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Two affine connections have exactly the same geodesics with the same parameters if and only if their difference tensor vanishes on every diagonal pair $(v,v)$. Indeed, their covariant accelerations differ by $\Delta (\dot{\gamma },\dot{\gamma })$; starting a geodesic at each arbitrary initial vector proves necessity. Polarization then says that the symmetric part of the difference is zero. This condition is stronger than agreement of unparametrized geodesic equations.

Let $\lambda$ be an affine parameter and $X=d{x}^{\mu }/d\lambda {\partial }_{\mu }$ the tangent to the geodesic. Since ${\nabla }_{{\partial }_{\mu }}{\partial }_{\nu }={\Gamma }^{\rho }{}_{\mu \nu }{\partial }_{\rho }$, the product rule for the covariant derivative givesIndependence of the coordinate basis vectors yields the coordinate geodesic equationThe affine parameter condition matters: an arbitrary reparametrization instead allows a multiple of the tangent on the right, as in the unparametrized geodesic equation.

For a timelike tangent $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2=-g(v,v)/2$, and use the Euler-Lagrange equations. For the quadratic geodesic Lagrangian, they reduce toHere the first equation has been multiplied by an irrelevant minus sign. For the length Lagrangian, its derivatives are $L^{-1}$ times those of $\mathcal{L}$, and differentiating that factor yields insteadThis is the unparametrized geodesic equation, the key to why timelike length and energy have the same geodesic images. Define proper time locally by $d\tau /du=L>0$ and put $w=dx/d\tau =v/L$. The product rule then gives ${\nabla }_{w}w=L^{-2}[{\nabla }_{v}v-(\dot{L}/L)v]=0$ and $g(w,w)=-1$. Conversely an affinely parametrized timelike geodesic has constant $L$, by metric compatibility, and therefore satisfies both equations.

Thus the two variational problems have the same oriented timelike geodesic images, after reparametrization. They do not have exactly the same parametrized solutions with an arbitrary fixed $u$: in Minkowski spacetime, $t(u)=u^2$, $\mathbf{x}=0$, $u>0$, satisfies the length equation but not the quadratic equation. The timelike assumption excludes $L=0$, where this argument and the length derivative would fail.