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 (2022)
| 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 nematic phase is head-tail symmetric (even when the molecules are not): \mathbf{u} and -\mathbf{u} are equivalent
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 | Density, temperature (no orientation) |
| Vector | 1 | Polarization or magnetization vector P / M (polar order) |
| Second‑rank tensor | 2 | Alignment tensor Q_{ij} (nematic order); its largest eigenvalue is \mathcal S |
| 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}{d-1}\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: the eigenvalues do not depend on the choice of axes, the eigenvectors give the principal directions
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 be confused with the entropy S.
In two dimensions, the director is simply characterised by the angle \psi expressed as \psi = \frac{1}{2} \operatorname{atan2}(\langle\sin{2\theta_i}\rangle, \langle\cos{2\theta_i}\rangle). 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)) |
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\\ &=\int_0^{\pi}P_2(\cos\theta)p(\theta)\sin\theta d\theta \end{align} where we used the Jacobian in spherical coordinates and normalised p(\theta) as \int_0^\pi p(\theta)\sin\theta d\theta=1.
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).
For historical reasons, in 3D the alignment tensor of a uniaxial nematic is usually written as (equivalent to our definition up to a multiplicative factor)
\mathbf{Q}=\mathcal{S}\left(\mathbf{n} \otimes \mathbf{n}-\frac{1}{3} \mathbf{I}\right)
If we choose z to be the axis along the 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)
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 transition:
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).