(adsbygoogle = window.adsbygoogle || []).push({}); 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 Euler-Lagrange equations reduce to the same thing.

The question I can't solve asks for a proof that the Euler-Lagrange (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

Euler-Lagrange 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.

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# Homework Help: Physical equivalence of Lagrangian under addition of dF/dt

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