Help a novice with EL equation derivation

Click For Summary
SUMMARY

The discussion focuses on the derivation of the Euler-Lagrange equation from the action principle as presented in Landau and Lifshitz's "Course of Theoretical Physics Volume 1: Mechanics." The key point is the expansion of the action S when the generalized coordinates q and their velocities \(\dot{q}\) are perturbed by small amounts \(\delta q\) and \(\delta \dot{q}\). The first variation of the action, \(\delta S\), must equal zero for the principle of least action to hold, leading to the integral expression involving partial derivatives of the Lagrangian L. The incorrect assumption made by the user was including \(\frac{\partial L}{\partial t}\) in the variation, which is unnecessary as time t serves merely as the integration variable.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with the principle of least action
  • Knowledge of Taylor series expansion
  • Basic concepts of functional derivatives
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equation in detail
  • Learn about functional derivatives and their applications
  • Explore the calculus of variations for deeper insights
  • Review examples of Lagrangian systems to apply these concepts
USEFUL FOR

Students of theoretical physics, particularly those studying mechanics, as well as educators and researchers looking to clarify the derivation of the Euler-Lagrange equation and the application of the principle of least action.

Alex Cros
Messages
28
Reaction score
1
Hello everyone,

Reading Landau and Lifshitz Course of Theoretical Physics Volume 1: Mechanics (page 3) I got suck in the following step (and I cite in italics):

The change in S when q is replaced by q+δq is
\int_{t_1}^{t_2} L(q+δq, \dot q +δ\dot q, t)dt - \int_{t_1}^{t_2} L(q, \dot q, t)dt

(So far so good)

When this difference is expanded in powers of δq and δ\dot q in the integrand, the leading terms are of the first order.

How do you expand that? Could anyone show me how explicitly if you don't know the explicit form of the Lagrangian?

The necessary condition for S (where S is the action) to have a minimum is that these terms (called the first variation, or simply the variation, of the integral) should be zero. Thus the principle of least action may be written in the form
δS = δ \int_{t_1}^{t_2} L(q, \dot q, t)dt = 0


(Which I'm fine with the above expression)

Or, effecting the variation,
\int_{t_1}^{t_2} (<br /> \frac{\partial L}{\partial q}δq+<br /> \frac{\partial L}{\partial \dot q}δ\dot q)<br /> dt = 0

Now my guess would have included \frac{\partial L}{\partial t} like:
\int_{t_1}^{t_2} (<br /> \frac{\partial L}{\partial q}δq+<br /> \frac{\partial L}{\partial \dot q}δ\dot q + \frac{\partial L}{\partial t}dt)<br /> dt = 0
To perform the total differential of all variables.
Explain me like if I'm five why this guess is wrong.

Thanks so much in advance! And sorry for my lack of elemental knowledge.
-Alex
 
Physics news on Phys.org
Hi,

##t## plays a different role here: it is the integration variable. We want to minimize ##S## by finding a path ##\Bigl ( q(t), \dot q(t) \Bigr )## that achieves this ##\delta S=0## condition. There is no value of ##t## to find; it just runs from ##t_1## to ##t_2## and nothing can be done with ##\partial L \over \partial t##. The time dependence of ##L## is present in the integration, though.
 
Thanks BvU!
Any ideas on how to do this and why is it relevant?
When this difference is expanded in powers of δq and δ\dot q in the integrand, the leading terms are of the first order.
 
Alex Cros said:
Thanks BvU!
Any ideas on how to do this and why is it relevant?

The expansion is basically a Taylor expansion about δq and δ{itex}\dot q{\itex}t keeping the first order powers in in the differentials. For a deeper understanding, read on functional derivatives and or the calculus of variations.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
671