Anisotropy and orientational order
Positional order: Regular arrangement of particle positions (lattice)
Orientational order: Angular alignment of particles
| Phase | Positional | Orientational |
|---|---|---|
| Crystal | Yes (3D) | Yes |
| Liquid | No | No |
| Liquid Crystal | Partial/No | Yes |
The strength of the ordering is determined by the decay of correlation functions:
Many natural and artificial systems have anisotropy
Golden nanorods, tobacco virus rods , ellipsoidal silica-coated hematite particles, hard platelets in the isotropic phase, and an arrangement of hard octahedra
Different liquid crystalline phases, from Doostmohammadi and Ladoux, Trends in Cell Biology (2002)
| Phase | Positional Order | Orientational Order |
|---|---|---|
| Isotropic | No | No |
| Nematic | No | Yes (long-range) |
| Smectic | Yes (1D: layered) | Yes |
| Columnar | Yes (2D: columns) | Yes |
| Cholesteric | No | Yes (helical) |
| Crystal | Yes (3D: lattice) | Yes |
Orientation and order in various phases, from Andrienko, Journal of Molecular Liquids (2018)
Examples of phases/textures: (a) Nematic with surface point defects. (b) Thin nematic film on isotropic substrate. (c) Thread-like nematic texture. (d) Cholesteric fingerprint (helical axis in plane). (e) Short‑pitch cholesteric (standing helix) with vivid colors. (f) Long-range cholesteric DNA alignment in a magnetic field. (g, h) Focal conic textures of chiral smectic A. (i) Focal conic texture of chiral smectic C. (j) Hexagonal columnar spherulitic texture. (k) Rectangular discotic phase. (l) Hexagonal columnar phase. From Andrienko, Journal of Molecular Liquids (2018)
Orientation of a rigid rod is described by a unit vector \mathbf{u} along its long axis
The particles are head-tail symmetric, so they have a centre of symmetry
Because \mathbf{u} and -\mathbf{u} are equiprobable, on average \langle \mathbf{u}\rangle=0
Vectorial order parameter is zero. The next non-trivial invariant is a second rank tensor
| Tensor | Rank | Example order parameter |
|---|---|---|
| Scalar | 0 | Scalar nematic S (degree of alignment), e.g., S = ⟨P2(cos θ)⟩ |
| Vector | 1 | Polarization or magnetization vector P / M (polar order) |
| Second‑rank tensor | 2 | Matrices, Alignment tensor Q_{ij} (nematic order) |
| Third‑rank tensor | 3 | Octupolar / tetrahedral order T_{ijk} |

For head-tail symmetric rods, we cannot simply average orientations.
Instead, construct the alignment tensor
\mathbf{Q}=\dfrac{d}{2}\left\langle\mathbf{u}_i \otimes \mathbf{u}_i-\dfrac{1}{d}\mathbf{I}\right\rangle
where d is dimensionality, \mathbf{u}_i is the unit orientation vector. For us, d=2,3
Diagonalisation: it extracts the invariants (the properties that do not change wrt a coordinate change or rotations)
Largest eigenvalue \rightarrow scalar order parameter \mathcal{S}
Corresponding eigenvector \rightarrow director \mathbf{n}
\mathcal{S} \in [0, 1]
Caution
The nematic order parameter \mathcal{S} should not nbe cconfused with the entropy S.
In two dimensions, the director is simply characterised by the angle \psi expressed as \psi = \frac{1}{2} \operatorname{atan2}(\sin{2\theta_i}, \cos{2\theta_i}). and the nematic order parameter is simply
\mathcal{S}=\sqrt{\left\langle\cos 2 \theta_i\right\rangle^2+\left\langle\sin 2 \theta_i\right\rangle^2}
In 3D, choosing coordinates along director \mathbf{n}:
\mathcal{S}=\dfrac{1}{2} \left\langle 3\cos ^2 \theta_i-1\right\rangle
where \theta_i is angle between particle i and director.
