Question 5.2.4: If the external magnetic field applied to a cholesteric is d......

If the twist axis of the cholesteric is directed along the z-axis, the distribution of the directors can be written as:

n_{x}(z) = \cos \phi(z), n_{y}(z) = \sin \phi(z), n_{z} = 0

Thus, if the external magnetic field is directed along the y-axis, the volumetric density of the elastic energy is equal, according to expressions (5.23)-(5.25), to

\mathcal{F}=K_1(\vec{∇} ⋅ \vec{n})^2+K_2(\vec{n}⋅(\vec{∇} ⋅ \vec{n}))^2\\ +K_{3}(\vec{n} × (\vec{∇} × \vec{n}))^{2} + α(\vec{n} · (\vec{∇} × \vec{n}))           (5.23)

\vec{∇} ⋅ \vec{n}=0,\ \ \vec{n}×(\vec{∇} × \vec{n})=0, \\ (\vec{n}⋅(\vec{∇} × \vec{n}))=-α/2K_2=t_0=const              (5.24)

\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)

\mathcal{F} = K_{2} \left(\frac{d\phi}{dz}  − t_{0}\right)^{2} − βB^{2} \sin^{2} \phi, t_{0} = \frac{α}{2K_{2}}        (5.28)

As seen from (5.28), if β > 0, the magnetic field tries to orient directors along the y-axis; hence, its effect is unwinding. In order to find out how it competes with the originally twisted structure of the cholesteric, one has to obtain the equation of equilibrium, which follows from the minimization of the total elastic energy

\mathcal{W} = \int dz \mathcal{F} = \int dz \left[K_{2} \left(\frac{d\phi}{dz}  − t_{0}\right)^{2} − βB^{2} \sin^{2} \phi\right]        (5.29)

Then, the respective Euler-Lagrange equation (the equlibrium equation) reads

\frac{d^{2}\phi}{d^{2}z} + \frac{βB^{2}}{K_{2}} \sin \phi \cos \phi = 0

Its first integral is

γ^{2} \left(\frac{d\phi}{dz} \right)^{2} + \sin^{2} \phi = const = κ^{−2},          (5.30)

where γ ≡ (K_{2}/βB^{2})^{1/2}, and the integration constant, κ^{−2}, must be greater than unity since \phi is a periodic function of z. Then, the dependence of \phi on z can be written implicitly as

\frac{z}{γκ} = \int_{0}^{\phi} dx(1 − κ^{2} \sin^{2} x)^{−1/2} ≡ F(κ, \phi),

where F(κ, \phi) is the elliptic integral of the first kind (see, e.g., M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, §17, Dover, New York).

The spatial period of the structure, L, which corresponds to the variation of \phi by 2π, is equal to L = 4γκK(κ), where

K(κ) = \int_{0}^{π/2} dx(1 − κ^{2} \sin^{2} x)^{−1/2} ≡ F(κ, π/2)

is the complete elliptic integral of the first kind. Hence, the elastic energy per unit volume, averaged over the period, is, according to expression (5.29), equal to

\tilde{\mathcal{W}} = \frac{1}{L} \int_{0}^{L} dz \left[K_{2} \left(\frac{d\phi}{dz}  − t_{0}\right)^{2} − βB^{2} \sin^{2} \phi \right] =

\frac{4βB^{2}}{L} \int_{0}^{π/2} \frac{d\phi}{d \phi /dz} \left[γ^{2}  \left(\frac{d\phi}{dz}  − t_{0}\right)^{2} − \sin^{2}\phi \right]        (5.31)

By substitution of d\phi /dz from (5.30), the expression (5.31) for \tilde{\mathcal{W}} can be transformed into

\tilde{\mathcal{W}} = βB^{2} \left(2κ^{−2}  \frac{E(κ)}{K(κ)} − \frac{γπt_{0}}{κK(κ)} − κ^{−2} + γ^{2}t^{2}_{0}\right)        (5.32)

where

E(κ) ≡ \int_{0}^{π/2} dx(1 − κ^{2} \sin^{2} x)^{1/2}

is the complete elliptic integral of the second kind (the identity

\int_{0}^{π/2} \frac{\sin^{2} xdx}{(1 − κ^{2} \sin^{2} x)^{1/2}} = \frac{K(κ) − E(κ)}{κ^{2}}

has been used here). The yet unknown value of κ has to be obtained by seeking the minimum of the energy (5.32) with respect to this parameter. By taking into account the following differentiation rules for the complete elliptic integrals:

\frac{dK}{dκ} = \frac{E}{κ(1 − κ^{2})} − \frac{K}{κ}, \frac{dE}{dκ} = \frac{E − K}{κ} ,

the required condition d \tilde{\mathcal{W}}/dκ = 0 yields

\frac{E(κ)}{κ} = \frac{π}{2} γt_{0}      (5.33)

This relation determines κ as a function of γ and, hence, its dependence on the applied magnetic field B. By using (5.33), one can get the following expression for the spatial period of the structure:

\frac{L}{L_{0}} = \frac{4}{π^{2}} K(κ)E(κ),        (5.34)

where L_{0} = 2π/t_{0} is the pitch length of the helical structure in the absence of the magnetic field. It follows from relations (5.33) and (5.34), that as the magnetic field is increasing, the cholesteric unwinds (the pitch length grows).

For a weak field, when (γt_{0})^{−1} = Bt_{0}(β/K_{2})^{1/2} ≪ 1, the parameter κ is small: κ ≈ (γt_{0})^{−1}, and, according to relation (5.34), L/L_{0} ≈ 1 + (γt_{0})^{−4}/32, so the pitch grows slowly with the applied magnetic field. The sought after transformation from the cholesteric helicoidal structure to the uniform field of directors corresponds to the infinite pitch length, which occurs at κ = 1. As seen from relation (5.33), at this point γ_{cr}t_{0} = 2/π, thus the critical magnetic field is equal to B_{cr} = (πt_{0}/2)(K_{2}/β)^{1/2}. If the magnitude of the external magnetic field exceeds this value, the cholesteric liquid crystal acquires a uniform structure.