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Analyzing a coordinate transformation

  1. Oct 6, 2012 #1
    In McCauley's book Classical Mchanics: Transformations, Flows, Integrable and Chaotic Dynamics we are analyzing a coordinate transformation in order to arrive at symmetry laws. A coordinate transformation is given by [itex]q_i(\alpha) = F_i(q_1,...,q_f, \alpha)[/itex]. Then, to the first order Mccauley states in equation 2.50 an infinitesimal shift in the coordinates can be given by

    [tex]
    \delta q_i = \left[ \frac{\partial q_i(\alpha)}{\partial \alpha} \right]_{\alpha=0} \delta \alpha
    [/tex]

    The variation in action is then given as

    [tex] \delta A = \int_{t_1}^{t_2} \frac{dL_{\alpha}}{d \alpha} \delta \alpha dt = \int_{t_1}^{t_2} \left(\frac{\partial L_{\alpha}}{\partial q_i(\alpha)}{\partial q_i(\alpha)}{\partial \alpha} + \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \frac{\partial \dot{q}_i(\alpha)}{\alpha} \right) \delta \alpha dt.[/tex]

    I understand up to here. The book then states that we can deduce the following from the above:

    [tex] \delta A = \left[ \left[ \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \delta q_i \right]^{t_2}_{t_1} \right]_{\alpha=0} = \left[ p_i(\alpha) \left[\frac{\partial q_i(\alpha)}{\partial \alpha} \right]_{\alpha = 0} \delta \alpha \right]_{t_1}^{t_2} .[/tex]

    I don't understand this line. For instance, shouldn't it be [itex] \delta \dot{q}_i[/itex] in the second expression? Or shouldn't the q be undotted? But then where does the second term from the above equation go? Again, any help would be appreciated.

    Thanks.
     
  2. jcsd
  3. Oct 6, 2012 #2

    vanhees71

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    According to the form of your transformation, i.e., a point transformation that is not explicitly time dependent (also [itex]\alpha[/itex] is not time dependent here, i.e., you have a global symmetry under consideration) you have
    [tex]\frac{\partial \dot{q}_i}{\partial \alpha}=\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial q_i}{\partial \alpha}.[/tex]
    Further, due to the equations of motion (Euler-Lagrange equations of the variational problem of Hamilton's principle) you have
    [tex]
    \frac{\partial L}{\partial q_i} = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}_i}.
    [/tex]
    Thus, for the trajectory of the particle the integrand is a total time derivative and thus you have
    [tex]\delta A=\delta \alpha \left [ \left (p_i \frac{\partial q_i}{\partial \alpha} \right )_{\alpha=0} \right]_{t=t_1}^{t=t_2}.[/tex]
    As written in the book.

    Further by assumption your infinitesimal transformation is a symmetry of the action, i.e., the actino doesn't change, and our derivation shows that along the trajectory of the particle the quantity
    [tex]Q=\left (p_i \frac{\partial q_i}{\partial_ \alpha}\right )_{\alpha=0}[/tex]
    is constant in time, i.e., it is the conserved quantity of the symmetry under consideration, which is one of Noether's theorem.

    You can also show the opposite: I.e., if you have a conserved quantity there must be a symmetry of the action, generated by the conserved quantity. In other words each conserved quantity implies a one-parameter symmetry group of the action. This is another of Noether's theorems.

    The latter theorem becomes much more elegant in the Hamiltonian formulation of Hamilton's principle, where you have everything in terms of the Lie algebra on the phase-space functions with the Poisson bracket as the Lie product. Then the infinitesimal symetry transformations form a subalgebra, and the finite symmetry transformations are given by the flow of the Lie derivatives of the phase-space variables with respect to the generator of the symmetry, which then turns out to be the corresponding conserved quantity.
     
  4. Oct 6, 2012 #3
    I think I understand now. The fact I think I overlooked is that everything we are working with must satisfy the Euler-Lagrange equation since we are dealing with real, physical particles. So essentially the integral representing the variation in action can be rewritten as

    [tex]
    \int_{t_1}^{t_2} \left( \frac{d}{dt} \left[ \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \right] \frac{\partial q_i(\alpha)}{\partial \alpha} + \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \frac{\partial \dot{q}_i(\alpha)}{\partial \alpha} \right) \delta \alpha dt = \int_{t_1}^{t_2} \frac{d}{dt} \left[ \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \frac{\partial q_i}{\partial \alpha} \right] \delta \alpha dt = \left[ \frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \frac{\partial q_i}{\partial \alpha} \delta \alpha \right]_{t_1}^{t_2}.
    [/tex]

    Then based on the equation given above for an infinitesimal shift in the coordinates this can all be rewritten as

    [tex]
    \left[\left(\frac{\partial L_{\alpha}}{\partial \dot{q}_i(\alpha)} \delta q_i \right)_{\alpha=0} \right]_{t_1}^{t_2},
    [/tex]

    and by definition of the canonical momentum you get, as you stated

    [tex]
    \delta A=\delta \alpha \left [ \left (p_i \frac{\partial q_i}{\partial \alpha} \right )_{\alpha=0} \right]_{t=t_1}^{t=t_2}.[/tex]

    Then if the action variation is 0, Q will be conserved across the transformation, correct? I don't fully understand how this leads to a conservation law in general, though. For instance, lets say we are restricted to motion on a circle of radius r. Then, we can say that [itex]\alpha = \theta[/itex]. So a change in alpha would be some small rotation of the coordinate system. As we rotate the coordinate system, this means that the canonical (angular) momentum would be conserved. But how does this translate into a conservation law when the particle is moving along the circle itself. Or have I misinterpreted the math?
     
    Last edited: Oct 6, 2012
  5. Oct 7, 2012 #4

    vanhees71

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    Of course, you have to make sure that your transformation really leaves the action invariant. Thus, in your example, the rotation is only a symmetry, if the Lagrangian doesn't depend on [itex]\theta[/itex]. Then the canonical momentum [itex]p_{\theta}[/itex] (which is an angular-momentum component along the axis of rotation, parametrized by [itex]\theta[/itex]). That's why generalized coordinates, on which the Lagrangian doesn't depend are called cyclic.
     
  6. Oct 7, 2012 #5
    So assuming the Lagrangian is independent of theta should make the angular momentum a conserved quantity when the coordinate system is rotated. But can the rotation of the coordinate system be reinterpreted as the particle rotating rather than the coordinate system rotating? Is this interpretation what makes the angular momentum a constant when it is the particle moving rather than the coordinate system rotating?
     
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