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The classical harmonic oscillator is commonly obtained from the canonical first order Lagrangian: $$L_1=\textstyle\frac{1}{2}m\dot{q}^2-\textstyle\frac{1}{2}kq^2$$ However, if you add the term (I do not understand the reason why it works) $L_2=d(-mq\dot{q}/2)/dt$, you get the Lagrangian (equivalent?):

$$L_3=L_1+L_2=-\dfrac{mq\ddot{q}}{2}-\dfrac{1}{2}kq^2$$

The equations of motion for $L_1$ and $L_3$ are the same since they only differ by a total derivative with respect to time. My questions are:

  1. What is the meaning of $L_3$? Is it really different to $L_1$?
  2. What are the physical applications of $L_3$? I think that it is related to superparticle models but I am not sure.
  3. Sometimes, with respect to the Noether procedure, some books define that equivalent Lagrangians are only those with $L_3=L_1+df(q)/dt$, but sometimes I have seen other refer to this "gauge" symmetry as generalized symmetry if you plug $f(q,\dot{q})$. Why does this difference matter?
  4. Why equivalent lagrangians $L_1, L_3$ do provide different "quantizations"? After all, the correspond to the same "classical" system. Of course, here I pressume there is some caveat due to the Grassmaniann variables or any other stuff I don't understand.
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Comment to the question (v2):

  1. The system has a second order eom, so 2 boundary conditions (BCs) are needed, 1 at initial time, and 1 at final time. Without BCs the variational principle is not well-defined.

    For the Lagrangian $L_1$ there are 2 consistent choices at each end-point: Dirichlet BC or Neumann BC, yielding a total of $2\times 2=4$ possible pairs of consistent BCs.

    For the Lagrangian $L_3$ there are no consistent choices of BCs. In other words, the Lagrangian $L_3$ is not suitable for a well-posed variational problem.

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  • $\begingroup$ And however I saw that lagrangian in superparticle actions, like this Revista Brasileira de Física, Vol. 20, no 2, 1990. On pseudodynamics of second order Lagrangian systems. sbfisica.org.br/bjp/download/v20/v20a09.pdf $\endgroup$ – riemannium Aug 9 '17 at 17:49
  • $\begingroup$ @riemannium the starting point of that paper is a Lagrangian that depends on 2nd order derivatives in time... $\endgroup$ – ZeroTheHero Aug 9 '17 at 18:11
  • $\begingroup$ Right, but why they chose eq. 3.11 or 3.14 for (accelerated) superparticle. I am puzzled...Since usually, we prefer to work out with first order lagrangians... $\endgroup$ – riemannium Aug 9 '17 at 18:16
  • $\begingroup$ Notice that the bosonic sector of the action in eqs. (3.1), (3.2), (3.11) & (3.14) in the link is OP's $\int \! dt~(L_3-L_2)=\int \! dt~ L_1$, which is OK; not OP's $\int \! dt~ L_3$, which is not allowed. $\endgroup$ – Qmechanic Aug 9 '17 at 18:26
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Two Lagrangians which differ by a total derivative in time (as is your $L_2$) do produce the same equations of motion. There is no physical distinction between them in the sense that their physics contents is equivalent (i.e. the produce the same equations of motion).

The situation is similar to that encountered with potentials: you can always add to the electric potential $V$ a constant, and to the vector potential $\vec A$ the gradient of a function. Neither of these change the fields, but adding such additional terms may simplify some calculations or provide additional insights into the solutions.

The possibility of adding a total time derivative is closely related to (in fact the starting point of) the method of generating functions for canonical transformations of the Hamiltonian.

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  • $\begingroup$ Yes, but...Is not a high derivative showing additional curvatures, like torsion, and is different? $\endgroup$ – riemannium Aug 9 '17 at 17:07
  • $\begingroup$ @riemannium I'm not sure I follow. $\endgroup$ – ZeroTheHero Aug 9 '17 at 18:07
  • $\begingroup$ I mean, when you have a lagrangian having more than first derivatives, it includes higher curvatures from the differential geometry viewpoint. What I can not understant (as well) is why they are equivalent if describing "different" curvature types. It is ... curious... $\endgroup$ – riemannium Aug 9 '17 at 18:31
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    $\begingroup$ @riemannium Ok then I had not properly understood your original question as the original Lagrangian as a function of $q$ and $\dot q$ only. $\endgroup$ – ZeroTheHero Aug 9 '17 at 18:39

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