This uses the Legendre polynomial P_2(\cos\theta) to enforce head-tail symmetry.
| n | Legendre polynomial P_n(x) | Head-tail symmetric? |
|---|---|---|
| 0 | P_0(x)=1 | Yes (even: P_0(-x)=P_0(x)) |
| 1 | P_1(x)=x | No (odd: P_1(-x)=-P_1(x)) |
| 2 | P_2(x)=\dfrac{1}{2}(3x^2-1) | Yes (even: P_2(-x)=P_2(x)) |
| 3 | P_3(x)=\dfrac{1}{2}(5x^3-3x) | No (odd: P_3(-x)=-P_3(x)) |
| 4 | P_4(x)=\dfrac{1}{8}(35x^4-30x^2+3) | Yes (even: P_4(-x)=P_4(x)) |
Using the tensorial order parameter one can express a Landau theory of the isotropic-to-nematic transition (the simplest form of orientational order appearing in liquid crystals).
Leveraging that \mathcal{S} is an eigenvalue of \mathbf{Q}, i.e. \mathbf{Qn} = S\mathbf{n} we have
\mathbf{Q}=\mathcal{S}\left(\mathbf{n} \otimes \mathbf{n}-\frac{1}{d} \mathbf{I}\right)
If we choose z to be the axis along teh director \mathbf{n} then explicitly
\mathbf{Q}=\mathcal{S}\left(\begin{array}{ccc} -\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{3} \end{array}\right)
We want to characterise the deviations from the director’s orientation.
We can think of compiling a probability distribution p(\Omega) of finding a rod with pair of polar angles \Omega=(\theta,\phi) in a coordinates system with polar axis along \mathbf{n}.
So, by definition of an average, the orientational order parameter reads
\begin{align}\mathcal{S}&=\dfrac{3}{2}\left\langle(\mathbf{u} \cdot \mathbf{n})^2-\frac{1}{3}\right\rangle = \dfrac{1}{2}\langle 3 \cos^2\theta-1\rangle=\langle P_2(\cos\theta)\rangle\\ &=2\pi \int_0^{\pi}P_2(\cos\theta)p(\theta)\sin\theta d\theta \end{align} where we used the Jacobian in spherical coordinates.
The phenomenological Landau-de Gennes approach is to construct a generic free energy density as an expansion in scalar combinations of the tensor \mathbf{Q} (of which \mathcal{S} is an example)
f=f_0+\frac{A}{2} Q_{i j} Q_{j i}-\frac{B}{3} Q_{i j} Q_{j k} Q_{k i}+\frac{C}{4}\left(Q_{i j} Q_{i j}\right)^2 or in more compact form
f=f_0+\frac{A}{2} \operatorname{tr} \mathbf{Q}^2-\frac{B}{3} \operatorname{tr} \mathbf{Q}^3+\frac{C}{4}\left(\operatorname{tr} \mathbf{Q}^2\right)^2
According to our earlier 3D definitions
\operatorname{Tr}\left(\mathbf{Q}^2\right)=\frac{2}{3} S^2 \quad,\quad \operatorname{Tr}\left(\mathbf{Q}^3\right)=\frac{2}{9} S^3
So f=f_0+\frac{A}{2} \operatorname{tr} \mathbf{Q}^2-\frac{B}{3} \operatorname{tr} \mathbf{Q}^3+\frac{C}{4}\left(\operatorname{tr} \mathbf{Q}^2\right)^2 becomes f=f_0+\frac{A}{3}S^2-\frac{2B}{27}S^3+\frac{C}{9}S^4
We can use the Landau phase transition approach.
We can expand A,B,C around a reference temperature T^\ast :
and obtain
f-f_0=\frac{a}{3}\left(T-T^{\ast}\right) S^2-\frac{2 b}{27} S^3+\frac{c}{9} S^4 .
