Question 5.2.6: Two parallel straight disclinations with the Frank indices m......

If the separation of the two parallel disclinations is equal to d, then at distances much larger than d, the field of directors should resemble a single disclination with the Frank index equal to the sum of the indices of both disclinations. Since, in this particular case, m_{1} + m_{2} = 2, the resulting diagram depends on the relative orientation of the disclinations, and Figure 5.10 depicts just one of the possibilities.

The sought after interaction force can be derived from the energy consideration. First, one should calculate the elastic energy of a single straight disclination with the Frank index m, which is equal to

\mathcal{W} = K \int dxdy(\vec{∇}\phi)^{2} = K \int \left(\frac{1}{r} \frac{d \phi}{dθ} \right)^{2} 2πrdr = \frac{πm^{2}K}{2} \int \frac{dr}{r}        (5.35)

As seen from expression (5.35), the resulting integral is logarithmically divergent both at small and large distances from the disclination line. Therefore, the global size, L, of the domain occupied by the liquid crystal should be used as the upper limit of integration in (5.35). As far as the lower limit is concerned, it has to be of the order of the molecular size r_{0}, because the macroscopic description used here is not applicable for r ≤ r_{0}. Hence, equation (5.35) can be written as

\mathcal{W} ≈ \frac{1}{2} πm^{2}K  ln \left(\frac{L}{r_{0}} \right)

For a macroscopic sample of a liquid crystal the ratio L/r_{0} ≫ 1, so the energy of the disclination is proportional to a large logarithm, the consequences of which are twofold. Firstly, it makes the disclination elastic energy practically insensitive to the particular value of L and r_{0}. Secondly, this energy greatly exceeds the elastic energy concentrated at the core of the disclination (at r ≤ r_{0}), and this very feature allows one to consider disclination as a macroscopic object.

One may notice from expression (5.35) that from the energy viewpoint such a disclination is similar to a straight uniformly charged line. Indeed, if the charge per unit length is κ, the respective energy of the electric field is

\mathcal{W}_{el} = \int \frac{E^{2}}{8π}2πrdr = κ^{2} \int \frac{dr}{r}

Thus, by comparing this with (5.35), one concludes that the “effective” charge of the disclination is equal to \tilde{κ} = m(πK/2)^{1/2}, and, therefore, the interaction force of the two parallel disclinations is equal to

T = \tilde{κ}_{1}(2\tilde{κ}_{2}/d) = πK m_{1}m_{2}/d

If m_{1} = m_{2} = 1, the disclinations repel each other with the force T = πK/d.