= 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 $\mathscr L=L^2/2$. The <Euler-Lagrange equations> of $\mathscr L$ give $\nabla_vv=0$, while those of $L$ give $\nabla_vv=(\dot L/L)v$. 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 $L=0$.
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