I What is "making the time integration redundant"?

[Happiness]

What does it mean by "making the time integration redundant" (5th line)? If I let $t_2=t_1$, I will only get $0=0-0$.

Source: http://www.phys.ufl.edu/~maslov/classmech/flannery.pdf

 

[haruspex]

I don't pretend to know what half of this is about, but it looks to me that (28) can be got from (50) by differentiating wrt time. That certainly avoids the integration.

 

[vanhees71]

I don't understand why many textbooks make nonholonomic constraints so unnecessarily complicated. It's a constraint on the "allowed variations" not a restriction to a submanifold as are the holonomic constraints. See the parallel thread on the subject:

https://www.physicsforums.com/threa...n-hamiltons-and-dalemberts-principles.861543/

 

[Happiness]

I don't pretend to know what half of this is about, but it looks to me that (28) can be got from (50) by differentiating wrt time. That certainly avoids the integration.

I think I figured it out.

$\delta q_j=0$ when $t=t_1$ and $t=t_2$ since the end points are fixed as we vary the trajectory.

That leaves us with $\int_{t_1}^{t_2}\delta L\,dt = -\int_{t_1}^{t_2}Q_j^{NP}\delta q_j\,dt$.

By "making the time integration redundant", we have $\delta L = -Q_j^{NP}\delta q_j$.

Then by expressing $\delta L$ in terms of $\delta q_j$, we obtain (14) and subsequently (28).