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Metric tensor

  1. Feb 19, 2016 #1
  2. jcsd
  3. Feb 19, 2016 #2

    Samy_A

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    While xμ = gμνxν and xμ=gμνxν are both correct, when you put them together you can't use v twice as the dummy summation index.
    This is correct, I think:
    xμ xμ = gμνxν gμσxσ
     
  4. Feb 19, 2016 #3

    haushofer

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    Indeed, the problem is you use the same dummy index twice, which is not part of the rules.
     
  5. Feb 19, 2016 #4

    Orodruin

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    Yes, that is correct. On the other hand, it is unclear why you would want to do that particular manipulation.

    To OP: It is important not to get your indices mixed up. This can happen by the same error displayed here or others. You should avoid using the same summation indices when inserting different expressions into a single one or using the same name for a summation index and a free one. As a rule of thumb, your expressions should contain at most two of each index (and then one up and one down). If you have free indices in an equality, you need to have the same free indices on both sides.
     
  6. Feb 19, 2016 #5

    Ibix

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    This is kind of a tensor equivalent of the simple algebraic proof that 2=1, isn't it? The trick is that you've done something illegitimate in the middle. One could interpret ##g^{\mu\nu}x_\nu g_{\mu\nu}x^\nu## in multiple ways. You initially write it to mean
    $$\sum_\mu\left(\left(\sum_\nu g^{\mu\nu}x_{\nu}\right) \left(\sum_\nu g_{\mu\nu}x^{\nu}\right)\right)$$where I do not intend to imply any summation convention. But you then use it to mean (again with no implied summation)
    $$\sum_\mu\left(\left(\sum_\nu g^{\mu\nu}g_{\mu\nu}\right) \left(\sum_\nu x^\nu x_\nu\right)\right)$$These are two different things. They only appear to be the same because you used ambiguous notation. As others have noted, dummy indices must appear twice and only twice in any summed term to avoid this kind of thing.
     
    Last edited: Feb 19, 2016
  7. Feb 19, 2016 #6

    pervect

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    Looks good to me. I believe that ##g^{\mu\nu} g_{\mu\sigma}## is ##\delta^\nu{}_\sigma## where ##\delta## is the kronecker delta function which equal one if the indices are equal and zero if they are unequal.
     
  8. Feb 20, 2016 #7
    Hi all,

    thank you very much, especially to Samy_A for a very quick answer. Now it is completely clear to me. To Orodruin: I did this manipulation exactly because I thought that I can. The result was 1=4 as Ibix said which sound weird to me and I was not able to tell what was wrong. Now I know.

    No more than two same indices in one multiplication.

    Maybe the equation (in the first post) is a good example for students. Shows where one can find himself (1 = 4) if not obey the discussed rule.

    Very nice forum by the way. :-)
     
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