# Infinite period of finite oscillation?

This is not-very-hard problem in my classical mechanics class, but it has a confusing result.

So the problem is about a particle moving in the following potential $$V(x)=\begin{cases}\infty & \mathrm{if }x\leq 0 \\ A[a^2-(x-a)^2]&\mathrm{if } 0<x\leq2a \\0 & \mathrm{if }x\geq2a\end{cases}$$

with $A>0$. Let's say the particle starts out at $x=0$ with some initial energy $E=\epsilon^2Aa^2$ where $\epsilon<1$. Because energy is smaller than $Aa^2$, then it can't cross the top of the hill, and will roll back down to $x=0$ then get reflected by the wall and repeat the motion.

The question asks you to find the action-angle variable, which I found to be $$J = \frac{a^2\sqrt{2mA}}{2\pi}\left[\epsilon-\frac12(1-\epsilon^2)\ln\left(\frac{1+\epsilon}{1-\epsilon}\right)\right]$$

which is confirmed to be correct.

Then it asks you to verify that for $\epsilon\rightarrow1$, the period blows up. Yes, indeed, I've verified it blows up.

This seems to be very confusing. It takes infinite time for the particle to get to the top if the energy is just equal to $Aa^2$?

I tried another approach, $$T = \int \frac{dx}{\sqrt{2E-2V}} \sim \frac{1}{x-a}.$$ which also indicates the period blows up...

But this can't be right, right? It must be some kind of coordinate singularity?

Somehow this reminds me of the infinite redshift at the event horizon of the black hole. Although the co-moving observer can cross the event horizon in a finite time, a distant observer can never see this happen. Are these two related?

• Worth drawing yourself a picture of the potential well... Oct 30 '16 at 0:14
• @Floris Of course I did! It's an inverted parabola. Oct 30 '16 at 0:28
• @Jamie What Floris means is that the question would benefit from actually having such a drawing included in the post itself. Oct 30 '16 at 0:36

This is perfectly reasonable behaviour, and it's in line with expectations. The reason for this is that if you take such a repulsive potential, the force looks like $$m\ddot x = F = +kx$$ (with a shifted origin for simplicity), which is easily solved to give solutions of the form $$x(t) = x_0 e^{\pm \gamma t}.$$ In principle, the general solution is a superposition of both of these, but the limit you're taking is precisely designed so that the trajectory will look like $x(t) = x_0 e^{- \gamma t}$ as it approaches the top of the hill, and like $x(t) = x_0 e^{+ \gamma t}$ on the outgoing leg.
• Yeah, you can make $\tau = e^{\gamma t}$ and you'll have a finite "period" in that unphysical "time", and it will mean somewhere between very little and nothing. The fact is that the particle takes unbounded physical time to reach the top at that energy, and until you realize it and make use of it you're mostly just wasting time and energy. Oct 30 '16 at 0:35