Statistical models of elasticity in main chain and smectic liquid ...

16 CHAPTER 2. HAIRPIN CHAIN ELASTOMERS=×n12∑n( n 2 −1)!(n 2 )! 2 Cs e −ikN N n −s−1( n)22 +s−1 !(−1) s(−2ik) n 2 +s.Thenumberofconfigurationscanbefoundbysummingtheresiduesatthetwopoles and calculating the integral transforms Eq. (2.15) and then summingtheresultsforstartingintheupanddowndirections. Firsttheintegral transformsof the residue at q = ik is calculated, followed by the corresponding result atq = −ik. Denoting these two results as ω + and ω − we haves=0ω + =×=×n2( n2 −1) ! ( ∑2 −1 (n n) 2 −1 nC s2 ! 2 +s (−1))!ss=0(2i) n 2 +s+1 ×{i(( n 2 +s)! (i(1+ z N ))n 2 +s +(i(1− z +s)} N ))n 2 N n 2 −112 ( n2 −1) ! ( n2)!( (1+ )nz ( 2N1− z N2 2)n2 −1 )n1−zN+( ( 2 1+ z )n−1) 2NN n−1 ,2 2and at the other poleω − =×=×n22∑ (( n2 −1) ! ( n n) 2n Cs2 ! 2 +s−1 (−1))!s(−2i) n 2 +s ×s=0{−i(2( n 2 +s−1)! (−i(1− z N ))n 2 +s−1 +(−i(1+ z +s−1)} N ))n 2 N n−112 ( n2 −1) ! ( n2)!( (1+ )nz ( 2N1− z )n2 −1 )nN 1−z+( ( 2N1+ z )n−1) 2NN n−1 .2 2 2 2Summing these two results gives the number of configurations for even nΩ (n) = ω − +ω + (2.23)2N (= N 2n(n−2)!! 2 −z 2)n 2 −1 . (2.24)A similar analysis can be done for the odd case. The result for the number ofconfigurations of the defects on the chain is{2Ω (n) N(N 2 −z 2 ) n−2n(n−2)!!=2 2 even n2(N 2 −z 2 (2.25)) n−1(n−1)!! 2 2 odd n.