I Momentum and Action: Understanding Lagrangian Mechanics

[happyparticle]

TL;DR

Action after a variation of $q_2$

Hi,
In my book I have and expression that I don't really understand.
Using the definition of action $\delta S = \frac{\partial L}{\partial \dot{q}} \delta q |_{t_1}^{t_2} + \int_{t_1}^{t_2} (\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}) \delta q dt$
Where L is the lagrangian and q the position.

For a variation of $q_2$

$\delta S = \frac{\partial L}{\partial \dot{q}} \delta q |_{t_2}$

I don't see why the integral on the right hand side is 0.
If I understand correctly, at $t_1$ I have 2 points, $q_1$ and $q_2$ and then at $t_2$, $q_2$ moves but not $q_1$
However, why we only have $t_2$ as lower limit for the first term on the right hand side and why the second term vanish?

 

[vanhees71]

I guess you talk about the least-action principle in the Lagrangian form. Then by assumption you vary within the space of all trajectories with fixed boundaries at the initial and final time, i.e., $\delta q(t_1)=0$ and $\delta q(t_2)=0$. That means that the boundary term in $\delta S$ vanishes. Then the action principle tells you that for the physical trajectory of the particle, i.e., for the solutions of the equations of motion, the integral must be vanish for otherwise arbitrary $\delta q$. Then the fundamental lemma of variational calculus says that for this the expression in the bracket under the integral must vanish, i.e., that the equations of motion are given by the Euler-Lagrange equations of the least-action variational principle,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}.$$
I don't know, what your secret book is discussing considering only one boundary term.

 

[happyparticle]

Is $\delta q(t_1)=0$ the distance between $q_1$ and $q_2$?

Basically, in my situation, $q_1$ is fixed and there's a variation of $q_2$. At the end I have $\frac{\partial S}{\partial q} = p$

 

[TSny]

The value of the action $S$ for the actual path depends on the initial and final space-time points of the path. I think your book is probably discussing how the value of $S$ for the actual path varies when you make a small change in the final position $q_2$ of the path (but no change in the final time $t_2$). See the first part of this article.

 

[happyparticle]

This is exactly what I have.

The author says:

"Both paths P,P′ are fully determined by their initial and final positions and times, so P,P′′ must correspond to slightly different initial velocities. The important point is that since both paths describe the natural dynamical development of the system from the initial conditions, the system obeys the equations of motion at all times along both paths, and therefore the integral term in the above equation is identically zero. "

However, I can see exactly why the term in the integral must be 0. I don't understand the link between the difference of velocity between both path and the integral term.

 

[TSny]

However, I can see exactly why the term in the integral must be 0.

Good

I don't understand the link between the difference of velocity between both path and the integral term.

The author points out that paths P and P' have different initial velocities. But, I don't think this fact is directly connected to why the integral term is zero. As I see it, the integral term is zero since P' represents a variation of P and P is an extremum path. So, the Euler-Lagrange equation is satisfied along P.

However, paths P' and P do not have the same end point $q_2$ at time $t_2$. The endpoints differ by a small amount $\delta q(t_2)$. This is accounted for by the term $\frac{\partial L}{\partial \dot q} \delta q|_{t_2} \equiv \frac{\partial L}{\partial \dot q}|_{t_2} \delta q(t_2)$.

In your first post you said

If I understand correctly, at $t_1$ I have 2 points, $q_1$ and $q_2$ and then at $t_2$, $q_2$ moves but not $q_1$

At $t_1$, paths P and P' have the same initial point $q_1$. At $t_2$, path P has $q= q_2$ and path P' has $q = q_2 + \delta q(t_2)$.