A liquid crystal is composed of molecules which are elongated (and often have flat segments). It behaves like a liquid because the locations of the center- of-mass of the molecules have no long-range order. It behaves like a crystal because the orientation of the molecules does have long-range order. The order parameter for a liquid crystal is given by the dyatic \(S=\eta(n n-1 / 3 I)\), where \(\boldsymbol{n}\) is a unit vector (called the director) which gives the average direction of alignment of the molecules. The free energy of the liquid crystal can be written. $$ \phi=\phi_{0}+\frac{1}{2} A S_{i j} S_{i j}-\frac{1}{3} B S_{i j} s_{j k} S_{k i}+\frac{1}{4} C S_{i j} S_{i j} S_{k l} S_{k l} $$ where \(A=A_{0}\left(T-T^{*}\right), A_{0}, B\) and \(C\) are constants, \(I\) is the unit tensor so \(\hat{x}_{i} \cdot I \cdot \hat{x}_{i}=\hat{x}_{i}\) \(\delta_{i i}, S_{i i}=\hat{x}_{i} \cdot S \cdot \hat{x}_{i, \text { and the summation is over repeated indices. The quantities are the unit vectors }}\) \(\hat{x}_{1}=\hat{x}, \hat{x}_{2}=\hat{y}\), and \(\hat{x}_{3}=\hat{z}\). (a) Perform the summations in the expression for \(\Phi\) and write \(\Phi\) in terms of \(\eta, A, B, C\). (b) Compute the critical temperature \(T_{\mathrm{c}}\) at which the transition from isotropic liquid to liquid crystal takes place, and compute the magnitude of the order parameter \(\eta\) at the critical temperature. (c) Compute the difference in entropy between the isotropic liquid \((\eta=0)\) and the liquid crystal at the critical temperature.

Step by step solution

01

Understanding the Structure of \(S\)

From the given problem, the order parameter \(S\) is defined as \(S = eta(n n-1 / 3 I)\). The unit vector \(\boldsymbol{n}\) gives the average direction of alignment of the molecules.

02

Express the Free Energy \(\Phi\)

The free energy \(\Phi\) is given by\[ \Phi = \Phi_0 + \frac{1}{2} A S_{ij} S_{ij} - \frac{1}{3} B S_{ij} s_{jk} S_{ki} + \frac{1}{4} C S_{ij} S_{ij} S_{kl} S_{kl} \]We need to express this in terms of \(\eta, A, B,\) and \(C\).

03

Compute \(S_{ij} S_{ij}\)

First, compute \(S_{ij} S_{ij}\) using \(S = \eta(\boldsymbol{n} \boldsymbol{n} - \frac{1}{3} I)\). Thus, \[ S_{ij} = \eta (n_i n_j - \frac{1}{3} \delta_{ij}) \]so\( S_{ij} S_{ij} = \eta^2 \left((n_i n_j - \frac{1}{3} \delta_{ij}) (n_i n_j - \frac{1}{3} \delta_{ij}) \right)\).

04

Simplify \(S_{ij} S_{ij}\)

Simplify using properties of \(\boldsymbol{n}\) and \(\delta_{ij}\): \[ S_{ij} S_{ij} = \eta^2 \left(1 - \frac{2}{3} n_j n_j + \left(\frac{1}{3}\right)^2 \delta_{ij} \delta_{ij}\right) = \eta^2 \left(1 - \frac{2}{3} + \frac{1}{9} \times 3\right) = \frac{2\eta^2}{3} \]

05

Compute \(S_{ij} S_{jk} S_{ki}\)

Next, compute \( S_{ij} S_{jk} S_{ki} = (\eta \boldsymbol{n} \boldsymbol{n_i} - \frac{1}{3} \delta_{ij}) (\eta \boldsymbol{n_j} \boldsymbol{n_k} - \frac{1}{3} \delta_{jk}) (\eta \boldsymbol{n_k} \boldsymbol{n_i} - \frac{1}{3}\delta_{ki}) \). Simplifying, \[ S_{ij} S_{jk} S_{ki} = \eta^3(n_i n_j n_j n_k n_k n_i - \frac{1}{9} \delta_{ij} \delta_{jk} \delta_{ki}) = \frac{1}{27}(1 -2 + 1) \eta^3 = \frac{4 \eta^3}{9}.\]

06

Expand Beyond Three-Term Product

Compute the product \[ S_{ij} S_{ij} S_{kl} S_{kl} \], which can be expanded using previously simplified expressions for individual products.

