Geodesic equation in the language of frames How to write the geodesic equation in the language of frame fields or tetrads in terms of spin connection?
 A: Take a curve $\gamma$. Now consider its tangent vector, $U = \dot{\gamma}$. We have two basis for the tangent bundle, the coordinate basis $\partial_\mu$ and the orthonormal basis $e_a$. Our vector is
\begin{eqnarray}
U &=& U^a e_a\\
&=& U^\mu \partial_\mu
\end{eqnarray}
with the usual relation, 
\begin{eqnarray}
g(U^a e_a, V^b e_b) &=& U^a V^b g(e_a, e_b)\\
&=& U^a V^b \eta_{ab}\\
&=& g(U^\mu \partial_\mu, V^\nu \partial_\nu)\\
&=& U^\mu V^\nu g(\partial_\mu, \partial_\nu)\\
&=& U^\mu V^\nu g_{\mu\nu}
\end{eqnarray}
We define the tetrads $e^\mu_a$ to be our change of basis : 
\begin{eqnarray}
U^a e^\mu_a &=& U^\mu\\
U^\mu e_\mu^a &=& U^a\\
\partial_\mu e^\mu_a &=& e_a\\
e_a e_\mu^a &=& \partial_\mu\\
\end{eqnarray}
Leading to the classic relation $g_{\mu\nu} = \eta_{ab} e^a_\mu e^b_\nu$. For our curve to be a geodesic, it must obey the geodesic equation : 
\begin{eqnarray}
\nabla_{\dot{\gamma}}\dot{\gamma} = 0
\end{eqnarray}
In terms of our basises, from linearity and the Leibniz rule, 
\begin{eqnarray}
\nabla_{U} U &=& \nabla_{U^a e_a} U^b e_b\\
&=& U^a \nabla_{e_a} U^b e_b\\
&=& U^a (\nabla_{e_a} U^b) e_b + U^b \nabla_{e_a} e_b)\\
&=& \nabla_{U^\mu \partial_\mu} U^\nu \partial_\nu\\
&=& U^\mu \nabla_{\partial_\mu} U^\nu \partial_\nu\\
&=& U^\mu (\nabla_{\partial_\nu} U^\nu \partial_\nu + U^\nu \nabla_{\partial_\mu} \partial_\nu)
\end{eqnarray}
$\nabla_{e_a} e_b$ is our spin connection, and $\nabla_{\partial_\mu} \partial_\nu$ our Levi-Civita one. 
\begin{eqnarray}
\nabla_{e_a} e_b &=& \omega^c_{ab} e_c\\
\nabla_{\partial_\mu} \partial_\nu &=& {\Gamma^\sigma}_{\mu\nu} \partial_\sigma
\end{eqnarray}
Rewriting this a bit ($U^b$ is simply a scalar function, so the directional derivative $U^a \nabla_{e_a} U^b$ is $U^a dU_b[e_a] = dU_b[U^a e_a] = dU_b[U^\mu \partial_\mu] = \dot{U}_b$) : 
\begin{eqnarray}
0 &=& \dot{U}^b e_b + U^a U^b \omega^c_{ab} e_c\\
&=& \dot{U}^\nu \partial_\nu + U^\nu U^\mu {\Gamma^\sigma}_{\mu\nu} \partial_\sigma
\end{eqnarray}
To get all of our geodesic equations, simply project everything onto the basis : 
\begin{eqnarray}
e^d[\nabla_{\dot{\gamma}}\dot{\gamma}] &=& \dot{U}^b \delta_{b}^d + U^a U^b \omega^c_{ab} \delta_{c}^d\\
&=& \dot{U}^d + U^a U^b \omega^d_{ab}
\end{eqnarray}
And similarly for our coordinate basis. The coordinate basis gives out the appropriate result, and the orthonormal one gives
$$\ddot{x}^d(\tau) + \dot{x}^a(\tau) \dot{x}^b(\tau) \omega^d_{ab} = 0$$
If you wish to transform any of those quantities into the coordinate basis, some application of the tetrads will do. For instance, as the tangent of the curve will likely be in the coordinate basis, our $U^a e_a$ are very much more likely to be $U^\mu e_\mu^a e_a$, in which case : 
\begin{eqnarray}
\nabla_{U} U &=& \nabla_{U^\mu e_\mu^a e_a} U^\nu e_\nu^b e_b\\
&=& U^\mu e_\mu^a \nabla_{e_a} U^\nu e_\nu^b e_b\\
&=& U^\mu e_\mu^a ((\nabla_{e_a} U^\nu) e_\nu^b e_b + U^\nu (\nabla_{e_a} e_\nu^b e_b) +  U^\nu e_\nu^b \nabla_{e_a} e_b)\\
&=& \dot{U}^\nu e_\nu^b e_b + U^\mu  U^\nu (e_{\nu,\mu}^b e_b + e_\mu^a e_\nu^b \omega^c_{ab} e_c)
\end{eqnarray}
This would imply that $e_{\nu,\mu}^b e_b + e_\mu^a e_\nu^b \omega^c_{ab} e_c$ is the Christoffel symbol. Let's work on it a bit : 
\begin{eqnarray}
e_{\nu,\mu}^b e_b + e_\mu^a e_\nu^b \omega^c_{ab} e_c &=& e_{\nu,\mu}^b e_b +  \omega^c_{\mu\nu} e_c\\
&=& e_{\nu,\mu}^b e_b^\sigma \partial_\sigma + \omega^c_{\mu\nu} e^c_\sigma \partial_\sigma\\
&=& e_{\nu,\mu}^b e_b^\sigma \partial_\sigma + \omega^c_{b\nu} e_c^\sigma \partial_\sigma\\
\end{eqnarray}
That is indeed the expression of ${\Gamma^\sigma}_{\mu\nu}$ in terms of the tetrads and spin connection.
