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

Local lorentz tranformations of fermion action

  1. Aug 18, 2010 #1
    The action for a fermion in curved spacetime is

    [itex]S = -\int d^4 x \sqrt{- \det(\eta^{ab} e_{a\mu}e_{b\nu})} \left[ i\overline{\psi} e^\mu_a \gamma^a D_\mu \psi + i m \overline{\psi}\psi \right][/itex]

    where [itex]g_{\mu\nu} = \eta^{ab} e_{a\mu} e_{b\nu}[/itex] and the derivative operator acting on fermions is given by

    [itex]D_\mu \psi = \partial_\mu \psi - \frac{i}{2} \omega_{ab\mu} S^{ab} \psi[/itex]

    where [itex]S^{ab} = -(i/4)[\gamma^a,\gamma^b][/itex].

    I am failing to show that [itex]D_\mu \psi[/itex] transforms covariantly under local Lorentz transformations.

    As far as I understand, the relation [itex]\omega_{ab\mu} = e_a^\nu \nabla_\mu e_{b\nu}[/itex] implies that the following transformation rules

    [itex]\psi(x) \mapsto \Lambda_{1/2}(x) \psi(x)[/itex]
    [itex]e_{a\mu}(x) \mapsto {\Lambda^a}_b(x) e_{b\mu}(x)[/itex]
    [itex]\omega_{ab\mu}(x) \mapsto {\Lambda^{\alpha}}_a(x) \omega_{\alpha\beta\mu}(x) {\Lambda^\beta}_b(x) + \eta_{\alpha\beta}{\Lambda^\alpha}_a(x)\partial_\mu {\Lambda^\beta}_b(x)[/itex].

    Plugging these transformations into [itex]D_\mu \psi[/itex] for an infinitesimal Lorentz transformation, for which
    [itex]{\Lambda^a}_b = \delta^a_b + \theta^a_b[/itex]
    [itex]\Lambda_{1/2} = 1 + (i/2)\theta_{ab} S^{ab}[/itex]
    I obtain

    [itex]D_\mu \psi \mapsto D_\mu' \psi' = (\partial_\mu \Lambda_{1/2})\psi + \Lambda_{1/2} \partial_\mu \psi - \frac{i}{2}\left[{\Lambda^{\alpha}}_a \omega_{\alpha\beta\mu} {\Lambda^\beta}_b + \eta_{\alpha\beta}{\Lambda^\alpha}_a \partial_\mu {\Lambda^\beta}_b \right]S^{ab}\Lambda_{1/2}\psi[/itex]

    In infinitesimal form:

    [itex]D_\mu' \psi' = \frac{i}{2} \partial_\mu \theta_{ab} S^{ab}\psi + (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2}\left[(\delta^\alpha_a + \theta^\alpha_a)\omega_{\alpha\beta\mu}(\delta^\beta_b + \theta^\beta_b) + \eta_{\alpha\beta} (\delta^\alpha_a + \theta^\alpha_a ) \partial_\mu \theta^\beta_b \right] S^{ab}(1 + i/2\, \theta_{cd}S^{cd})\psi[/itex]

    Ignorning quadratic terms and using the anti-symmetry of [itex]\theta_{ab}[/itex]:

    [itex]D_\mu' \psi' = (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2} \left( \omega_{ab\mu} S^{ab} + \frac{i}{2} \omega_{ab\mu} \theta_{cd} S^{ab} S^{cd}\right)\psi - \frac{i}{2}\left( \omega_{a\beta\mu} \theta^\beta_b + \omega_{\alpha b \mu} \theta^\alpha_a\right) S^{ab} \psi[/itex]

    The first two terms combine to give [itex](1+ i/2 \, \theta_{cd}S^{cd})(\partial_\mu - i/2\, \omega_{ab\mu}S^{ab})\psi = \Lambda_{1/2} D_\mu \psi[/itex] as required, but the last term does not have anything to cancel with. Does anyone have any suggestions about where I might be going wrong?
     
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 18, 2010 #2
    I think the error lies in the fact that I should be considering the transformation of [itex]{\omega^a}_{b}[/itex] rather than [itex]\omega_{ab}[/itex]. The transformation rule for [itex]{\omega^a}_{b}[/itex] is

    [itex]{\omega^a}_{b\mu}(x) \mapsto {\Lambda^\beta}_b(x) {\omega^\alpha}_{\beta\mu}(x) {\Lambda_{\alpha}}^a(x) + [\partial_\mu {\Lambda^\alpha}_b(x)]{\Lambda_\alpha}^a(x)[/itex]

    expanding this into infinitesimal form gives extra minus signs because of [itex]{\Lambda^a}_b = \delta^a_b + \theta^a_b[/itex] and [itex]{\Lambda_a}^b = \delta^b_a - \theta^b_a[/itex]. These cause the last term in the expression for [itex]D_\mu' \psi'[/itex] to cancel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Local lorentz tranformations of fermion action
  1. Local lorentz (Replies: 1)

  2. Locally Lorentz (Replies: 40)

  3. Local Lorentz Frames (Replies: 47)

Loading...