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Is This Contraction of a Tensor Allowed?

  1. Jul 17, 2015 #1
    Say you have a scalar ##S=A^{\alpha}_{\beta}B^{\beta}_{\alpha}## . Since this just means to sum over ##{\alpha}## and ##{\beta}## , is it allowable to rewrite it as ##S=A^{\alpha}_{\alpha}B^{\beta}_{\beta}## . I don't see anything wrong with this, I simply rewrote the dummy indices, but since I am not by far an expert I would like some confirmation.
     
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  3. Jul 17, 2015 #2

    ShayanJ

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    They're both correct contractions but not equal to each other.
    More precisely, ## A^a_a B^b_b ## is the product of two scalars.
    But ## A^a_bB^b_a ## is different. Consider the tensor ## S^{ad}_{bc}=A^a_b B^d_c ##. Now if I contract a with c and b with d, I get## A^a_bB^b_a ##. So here you're contracting a fourth order tensor twice, instead of multiplying two scalars where each is a contracted 2nd rank tensor.
     
  4. Jul 17, 2015 #3

    Fredrik

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    Suppose e.g. that the indices run from 0 to 1. Then the former means ##A^0_0 B^0_0+ A^0_1B^1_0 +A^1_0B^0_1+A^1_1B^1_1## and the latter means ##(A^0_0+A^1_1)(B^0_0+B^1_1)##, which is equal to ##A^0_0B^0_0+A^0_0B^1_1+A^1_1B^0_0+A^1_1B^1_1##. The difference is ##A^0_1B^1_0+A^1_0B^0_1-A^0_0B^1_1-A^1_1B^0_0##. This isn't always zero.
     
  5. Jul 17, 2015 #4
    Thanks for the quick replies. I understand now the mistake I made when I assumed the indices could be exchanged just because they were dummy indices.
     
  6. Jul 17, 2015 #5

    robphy

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    In matrix language,
    the first is Trace(AB), and the second is Trace(A)Trace(B), which are generally not equal.
     
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