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\begin{document}
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\heading{E5010: One dimensional hard sphere gas}
\auname{Eyal Gavish}
{\bf The problem:}
\Dn
N beads of diameter $a$ and mass $m$ are threaded over a wire of length $L$. Assume $N>>1$, $Na<>Na$ and for $N>>1$ we get:
\[
\ln Z \simeq \ln [(\frac{1}{\lambda_T})^N \frac{1}{N!}] + N \ln L - \frac{N^2a}{L}
\]
Now we calculate the force that is acting on the edge of the wire:
\[
F = T \frac{\partial \ln Z}{\partial L} = \frac{NT}{L^2/(L+Na)}
\]
For $L >> Na$ we can write: $\frac{L^2}{L+Na} \simeq L-Na$ and the expression for the force becomes:
\[
F \simeq \frac{NT}{L-Na} = \frac{NT}{L_{eff}}
\]
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\Dn
(2)
The result we have got is exactly the result of the kinetic picture. In the kinetic picture, in 1D, the force is given by $F = \frac{Nmv^2}{L}$ and for $mv^2=T$ it becomes $F=\frac{NT}{L}$. This is not surprising because all of the assumptions we made here are corresponding to the kinetic theory postulates.
The expression for $L_{eff}$ is the part of the wire that is not occupied by the beads.
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\end{document}