After 1 and 2 collide, 1 and 3 must collide if we are considering a process of the form (32) n or (2)(32) $^{n}$. Now if the previous collision between 1 and 2 was preceded by one between 1 and 3, 3 is moving radially away from 1 at the moment of the collision between 1 and 2. So for another collision to occur between 1 and 3, the velocity of 3 after the collision, q ′ = q + p ( x − 1) / ( a + 1) , must be directed toward the origin and so be antiparallel to its velocity before the collision, q. But if q and q + p ( x − 1) / ( a + 1) are antiparallel, the same is true of q and p ( x − 1) / ( a + 1) . The phase of x is α and by a trigonometric identity, the phase of $^{x}$− 1 is ( α + π ) / 2 . Therefore, for q and p ( $^{x}$− 1) / ( a + 1) to be antiparallel, we need: The above derivation was for the required scattering angle of a collision between 1 and 2 in order for a collision between 1 and 3 to follow. But since the final result is independent of any mass ratios, it also works for the case of a collision between 1 and 3. However for that case, by φ we mean the angle of incidence of particle 3 and by ψ we mean the initial angle of 2’s velocity. Following the collision between 1 and 2 described above, the phase of 2’s velocity is: ψ ′ = φ + α = 2 ψ − φ + π and the phase of 3’s is φ ′ = ψ + π. Now if these are plugged into the formula for scattering angle, we obtain: This means that if four collisions occur in a row (either (2323) or (3232)) the second and third scattering angles are equal. The first and fourth collisions are required because the collisions discussed above were assumed to be preceded and followed by collisions. Processes which lead to capture leave 1 and 2 with the same velocity. Two particles with the same velocity may be said to collide after an infinite time so the process concludes with a virtual collision (3). Since this may be counted as the fourth collision, all collisions have the same 34 α ′ = 2( ψ ′ − φ $^{′}$) + π = 2( ψ − φ ) + π = α. (B4) ψ + π = φ + ( α + π ) / 2 ⇒ (B2) α = 2( ψ − φ ) + π (B3)   p ′ q ′   =   x 0 ( x − 1) / ( a + 1) 1     p q   =   xp q + p ( x − 1) / ( a + 1)   (B1) are given by: