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More index gymnastics question

  1. Sep 18, 2015 #1

    dyn

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    Hi. Got 2 more questions on index placements.

    1 - I found the following on an exam paper and it seems wrong to me. It concerns a weak perturbation to the Minkowski metric gik = ηik + hik. It then states that to first order in hik the contravariant metric tensor is gik = ηik - ηinηkmhnk. This seems wrong to me as the RHS has a free m index but the LHS does not. I think 2nd term on the RHS should be -ηaiηbkhab which I think is equal to -hik. Am I right ?

    2 - In some notes I found nuvTuv = Tuv. This doesn't seem right either but i'm not sure exactly what it should be ? If I contract the u I get Tvv but if I contract the v first I get Tuu. Are these different ? Do they both equal scalar T ?
     
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  3. Sep 19, 2015 #2

    Orodruin

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    You are correct on both accounts. The two contractions at the end of 2 are equivalent (they must be as they come from the same expression). This is called the trace of T, which is a scalar.
     
  4. Sep 20, 2015 #3

    dyn

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    Thanks. Just to confirm for me ; ηuvTuv does not equal Tuv ? It actually equals Tvv which equals Tuu and these are both equal to T ?
    For a general tensor does Auu always equal Avv ?
     
  5. Sep 21, 2015 #4

    Orodruin

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    Two index expressions with different free indices will in general not be equal.

    Correct.
     
  6. Sep 21, 2015 #5

    bcrowell

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    I would say there are several different cases to consider if you want to say this as a general rule.

    If the number or placement of the free indices differ (as in #3), then they are different types of mathematical objects, e.g., a vector and a scalar.

    If the number and placement of free indices are the same, and it's concrete index notation, but the indices are different letters, e.g., [itex]v^\mu[/itex] and [itex]v^\nu[/itex], then there is some ambiguity. We could be talking about components in different coordinate systems, which could be equal but probably wouldn't be. Or comparing them could be a notational mistake.

    For the same case in abstract index notation, [itex]v^a[/itex] and [itex]v^b[/itex], these represent exactly the same mathematical object, and they are guaranteed to be equal to one another. However, we have a notational rule that when we mix them in the same equation, we should not use different letters, e.g., we write [itex]v^a=v^a[/itex], not [itex]v^a=v^b[/itex], and [itex]v^a+v^a[/itex], not [itex]v^a+v^b[/itex]. This is because the whole purpose of using different letters in abstract index notation is to keep straight how we're hooking up the "plumbing."
     
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