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

112 CHAPTER 4. THE ELASTICITY OF SMECTIC-A ELASTOMERSdeformation is tanγ = −λ xz . The following deformation matrix is then obtained⎛ ⎞α 0 0λ = ⎝ 10αβ0 ⎠ (4.137)λ 0 βwhereα = √ λ 2 xx −λ 2 ; β = λ xxλ zz√λ 2 xx −λ 2 ; λ = λ xxλ xz√1+λ 2 xz. (4.138)The free energy is then of the same form as Eq. (4.133) and we have thedecomposition of the mode.A decomposition can be performed on the imposed λ zz deformation. Thestarting point is again Eq. (4.135), but this time setting tanγ = −λ xz . Theresulting matrix is then compared to that of imposed λ zz so that the identificationλ = λ zz√1+λ 2 xz can be made. This decomposition gives a geometricreason for the threshold observed when λ zz is imposed. Suppose that theelastomer deforms with only the λ zz component. The free energy density isthenf zz = 1 2 B(λ−1)2 . (4.139)Alternatively the sample could deform by a shear λ xz , which leaves the layerspacing unchanged, and then rotate to accomplish the same λ zz value. In thiscase the free energy density isf xz = 1 2 µrλ2 xz = 1 2 µr(λ2 −1) . (4.140)Comparing these two energy densities for small ǫ where λ = 1+ǫ, it is clearthat that f zz ∼ 1 2 Bǫ2 < f xz ∼ µrǫ, provided that ǫ < 2µr/B. The latterenergy is first order rather than second order in the strain and explains whyit is so costly and unphysical (it is second order finally where it intercedesafter λ cr ). This decomposition also shows that an imposed λ zz is equivalentto an imposed λ xz deformation plus a rotation and a fixed stretch along thelayer normal such that d/d 0 = λ cr . The decomposition explains why themodulus of the sample after the threshold is the same as that for an imposedλ xz deformation.4.4 Comparison with experimentThe model outlined in §4.2 can be compared with numerous pieces of experimentaldata. Here the Poisson ratios, the elastic moduli and the x-rayscattering are considered.