Lagrangian for velocity dependent potential

Click For Summary

Homework Help Overview

The discussion revolves around the implications of velocity-dependent potentials in the context of Lagrangian mechanics, specifically focusing on the canonical momentum associated with rotational coordinates. Participants explore the relationship between the Lagrangian, potential energy, and the Euler-Lagrange equations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants attempt to derive the canonical momentum from the Lagrangian and question the necessity of satisfying the Euler-Lagrange equations. There is a discussion on the definition of canonical momenta and the implications of using a velocity-dependent potential.

Discussion Status

Some participants express confusion about the foundational principles behind the Euler-Lagrange equations and their applicability to systems with velocity-dependent potentials. Others suggest that the choice of generalized potential is crucial for ensuring that the principle of stationary action leads to the correct dynamics.

Contextual Notes

There are inquiries regarding the generalizability of finding a Lagrangian for various systems and whether a theorem exists that guarantees the derivation of force laws from the Euler-Lagrange equations in all cases.

aaaa202
Messages
1,144
Reaction score
2

Homework Statement


Show that if the potential in the Lagrangian contains-velocity dependent terms, the canonical momentum corresponding to the coordinate of rotation θ, is no longer the mechanical angular momentum but is given by:

p = L - Ʃn\bulletri x ∇viU


Homework Equations





The Attempt at a Solution


Setting: L = T - V(qi,qi')
Lagranges equation must be satisfied***:
d/dt(∂L/∂qi') - ∂L/∂qi = 0
=>
d/dt(∂T/∂qi' - ∂V/∂qi') - ∂V/∂qi = 0
Am I on the right track?
I know I am supposed to use ∂ri/∂qi = nxr somewhere.

** Why is it that it MUST be satisfied?
 
Physics news on Phys.org
aaaa202 said:
Setting: L = T - V(qi,qi')
Lagranges equation must be satisfied***:
d/dt(∂L/∂qi') - ∂L/∂qi = 0
=>
d/dt(∂T/∂qi' - ∂V/∂qi') - ∂V/∂qi = 0
Am I on the right track?
I know I am supposed to use ∂ri/∂qi = nxr somewhere.

You are being asked to find the canonical momentum associated with the rotation coordinate \theta, so I would start with the definition of canonical momenta...which would be?

** Why is it that it MUST be satisfied?

If you are referring to the Euler-Lagrange equations, the we choose the generalized potential (and hence the Lagrangian) so that a stationary action \int_{t_1}^{t_2}Ldt produces the correct physics. When the Lagrangian is defined as such, the Euler-Lagrange equations mathematically follow.
 
okay I solved it. But I'm still confused as to why the Euler Lagrange equations MUST be satisfied.
You can show that the principle of stationary action leads to the euler lagrange equation and for simple systems where the potential is conservative you can show directly that L=T-V where T is the kinetic energy and V the potential which again leads to Newtons laws of motion.
But how do you know that L=T-U for this weird velocity-dependent will also generate Newtons laws for the system it describes- without assuming that the principle of stationary action is actually the deeper principle. You must somehow be able to show that the principle of least action always leads to Newtons laws of motion, else I don't see why you can assume it is as general as it is.
 
aaaa202 said:
okay I solved it. But I'm still confused as to why the Euler Lagrange equations MUST be satisfied.
You can show that the principle of stationary action leads to the euler lagrange equation and for simple systems where the potential is conservative you can show directly that L=T-V where T is the kinetic energy and V the potential which again leads to Newtons laws of motion.
But how do you know that L=T-U for this weird velocity-dependent will also generate Newtons laws for the system it describes- without assuming that the principle of stationary action is actually the deeper principle. You must somehow be able to show that the principle of least action always leads to Newtons laws of motion, else I don't see why you can assume it is as general as it is.

The point is that you choose your generalized potential in a way that makes the principle of stationary action lead to the correct forces (dynamics). Your generalized potential is not just the path integral of some conservative force field, but rather it must be deduced from the appropriate force laws that describe your system.
 
Okay, but is it always possible to find the lagrangian for the system? I.e. can you always find a generalized potential such that the force laws can be derived from the euler lagrange equations?
I mean you can deduce Newtons laws and the lorentz force law, but is there a general theorem which states in what situations you can find a lagrangian that describes the system completely?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
25
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K