\documentclass{article}

\begin{document}

\title{Compressing a One-dimensional One-atom Gas}
\author{Drew Dolgert for Michael Fowler}
\date{\today}
\maketitle

Our gas is just a single atom bouncing classically in a one-dimensional
chamber with a moveable piston on one end.  We want to know what happens
when the piston pushes in towards the atom.  We constructed this thought
experiment to give a kinetic reasoning for the familiar $PV=nkT$ and hope
to recover that equation.

We will denote the speed of the atom by $v_a$ and its initial speed
before compression by $v_0$.  The speed of the piston is $v_p$.  The cavity
begins at a length $l_0$ so that the moving piston makes the cavity length
$l=l_0-v_p t$ at any point in time.

As the piston compresses, the atom gains velocity by colliding classically
with the piston. In a thermodynamic system, the velocity of the piston would not matter.
The more slowly it moves, the more times the atom will collide so that the
speed of the atom depends only on the cavity length.  If the piston speed is much
slower than the atom speed, then the velocity of the atom depends only on the
length of the cavity.

What are $P$, $V$, and $T$ for a one-dimensional one-atom gas?  Our new equivalent to
volume is the cavity length, $l$.  The temperature can be found from the
equipartition of energy
\begin{equation}
	\frac{1}{2}mv_a^2=\frac{1}{2}k_BT
\end{equation}
where $k_B$ is Boltzmann's constant and $m$ is the atom's mass.  The factor
on the right is only one half because the atom moves in one dimension (three
dimensions, three halves).  Solving the above equation for the temperature gives
\begin{equation}
	T = \frac{mv^2}{k_B}.
\end{equation}
The last variable of concern is the pressure.  Pressure is normally the force
per unit area exerted on the piston.  In a two-dimensional gas, pressure would
be a force per unit length.  Here, pressure is just the average force the atom
exerts on the piston.  We can find an average force by averaging the familiar
Newton's law
\begin{equation}
	F = ma = m\frac{\Delta v}{\Delta t}.
\end{equation}
During a collision with the piston, the atom velocity changes from $+v_a$ to $-v_a$ for
a total change of $\Delta t= 2v_a$.  The total time per collision is the time it takes
the atom to travel down the chamber and back or $t = 2l/v_a$.  Our pressure is then
\begin{equation}
	P = F = m\frac{2v_a}{2l/v_a} = \frac{mv_a^2}{l}.
\end{equation}
Do these variables, $P$, $V$, and $T$, fit into an equation of state?
\begin{equation}
  \left(\frac{mv_a^2}{l}\right)\cdot l = 1\cdot k\cdot\left(\frac{mv_a^2}{k}\right).
\end{equation}
It is exactly $PV=nkT$ where $n=1$ for one atom.

We can start from examining the simple kinetic collision and show that, for a
slowly moving piston, the velocity is a function of the cavity length.  When
the atom hits the fixed end, it just bounces, but when it hits the compressing
piston, it gains the speed of the piston.
\begin{equation} v_a' = -v_a+2v_p \end{equation}
At any point in time, the velocity of the atom is equal to its initial velocity
plus additions from every piston collision
\begin{equation}v_a=v_0+2\sum_{\mbox{piston hits}}v_p.\end{equation}
Instead of summing over the number of times the piston hit, we could sum over the
total time and factor in the number of hits per time.
\begin{equation}
v_a=v_0+2\sum_{\mbox{time}}v_p\times\frac{\mbox{hits}}{\mbox{time}}
\end{equation}
The frequency of hits is determined by the atom's speed and path length
\begin{equation}
2l=v_at\quad\Rightarrow\quad\frac{1}{t}=\frac{v_a}{2l}=\frac{v_a}{2(l_0-v_pt)}
\end{equation}
We can put this back in the sum to find
\begin{equation}
v_a=v_0+\sum_{t}2v_p\frac{v_a}{2(l_0-v_pt)}=v_0+\int \frac{v_pv_a}{(l_0-v_pt)}dt.
\end{equation}
We can't do that integral because $v_a=v_a(t)$, and it is unknown.  We can
rewrite the equation in terms of length using $v_pdt = dl$
\begin{equation}
v_a=v_0-\int_{l_0}^l \frac{v_a}{l}dl
\end{equation}
Taking the derivative of both sides gives a tractable differential equation
\begin{equation}
\frac{dv_a}{v_a}=-\frac{dl}{l}
\end{equation}
whose solution is
\begin{equation}
	\frac{v_a}{v_0}=\frac{l_0}{l}.
\end{equation}
Squaring both sides, we could write it as
\begin{equation}
	\left(\frac{mv_a^2}{l}\right)\cdot l^3=\left(\frac{mv_0^2}{l_0}\right)\cdot l_0^3.
\end{equation}
Looking above at the definitions for pressure and volume in this system, we see
our relation between cavity length and velocity is equivalent to
\begin{equation}
	PV^3 = \mbox{constant}.
\end{equation}
This is not the standard relation for adiabatic compression but is
of similar form.

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