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Emmy Noether theorem

  1. Mar 21, 2012 #1
    If we watch some translation in space.

    [tex]L(q_i+\delta q_i,\dot{q}_i,t)=L(q_i,\dot{q}_i,t)+\frac{\partial L}{\partial q_i}\delta q_i+...[/tex]

    and we say then
    [tex]\frac{\partial L}{\partial q_i}=0[/tex]

    But we know that lagrangians [tex]L[/tex] and [tex]L'=L+\frac{df}{dt}[/tex] are equivalent. How we know that [tex]\frac{\partial L}{\partial q_i}\delta q_i[/tex] isn't time derivative of some function [tex]f[/tex]?
  2. jcsd
  3. Mar 21, 2012 #2


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    For more general symmetries than translation symmetries, you need to take into account the possibility that the Lagrangian changes by the total time derivative of a function that is only a function of the generalized coordinates (and perhaps explicitly on time) but not the generalized velocities. An example is invariance under Galileo transformations in Newtonian mechanics (Galileo boosts) or Lorentz boosts in Special Relativity, which both lead to the constant velocity of the center of mass of a closed system of point particles.
  4. Mar 21, 2012 #3
    I don't want more general symmetries than translation symmetries. I asked my question about invariance under translation symmetries. Can you answered the question about this problem which I asked?
  5. Mar 22, 2012 #4


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    But for this problem, you've already given the answer yourself: A generalized translation is a symmetry if the Lagrangian doesn't depend on the corresponding generalized coordinate [itex]q_1[/itex], and then from the Euler-Lagrange equation, you get

    [tex]\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}_1} = \frac{\partial L}{\partial q_1}=0,[/tex]

    which means that the conserved quantity is given by the canonical momentum of this variable, i.e.

    [tex]p_i=\frac{\partial L}{\partial \dot{q}_1}.[/tex]
  6. Mar 22, 2012 #5
    I think we don't understand each other. My question is. If two lagrangians [tex]L'=L+\frac{df}{dt}[/tex] and [tex]L[/tex] gives us same dynamics. Why can't be that

    [tex]\frac{\partial L}{\partial q}\delta q=\frac{df}{dt}[/tex]?
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