Why don't dislocations go away?

Take a unit vector n and move it continuosly along the rod boundary so that it remains everywhere parallel to the tooth picks. In case (a), the vector n turns on 360 degrees after returning to it's original position, while in case (b), the vector n turns on 180 degrees. In other words, the Frank index of the dislocation is 1 in case (a) and is 1/2 in case (b).

Direction of the vector n can be represented by a point on a sphere. When the vector n goes around a closed contour, this point makes a closed loop on the sphere if the Frank index is integer. If the Frank index is semi-integer, the end point will be always opposite to the starting point. For these reasons, dislocations with integer Frank indices can be eliminated by continuous deformations, and dislocations with semi-integer Frank indices do not go away.

By deforming the closed contour smoothly, we also deform trajectories on the sphere. The closed loop on the sphere representing case (a) can be smoothly contracted into a point, removing the singularity. Physically this means that the tooth picks start orienting along the rod axis as we approach the axis. The trajectory on the sphere representing case (b) always connects opposite points on a sphere. Hence, the singularity will be always somewhere inside the rod for boundary conditions (b).

For rigorous calculations one can use the expression for the free energy of liquid crystals, 2 F = (K_1) (∇ * n)^2 + (K_2) (n * curl n)^2 + (K_3) (n x curl n)^2, and determine the elastic moduli from the specified interaction energy of the tooth picks

(K_j≅1). By the order of magnitude, the energy per layer in case (a) will be equal to the number of cells in the rod cross-section π N^2 (N =10^9) times the interaction energy per cell. The tooth picks change their orientation on angle π/4 at distance N. The angle between neigbours is about π/(4N). Their interaction energy is π^2/(16 N^2). Each tooth pick interacts with four neigbours (in z-direction there is symmetry). The result should be divided by 2, since each interaction is taken into account twice. In total we have E ≅π^3/8≅ 5.

In case (b), we have a singularity. Assue for an estimation the radial symmetry. The divergence is ∇ * n = (1/r) (d/dr) (r n) ≅1/r. The energy E ≅ ∫ F dr dφ ≅ 2π K_1 ∫ (∇ * n)^2 r dr ≅ 2π∫ dr/r = 2π ln(R/a) = 2π ln N (R is the rod radius and a is the cell size). In reality, the boundary condition are not axially symmetric and the tooth picks will rearrange to minimize the energy. So the actual energy might be smaller than 2π ln N. Still, it will be somewhere about 50 (several tens) , which is an order of magnitude larger than in case (a).