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Understanding Kronecker Delta

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data

    I am a bit confused with how to deal with Kronecker Delta.
    I need to show that P,i = ( P*δij ),j

    The i and j are subscripts.

    2. Relevant equations



    3. The attempt at a solution

    P,i = ( P*δij ),j = P,j*δij + P*δij,j
    I assumed I could get rid of P*δij,j leaving me with P,j*δij.
    After this I am stuck. I'm not sure if I am supposed to consider Kronecker Delta as the Identity Matrix now or not. If I do, I end up with P,j which technically would equal P,i.
     
  2. jcsd
  3. Sep 12, 2011 #2

    vela

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    What exactly is P here?
     
  4. Sep 13, 2011 #3
    is it like this?

    we have
    {P * dij }j

    since dij is 0 for all i,j except i=j where it is 1,

    then {P * dij }j
    = {P * 1 }i ==> since i = j
    = Pi


    i was taught to think of the kronecker delta as a sieving function, i.e, it 'sieves' out the index 'j' and replace it by 'i' , as per above
     
  5. Sep 13, 2011 #4

    vela

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    There's no need to assume the second term is 0. δij is equal to 0 or 1, so if you differentiate it, you get 0.

    The second term isn't equal to P,j. The index j is a dummy index you're summing over. Just expand the summation out to see what's happening:
    [tex]P_{,j}\delta_{ij} = P_{,1}\delta_{i1} + P_{,2}\delta_{i2} + \cdots + P_{,n}\delta_{in}[/tex]In only one term, the ith term, does the Kronecker delta not vanish, so you end up with P,i.
     
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