[itex](1,a^2,a^2,a^2)[/itex]) from the action; [itex] \mathcal{S}_{D}[\phi,\psi,e^{\alpha}_{\mu}] = \int d^4 x \det(e^{\alpha}_{\mu}) \left[ \mathcal{L}_{KG} + i\bar{\psi}\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi} + g\phi)\bar{\psi}\psi \right](adsbygoogle = window.adsbygoogle || []).push({});

[/itex] I can show that, [itex]i\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi}+g\phi)\psi =0 [/itex] by varying the action. I know that [itex]D_{\mu}=\partial_{\mu}+\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}[/itex] and I know how to relate [itex]\bar{\gamma}^{\mu}[/itex]to the flat space-time gamma matrices [itex]\gamma[/itex], I am just stuck trying to prove that [itex]\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}=\frac{3}{2}\frac{\dot{a}}{a}[/itex] I think this term is equal to [itex] \frac{1}{4}\left( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}\right) \left( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma \mu}\right)[/itex] but if it is I can't get the above result, can anyone help?

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# Curved Dirac equation, Spin connection

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