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

68 CHAPTER 3. POLARISATION OF CHIRAL ELASTOMERSwhere de Gennes elastic constants D 1 = µ 2r (1 − r)2 and D 2 = µ r (1 − r2 )[33] have been substituted in. Note the separation of the symmetric and theantisymmetric parts together with the appropriatede Gennes elastic constant.For the case of an oscillating λ xz component⎛λ = ⎝1 0 ǫ0 1 00 0 1⎞⎠, (3.50)the resulting perturbation can be calculated by breaking up the above λ intoits symmetric and antisymmetric parts, and then using the expression for l ′ .The result is⎛l ′ = ⎝0 0 ǫr0 0 0ǫr 0 0⎞⎠. (3.51)The initial orientation of the director is n 0 = (0,0,1) and the current directororientation can be written as n = (sinθ,0,cosθ), then all the components ofthe equation of motion reduce toγ ˙θ = −D 1 sin2θ − ( D 22−D 1)ǫ(t)cos2θ (3.52)where the de Gennes elastic constants have again been substituted in. Forsmall oscillations this reduces to a linear, driven, over-damped oscillator equation.The driving deformation is assumed to be ǫ(t) = ǫe iωt . The amplitudein complex notation is then( )ǫ21− D 22D 1A(ω) = , (3.53)1+iωτ cwhere τ c = γ2D 1. The polarisation of the liquid crystal elastomer can becalculated by using the orientation of the director and Eq. (3.8). Althoughthis formula is derived entirely from equilibrium statistical mechanics, it isused here for a system that is considered locally close to equilibrium at eachstage of its relaxation)P y = 1 2 n sd(b/a)((r −1)ǫe iωt − (r−1)2rA(ω)e iωt . (3.54)Denoting P ∞ = 1 2 n sd(b/a)(r − 1)ǫ (the ∞ denoting the large frequency response)and use the explicit forms above for D 1 and D 2 , then this simplifiestoiωτ cP y = P ∞ e iωt . (3.55)1+iωτ cThe amplitude of this response is given by∣ P y ∣∣∣ ωτ∣ = √ c(3.56)P ∞ 1+(ωτc ) 2.