Independence of Position and Velocity in Lagrangian Mechanics

[quickAndLucky]

In Lagrangian mechanics, both q(t) and dq/dt are treated as independent parameters. Similarly, in Hamiltonian mechanics q and p are treated as independent. How is this justified, considering you can derive the generalized velocity from the q(t) by just taking a time derivative. Does it have anything to do with a freedom in choosing boundary conditions?

 

[BvU]

It has to do with the variational calculus of the mechanics: in a way of speaking the time is frozen and variation of variables is used to derive the L and/or H equations.

 

[PeroK]

In Lagrangian mechanics, both q(t) and dq/dt are treated as independent parameters. Similarly, in Hamiltonian mechanics q and p are treated as independent. How is this justified, considering you can derive the generalized velocity from the q(t) by just taking a time derivative. Does it have anything to do with a freedom in choosing boundary conditions?

They are independent in the sense that, in general, a particle can have any position and any velocity at a given time. The Lagrangian itself is then a function of those variables. For example, for 3D unconstrained motion, the Lagrangian is a function of 6 variables. You should think of this as an abstract mathematical object.

If you have the specific trajectory of a particle, then clearly there is a relationship between the velocity at one time and the change in position. You can get this from Newton's laws. But, you can also get this specific trajectory by looking at the properties of the Lagrangian. This, of course, depends on the initial conditions of the particle as well as the Lagrangian.