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

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66 CHAPTER 3. POLARISATION OF CHIRAL ELASTOMERSThese two quantities can be related to the quantities at the second cross-linkstage as followsg 0 = − j n λ·λ−1 f·r f n0 −λ·λ−1 f·z f +tλ·λ −1f cf −tcg n = n−jn λ·λ−1 f·r f n0 −λ·λ−1 f·z f +tλ·λ −1f cf −tcThe binormal average for the two strands g 0 and g n can be evaluated bysubstituting these expressions into the formula for the binormal vector. Theaverage is slightly simpler than that of the free energy density because noannealed terms have to be expanded. Note that z f and r f are un-correlated;also c·(−)·c terms are ignored as are c·(−)·z f , c f ·(−)·c and c f ·(−)·r n0 .After using these averages the following expressions for the binormals of eachstrand are obtained〈V g0 〉 f = b2a ǫ ijk+ t2[jn λ·l 0 ·λ T ·l −1 + n−j λ·( )λ −1nf·l f ·λ −Tf·λ T ·l −1]f·C f ·λ −Tf·λ T ·l −1ja 2λ·λ−1[〈V gn 〉 f = b2a ǫ n−jijkn λ·l 0 ·λ T ·l −1 + j λ·( )λ −1nf·l f ·λ −Tf·λ T ·l −1t 2+(n−j)a 2λ·λ−1 f·C f ·λ −Tf·λ T ·l].−1Summing the contributions from each chain and then average over the position,j, to which the rod molecule is bonded, with equal probability for eachsite, andsummingtheresultover all strandsresultsin thefollowing expressionP i = n sd(b/a)ǫ ijk[λ·A ′ ·λ T ·l −1] (3.44)2This is a polarisation formula but with a different matrix compared to the freeenergy density. The minimum of the free energy density can be found as in§3.2, assuming that l ∗ has the same principal axes as l. This follows becauseC and l have the same principal axes as a result of being in the same nematicenvironment. In order to develop a polarisation it is required that A ′ andB ′ have different principal axes. However, this cannot be the case becausethe anisotropy tensor l f is related to the original anisotropy tensor by theconventional trace formula so l o and λ −1f· l f · λ −Tfhave the same principalaxes and hence A ′ and B ′ have the same principal axes. As a result the traceformula is still the contraction of a symmetric with an antisymmetric tensorand so again the polarisation is zero.
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)
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iStatistical models of elasticity i
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AcknowledgementsDuring the course o
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viiiCONTENTS5 Soft elasticity of sm
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2 CHAPTER 1. INTRODUCTION1.2 Neo-cl
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4 CHAPTER 1. INTRODUCTIONticular in
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Chapter TwoThe elasticity of hairpi
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2.1. INTRODUCTION 9Fig. 2.1. This b
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2.2. MODEL OF HAIRPIN CHAINS 11nFig
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2.2. MODEL OF HAIRPIN CHAINS 13za)
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Page 45 and 46: 2.5. CONCLUSIONS 35to their limit
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Page 95 and 96: 4.1. INTRODUCTION 85HzyxGlobal rota
Page 97 and 98: 4.1. INTRODUCTION 87Figure 4.3: The
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4.5. CONCLUSIONS 117formation in th
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120 CHAPTER 5. SOFT ELASTICITY OF S
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Bibliography[1] P. J. Flory. Statis
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139[27] J-.C. Meiners and S. R. Qua
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141[58] W. L. McMillan. Measurement
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143[86] L. Golubovic and T. C. Lube
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