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Summation Convention – Substitution Rule

  1. Feb 16, 2016 #1
    Hello. I'm new to this forum. I'm starting a PhD – it's going to be a big long journey through the jungle that is CFD. I would like to arm myself with some tools before entering. The machete is Cartesian Tensors.

    I know the rules regarding free suffix's and dummy suffixes, but I'm having trouble proving the substitution rule:
    δi j δj k = δi k

    Assuming a range of 3 for each component, I will choose that the free suffix's have the following values: i=3, k=1. Therefore, using the implicit summation rule for the dummy index:
    δi j δj k = δ3 1 δ1 1 + δ3 2 δ2 1 + δ3 3 δ3 1 = ?

    Is my expansion correct?

    Many thanks.
     
    Last edited by a moderator: Feb 17, 2016
  2. jcsd
  3. Feb 16, 2016 #2

    blue_leaf77

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    You can visualize this equation by remembering that kronecker delta symbol actually represents the matrix elements of an identity matrix.
     
  4. Feb 17, 2016 #3

    HallsofIvy

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    No, It is not because, since "i" and "k" are NOT repeated your result should have "i" and "k" in it, only "j" is replaced by numbers:
    δijδjk= δi1δ1k+ δi2δ2k+ δi3δ3k

     
  5. Feb 24, 2016 #4
    Ok, that makes sense (since the index which appears once simultaneously represents all the possible values of that index).

    However, from your expansion, I do not understand how one could end up with δik. This was really the point in my question – I would like to see step-wise how δijδjk = δik.
     
  6. Feb 24, 2016 #5

    blue_leaf77

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    What do you get when multiplying an identity matrix with another identity matrix?
     
  7. Feb 24, 2016 #6
    I * I = I
     
  8. Feb 24, 2016 #7

    blue_leaf77

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    Now, imagine you have a 3x3 matrices ##A##, ##B##, and ##C## such that ##AB=C##. How do you write the element ##c_{ik}## of ##C## in terms of the elements of ##A## and ##B##?
     
  9. Feb 24, 2016 #8
    cik = aiibik + aijbjk + aikbkk
     
  10. Feb 24, 2016 #9

    blue_leaf77

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    Almost, just that the index to be summed must be written as numbers, not as indices as you did there.
     
  11. Feb 24, 2016 #10
    Do you mean:

    cik = a11b13 + a12b23 + a13b33 ?
     
  12. Feb 24, 2016 #11

    blue_leaf77

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    No. The left most and right most indices in each term in right hand side should be equal to the indices in the left hand side, which is ##ik##. Only the two two middle indices changes due to the summation.
    Having fixed this, you should be able to see how that equation becomes of when ##A=B=C=I##.
     
  13. Feb 24, 2016 #12

    HallsofIvy

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    I presume this is the "Kronecker delta"- [itex]\delta_{ii}= 1[/itex], [itex]\delta_{ij}= 0[/itex] if [itex]i\ne j[/itex].
    Then [itex]\delta_{ij}\delta_{jk}= \delta_{i1}\delta_{1k}+ \delta_{i2}\delta_{2j}+ \delta_{i3}\delta_{3j}[/itex]
    If i= 1, k= 1, then only the first term is non-zero and it is 1: [itex]\delta_{11}= 1+ 0+ 0= 1[/quote].
    If i= 1, k= 2, then at least one of the factors in every term is 0: [itex]\delta_{12}= 0+ 0+ 0= 0[/quote].
    If i= 1, k= 3, then at least one of the factors in every term is 0: [itex]\delta_{13}= 0+ 0+ 0= 0.

    If i= 2, k= 2, then only the second term is non-zero and it is 1: [itex]\delta_{22}= 0+ 1+ 0= 1.
    etc.
     
  14. Feb 24, 2016 #13
    Ok, so then cik = ai1b1k + ai2b2k + ai3b3k?

    In this way I can see that A=B=C=I. However, I still don't see how this leads to the proof that δijδjk= δik. If you want to expand on this for others visiting the thread, feel free. However, I can see it clearly from this:

    In all the cases where i = k, we end up with 1, while in all the cases where i ≠ k, we end up with 0. In other words, we end up with δik. Very good, thank you both for your help.
     
  15. Feb 24, 2016 #14

    blue_leaf77

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    That the Kronecker delta is defined as ##\delta_{ik}=1## for ##i=k## and zero for ##i\neq k##, doesn't it remind you of the elements of the identity matrix?
     
  16. Feb 24, 2016 #15
    Yes, it does. I can see that the Kronecker Delta in matrix form is simply an identity matrix. But how does this help me to prove:
    ?
     
  17. Feb 24, 2016 #16

    blue_leaf77

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    $$\delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik}$$
    Now you already have
    for the matrix equation ##AB=C##. If A=B=C=I, you will only need to replace ##a##, ##b##, and ##c## with ##\delta## in that quoted equation.
     
  18. Feb 24, 2016 #17
    I see that replacing each with δ gives the required proof. But is it "ok" to replace a, b and c with δ? Aren't a, b and c simply elements of matrices? How can we just replace them with δ?
     
  19. Feb 24, 2016 #18
    At the risk of repetition: To put the essential point in one place:

    Let A = (aij), B = (bij), and C = (cij) be n-dimensional matrices such that

    AB = C.​

    Then you know from the definition of matrix multiplication that

    cij = Σ aik bkj

    for any i, j in the range 1 ≤ i, j ≤ n, where the sum is over k = 1,...,n.

    Specializing to the case of n = 3, for the case of A = B = C = I3 (the 3D identity matrix), and then using the repeated-index summation convention, you get

    δik δkj = δij

    for any i, j in the range 1 ≤ i, j ≤ 3.
     
  20. Feb 24, 2016 #19

    blue_leaf77

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    Note that the last sentence in post #16 is a conditional statement, look what the condition is in order for the conclusion to be true.
     
  21. Feb 24, 2016 #20

    PeroK

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    Here's an alternative proof (for which I'll abandon the summation convention):

    ##\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} = \delta_{ik} \ \ ## (as ##\delta_{ij} = 0 \ (i \ne j)##)

    Or, more laboriously:

    ##\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} + \Sigma_{j \ne i} \delta_{ij} \delta_{jk} = \delta_{ik} + 0 \ \ ## (as ##\delta_{ii} = 1, \ \delta_{ij} = 0 \ (i \ne j)##)
     
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