A How to understand when surface terms go to zero

AI Thread Summary
Surface terms in Lagrangian mechanics go to zero at fixed endpoints, as noted in Liboff's Kinetic Theory. The discussion clarifies that while the Lagrangian may have a differential with respect to qdot, the variations (δq) only yield zero at the endpoints, not throughout the entire interval. Understanding when terms become surface terms involves recognizing that they are evaluated at these fixed endpoints. The conversation also touches on identifying surface terms in integrals, emphasizing that variations along a path between fixed endpoints are crucial. Overall, the key takeaway is that surface terms vanish at the boundaries, which is essential for applying Lagrange's equations correctly.
Ebarval
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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?
 

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If ##\delta q = 0## then ##\delta q \times X = 0##, whatever ##X## is.
 
PeroK said:
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?
 
Ebarval said:
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.
 
PeroK said:
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?
 
Ebarval said:
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.
 
How about for the attached section? They mention the surface terms go to zero which I assume are the uk*ul
 

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