Question 5.2.5: Determine the field of directors n(r) for a straight disclin......
Determine the field of directors \vec{n}(\vec{r}) for a straight disclination with the Frank index m in a nematic, when the moduli K_{1} and K_{3} are equal to each other: K_{1} = K_{3} = K.
If the disclination line is directed along the z-axis, the field of directors is plane, with the non-zero components n_{x} = \cos \phi(x, y), n_{y} = \sin \phi(x, y).
Clearly, for a given field of directors \vec{n}(x, y) the value of the angle \phi(x, y) is not specified uniquely: any quantity of the form lπ, where l is either a positive or a negative integer, can be added to \phi without making any difference in \vec{n}(x, y). The very presence of the disclination means that while passing along any closed path in the (x, y) plane around the origin of the coordinate system (which is the projection of the disclination line), the angle \phi acquires the increment equal to mπ, with the integer m called the Frank index of the disclination: \int \vec{∇} \phi · d\vec{l} = mπ.
The equilibrium equation for the function \phi(x, y) can be obtained in the same way as it has been done in Problem 5.2.2. In this case there is no twisting deformation, so the free energy density takes the simple form: \mathcal{F} = K(\vec{∇}\phi)^{2}.
Therefore, the total energy per unit length of the disclination is equal to \mathcal{W} = \int K(\vec{∇}\phi)^{2} dxdy, and the required minimum of \mathcal{W} yields the equilibrium equation \vec{∇}^{2}\phi = 0. By introducing the polar coordinates (r, θ) in the (x, y) plane, one concludes that its relevant solution must not depend on r (otherwise the angle increment \int \vec{∇} ·d\vec{l} would be different for the circles with the different radii). Thus, the equilibrium equation reduces to d^{2}\phi /d^{2}θ = 0, so that \phi(θ) = Aθ + \phi_{0}. The constant A is related to the Frank index as \int \vec{∇} \phi · d\vec{l} = mπ, hence
\phi(θ) = \frac{m}{2} θ + \phi_{0}
The resulting field of directors \vec{n} can be sketched by plotting the “streamlines” of the directors, which are tangential to \vec{n} at each point in the (x, y) plane.
Thus, their differential equation is (see Figure 5.8):
\frac{1}{r} \frac{dr}{dθ} = cot α = cot \left[ \left( \frac{m}{2} − 1 \right) θ + \phi_{0}\right]
If the Frank index of the disclination is such that m ≠ 2, the constant \phi_{0} can be eliminated by the appropriate choice of the line θ = 0. The field of directors for the cases m = 1 and m = −1 are shown in Figure 5.9. When m = 2, the result depends on \phi_{0}; hence some of these possibilities are also depicted there.
