The order parameter and the hard-particle model of nematics

One number, S, says how well the rods line up

Intuition

Picture a box of pencils. Shake them loose and they point every which way. Tap the box on the table and they start to line up. You want one number that says “how lined up” without listing every pencil. That number is the order parameter.

Slide 4 shows two pressure against temperature phase diagrams for a cyano-biphenyl compound with an alkyl tail. One has Solid, Nematic and Isotropic fields. The other adds a Smectic A field. Slide 5 shows the isotropic to nematic transition happening in a nematic mixture on cooling from about 78 to 73 degrees.

On the microscopic scale, $\theta$ is the angle between the long axis of a chosen molecule and the average direction of orientation, the director $\bar n$. The brackets $\langle\rangle$ average over all the molecules in the sample.

$S = \frac{1}{2}\left\langle 3\cos^2\theta - 1\right\rangle$

This is the order parameter. It is 1 for perfect alignment and 0 for isotropic. As temperature rises, $S$ decreases. When the temperature is high enough, the orientational order is suddenly and completely lost ($S = 0$). That is a first-order phase transition into the isotropic state.

Exam detail

Know the formula for $S$, what $\theta$ and the brackets mean, and the two end values (1 and 0). Know that the nematic to isotropic transition is first order, meaning $S$ jumps to 0.

Hard rods are a good enough model because molecular detail hardly matters

The lecture’s good news is that molecular detail does not matter much. Inflexible molecules and colloidal rods behave in a similar way. So we can try a collection of hard rods. In a hard-particle system, particles are not allowed to overlap. No other interactions are considered. Increasing the pressure or lowering the temperature is equivalent to raising the density $\rho$.

Once a hard-particle problem is solved, the answer converts quasi-universally into energy, length and time units. These reduced units are

$e_0 = k_B T \qquad l_0 = \rho^{-1/3} \qquad t_0 = \rho^{-1/3}\sqrt{\frac{m}{k_B T}}$

In plain words, the energy unit is the thermal energy, the length unit is the typical gap between particles, and the time unit is built from the mass and the thermal energy. The slide does not define $m$, but it is the particle mass.

With no interactions, entropy alone picks the winner

To find the stable phase of an open system with $P$ and $T$ fixed by the lab, minimise the Gibbs free energy.

$G = H - TS = U + PV - TS$

If the system has fixed volume (fixed density), minimise the Helmholtz free energy instead.

$F = U - TS$

Hard particles have no interactions, so $U = 0$. At the temperature of interest, the only thing left to do is to maximise $S$. The lecture says it is this feature that makes hard-particle systems interesting. The next topic shows how maximising entropy can make rods line up.

Pitfall

Do not say hard rods “attract” each other to align. They do not attract at all. There is no energy term. Order appears because entropy can rise when the rods line up (next topic).

Optional aside

Slide 4 gives only the shape of the phase diagrams. The lecture does not read off temperatures from them in text, so use the numbers on the slide itself if a question asks for them.

Recall

For hard particles, why is the stable phase just the one with the biggest entropy?

With no interactions, the internal energy $U$ is zero. The free energy then reduces to $-TS$, so the lowest free energy at a given $T$ means the highest entropy.