# When does a Lagrangian dynamical system have an equivalent Hamiltonian description?

Let a Lagrangian dynamical system with $n$ degrees of freedom and configuration space $\mathbb{R}^n$ (i.e. phase space $\mathbb{R}^{2n}$), which is described by $L=L(q_{i},\dot{q}_{i},t)$, $i=1,2,...,n$.

It is well known from classical mechanics, that a sufficient condition for a Lagrangian dynamical system to be equivalently described as a Hamiltonian system, is the Hessian of the Lagrangian w.r.t. the generalized velocities to be non-degenerate: $$\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0$$ Now the question is the following: It is frequently mentioned in the literature of classical mechanics (but I have not seen a direct proof of this), that the above sufficient condition is independent of the choice of generalized coordinates and depends only on the dynamical system itself (i.e. only on the Lagrangian function and the underlying configuration space). What kind of direct proof could be provided for this ?

P.S.: the motivation for posting this question was a student's (not my student) question on whether the property of a Lagrangian system to be Hamiltonian as well, is a purely geometrical property. I am not an expert in the subject but I think that essentially, it suffices to answer the question for the case that the configuration space is $\mathbb{R}^n$. In the sense that the independence of the property of the particular coordinate system used, means that in the general case, it will be a property of the underlying differentiable structure on the configuration space manifold.

$\bullet$ Let the Lagrangian $L(q_{i},\dot{q}_{i},t)$, which under the point transformations
$$\{q_{i}\}\leftrightsquigarrow\{Q_{i}\}$$ given by the invertible relations $Q_{i}=Q_{i}\big(q_{j},t\big)\Leftrightarrow q_{j}=q_{j}\big(Q_{i},t\big)$, $\ \ i,j=1,2,...,n$, (i.e. $\det\Big|\frac{\partial Q_{j}}{\partial q_{i}}\Big|\neq 0$), can be written as: $$L(q_{i},\dot{q}_{i},t)\stackrel{}{\leftrightsquigarrow}L(Q_{i},\dot{Q}_{i},t)$$ $\bullet$ Differentiating the point transformations $Q_{i}=Q_{i}\big(q_{j},t\big)$, we get that: $$\frac{dQ_i}{dt}\equiv\dot{Q}_i=\frac{\partial Q_i}{\partial q_j}\dot{q}_i+\frac{\partial Q_i}{\partial t}\ \ \ \ \ \Rightarrow \ \ \ \ \ \frac{\partial\dot{Q}_i}{\partial\dot{q_j}}=\frac{\partial Q_i}{\partial q_j} \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ $\bullet$ On the other hand, differentiating the Lagrangian we get that: $$\frac{\partial L}{\partial\dot{q}_i}=\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i} \Rightarrow$$ $$\Rightarrow \frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\frac{\partial}{\partial\dot{q}_j}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)= \sum_{m=1}^{n}\frac{\partial}{\partial\dot{Q}_m}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}=$$ $$=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}\stackrel{(1)}{\Rightarrow}$$ $$\stackrel{(1)}{\Rightarrow} \frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial Q_k}{\partial q_i}\frac{\partial Q_m}{\partial q_j}$$ Thus: $$\Big[\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big]=\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]^{t} \Big[\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big]\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]$$ where $[..]$ stands for the respective Jacobian and hessian matrices and $[..]^t$ stands for the transpo-se matrix. The last relation clearly tells us that the non-vanishing of the hessian determinant is invari-ant under the point transformations, i.e. $$\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0 \Leftrightarrow \det\Big|\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big|\neq 0$$ which concludes the proof.
P.S.: We can easily see why the above, is a sufficient condition for a Hamiltonian description to exist: Generalized momenta $p_{i}$, are defined by $$\label{7.55} p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}=p_{i}\big(q_{j},\dot{q}_{j},t\big), \ \ \ i,j=1,2,...,n$$
If the Jacobian of the generalized momenta w.r.t. the generalized velocities, or equivalently the Hessian of the Lagrangian w.r.t. the generalized velocities $$\label{7.57} \det\Big|\frac{\partial p_{i}}{\partial\dot{q}_{j}}\Big|=\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i} \partial\dot{q}_{j}}\Big|\neq 0$$ is non-zero, then, due to the inverse function theorem in $\mathbb{R}^{n}$, the above relations can be solved w.r.t. the generalized velocities: $$\label{7.58} \dot{q}_{j}=\dot{q}_{j}\big(q_{i},p_{i},t\big), \ \ \ i,j=1,2,...,n$$ Then, the Hamiltonian function is defined through a Legendre transform: $$\label{7.59} H=H\big(q_{i},p_{i},t\big)=\sum_{i=1}^{n}p_{i}\dot{q}_{i}-L\big(q_{i},\dot{q}_{i},t\big)$$ with $i=1,2,...,n$. The Hamiltonian function $H:\mathbb{R}^{2n+1}\rightarrow\mathbb{R}$, is a function, of generalized coordinates, generalized momenta and time.
• Indeed, your calculation shows that the condition that the determinant is nonzero is independent of coordinate transformations, thus on a manifold is defined in terms of the function $L$ and differentiable structure only. Dec 28 '16 at 14:06