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For a classical free 1-d particle, the conserved quantities of the dynamics are:

$Q_1=p$
$Q_2=q-\frac{pt}{m}$.

The symmetry associated with $Q_1$ is translation symmetry, as I know.
What is the symmetry associated with $Q_2$?
Why is this symmetry (and conserved quantity) neglected in comparison to time and space translation symmetries?

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  • $\begingroup$ Do you mean by $q$ the position? If so, how can $q$ and $\frac{pt}{m}$ have the same units? $\endgroup$ – Deschele Schilder Aug 25 '20 at 17:59
  • $\begingroup$ They have the same units as $p=mv$. $\endgroup$ – Valter Moretti Aug 25 '20 at 18:16
  • $\begingroup$ @ValterMoretti I mean the unit of $q$ (generalized position). You can't subtract $\frac{pt}{m}$ from $q$ if they don't have the same units. $\endgroup$ – Deschele Schilder Aug 25 '20 at 18:49
  • $\begingroup$ Yes, that is true in general, but here $q$ is the true position and $p$ the true momentum, they are not generalized coordinates...otherwise the exercise is pointless. $\endgroup$ – Valter Moretti Aug 25 '20 at 18:52
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I prefer to use $x$ to denote your coordinate $q$ since it is evidently the Cartesian position of the material point (otherwise the exercise does not make sense).

The symmetry associated to $mQ_2$ is the boost: $$x \to x-vt =: x'\:,$$ where the real $v$ is the parameter of the group of transformations. Under this transformation the free Lagrangian is invariant at the first order in $v$ up to a total derivative. $$L\to L- v \frac{dmx}{dt} +O(v^2)$$ Applying Noether's theorem, you see that $m Q_2$ is the corresponding constant of motion: $$ I := \frac{\partial L}{\partial \dot{x}} \frac{\partial x'}{\partial v}|_{v=0} - mx =m Q_2\:.$$ This symmetry is sometimes neglected because it needs a formulation of the Noether theorem a bit more sophisticated, and it can be also obtained indirectly from other constants of motion. However it is of crucial relevance as it relies upon one of the most important notions of invariance in classical physics: the classical (Galileian) relativity principle.

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$Q_2$ is the initial position, which is a constant of motion$^1$. As always in an Hamiltonian formulation, the corresponding infinitesimal symmetry is generated by the Poisson bracket $\{Q_2,\cdot\}$, cf. e.g. this Phys.SE post.

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$^1$ $Q_2$ is by definition not an integral of motion, since it depends explicitly on time, cf. e.g. this Phys.SE post.

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