For a double pendulum we can consider 2 generalised coordinates $\theta_1$ (angle between first mass and vertical axis) and $\theta_2$ (angle between second mass and vertical axis).

The Lagrangian to this system is:


I found here , that for small oscillations we can assume the following approximations:

For $T$: $\cos(\theta_1+\theta_2)\approx 1 $ (working in zeroth order)

For $V$: $\cos(\theta_1)\approx 1-\theta_1^2/2$, as for $\cos(\theta_2)$ (working in second order)

Why can we work with different orders on the same system for small oscillations?

If we assume an $n$ order, shouldn't we maintain that order independently if it's $T$ or $V$?

  • $\begingroup$ I don't believe there is a rigorous reason, or any reason at all except "this is easier to solve, and the result isn't that wrong". $\endgroup$
    – ACuriousMind
    Apr 7, 2015 at 14:21
  • $\begingroup$ $\uparrow$ Which book? $\endgroup$
    – Qmechanic
    Apr 7, 2015 at 15:08
  • $\begingroup$ steadyserverpages.com/steadystate/CourseWork/721/hw6cm.pdf $\endgroup$ Apr 7, 2015 at 15:12
  • $\begingroup$ This is a problem from Goldstein. I don't know who wrote the resolution, but he/she shows credibility $\endgroup$ Apr 7, 2015 at 15:15

1 Answer 1


The small oscillation approximation considers terms in the Lagrangian to quadratic order in $\theta_i$ and $\dot{\theta}_i$. The reason to only work to zeroth order in $\theta_i$ in one term is because the pertinent term in the Lagrangian is already quadratic in $\dot{\theta}_i$.


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