Lagrangian for velocity dependent potential


by aaaa202
Tags: dependent, lagrangian, potential, velocity
aaaa202
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#1
Sep16-12, 08:07 AM
P: 992
1. The problem statement, all variables and given/known data
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[itex]\bullet[/itex]ri x ∇viU


2. Relevant equations



3. 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?
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gabbagabbahey
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#2
Sep16-12, 11:23 PM
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P: 5,004
Quote Quote by aaaa202 View Post
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 [itex]\theta[/itex], 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 [itex]\int_{t_1}^{t_2}Ldt[/itex] produces the correct physics. When the Lagrangian is defined as such, the Euler-Lagrange equations mathematically follow.
aaaa202
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#3
Sep17-12, 04:55 AM
P: 992
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.

gabbagabbahey
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#4
Sep17-12, 07:06 PM
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P: 5,004

Lagrangian for velocity dependent potential


Quote Quote by aaaa202 View Post
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.
aaaa202
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#5
Sep18-12, 02:29 AM
P: 992
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?


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