On the theory of the shear-induced isotropic-to-nematic phase transition of side chain liquid-crystalline polymers

Abstract

The shear-induced isotropic-to-nematic phase transition of side chain liquid-crystalline polymers is studied theoretically. A modification of the previous models of main-chain liquid crystals to the case of side chain liquid-crystalline polymers is proposed. Orientational and rheological properties of the model are studied in plane-shear flow. Predictions of the present model agree qualitatively with experimental results (Pujolle-Robic, Noirez in Nature 409:167, 2001).

Appendix

Coefficients of low and high shear expansion

In the stationary state, the values of b _j are determined by those of a _j . In the symmetry-adapted case a₃=a₄=0 and b₃=b₄=0, the components in the plane of shear are given by

$$ \begin{aligned} b_0 & = \frac{1} {N}\left\{ { - \frac{{\nu _{{\text{ab}}} }} {{\vartheta _{\text{b}} }}\left( {1 + \Gamma ^2 } \right)a_0 - \bar \kappa _{\text{b}} \Gamma \left( {\Lambda _{\text{K}}^b \Gamma + \frac{{\nu _{{\text{ab}}} }} {{\vartheta _{\text{b}} }}\left( {\Gamma a_1 - a_0 } \right)} \right)} \right\} \\ b_1 & = \frac{1} {N}\left\{ {\Lambda _{\text{K}}^b \Gamma ^2 - \frac{{\nu _{{\text{ab}}} }} {{\vartheta _{\text{b}} }}\left( {\Gamma a_2 + \left( {1 - \bar \kappa _{\text{b}}^2 \Gamma ^2 } \right)a_1 - \bar \kappa _{\text{b}} \Gamma a_0 } \right)} \right\} \\ b_2 & = \frac{1} {N}\left\{ {\Lambda _{\text{K}}^b \Gamma - \frac{{\nu _{{\text{ab}}} }} {{\vartheta _{\text{b}} }}\left( {\Gamma a_1 - a_2 + \bar \kappa _{\text{b}} \Gamma a_0 } \right)} \right\} \\ \end{aligned} $$

(46)

with $\bar \kappa _{\text{b}} \equiv \kappa _{\text{b}} /\sqrt 3$ and $\Gamma \equiv \dot \gamma /(\alpha \vartheta _{\text{b}} )$ and $N \equiv 1 + (1 - \bar \kappa _{\text{b}}^2 )\Gamma ^2 .$ Furthermore, the abbreviations $\Lambda _{\text{K}} = (\sqrt 3 /2)\lambda _{\text{K}} ,\;\Lambda _{\text{K}}^b = (\sqrt 3 /2)\lambda _{\text{K}}^b$ were used. In equilibrium, Γ=0 and above equations reduce to the equilibrium conditions b _j =−(ν_ab/ϑ_b)a _j .

Inserting Eqs. 46 into Eqs. 31 results, for the stationary case $\dot a_j = 0,$ in a set of three nonlinear algebraic equations for a₀, a₁ and a₂. For the special case κ=κ_b=0 one obtains

$$ \begin{aligned} 0 = & - (\vartheta _{{\text{eff}}} - 3a_0 + 2a^2 )a_0 - 3(a_1^2 + a_2^2 ), \\ 0 = & - (\vartheta _1 + 6a_0 + 2a^2 )a_1 + \dot \gamma _{{\text{eff}}} a_2 - f, \\ 0 = & - (\vartheta _1 + 6a_0 + 2a^2 )a_2 - \dot \gamma _{{\text{eff}}} a_1 + \Lambda _{\text{K}}^{{\text{eff}}} \dot \gamma _{{\text{eff}}} , \\ \end{aligned} $$

