Does the Euler-Lagrange equation for action always extremize the abbreviated action as well?

Question

In continuation of my previous Phys.SE post, the action functional $$S = \int_{}{}{L}{\, \mathrm{d}t}$$ for the standard Lagrangian $L = \dfrac{1}{2}\dot{q}^2 - V(q)$ is extremized by the Euler-Lagrange equation: $$\frac{\mathrm{d}^2 q}{\mathrm{d}t^2} = - \frac{\partial V}{\partial q}.$$ In S. Coleman's Fate of the False Vacuum I, this same Euler-Lagrange equation is used to extremize the abbreviated action of the system:$$\int \sqrt{2(E - V) } \, \mathrm{d}s, \qquad ds = \sqrt{(dq)^2} = |\dot{q}| dt. $$ where $E$ is given as $$ E = \frac{1}{2} \dot{q}^2 + V.$$

Does the Euler-Lagrange equation that extremizes the standard action always extremize the abbreviated action as well? Or is this a special case?

Answer 1

The claim is that if we evaluate the "reduced action" $$ I[q]=\int_{q_1}^{q_2} p dq $$ along a path in $q$-space where $p$ is determined by $q$ via the constraint $H(p,q)=E$ for a fixed value of $E$, then the path from $q_1$ to $q_2$ that makes $I$ stationary under variations that respect the constraint coincides with that given by the usual equations of motion. What is not determined, is the rate at which the path is traversed.

To see that this is so, and in passing explain what $s$ is, we obsrve can enforce the constraint in two distinct ways: (1) we can explicitly enforce the constraint by solving for $p$ as a function of $q$, or (2) we can introduce a Lagrange multiplier $\lambda(s)$ and make stationary the functional $$ J[p,q]= \int \left\{p \frac{\partial q}{\partial s} -\lambda(s) (H(p,q)-E)\right\}ds $$ where $s$ parameterizes the path. We can now vary $p$ and $q$ independently and appeal to the usual Lagrange multiplier arguments that the existence of $\lambda(s)$ locates the constrained stationary path. The stationarity conditions for $J$ are $$ \frac {dp}{ds} =-\lambda(s) \frac{\partial H}{\partial q}\\ \frac {dq}{ds} =+\lambda(s) \frac{\partial H}{\partial p} $$ These keep $dH/ds=0$ for any $\lambda(s)$ and coincide with Hamilton's equations if we make $dt/ds=\lambda(s)$.