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The problem is the following:

Set up:

A sealed box contains a collisionless classical gas of particles, each of mass $m$. The box is coupled to a thermostat such that interactions between particles and the box walls maintain the temperature of the gas but which can otherwise be neglected.

The single particle distribution function is given by $$f(\vec{r} , \vec{p} ,t) = \frac{\rho}{(2 \pi m k_B T)^{\frac{3}{2}}} \exp \left(-\frac{p^2}{2mk_B T} \right)$$

The question:

The box is divided into two compartments by an internal wall in the $yz$-plane which has a small hole. Using $f$ as given above, calculate the flux $j_x$ of particles incident on the hole moving in the positive $x$ direction, as a function of $T, \rho $

What I'm struggling is how to use the fact that there is a small hole. Do I just compute $j_x$ as usual? In that case, I find it to be just $\int \frac{p_x}{m} f(\vec{r},\vec{p},t) d\vec{p}$

When I switch to polar coordinates, using $p_x = p \sin(\theta) \cos(\phi)$ I get the integral of $\cos (\theta)$ from $0$ to $2\pi$ which vanishes, giving me that $j_x =0$, which I am not sure if it is correct since it doesn't use the fact that there is a wall there and a small hole.

Any help would be appreciated.

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Physical reasoning

You clearly get zero because you are integrating on all the values of $p_x$, from $-\infty$ to $+\infty$, which means you are summing the two fluxes of all the particles going from the left to the right and vice versa, but since the gases in the two sub-boxes are at equilibrium, the fluxes are equal and opposite in sign.

Suggestion

If you want to study only the flux of particles going from the left box to the right box, you simply need to take positive values of $p_x$. In polar coordinates the angle $\phi$ goes from $-\pi/2$ to $+\pi/2$ (not from $0$ to $2\pi$).

Solution

However the integral is much simpler in cartesian coordinates: $$ j_x = \rho \frac{\alpha}{\sqrt{\pi}} \int_0^{\infty} dp_x p_x e^{-\alpha p_x^2} \int_{-\infty}^{+\infty} dp_y e^{-\alpha p_y^2} \int_{-\infty}^{+\infty} dp_y e^{-\alpha p_z^2}, $$ where I set $\alpha = (2mk_BT)^{-1/2}$. This is much simpler because you have two one dimensional gaussian integrals that give you a factor $\pi/\alpha$, and the trivial integral $$ \int_0^{\infty} dp_x p_x e^{-\alpha p_x^2} = \frac{1}{2\alpha}. $$ All together your final result is $$ j_x = \frac{\rho}{\sqrt{4 \pi \alpha^2}} = \frac{\rho}{\sqrt{8\pi m k_BT}} $$

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