Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Varying with respect to vierbein

  1. Feb 22, 2013 #1
    Hi
    Can anyone explain how to vary an action e.g [itex]\int d^4 x e [\frac{1}{2}\partial_{\mu}\partial^{\mu} \phi + i\bar{\psi}\bar{\gamma}^{\mu}D_{\mu}\psi] [/itex] w.r.t the vierbein?

    Where e here is the determinant of the vierbein, and [itex]D_{\mu} [/itex] is = to [itex]\partial_\mu + \frac{1}{4}\gamma_{\alpha \beta} \omega_{\mu}^{\alpha \beta} [/itex]
     
  2. jcsd
  3. Feb 22, 2013 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    For the first term, I think you actually mean ##\partial_\mu \phi \partial^\mu \phi = g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi##. So remember that the metric is made of vierbeins.

    For the second term, are you assuming the spin connection is a function of the vierbein, or is it an independent field? If the spin connection depends on the vierbein, then write down the expression for it in terms of the vierbein, and vary that.

    For the determinant, same technique applies as for varying the metric determinant.
     
  4. Feb 23, 2013 #3
    I know that the answer should be [itex] T^{\mu \nu} = i/2 [\bar{\psi}\bar{\gamma}(\mu D_\nu)\psi - \bar{\psi}D(\mu \bar{\gamma}_\nu)\psi [/itex], but i just don't see why
     
  5. Feb 27, 2013 #4

    haushofer

    User Avatar
    Science Advisor

    I don't have the time to do the calculation, but you should check Van Proeyen's book on SUGRA. He treats in great detail these kind of calculations. Also, you didn't answer Ben's question: do you treat the spin connection as dependent field? The different treatments (first, second, one and a half order) can be confusing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook