jdstokes
Aug18-10, 08:46 AM
The action for a fermion in curved spacetime is
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]
where g_{\mu\nu} = \eta^{ab} e_{a\mu} e_{b\nu} and the derivative operator acting on fermions is given by
D_\mu \psi = \partial_\mu \psi - \frac{i}{2} \omega_{ab\mu} S^{ab} \psi
where S^{ab} = -(i/4)[\gamma^a,\gamma^b].
I am failing to show that D_\mu \psi transforms covariantly under local Lorentz transformations.
As far as I understand, the relation \omega_{ab\mu} = e_a^\nu \nabla_\mu e_{b\nu} implies that the following transformation rules
\psi(x) \mapsto \Lambda_{1/2}(x) \psi(x)
e_{a\mu}(x) \mapsto {\Lambda^a}_b(x) e_{b\mu}(x)
\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_\m u {\Lambda^\beta}_b(x).
Plugging these transformations into D_\mu \psi for an infinitesimal Lorentz transformation, for which
{\Lambda^a}_b = \delta^a_b + \theta^a_b
\Lambda_{1/2} = 1 + (i/2)\theta_{ab} S^{ab}
I obtain
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
In infinitesimal form:
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^\be ta_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
Ignorning quadratic terms and using the anti-symmetry of \theta_{ab}:
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
The first two terms combine to give (1+ i/2 \, \theta_{cd}S^{cd})(\partial_\mu - i/2\, \omega_{ab\mu}S^{ab})\psi = \Lambda_{1/2} D_\mu \psi 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?
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]
where g_{\mu\nu} = \eta^{ab} e_{a\mu} e_{b\nu} and the derivative operator acting on fermions is given by
D_\mu \psi = \partial_\mu \psi - \frac{i}{2} \omega_{ab\mu} S^{ab} \psi
where S^{ab} = -(i/4)[\gamma^a,\gamma^b].
I am failing to show that D_\mu \psi transforms covariantly under local Lorentz transformations.
As far as I understand, the relation \omega_{ab\mu} = e_a^\nu \nabla_\mu e_{b\nu} implies that the following transformation rules
\psi(x) \mapsto \Lambda_{1/2}(x) \psi(x)
e_{a\mu}(x) \mapsto {\Lambda^a}_b(x) e_{b\mu}(x)
\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_\m u {\Lambda^\beta}_b(x).
Plugging these transformations into D_\mu \psi for an infinitesimal Lorentz transformation, for which
{\Lambda^a}_b = \delta^a_b + \theta^a_b
\Lambda_{1/2} = 1 + (i/2)\theta_{ab} S^{ab}
I obtain
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
In infinitesimal form:
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^\be ta_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
Ignorning quadratic terms and using the anti-symmetry of \theta_{ab}:
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
The first two terms combine to give (1+ i/2 \, \theta_{cd}S^{cd})(\partial_\mu - i/2\, \omega_{ab\mu}S^{ab})\psi = \Lambda_{1/2} D_\mu \psi 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?