I would like to show that the time it takes for a pendulum to leave the unstable equilibrium at the top is logarithmic; i.e. if the starting position of the mass is $\theta_0 = \delta$ for some small $\delta > 0$, then $t(\theta = \pi/2) \sim \log(1/\delta)$ (here the angle is measured from the positive $y$-axis, so $\theta = 0$ is the unstable equilibrium point). Here's what I have so far:

Since there are no external forces in the system, a loss in potential energy is precisely the gain in kinetic energy, so $$\frac{1}{2}mv^2 = mgh$$ where $v = l\dot{\theta}$ and $h$ denotes the height fallen. Hence, $$\dot\theta = \frac{\sqrt{2gh}}{l} = \sqrt{2\frac{g}{l}(\cos\delta - cos\theta)}$$ since $h = l(\cos\delta - \cos\theta)$. Ignoring constants, we have $$\frac{d\theta}{dt} = \sqrt{\cos\delta - \cos\theta}$$ hence $$dt = \frac{d\theta}{\sqrt{\cos\delta - \cos\theta}}$$ so $$t = \int_\delta^{\pi/2} \frac{1}{\sqrt{\cos\delta - \cos\theta}}d\theta.$$ Away from $\theta = \delta$, the integral is bounded above on a finite interval and hence is finite. Near $\theta = \delta$, there are two reasonable options: the first is to expand $\cos\theta$ as $$\cos\theta = \cos\delta - \sin(\delta)(\theta - \delta) + O(\theta - \delta)^2,$$ so the denominator locally looks like $\frac{1}{\delta\sqrt{\theta - \delta}}$ using the small angle approximation $\sin\delta \approx\delta$. This gives $$t \sim \int_\delta^{\delta + \epsilon} \frac{1}{\delta\sqrt{\theta - \delta}}d\theta = \frac{1}{\delta}\sqrt{\theta - \delta}\vert_{\delta}^{\delta + \epsilon} = \sqrt{\frac{\epsilon}{\delta}}$$ For fixed $\epsilon$, this gives square-root divergence.

The other option that may be reasonable is to approximate $$\cos\delta = 1-\frac{1}{2}\delta^2 + O(\delta)^4$$ and Taylor expand $\cos\theta$ around zero (as both $\delta$ and $\epsilon$ are small) as $$\cos\theta = 1 - \frac{1}{2}\cdot\theta^2 + O(\theta)^4$$ then, ignoring constants $$t \sim \int_\delta^{\delta + \epsilon}\frac{1}{\sqrt{\theta^2 - \delta^2}}d\theta = \log \vert \sqrt{\theta^2 - \delta^2} + \theta \vert \big\vert_\delta^{\delta + \epsilon} = \log(\sqrt{2\delta\epsilon + \epsilon^2} + \delta + \epsilon) - \log\delta.$$ As $\delta \rightarrow 0$, the first term is a constant $\log(2\epsilon)$, and the second term gives the logarithmic divergence.

Which of these two approaches is correct? They both seem reasonable to me but give very different divergence.


2 Answers 2


The logarithmic divergence is correct. To make things simple, I will consider the dimensionless system: $$ \ddot \theta=\sin\theta $$ with energy: $$ H = \frac12\dot\theta^2+\cos\theta $$

This is pretty intuitive by linear approximation. At the unstable point, your system is in a bottle neck where the equation is roughly linear: $$ \ddot \theta=\theta $$ with quadratic energy: $$ H = \frac12\dot\theta^2-\frac12\theta^2 $$ You therefore have an exponential time dependence about this point, which translates into a logarithmic divergence of the sojourn time.

