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Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as: $$ ma = -mg $$ From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian: $$ L = \frac{mv^2}{2} - mgx $$ because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see).

Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?

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Well, it seems to me that under a translation $x(t) \to x(t) + c$, the Lagrangian goes to $$\mathcal{L} \to \mathcal{L}' = \frac{1}{2}m v^2 - mg(x+c) = \mathcal{L} -mgc.$$

So yes, the Lagrangian may appear to be different, however since it only shifts by a constant, these two Lagrangians ($\mathcal{L}$ and $\mathcal{L}'$) are equivalent and produce the same Euler-Lagrange Equations. Indeed, more generally, two Lagrangians are equivalent if their difference is a total time derivative. i.e. $\mathcal{L}$ and $$\mathcal{L}' = \mathcal{L} + \frac{\text{d}f}{\text{d}t}$$ are equivalent for any $f(t)$.

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  • $\begingroup$ Thank you. But do we have a conserved quantity? $\endgroup$
    – RedGiant
    Aug 31 at 14:06
  • $\begingroup$ @RedGiant Ah yes, I forgot to address that. I believe (though it's been a while since I actually did the calculations) that the conserved quantity is just the energy. This is not strictly a symmetry of the Lagrangian, but a "quasi"-symmetry, but happily I believe Noether's theorem still holds for it. See the answers to this question: Invariance of Lagrangian in Noether's theorem. $\endgroup$
    – Philip
    Aug 31 at 14:11
  • $\begingroup$ The second answer in the question I linked above seems to corroborate that the conserved quantity is the Hamiltonian, but I feel @MichaelSeifert's answer below showing that it's the initial momentum is much more obvious. I'm try to reproduce the steps in the other answer. Nevertheless, this at least drives home the point that not all "conserved" quantities are useful quantities! :) $\endgroup$
    – Philip
    Aug 31 at 14:28
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You can do an integration by parts on the last term (and discard the resulting boundary term) to yield an action with equivalent EOMs:$$ S'= \int \left( \frac{1}{2} m \dot{x}^2 + m g t \dot{x} \right) \, dt $$ In this context, the symmetry $x \to x + C$ is obvious at the level of the Lagrangian. Moreover, the Euler-Lagrange equations become $$ \frac{d}{dt} \left( m \dot{x} + m g t \right) = \frac{\partial \mathcal{L}'}{\partial \dot{x}} = 0 $$ and thus the quantity $m \dot{x} + m g t$ is a constant of the motion. Specifically, it's the initial momentum of the particle.

(This seems "cheap", somehow, and I'm not 100% sure whether it's a legitimate move. Comments are welcome.)

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  1. The (infinitesimal) translation $$\delta x~=~\epsilon$$ changes OP's Lagrangian with a total time-derivative $$\delta L~=~mg \epsilon~=~ \frac{d}{dt}(mg \epsilon t).$$ This is known as a quasi-symmetry. Noether's theorem does also hold for quasi-symmetries.

  2. Concerning symmetries of action vs. EOM, see also e.g. this related Phys.SE post.

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