# Difference between revisions of "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

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:

• External Potential; the whole length of the rod must be inside the range:
$V_{0}(x_i) = \left\{ \begin{array}{lll} 0 & ; & \sigma/2 < x < L - \sigma/2 \\ \infty &; & {\rm elsewhere}. \end{array} \right.$
• Pair Potential:
$V (x_i,x_j) = \left\{ \begin{array}{lll} 0 & ; & |x_i-x_j| > \sigma \\ \infty &; & |x_i-x_j| < \sigma \end{array} \right.$

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:
$\frac{ Z \left( N,L \right)}{N!} = \int_{\sigma/2}^{L+\sigma/2-N\sigma} d x_0 \int_{x_0+\sigma}^{L+\sigma/2-N\sigma+\sigma} d x_1 \cdots \int_{x_{i-1}+\sigma}^{L+\sigma/2-N\sigma+i \sigma} d x_i \cdots \int_{x_{N-2}+\sigma}^{L+\sigma/2-N\sigma+(N-1)\sigma} d x_{N-1}.$

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

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

## 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

From the basic thermodynamics, the pressure [linear tension in this case] $\left. p \right.$ can be written as:

$p = - \left( \frac{ \partial A}{\partial L} \right)_{N,T} = k_B T \frac{ N}{L - N \sigma};$

$Z = \frac{p L}{N k_B T} = \frac{1}{ 1 - \eta},$

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

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