Help with connection term in curved Dirac equation

Hi when trying to find the Dirac equation (in the metric $\displaystyle (1,a^2,a^2,a^2)$) from the action; $\displaystyle \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]

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