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Lagrangian |
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| Apr23-11, 01:25 AM | #1 |
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Lagrangian
1. The problem statement, all variables and given/known data
I am trying to prove that Lagrangian L is not uniquely defined, but only up to a time derivative of a function: [tex]\frac{d\Lambda}{dt}, \Lambda(\vec{q}, t)[/tex] So [tex] L > L+\frac{d\Lambda}{dt} = L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t}[/tex] But when I put it in the E-L eqns they definitely aren't as before. Where have I gone wrong? 2. Relevant equations 3. The attempt at a solution |
| Apr23-11, 02:24 AM | #2 |
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Show us what you got when you tried to crank out the Euler-Lagrange equations.
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| Apr23-11, 03:07 AM | #3 |
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Alright:
[tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}[/tex] [tex]\frac{d}{dt}\frac{\partial}{\partial \dot{q}}(L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t})=\frac{\partial}{\partial q}(L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t})[/tex] [tex]\frac{d}{dt}(\frac{\partial L}{\partial q}+\frac{\partial \Lambda}{\partial q})=\frac{\partial L}{\partial q}+\frac{\partial^2 \Lambda}{\partial q^2}~\dot{q}+\frac{\partial^2 \Lambda}{\partial q \partial t}[/tex] |
| Apr23-11, 03:21 AM | #4 |
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Lagrangian
Now calculate what
[tex]\frac{d}{dt}\left(\frac{\partial \Lambda}{\partial q}(q,t)\right)[/tex] is equal to. (Do you only have one coordinate, or should you have qi's?) |
| Apr23-11, 03:37 AM | #5 |
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Well I can't, so I am asking for help.
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| Apr23-11, 03:50 AM | #6 |
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Don't be intimidated by the notation. The partial of Λ with respect to q is just another function of q and t. You find the total time derivative of it the same way you found the total time derivative of Λ(q,t).
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| Apr23-11, 04:00 AM | #7 |
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I see. Thank you a lot.
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