Are Lagrangian and Hamiltonian Hessians Always Non-Singular?

[espen180]

When going from the Lagrangian to the Hamtiltonian, we define

$$p_i=\frac{\partial L}{\partial \dot{q}_i}$$

as the independent variables in place of $$\dot{q}$$. This change of variables is possible if and only if the Hessian matrix

$$\frac{\partial^2L}{\partial \dot{q}_i\partial\dot{q}_j}$$

is non-singular. The Hamiltonian is then defined by performing a Legendre transformation on the Lagrangian:

$$H(q,p,t)\equiv \sum_{i=1}^n \dot{q}_i p_i - L(q,\dot{q},t)$$.

Since the first Hamilton's equation is

$$\dot{q}_i=\frac{\partial H}{\partial p_i}$$

, this implies that we can perform the inverse change of coordinates if and only if the Hessian matrix

$$\frac{\partial^2 H}{\partial p_i \partial p_j}$$

is non-singluar, and we would re-obtain $$L$$ as

$$L(q,\dot{q},t)\equiv \sum_{i=1}^n p_i \dot{q}_i - H(q,p,t)$$

Now for my questions.

1) Does there exist a physical system with a Lagrangian for which $$\frac{\partial^2L}{\partial \dot{q}_i\partial\dot{q}_j}$$ is singular, or a physical system with a Hamiltonian such that $$\frac{\partial^2 H}{\partial p_i \partial p_j}$$ is singular?

2) Does the non-singularity of $$\frac{\partial^2L}{\partial \dot{q}_i\partial\dot{q}_j}$$ immediately imply the non-singularity of $$\frac{\partial^2 H}{\partial p_i \partial p_j}$$? If so, how? How about the opposite case (singular implies singular)? Thanks for any help.

 

[gtoguy97]

A:1) Yes. Consider the Lagrangian $$L~=~\frac{1}{2}m\dot{q}^2. \tag{A}$$ Then the Hessian matrix is $$\left(\frac{\partial^2 L}{\partial \dot{q}_i\partial\dot{q}_j}\right) ~=~m\delta_{ij}. \tag{B}$$ So the Hessian matrix (B) is singular when $m=0$.2) No. The non-singularity of the Hessian matrix (B) does not immediately imply the non-singularity of the Hessian matrix $$\left(\frac{\partial^2 H}{\partial p_i \partial p_j}\right) ~=~\left(\frac{\partial^2 L}{\partial q_i \partial q_j}\right). \tag{C}$$ Consider again the Lagrangian (A). Then the Hessian matrix (B) is non-singular, but the Hessian matrix (C) is singular.