You weren't careful with the estimation of the integral. You want to calculate the time to get from $\delta$ to $\theta_0$ as $\delta\to0$. The time is indeed: $$ T = \int_\delta^{\theta_0}\frac{d\theta}{\sqrt{2(\cos\delta-\cos\theta)}} $$ On a side note, you made a minor mistake in the first method, which was corrected by Michel Seifert. You are using: $$ \cos\delta-\cos\theta = \sin\delta(\theta-\delta)+o(\theta-\delta) $$ so: $$ t\sim \frac1{\sqrt{2\sin\delta}}\int_\delta^{\delta+\epsilon}\frac{d\theta}{\sqrt{\theta-\delta}} \sim \sqrt{\frac\epsilon{2\delta}} $$

The problem with your approach and Michael Seifert's is that you are assuming that the contribution of the integral comes from the singularity as $\theta\to\delta$ and just the leading order of the integrand at the singularity gives the leading order contribution. This is false. You need to go to second order in expansion of the potential. By Taylor expansion, $$ \cos\delta-\cos\theta = \sin\delta(\theta-\delta)+\frac{\cos\delta}2(\theta-\delta)^2+o((\theta-\delta)^2) $$ You get the $1/\sqrt{\delta}$ by keeping only the first term and neglecting the second term. However, for the integral, it is the second term that is responsible for the leading order contribution to the integral as your second method suggests.

Heuristically, the region of validity for the linear term dominating the quadratic term is: $$ \theta-\delta\ll \frac{\tan\delta}2 $$ i.e. $\theta-\delta\ll \delta$, so the region of validity shrinks. This has two effects. First off, it means that you should not take your $\epsilon$ to be fixed but rather $\epsilon\sim \delta$. This reduces the effect of the contribution of the singularity to a constant. Thus, rigorously, your argument does not even predict a divergence of sojourn time, but a convergence to a finite value.

On the other hand, this allows the integral to "see for a longer time" second order term. This leads to an apparent $1/(\theta-\theta_1)$ singularity, which has a stronger, logarithmic divergence. It is therefore the latter that dominates as $\delta\to0$.

You can translate this mathematically. You can get rid of the dependence of the bound by translation, writing $\Delta\theta = \theta_2-\delta$: $$ T = \int_0^{\Delta\theta}\frac{d\phi}{\sqrt{2(\cos\delta-\cos(\phi +\delta))}} $$ Now, it is easier to see that the integrand does converge to: $$ \frac1{\sqrt{2(\cos\delta-\cos(\phi +\delta))}}\to \frac1{\sqrt{2(1-\cos\phi)}}\sim\frac1\phi $$ Even if for every $\delta>0$ the singularity is in $1/\sqrt\phi $, the limiting function has a stronger singularity. This is because the convergence is not uniform. The region of validity is given by the previous calculation: $$ \phi \gg \delta $$ which excludes the singular behaviour at $\phi=0$. You can see this non uniform convergence graphically: graph

Therefore, the leading order contribution for $T$ is given by integrating the limiting function up to its region of validity: $$ T = \int_\delta^{\Delta\theta}\frac{d\theta}{\sqrt{2(1-\cos\theta)}}\sim -\ln\delta $$

Note that this is not specific to the inverted pendulum. This is true for any linearly unstable point. It even generalises when the leading order is not quadratic, but polynomial, in which case the can diverge or converge as a power law by simple scaling arguments. The linear case is in some sense marginal as the logarithmic dependence is a signature of exponent given by dimensional analysis being $0$.

Hope this helps.


Are you sure that the period diverges logarithmically? I note that since $\sin \delta \neq 0$ so long as $\delta \neq 0$, the leading order term in your expansion of $\cos \theta$ should be $$ \cos \theta \approx \cos \delta - \sin \delta (\theta - \delta) + \mathcal{O}(\theta - \delta)^2 $$ which (following your logic with this correction) implies a divergence proportional to $(\sin \delta)^{-1/2} \sim \delta^{-1/2}$.

  • $\begingroup$ I am not confident about this, but this is an exercise in Chapter 2 of Hall's "Quantum Theory for Mathematicians" and he claims the divergence should be logarithmic. $\endgroup$
    – SescoMath
    Commented Nov 21, 2022 at 0:15
  • $\begingroup$ I agree that (assuming the prior reasoning is correct) the divergence seems like it goes like $\delta^{-1/2}$. I'm wondering though if theres an argument to the tune of taking $\delta$ small means that we should instead taylor expand around $0$, approximately cancel the first terms ($\cos\delta \approx 1-\delta$), and and we are in fact left with an integral that looks like $(\theta - \delta)^{-1}$. $\endgroup$
    – SescoMath
    Commented Nov 21, 2022 at 0:40

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