Principle of least action to prove that conservation of momentum results from translational symmetry

Question

In an article that I am reading http://go.owu.edu/~physics/StudentResearch/2005/LauraBecker/SymmetrytoConservation.html - the author proves, firstly, why translational symmetry in space results from the principle of least action and, secondly, why conservation of momentum results from translational symmetry, using the same principle of least action. In the first proof, the author defines the action as $$S=T_{\rm avg}(t_2-t_1)$$ rather than the conventional definition of action as $$\int_{t_1}^{t_2}(T-V)dt, $$ which has confused me. (I am very new to this topic and self-taught). I am also struggling the understand the logic for the second proof of the conservation of momentum, which can be found in the link above. I would appreciate an explanation of both of these misunderstandings and, perhaps, a walkthrough of the steps of the proof of conservation of momentum.

Answer 1

The author Laura Becker of OP's linked webpage is considering 2 types of actions:

1. The off-shell action functional $$ I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t), \tag{1} $$

2. The Dirichlet on-shell action function $$ S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl};t_i,t_f], \tag{2} $$ where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the extremal/classical path.

Noether's theorem is usually/originally formulated with (1). However e.g. Feynman sometimes uses (2), cf. this related Phys.SE post, which also provides the sought-for proofs.