07

Express Free Energy in Terms of \(\eta\), \(A\), \(B\), and \(C\)

Combine earlier results to express \(\Phi\):\[ \Phi = \Phi_0 + \frac{1}{2} A \times \frac{2 \eta^2}{3} - \frac{1}{3} B \times \frac{4 \eta^3}{9} + \frac{1}{4} C \times \left( \frac{2 \eta^2}{3} \right)^2 \]Simplify each term:\[ \Phi = \frac{A \eta^2}{3} - \frac{4 B \eta^3}{27} + \frac{C \eta^4}{9} \]

08

Determine Critical Temperature \(T_c\)

Set first derivative \( \frac{d \Phi}{d \eta} = 0 \) for equilibrium conditions, and solve for \(\eta\). This determines the critical temperature \(T_c\): \[ T_c = T^* + \frac{32 B^2}{81 A_0 C} \].

09

Magnitude of Order Parameter \(\eta(T_c)\)

Evaluate the magnitude of \(\eta\) at the critical temperature.\[ \eta_c = \pm \left(\frac{8 B}{9 C}\right)^{1/2} \]. For positive temperature assumption, choose positive value.

10

Compute Entropy Difference

The entropy difference between isotropic phase \((\eta=0)\) and liquid crystal at \(T_c\) can be determined from the nature of free energy reduction:\[ \Delta S = - \frac{\partial \Phi}{\partial T} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Liquid Crystals

Liquid crystals are fascinating states of matter, existing between solid and liquid. Unlike typical liquids where molecules move freely, liquid crystal molecules have an average direction, offering them a unique combination of fluidity and organized structure. This dual behavior is evident as liquid crystals flow like a liquid but exhibit ordered molecular orientation similar to crystals.
Applications of liquid crystals can be found in devices like LCD screens where their unique properties are exploited. Understanding these properties helps in manipulating their states effectively, ensuring better performance in practical applications.
Furthermore, their structural properties are strongly influenced by temperature, making them sensitive to changes in thermal states. This learning is crucial for their use in temperature-responsive devices.

Order Parameter

In statistical mechanics, the order parameter is a critical concept for defining the state of a system. For liquid crystals, it quantifies the degree of molecular alignment. Mathematically, for liquid crystals, it is given by the dyadic form: \(S = \eta(\boldsymbol{n} \boldsymbol{n} - \frac{1}{3} I)\)
Here, \(\eta\) is the magnitude of the order, \(\boldsymbol{n}\) is a unit vector indicating the average direction of molecular alignment, and \(I\) is the identity tensor. This expression reveals how the arrangement varies, pinpointing the extent of alignment.
The order parameter becomes zero in isotropic phases where molecules have no preferred orientation, and it reaches maximum values in fully aligned states. It bridges microscopic configurations to macroscopic observable properties.

Free Energy

The free energy \(\Phi\) of a liquid crystal determines its thermodynamic stability and can be formulated as:
\[ \Phi = \Phi_0 + \frac{1}{2} A S_{ij} S_{ij} - \frac{1}{3} B S_{ij} s_{jk} S_{ki} + \frac{1}{4} C S_{ij} S_{ij} S_{kl} S_{kl} \] where constants \(A\), \(B\) and \(C\) vary with temperature and system properties.
The expression for \(A\), specifically given by \(A = A_0(T-T^*)\), highlights the temperature dependence. Free energy encompasses contributions from molecular interactions and alignment, balancing various forces within the system. By minimizing free energy, we derive stable configurations and phase transitions, hinting at equilibrium positions based on interactions modeled by these constants.

Critical Temperature

Critical temperature \(T_c\) marks the transition point from an isotropic liquid to an ordered liquid crystal phase. Understanding this is vital for applications requiring precision control over these phases.
Using thermodynamic formulations, \(T_c\) can be determined by setting the first derivative of \(\Phi\) with respect to the order parameter \(\eta\) to zero, ensuring equilibrium:
\[ T_c = T^* + \frac{32 B^2}{81 A_0 C} \]
Such transitions are crucial in devices where phase stability and dynamic switching between states are needed for functionality, influencing how these devices are designed and operated, ensuring they perform optimally at desired operating conditions.

Entropy Difference

Entropy difference, \(\Delta S\), between the isotropic liquid and liquid crystal phase provides insights into the disorder and thermal fluctuations within the system. It can be computed using the derivative of free energy with respect to temperature: \(\Delta S = - \frac{\partial \Phi}{\partial T}\)
This difference gives a quantitative reflection of energy dispersal and the degree of molecular disorder between two phases. Higher entropy in isotropic phases suggests greater molecular freedom, while lower entropy in the ordered phase indicates restricted molecular movement.
Such calculations are useful to understand the thermal behavior and energy requirements for phase transitions in liquid crystal systems, aiding in the design of temperature-managing components.