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

2.B. UNDULATING CHAIN STATISTICS 39The result isZ ∝L2l hsinh( L2l h)The free energy associated with this partition function is then[F = F 0 −k B T ln L ( )] L−lnsinh2l h 2l h(2.114)(2.115)If L/l h is large then this expression shows that the free energy is approximatelylinear in L and hence that the free energy of an undulating chain isproportional to its length. The alternative way to evaluate Eq. (2.112) is towrite the product asZ = Aexp{−∫ qmaxq minLdq2π(ln ( Bq 2 +J ) +ln βL 2)}. (2.116)This integral can be evaluated by substituting q min = 0 and q max = 2πl h. Theassociated free energy isF = F 0 +k B T( Ll hln JβL2− 2Ll h+ L l hln ( 1+4π 2)) (2.117)This expression is also essentially linear in L. This justifies the assumptionthat the hairpin defects and the undulating chain sections are very weaklycoupled. An hairpin defect essentially acts as a node as regards undulationsof a chain because of the extreme cost of additional bend around a hairpindefect. Thus if an undulating chain is divided up into segments separated byhairpins then the free energy of the undulating sections is the same as that ofthe reassembled sections.The end-to-end distribution of undulating chains.Using the same small angle approximation as above the probability distributionfor end-to-end vectors projected onto the perpendicular plane, R ⊥ canbe calculated.〈 ( ∫ L)〉P(R ⊥ ) = δ R ⊥ − dsσ(s)0∫ ∫ d 2 {p= Dσ(s)(2π) 2 exp− β ∫ (L)}ds B∂σ(s)22 ∣ ∂s ∣ +Jσ(s) 20∫ip·R ⊥ −ip·dsσ(s)(2.118)(2.119)This integral can be evaluated by using the Fourier representation. The resultisP(R ⊥ ) ∝ e −βJ 2L R2 ⊥ (2.120)