Question 5.2.3: Determine the equilibrium field of directors for the setup o......

If (B −B_{cr}) ≪ B_{cr}, the deformation of the otherwise uniform distribution of \vec{n} is weak; therefore, the equilibrium equation (5.26) can be approximated as

\frac{d^{2}\phi}{d^{2}y} + \frac{βB^{2}}{K} \sin \phi \cos \phi = 0        (5.26)

\frac{d^{2}\phi}{d^{2}y} ≈ −\frac{βB^{2}}{K} \left(\phi  − \frac{2}{3} \phi^{3} \right)

Its first integral (which can be easily found with the help of the mechanical analogy, as in Problem 3.5.7) reads:

\frac{1}{2} \left(\frac{d\phi}{dy} \right)^{2} + \frac{βB^{2}}{2K} \left(\phi^{2}  −  \frac{\phi^{4}}{3}\right) = const = \frac{βB^{2}}{2K} \left(\phi^{2}_{m} − \frac{\phi^{4}_{m}}{3}\right) ,

where \phi_{m} is the maximum tilting angle of directors, corresponding to y = d/2 as shown in Figure 5.7. Another integration with the boundary condition \phi(0) = 0 yields

B(β/K)^{1/2}y = \int_{0}^{\phi} dξ \left(\phi^{2}_{m} − ξ^{2} − \frac{1}{3} \phi^{4}_{m} + \frac{1}{3} ξ^{4} \right)^{−1/2}

Then, the maximum tilt angle \phi_{m} is determined by the following relation:

B \left( \frac{β}{K}\right)^{1/2} \frac{d}{2} = \int_{0}^{\phi_{m}} dξ \left(\phi^{2}_{m} − ξ^{2} − \frac{1}{3} \phi^{4}_{m} + \frac{1}{3} ξ^{4} \right)^{−1/2} =

\int_{0}^{1} dx \left[1 − x^{2} − \frac{1}{3} \phi^{2}_{m} (1 − x^{4})\right]^{−1/2}        (5.27)

Under \phi_{m} ≪ 1, the integral in equation (5.27) can be calculated approximately as

\int_{0}^{1} dx \left[1 − x^{2} − \frac{1}{3} \phi^{2}_{m} (1 − x^{4})\right]^{−1/2} ≈

\int_{0}^{1} \frac{dx}{(1 − x^{2})^{1/2}} \left[1 + \frac{\phi^{2}_{m}}{6} (1 + x^{2}) \right] = \frac{π}{2} + \frac{π}{8} \phi^{2}_{m}

Therefore,

\phi_{m} ≈ 2 \left[ \frac{B}{B_{cr}} − 1\right]^{1/2} ,

so that when the external magnetic field slightly exceeds the critical one, the respective deformation in the distribution of directors is proportional to a square root of the supercriticality.