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Question regarding Kronecker Delta

  1. Aug 29, 2013 #1
    Question regarding Kronecker Delta and index notation

    I am reading a book which covers the Kronecker delta and shows some examples of how it works. One of the examples confuses me, because it seems to be impossible.

    This book uses the notation that a repeated index is a summation over the range of that index. In this case, d and a are both 3x3 matrices.

    The statement I have been mulling over is:
    [itex]\delta_{jk} a_{ik} = a_{ij}[/itex]

    I interpret it this way: since the k is the repeated index, then the summation is over the range of k, like so:

    [itex]\sum\limits_{k=1}^3 \delta_{jk} a_{ik} = \delta_{j1} a_{i1} +\delta_{j2} a_{i2} +\delta_{j3} a_{i3} [/itex]

    Aren't these column vectors multiplied by column vectors? This makes no sense to me and seems impossible. My professor told me that this statement of mine is wrong, and that all the terms are scalars. I am sure he is right, but I simply do not see how. Could anyone offer some assistance?
  2. jcsd
  3. Aug 29, 2013 #2


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    Hi TheFerruccio! :smile:
    No, a vector would have only one index. :wink:

    Compare your equation with D = ABC, where they're all matrices.

    Then Dij = AikBklClj … and all the indices are subscripts!

    Think of the Kronecker delta as a matrix. :smile:
  4. Aug 29, 2013 #3

    Sorry, I wasn't clear enough. I believe you can make one column out of two indices, if one of the indices is fixed. This is why I was saying vector, because the index is already a fixed one. Let me explain.
    [itex]a_{ij} = \left[\begin{array}{ccc}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{array}\right][/itex]


    [itex]a_{i1} = \left[\begin{array}{ccc}1\\4\\7\end{array}\right][/itex]

    which is why I was calling it a vector, since it only had 3 elements. I am sorry for the confusion.

    I am aware that the Kronecker delta can be expressed as the identity matrix. That is why I am getting this for a term, for example:

    [itex]\delta_{j1}a_{i1} = \left[\begin{array}{ccc}1 \\ 0 \\ 0\end{array}\right] \left[\begin{array}{ccc}1 \\ 4 \\ 7\end{array}\right][/itex]

    which can't be done.

    I'm sorry, I do not understand your example and how that works. I have tried expanding out but I end up with the same problem that I am showing above. I am sure it is something very, very simple, but I am not getting it.
  5. Aug 29, 2013 #4


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    ah, i think i see what's worrying you …

    remember that δ is symmetric, so δij = δji, and you can write that vector horizontally: [1,0,0] :wink:
    my example does not work, if you mean can we split the equation into three vectors A B and C …

    i gave that example to show you that trying to interpret everything in terms of vectors doesn't work!
  6. Aug 29, 2013 #5
    Thanks. I think I am beginning to understand the problem. In grouping indexed things as vectors, I am limiting myself in what I can do with them, so I am ending up with the problem shown above. Is this roughly right in what I am thinking?

    Would the appropriate technique be to find the number of free indices, make a blank "final matrix" and run through the summation equation by fixing i and j at specific values? I guess I am slightly baffled as to why this method breaks down the way it does, since the book itself uses notation such as...

    [itex]b_i = \left[\begin{array}{ccc}2\\4\\0\end{array}\right][/itex]
  7. Aug 29, 2013 #6


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    Yes. :smile:
    Not really understanding you. :confused:

    The technique for multiplying something by Kronecker delta δij is to replace i by j (or vice versa).
  8. Aug 29, 2013 #7
    I mean I would simply draw the brackets and fill in each value as I go through the indices of i and j. I'm not sure what you mean by replacing i by j.
  9. Aug 29, 2013 #8


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    eg in [itex]\delta_{jk} a_{ik} = a_{ij}[/itex],

    δjk replaces k by j

    ie you replace the k in aik by j, making it aij :wink:
    Still not following you. :redface:
  10. Aug 29, 2013 #9
    Ah, I see what you mean there. That makes sense. I still have zero intuition for this, though. Is this just one of those things that I need to practice (much like the Levi-Civita tensor) in order to get a feel of what kind of patterns this makes? I can't seem to just look at the matrices and immediately see what the presence of the Kronicker Delta is going to do to the indices, or what it would equate to.

