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

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?

  • $\begingroup$ Your language seems imprecise: There are many quantities invariant under Galilean transformations (a large class of examples being any quantity that depends only on second or higher time derivatives of the $x_i$)), but I suspect that you are not looking for simply any such quantity. Are you looking for a quantity that is conserved under the dynamics of that Lagrangian and additionally Galilean invariant? If yes, why would you think there needs to be an explanation why it doesn't exist, rather than an explanation why it should (as you seem to think, but you don't explain why)? $\endgroup$ – ACuriousMind Mar 31 '18 at 18:44
  • $\begingroup$ Yes that is what I am looking for. I just have no idea whether there exists such a quantity. How can I find a physical quantity that is conserved under the Lagrangian and also Galilean invariant? $\endgroup$ – Keith Mar 31 '18 at 18:54
  • $\begingroup$ Why would you expect such a quantity to exist if the Lagrangian itself is not Galilean invariant? With the linear and angular momenta you've already found $2N$ independent conserved quantities, where $N$ is the number of particles, so there are no more independent ones, what do you hope to gain by finding such a quantity? $\endgroup$ – ACuriousMind Mar 31 '18 at 18:57
  • $\begingroup$ Bexause the equations of the motion are invariant. I just hope there might exist some other invariant ones. What do you mean by there are no more independent ones? $\endgroup$ – Keith Mar 31 '18 at 19:02
  • $\begingroup$ Related: physics.stackexchange.com/q/14875/2451 $\endgroup$ – Qmechanic Mar 31 '18 at 19:25

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.

  • $\begingroup$ I managed to get this answer. But, clearly the quantity $Q$ you presented does seem to have a nonzero time derivative. That, is $\frac{dQ}{dt}=m \ddot x t$ $\endgroup$ – Keith Apr 2 '18 at 18:24
  • $\begingroup$ Oh, you assumed free Lagrangian. So, in the case there is interaction, then there is no hope of finding a conserved quantity? $\endgroup$ – Keith Apr 2 '18 at 18:25
  • $\begingroup$ @Keith Of course there is hope - if the Lagrangian is still quasi-symmetric under the transformation, then Noether's theorem still works. You just need to apply it. $\endgroup$ – ACuriousMind Apr 2 '18 at 18:27
  • $\begingroup$ So, in the above original question of mine, if the potential $V$ is given to make the Lagrangian quasi-symmetric, then according to you there is hope...I see. $\endgroup$ – Keith Apr 2 '18 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.