 Timelike length and energy have the same geodesic images

For a Lorentzian metric and timelike $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2$. The Euler-Lagrange equations of $\mathcal{L}$ give ${\nabla }_{v}v=0$, while those of $L$ give ${\nabla }_{v}v=(\dot{L}/L)v$. Taking $ds/du=L$ removes the latter tangential acceleration and gives a unit timelike tangent. Thus the two actions have the same geodesic images; the quadratic action singles out an affine parameter. They do not have the same solutions for an arbitrary fixed parameter: $t=u^2$, $\mathbf{x}=0$, $u>0$, in Minkowski spacetime solves the length equation but not the energy equation. Null curves are excluded since $L=0$.