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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 $N>>1$, $Na<<L$. The system is in a thermal equilibrium and its temperature is $T$.

\Dn

(1) Find the force $F$ that is acting on the edges of the wire. Write the result using the form $F=N\frac{T}{L_{eff}}$. Express $L_{eff}$ using the variables $T,m,N,a$.

(2) Explain the physical meaning of the result.

\Dn

Hints:

(a) When calculating the partition function, notice that if permutations were allowed it would have caused $Z \rightarrow N!Z$.

(b) Assume that a typical distance between two beads is much larger than $a$.

(c) To calculate the product ${A=\prod\limits_{n=1}^{N}a_{n}}$ look at the sum $\ln(A)$ and use reasonable approximations to calculate it.
\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
We begin by calculating the partition function. We denote by $x_{_1}...x_{_N}$ the positions of the beads on the wire where $x_{_1}$ is the position of the bead closest to the edge of the wire at $x=L$. The partition function is:

\[
Z = \frac{1}{(2\pi)^N} \int\limits_{-\infty}^{\infty} \! \mathrm{d}p_{_N}  e^{-\beta \frac{p_{_N}^2}{2m}} \cdot ... \cdot \int\limits_{-\infty}^{\infty} \! \mathrm{d}p_{_1} e^{-\beta \frac{p_{_1}^2}{2m}} \cdot \int\limits_{0}^{L-(N-1)a} \! \mathrm{d}x_{_N} \cdot ... \cdot \int\limits_{x_{_3} + a}^{L-a} \! \mathrm{d}x_{_2} \cdot \int\limits_{x_{_2} + a}^{L} \! \mathrm{d}x_{_1}
\]

Where in this calculation we neglect edge effect. We calculate the momentum integrals and get:

\[
Z = (\frac{1}{\lambda_T})^N \cdot \int\limits_{0}^{L-(N-1)a} \! \mathrm{d}x_{_N} \cdot ... \cdot \int\limits_{x_{_3} + a}^{L-a} \! \mathrm{d}x_{_2} \cdot \int\limits_{x_{_2} + a}^{L} \! \mathrm{d}x_{_1}
\]

We denote by $I_n$ the product: $\int\limits_{0}^{L-(n-1)a} \! \mathrm{d}x_{_n} \cdot ... \cdot \int\limits_{x_{_2} + a}^{L} \! \mathrm{d}x_{_1}$

We calculate $I_2$:

\[
I_2 = \int\limits_{x_{_3} + a}^{L - a} \! \mathrm{d}x_{_2} \cdot \int\limits_{x_{_2} + a}^{L} \! \mathrm{d}x_{_1} = \int\limits_{x_{_3} + a}^{L -a} \! \mathrm{d}x_{_2} \cdot (L- x_{_2} -a)
\]
We change the variable of integration to be $y \equiv L - x_{_2} - a$  and get:

\[
I_2 = \int\limits_{L - x_{_3} - 2a}^{0} \! (-y) \mathrm{d}y = \frac{1}{2} (L- (x_{_3} + 2a))^2
\]

Now for $I_n$ we assume:

\[
I_n = \int\limits_{x_{_{n+1}}+a}^{L-(n-1)a} \mathrm{d}x_{_n} I_{n-1} =  \frac{(L-(x_{_n+1} + na))^n}{n!}
\]

and show that if the formula for $I_n$ is true then the formula is true for $I_{n+1}$:

\[
I_{n+1} = \int\limits_{x_{_{n+2}} + a}^{L -na} \! \mathrm{d}x_{_{n+1}} I_n = \frac{1}{n!} \int\limits_{x_{_{n+2}} + a}^{L-na} \! \mathrm{d}x_{_{n+1}}(L- (x_{_{n+1}} + na))^n = \frac{(L-(x_{_{n+2}} + (n+1)a))^{n+1}}{(n+1)!}
\]

Hence, by induction the formula for $I_n$ is true. Now for $N$ beads, the position of the bead $N+1$ is the position of the edge of the wire ($x=0$) minus the diameter of the bead. So we get:

\[
I_N = \frac{(L-(N-1)a)^N}{N!}
\]

So the partition function is:

\[
Z = (\frac{1}{\lambda_T})^N \cdot \frac{ (L-(N-1)a)^N}{N!}
\]

We now calculate $\ln Z$:

\[
\ln Z = \ln [(\frac{1}{\lambda_T})^N \frac{1}{N!}] + N \ln (L-(N-1)a)
\]

For $L>>Na$ 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:

\[
F \simeq \frac{NT}{L-Na} = \frac{NT}{L_{eff}}
\]

%%%%%%%%%%%%%%%
\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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