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ShayanJ

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Thanks

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- Thread starter ShayanJ
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- #1

ShayanJ

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Thanks

- #2

jedishrfu

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- #4

ShayanJ

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But sometimes we just dismiss some terms in the equation of motion because they're negligible compared to other terms. Your objection is applicable there too. But in those cases we're considering the equation in only a time interval which is shorter than the full time interval that the equation is applicable in. So here, we can take the action integral only in that subinterval where the approximation is a good one.

Well, I can take only terms which contain a dynamical variable!

- #5

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Well, I can take only terms which contain a dynamical variable!

The instead of a constant term, consider a large but very slowly varying term.

- #6

ShayanJ

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So you're saying that we may have a term in the Lagrangian that is negligible compared to other terms but is varying fast. Then it doesn't matter that its magnitude is small?The instead of a constant term, consider a large but very slowly varying term.

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I would express that differently. It is not magnitude *alone* that matters.

- #8

ShayanJ

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There are many famous approximations in physics. Of course some of them are in parts of physics that have a well established variational formulation. I think if I see one example of an approximation implemented in the variational language, I'll be OK. Do you know such an example?(I mean applying the approximation before getting the equations of motion, applying it to the Lagrangian!)I would express that differently. It is not magnitudealonethat matters.

- #9

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Say kinetic energy is ## {f(q) \over 2} \dot q^2 ## and potential energy is ## g(q) ##. Then kinetic energy is approximated with ## {f(q_0) \over 2} \dot q^2 ##, and potential energy with ## {g''(q_0) \over 2} q^2 ## (remember, ## g'(q_0) = 0 ## because ##q = q_0## is an equilibrium). So the approximated Lagrangian is ## {f(q_0) \over 2} \dot q^2 - {g''(q_0) \over 2} q^2 ##, whose Euler-Lagrange equation is ## f(q_0) \ddot q + g''(q_0) q = 0 ##, i. e., that of a harmonic oscillator.

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