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Free action invariant under galliean boosts?

  1. Apr 29, 2013 #1
    Is the non-relativistic Lagrangian:

    [tex]\mathcal L=\frac{1}{2}m \dot{x}^2 [/tex]

    invariant under boosts x'=x+vt?

    It doesn't seem like it is. Surely something must be wrong?
     
  2. jcsd
  3. Apr 29, 2013 #2

    dextercioby

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    Well, the action must be invariant, not the Lagrangian.
     
  4. Apr 29, 2013 #3
    This is purely a transformation on the field x, and not on the coordinate t, so I think we can consider just the Lagrangian rather than the action.

    In any case, the change in the Lagrangian:

    [tex]\Delta \mathcal L=m \dot{x}v+m\frac{v^2}{2} \approx m \dot{x}v [/tex]

    can't be written as a total time derivative as far as I can tell, so the Lagrangian's are inequivalent. O damn, yes it can:

    [tex]\Delta \mathcal L=\frac{d}{dt}\left(mvx \right) [/tex]

    So the conserved quantity is

    [tex]Q=m\dot{x}t-mx [/tex]

    which is clearly conserved for a free particle.

    So adding a potential where the forces between particles only depends on relative distances doesn't add anything to ΔL, so Q is still conserved for that case.

    So this is just saying that in the absence of external forces, the acceleration of the center of mass is zero, expressed as the conservation of Q. Neat.

    Thanks.
     
  5. Aug 25, 2013 #4
    How did you get from Delta L to Q?
     
  6. Aug 25, 2013 #5

    DrDu

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    Under an infinitesimal transformation
    ##\dot{q} \rightarrow\dot{q}+v## where ##v## is infinitesimally small, the lagrangian transforms as
    ##L=m\dot{q}^2/2 \rightarrow m\dot{q}^2/2+m\dot{q}v##.
    The second term clearly is a time derivative.
    It is sufficient to consider infinitesimal variations of the action.
    You are right about the conservation of the center of mass.
     
  7. Aug 26, 2013 #6

    Bill_K

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    It's done in this paper, Eq.(4.40).
     
    Last edited: Aug 26, 2013
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