I have some questions about the Legendre Transformation of a Lagrangian in Classical Mechanics to the Hamiltonian:

We start with a Lagrangian $L(q,\dot{q})=\frac{\langle \dot{q} , \dot{q}\rangle }{2} - V(q,\dot{q})$ and to be able to have a unique Legendre Transformation we require that the partial derivative $ p:= \frac{\partial L}{\partial \dot{q}}$ exists and the Lagrangian is convex with respect to $\dot{q}$ or are there any further things that we need? Now I was wondering about the following:

If our potential $V(q,\dot{q})$ also depends in some weird way of the velocities, this could mean that our Lagrangian is no longer convex with respect to the velocity, so the Legendre Trafo would be no longer necessarily unique, right? has this any consequences on classical mechanics?

If the Lagrangian would be convex with respect to $q$, could we also substitute this variable by using the Legendre Transformation and keep $\dot{q}$, just out of curiosity?

Can we conclude from L being convex with respect to $q$ and $\dot{q}$ separately, that it is convex function or does this require more? Probably one could say this by investigating the Hessian matrix somehow?


How I thought of it, was that the Legendre transform is unique to each function $L$ on the tangent bundle $T M$. Here $q$ is representing standard coordinates and the $\dot q$ are the induced coordinates. No matter what the Lagrangian is you can always define a Legendre transform associated to it, it just might not have nice properties.

If the Lagrangian is convex, then the associated Legendre transform is also convex. If the Lagrangian has positive definite Hessian at each point, then the associated Legendre transform is a diffeomorphism onto its image.

If the Lagrangian has quadratic growth at infinity (meaning has positive definite Hessian at each point, and there exists some positive definite quadratic form $Q$ and constant $K$ such that $L( p)\geq Q(p )+K$) then the Legendre transform is onto the entire cotangent bundle $T^\ast M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy