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Physical equivalence of Lagrangian under addition of dF/dt 
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#1
Feb1812, 06:18 AM

P: 14

1. The problem statement, all variables and given/known data
This isn't strictly a homework question as I've already graduated and now work as a web developer. However, I'm attempting to recover my ability to do physics (it's been a few months now) by working my way through the problems in Analytical Mechanics (Hand and Finch) in my free time and have got stuck on a question about physically invariant Lagrangians. I understand that the Lagrangian of a physical system is not unique because there are many Lagrangians for which the EulerLagrange equations reduce to the same thing. The question I can't solve asks for a proof that the EulerLagrange (EL) equations are unchanged by the addition of [itex]dF/dt[/itex] to the Lagrangian, where [itex]F \equiv F(q_1, ..., q_2, t)[/itex]. 2. Relevant equations EulerLagrange equations: [itex]\frac{d}{dt}\frac{∂L}{∂\dot{q_k}}  \frac{∂L}{∂q_k} = 0[/itex] Change of Lagrangian [itex]L \rightarrow L' = L + \frac{dF}{dt}[/itex] Chain rule: [itex]\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\frac{∂q_k}{∂t}}[/itex] 3. The attempt at a solution I guess the solution is to substitute the new Lagrangian, [itex]L'[/itex], into the EL equations and somehow show that it reduces to exactly the EL equations for [itex]L[/itex]. The substitution gives: [itex]\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\left(L + \frac{dF}{dt}\right)  \frac{∂}{∂q_k}\left(L + \frac{dF}{dt}\right) = 0[/itex] This can be rewritten as: [itex]\left(\frac{d}{dt}\frac{∂L}{∂\dot{q_k}}  \frac{∂L}{∂q_k}\right) + \left(\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{dF}{dt}  \frac{∂}{∂q_k}\frac{dF}{dt}\right) = 0[/itex] where the first set of bracketed terms are just the left hand side of the EL equation in L. Therefore, I now need to show that the other bracketed terms equate to zero also. i.e. [itex]\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{dF}{dt}  \frac{∂}{∂q_k}\frac{dF}{dt} = 0[/itex] All I can think of to try next is apply the chain rule to the differentiation of [itex]F[/itex] and then hope that everything cancels nicely. Applying the chain rule gives: [itex]\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{∂F}{∂q_k} \frac{∂q_k}{∂t} + \frac{∂}{∂q_k}\frac{∂F}{∂q_k}\frac{∂q_k}{∂t} = 0[/itex] So I guess my proof works if [itex]\frac{d}{dt}\frac{∂}{∂\dot{q_k}} = \frac{d}{dt}\frac{∂}{∂q_k/∂t}[/itex] is equivalent to [itex]\frac{∂}{∂q_k}[/itex]. Is this correct? I feel like you can't just cancel the [itex]dt[/itex] and [itex]∂t[/itex] like this, but I can't see how else this proof can be done. 


#2
Feb1812, 09:23 AM

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that is [tex]\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\dot q_k}+ \frac {∂F}{∂t}[/tex] Substitute equation [itex]\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\dot q_k}+ \frac {∂F}{∂t}[/itex] for dF/dt, derive it with respect of [itex]\dot q_k[/itex] and apply chain rule again when determining the time derivative. ehild 


#3
Feb1812, 09:56 AM

P: 615

You can do it without going all that dirty work by looking at the action
[itex]S=\int L dt = \int (L' + \frac{dF}{dt})dt [/itex] That method seems much neater to me Mod note: removed remaining steps. 


#4
Feb1812, 10:25 AM

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Physical equivalence of Lagrangian under addition of dF/dt



#5
Feb2112, 04:08 PM

P: 14

Indeed, I was aware of the action as the simpler way to prove it, but I wanted to do it with derivatives too as the book I'm working from doesn't discuss action until later on. Thought it made a good exercise in partial differentiation and chain rule, which it seems I needed! In fact I'm still struggling a little. Applying the chain rule gives me:
[itex] \frac{d}{dt}\frac{\partial}{\partial \dot{q_k}}\frac{dF}{dt} = \left(\frac{\partial^2 F}{\partial q_k^2} + \frac{\partial^3F}{\partial q_k \partial \dot{q_k}\partial q_k} + \frac{\partial^3F}{\partial q_k\partial \dot{q_k}\partial t}\right) \dot{q_k} + \frac{\partial^2F}{\partial t\partial q_k} + \frac{\partial^3 F}{\partial t\partial\dot{q_k}\partial q_k} + \frac{\partial^3F}{\partial t\partial\dot{q_k}\partial t} [/itex] and [itex] \frac{\partial}{\partial q_k}\frac{dF}{dt} = \frac{\partial}{\partial q_k}\left(\frac{\partial F}{\partial q_k}\dot{q_k} + \frac{\partial F}{\partial t}\right) [/itex] Now, it seems to me that the only way this doesn't end with a lot of second and third order partial derivatives that don't cancel out is if I've got a sign wrong. For instance, if the chain rule had a minus [itex]\frac{\partial F}{\partial t}[/itex] at the end instead of plus, then I think everything would cancel out perfectly with each of the above becoming equal to zero, as required. But I don't think the chain rule should have a negative term like that. Have I missed something? 


#6
Feb2212, 10:05 AM

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Have you read my post #2?
[tex]\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\dot q_k}+ \frac {∂F}{∂t}[/tex] First determine he derivative of dF/dt with respect to one of the velocity components, [itex]\dot q_j[/itex] As F itself does not depend on the velocities, [tex]\frac{\partial}{\partial \dot q_j}\frac{dF}{dt}=\frac{∂F}{∂q_j} [/tex] Then take the time derivative of [itex]\frac{\partial}{\partial \dot q_j}\frac{dF}{dt}=\frac{∂F}{∂q_j}[/itex]: [tex]\frac{d}{dt} ( \frac{∂F}{∂q_j} )[/tex] and you have to subtract [itex] \frac{∂}{∂q_j}\frac{dF}{dt}[/itex] at the end. ehild 


#7
Feb2212, 04:22 PM

P: 14

Ah ha, it was the [itex]\frac{\partial }{\partial \dot{q_j}}\frac{dF}{dt} = \frac{\partial F}{\partial q_j}[/itex] that I missed. That got rid of all the 3rd order derivatives, leaving only 2nd order one's which all cancel each other out in the end. Thanks very much for the help ehild!



#8
Feb2212, 04:58 PM

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It was a great battle that ended well...
ehild 


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