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$$ L = \sum _ { i = 1 } ^ { N } \frac { 1 } { 2 } m _ { i } \left| \dot { \vec { x } _ { i } } \right| ^ { 2 } - \sum _ { i < j } V \left( \vec { x } _ { i } - \vec { x } _ { j } \right) $$

This is just a typical classical Lagrangian for $N$ particles. Since the Lagrangian does not explicitly depend on time, the energy must be conserved. Also, the linear and angular momentum seem to be conserved too.

However, if there is a change in the coordinate by the Galilean transformation $\overrightarrow{x}_i(t) \to \overrightarrow{x}_i(t) +\overrightarrow{v}t$, then the aforementioned quantities seeem clearly "variant". So, my question is that whether there exists a quantity that is invariant under this Galilean transformation. Could anyone please present me one? Or if there is no such quantity, could anyone please explain why?

Wasserwaage
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Keith
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1 Answers1

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In general, there is no reason to expect that there exist conserved quantites for a symmetry which is not a symmetry of the action, but merely of the equations of motion.

The case of the non-relativistic Lagrangian and Galilean transformations is a special case. As Qmechanic works out in this answer, the Galilean transformations are quasi-symmetries of the Lagrangian, i.e. only change it by a total time derivative. In this case, Noether's theorem still applies and yields a conserved quantity (for the free Lagrangian) $$ Q = m(\dot{x}t - x),$$ which is Galilean invariant. Note that Qmechanic's third example shows that a symmetry of the equation of motion does not always imply a quasi-symmetry of the Lagrangian, and therefore there is no conserved quantity associated to it in the general case.

ACuriousMind
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