Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have a Riemannian surface $(M,g)$. It's clear that closed geodesics are critical points of the length functional $l(\gamma)=\int\left|\gamma(t)^{\prime}\right|_{g(\gamma(t))}dt$ over the space $Imm(S^{1},M)=\left\{\gamma\in C^{2}(S^{1},M)\left.\right|\gamma^{\prime}(t)\neq0\; \forall t\in S^{1} \right\}$ or equivalently (in the case of constant speed) minimisers of the Energy functional $E(\gamma)=\frac{1}{2}\int\left|\gamma(t)^{\prime}\right|^{2}_{g(\gamma(t))}dt$. And we can also view them as solutions to Euler-Lagrange equations or Hamiltonian equations with a particular Hamiltonian and Lagrangian. My question is given a second order Lagrangian $L(\gamma,\dot{\gamma},\ddot{\gamma})=\left|\ddot{\gamma}\right|^{2}+K(\gamma,\dot{\gamma})$ does there exist a metric $\tilde{g}$ for which the action functional for $L$ yields closed geodesics as its minimisers?

One thing I had in mind was to write: $A(\gamma)=\int\left|\ddot{\gamma}\right|^{2}+K(\gamma,\dot{\gamma})dt=\int\left|\dot{\gamma}\right|^{2}+\left|\ddot{\gamma}\right|^{2}-\left|\dot{\gamma}\right|^{2}+K(\gamma,\dot{\gamma})=\int\left|\dot{\gamma}\right|^{2}+\left|\ddot{\gamma}\right|^{2}+\tilde{K}(\gamma,\dot{\gamma})$ Now let $x=(x_{1},x_{2})$ and define another action $S(x,\dot{x})=\int\left|\dot{x_{1}}\right|^{2}+\left|\dot{x_{2}}\right|^{2}+\tilde{K}(x)=\int\left|\dot{x}\right|^{2}+\tilde{K}(x)$ in the case where $\gamma=x_1,\dot{\gamma}=x_{2}$, $A$ and $S$ coincide so closed geodesics for A are closed geodesics for S in the $(\gamma,\dot{\gamma})$ plane. And S is a "pertubation" of the energy action.

share|cite|improve this question
equivalently? A minimizer of length need not minimize the energy (it never does unless parametrized by length). – user31373 Jul 18 '12 at 4:51
Right, though assuming $\left|\gamma^{\prime}\right|$ is constant in time minimizing the energy of the parameterized curve is the same as minimizing the length. – Paul Jul 18 '12 at 5:01
@Paul: "assuming $\left|\gamma^{\prime}\right|$ is constant in time" is the same as "parametrizing by length". – joriki Jul 18 '12 at 5:45
I don't see this at all. The geodesic equations are 2nd order, but the Euler-Lagrange equations associated to a 2nd order Lagrangian are of 3rd order. So these can't be the same thing. You are trying to lower the order by making $\dot\gamma $ another variable, but now you don't have a manifold on which to consider geodesics. – user31373 Jul 18 '12 at 23:58
@Leonid A second order lagrangian $L(u,\dot{u},\ddot{u})$ yields a 4 dimensional Hamiltonian system, closed characteristics of that Hamiltonian system (for the moment consider the ones on a regular energy surface) project down to closed characteristics in the $(u,\dot{u})$ plane, now I'm trying to show that these periodic orbits are related to the geodesics of a manifold in the $(u,\dot{u})$ plane. If $\gamma$ is a minimizer of $A$ then $(\gamma,\dot{\gamma})$ is a minimizer of S in the space of curves $\left\{(\gamma,\dot{\gamma})\right\}$ for $\gamma$ an immersed closed curve. – Paul Jul 19 '12 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.