Question 5.2.2: The domain between two parallel glass plates is filled with ......
The domain between two parallel glass plates is filled with a nematic.
Due to its relatively strong interaction with the glass, all directors are constrained to lie parallel to the glass surfaces at the boundaries (see Figure 5.7). Derive the magnitude of an external magnetic field which makes the uniform field of directors unstable (the Fredericksz transition).
Consider, for simplicity, that the moduli K_{1} and K_{3} are equal: K_{1} = K_{3} = K.
The presence of an external magnetic field affects the free energy of the nematic. For a sufficiently weak field it can be accounted for with an additional term in the free energy:
\mathcal{F} = K_{1}(\vec{∇} · \vec{n})^{2} + K_{2}(\vec{n} · (\vec{∇} × \vec{n}))^{2} +
K_{3}(\vec{n} × (\vec{∇} × \vec{n}))^{2} − β(\vec{n} · \vec{B})^{2} (5.25)
In the absence of the field the directors are oriented uniformly between the plates (by using the coordinates chosen in Figure 5.7, one has n_{x} = 1, n_{y} = 0).
However, if the constant β in (5.25) is positive, the magnetic field tries to make \vec{n} parallel to \vec{B}, which brings about a competition between the orientational actions of the field and of the glass. The result is as follows. If the strength of the external magnetic field is below some critical value B_{cr}, the orientation of the directors remains uniform (parallel to the glass surfaces) in the entire volume. Then, when B exceeds B_{cr}, such a uniform orientation of directors becomes unstable, which yields a non-uniform stable equilibrium.
In order to determine the value of B_{cr}, one has to obtain the equilibrium equation for the field of directors \vec{n}(\vec{r}). In this particular case there are only n_{x}(y), n_{y}(y), which can be represented as n_{x} = \cos \phi, n_{y} = \sin \phi. Thus, there is no twist component of the deformation, and the total elastic energy, according to expression (5.25), takes the form:
\mathcal{W} = \int_{0}^{d} \mathcal{F}dy = \int_{0}^{d} dy \left[K_{1} \cos^{2} \phi \left(\frac{d\phi}{dy} \right)^{2} + K_{3} \sin^{2} \phi \left( \frac{d\phi}{dy} \right)^{2} − βB^{2} \sin^{2} \phi \right]
= \int_{0}^{d} dy \left[ K \left(\frac{d\phi}{dy} \right)^{2} − βB^{2} \sin^{2} \phi\right]
The equilibrium state corresponds to a minimum of \mathcal{W} as a functional of \phi(y).
Therefore, the respective Euler-Lagrange equation for the \phi(y) reads
\frac{d^{2}\phi}{d^{2}y} + \frac{βB^{2}}{K} \sin \phi \cos \phi = 0 (5.26)
Whatever the value of B, it has the solution \phi(y) = 0, which corresponds to the uniform orientation of directors parallel to the glass surface. However, this solution becomes unstable at such a field B = B_{cr}, when another solution, with a small deformation of the nematic, \phi ≪ 1, does exist (note the analogy with the Euler’s instability of a compressed rod considered in Problem 5.1.6).
Then, for small \phi equation (5.26) reduces to
\frac{d^{2}\phi}{d^{2}y} + \frac{βB^{2}}{K} \phi = 0
It has the solution with the required boundary conditions \phi(0) = \phi(d) = 0, when (βB^{2}/K)^{1/2}d = π, therefore
B_{cr} = \frac{π}{d} \left( \frac{K}{β}\right)^{1/2}