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[edit]

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 of a system of $N$ particles as:

{\begin{aligned}{\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{aligned}}

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

{\begin{aligned}{\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{aligned}}

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[edit]

Helmholtz energy function

\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)=Nk_{B}T\left[\log \left({\frac {N\Lambda }{L-N\sigma }}\right)-1\right].

Equation of state[edit]

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 {Nk_{B}T}{L-N\sigma }};

The compressibility factor is

Z={\frac {pL}{Nk_{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[edit]

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[edit]

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}}dx_{0}\int _{0}^{x_{2}}dx_{1}\cdots \int _{0}^{L}dx_{N-1}f(x_{1}-x_{0})f(x_{2}-x_{1})\cdots f(x_{0}+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 }dy_{0}\int _{0}^{\infty }dy_{1}\cdots \int _{0}^{\infty }dy_{N-1}f(y_{0})f(y_{1})\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 }dy_{0}\int _{0}^{\infty }dy_{1}\cdots \int _{0}^{\infty }dy_{N-1}f(y_{0})f(y_{1})\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 }dse^{Ls}\left\{\int _{0}^{\infty }dyf(y)e^{-sy}\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 }dyf(y)e^{-sy}\right\}^{N},

so that

Z'(s)=\int _{0}^{\infty }dse^{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 $G=-kT\log Z'(p/kT)$ , 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.