# I What is "making the time integration redundant"?

1. Mar 11, 2016

### Happiness

Last edited: Mar 11, 2016
2. Mar 12, 2016

### 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.

3. Mar 13, 2016

### vanhees71

4. Mar 13, 2016

### Happiness

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).