aligned

An entropic slip

After Daan Frenkel gave a fantastic seminar on nucleation of solid crystals in a hard-sphere liquid, I thought it would be educational to make a toy example explaining how crystallization, often viewed as an entropy-reducing transformation, can actually be entropy-increasing.

I decided to study a simple `lattice nematic' model, which I thought would show an analogous effect, but I was wrong.

Consider K elongated particles of length L on a NxN grid with periodic boundary conditions. Define the density to be (KL/N2). Use Monte Carlo moves to sample uniformly from all possible states. At sufficiently high density I expected that almost all the particles would align as illustrated in the right hand figure1. An aligned state, I reasoned, would allow all the gaps between particles to be accessible and thus useful for creating lots and lots of configurations. In contrast, a `liquid' state with a mixture of orientations would have traffic jams with `wasted' gaps. Thus as the density increased, the particles were expected to `crystallize' into an aligned state.

I hoped to estimate the free energy of this system in its liquid and aligned states, and make simulations to confirm these ideas.

In fact, two typical states of a system with L=5 and density 0.85 are shown below, and it's evident that alignment is not the norm.

twostates

These two states were created by starting the system in an all-vertical and a 50:50 mixed state respectively. The time-history of the number of horizontal particles is shown below. (150 corresponds to the 50:50 state.)

twohistories

The clinching argument is a theoretical one. At density 1, an aligned system can be in a number of states equal to LN, where N is the number of rows in the grid, which scales as the square root of the number of particles, K. Thus the entropy of an aligned system does not grow linearly with K. In contrast, consider a 50:50 ensemble in which the grid is tiled with square LxL patches, each having a randomly chosen orientation. The entropy of this ensemble is K/L log(2), which grows linearly with K. Therefore mixed states dominate over aligned states at density 1.

Oops!

I thank Sanjoy Mahajan for programming help, and Mike Cates and Peter Sollich for helpful discussions.

Footnotes
1. Periodic boundary conditions were used correctly in the simulations but the figures do not render these situations correctly.


Peter Sollich raises this question regarding the relationship to the continuum model: Suppose you make the particles have width W and length L*W, i.e. divide each old lattice square in to W^2 new ones. As one makes W^2 large, the behaviour of the continuum model should be approached, which (at least in 3d) does have an isotropic-nematic phase transition. On the other hand, at density = 1 your argument should apply again. If both are true, then for sufficiently large W one would expect an I-N phase separation first and then a re-entrance to I (isotropic) at some density below 1 (but which presumably -> 1 as W->infty)? Which would seem rather odd, so there ought to be a flaw somewhere. (One possibility being that the density=1 limit becomes singular in the continuum version, although for W large but finite that doesn't matter.)

Here's the answer. Let x be how big the particles are compared with the grain-size of the state space; let all particles be $x$ by $Lx$ in size. Define the undensity $u$ to be the fraction of the volume not covered by the particles. We are interested in the entropy of various ensembles as $u \rightarrow 0$. Let's introduce a useful quantity $v$, which is the number of translational positions each particle can take if each is given a rectangular hole (of size $(Lx+v) \times x$) as its own territory. These holes completely fill the volume, and \[ v = Lx \frac{u}{1-u} \]

nematic and parquet entropies

We now consider two ensembles, whose entropies we can compute exactly. In the NEMATIC ensemble, all particles are oriented horizontally and are arranged in layers, with zero gap between layers. Assuming identical number of particles, $M$, per row, and that the width of the box equals $M(Lx+v)$, the entropy is exactly \[ N [ (1+v) \log (1+v) - v \log v ] \]

In the PARQUET ensemble, particles are grouped in square collections of $L$. Each square tile can have either orientation, 50:50. The tiles are packed tight in the vertical direction and can slip horizontally in the same way as the particles in the nematic ensemble. The entropy is exactly \[ \frac{N}{L} [ (1+v) \log (1+v) - v \log v + \log 2 ] \] Legal values of $v$ are $\{ 0 , 1/M , 2/M, 3/M, \ldots \}$. An interesting value of $v$ is $v=1$, which corresponds to the density when it's easy to arrange for each particle to have two translational positions. We could call this the Planck density.

The figure at the right shows the two entropies for the case of aspect ratio $L=5$.

We conclude: For sufficiently high densities -- a little beyond the Planck density -- that the nematic, all-aligned, phase is not the dominant state.

A reason for the difference of this system from a hard-sphere liquid, I guess, is that for the hard-sphere liquid, the maximum density achievable by the crytalline solid is greater than the maximum achievable by the "liquid" state. Here, a "disordered" state can achieve the same density.


David MacKay
Last modified: Fri May 28 18:25:25 2004