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Action and equations of motion

  1. Aug 30, 2007 #1
    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. jcsd
  3. Aug 30, 2007 #2
    More details would certainly be appreciated :)
  4. Aug 30, 2007 #3


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    Okay [itex] L'=L+\frac{d}{dt}f(q,t) [/itex]. Compute the E-L equations for S' defined in terms of L'. Can you do it ?
  5. Aug 30, 2007 #4
    action.jpg (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...
  6. Aug 30, 2007 #5


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    Perfect so far. Now who's

    [tex] \frac{d}{dt}f(q(t),t) [/tex] equal to ? HINT: Chain rule, you've got both implicit and explicit time dependence.
  7. Jan 2, 2011 #6
    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.
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