Case Study 3: Hard-Disk Dumbbells in 2D

In this case study, we consider a slightly different system than the hard-disk system in the previous case study. Here, we imagine that pairs of disks are tethered together to form dumbbells. The ``bond-length'' of a dumbbell is a constant parameter, $r_0$. We will again confine the dumbbells to a circle.

Configuration snapshot of a system of hard-disk-dumbbells. $N$ = 200 particles, $rho$ = 0.5, and $R$ = 11.3.

What is new is how we have to consider the trial moves for this system. We cannot simply select a random particle and try to displace it, because this is likely to violate the constant bond-length of the dumbbell that particle belongs to. How then do we generate new configurations? A simple idea is to use two kinds of trial moves, translation of entire dumbbells and rotation of dumbbells around their centers of mass. This was originally presented in Sec. 3.3.3. In order to implement an MC code with more than one trial move, we must include a ``trial move selection rule'' which randomly selects a trial move based on their user-defined ``weights''.

The code hdb.c implements a MC simulation of hard-disk dumbbells. Consider the following question: Does the acceptance ratio of rotational moves depend upon the weight given to displacement moves? Why or why not? Below is a plot of the acceptance ratio vs. the maximum displacement for a system of 100 dumbbells at a density of 0.5, for various displacement move weights between 0.1 and 0.9. As you can see, there appears to be no effect on the acceptance of trial displacements if we change how frequently we perform them relative to trial rotations.

Acceptance ratio vs. maximum dumbbell displacement for various displacement trial move weights between 0.1 and 0.9. $N$ = 200 (100 dumbbells), $rho$ = 0.5, and $R$ = 11.3.

As a second suggested exercise (one of an advanced nature, suitable for a course project), we can use this code to explore the liquid crystalline nature of the dumbbell fluid. Because the dumbbells are slightly elongated along one direction, they may tend to line up in a dense liquid. For dumbbell $i$, let the quantity $\theta_i$ represent the angle made by the interparticle segment and some global coordinate frame axis (say the $x$ axis). We could use this code to compute the average orientation, $\left<\theta\right>$ as a function of density and temperature. The problem is that if the system is truly a liquid, even if large numbers of dumbbells are lined up at any one time, the system will slowly evolve so that all dumbbell orientations are eventually realized. So $\left<\theta\right>$ is probably not so interesting to calculate. What would be interesting would be to assess the effect of the confining circle on the spatially resolved orientation field, $\left<\theta\right>(x,y)$. You could construct a 2D histogram of orientation and perform MC to populate it. But because the system is nominally cylindrically symmetric, it would suffice to consider a one-dimensional field, $\left<\theta\right>(r)$. Plot $\left<\theta\right>(r)$ vs. $r$ for various densities and temperatures.

A third suggested (advanced) exercise would be to measure an orientational correlation function. Here, we let the quantity $\theta_{ij}$ be the relative angle between the segments of dumbbells $i$ and $j$. You can use the code to accumulate statistics on $\theta_{ij}$ as a function of $r_{ij}$, where $r_{ij}$ is the center-of-mass-to-center-of-mass distance of the two dumbbells. It is actually better to accumulate statistics of the following polynomial of the cosine of $\theta_{ij}$:

$$P\left(\cos\theta_{ij}\right) = \cos^2\theta_{ij}-\frac12.$$ (88)

$\left<P\right>$ as a function of $r$ will reveal the character of the liquid-crystalline-like ordering of the dumbbells; this is the orientational correlation function. When two dumbbells are aligned, $\cos\theta_{ij}$ is unity, and therefore $P$ approaches unity. When two dumbbells are perpendicular, $P$ is -0.5. The average, $\left<P\right>(r)$ at a particular distance $r$, will range between -0.5 and 1.0, for perfectly perpendicular to perfectly aligned, and 0 implies no preferred orientation. Try to hypothesize how $\left<P(r)\right>$ behaves as one changes temperature and density.

A final note: Since this is a 2D system, we don't use the second Legendre polynomial of $\cos\theta$:

$$P_2(x)=\frac12(3x^2-1)$$ (89)

The reason for this is clear. Suppose $f(\theta)$ is the probability distribution of the angle $\theta$. Now, since the molecules are not polar (they have no head or tail), $\theta$ is meaningful only on the domain $[0,\frac{\pi}{2}]$. Suppose further that there is no preferred angle; i.e., $f$ is a constant. In 2D, normalization requires

$$1=f\int_0^{\pi/2}d\theta \rightarrow f=\frac{2}{\pi},$$ (90)

and in this case, with no preferred orientation, the average of $\cos^2\theta$ is

$$\left<\cos^2\theta\right> = \frac{2}{\pi}\int_0^{\pi/2} \cos^2\theta d\theta = \frac{2}{\pi}\frac{\pi}{4} = \frac12.$$ (91)

And Eq. 88 will have the described behavior. The second Legendre polynomial evaluates to zero when $\cos^2\theta$ is $\frac13$, which is indeed what $\left<\cos^\theta\right>$ evaluates to when $f$ is a constant in three dimensions.