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I Help a novice with EL equation derivation

  1. Jun 15, 2017 #1
    Hello everyone,

    Reading Landau and Lifshitz Course of Theoretical Physics Volume 1: Mechanics (page 3) I got suck in the following step (and I cite in italics):

    The change in S when q is replaced by q+δq is
    [tex]\int_{t_1}^{t_2} L(q+δq, \dot q +δ\dot q, t)dt - \int_{t_1}^{t_2} L(q, \dot q, t)dt[/tex]

    (So far so good)

    When this difference is expanded in powers of δq and [itex]δ\dot q[/itex] in the integrand, the leading terms are of the first order.

    How do you expand that? Could anyone show me how explicitly if you don't know the explicit form of the Lagrangian?

    The necessary condition for S (where S is the action) to have a minimum is that these terms (called the first variation, or simply the variation, of the integral) should be zero. Thus the principle of least action may be written in the form
    [tex]δS = δ \int_{t_1}^{t_2} L(q, \dot q, t)dt = 0[/tex]


    (Which I'm fine with the above expression)

    Or, effecting the variation,
    [tex]\int_{t_1}^{t_2} (
    \frac{\partial L}{\partial q}δq+
    \frac{\partial L}{\partial \dot q}δ\dot q)
    dt = 0[/tex]

    Now my guess would have included [itex]\frac{\partial L}{\partial t}[/itex] like:
    [tex]\int_{t_1}^{t_2} (
    \frac{\partial L}{\partial q}δq+
    \frac{\partial L}{\partial \dot q}δ\dot q + \frac{\partial L}{\partial t}dt)
    dt = 0[/tex]

    To perform the total differential of all variables.
    Explain me like if I'm five why this guess is wrong.

    Thanks so much in advance! And sorry for my lack of elemental knowledge.
    -Alex
     
  2. jcsd
  3. Jun 15, 2017 #2

    BvU

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    Science Advisor
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    Hi,

    ##t## plays a different role here: it is the integration variable. We want to minimize ##S## by finding a path ##\Bigl ( q(t), \dot q(t) \Bigr )## that achieves this ##\delta S=0## condition. There is no value of ##t## to find; it just runs from ##t_1## to ##t_2## and nothing can be done with ##\partial L \over \partial t##. The time dependence of ##L## is present in the integration, though.
     
  4. Jun 15, 2017 #3
    Thanks BvU!
    Any ideas on how to do this and why is it relevant?
     
  5. Jun 15, 2017 #4
    The expansion is basically a Taylor expansion about δq and δ{itex}\dot q{\itex}t keeping the first order powers in in the differentials. For a deeper understanding, read on functional derivatives and or the calculus of variations.
     
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