1-dimensional hard rods

Hard Rods, 1-dimensional system with hard sphere interactions.

The statistical mechanics of this system can be solved exactly (see Ref. 1).

Canonical Ensemble: Configuration Integral

This part could require further improvements

Consider a system of length $\left.L\right.$ defined in the range $\left[0,L\right]$ .

Our aim is to compute the partition function of a system of $\left.N\right.$ hard rods of length $\left.\sigma \right.$ .

Model:

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

V(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.

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 parttion function (configuration integral) of a system of $N$ particles as:

{\frac {Z\left(N,L\right)}{N!}}=\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}.

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

{\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}.

Therefore: ${\frac {Z\left(N,L\right)}{N!}}={\frac {(V-N)^{N}}{N!}}.$

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