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Tensor identity - Help

  1. Jul 31, 2004 #1
    I'm having some trouble to prove the following tensor identity shown below in Einstein's summation convention:


    I expanded the terms but when I did group them I didn't get the identity. The only way I could get the identity is if


    and I don't see a reason why this would be so.

    Obviously I'm missing something. Can somebody tell what is it that I'm doing wrong?
  2. jcsd
  3. Jul 31, 2004 #2
    It's true that [tex]a_{ij}x_{i}x_{j} = a_{ji}x_{i}x_{j}[/tex]. Putting in the summation signs may help you to see this.
    Last edited: Jul 31, 2004
  4. Jul 31, 2004 #3


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    To elaborate on Lonewolf's reply...
    &= a_{ji}x_{j}x_{i}&&\text{relabel repeated [dummy] indices}\\
    &= a_{ji}x_{i}x_{j}&&\text{reorder the writing of the x factors}\\

    You'll learn that [itex]\frac{1}{2}(a_{ij} + a_{ji})[/itex]
    is called "the symmetric part of [itex] a_{ij} [/itex]", and is written as [itex] a_{(ij)} [/itex].
    Hence, your identity can be written
    [tex] 2a_{(ij)}x_{i}x_{j} =2a_{ij}x_{i}x_{j}[/tex].

    Here is an "index gymnastics" proof, starting with half of your right-hand-side:
    &= a_{ij}x_{(i}x_{j)}&&\text{since } x_{i}x_{j} = x_{(i}x_{j)} \\
    &= a_{(ij)}x_{i}x_{j}&&\text{since i and j are being symmetrized}
  5. Jul 31, 2004 #4
    Guys, thanks to you help I got it right this time after some thinking and some calculations. I realized one of my problems was the fact that when I was looking at
    I was thinking of it as when we say
    The way I should look at that expression is not isolated from the rest of the other terms. The important thing, I think, is not the term itself isolated but how the summation terms comes up after the entire summation is expanded.
    I can continue now with the next page of my book :smile:
    Last edited: Jul 31, 2004
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