# 1-dimensional hard rods

1-dimensional hard rods (sometimes known as a Tonks gas [1]) consist of non-overlapping line segments of length $\sigma$ who all occupy the same line which has length $L$. One could also think of this model as being a string of hard spheres confined to 1 dimension (not to be confused with 3-dimensional hard rods). The model is given by the intermolecular pair potential:

$\Phi_{12}(x_{i},x_{j})=\left\{ \begin{array}{lll} 0 & ; & |x_{i}-x_{j}|>\sigma\\ \infty & ; & |x_{i}-x_{j}|<\sigma \end{array}\right.$

where $\left. x_k \right.$ is the position of the center of the k-th rod, along with an external potential. Thus, the Boltzmann factor is

$e_{ij}:=e^{-\beta\Phi_{12}(x_{i},x_{j})}=\Theta(|x_{i}-x_{j}|-\sigma)=\left\{ \begin{array}{lll} 1 & ; & |x_{i}-x_{j}|>\sigma\\ 0 & ; & |x_{i}-x_{j}|<\sigma \end{array}\right.$

The whole length of the rod must be inside the range:

$V_{0}(x_i) = \left\{ \begin{array}{lll} 0 & ; & \sigma/2 < x_i < L - \sigma/2 \\ \infty &; & {\mathrm {elsewhere}}. \end{array} \right.$

## Canonical Ensemble: Configuration Integral

The statistical mechanics of this system can be solved exactly. Consider a system of length $\left. L \right.$ defined in the range $\left[ 0, L \right]$. The aim is to compute the partition function of a system of $\left. N \right.$ hard rods of length $\left. \sigma \right.$. 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:

\begin{align} \frac{Z\left(N,L\right)}{N!} & =\int_{\sigma/2}^{L-\sigma/2}dx_{0}\int_{\sigma/2}^{L-\sigma/2}dx_{1}\cdots\int_{\sigma/2}^{L-\sigma/2}dx_{N-1}\prod_{i=1}^{N-1}e_{i-1,i}\\ & =\int_{\sigma/2}^{L+\sigma/2-N\sigma}dx_{0}\int_{x_{0}+\sigma}^{L+\sigma/2-N\sigma+\sigma}dx_{1}\cdots\int_{x_{i-1}+\sigma}^{L+\sigma/2-N\sigma+i\sigma}dx_{i}\cdots\int_{x_{N-2}+\sigma}^{L+\sigma/2-N\sigma+(N-1)\sigma}dx_{N-1}. \end{align}

Variable change: $\left. \omega_k = x_k - \left(k+\frac{1}{2}\right) \sigma \right.$ ; we get:

\begin{align} \frac{Z\left(N,L\right)}{N!} & =\int_{0}^{L-N\sigma}d\omega_{0}\int_{\omega_{0}}^{L-N\sigma}d\omega_{1}\cdots\int_{\omega_{i-1}}^{L-N\sigma}d\omega_{i}\cdots\int_{\omega_{N-2}}^{L-N\sigma}d\omega_{N-1}\\ & =\int_{0}^{L-N\sigma}d\omega_{0}\cdots\int_{\omega_{i-1}}^{L-N\sigma}d\omega_{i}\frac{(L-N\sigma-\omega_{i})^{N-1-i}}{(N-1-i)!}=\int_{0}^{L-N\sigma}d\omega_{0}\frac{(L-N\sigma-\omega_{0})^{N-1}}{(N-1)!} \end{align}

Therefore:

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

## Thermodynamics

$\left. A(N,L,T) = - k_B T \log Q \right.$

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

$A \left( N,L,T \right) = N k_B T \left[ \log \left( \frac{ N \Lambda} { L - N \sigma }\right) - 1 \right].$

## Equation of state

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

$p = - \left( \frac{ \partial A}{\partial L} \right)_{N,T} = \frac{ N k_B T}{L - N \sigma};$
$Z = \frac{p L}{N k_B T} = \frac{1}{ 1 - \eta} = \underbrace{1}_{Z_{\mathrm{id}}}+\underbrace{\frac{\eta}{1-\eta}}_{Z_{\mathrm{ex}}},$

where $\eta \equiv \frac{ N \sigma}{L}$; is the fraction of volume (i.e. length) occupied by the rods. 'id' labels the ideal and 'ex' the excess part.

It was shown by van Hove [2] that there is no fluid-solid phase transition for this system (hence the designation Tonks gas).

## Chemical potential

The chemical potential is given by

$\mu=\left(\frac{\partial A}{\partial N}\right)_{L,T}=k_{B}T\left(\ln\frac{\rho\Lambda}{1-\rho\sigma}+\frac{\rho\sigma}{1-\rho\sigma}\right)=k_{B}T\left(\ln\frac{\rho\Lambda}{1-\eta}+\frac{\eta}{1-\eta}\right)$

with ideal and excess part separated:

$\beta\mu=\underbrace{\ln(\rho\Lambda)}_{\beta\mu_{\mathrm{id}}}+\underbrace{\ln\frac{1}{1-\eta}+\frac{\eta}{1-\eta}}_{\beta\mu_{\mathrm{ex}}}$

## Isobaric ensemble: an alternative derivation

Adapted from Reference [3]. 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 as:

$Z= \int_0^{x_1} d x_0 \int_0^{x_2} d x_1 \cdots \int_0^{L} d x_{N-1} f(x_1-x_0) f(x_2-x_1) \cdots f(L-x_{N-1}),$

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

$Z = \int_0^{\infty} d y_0 \int_0^{\infty} d y_1 \cdots \int_0^{\infty} d y_{N-1} f(y_1) f(y_2) \cdots f(y_{N-1}) \delta \left( \sum_{i=0}^{N-1} y_i-L \right):$

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:

$Z = \int_0^{\infty} d y_0 \int_0^{\infty} d y_1 \cdots \int_0^{\infty} d y_{N-1} f(y_1) f(y_2) \cdots f(y_{N-1}) \frac{1}{2\pi i } \int_{-\infty}^{\infty} ds \exp \left[ - s \left(\sum_{i=0}^{N-1} y_i-L \right)\right].$

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

$Z = \frac{1}{2\pi i } \int_{-\infty}^{\infty} ds e^{ L s } \left\{ \int_0^{\infty} d y f(y) e^{ - s y } \right\}^N.$

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,

$Z'(s)= \left\{ \int_0^{\infty} d y f(y) e^{ - s y } \right\}^N,$

so that

$Z'(s) = \int_0^{\infty} ds e^{ L s } Z(L).$

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 $s=p/k T$. Therefore, the Gibbs energy function is simply $G=-kT\log Z'(p/kT)$, which easily evaluated to be $G=kT N \log(p/kT)+p\sigma N$. 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.