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I can't do a differentiation during the proof of Noether's theorem.

  1. Jan 20, 2010 #1
    In Wikipedia,

    http://en.wikipedia.org/wiki/Noether's_theorem#One_independent_variable

    You can see the proof of Noether's theorem for the system that has only one symmetry.
    I can't do the calculation of this, for

    [tex]\frac{dI'}{d\epsilon} = \frac{d}{d\epsilon} \int_{t_1+\epsilon T}^{t_2+\epsilon T} L(\phi(q(t'-\epsilon T),\epsilon), \frac{\partial \phi}{\partial q} (q(t'-\epsilon T), \epsilon) \dot{q} (t' - \epsilon T), t') dt'[/tex]

    it becomes

    [tex]L(q(t_2) , \dot{q}(t_2), t_2)T - L(q(t_1),\dot{q}(t_1) , t_1)T +\int_{t_1}^{t_2} \frac{\partial L}{\partial q} \left( -\frac{\partial \phi}{\partial q} \dot{q} T + \frac{\partial \phi}{\partial \epsilon} \right) + \frac{\partial L}{\partial \dot{q}} \left( -\frac{\partial^2 \phi}{(\partial q)^2} \dot{q}^2 T + \frac{\partial^2 \phi}{\partial \epsilon \partial q} \dot{q} - \frac{\partial \phi}{\partial q} \ddot{q} T \right) dt[/tex]

    How it becomes like this?
     
  2. jcsd
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