- #1
observer1
- 82
- 11
The form of the Lagrangian is: L = K - U
When cast in terms of generalized coordinates, the kinetic energy (K) can be a function of the rates of generalized coordinates AND the coordinates themselves (velocity and position); a case would be a double pendulum.
However, the potential energy (U) is ONLY a function of the generalized coordinates and NOT the rates: position, NOT velocity.
Yes, I understand that the potential energy is the potential function whose negative gradient delivers the force (from the path independence of the work done) and I can anticipate this will only be function of position and not velocity.
However, could anyone extend a few descriptive words - not equations: I got those - on why it is natural to expect the potential energy NOT to depend on velocities. Because... I am not really content with my explanation and am looking for something... just a few more words so I can rest more assured.
When cast in terms of generalized coordinates, the kinetic energy (K) can be a function of the rates of generalized coordinates AND the coordinates themselves (velocity and position); a case would be a double pendulum.
However, the potential energy (U) is ONLY a function of the generalized coordinates and NOT the rates: position, NOT velocity.
Yes, I understand that the potential energy is the potential function whose negative gradient delivers the force (from the path independence of the work done) and I can anticipate this will only be function of position and not velocity.
However, could anyone extend a few descriptive words - not equations: I got those - on why it is natural to expect the potential energy NOT to depend on velocities. Because... I am not really content with my explanation and am looking for something... just a few more words so I can rest more assured.