# Action and equations of motion

1. Aug 30, 2007

### TeTeC

Hello !

I'm reading Landau/Lifchitz's mechanics book.

At equation (2.8), the author explains that when I add a time derivative of any function of time and coordinates f(q,t) to the lagrangian, the equations of motion are unchanged.

I understand the mathematical development leading to S' = S + f(q(2),t_2) + f(q(1),t_1), but I can't see why the equations of motion don't change.

I've tried to substitude the lagrangian of equation (2.8) in the Euler-Lagrange equations to convince myself it works, but that doesn't seem to be a good idea.

If you don't have this book and can't see what I'm talking about, I can provide you with more details.

Thank you !

2. Aug 30, 2007

### Nesk

More details would certainly be appreciated :)

3. Aug 30, 2007

### dextercioby

Okay $L'=L+\frac{d}{dt}f(q,t)$. Compute the E-L equations for S' defined in terms of L'. Can you do it ?

4. Aug 30, 2007

### TeTeC

(click on it, obviously my attachment becomes very small...)

Like this ? If this is correct, I don't see why the two last terms should disappear...

5. Aug 30, 2007

### dextercioby

Perfect so far. Now who's

$$\frac{d}{dt}f(q(t),t)$$ equal to ? HINT: Chain rule, you've got both implicit and explicit time dependence.

6. Jan 2, 2011

### sijokjoseph

That simply imply that you can sit in a different frame of reference /coordinate and see the system.
But you can prove that simply,
When the perfect time derivative dF is added to Lagrangian, to calculate the action you have to integrate a perfect differential dF. Then you get a function F which is the functions of position and momentum coordinates and their variation vanishes at the end points. So the variation of the function F also vanishes.