# Is the Legendre transformation a unique choice in analytical mechanics?

Consider a Lagrangian $L(q_i, \dot{q_i}, t) = T - V$, for kinetic energy $T$ and generalized potential $V$, on a set of $n$ independent generalized coordinates $\{q_i\}$. Assuming the system is holonomic and monogenic, it follows from Hamilton's principle that $L$ satisfies the Euler-Lagrange equations:

$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0.$$

The Lagrangian is convex with respect to the generalized velocities $\{\dot{q_i}\}$, so it is natural to consider its associated Legendre transform:

$$H(q_i, p_i, t) = \sup_{\dot{q}_i} [\dot{q}^i p_i - L(q_i, \dot{q_i}, t)],$$

where the $p_i = \partial L / \partial \dot{q_i}$ are the conjugate momenta. One can show that the equations of motion of $H$ are:

$$\dot{p_i} = - \frac{\partial H}{\partial q_i}, \quad \dot{q_i} = \frac{\partial H}{\partial p_i}, \quad \frac{dH}{dt} = - \frac{\partial L}{\partial t}$$

In this sense, $H$ is none other than the Hamiltonian of the system. While it is apparent that $H$ leads to a satisfactory picture of classical mechanics, could there exist another function of the canonical variables $(q_i, p_i)$ with its own equations of motion?

Question: Does there exist a nontrival non-Legendre transformation $\mathcal{T}$ such that the function defined by $F(q, p, t) = \mathcal{T}[L(q, \dot{q}, t)]$ contains the full dynamics of the system?

• Another associated and interesting question is, if other non-Legendre transforms produce the dynamics, what if anything is special about the Legendre transform? – DanielSank Jul 4 '15 at 1:47
• @DanielSank: There is something "special" about the Legrendre transform: It's the switch between describing a function by its graph or the tangents to its graph, and thus the switch between describing a system's evolution "globally" or "infinitesimally". – ACuriousMind Jul 4 '15 at 9:52
• For the question: Is it not obvious that any invertible transform will yield a function that contains the same information and obeys the transformed version of the E-L equations? – ACuriousMind Jul 4 '15 at 9:53
• The question (v2) seems like a list question, starting with the identity transformation, the Legendre transformation, and various Routhian hybrids. – Qmechanic Jul 4 '15 at 10:57
• @Qmechanic: Transforming from a Lagrangian to some type of Routhian may not answer my question, since the transformation must yield a function only of $(q,p)$ (and time) and none of their derivatives. – Ultima Jul 4 '15 at 18:27

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system?

Answer: any function that produces the equations of motion under some sort of rules that you state is an allowed function to describe the dynamics. In particular any function that you can place into an action principle and take derivatives thereof giving rise, after some steps, to the equations of motion.

This said, the Lagrangian and the Hamiltonian descriptions are not equivalent. They only are if the Hessian matrix with respect to positions and velocities is invertible in any point of the domain of the configuration space, which in general does not happen for field theories and interacting particles theories (it does happen for the point particle, though). Any always invertible function $\tau[\mathcal{L}(q,\dot{q},t)]$ can equivalently describe the dynamics of the system because, due to invertibility, you can always express $q,\dot{q}$ in terms of the other variables that you introduce by using Dini's theorem on implicit functions (provided its assumptions hold).

Nevertheless, there is something that makes you want to use the Hamiltonian-Lagrangian Legendre transformation, which is exactly what @ACuriousMind states above on describing the graph of a function by its tangent in any point (velocity). Other than that, you can easily introduce any other function you want and take derivatives thereof.

• The Lagrangian and Hamiltonian descriptions are equivalent (or rather, can be made equivalent) in the case of non-invertible Hessian if one remembers the data lost due to the non-invertibility in the form of (primary) constraints. – ACuriousMind Jul 4 '15 at 23:05
• Which, in turn, must be invertible to recover back positions and velocities. Equivalent to ask that the entire Hessian is. – gented Jul 4 '15 at 23:08
• I wonder if there is a typo in the following sentence: "... does not happen for field theories and interacting particles theories (it does happen for the point particle, though)". Its seems a bit unclear as obviously point particles can also be interacting. – Virgo Jul 6 '15 at 3:50
• Yes, sorry, I was unclear, I meant free point particles. – gented Jul 6 '15 at 9:28

Building on the responses from ACuriousMind and Gennaro Tedesco, I will make an attempt to provide a satisfactory, though not mathematically rigorous, answer.

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system?

Yes, any invertible transformation that takes a function of $(q,\dot{q}, t)$ to a function of $(q, p, t)$ will yield the correct dynamics of the system. Consider an invertible transformation $\mathcal{T}$ of the Lagrangian $L(q, \dot{q}, t)$ such that

$$A(q, p, t) = \mathcal{T}[L(q, \dot{q}, t)],$$

where $p$ is the momentum conjugate to $q$. Hamilton's principle requires that the first order variation of the action assume zero, i.e.

$$\delta \int_{t_1}^{t_2}L(q, \dot{q}, t) dt = 0.$$

Substituting $A$ into the integral yields

$$\delta \int_{t_1}^{t_2} \mathcal{T}^{-1}[A(q,p,t)] dt = 0.$$

Let $F(q,p,t) = \mathcal{T}^{-1}[A]$. The integrand in consideration depends independently on $q$ and $p$, therefore zero variation of the integral implies that the Euler-Lagrangian equations of motion are independently satisfied for both $q$ and $p$:

$$\frac{d}{dt}\frac{\partial F}{\partial \dot{q}} - \frac{\partial F}{\partial q} = 0,$$ $$\frac{d}{dt}\frac{\partial F}{\partial \dot{p}} - \frac{\partial F}{\partial p} = 0.$$

As ACuriousMind and Phoenix87 correctly point out, the Hamiltonian generated by the Legendre transformation is a natural choice. This is due to the fact that it provides a systematic way to go from a convex function $f(x)$ to its slope representation $g(s)$, where $s = f'(x)$.

• I'm not sure this is right: $F$ is a function of $\dot{q}$, not $p$, with $\dot{q}$ being a parameter of a functional of $H_0(p) = H(p,\,q_0,\,t_0)$, so $p$ disappears. Hence, how can you write down the last pair of equations? – Selene Routley Jun 6 '17 at 1:19