1-dimensional hard rods: Difference between revisions

Revision as of 11:55, 27 February 2007

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

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

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 {(L-N\sigma )^{N}}{N!}}.

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

\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].

@@ Line 58: / Line 58: @@
 : <math> \left. A(N,L,T) = - k_B T \log Q \right. </math>
-In the thermodynamic limit (i.e. <math> N \rightarrow \infty; L \rightarrow \infty</math> with <math> \rho = N/L </math> remaining finite(:
+In the thermodynamic limit (i.e. <math> N \rightarrow \infty; L \rightarrow \infty</math> with <math> \rho = \frac{N}{L} </math>,  remaining finite):
-:
+:<math>  A \left( N,L,T \right) = - N k_B T \left[ \log \left( \frac{ N \Lambda} { L - N \sigma }\right)  - 1 \right]. </math>
 ==References==