The cubic term implies that there is a temperature range where two minima coexist:
First order phase transition. The cubic terms prevents the free energy from having an inflection at S=0 (the free energy nver gets truly flat)
The transition temperature is
T_{N I}=T^{\ast}+\frac{b^2}{27 a c}
while T^{\ast} is a material dependent temperature, identifying instability (spinodal).
Many physics problems can be cast in terms of constraiend optimisation
| Area | Problem | Constraint |
|---|---|---|
| Mechanics | Minimize action S=\int L dt | Energy conservation |
| Thermodynamics | Maximize entropy S | Fixed total energy E |
| Electromagnetism | Minimize energy | Gauss’s law \nabla \cdot \mathbf{E} = \rho/\epsilon_0 |
| General Relativity | Minimize Einstein-Hilbert action | Metric signature constraint |
| Liquid Crystals | Minimize Frank energy | Director normalization \|\mathbf{n}\|=1 |
| Statistical Mechanics | Maximize partition function | Micro/Gran/canonical ensembles |
| Optics | Minimize optical path (Fermat) | Snell’s law at interfaces |
Method: Use Lagrange multipliers \lambda to convert constrained into unconstrained problem: \mathcal{L}=f(\mathbf{x})-\lambda g(\mathbf{x}) then solve \nabla\mathcal{L}=0.
Take a function f(x,y).
A constraint is expressed by fixing the value of a second function g(x,y)=c
We want to find the extrema of f(x,y) subject to the constraint g(x,y)=c: It has to be a curve
This means find the extremal values of f on the curve \mathcal{C}: g(x,y)=c
We could parametrise \mathcal{C} but we can also simply think in terms of its geometrical features.
Take the example of a minimum: - at every point of \mathcal{C} there is a tangent vector \vec{t} - we want to find the points on \mathcal{C} where f is miniimal. This means that changes of f along \mathcal{C} are minimal. - we can calculate such changes as the projection of the local gradient with the tangent vector and state that \vec{\nabla}{f} \perp \vec{t} - Notice that, by definition, the \vec{\nabla}{g}\perp \vec{t} - So we are simply saying that
\vec{\nabla}{f}\parallel \vec{\nabla}{g}\to \vec{\nabla}{f}=\lambda\vec{\nabla}{g}
for some scalar \lambda
Discrete case: Find extremum of entropy S = -\sum_i p_i \ln p_i with constraints \sum_i p_i = 1 \quad \sum_i p_i E_i = \langle E \rangle
Using Lagrange multipliers: \mathcal{L} = -\sum_i p_i \ln p_i - \lambda_1 \left(\sum_i p_i - 1\right) - \lambda_2 \left(\sum_i p_i E_i - \langle E \rangle\right)
Setting \frac{\partial \mathcal{L}}{\partial p_i} = 0: -\ln p_i - 1 - \lambda_1 - \lambda_2 E_i = 0 \quad \Rightarrow \quad p_i = \exp(-\lambda_2 E_i - \lambda_1 - 1)
This is the Boltzmann distribution: p_i \propto \exp(-\beta E_i) with \beta = \lambda_2.
Replace sums with integrals. Let p_i \to p(x) where x is a continuous variable.
Constraints become: \int p(x) dx = 1, \quad \int p(x) E(x) dx = \langle E \rangle
Entropy functional: S[p] = -\int p(x) \ln p(x) dx
Lagrangian: \mathcal{L} = -\int p(x) \ln p(x) dx - \lambda_1 \left(\int p(x) dx - 1\right) - \lambda_2 \left(\int p(x) E(x) dx - \langle E \rangle\right)
Functional derivative \frac{\delta \mathcal{L}}{\delta p(x)} = 0: -\ln p(x) - 1 - \lambda_1 - \lambda_2 E(x) = 0 \quad \Rightarrow \quad p(x) = \exp(-\beta E(x) - c)