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A Time-reparameterization invariance

  1. Apr 17, 2017 #1
    This post considers an aspect of time-reparametization invariance in classical Hamiltonian mechanics. Specifically, it concerns the use of Lagrange multipliers to rewrite the action for a classical system in a time-reparametization-invariant way.

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    Prelude:

    Suppose we have a system with a single degree of freedom ##q(t)## with conjugate momentum ##p##, and action
    $$I = \int dt\ L.$$
    The Hamiltonian is the Legendre transform
    $$H(p,q) = p\dot{q}-L(q,\dot{q})|_{p=\partial L/\partial\dot{q}}.$$
    The independent variable ##t## is special. It labels the dynamics but does not participate as a degree of freedom.

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    Time-reparametrization symmetry:

    Let us introduce a fake time-reparameterization symmetry by labelling the dynamics by an arbitrary parameter ##\tau## and introducing a physical 'clock' variable ##T##, treating it as a dynamical degree of freedom. So we consider the system of variables and conjugate momenta
    $$q(\tau),\qquad p(\tau),\qquad T(\tau), \qquad \Pi(\tau)$$
    where ##\Pi## is the momentum conjugate to ##T##. This is equivalent to the original original system if we use the 'parameterized' action
    $$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),$$
    where prime ##= d=d/d\tau##. Here ##N(\tau)## is a Lagrange multiplier, which enforces the 'constraint equation'
    $$\Pi + H(p,q) = 0.$$

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    My difficulty lies with the introduction of the Lagrange multipliers. How do you show that the action

    $$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),$$

    with Lagrange multiplier ##N(\tau)##, reduces to the action

    $$I' = \int d\tau\ (pq' - H(p,q)T')?$$
     
  2. jcsd
  3. Apr 18, 2017 #2

    stevendaryl

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    Well, they certainly are the same once you plug in [itex]R=0[/itex] (which implies that [itex]\Pi = -H[/itex]).
     
  4. Apr 18, 2017 #3
    No, you solve the Euler-Lagrange equation for ##N## to find that ##R=0##. This is your constraint.
     
  5. Apr 19, 2017 #4

    stevendaryl

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    I think you have it backwards. The constraint is not something you prove is true. It's what you assume is true. Then you use the constraint to find the value of the Lagrange multiplier (if you care about that--it's usually not necessary).

    The Euler-Lagrange equations do not uniquely determine [itex]\Pi[/itex], they only imply that [itex]\Pi[/itex] is a constant. And they also imply that [itex]H[/itex] is constant. So the combination [itex]\Pi + H[/itex] is some constant, but the Euler-Lagrange equations don't determine what that constant is. They give the following equations (varying all 4 quantities: [itex]q, p, \Pi, T[/itex]:

    1. [itex]\frac{dq}{d\tau} -N \frac{\partial H}{\partial p} = 0[/itex]
    2. [itex]-N \frac{\partial H}{\partial q} = \frac{dp}{d\tau}[/itex]
    3. [itex]\frac{dT}{d\tau} - N = 0[/itex]
    4. [itex]\frac{d\Pi}{d\tau} = 0[/itex]
    There is no information about the value of [itex]\Pi[/itex] other than that it's constant. The constraint provides another equation: [itex]R = 0[/itex]. With that additional equation, you have 5 equations and 5 unknowns ([itex]p,q,\Pi,T, N[/itex].

    Here's a much simpler example for using Lagrange multipliers: Suppose you want to maximize [itex]x+y[/itex] subject to the constraint that [itex]x^2 + y^2 = 1[/itex]. Then you create the variational problem: Maximize:

    [itex]x + y - \lambda R[/itex] (where [itex]R = x^2 + y^2 - 1[/itex]

    Varying x gives: [itex]1 - 2 \lambda x = 0[/itex]
    Varying y gives: [itex]1 - 2 \lambda y = 0[/itex]

    So the solution is [itex]x = \frac{1}{2\lambda}, y = \frac{1}{2\lambda}[/itex]. That doesn't determine the value of [itex]R[/itex] (and it also doesn't determine [itex]\lambda[/itex]). You just get: [itex]R = \frac{1 - 2 \lambda^2}{2\lambda^2}[/itex]. You have to impose [itex]R=0[/itex] as an additional equation, you can't derive it. Once you impose the constraint [itex]R = 0[/itex], you uniquely determine [itex]\lambda = \frac{1}{\sqrt{2}}[/itex], which allows you to solve for [itex]x[/itex] and [itex]y[/itex].
     
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