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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?

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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 $L$.

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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 $\dot{\gamma}(t)$.

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  • $\begingroup$ Oups, it seems I was writing my answer as Robert was posting his. One thing to keep in mind is a sort of general principle that says that if you have two truly different (functionally independent) Lagrangians for the same variational problem, you have some invariants of motion. Thus, it is often the case that the Lagrangian you were given is the only one. $\endgroup$
    – alvarezpaiva
    Commented Mar 12, 2014 at 19:25
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    $\begingroup$ Actually, this statement has to be taken with a grain of salt; one can add a divergence term to any Lagrangian without changing the the extrema. Thus, for example, if $f:M\to \mathbb{R}$ is any smooth function and $L:TM\to\mathbb{R}$ is any Lagrangian, then $\bar L = L + \mathrm{d}f$ will have the same extrema as $L$, and, for generic $f$, the two Lagrangians $L$ and $\bar L$ will not be functionally dependent. (Of course, this divergence ambiguity disappears when you go over to the Hamiltonian side.) $\endgroup$
    – Robert Bryant
    Commented Mar 15, 2014 at 10:59

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