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Variation of the lagrangian

  1. Aug 27, 2011 #1
    Given an action:

    [tex]S = \int L(q,\dot{q},t) \,dt[/tex]

    The variation is:

    [tex]\delta S = \int \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta\dot{q}\right)\,dt[/tex]

    I'm guessing this is some type of chain rule, but I haven't been able to derive it... how is it justified?
     
  2. jcsd
  3. Aug 27, 2011 #2

    WannabeNewton

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    Let [itex]x_{l}(t)[/itex] be the path of least action. Consider the path [itex]x_{l}(t) + \zeta (t)[/itex] where [itex]\zeta (t) << 1 [/itex]. So, [tex]S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}L(x_{l}(t) + \zeta (t); \dot{x_{l}}(t) + \dot{\zeta }(t))dt [/tex] After a taylor expansion to first order in [itex]\boldsymbol{\zeta }[/itex], [tex]S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}[L(x_{l}(t), \dot{x_{l}}(t)) + \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt[/tex] (with the expansion terms along the path of least action) so the variation is given by [tex]\delta S = \int_{t_{i}}^{t_{f}}[ \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt[/tex]
     
  4. Aug 27, 2011 #3
    Ah, thanks for that and the fast reply :)
     
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