Geodesic equation and potential energy It is given here that the Geodesics equation is formulated as
$$\frac{d^2x^a}{ds^2} + \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} = 0$$
However, it is also given that one can derive the Geodesics equation using the Lagrangian $= T -V$ 
$$L = g_{\mu \nu} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s}$$
which, to me, looks like a generalized version of $\frac{m}{2}\dot{x}\dot{x}=T$. That is, no contribution from the potential energy. 
How does potential energy enter into the Geodesics equation? Is this information somehow embedded in the manifold? 
 A: The formula $L = T - V$ is not valid in general relativity theory.  It's a classical approximation.  The "gravitational potential" is hidden inside the metric components $g_{\mu \nu}$.  For weak gravitational fields, the metric in cartesian coordinates can be expressed as $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, where $\eta_{\mu \nu} = \mathrm{diag}(1, -1, -1, -1)$ is the Minkowski metric of special relativity and $|\, h_{\mu \nu}| \ll 1$.  Then, to the lowest order (neglecting the components $g_{0 i}$, writing $g_{00} = 1 + 2 \phi/c^2$ and $g_{ij} \approx \eta_{ij} =  -\, \delta_{ij}$ for small velocities : $v^2 \ll c^2$) :
\begin{align}
L = \frac{m}{2} \, g_{\mu \nu} \, \frac{d x^{\mu}}{d \tau} \, \frac{d x^{\nu}}{d \tau} &\approx \frac{m}{2} \, g_{00} \, c^2 + \frac{m}{2} \, g_{ij} \, \frac{d x^i}{d \tau} \, \frac{d x^j}{d \tau}, \\[12pt]
&\approx \frac{1}{2} \, m \, c^2 + m \, \phi - \frac{1}{2} \, m \, v^2. \tag{1}
\end{align}
The first term is just a constant.  The rest is your usual $T - V$ with a global sign change (which doesn't change the equations of motion).
The metric global signature is arbitrary.  Some prefer to use $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$, so you'll get the usual $T - V$ without a global sign change.
