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Let us assume we are doing classical one point particle mechanics. Assume that the least action principle holds. Also, assume that Lagrangian $L$ is a function only of coordinate $x$, its derivative $\dot{x}$ and time $t$. Then by the principle, we know that the following differential equation gives the equation of motion for a particle.

$$ \frac{\partial}{\partial x} L(x, \dot{x}, t) = \frac{d}{dt} \frac{\partial}{\partial \dot{x}} L(x, \dot{x}, t) $$

Now assume that whenever $\phi(t)$ is a solution for this differential equation then $\phi(t) + vt$ is a solution as well. Here, $v$ is assumed to be constant but arbitrary.

What can be said about functional dependence of $L$ on $x, \dot{x}, t$?

I would appreciate your help!

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  • $\begingroup$ Referring to the title, the Lagrangian is different in different frames of reference, in general you can say nothing from first principles, but if you go to an inertial frame of reference you can use symmetry principles (e.g. isotropy/homogeneity) and invoke either Galilean or Einsteinian relativity to fix the Lagrangian for free particles, and then add interactions in an inertial frame, and then go to non-inertial frames, c.f. Landau Mechanics sec. 3, 4, 5 and beyond. $\endgroup$ – bolbteppa Jun 9 at 5:24
  • $\begingroup$ @bolbteppa Thanks for your comment. I agree that Lagrangian depends on a frame of reference. Unfortunately, I do not see connection of that fact to my question. Would you agree that not all Lagrangians satisfy the property I mentioned? If that is the case, I feel that there should be something that we can conclude about what Lagrangian must have to satisfy the property. Do I correctly understand that you think this question is too difficult to answer or maybe meaningless as the result will not be practically useful? I feel that Galilean relativity is the most general principle so I am curious $\endgroup$ – Daniels Krimans Jun 9 at 5:27
  • $\begingroup$ @Daniels Krimans have transformation from cartesian coordinates $(x,\dot{x},t)$ in an inertial frame to some general coordinates $(x'(x,\dot{x},t),\dot{x}'(x,\dot{x},t),t)$. The Galilean relativity tells you that lagrangian of boosted inertial frame differs from not boosted at most by total time derivative. Does this imply $L(x',\dot{x}'+v',t)-L(x',\dot{x}',t)$ is still total time derivative? I dont think so, because when you transfer back, the $v'$ term will not transfer back to simple added constant. So you need to start from an inertial frame in which (cont.) $\endgroup$ – Umaxo Jun 9 at 5:57
  • $\begingroup$ @Daniels Krimans (cont.) space is homogeneous and isotropic and time is homogenenous. So you need lagrangian with this symmetry in an appropriate coordinates to use galilean relativity in a form of addition of constant to velocity. But this symmetry holds only for free particle, not in general from which you get free particle lagrangian to be proportional to $\dot{x}^2$ and nothing else (check mentioned landaus mechanics). $\endgroup$ – Umaxo Jun 9 at 6:04
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/23098/2451 and links therein. $\endgroup$ – Qmechanic Jun 9 at 6:55

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