How to understand when surface terms go to zero

[Ebarval]

TL;DR

When do surface terms go to zero?

Hi all,
I'm trying to understand when surface terms go to zero. I'm not really getting a connection other than many textbooks just saying surface terms go to zero.
I have added a photo of Liboff's Kinetic Theory page 3 on Lagrange's equations. Before equation 1.7, he says the surface terms go to zero because the end points 1 &2 are fixed. But can't the Lagrangian still have a differential with respect to qdot?

 

[PeroK]

If $\delta q = 0$ then $\delta q \times X = 0$, whatever $X$ is.

 

[Ebarval]

If $\delta q = 0$ then $\delta q \times X = 0$, whatever $X$ is.

But then why can't that be done on the step after eqn 1.6 or 1.6 itself?

 

[PeroK]

But then why can't that be done on the step after eqn 1.6 or 1.6 itself?

Because it's not zero everywhere. Only at the end points where the Parts terms are evaluated.

 

[Ebarval]

Because it's not zero everywhere. Only at the end points where the Parts terms are evaluated.

Ah I see! Thank you!
But now I have more questions. How can I tell from integral if something will end up as a surface term and that it goes to zero if there is no such delta term explicitly multiplying?

 

[PeroK]

Ah I see! Thank you!
But now I have more questions. How can I tell from integral if something will end up as a surface term and that it goes to zero if there is no such delta term explicitly multiplying?

In this case, $\delta q$ is any variation on the path between two fixed endpoints. That's all there is to it.

 

[Ebarval]

How about for the attached section? They mention the surface terms go to zero which I assume are the u_k^*u_l