When is a curve “straight”?

Consider the velocity vector associated with a curve, given by \begin{equation} \vv = \frac{d\rr}{dt} \end{equation} so that \begin{equation} \vv\,dt = d\rr = \sigma^i \ee_i \end{equation} Thus, \begin{equation} \sigma^i = v^i\,dt \end{equation} Taking the exterior derivative of $\vv$, we obtain \begin{align} d\vv = d(v^i \ee_i) &= dv^i \ee_i + v^i d\ee_i \nonumber\\ &= (dv^j + \omega^j{}_i v^i) \ee_j \nonumber\\ &= (dv^j + \Gamma^j{}_{ik} v^i\,\sigma^k) \ee_j \nonumber\\ &= (dv^j + \Gamma^j{}_{ik} v^i\,v^k\,dt) \ee_j \end{align} Thus, \begin{equation} \frac{d\vv}{dt} = \left(\frac{dv^j}{dt}+\Gamma^j{}_{ik} v^i\,v^k \right) \ee_j \end{equation} or equivalently \begin{equation} \dot\vv = \left(\dot{v}^j+\Gamma^j{}_{ik} v^i\,v^k \right) \ee_j \end{equation} Intuitively, a curve is straight if there is no acceleration, so we require \begin{equation} \dot\vv=0 \label{straight} \end{equation} or equivalently \begin{equation} \dot{v}^j+\Gamma^j{}_{ik} v^i\,v^k = 0 \label{geodesic} \end{equation} A curve satisfying ($\ref{straight}$), and hence also ($\ref{geodesic}$), is called a geodesic.