- #1

happyparticle

- 442

- 20

- TL;DR Summary
- 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

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?

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?