# Most general form of Lagrangian only with respect to Galilean invariance

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 $$x, \dot{x}, t$$?

• @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.) – Umaxo Jun 9 at 5:57
• @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). – Umaxo Jun 9 at 6:04