Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 15C / c / Solution

Between collisions the momentum is $p_0=\sqrt{2mE}$ or $-p_0$. In phase space the cycle runs right along the upper horizontal segment from $q=0$ to $q=b$, jumps vertically to $-p_0$ at the right wall, runs left along the lower segment, and jumps back at the left wall. The jumps have $dq=0$. ThereforeDifferentiation gives $dI/dE=2bm/\sqrt{2mE}=2b/v$, exactly the out-and-back period. Under slow wall motion the action $I$ is the adiabatic invariant; a convention using $I/(2\pi )$ is equivalent. HenceCompression raises the energy and expansion lowers it, provided the wall speed is small compared with the particle speed and the change per cycle is small.