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Lagrangian for a free particle

  1. Nov 16, 2011 #1
    In Landau's Mechanics, if an inertial frame [itex]\textit{K}[/itex] is moving with an infinitesimal velocity [itex]\textbf{ε}[/itex] relative to another inertial frame [itex]\textit{K'}[/itex], then [itex]\textbf{v}'=\textbf{v}+\textbf{ε}[/itex]. Since the equations of motion must have the same form in every frame, the Lagrangian [itex]L(v^2)[/itex] must be converted by this transformation into a function [itex] L'[/itex] which differs from [itex]L(v^2)[/itex], if at all, only by the total time derivative of a function of co-ordinates and time. Then he gave the formula [itex]L'=L(v'^2)=L(v^2+2\textbf{v}\bullet\textbf{ε} + \textbf{ε}^2)[/itex].
    So my question is what does the sentence 'the equations of motion must have the same form in every frame' mean? Whether [itex]L'(v'^2)=L(v'^2)[/itex] or [itex]L'(v'^2)=L(v^2)[/itex]? Why?
    And what is the variable in the two Lagrangians,[itex]\textbf{v}[/itex] or [itex]\textbf{v}'[/itex]?
  2. jcsd
  3. Nov 16, 2011 #2


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    The most general symmetry transformation is one that leaves the first variation of the action invariant since the Hamilton principle of stationary action simply says that the first variation of the action vanishes for the solutions of the equation of motion.

    The action itself stays invariant, if the Langrangian in terms of the transformed variables differs from the original Lagrangian only by a total time derivative. Then of course also the first variation is invariant and thus the transformation describes a symmetry of the system.
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