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Index Notation, taking derivative

  1. Sep 14, 2015 #1
    Can anyone explain how to take the derivative of (Aδij),j? I know that since there is a repeating subscript I have to do the summation then take the derivative, but I am not sure how to go about that process because there are two subscripts (i and j) and that it is the Kronecker's Delta (not sure if this would change anything).
     
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  3. Sep 15, 2015 #2

    Orodruin

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    You really should be using sub and super scripts because it becomes difficult to imagine what you intend, I am going to assume that you are meaning ##(A\delta_{ij})_{,j} = \partial_j A\delta_{ij}##.

    The wonderful thing about index notation is that you can treat each term as if it was just a number and in the end you sum over repeated indices. So what you need to think about is what is the partial derivative ##\partial_k (A \delta_{ij})##. Once you have done that you can let ##k = j## and perform the sum.
     
  4. Sep 15, 2015 #3
    How would the partial derivative look like? This is what I got ##(dp/dx_{j}) \delta_{ij} + p (d \delta_{ij}/dx_{j})##
     
  5. Sep 15, 2015 #4

    Avodyne

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    The second term is zero, since ##\delta_{ij}## is a constant.
     
  6. Sep 15, 2015 #5
    Okay so after I do the summation, I would get ##(dp/dx_{1})A\delta_{i1} + (dp/dx_{2})A\delta_{i2} + (dp/dx_{3})A\delta_{i3}##

    Then if I set i=1, I would get just ##dp/dx_{1}## ?
     
  7. Sep 15, 2015 #6

    Orodruin

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    In general you can use the property ##\delta_{ij}A_j = A_i## as long as ##A_i## is any indexed expression.
     
  8. Sep 15, 2015 #7

    Avodyne

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    OP, I'm not sure what you're doing with the ##p## and the ##A##, so let's go back to basics:

    Suppose ##A## is a scalar and ##T_{ij}## is a two-index tensor. Consider ##(AT_{ij})_{,k}##. (Later we will set ##T_{ij}=\delta_{ij}## and ##k=j##.) Then, using the product rule for derivatives, we have ##(AT_{ij})_{,k}=A_{,k}T_{ij}+A(T_{ij})_{,k}##. (As a side comment, the standard notation would be to write ##T_{ij,k}## instead of ##(T_{ij})_{,k}##.)

    In the case of interest, we have ##T_{ij}=\delta_{ij}##. This is constant, and so its derivatives vanish: ##(\delta_{ij})_{,k}=0##. We are left with ##(A\delta_{ij})_{,k}=A_{,k}\delta_{ij}##.

    Now let ##k=j##, with an implicit sum over the repeated ##j## index. Then we have ##(A\delta_{ij})_{,j}=A_{,j}\delta_{ij}##. We can perform the sum over ##j## using the general rule given by Orodruin: ##A_{,j}\delta_{ij}=A_{,i}##. So the final result is ##(A\delta_{ij})_{,j}=A_{,i}##.

    Now you can set ##i## to a particular value (1, 2, or 3) if you like.
     
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