Monte Carlo simulation of hard-, square-well, and square-shoulder disks in narrow channels

Abstract

We perform Monte Carlo simulation of the thermodynamic and structural properties of hard-, square-well, and square-shoulder disks in narrow channels. For the thermodynamics, we study the internal energy per particle and the longitudinal and transverse compressibility factor. For the structure, we study the transverse density and density of pairs profiles, the radial distribution function and longitudinal distribution function, and the (static) longitudinal structure factor. We compare our results with a recent exact semi-analytic solution found by Montero and Santos for the single file formation and first nearest neighbor fluid, and explore how their solution performs when these conditions are not fulfilled making it just an approximation.

Graphical abstract

Appendices

Appendix A: On the longitudinal pressure of HD from the LDF

Using the notation of Refs. [22, 23], we have for the Equation Of State (EOS)

$$\begin{aligned} Z_\mathrm{{L}}=\frac{\beta p}{\lambda }= 1+A^2\sum _{i,j}\phi _i\phi _ja_{ij}e^{-\beta p a_{ij}}, \end{aligned}$$

(A1)

where $A^2$ and $\phi _i$ are the solutions to

$$\begin{aligned} \sum _je^{-\beta p a_{ij}}\phi _j=\frac{\beta p}{A^2}\phi _i. \end{aligned}$$

(A2)

The LDF in the range $a(\epsilon )<x<2a(\epsilon )$ is

$$\begin{aligned} g(x)=\frac{A^2}{\lambda }\sum _{i,j}\phi _i\phi _ja_{ij}e^{-\beta p x} \Theta (x-a_{ij}). \end{aligned}$$

(A3)

Our aim is to express the EOS in terms of the integrals

$$\begin{aligned} I_n=\int _{a(\epsilon )}^1\mathrm{{d}}x\,x^ng(x), \quad n=0,1 \end{aligned}$$

(A4)

Inserting Eq. (A3) into Eq. (A4)

$$\begin{aligned} I_0= & {} \frac{A^2}{\beta p\lambda }\sum _{i,j}\phi _i\phi _j \left( e^{-\beta p a_{ij}}-e^{-\beta p}\right) ,\end{aligned}$$

(A5a)

$$\begin{aligned} I_1= & {} \frac{A^2}{(\beta p)^2\lambda }\sum _{i,j}\phi _i\phi _j \left[ e^{-\beta p a_{ij}}(1+\beta p a_{ij})\right. \nonumber \\{} & {} \left. -e^{-\beta p}(1+\beta p)\right] . \end{aligned}$$

(A5b)

From Eq. (A2), we have $\sum _{i,j}\phi _i\phi _je^{-\beta p a_{ij}}=\beta p/A^2$. Therefore,

$$\begin{aligned} \lambda I_0= & {} 1-\frac{A^2}{\beta p}e^{-\beta p}\sum _{i,j}\phi _i\phi _j,\end{aligned}$$

(A6a)

$$\begin{aligned} \beta p\lambda I_1= & {} 1+A^2\left[ \sum _{i,j}\phi _i\phi _ja_{ij}e^{-\beta p a_{ij}} \right. \nonumber \\{} & {} \left. -e^{-\beta p}\left( 1+\frac{1}{\beta p}\right) \sum _{i,j}\phi _i\phi _j\right] . \end{aligned}$$

(A6b)

Comparison with Eq. (A1) yields

$$\begin{aligned} \beta p\lambda I_1=Z_\mathrm{{L}}-(1+\beta p)(1-\lambda I_0). \end{aligned}$$

(A7)

This is a linear equation in $Z_\mathrm{{L}}$ which is solved by Eq. (2.13a) in the main text. From which immediately follows that for the pure 1D (Hard Rods) case, we find $Z_\mathrm{{L}}=1/(1-\lambda )$, since $\epsilon \rightarrow 0$ and $a(\epsilon )\rightarrow 1$ so that $I_n=0$, as it should be [16].

Note also that from Appendix C of Ref. [23] follows that in the $p\rightarrow \infty$ limit or equivalently in the $\lambda \rightarrow \lambda _{\textrm{cp}}$ limit one finds $\lim _{\lambda \rightarrow \lambda _{\textrm{cp}}}\lambda I_0=\lim _{\lambda \rightarrow \lambda _{\textrm{cp}}}\lambda ^2 I_1=1$. In the continuum limit, one has from Eq. (A6a)

$$\begin{aligned} \lambda I_0= & {} 1-\frac{e^{-\beta p}}{\ell }J^2, \end{aligned}$$

(A8a)

$$\begin{aligned} J= & {} \int _{-\epsilon /2}^{\epsilon /2}\phi (y)\,\mathrm{{d}}y. \end{aligned}$$

(A8b)

