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In Lagrangian formalism, we consider point transformations $Q_i=Q_i(q,t)$ because the Euler-Lagrange equation is covariant only under these transformations. Point transformations do not explicitly depend on $\dot{q}_i$ (except when we include change of time $t$). A point transformation that satisfies $$L(Q,\dot{Q},t)=L(q,\dot{q},t)+\frac{d}{dt}\chi(q,t)$$ is a symmetry of the system. When it is continuous its infinitesimal version implies the existence of a conserved quantity (Noether theorem).

In Hamiltonian formalism, we consider a larger class of transformations, called canonical transformations (definition is summarized in this post), $Q_i=Q_i(q,p,t)$, $P_i=P_i(q,p,t)$. A generator $G(q,p,t)$ that satisfies $$ \{G,H\}+\frac{\partial G}{\partial t}=0 $$ is a symmetry generator and is conserved (Noether theorem).

Does this mean we could get a larger class of conserved charges in Hamiltonian formalism? Are there any concrete examples? I was thinking about the Runge–Lenz vector in the Kepler-type problem but not sure if it can be discussed in the Lagrangian when the change of time is allowed.

---edits---

I now think we need to make the following clear distinction.

[A] A coordinate transformation that leaves the Euler-Lagrange equation covariant is point transformation $Q_i(q,t)$. The ambiguity in the Lagrangian that does not affect EL eq takes the form $\frac{d}{dt}\chi(q,t)$.

[B] To derive a conserved quantity, we can consider wider class of transformation $Q_i(q,\dot{q},t)$. The change in the Lagrangian is allowed to take the form $\frac{d}{dt}\chi(q,\dot{q},t)$. This is not usually considered as a symmetry because the EL eq is not covariant (at least without modification).

Examples of type [B] include the Runge-Lenz vector.

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    $\begingroup$ Related: physics.stackexchange.com/q/352393/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 15 at 22:23
  • $\begingroup$ I believe this is different. The link asks the same transformation in two formalisms, and tries to verify the results are the same. Here I'm asking a possible transformation not allowed in Lagrangian formalism that gives a conserved quantity. $\endgroup$
    – watahoo
    Jan 15 at 22:28
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    $\begingroup$ The classic example that comes to mind is the $U(1)$ symmetry $q \to q\cos\theta - p\sin\theta; p \to q\sin\theta + p\cos\theta$ of the Hamiltonian $H = \frac{1}{2}q^2 + \frac{1}{2}p^2$ (the harmonic oscillator with a convenient choice of units), which corresponds to conservation of energy. The Lagrangian formalism doesn't know about this symmetry in a direct way (there is no corresponding transformation of the positions and velocities that leaves the Lagrangian invariant), but of course does in the indirect way that it knows that the Hamiltonian is a constant of motion. $\endgroup$ Jan 16 at 5:05
  • $\begingroup$ Yes, I thought about this. This U(1) symmetry is generated by H itself. In other words, the Noether charge associated with this symmetry is H. Then the question is whether the Lagrangian formalism can deal with the time-translation symmetry, and the answer is yes if we allow for the change of time: $t'=t+\epsilon$ and $Q(t')=q(t)$. $\endgroup$
    – watahoo
    Jan 16 at 5:39
  • $\begingroup$ I found that higher dimensional generalizations of harmonic oscillator (e.g., 2D version) are widely discussed as an example of hidden symmetries. In this case conserved charges are more than the Hamiltonian itself. $\endgroup$
    – watahoo
    Jan 19 at 6:43

1 Answer 1

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In the Lagrangian formulation, an (infinitesimal) quasi-symmetry transformation does not have to be a point transformation; it may also depend on velocities.

With this in mind, one may argue that there is a bijective correspondence between quasisymmetries in the Lagrangian and Hamiltonian formalisms, cf. e.g. this Phys.SE post.

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  • $\begingroup$ I looked at J.V. Jose & E.J. Saletan, Classical Dynamics: A Contemporary Approach, 1998; p. 565. It seems to me that it just refers to $\chi$ in $L'=L+\frac{d}{dt}\chi$. Why does this imply "it may also depend on velocities"? $\endgroup$
    – watahoo
    Jan 15 at 23:10
  • $\begingroup$ J&S on p. 565 generalizes the Lagrangian formalism to field theory (rather than point mechanics), and does seemingly not stress the possible dependence on velocity fields (i.e. time derivatives of fields). $\endgroup$
    – Qmechanic
    Jan 16 at 7:54
  • $\begingroup$ If a transformation depends on velocities, the Euler-Lagrange equation may not be left invariant. Usually such a transformation is not called a symmetry. I thought the idea of Noether theorem is the correspondance between a continuous symmetry and a conserved charge. In this sense, I think this treatment is still out of the standard Noether theorem. $\endgroup$
    – watahoo
    Jan 19 at 6:57
  • $\begingroup$ Or are you saying that the Hamiltonian-Lagrangain approach allows us to include the velocity dependence in the transformation rule without changing the Euler-Lagrange equation? $\endgroup$
    – watahoo
    Jan 19 at 6:58
  • $\begingroup$ Hi @watahoo. Thanks for the feedback. This Phys.SE post may partially(?) answer you comment. $\endgroup$
    – Qmechanic
    Jan 19 at 7:53

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