Question 5.2.8: A nematic liquid crystal with the equal elastic moduli (K1 =......

It follows from the definition of the problem, that the distribution of directors should be a plane one, with n_{z} = 0 (the z-axis is directed along the axis of the tube). Furthermore, since at the boundary circle (at r = R) the directors are parallel to it, the increment of the angle \phi while moving along the boundary circle is equal to 2π. It means that the disclinations must be present in the nematic, with the sum of their Frank indices equal to 2. An obvious solution is just one disclination with m = 2, located at the center, with the circular pattern of directors orientation (see Figure 5.9). However, such a configuration does not provide a minimum of the elastic energy, and, therefore, it is unstable. Indeed, suppose that this single disclination with m = 2 splits into two very close to each other disclinations with m_{1} = m_{2} = 1. Then, according to Problem 5.2.6, they would repel each other trying to increase their separation. On the other hand, as follows from Problem 5.2.7, they cannot become too close to the boundary with a fixed orientation of directors due to another respective repulsion. Therefore, in the ground energy equilibrium state their separation x from the center should be such that the two repulsive actions balance each other. The respective solution can be easily found by using again the electrostatic analogy, so that the two “reflected” disclinations with m = 1, located outside the tube at a distance l = R^{2}/x from the center, are added. Then, the balance of forces reads:

\frac{1}{(l − x)} = \frac{1}{2x} + \frac{1}{(l + x)} ,

yielding x = R/5^{1/4}. The resulting pattern of directors is shown in Figure 5.12.