1-dimensional hard rods

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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; 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 &; & {\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:

\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}.


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

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


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) = 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},

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

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

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:

\int_0^{x_1} d x_0
\int_0^{x_2} d x_1
\int_0^{L} d 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
\int_0^{\infty} d y_{N-1}
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
\int_0^{\infty} d 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 }
\int_0^{\infty} d y f(y) e^{ - s y }

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.

Confined hard rods



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