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About calculus of variation and lagrangian formulation

  1. Nov 9, 2008 #1
    I was reading about the principle of least action and how to derive Newton's second out of it.

    at a certain point I didn't follow the calculations,

    so the author defines a variation in the path, [tex]x(t) \longrightarrow x'(t) = x(t) + a(t), a \ll x[/tex]

    [tex]a(t_1) = a(t_2) = 0[/tex]

    Now, [tex]S \longrightarrow S' = \int_{t_1}^{t_2} (m/2 (\dot{x} +\dot{a})^2 - V(x +a)) dt[/tex]

    [tex]= \int_{t_1}^{t_2} {1/2 m\dot{x}^2 + m\dot{x}\dot{a} - [V(x) + aV'(x)]} dt + O(a^2)[/tex]

    (what happened exactly here? could anybody tell me??)

    then

    [tex]= S + \int_{t_1}^{t_2} [m\dot{x}\dot{a} - aV'(x)] dt[/tex]

    [tex]\equiv S + \delta{S}[/tex]
     
  2. jcsd
  3. Nov 9, 2008 #2

    Hurkyl

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    I assume you meant

    [tex]
    S \longrightarrow S' = \int_{t_1}^{t_2} (m/2 (\dot{x}(t) +\dot{a}(t))^2 - V(x(t) + a(t))) dt
    [/tex]

    ? It looks like they simply used a differential approximation. (i.e. a first-order Taylor series)
     
  4. Nov 9, 2008 #3
    Yes, I got it. Thanks
     
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