From the Lagrangian I've got the following equations of motion for the double pendulum in 2D. (The masses are different but the lengths of the two pendula are equal.) Let $m_2$ be the lowest-hanging mass.

$$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ -2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-(m_1+m_2)g/l\sin(\theta_1)$$


$$m_2\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ 2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-m_2g/l\sin(\theta_1)$$

In the small angle approximation these become, respectively

$$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2= -2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1(m_1+m_2)g/l$$


$$m_2\ddot{\theta_1}+2m_2\ddot\theta_2= 2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1m_2g/l$$.

Most sources don't have the terms of order $\dot\theta$. This is because they apply the small angle approximation to the Lagrangian before taking the derivatives, thereby ignoring terms of order $\theta.$ What justification do we have for getting rid of these terms?


2 Answers 2


I think the issue here is that you need to keep a consistent level of approximation in your "small angle approximation." By small angles, we typically mean $\theta_1$ and $\theta_2$ are both of order $\epsilon$, where $\epsilon \ll 1$. Then the question is - to what order in $\epsilon$ do you want to write down the equations of motion?

When you neglect the term $\frac{3}{2} \dot\theta_1 \dot \theta_2 (\theta_1 - \theta_2)^2$ in the Lagrangian, you are saying that terms of size $\epsilon^4$ are small compared to terms like $\dot \theta_1^2$, which is of size $\epsilon^2$. In the equation of motion, you get terms that are $\dot \theta_1 \dot \theta_2 (\theta_1 - \theta_2)$, which are of size $\epsilon^3$, compared to $\theta_1$, which is size $\epsilon$.

So neglecting the additional term in the Lagrangian gets you the same equation of motion as keeping the whole Lagrangian, and then dropping terms that are of size $\epsilon^3$.

This kind of argument is a little handwavy, and (in principle) could blow up if the time derivatives of $\theta_{1,2}$ were large - at some point, you might want to check out some books on perturbation theory in a more formal sense.

  • $\begingroup$ In the limit of large $g$ I find it hard to buy that $\dot\theta^2$ is of order $\epsilon^2$. (Basically what you say at the end of your answer.) I guess, I have a hard time seeing that if $f(t)$ is of order $\epsilon$ then so must $\dot f(t)$. $\endgroup$ Commented Sep 21, 2013 at 19:40
  • $\begingroup$ Surely one cannot generally conclude from $f(t)$ being of order $\epsilon$ that the same holds for the derivative. In case of a pendulum, however, things are different. Just take a simple pendulum. The height is related to $\cos \theta$. It is, hence, of order $\epsilon^2$. The corresponding potential energy is then transformed into kinetic energy. Consequently, $\dot\theta^2$ must be of order $\epsilon^2$. This is independent of the magnitude of $g$. This is more complicated in the double pendulum, but becomes approximately correct again in the small angle case. $\endgroup$ Commented Jan 28, 2014 at 8:45

I think that doing the approximation at the Lagrangian level is the correct method.

The Lagrangian is important, because it has a direct impact about conserved quantities by virtue of the Noether theorem.

Making the approximation at the Lagrangian level, ensures you, that the equations of movement, and the Lagrangian are compatible and coherent, and that the conserved Noether global conserved quantities, are in fact, easily obtained from the equations of movement.


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