In the high-pressure regime

$$\begin{aligned} \phi (y)\rightarrow & {} \frac{1}{\sqrt{{\mathcal {N}}}}[\phi _+(y)+\phi _-(y)],\end{aligned}$$

(A9a)

$$\begin{aligned} \phi _{\pm }(y)= & {} e^{-a(y\pm \epsilon /2)\beta p},\end{aligned}$$

(A9b)

$$\begin{aligned} {{\mathcal {N}}}\rightarrow & {} \frac{a(\epsilon )}{\epsilon \beta p}e^{-2a(\epsilon )\beta p},\end{aligned}$$

(A9c)

$$\begin{aligned} \ell\rightarrow & {} \frac{a(\epsilon )}{2\epsilon \beta p}e^{-a(\epsilon )\beta p}. \end{aligned}$$

(A9d)

Thus,

$$\begin{aligned} J= & {} \frac{2}{\sqrt{\mathcal {N}}}\int _{-\epsilon /2}^{\epsilon /2}\phi _+(y)\,\mathrm{{d}}y. \end{aligned}$$

(A10)

By expanding $a(y+\epsilon /2)$ around $y=\epsilon /2$

$$\begin{aligned} a(y+\epsilon /2)\rightarrow a(\epsilon )+\frac{\epsilon }{a(\epsilon )}(\epsilon /2-y) +\cdots \end{aligned}$$

(A11)

Therefore,

$$\begin{aligned} J\rightarrow & {} \frac{2}{\sqrt{{\mathcal {N}}}}e^{-a(\epsilon )\beta p} \int _{-\epsilon /2}^{\epsilon /2}e^{-\frac{\epsilon \beta p}{a(\epsilon )}(\epsilon /2-y)}\,\mathrm{{d}}y\nonumber \\\rightarrow & {} \frac{2}{\sqrt{{\mathcal {N}}}}e^{-a(\epsilon )\beta p} \frac{a(\epsilon )}{\epsilon \beta p}=2\sqrt{\frac{a(\epsilon )}{\epsilon \beta p}}. \end{aligned}$$

(A12)

Consequently

$$\begin{aligned} \lambda I_0\rightarrow 1-8e^{-\beta p[1-a(\epsilon )]}. \end{aligned}$$

(A13)

Consistency between this result and Eq. (2.15) gives Eqs. (2.14a)–(2.14b) in the main text.

Appendix B: RDF of the ideal gas in a narrow channel

We arrive at the analytically exact Eq. (3.1) for the RDF of the ideal gas confined in the narrow channel with the following steps

$$\begin{aligned} g_{\textrm{id}}(r)= & {} \frac{\lambda }{2N}\int _{0}^{L}\mathrm{{d}}x_1\int _{0}^{L}\mathrm{{d}}x_2 \int _{-\epsilon /2}^{\epsilon /2}\mathrm{{d}}y_1\,\frac{1}{\epsilon } \int _{-\epsilon /2}^{\epsilon /2}\mathrm{{d}}y_2\, \frac{1}{\epsilon }\nonumber \\{} & {} \times \delta (r-\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}). \end{aligned}$$

(B1)

Since the integrand depends only on $x=|x_2-x_1|$, we have $\int _0^L\mathrm{{d}}x_1\int _0^L\mathrm{{d}}x_2\ldots =2\int _0^L\mathrm{{d}}x\,(L-x)\ldots$. Moreover,

$$\begin{aligned} \delta (r-\sqrt{x^2+s^2})=\frac{r}{x}\delta (x-\sqrt{r^2-s^2}). \end{aligned}$$

(B2)

Therefore,

$$\begin{aligned}{} & {} g_{\textrm{id}}(r)\nonumber \\{} & {} =\frac{\lambda }{N\epsilon ^2}r\int _{-\epsilon /2}^{\epsilon /2}\mathrm{{d}}y_1 \int _{-\epsilon /2}^{\epsilon /2}\mathrm{{d}}y_2\, \left( \frac{L}{\sqrt{r^2-(y_2-y_1)^2}}-1\right) \nonumber \\{} & {} =\frac{2}{\epsilon ^2}r\int _0^{\min (\epsilon ,r)} ds\,(\epsilon -s) \left( \frac{1}{\sqrt{r^2-s^2}}-\frac{1}{L}\right) \nonumber \\{} & {} \approx \frac{2}{\epsilon ^2}r\int _0^{\min (\epsilon ,r)} ds\, \frac{\epsilon -s}{\sqrt{r^2-s^2}}. \end{aligned}$$

(B3)

Where in the first step, we have assumed that $\sqrt{r^2-(y_2-y_1)^2}<L$ and in the third step we have taken the limit $L\rightarrow \infty$. In the limit $r\gg 1$, $\sqrt{r^2-s^2}\approx r$, so that $g_{\textrm{id}}(r)\approx 1$ as expected.

The integral in Eq. (B3) can be analytically performed and the result is given by Eq. (3.1) in the main text.