B. Collisions of type (32) n Since the determinant is the product of the eigenvalues, the eigenvalues may be expressed as ϵ$_{1}$ = y/g and ϵ$_{2}$ = y · g, where g is to be determined. Furthermore, since the trace is the sum of these, we have: The trace and determinant of the matrix are given by: To exponentiate S$_{2}$S$_{3}$, we need to diagonalize it. Now there is only one unknown variable in the problem, namely x. It is found by insisting that, in the particle-1 frame, 2 have zero velocity finally. To find 2’s final velocity, we need to multiply the string of scattering matrices. Since only the first collision has a unique scattering angle, the latter 2 n − 2 collision matrices may be expressed as a power to which the first two scattering matrices are appended: The first step in analyzing (32) n or (2)(32) n is to find the relation between the first and second scattering angles. Let the first and second scattering phasors be x = exp( iα ) and y = exp( iβ ) . Suppose the particles 2 and 3 initially have unit velocity. Then following the first collision, the velocities of 2 and 3 are x and ( a + x ) / ( a + 1) . The second collision is between 3 and 1 and we should like to know its scattering angle, φ. If we exponentiate the equation derived above for scattering angle, we get: y = exp( iβ ) = − exp(2 iψ ) / exp(2 iφ ) . In the present case, exp(2 iφ ) = [( a + x ) / ( a + 1)] / [( a + x ) / ( a + 1)] ∗ = x ( a + x ) / ( ax + 1) and exp( iψ ) = x, so that: y = − x ( ax + 1) / ( a + x ) . (14) S = ( S$_{2}$S$_{3}$ ) ( n − $^{1)}$S ( β/π ) 2 S ( α/π ) 3 = ( S$_{2}$S$_{3}$ ) $^{n}$S ( α − β ) /π 3 . (15) S$_{2}$S$_{3}$ =   y + b ( y − 1) $^{2}$/ ( a + b )( a + 1) b ( y − 1) / ( a + b ) y ( y − 1) / ( a + 1) y   . (16) Trace( S$_{2}$S$_{3}$ ) = 2 y [ 1 + b ( y − 1) 2 2 y ( a + 1)( a + b ) ] (17) det( S$_{2}$S$_{3}$ ) = y $^{2}$. (18) 2 y · 1 2 ( g + 1 /g ) = 2 y [ 1 + b ( y − 1) 2 2 y ( a + 1)( a + b ) ] . (19) 11