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The objective is to prove that the Lagrangian: $$L'=\frac{2\dot x+\lambda x}{2\Omega x}\tan^{-1}(\frac{2\dot x+\lambda x}{2\Omega x})-\frac{1}{2}\ln(\dot x^2+\lambda \dot{x } x + \omega^2x^2), \qquad \Omega=\sqrt{\omega^2-\lambda^2/4},$$ is equivalent to the lagrangian of the damped harmonic oscillator: $$L=e^{\lambda t}(\frac{m}{2}\dot x^2 -\frac{m\omega^2}{2}x^2), $$ but I dont know how to show that there is a time derivative of a function that differs from one Lagrangian to the other;

(It's exercise 2.14 from Nivaldo Lemos, Analytical Mechanics, 2018.)

Qmechanic
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Acephalus
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  • Since you only have one Lagrangian in the current version of your question it's difficult to show it differs from a second one. – ZeroTheHero Sep 17 '20 at 14:26

1 Answers1

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Hints:

  1. Two Lagrangians, whose difference is not a total derivative, can still yield the same EOM, cf. e.g. this & this Phys.SE posts.

  2. Check that both Lagrangians lead to the same EOM $\ddot{x}+\lambda \dot{x}+\omega^2x~=~0$ of the damped harmonic oscillator.

Qmechanic
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