Physical equivalence of Lagrangian under addition of dF/dt

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Homework Statement

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 $dF/dt$ to the Lagrangian, where $F \equiv F(q_1, ..., q_2, t)$.

Homework Equations

Euler-Lagrange equations: $\frac{d}{dt}\frac{∂L}{∂\dot{q_k}} - \frac{∂L}{∂q_k} = 0$

Change of Lagrangian $L \rightarrow L' = L + \frac{dF}{dt}$

Chain rule: $\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\frac{∂q_k}{∂t}}$

The Attempt at a Solution

I guess the solution is to substitute the new Lagrangian, $L'$, into the EL equations and somehow show that it reduces to exactly the EL equations for $L$. The substitution gives:

$\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\left(L + \frac{dF}{dt}\right) - \frac{∂}{∂q_k}\left(L + \frac{dF}{dt}\right) = 0$

This can be rewritten as:

$\left(\frac{d}{dt}\frac{∂L}{∂\dot{q_k}} - \frac{∂L}{∂q_k}\right) + \left(\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{dF}{dt} - \frac{∂}{∂q_k}\frac{dF}{dt}\right) = 0$

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.

$\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{dF}{dt} - \frac{∂}{∂q_k}\frac{dF}{dt} = 0$

All I can think of to try next is apply the chain rule to the differentiation of $F$ and then hope that everything cancels nicely.

Applying the chain rule gives:

$\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{∂F}{∂q_k} \frac{∂q_k}{∂t} + \frac{∂}{∂q_k}\frac{∂F}{∂q_k}\frac{∂q_k}{∂t} = 0$

So I guess my proof works if $\frac{d}{dt}\frac{∂}{∂\dot{q_k}} = \frac{d}{dt}\frac{∂}{∂q_k/∂t}$ is equivalent to $\frac{∂}{∂q_k}$. Is this correct? I feel like you can't just cancel the $dt$ and $∂t$ like this, but I can't see how else this proof can be done.

 

[ehild]

Chain rule: $\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\frac{∂q_k}{∂t}}$

Correctly: $$\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\frac{dq_k}{dt}}+ \frac {∂F}{∂t}$$,

that is

$$\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\dot q_k}+ \frac {∂F}{∂t}$$

Therefore, I now need to show that the other bracketed terms equate to zero also. i.e.

$\frac{d}{dt}\frac{∂}{∂\dot{q_k}}\frac{dF}{dt} - \frac{∂}{∂q_k}\frac{dF}{dt} = 0$

Correct.

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


ehild

 

[genericusrnme]

You can do it without going all that dirty work by looking at the action

$S=\int L dt = \int (L' + \frac{dF}{dt})dt$

That method seems much neater to me

Mod note: removed remaining steps.

 

[ehild]

You can do it without going all that dirty work by looking at the action
That method seems much neater to me

Yes, the variation of F disappears between the endpoints of the path... That was as we learnt, but it works with the derivatives, too.

 

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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:

$\frac{d}{dt}\frac{\partial}{\partial \dot{q_k}}\frac{dF}{dt} = \left(\frac{\partial^2 F}{\partial q_k^2} + \frac{\partial^3F}{\partial q_k \partial \dot{q_k}\partial q_k} + \frac{\partial^3F}{\partial q_k\partial \dot{q_k}\partial t}\right) \dot{q_k} + \frac{\partial^2F}{\partial t\partial q_k} + \frac{\partial^3 F}{\partial t\partial\dot{q_k}\partial q_k} + \frac{\partial^3F}{\partial t\partial\dot{q_k}\partial t}$

and

$\frac{\partial}{\partial q_k}\frac{dF}{dt} = \frac{\partial}{\partial q_k}\left(\frac{\partial F}{\partial q_k}\dot{q_k} + \frac{\partial F}{\partial t}\right)$

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 $\frac{\partial F}{\partial t}$ 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?

 

[ehild]

Have you read my post #2?

$$\frac{dF}{dt} = \sum\limits_k{\frac{∂F}{∂q_k}\dot q_k}+ \frac {∂F}{∂t}$$

First determine he derivative of dF/dt with respect to one of the velocity components, $\dot q_j$
As F itself does not depend on the velocities,
$$\frac{\partial}{\partial \dot q_j}\frac{dF}{dt}=\frac{∂F}{∂q_j}$$

Then take the time derivative of $\frac{\partial}{\partial \dot q_j}\frac{dF}{dt}=\frac{∂F}{∂q_j}$:
$$\frac{d}{dt} ( \frac{∂F}{∂q_j} )$$

and you have to subtract $\frac{∂}{∂q_j}\frac{dF}{dt}$ at the end.


ehild

 

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Ah ha, it was the $\frac{\partial }{\partial \dot{q_j}}\frac{dF}{dt} = \frac{\partial F}{\partial q_j}$ 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!

 

[ehild]

It was a great battle that ended well...

ehild