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

3.4. SYSTEMS WHERE A POLARISATION RESULTS 673.4 Systems where a polarisation results3.4.1 Oscillating shearThe dynamics of liquid crystal elastomers are complicated because of thecoupling between the liquid crystalline degrees of freedom and the underlyingpolymer matrix. As stated already, it is possible to identify several time scalesin the elastomer. A single free polymer strand has an associated relaxationtime called theRousetime τ p ∼ 10 −4 −10 −6 s, and themesogenic unitshave anassociated relaxation time τ LC ∼ 10 −1 −10 −2 s. However the polymer networkas a whole relaxes according to a power law and so has no associated scale.The polarisation associated with the dynamics of a liquid crystal elastomer isnow calculated under the assumption that the polymer network responds veryquickly and then the director relaxes much more slowly into its equilibriumorientation. For an elastomer that displays suchaseparation of timescales therelaxation of the polymer can be neglected (it is assumed to be instantaneous)and the slower dynamics of the director focused upon [52]. In this case thesystem can bemodelled as an over-damped oscillator with a dampingconstantγ associated with the return to equilibrium of the director. The force causingthe return to equilibrium is governed by the gradient of the equilibrium freeenergy density∂(f −χn·n)γṅ = − , (3.45)∂nwhere the Lagrange multiplier has again been included to prevent changingof the length of the director during the motion. Using the the small deformationform for the free energy density given previously and then applying theconstraint n·n = 1 to find χ as before, then in the notation of section §3.3.1,the resulting equation isγṅ = −µ ( )1r −1)( (l 0 +l ′ )·n−n(n·(l 0 +l ′ )·n)(3.46)To pick out the symmetric and antisymmetric contributions to this expressionconsider a deformation of the form⎛ ⎞1 0 η +δλ = ⎝ 0 1 0 ⎠. (3.47)η −δ 0 1On substituting this deformation into Eq. (3.46) the following expression forthe angle of the director is obtained( ) 1−rγ˙θ = µ (((r −1)δ −(r +1)η)cos2θ +(r −1)sin2θ) (3.48)r= −D 1 sin2θ −(2D 1 δ(t)+D 2 η(t))cos2θ, (3.49)