Lagrangians with the same extremal curves It is well know that (in the sense having the same image not parametrization) the extremal curves of the energy functional on a Riemanian manifold $(M,g)$:
$E[\gamma(t)] = \int g_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) dt$
are the same as the length functional:
$L[\gamma(t)] = \int \sqrt{g_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t))}dt$.
Is there a more general way to see which functions $F: \mathbb{R} \rightarrow \mathbb{R}$ preserve the extremal curves of a Lagrangian on $M$. I.e. $\mathcal{L} \in C^{\infty}(TM)$?
I mean $F$ such that:
$S[\gamma(t)] = \int \mathcal{L}(\gamma(t), \dot{\gamma}(t)) dt$
has the same extremal curves as:
$\hat{S}[\gamma(t)] = \int F(\mathcal{L}(\gamma(t), \dot{\gamma}(t))) dt$.
Is it known which $F$ would preserve the geodesics of a Finsler metric specifically?
 A: If you have a nondegenerate Lagrangian $L:TM\to\mathbb{R}$ (such as the energy Lagrangian of a pseudo-Riemannian metric or the square of a Finsler metric, though these are not the only cases) with the property that $R(L)$ is a function of $L$ (where $R$ is the 'radial' vector field on $TM$, i.e., it is tangent to the fibers of $TM\to M$ and is the usual radial vector field on each of the vector spaces $T_pM$), then the extrema of $L$ will also be extrema of $F(L)$ for any $F$ with $F'\not=0$.  This follows from the Euler-Lagrange equations (and the nondegeneracy of $L$), since they imply that $L$ is constant on any extremal curve.  
More generally, this property will hold as long as $L$ (assumed nondegenerate) has the property that it is constant along all extrema of $L$.
A: I don't know if this will be useful, but if $L$ is homogeneous of degree one (like a Finsler metric, for instance) and $F$ is any continuous differentiable function with $F'(t) > 0$ for all $t > 0$, then if you take an extremal $\gamma(t)$ for $L$ and parametrize it in such a way that $L(\dot{\gamma}(t)) \equiv 1$, then it will also be an extremal for $F \circ L$. Indeed, write down the Euler-Lagrange equations:
$$
\frac{d}{dt} \frac{\partial (F\circ L)}{\partial \dot{q}} - 
\frac{\partial (F\circ L)}{\partial q} = 0
$$
and use that $F' \circ L$ is non-zero and constant along $\dot{\gamma}(t)$.
