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Q. Does Galileo's principle of relativity I imply Galileo's principle of relativity II?

Galileo's principle of relativity I: Newton’s equations $\ddot{x} = F(x, \dot{x}, t)$ are invariant under the Galilean transformation group in an inertial frame of reference. In other words, if the world lines of all the points of any mechanical system in an inertial frame of reference are subjected to one and the same Galilean transformation, then the resulting world lines belong to the same system (with new initial conditions).

Galileo's principle of relativity II: there is some Lagrangian $L = L(x, \dot{x}, t)$ such that Galilean transformations are quasi-symmetries of the action $S = \int^{t_2}_{t_1} L(x, \dot{x}, t) dt$ in an inertial frame of reference.

The second statement implies the first one by this post or this post. I am wondering if the converse is true. That is, given a sytem in an inertial frame of reference, if any Galilean transformation sends every solution to another solution of the same system, then can we conclude that there exists a Lagrangian $L$ of the system such that any Galilean transformation is a quasi-symmetry of its corresponding action $S = \int^{t_2}_{t_1} L dt$? If this is not the case, please give some counterexamnples.

PS. Here, of course, we only consider mechanical systems which can be described by Lagarangians.

Background: The first one is stated by Arnold in Mathematical Methods of Classical Mechanics, while the second one is implicitly used in Mechanics by Landau and Lifshitz.

Zephyr
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  • Symmetries of the Euler-Lagrange equations generically need not be symmetries of the action. – Richard Myers Mar 30 '21 at 21:26
  • @RichardMyers You are correct. But in this setting I only consider symmetries induced by Galilean transformations, the counterexamples I know are all non-Galilean transformations. Also note that I don’t fix a Lagrangian here. – Zephyr Mar 30 '21 at 21:40
  • Oh fair enough. Tempted to say it can't be done in point mechanics because every function you can write down is already a good density on $0+1$ dimensions. If that's wrong though, looking for a function which does not transform like a density would probably be the way to find an example. – Richard Myers Mar 30 '21 at 22:56

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