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Torsion Scalar and Symmetries of Torsion Tensor

  1. Jul 13, 2015 #1
    I've started f(T) theory but I have a simple question like something that i couldn't see straightforwardly.
    In Teleparallel theories one has the torsion scalar:

    upload_2015-7-13_15-26-12.png
    upload_2015-7-13_15-27-27.png
    And if you take the product you should obtain
    upload_2015-7-13_15-29-31.png
    But there seems to be the terms like
    upload_2015-7-13_15-31-56.png upload_2015-7-13_15-32-16.png upload_2015-7-13_15-32-48.png .
    How does this one vanish?

    because we know that Torsion tensor is not symmetric in first two indices but antisymmetric in the last two indices.
     
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  3. Jul 13, 2015 #2

    bcrowell

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    Often one is interested in torsion tensors that are totally antisymmetric. Is that the case here?
     
  4. Jul 14, 2015 #3
    Oscillatory behavior
     
  5. Jul 14, 2015 #4
    Nope. These are defined in terms of Wietzenböck konnections:
    upload_2015-7-14_9-59-32.png and are not totally antisymmetric.
     
  6. Jul 14, 2015 #5

    samalkhaiat

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    Will you buy me strong black coffee if I do it for you? Okay, start by multiplying the two tensor
    [tex]S^{\rho\mu\nu}T_{\rho\mu\nu} = A + B , \ \ \ \ \ (1)[/tex]
    where
    [tex]A = \frac{1}{4} T^{\rho\mu\nu}T_{\rho\mu\nu} + \frac{1}{4} T^{\mu\rho\nu}T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu}T_{\rho\mu\nu} , \ \ \ \ (2)[/tex] and [tex]B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} + \frac{1}{2}T_{\alpha}{}^{\nu\alpha} \ T^{\rho}{}_{\rho\nu} ,[/tex] or
    [tex]B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} - \frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} = - T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} ,[/tex] which is the same as
    [tex]B = T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\rho\mu} = T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} . \ \ \ \ \ (3)[/tex] That is all for [itex]B[/itex]. To work on [itex]A[/itex], we start with the identity
    [tex]T^{\mu\nu}{}_{\rho} \left( T^{\rho}{}_{\nu\mu} + T^{\rho}{}_{\mu\nu} \right) = 0 .[/tex] In the first term, if we make [itex]\rho \to \nu[/itex], [itex]\mu \to \rho[/itex] and [itex]\nu \to \mu[/itex], we find
    [tex]T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\mu\rho} + T^{\mu\nu}{}_{\rho} \ T^{\rho}{}_{\mu\nu} = 0 .[/tex] Let us rewrite this as
    [tex]T^{\rho\mu\nu} \ T_{\nu\mu\rho} + T_{\mu\nu\rho} \ T^{\rho\mu\nu} = 0 .[/tex] Using the antisymmetry in the second term, we get
    [tex]T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\rho\mu\nu} \ T_{\mu\rho\nu} .[/tex] Now, on the RHS we make [itex]\rho \leftrightarrow \mu[/itex] and get
    [tex]T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\mu\rho\nu} \ T_{\rho\mu\nu} .[/tex] Add [itex]T^{\rho\mu\nu}T_{\nu\mu\rho}[/itex] to both sides,
    [tex]T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T^{\rho\mu\nu} \ T_{\nu\mu\rho} \right) .[/tex] In the second term on the RHS, lower the indices on the first tensor and raise them on the second tensor
    [tex]T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T_{\rho\mu\nu} \ T^{\nu\mu\rho} \right) .[/tex] Using the antisymmetry and dividing by 2, we obtain
    [tex]\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = \frac{1}{2} T^{\rho\mu\nu}T_{\nu\mu\rho} = - \frac{1}{2} T^{\rho\mu\nu} T_{\nu\rho\mu}.[/tex] And finally, we rewrite this identity in the form
    [tex]\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} .[/tex] Now, substituting this identity in (2) leads to [tex]A = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} . \ \ \ (4)[/tex] Now, put (3) and (4) in (1), you get
    [tex]S^{\rho\mu\nu}T_{\rho\mu\nu} = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} + T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} .[/tex]
    Okay, don’t forget my coffee.
    Sam
     
  7. Jul 28, 2015 #6
    I owe you a black coffee :)
     
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