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Homework Help: Transformation that Preserves Action (Lagrangian Mechanics)

  1. Sep 6, 2015 #1
    Hello, I am stuck on the following problem.

    1. The problem statement, all variables and given/known data

    Consider the continuous family of coordinate and time transformations (for small ##\epsilon##).

    [tex]Q^{\alpha}=q^{\alpha}+\epsilon f^{\alpha}(q,t)[/tex]
    [tex]T= t+\epsilon \tau (q,t)[/tex]

    Show that if this transformation preserves the action
    [tex]\frac{\partial L}{\partial \dot{q}^{\alpha}}(\dot{q}^{\alpha}\tau-f^{\alpha})-L\tau[/tex]
    is a constant of motion.

    2. Relevant equations

    I am guessing as to which equations I may need

    Extended Noether's Theorem:
    [tex]\Gamma=\sum_{\alpha}\frac{\partial L}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]
    Where ##\Psi## is a function in which it's time derivative is the difference between the Lagrangian before and after the transformation (Note that it is possible that the extended Noether's theorem doesn't apply and the normal Noether's theorem is sufficient, I do not know).

    The Energy Function:
    [tex]E=\sum_{\beta}\dot{x}_{\beta}\frac{\partial L}{\partial \dot{x}_{\beta}}-L[/tex]

    3. The attempt at a solution

    So the first thing I think I should do is apply the transformation to the Action. First I need to find ##dT##.
    [tex]T= t+\epsilon \tau (q,t) \rightarrow dT=(1+\epsilon \dot{\tau})dt \rightarrow dt=\frac{1}{1-(-\epsilon \dot{\tau})}dT \approx (1-\epsilon \dot{\tau})dT[/tex]
    Thus the action is,
    [tex]S'=\int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}L(Q,\dot{Q},T)dT-\epsilon \int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}\tau L(Q,\dot{Q},T)dT.[/tex]
    [tex]S'=\int_{T_1}^{T_2}L(Q,\dot{Q},T)dT-\epsilon \tau L[/tex]
    [tex]S'=S-\epsilon \tau L[/tex]

    Thus when applying this transformation, the two actions differ by ##\epsilon \tau L##. This seems very similar to Noether's theorem except for that theorem we are talking about Lagrangians and not Actions. However, since the action is just the integral of the Lagrangian, can I assume the same principles apply? In Noether's theorem, we had
    [tex]\frac{\partial L'}{\partial \epsilon}=\frac{d \Psi}{dt}[/tex]
    In this case it appears that we have the following,
    [tex]\frac{\partial S'}{\partial \epsilon}=\frac{d \Psi}{dT}=-L\tau[/tex]

    Can I therefore make the conjecture that,
    [tex]\Gamma=\sum_{\alpha}\frac{\partial S}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]

    However evaluating this I do not get the required answer. :(
  2. jcsd
  3. Sep 6, 2015 #2
    I have tried a different approach and am still getting stuck.

    The change in action up to first order is given by,

    [tex]S'-S=\int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial t}\delta t + \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q}\right][/tex]
    [tex]S'-S=\int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial t}\epsilon \tau + \frac{\partial L}{\partial q}\epsilon f+ \frac{\partial L}{\partial \dot{q}}\frac{d}{dt}\left( \epsilon f \right)\right][/tex]

    I have tried doing various integration by parts and still cannot seem to get to the correct answer. :(


    A little more headway. I can re-write the integrand as the following.
    [tex]\left[ \frac{\partial L}{\partial t}\epsilon \tau + \frac{\partial L}{\partial q}\epsilon f+ \frac{\partial L}{\partial \dot{q}}\frac{d \epsilon}{dt} f + \frac{\partial L}{\partial \dot{q}}\frac{df}{dt} \epsilon \right][/tex]

    [tex]\left[ \left( \frac{\partial L}{\partial t} \tau + \frac{\partial L}{\partial q} f + \frac{\partial L}{\partial \dot{q}}\frac{df}{dt}\right)\epsilon + \frac{\partial L}{\partial \dot{q}}\frac{d \epsilon}{dt} f \right][/tex]

    [tex]\left[ \frac{dL}{d\epsilon}\epsilon + \frac{\partial L}{\partial \dot{q}}\frac{d \epsilon}{dt} f \right][/tex]

    I am gettin' there! (I think). The goal is to write the integrand of the same form as the constant of motion. The integrand must be zero for the action to be invariant under the transformation.
    Last edited: Sep 6, 2015
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