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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 $\int \mathcal{L}^*$?

Andrew Bacon
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It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian.

Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as constraints on the system. In this way we can use the Dirac procedure for constrained Hamiltonians. This leads to a modified Poisson bracket called the Dirac bracket which describes the allowed region of phase space a system is allowed in.

AngusTheMan
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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:

$$ \mathcal{L}(q, \dot{q}, t) \to \mathcal{L}(q, \dot{q}, t) + \frac{d}{dt} \big( \frac{1}{2}mv^2 t + mvq \big). $$

Ultima
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Yes, it is, but it is not easy to know the invariant quantities beforehand, so the usual way to use it is to write down the Lagrangian in terms of any (non-necessarily invariant) quantities, and then use the principle of least action to find the invariant quantities, Nother currents, etc.

fredwhileshavin
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