Landau/Lifshitz action as a function of coordinates [duplicate]

Question

In Landau/Lifshitz' "Mechanics", $\S43$, 3ed, the authors consider the action of a mechanical system as a function of its final time $t$ and its final position $q$. They consider paths originating at some point $q^{(1)}$ at a time $t_1$, and terminating at the same point $q$ after different times $t$. The total derivative of the action is then

$$ \frac{dS}{dt}=\frac{\partial S}{\partial t}+\sum_i\frac{\partial S}{\partial q_i}\dot{q}_i\tag{p.139}. $$

The authors then claim that $\frac{\partial S}{\partial q_i}$ can be replaced by $p_i$. I don't understand this step. L+L derived this formula for $p_i$ by considering paths starting from the same point $q^{(1)}$, but passing through different endpoints after a common time interval. Why should $$p_i=\frac{\partial S}{\partial q_i}\tag{43.3}$$ continue to hold when the action is varied in a different way?

Answer 1

As defined in your question $p(q,t) = \frac{\partial S(q,t)}{\partial q}$ is the momentum at time $t$ of the classical (least action) path which passes through the points $(q_1,t_1)$ and $(q,t)$. Changing the end point $(q,t)$ will generally change the corresponding path but $\frac{\partial S}{\partial q}$ will always be the momentum at time $t$ for whatever path that happens to be.

That being said, in your total derivative equation the action is being evaluated on a classical path uniquely specified by certain end points, $q(t) \equiv q_{(q_1,t_1,q_2,t_2)}(t)$, to make it a function of only time $S(q(t), t)$. So by definition the classical path passing through the points $(q_1,t_1)$ and $(q(t), t)$ will be this same path for all $t$. This means that for all times $t$, $p(t) = \frac{\partial S(q(t),t)}{\partial q}$ is the momentum of the classical path at time $t$.