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Simplifying Lagrangian's Equations (Classial Dynamics)

  1. Aug 14, 2010 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img96.imageshack.us/img96/9288/lagrangiansimplifcation.jpg [Broken]
    "[URL [Broken]

    If you require more info in the derivation, it's on page 1:
    http://www.ph.qmul.ac.uk/~phy304/Homework/HW3sol.pdf


    2. Relevant equations
    L = T - V


    3. The attempt at a solution

    I understand exactly the process of obtaining the Lagrangian... But I do not understand his simplifying process at all.

    Could somebody please help/forward me into the right direction?

    Thanks in advance!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 14, 2010 #2

    cristo

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    If you think in terms of the action, you can always add a constant such that, when varied, the action will give the same equations of motion. This translates into meaning that you can always add to the Lagrangian (a) a constant, (b) a function that depends on time but not on the coordinates in the Lagrangian or (c) a total time derivative of a function.

    The first two in the list fall into category (b). If we change the Lagrangian so that

    [tex]\tilde{L}=L+F(t)[/tex],

    then the action is

    [tex]\tilde{S}=\int_{t_1}^{t_2}{\tilde{L}}dt=\int_{t_1}^{t_2}{L}dt+\int_{t_1}^{t_2}F(t)dt=S+\int_{t_1}^{t_2}F(t)dt=S+{\rm const.}[/tex]

    In the third point on that list, a term is rewritten in terms of a total time derivative plus some other term; the total time derivative then being discarded. To see why, let's look at the change of the action due to the Lagrangian changing to

    [tex]\tilde{L}=L+\frac{d}{dt}G(\varphi,t)[/tex]

    which gives

    [tex]\tilde{S}=\int_{t_1}^{t_2}Ldt+\int_{t_1}^{t_2}\frac{d}{dt}G(\varphi,t)dt
    =S+G(\varphi(t_1),t_1)-G(\varphi(t_2),t_2)=S+{\rm const.}[/tex]
     
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