This post considers an aspect of time-reparametization invariance in classical Hamiltonian mechanics. Specifically, it concerns the use of Lagrange multipliers to rewrite the action for a classical system in a time-reparametization-invariant way.
Prelude:
Suppose we have a system with a single degree of freedom $q(t)$ with conjugate momentum $p$, and action $$I = \int dt\ L. \tag{8.1}$$ The Hamiltonian is the Legendre transform $$H(p,q) = p\dot{q}-L(q,\dot{q})|_{p=\partial L/\partial\dot{q}}.\tag{8.2}$$ The independent variable $t$ is special. It labels the dynamics but does not participate as a degree of freedom.
Time-reparametrization symmetry:
Let us introduce a fake time-reparameterization symmetry by labelling the dynamics by an arbitrary parameter $\tau$ and introducing a physical 'clock' variable $T$, treating it as a dynamical degree of freedom. So we consider the system of variables and conjugate momenta $$q(\tau),\qquad p(\tau),\qquad T(\tau), \qquad \Pi(\tau)\tag{8.4}$$ where $\Pi$ is the momentum conjugate to $T$. This is equivalent to the original original system if we use the 'parameterized' action $$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),\tag{8.5}$$ where prime $= d=d/d\tau$. Here $N(\tau)$ is a Lagrange multiplier, which enforces the 'constraint equation' $$\Pi + H(p,q) = 0.\tag{8.6}$$
My difficulty lies with the introduction of the Lagrange multipliers. How do you show that the action
$$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),\tag{8.5}$$
with Lagrange multiplier $N(\tau)$, reduces to the action
$$I' = \int d\tau\ (pq' - H(p,q)T')~?\tag{8.8}$$
Edit to question:
OkThen's answer suggests that, when an action is extremised under a set of constraints, and the constraints are implemented using Lagrange multipliers, each Lagrange multiplier becomes a dynamical variable which satisfies its own equation of motion.
Why must each Lagrange multiplier be a dynamical variable which satisfies its own equation of motion?
