Excluded volume, entropy and Onsager theory of the nematic

Entropy counts room to move, so a particle’s accessible volume matters

Intuition

Think of a bus. When it is nearly empty, everyone has plenty of room and can stand anywhere. Entropy is a count of the ways things can be arranged, and more room means more arrangements. A crowded bus has few. But a crowded bus where everyone queues neatly in rows can leave each person a bit more wiggle room than a jumble where some are squeezed and a few have a lot. That is the surprising idea of this lecture.

For one particle, every microstate that meets the constraints is equally likely. The accessible volume is the region available to the centre of the particle. Slide 10 shows a particle in two chambers joined by a small aperture. The accessible area on each side is the same, so the particle is equally likely to be on either side. In general the probability of finding a particle in a region is proportional to its accessible volume.

With many particles, each one reduces the accessible volume of the others. That lost room is the excluded volume. The translational entropy of the particles is

$S = k_B \ln W$

and $W$ is roughly the product of the accessible volumes of each particle.

Entropy can drive order, because equal sharing makes the product biggest

At high density, every arrangement leaves much less accessible volume. In a crystal, the accessible volume is shared equally between particles. In a disordered arrangement, it is shared unequally. $W$ is largest when the volume is shared equally. So the crystal has higher entropy.

For hard spheres the slide 13 diagram plots pressure against volume fraction $\phi$. Freezing sits at $\phi = 0.49$ and melting at $\phi = 0.55$, joined by a horizontal coexistence line. The pressure climbs steeply towards “To FCC (Close Packed)” at $\phi = 0.74$. A dashed metastable branch runs towards RCP, random close packed, near $\phi = 0.64$. At high densities the crystalline FCC phase has higher entropy than the amorphous RCP phase, so the crystal becomes the stable phase and there is a phase transition.

Exam detail

Know that $S = k_B\ln W$, that $W$ is about the product of accessible volumes, that equal sharing maximises $W$, and the four volume fractions 0.49, 0.55, 0.64 and 0.74 on the hard-sphere diagram.

Rods can line up to free up room, and Onsager made that quantitative

Rod-like particles have length $L$ and diameter $D$. They can point in different directions, so they also have orientational entropy. At low concentrations they use it and form an isotropic phase. But the excluded volume now depends on the relative orientation of the rods.

Compare

Isotropic rods

The average excluded volume is much larger than the volume of a rod, about $DL^2$.

Aligned rods

The excluded volume is only about 4 times the volume of a rod, about $D^2L$.

As with spheres, these exclusions grow linearly with concentration. At high concentration, the accessible volume, and so the translational entropy, is far greater in the nematic than in the isotropic phase. To maximise total entropy, the nematic gives up orientational entropy to gain translational entropy.

Lars Onsager put these qualitative arguments into a quantitative theory. He assumed a distribution $f(\theta)$ of the angle $\theta$ between each rod and the director. At each concentration he then minimised the free energy over all variations of $f(\theta)$. It predicted that

$\text{concentration} < \frac{4.2}{DL^2}\quad\text{(isotropic stable)}$

$\text{concentration} > \frac{7.0}{DL^2}\quad\text{(nematic stable)}$

with phase coexistence in between. The volume of a rod is

$V_{\text{rod}} = \frac{\pi D^2 L}{4}$

so the packing fraction at which the transitions happen decreases as $L/D$ increases.

Slide 16 compares theory with simulation. The reduced concentration on its axis is

$c = \frac{1}{4}\pi L^2 D \rho$

plotted against $D/L$. The isotropic phase lies below a shaded two-phase band and the nematic above it. Simulation points lie along both edges. The Onsager limit is $D/L = 0$.

Pitfall

Do not confuse the two directions of the argument. Aligned rods lose orientational entropy but gain translational entropy, and the gain wins at high concentration. They do not align because of any attraction.

Optional aside

Slide 15 quotes the boundaries as 4.2 and 7.0 in units of $1/DL^2$, while the slide 16 plot uses the reduced concentration $c$ with edges near 4.2 and 3.3 (values read from the figure). The slides do not show how these two sets of numbers connect, so they are reported as given and not reconciled here. If a question asks for the Onsager boundaries, use slide 15.

Recall

Why would rods in a crowded box choose to line up?

Lined-up rods exclude much less volume from each other. That frees up room for every rod to move around, which raises translational entropy by more than the loss of pointing-direction entropy.