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Varying the metric with respect to the Veirbein

  1. Jun 25, 2015 #1
    Hello! Given a metric in terms of the Veirbein, ##g_{\mu\nu}=e^{a}_{\mu}e^{b}_{\nu}{\eta}_{ab}## , how would you go about varying it with respect to ##e^{a}_{\mu}## ? I know that ##{\delta}g_{\mu\nu}={\delta}e^{a}_{\mu}e^{b}_{\nu}{\eta}_{ab}+e^{a}_{\mu}{\delta}e^{b}_{\nu}{\eta}_{ab}## , with ##{\delta}{\eta}_{ab}=0## . Then I would divide both sides by ##{\delta}e^{a}_{\mu}## , but that leaves me with the term ##{\frac{{\delta}e^{b}_{\nu}}{{\delta}e^{a}_{\mu}}}## . What would I do with that? Thanks for any help :)
     
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  3. Jun 25, 2015 #2

    fzero

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    Typically you'd assume that the components of the vierbein are independent variables, so that ##\delta e^b_\nu / \delta e^a_\mu = \delta^b_a \delta^\mu_\nu##.
     
  4. Jun 25, 2015 #3
    Thank you! I have another question, assuming there is torsion, how would you go about taking the variation of the christoffel symbol with respect to the veirbein, ##{\delta}{\Gamma}^{\sigma}_{\mu\nu}/{\delta}e^{a}_{\tau}## ? If it was torsion-free I would just take the christoffel symbols in terms of the metric, and do a very long, algebraically tedious calculation. But I can't seem to find an equation for the christoffel symbols when there is torsion.
     
  5. Jun 25, 2015 #4

    fzero

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    Are you working from a particular reference? In Einstein-Cartan theory, I believe that the metric and torsion tensor are taken to be independent variables. In this theory, the covariant derivative is

    $$ \nabla_\mu V_\nu = \partial_\mu V_\nu - ({\Gamma^\rho}_{\mu\nu} + {K^\rho}_{\mu\nu} )V_\rho,~~~(*)$$

    where ##{\gamma^\rho}_{\mu\nu}## is the usual Christoffel symbol expressed in terms of the metric and ##{K^\rho}_{\mu\nu} ## is the contorsion tensor, related to the torsion tensor ##{T^\rho}_{\mu\nu} ## by

    $$ K_{\rho\mu\nu} = \frac{1}{2} ( T_{\rho\mu\nu} - T_{\mu\nu\rho} +T_{\nu\rho\mu} ).$$

    In this way, ##{\Gamma^\rho}_{\mu\nu}= {\Gamma^\rho}_{\nu\mu}## and all of the torsion is contained in ##{T^\rho}_{\mu\nu}##. Because of the added term in (*) we say that the connection is no longer compatible with the metric.

    I believe that you can also write this theory in terms of a vierbein and a spin connection ##{\omega_\mu}^{ab}+{\gamma_\mu}^{ab}##. Here ##{\omega_\mu}^{ab}## is the parti of the spin connection that can be related to the Christoffel symbol (as in the theory without torsion), while ##{\gamma_\mu}^{ab}## contains the torsion via something like

    $$ {\gamma_\mu}^{ab} = K_{\rho\sigma \mu} ( e^{a\rho} e^{b\sigma} - e^{a\sigma} e^{b\rho}).$$

    Presumably you can take ##e^a_\mu, {\omega_\mu}^{ab} ##, and ##{\gamma_\mu}^{ab}## as independent variables.

    I referred to this review by Shapiro to gather the formulas together, but I don't think he addresses the variational principle directly in this formalism.
     
  6. Jun 25, 2015 #5
    Thank you again. No I'm not working from any reference. I'm trying to self study GR from Sean Carroll's online notes, and am working with Veirbeins, Spin Connection, etc. because i find them interesting, and to get practice with these things. Your explanation cleared a lot of things up, I appreciate the help.
     
  7. Jun 26, 2015 #6

    naima

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    You wrote twice veirbein instead of vierbein. The word comes from Germany where vier = 4 (ein, zwei, drei,vier). Tetrad comes Greece (tetra = 4)
    Greetings from France.
     
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