# Formulating the Lagrangian in terms of invariant quantities

Consider a closed system consisting of $N$ point particles, whose Lagrangian is given in the standard way, by the total kinetic energy minus the potential energy: $\mathcal{L}(\dot{q},q):= T(\dot{q}) - U(q)$.

This definition of the Lagrangian is invariant under some symmetries but not others. For example, it is preserved under spatial translations, but a Galillean boost -- in which the velocity of every particle is uniformly shifted -- does not preserve the Lagrangian because it doesn't preserve kinetic energy. A single particle moving under no external forces can have any kinetic energy greater or equal to 0.

My question is: Is it possible to formulate the principle of least action and the Euler-Lagrange equations using quantities that are invariant under boosts? For example, suppose I were to define the invariant kinetic energy, $T^*$, as the kinetic energy in the center of momentum frame (or the total kinetic energy minus the velocity of the center of mass times the total mass), and introduce a Lagrangian $\mathcal{L^*}= T^* - U$. What's wrong with the principle that legitimate motions minimize $\int \mathcal{L}^*$?

• Concerning Galilean symmetry of the Lagrangian, see also this related Phys.SE post. – Qmechanic Jul 29 '15 at 13:02
• Thanks for the link to the post -- that was helpful. If I've got it right, then the answer to my question is that minimizing $\int\mathcal{L}^*$ does not result in the laws of motion. But a closely related trick can be employed where we try to find a path $q$ and a frame of reference $v$ that minimizes quantity "the value T - V in frame v". (So the frame of reference gets an explicit extra argument place in the Lagrangian). As it happens it's miminized when $v$ is the center of momentum frame, and $q$ satisfies the laws of motion. Is that about the jist of it? – Andrew Bacon Jul 31 '15 at 22:39

While neither the Lagrangian $\mathcal{L}$ nor the action $S$ are invariant under boosts of the form $$\dot{q}(t) \to \dot{q}(t) + v, \quad v \in \mathbb{R},$$ the Euler-Lagrange equations are. The dynamics of the systems are unchanged for any transformation that preserves $\delta S = 0$, i.e. a transformation of the form $$\mathcal{L}(q, \dot{q}, t) \to \mathcal{L}(q, \dot{q}, t) + \frac{d}{dt}f(q,t).$$ Notice that the boost in question merely adds to a total derivative to the standard Lagrangian $\mathcal{L} = T - V$, where:
$$\mathcal{L}(q, \dot{q}, t) \to \mathcal{L}(q, \dot{q}, t) + \frac{d}{dt} \big( \frac{1}{2}mv^2 t + mvq \big).$$