# Lagrangians, Hamiltonians and the Legendre Transform [duplicate]

I have been studying the Lagrangian formulation of classical mechanics. I have been introduced to the Hamiltonian formulation via the Legendre transform, and studied the transform from this excellent introductory paper

1. Are the Lagrangian and Hamiltonian always Legendre transforms of each other? If so, does this apply to physics in general, or just to classical mechanics?

2. What happens if either of the Lagrangian or Hamiltonian is not convex? For example, if the former is not convex does this mean the Hamiltonian does not exist, and vice versa? The first answer to this question says "At the classical level, there is no convexity condition ...", but the meaning of this phrase is obscure and I feel the gist is really lost in the details. From my limited understanding of the Legendre transform it could only be applied to (strictly) convex functions.

3. If either of the Lagrangian or Hamiltonian is not strictly convex this would mean non-unique values for the derivative that we need for the Legendre transformation. In this case, am I right in assuming that one would need to impose further conditions on the problem in order to restrict interest to a subset of the domain where the function in question is convex?

This question and this other question have some interesting material in their answers, but .

• Possible duplicates: physics.stackexchange.com/q/336818/2451 , physics.stackexchange.com/q/324120/2451 and links therein. – Qmechanic Apr 9 '18 at 20:15
• I have always taken the Hamiltonian to be defined as the Legendre transform of the Lagrangian, where "Lagrangian" refers to the functional that one is interested in minimizing. This is occasionally awkward, for example when the energy of an object is a functional of its shape; then the Hamiltonian is decidedly not the energy, which is often confusing. – ZachMcDargh Apr 9 '18 at 21:47