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’ z = D z = 0 b A 1, γ 1 A 2, γ 2 x z Burgers vectors Surface Tension

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 W E (per unit-length of dislocation) (1) The surface energies W 1 and W 2 (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 = W E + W 1 + W 2

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

7. -z’ z’+2D -z’+2D z’-2D -z’-2D z’-4D -z’+4D z’+4D z = 5D z = 4D z = 3D z = 2D z = D z = 0 z = -D z = -2D z = -3D z = -4D z A1bA1b b A2bA2b (A 1 A 2 )b (A 1 A 2 )A 2 b (A 1 A 2 ) 2 b (A 1 A 2 )b (A 1 A 2 )A 1 b (A 1 A 2 ) 2 b A 1, γ 1 A 2, γ 2 Burgers vectors x Surface Tension z’ In an Infinite medium

8. (5)

9. Error function :

10. Interaction between two parallel edge dislocations The interaction energy is equal to the work to create the first dislocation [b 1, (x 1, z 1 )] in the stress field of the second one [b 2, (x 2, z 2 )]. (6)

11. (7)

12. Interaction of a single dislocation with surfaces Put b 1 = b 2 = b, x 1 = x 2 and z 1 = z 2 = z 0 Rewrite equ(7) as (8)

13. In a symmetric case Polylogarithm function

14. Minimize Equ. (8)

15. In our case AIR H2OH2O 8CB Trilayer thicker layers (1+1/2) BILAYER (n+1+1/2) BILAYER

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

17. AIR H2OH2O 8CB θ EXAMPLE: If 10 bilayers on top of trilayer, (n = 10) Then, D = 375 Ǻ ξ= 173 Ǻ θ≈ 44 o D Obviously,θ with n

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

19. worksheet AIR H2OH2O 8CB