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Lagrangian

 
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Apr23-11, 01:25 AM   #1
 

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
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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.
Apr23-11, 03:07 AM   #3
 
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
 
Well I can't, so I am asking for help.
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).
Apr23-11, 04:00 AM   #7
 
I see. Thank you a lot.
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