    I mean, I would literally draw a [ ], with 9 blank spaces (since I know the final term is going to be described by two free indices), point to each blank spot and evaluate this equation:

    [itex]\sum\limits_{k=1}^3 \delta_{jk} a_{ik} = \delta_{j1} a_{i1} +\delta_{j2} a_{i2} +\delta_{j3} a_{i3} [/itex]

    termwise as I cycle through the different locations in the final matrix. "At this point, i = 1, j = 2, evaluate the sum for these indices." This seems kind of cumbersome, but I do not see a faster way to do this, and I still see zero way to get an intuition for this.
  11. Aug 29, 2013 #10


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    There's no pattern.

    You really do just replace one letter by the other.
  12. Aug 29, 2013 #11
    I mean, there's a reason why that happens. I can always arrive at the solution by working it out, but I have no intuition for why one letter is replaced by the other through that relation. I have no quick way of being able to look at these different relations and intuitively figuring out when they're right or wrong.
  13. Aug 30, 2013 #12


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    sorry, but i don't think there is anything intuitive behind it, in general :redface:

    can you give an example of a physical process for which the equation includes a Kronecker delta, and for which you'd prefer an intuitive explanation?​
  14. Sep 1, 2013 #13
    Sure. Sorry for the late reply. Really, I've just been thinking up some different examples that use the Kronecker Delta so I can get some good practice for it.

    For instance, suppose I wanted to do an element-wise multiplication of two tensors Aij and Bkl. How would the Kronecker delta help with this?

    My initial guess would be


    but that doesn't work, since that would contract everything down into a 0th order tensor. In this hypothetical example (which I can't imagine ever needing) would I just have to specify "no sum" with an underline somewhere?
  15. Sep 1, 2013 #14


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    Hi TheFerruccio! :smile:
    Do you mean a tensor C with Cij = AijBij ?

    Why would you want that?

    (Why would you want eg a vector C with Ci = AiBi ?)

    Kronecker deltas can be inserted where they're really not needed:

    eg the inner product (dot product) of two vectors is ∑AiBi, but you could write it ∑∑AiBjδij

    but why would you do so? :confused:

    Instead of trying to invent usages of δ, concentrate on finding a book with an actual usgae of δ that you would like explained. :wink:
  16. Sep 1, 2013 #15
    Like I said, it probably wouldn't be very useful. Also, doesn't that equation not work? The i and j subscripts appear twice, so the whole thing would be a summation across i and j.

    I think you don't need the summation symbols in the AiBi term and the subsequent term, since i appears twice, so the summation is implied.

    It helps me to be able to use δ in a way that is different from what the book provides, since that can give me a better understanding of the concept. I understand that you do not understand the usefulness of it, but it works for me. I like playing around with math and finding the outer boundaries of a convention, so I feel very comfortable with the traditional usage. I do not want to be faced with a test or an engineering problem that uses a notation that feels strange to me, because it didn't fall within the parameters that book examples or problems used. I have done the same thing with other concepts and it has helped me greatly.

    Here's an example that I saw across two books where I do not understand the usage of [itex]\delta[/itex], however.

    One book mentions that [itex]\delta_{ij}[/itex] can be seen as an identity matrix. However, another book states that... [itex]\delta_{ij} = \textbf{e}_i\cdot\textbf{e}_j[/itex]. I understand both concepts independently. However, in the second example, the rank of the delta is 0, since it is a dot product. The rank of the delta in the first example is 2. Why is there this discrepancy? Why is there not a discrepancy (which I anticipate will be said)? I'm guessing the answer is "A matrix is not a tensor."
  17. Sep 1, 2013 #16


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    δij is not the identity matrix

    δ is the identity matrix

    δij is one component (a typical component) of the identity matrix

    the statement δij = ei·ej is not a matrix (or tensor) equation, it is an equation about components of a matrix (or a tensor)

    the matrix (or tensor) equation would be δ = E where Eij = ei·ej
  18. Sep 2, 2013 #17
    I think you've kickstarted my brain about this concept, now. I'm starting to understand what is being discussed here. Thanks for the help!
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