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.[/color]

 

[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:

<br /> \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}<br />

and

<br /> \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)<br />

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}<br />

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