(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Physical equivalence of Lagrangian under addition of dF/dt

**Physics Forums | Science Articles, Homework Help, Discussion**