# Phase space area preservation

What is wrong with the following argument?

Suppose the initial configuration $(x,p)$ of a system of many non-interacting particles each of mass $m$ in phase space is given by a rectangle $x_0\in[-a,a]$ and $p_0\in[-b,b]$.

Then they are subjected to a constant acceleration in the $x$-direction, $c$.

I wish to find the image of this rectangle in phase space after time $t$ (under such acceleration).

Hint: The answer should have the same area -- $4ab$ by Liouville Theorem.

By the SUVAT equations, $$x=x_0+{p_0\over m} t+{1\over 2}ct^2\\ p=p_0+mct$$

So the image should be $$x\in \left[-a-{b\over m}t+{1\over 2}ct^2\,\,\,\,\,\,,\,\,\,\,\,\,a+{b\over m}t+{1\over 2}ct^2\right]\\ p\in \left[-b+mct\,\,\,,\,\,\,b+mct\right]$$

So the area would be $$\left(2a+2{b\over m}t\right)\left(2b\right)\neq 4ab$$

What is wrong here?

Thank you.

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Are you sure the image remains a rectangle? – Rahul Jan 2 '13 at 21:58
@RahulNarain: Ah thank you. My assumption was totally unjustifiable! – Kurt Jan 2 '13 at 22:30

The system does not remain a rectangle for $t\ne0$. The transformation from $x_0,p_0$ to $x,p$ is affine and can be written as $$\begin{bmatrix}x \\ p\end{bmatrix} = \begin{bmatrix}1 & \frac tm \\ 0 & 1\end{bmatrix} \begin{bmatrix}x_0 \\ p_0\end{bmatrix} + \begin{bmatrix}\frac12ct^2 \\ mct\end{bmatrix}.$$ The matrix in there has determinant $1$, so it preserves area. But it is not a scalar multiple of an orthogonal matrix, so it may not preserve rectangularity. (Nor is it diagonal, which would preserve axis-aligned rectangles.)

Alternatively, you can compute the images of the vertices of the rectangle, and find that they form a parallelogram with one pair of edges of length $2a$ and corresponding perpendicular height $2b$.

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