- #1
jostpuur
- 2,116
- 19
Here's a simple question. Anyone who knows the answer can save me from some effort by telling the answer.
For arbitrary canonical transform [itex](q,p)\mapsto (Q,P)[/itex] in the Hamilton's formulation, does there always exist a corresponding transformation [itex]q\mapsto Q[/itex] in the Lagrange's formulation? By corresponding transformation, I mean such transformation, that when we move from the Lagrange's system [itex]L(Q,\dot{Q})[/itex] to the Hamilton's formulation, we get the same [itex]K(Q,P)[/itex] as from the original canonical transform.
Another question (or perhaps the same): I know that we can always move from Lagrange's formulation to the Hamilton's formulation, but is the converse true. Do such Hamiltonian systems exist, that you cannot achieve by starting from some Lagrangian system?
For arbitrary canonical transform [itex](q,p)\mapsto (Q,P)[/itex] in the Hamilton's formulation, does there always exist a corresponding transformation [itex]q\mapsto Q[/itex] in the Lagrange's formulation? By corresponding transformation, I mean such transformation, that when we move from the Lagrange's system [itex]L(Q,\dot{Q})[/itex] to the Hamilton's formulation, we get the same [itex]K(Q,P)[/itex] as from the original canonical transform.
Another question (or perhaps the same): I know that we can always move from Lagrange's formulation to the Hamilton's formulation, but is the converse true. Do such Hamiltonian systems exist, that you cannot achieve by starting from some Lagrangian system?