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Question on Lagrangian Mechanics

  1. Apr 7, 2005 #1
    "There is some freedom as to what we choose for the Lagrangian in a given problem: We can add a constant, multiply by a constant, change the time scale by a multiplicative constant, or add the total time derivative ........... Any of these transformations will lead to a Lagrangian that is perfectly satisfactory for describing the motion."

    I could not verify the 3rd one. Why is it possible to change the time scale by a constant factor?

    For example:
    [tex]L = \frac{1}{2}m\dot{q}^2 - kqt[/tex]

    The E-L equation is
    [tex]m\ddot{q} + kt = 0[/tex]

    If we modify the Lagrangian, multiplying all time by constant c
    [tex]L = \frac{1}{2c^2}m\dot{q}^2 - ckqt[/tex]

    The E-L equation is
    [tex]m\ddot{q} + c^3kt = 0[/tex]
  2. jcsd
  3. Apr 8, 2005 #2
    http://www.physik.fu-berlin.de/~kleinert/b6/psfiles/conslaw.pdf [Broken]

    I refer to section 6.2 on page 4 of the above link.

    If the Lagrangian does NOT depend on time explicitely (which is NOT the case in your example), then this L will be invariant under time-transformations t-->t + e where e is a given constant. This symmetry yields the energy conservation law.

    Last edited by a moderator: May 2, 2017
  4. Apr 8, 2005 #3
    Yeah, but that's translating time, and the excerpt he quoted was about scaling time.

    Something's not quite right in kakarukeys' example, but I can't put my finger on it. Certainly it shouldn't matter whether you measure time in seconds or milliseconds. And in your [itex] L = T - V [/itex] example, it doesn't make any sense that [itex]V[/itex] became bigger but [itex]T[/itex] became way smaller (assuming [itex] c > 1[/itex]) when you changed the time scale. They should stay at the same ratio.

    OK, I think I know what the problem is. I think [itex] k [/itex] needs to be in units of mass * length / time^3 for the units to come out right. So when you did your scaling, you also needed to scale [itex] k [/itex], by dividing it by [itex]c^3[/itex]. Then everything has [itex] c^2 [/itex] on the bottom and everyone's happy again. Does that sound right?
    Last edited by a moderator: May 2, 2017
  5. Apr 9, 2005 #4


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    Okay.It doesn't really matter.Rescaling

    [tex] t'=:ct [/tex]

    (adimensional constant,dfan!!).

    [tex] L\left(q,\frac{dq}{dt'},t'\right)=\frac{m}{2}\left(\frac{dq}{dt'}\right)^{2}-kqt' [/tex]

    It can be easily proven that a time rescaling doesn't modify the Lagrange equation(s),so it looks (for this 1D-case)

    [tex]\frac{\partial L}{\partial q}-\frac{d}{dt'}\left(\frac{\partial L}{\partial \left(\frac{dq}{dt'}\right)}\right) =0 [/tex]

    and so,in the rescaled variables,the equation of motion is the same...

    We all know that time inversion is a weird (for relativistic systems,classical & quantum fields) example of a time-rescaling.

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