Euler Lagrange Equation trough variation

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SUMMARY

The discussion focuses on deriving the Euler-Lagrange equations of motion through the principle of variation. Participants highlight the challenge of ensuring that the variation of the action, represented as \( \delta S = \int dt \delta q \), equals zero for all variations \( \delta q \). A key insight is the necessity to define the function as \( q = q(t) + \delta q(t) \) while maintaining fixed endpoints, leading to the condition \( \delta q(t_1) = \delta q(t_2) = 0 \). The conversation emphasizes the importance of correctly applying the variation principle to avoid contradictions in the equations.

PREREQUISITES
  • Understanding of the calculus of variations
  • Familiarity with the concept of action in physics
  • Knowledge of differential equations
  • Basic grasp of functional derivatives
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equations in detail
  • Explore examples of action principles in classical mechanics
  • Learn about fixed endpoint conditions in variational problems
  • Investigate the role of functional derivatives in physics
USEFUL FOR

Students of physics, particularly those studying classical mechanics and the calculus of variations, as well as educators seeking to clarify the application of the Euler-Lagrange equations in problem-solving.

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Homework Statement



"Vary the following actions and write down the Euler-Lagrange equations of motion."

Homework Equations



[itex]S =\int dt q[/itex]

The Attempt at a Solution



Someone said there is a weird trick required to solve this but he couldn't remember. If you just vary normally you get [itex]\delta S=\int dt \delta q=0[/itex]
and that's not helpful. Any suggestion on how to avoid this problem?
 
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I believe you need to make your function [itex]q = q(t) + \delta q(t)[/itex] and then observe that since this new function needs to have the same ending points it implies that
[tex] \delta q(t_1) = \delta q(t_2) = 0[/tex]
And follow that with what [itex]\delta S[/itex] becomes :)
 
thanks but isn't this just the general way of variation?

[itex]\delta S = \int dt (f(q+\delta q) - f(q))= \int dt \frac{\partial f}{\partial q} \delta q = \int dt \delta q[/itex]

and there still the same problem remains that I can't find any function that makes [itex]\delta S = 0[/itex] for every [itex]\delta q[/itex] because here I would get [itex]1=0[/itex].
Or did i miss your point?
 

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