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Dynamical definition of spin tensor- Einstein Cartan theory

  1. Nov 3, 2015 #1
    Hi, this is my first message on thi forum :D
    I apologize in advance for my english.

    I'm doing my thesis work on the theory of relativity of Einstein-Cartan.
    I'm following the article of Hehl of 1976; it's title is "General relativity with spin and torsion: Foundations and prospects".

    I can't understand the equations (3.7) at page 399:
    $$\mu^{\lambda\nu\mu}=-\tau^{\lambda\nu\mu}+\tau^{\nu\mu\lambda}-\tau^{\mu\lambda\nu}\qquad \tau^{\nu\mu\lambda}= \mu^{[\mu\nu]\lambda} $$
    Where $\tau$ (spin tensor) e $\mu$ ( pseudo spin tensor) are defined as:
    $${\tau_{\lambda}}^{\nu\mu}=\frac{1}{\sqrt{-g}}\frac{\delta\left(\sqrt{-g} \mathcal{L}\right)}{\delta {K_{\mu\nu}}^{\lambda}}\qquad
    {\mu_{\lambda}}^{\nu\mu}=\frac{1}{\sqrt{-g}}\frac{\delta\left(\sqrt{-g} \mathcal{L}\right)}{\delta {\Theta_{\mu\nu}}^{\lambda}} $$
    Where $\Theta$ and $K$ are respectively torsion and contorsion. For them we have also:
    $${K_{\mu\lambda}}^\nu=-{\Theta_{\mu\lambda}}^\nu+{{\Theta_{\lambda}}^{\nu}}_\mu-{\Theta^{\nu}}_{\mu\lambda}\qquad {\Theta_{\lambda\mu}}^\nu={K_{[\mu\lambda]}}^\nu $$
    I managed to get the formulas in question in a way that implies the antisymmetry of the spin tensor in the last two indices; but i know that, in general, this is not true. On the other hand, i can't find my mistake; it's look like as all is good.
    My reasoning is based on the previous equations and on the chain rule:
    {\mu_{\lambda}}^{\nu\mu}&=\frac{1}{\sqrt{-g}}\frac{\delta\left(\sqrt{-g} \mathcal{L}\right)}{\delta {\Theta_{\mu\nu}}^{\lambda}}\\
    &=\frac{1}{\sqrt{-g}}\frac{\partial\left(\sqrt{-g} \mathcal{L}\right)}{\partial {\Theta_{\mu\nu}}^{\lambda}}\\
    &=\frac{1}{\sqrt{-g}}\frac{\partial\left(\sqrt{-g} \mathcal{L}\right)}{\partial {K_{\rho\sigma}}^{\epsilon}}\frac{\partial{K_{\rho\sigma}}^{\epsilon}}{\partial {\Theta_{\mu\nu}}^{\lambda}}\\
    &={\tau_\epsilon}^{\sigma\rho}\left[-\frac{ \partial{\Theta_{\rho\sigma}}^{\epsilon}}{\partial {\Theta_{\mu\nu}}^{\lambda}}+\frac{\partial {\Theta_{\sigma}}^{\epsilon}\,_\rho}{\partial {\Theta_{\mu\nu}}^{\lambda}}-\frac{\partial {\Theta^{\epsilon}}_{\rho\sigma}}{\partial {\Theta_{\mu\nu}}^{\lambda}}\right]\\
    &={\tau_\epsilon}^{\sigma\rho}\left[-\frac{\partial{\Theta_{\rho\sigma}}^{\epsilon}}{\partial {\Theta_{\mu\nu}}^{\lambda}}+g^{\epsilon\gamma}g_{\rho k}\frac{\partial {\Theta_{\sigma\gamma}}^{k}}{\partial {\Theta_{\mu\nu}}^{\lambda}}-g^{\epsilon\gamma}g_{\sigma k}\frac{\partial {\Theta_{\gamma\rho}}^{k}}{\partial {\Theta_{\mu\nu}}^{\lambda}}\right]\\
    Which is (Formula 1)
    Up to this point all it's ok. The problem is that if i remember the definition of the spin tensor and follow a similar reasoning:
    {\tau_{\lambda}}^{\nu\mu}&={\mu_\epsilon}^{\sigma\rho}\frac{\partial {\Theta_{\rho\sigma}}^\epsilon}{\partial {K_{\mu\nu}}^\lambda}\\
    &={\mu_\epsilon}^{\sigma\rho}\frac{\partial K_{[\sigma\rho]}\,^\epsilon }{\partial {K_{\mu\nu}}^\lambda}\\
    &={\mu_\epsilon}^{\sigma\rho}\frac{1}{2}\frac{\partial }{\partial {K_{\mu\nu}}^\lambda}\left[ {K_{\sigma\rho}}^\epsilon-{K_{\rho\sigma}}^\epsilon\right]
    we find the following result:
    \tau^{\lambda\nu\mu}= \mu^{\lambda[\mu\nu]}
    Where is the mistake?
    Following the previous reasoning and using the antisymmetry of the spin tensor in the last two indices in the formula (1) we arrive easily to:
    \tau^{\lambda\nu\mu}= \mu^{[\mu\nu]\lambda}
    Therefore we have demonstrated the antisymmetry of the spin tensor in the first two indices. Using this asymmetry another time in the formula (1) we obtain finally:
    \tau^{\nu\mu\lambda}= \mu^{[\mu\nu]\lambda}
    Where is my mistake? And in which way i can demonstrate the last formula without the antisymmetry of $\tau$ in the last two indices?

    thank you very much, bye!!!
  2. jcsd
  3. Nov 8, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Nov 10, 2015 #3
    i haven't yet reach the right answer..
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