1-dimensional hard rods

1-dimensional hard rods are basically hard spheres confined to 1 dimension (not to be confused with 3-dimensional hard rods). The model is given by the intermolecular pair potential:

$\mathrm{\Phi }(x_i,x_j)=\{\begin{array}{ccc}0& ;& |x_i-x_j|>\sigma \\ \mathrm{\infty }& ;& |x_i-x_j|<\sigma \end{array}$

where $x_k$ is the position of the center of the k-th rod, along with an external potential; the whole length of the rod must be inside the range:

$V_0(x_i)=\{\begin{array}{ccc}0& ;& \sigma /2<x<L-\sigma /2\\ \mathrm{\infty }& ;& .\end{array}$

Canonical Ensemble: Configuration Integral

The statistical mechanics of this system can be solved exactly (see Ref. 1). Consider a system of length $L$ defined in the range $[0,L]$. The aim is to compute the partition function of a system of $N$ hard rods of length $\sigma$. Consider that the particles are ordered according to their label: $x_0<x_1<x_2<\cdots <x_{N-1}$; taking into account the pair potential we can write the canonical partition function (configuration integral) of a system of $N$ particles as:

$\frac{Z(N,L)}{N!}=\underset{\sigma /2}{\overset{L+\sigma /2-N\sigma }{\int }}d{x}_0\underset{x_0+\sigma }{\overset{L+\sigma /2-N\sigma +\sigma }{\int }}d{x}_1\cdots \underset{x_{i-1}+\sigma }{\overset{L+\sigma /2-N\sigma +i\sigma }{\int }}d{x}_i\cdots \underset{x_{N-2}+\sigma }{\overset{L+\sigma /2-N\sigma +(N-1)\sigma }{\int }}d{x}_{N-1}.$

Variable change: ${\omega }_k=x_k-(k+\frac{1}{2})\sigma$ ; we get:

$\frac{Z(N,L)}{N!}=\underset{0}{\overset{L-N\sigma }{\int }}d{\omega }_0\underset{{\omega }_0}{\overset{L-N\sigma }{\int }}d{\omega }_1\cdots \underset{{\omega }_{i-1}}{\overset{L-N\sigma }{\int }}d{\omega }_i\cdots \underset{{\omega }_{N-2}}{\overset{L-N\sigma }{\int }}d{\omega }_{N-1}.$

Therefore:

$\frac{Z(N,L)}{N!}=\frac{(L-N\sigma {)}^N}{N!}.$

$Q(N,L)=\frac{(L-N\sigma {)}^N}{{\mathrm{\Lambda }}^{N}N!}.$

Thermodynamics

Helmholtz energy function

$A(N,L,T)=-k_{B}T\log Q$

In the thermodynamic limit (i.e. $N\to \mathrm{\infty };L\to \mathrm{\infty }$ with $\rho =\frac{N}{L}$, remaining finite):

$A(N,L,T)=N{k}_{B}T[\log (\frac{N\mathrm{\Lambda }}{L-N\sigma })-1].$

Equation of state

Using the thermodynamic relations, the pressure (linear tension in this case) $p$ can be written as:

$p=-{(\frac{\partial A}{\partial L})}_{N,T}=\frac{N{k}_{B}T}{L-N\sigma };$

$Z=\frac{pL}{N{k}_{B}T}=\frac{1}{1-\eta },$

where $\eta \equiv \frac{N\sigma }{L}$; is the fraction of volume (i.e. length) occupied by the rods.

Isobaric Ensemble: an alternative derivation

Adapted from Reference [4]. If the rods are ordered according to their label: $x_0<x_1<x_2<\cdots <x_{N-1}$ the canonical partition function can also be written:

$Z=\underset{0}{\overset{x_1}{\int }}d{x}_0\underset{0}{\overset{x_2}{\int }}d{x}_1\cdots \underset{0}{\overset{L}{\int }}d{x}_{N-1}f(x_1-x_0)f(x_2-x_1)\cdots f(L-x_{N-1}),$

where $N!$ does not appear one would have $N!$ analogous expressions by permuting the label of the (distinguishable) rods. $f(x)$ is the Boltzmann factor of the hard rods, which is $0$ if $x<\sigma$ and $1$ otherwise.

A variable change to the distances between rods: $y_k=x_k-x_{k-1}$ results in

$Z=\underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_0\underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_1\cdots \underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_{N-1}f(y_1)f(y_2)\cdots f(y_{N-1})\delta (\sum _{i=0}^{N-1}{y}_i-L):$

the distances can take any value as long as they are not below $\sigma$ (as enforced by $f(y)$) and as long as they add up to $L$ (as enforced by the Dirac delta). Writing the later as the inverse Laplace transform of an exponential:

$Z=\underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_0\underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_1\cdots \underset{0}{\overset{\mathrm{\infty }}{\int }}d{y}_{N-1}f(y_1)f(y_2)\cdots f(y_{N-1})\frac{1}{2\pi i}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}ds\exp [-s(\sum _{i=0}^{N-1}{y}_i-L)].$

Exchanging integrals and expanding the exponential the $N$ integrals decouple:

$Z=\frac{1}{2\pi i}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}ds{e}^{Ls}{\{\underset{0}{\overset{\mathrm{\infty }}{\int }}dyf(y)e^{-sy}\}}^N.$

We may proceed to invert the Laplace transform (e.g. by means of the residues theorem), but this is not needed: we see our configuration integral is the inverse Laplace transform of another one,

$Z^{\prime }(s)={\{\underset{0}{\overset{\mathrm{\infty }}{\int }}dyf(y)e^{-sy}\}}^N,$

so that

$Z^{\prime }(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}ds{e}^{Ls}Z(L).$

This is precisely the transformation from the configuration integral in the canonical ($N,T,L$) ensemble to the isobaric ($N,T,p$) one, if one identifies $s=p/kT$. Therefore, the Gibbs energy function is simply $G=-kT\log Z^{\prime }(p/kT)$, which easily evaluated to be $G=kTN\log (p/kT)+p\sigma N$. The chemical potential is $\mu =G/N$, and by means of thermodynamic identities such as $\rho =\partial p/\partial \mu$ one arrives at the same equation of state as the one given above.

References

1. Lewi Tonks "The Complete Equation of State of One, Two and Three-Dimensional Gases of Hard Elastic Spheres", Physical Review 50 pp. 955- (1936)
2. L. van Hove "Quelques Propriétés Générales De L'intégrale De Configuration D'un Système De Particules Avec Interaction", Physica, 15 pp. 951-961 (1949)
3. L. van Hove, "Sur L'intégrale de Configuration Pour Les Systèmes De Particules À Une Dimension", Physica, 16 pp. 137-143 (1950)
4. J. M. Ziman Models of Disorder: The Theoretical Physics of Homogeneously Disordered Systems. ISBN 0521292808. Cambridge University Press (1979)