Edge Dislocation in Smectic A Liquid Crystal (Part II)

1. Edge Dislocation in Smectic A Liquid Crystal(Part II) Lu Zou Sep. 19, ’06 For Group Meeting

2. Reference and outline • General expression • “Influence of surface tension on the stability of edge dislocations in smectic A liquid crystals”, L. Lejcek and P. Oswald, J. Phys. II France, 1 (1991) 931-937 • Application in a vertical smectic A film • “Edge dislocation in a vertical smectic-A film: Line tension versus film thickness and Burgers vector”, J. C. Geminard and etc., Phys. Rev. E, Vol. 58 (1998) 5923-5925

3. z Surface Tension Burgers vectors A1, γ1 z = D z’ b x z = 0 A2, γ2

4. Notations • K  Curvature constant • B  Elastic modulus of the layers • γ Surface tension • b  Burgers vectors • u(x,z)  layer displacement in z-direction • λ  characteristic length of the order of the layer thickness λ= (K/B) 1/2

5. The smectic A elastic energy WE (per unit-length of dislocation) • (1) • The surface energies W1 and W2 (per unit-length of dislocation) • (2) u = u (x, z) the layer displacement in the z-direction The Total Energy W of the sample (per unit-length of dislocation) W = WE + W1 + W2

6. Minimize W with respect to u, Equilibrium Equation (3) Boundary Conditions at the sample surfaces (Gibbs-Thomson equation) (4)

7. In an Infinite medium z Burgers vectors z = 5D (A1A2)2b z’+4D z = 4D Surface Tension (A1A2)A1b -z’+4D z = 3D z’+2D (A1A2)b z = 2D A1b -z’+2D A1, γ1 z = D z’ b x z = 0 A2, γ2 A2b -z’ z = -D z’-2D (A1A2)b z = -2D -z’-2D (A1A2)A2b z = -3D z’-4D (A1A2)2b z = -4D

8. (5)

9. Error function :

10. Interaction between two paralleledge dislocations • The interaction energy is equal to the work to create the first dislocation [b1, (x1, z1)] in the stress field of the second one [b2, (x2, z2)]. (6)

11. (7)

12. Interaction of a single dislocation with surfaces • Put b1 = b2 = b, x1= x2 and z1 = z2 = z0 Rewrite equ(7) as (8)

13. In a symmetric case Polylogarithm function

14. Minimize Equ. (8)

15. In our case AIR thicker layers 8CB (n+1+1/2) BILAYER Trilayer (1+1/2) BILAYER H2O

16. Calculation result with γ, λ, B, K for both AIR/8CB and 8CB/Water, t = 0.54 ≈ 0.5

17. AIR EXAMPLE: If 10 bilayers on top of trilayer, (n = 10) Then, D = 375 Ǻ ξ= 173 Ǻ θ≈ 44o θ 8CB D H2O Obviously,θ with n

18. Because of the symmetry, In our case, b = n d = ΔL d is the thickness of bilayer. } ΔL cutoff energy  γc = 0.87 mN/m

19. worksheet AIR 8CB H2O