# How do we know the Hamiltonian is well-defined

1. Sep 5, 2011

### andrewkirk

I have been reading about the derivation of the Hamiltonian from the Lagrangian using a Legendre transform. The Lagrangian is a variable whose value, by definition, is independent of the coordinates used to express it. (The Lagrangian is defined by means of a formula in one set of coordinates, and the formula for its value in any other set of coordinates is simply what you get from substituting the coordinate-transformation functions into the original formula).

The Hamiltonian is defined as the Legendre transform of the Lagrangian, with respect to a particular set of coordinates. The formula for that transform uses the chosen coordinates. So the Hamiltonian is not well-defined unless we can be certain that the value will be the same if we use any different set of coordinates to perform the Legendre transformation.

The derivations I have seen have not addressed this point. They just seem to assume the Hamiltonian will be well-defined.

Am I missing something obvious here? Is there a simple reason why the Hamiltonian's value will not depend on the coordinates used to derive it?

2. Sep 6, 2011

### Bill_K

When going from a Lagrangian formulation to a Hamiltonian one, the equations of motion remain equivalent. Consequently, if the equations of motion derived from L1 and L2 are equivalent, then the ones derived from H1 and H2 will also be equivalent.

3. Sep 6, 2011

### vanhees71

The Lagrangian formulation of Hamilton's principle is invariant under point transformations (diffeomorphisms in configuration space), and the Hamiltonian formulation even under the larger group of canonical transformations (symplectomorphisms on phase space). This implies what Bill_K said about the equations of motion.