(47)

where the abbreviation a²=a ²₀ +a ²₁ +a ²₂ is used. Equations 47) are of the same form as in the one alignment tensor theory with no coupling to the polymer backbone (ν_ab=0) except for an additional constant forcing f=ν_abϑ_bΛ ^b_K Γ²/(1+Γ²). Effective temperatures, shear rate and ratio of relaxation times were used in Eqs. 47 which are defined by ϑ_eff=ϑ−ν ²_ab /ϑ_b, ϑ₁=ϑ−ν ²_ab /(ϑ_b(1+Γ²)), $\dot \gamma _{{\text{eff}}} = g\dot \gamma$ and $\Lambda _{\text{K}}^{{\text{eff}}} = (\tilde g/g)\Lambda _{\text{K}} ,$ respectively, with g=1+(ν_ab/ϑ_b)²/(ζ(1+Γ²)), $\tilde g = 1 - \left( {v_{{\text{ab}}} /\vartheta _{\text{b}} } \right)/\left( {\zeta _{\text{p}} \left( {1 + \Gamma ^2 } \right)} \right),$ ζ≡τ_a/τ_b and ζ_p≡τ_ap/τ_bp. Note, that these effective quantities are highly nonlinear in the shear rate $\dot \gamma .$

Expanding Eqs. 46 and 47 up to second order in the shear rate one obtains the coefficient of the low shear expansion (Eqs. 39, 40),

$$ \alpha _2 = \frac{{\Lambda _{\text{K}} }} {{\sqrt 2 \vartheta _{{\text{eff}}} }}\left( {1 - \tilde v_{\text{p}} } \right),\quad \beta _2 = \frac{1} {{\vartheta _{\text{b}} }}\left( {\frac{{\Lambda _{\text{K}} }} {{\sqrt 2 \zeta _{\text{p}} }} - v_{{\text{ab}}} \alpha _2 } \right) $$

(48)

$$ \alpha _1 = \frac{1} {{\vartheta _{{\text{eff}}} }}\left( {\alpha _2 - \tilde v\beta _2 } \right),\quad \beta _1 = \frac{1} {{\vartheta _{\text{b}} }}\left( {\frac{1} {\zeta }\beta _2 - v_{{\text{ab}}} \alpha _1 } \right) $$

(49)

$$ \alpha _0 = - \frac{1} {{\vartheta _{{\text{eff}}} }}\left( {3\alpha _2^2 + \frac{\kappa } {{\sqrt 3 }}\alpha _2 - \frac{{\kappa _{\text{b}} \tilde v\beta _1 }} {{\sqrt 3 }}} \right),\quad \beta _0 = - \frac{1} {{\vartheta _{\text{b}} }}\left( {\frac{{\kappa _{\text{b}} }} {{\sqrt 3 \zeta }}\beta _2 + v_{{\text{ab}}} \alpha _0 } \right) $$

(50)

with $\tilde v \equiv v_{{\text{ab}}} /(\vartheta _{\text{b}} \zeta )$ and $\tilde v_{\text{p}} \equiv v_{{\text{ab}}} /(\vartheta _{\text{b}} \zeta _{\text{p}} ).$ Note, that ϑ_b>0 by definition.

In the limit of high shear rates one obtains from Eqs. 46 and 47 that a₂ and b₂ decrease as $\dot \gamma ^{ - 1}$ where the coefficients are given by

$$ a_2^\infty = (\vartheta + 6a_0^\infty + 2(a_0^\infty )^2 + v_{{\text{ab}}} \Lambda _{\text{K}}^b + 2\Lambda _{\text{K}}^2 )\Lambda _{\text{K}} , $$

and b ^∞₂ =ζ(ϑ_bΛ ^b_K +ν_abΛ_K) are used. The asymptotic value of a₀ is

$$ a_0^\infty = \frac{1} {2} + \frac{E} {{3(2)^{2/3} F^{1/3} }} - \frac{{F^{1/3} }} {{6(2)^{1/3} }}, $$

where $F = G + \sqrt {4E^3 + G^2 } ,$ E=−9+12Λ ²_K +6ϑ_eff, G=−54+432Λ ²_K +54ϑ_eff.