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Tensor calculus problem

  1. Mar 11, 2015 #1

    Mentz114

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    If you don't like indexes, look away now. I got these terms from a tensor calculus program as part of a the transformed F-P Lagrangian.
    [tex]
    \begin{align}
    {g}^{b a}\,{g}^{d e}\,{g}^{f c}\,{X}_{a,b c}\,{X}_{d,e f}\\
    +{g}^{b a}\,{g}^{c f}\,{g}^{e d}\,{X}_{a,b c}\,{X}_{d,e f}\\
    +{g}^{b a}\,{g}^{c e}\,{g}^{d f}\,{X}_{a,b c}\,{X}_{d,e f}\\
    +{g}^{a b}\,{g}^{c e}\,{g}^{d f}\,{X}_{a,b c}\,{X}_{d,e f}
    \end{align}
    [/tex]

    I think I can substitute ##g^{pq}## with ##g^{qp}## without harm. Also ##,{X}_{p,q r}={X}_{p,r q}## so I can exchange ##q## and ##r##. But can I do this if ##q## and ##r## are in different ##g##'s (like swapping ##e## and ##f## in the fourth term) ?

    If these gymnastics are allowed then the terms are equal and there is a good simplification.
     
  2. jcsd
  3. Mar 11, 2015 #2

    Matterwave

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    The metric tensor is symmetric, so there's never any harm with changing ##g^{ab}\leftrightarrow g^{ba}##. And the same goes for partial derivatives (important to note though that this is not true for covariant derivatives, so if it was ##X_{p;qr}## instead, you can't just arbitrarily make this swap). In addition, every index seems to be summed over, so they are all dummy indices anyways, so within each term you are allowed to make arbitrary index substitutions (as long as you replace both instances of said index simultaneously).
     
  4. Mar 11, 2015 #3

    Mentz114

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    Thank you. I thought it would be OK but not certain. I should have mentioned that ##g## is ##\eta##, the Minkowski metric.

    It's a pity there isn't a change of sign so some of these pesky things could cancel ...

    (you wouldn't like to look over the other 80 terms, by any chance ? :-)
     
  5. Mar 11, 2015 #4

    Matterwave

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    Nope, no change in sign for either the metric or the derivatives term. Both are symmetric. Although, since they are all added together...and they are all summed over every index...my suspicion is that all 4 terms are the same term...

    Certainly the first two terms are identical, and the bottom two are identical. I'm not sure if the top and bottom are identical though.

    I hope someone can check this result. It's been a while since I've done much index gymnastics.
     
  6. Mar 11, 2015 #5

    Mentz114

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    The iTensor program agrees with you. Those terms got amalgamated into 2 after I tidied the symmetry declarations. In fact The four terms in the massless Lagrangian only have 28 terms after canonicalising (?). I can make about 8 cancel, but the program disagrees.

    The problems are cause by the programs inabilty to handle a contravariant derivative index. So I have to write ##\partial^\lambda \phi^{\mu\nu}## as ##g^{k\lambda}\partial_k \phi^{\mu\nu}##. When the gauge transformation done the humber of dummy indexes rises to 10. In the canonical form though it drops to 6 which is the same as the untransformed Lagrangian.

    This is what I get for the four terms in the first post

    ##2{g}^{\%1 \%2}\,{g}^{\%3 \%5}\,{g}^{\%4 \%6}\,{X}_{\%1,\%2 \%3}\,{X}_{\%4,\%5 \%6}+2{g}^{\%1 \%2}\,{g}^{\%3 \%6}\,{g}^{\%4 \%5}\,{X}_{\%1,\%2 \%3}\,{X}_{\%4,\%5 \%6}##

    (yes, it looks horrible). If we swap ##\%4## and ##\%5## in the first term it is the same as the second. This is the same procedure used to amalgamate the 4 into 2, isn't it ?

    I don't know why the program can't see this. If it was legal the first time, why not now ?

    All good fun.
     
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