I Proving that the Lagrangian of a free particle is independent of q

[okaythanksbud]

TL;DR Summary

Authors typically cite "symmetry" when allowing a Lagrangian to be independent of its position or direction, but how can we prove this?

One of the first things Landau does in his Mechanics book is give an argument as to why the Lagrangian of a free particle must be our conventional kinetic energy. Heuristically, he justifies it, but leaves out the details, perhaps being too obvious. They aren't obvious to me. While in free space we will see a particle travel in the same fashion regardless of when, where, and the angle we shoot it at, translating this to time, location, and directional independence seems like way too big of a leap.

If we start at Hamiltons principle and claim there to be a function/lagrangian capable of maximizing the action whenever we supply a path a free particle will take, Id imagine we should be able to use the previous claims to actually prove this invariance. For example, for any path q, translational invariance should allow us to say the integral of L(q+a, q',t) over a time interval is maximized for any value of a. If we assume L to be analytic in some region we should be able to argue d^k/dq^k L is also maximized. I dont know where to go from here but imagine one could produce a rigorous argument as to why we can say that L(q+a,q',t)-L(q,q',t) is 0 for any value of a. From the previous observation we can see that the difference should be a total time derivative, however I do not know how to show independence from q', and do not know if this would necessarily advance the argument. Any help is appreciated.

 

[vanhees71]

The argument is based on Noether's theorem. A transformation
$$t'=t'(q,\dot{q},t), \quad q'=q'(q,\dot{q},t)$$
is called a symmetry transformation, if the Lagrangian wrt. the new time and configuration variables is an equivalent Lagrangian to the original one. That means that there is a function $\Omega(q,t)$ such that
$$\frac{\mathrm{d} t'}{\mathrm{d} t} L[q'(q,\dot{q},t),\dot{q}'(q,\dot{q},t),t']=L(q,\dot{q},t)+\frac{\mathrm{d}}{\mathrm{d} t} \Omega(q,t).$$
If you exploit this for the given symmetry group of Newtonian mechanics, i.e., the 10-parametric Galilei group (generated by time translations, spatial translations, rotations, and Galilei boosts) you get, for a single particle
$$L=\frac{m}{2} \dot{\vec{x}}^2$$
with Cartesian coordinates $\vec{x}$ for the particle's position, modulo an arbitrary function $\Omega$, which is physically unimportant, because all these Lagrangians lead to the same equations of motion. In the case of a free particle of course that $\ddot{\vec{x}}=\text{const}$. It is sufficient to consider "infinitesimal transformations" to derive this.

The detailed treatment is too long for a newsgroup posting. Unfortunately I have it only in my German manuscript on mechanics (Sect. 3.5):

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf