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Tensors Notation - Summation Convention - meaning of (a_ij)*(a_ij)

  1. Jan 4, 2013 #1
    The summation convention for Tensor Notation says, that we can omit the summation signs and simply understand a summation over any index that appears twice.

    So consider a 3X3 matrix A whose elements are denoted by aij, where i and j are indices running from 1 to 3.

    Now consider the multiplication aa.

    Using the summation convention described above, the summation here would be over the index i since it occurs twice.

    Now if the matrix A is an orthogonal matrix, then it has the property that elements of any row or column can be thought of as components of a vector whose magnitude is 1, and that they are all mutually orthogonal.

    So, aaαβ

    Where δ is the dirac delta function.

    Now what if α=β?

    According to the above equation, aa should equal 1 since δαβ=1 for α=β.

    But if we write it as aa, by summation convention, this means a summation over both i and α(or β).

    First summing over α, this means multiplication of each element of the i th row with itself.
    This will equal 1, as a result of A being orthogonal.

    Now summing over i, we'll get i*1=i.

    Also, if we had summed over i first and then α, we would have got α*1=α.

    Where am I going wrong??
  2. jcsd
  3. Jan 4, 2013 #2


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    If you say "δαβ=1 for α=β" you don't sum; you consider just an element on the diagonal regarding the delta as a matrix!

    However, you are summing. So:

    \delta^{\alpha}_{\alpha} = 1 + 1 + 1 = 3

    It's just the trace of the identity matrix, and I think you'll agree that that's equal to the dimension of the space you're working in (3) ;)

    Btw, it's a good habit to write upper and lower indices, even though in flat space and Euclidean coordinates these are equivalent.
  4. Jan 4, 2013 #3
    What you seem to have done is this :


    However, my question is regarding

  5. Jan 4, 2013 #4


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    That would be the sum of the squares of all elements of the tensor.
  6. Jan 4, 2013 #5
    I missed out on a subtle point in my book. Hence the confusion.

    I think i am clear now. Thanks for the replies anyway.
  7. Jan 4, 2013 #6


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    No, i didn't. Check your expression for a*a-transpose.
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