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Influence of Surface Tension on Edge Dislocations in Smectic A Liquid Crystals

This document explores the impact of surface tension on the stability of edge dislocations in smectic A liquid crystals. It discusses general expressions and energy equations related to dislocations, incorporating the effects of Burgers vectors and layer displacement. The analysis includes applications in a vertical smectic A film, noting the relationship between line tension, film thickness, and the interaction energies among dislocations. Detailed calculations involving characteristic lengths, elastic modulus, and surface energies provide insights into the complex behavior of these materials.

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Influence of Surface Tension on Edge Dislocations in Smectic A Liquid Crystals

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  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

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