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Lagrangian for velocity dependent potential |
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| Sep16-12, 08:07 AM | #1 |
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Lagrangian for velocity dependent potential
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? |
| Sep16-12, 11:23 PM | #2 |
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| Sep17-12, 04:55 AM | #3 |
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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. |
| Sep17-12, 07:06 PM | #4 |
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Recognitions:
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Lagrangian for velocity dependent potential |
| Sep18-12, 02:29 AM | #5 |
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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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