newtonian mechanics - Galilean invariance of Lagrangian for non-relativistic free point particle?

In QFT, the Lagrangian density is explicitly constructed to be Lorentz-invariant from the beginning. However the Lagrangian

$$L = \frac{1}{2} mv^2$$

for a non-relativistic free point particle is not invariant under Galilean transformation. This does not ultimately matter because the difference is a total time derivative.

However, is it possible to exhibit a Galilean invariant Lagrangian for a non-relativistic free point particle?

Answer

The answer is negative. There is no action of the free particle invariant under the Galilean group. In the following, a heuristic explanation will be given and in addition a reference where a more detailed proof is provided.

The basic reason is that the Galilean group cannot be realized on the Poisson algebra of functions on the phase space of the free particle $T^{*}\mathbb{R}^3$ (equipped with the canonical symplectic form). It is only its central extention (Please, see the following Wikipedia page) which can be realized in terms of Poisson brackets. For this central extension, the Poisson brackets between the generators of the boosts $B_i$ and the translations (i.e., the components of the momentum) $P_i$ no longer vanishes but depends on the mass of the particle:

$\{B_i, P_j\} = m\delta_{ij}$.

Since boosts must generate the transformation: $ P_i \rightarrow P_i + m v_i $ on the momentum coordinates via the canonical Poisson brackets, the Boost generators have to be realized as multiples of the position coordinates $Q_i$.

$B_i = m Q_i$

The transformation law of the Boosts on the phase space (which is the manifold of the initial data, thus this realization does not involve time):

$ Q_i \rightarrow Q_i $

$ P_i \rightarrow P_i + m v_i $

$ H(\vec{P}) \rightarrow H(\vec{P}+m\vec{v}) - \vec{P}.\vec{v}-\frac{1}{2}m v^2 $

It is easy to verify that the free particle Hamiltonian is invariant and its transformation satisfies a group law. But this realization still does not make the Lagrangian $ L = \vec{P} .\dot{\vec{Q}} - H$ invariant, because the Cartan-Poincare form: $ \vec{P}.d\vec{Q} $ is not invariant and changes by a total derivative: $m d \vec{v}.\vec{Q}$. Thus the existence of mass prevents the action from being invariant, because of the canonical Poisson bracket and not because of the choice of the dynamics through the particular choice of the Hamiltonian.

The noninvariance of the Cartan-Poincare form under the boosts is referred to as non-equivariance of the momentum maps associated to the Boosts, which indicates that we cannot redefine the group generators so the Poisson bracket between the Boosts and translation generators vanishes. Please see pages 430-433 and exercise 12.4.6 in "Introduction to mechanics and symmetry", by Marsden and Ratiu for a rigorous proof.