delta problem

  • Thread starter bossman007
  • Start date
  • #1
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Homework Statement



[PLAIN]http://postimage.org/image/s7m1kohst/ [Broken][/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/ [Broken][/PLAIN]
 
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Answers and Replies

  • #2
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Still no luck on my own :(
 
  • #3
Zondrina
Homework Helper
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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.
 
  • #4
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Thats the entire problem...im still confused from ur response, but I feel u gave me a good nudge in the right direction. thx
 
Last edited:
  • #5
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That's the entire problem...hmmm
 
  • #6
60
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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
 

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