OPhO 2022, Теория — Задача 4. Magical Box

a.

Let us consider two particles of mass $m$, both starting at the center of the box. Particle A travels in the $x$-direction to the conductive wall, while particle B similarly goes to the opposite insulated one, both travelling at a speed $v$. We neglect the impact of the other walls, as due to symmetry, any effects will cancel out. Furthermore, we can take such an approximation because Maxwell’s distribution shows that half the particles would travel to the opposite wall and vice versa.

Due to the given conditions, the momentum of the system must initially be zero. As the velocity of a particle is proportional to the square root of its temperature, $v\propto\sqrt T$, then the velocity of A after rebounding will be slightly greater as $v+\Delta v$ in the negative $x$-direction. The velocity of B after rebounding will be the same in the positive $x$-direction. As such, the net momentum of both particles is $-m\Delta v$. This means that the box must start moving in the positive $x$-direction, or in the direction of the conductive wall, to conserve the momentum of the system.

b.

Inside the box, the particles will be moving randomly at different speeds. We want to find the number of particles approaching a wall in the given speeds $[v,v+dv]$ and angles $[\theta,\theta+d\theta]$. If all molecules are moving in equal directions, the fraction of particles within a solid angle $d\Omega$ is $d\Omega/4\pi$. If we consider the angles between $\theta$ and $\theta+d\theta$, we can relate $\Omega$ as

The number of particles in a unit volume is then

where $f(v)$ is the Maxwell speed distribution function. For molecules approaching a wall of area $A$ at angle $\theta$, the volume encapsulated within a unit time $dt$ is

where $A$ is the area of the wall, or in this case, $L^2$. Therefore, the number of particles approaching the wall is

Suppose the velocity of particles after hitting the insulated wall is $v_b$. Then, by momentum conservation in the parallel direction of the wall, impulse on the box per collision is:

Thus, the net impulse imparted for collisions at speed $v$ and at angle $\theta$ is:

We can then find the force, and hence our acceleration, imparted on the wall by integrating over impulse and dividing by unit time:

For the insulated wall, $v_b=v$, so

The net force imparted on the box is thus

Since $\langle vv_b\rangle/\langle v^2\rangle\approx\frac{\sqrt{T_0T_b}}{T_0}=\sqrt{\frac{T_b}{T_0}}$, we can write our final answer as

Remarks on the Net Momentum of the System

Though analyzing the impulse imparted at the conductive and insulated walls clearly indicates that the box moves from its initial position, most solvers, including the problem writer, overlooked the rather obvious fact that the net force on the system is zero. If we set the system to include the box, the gas, and the heating element, all interactions that occur in the consequent motion of the box are undoubtedly adiabatic; in other words, it doesn’t make physical sense for the final momentum of the system to be nonzero – the box’s COM must return to rest, obeying the laws of Newtonian mechanics. Let’s divide the problem into two cases based on the size of $L$ relative to the gas’s mean free path $\lambda$.

$L\ll\lambda$

$L\ll\lambda$ implies that particles interact solely with the walls and not with themselves – once a particle collides with the conductive wall, gains momentum, and moves straight towards the insulated wall and collides with it. It’s pretty easy to imagine that the box won’t move all that much from its original position. In the first few collisions, the box will surely be accelerated towards the conductive wall, but it will come to a stop when the “hot” particles collide with the insulated wall, delivering their momentum. Once all particles collide with the conductive wall, the collision frequency and average impulse per collision at the two walls will be the same, resulting in zero net force.

$L\gg\lambda$

For this case, particles do collide with each other, and it is possible for the box to be accelerated for a sustained period of time. The temperature of the gas will gradually increase from the conductive side through gas particle collisions, eventually reaching thermal equilibrium at $T_b$.

The box will certainly accelerate from its initial position. There are multiple factors that causes the box to slow down, as it has been noted by several solvers. For instance, when the box’s speed becomes greater than the average velocity of gas particles, there will be a rapid decline in the collision rate of gas particles at the conductive wall. This is supported by some quick calculations.

When the box is moving with velocity $v_b$, particles that collide with the conductive wall have a horizontal component of velocity greater than or equal to $v_b$. The number of particles per area that collide with the conductive wall can therefore be found as:

where f(v) is the Maxwell distributuon function of the gas in the vicinity of the conductive wall. If we evaluate the integral for different values of $v_b$, we find that the collision frequency per unit area decreases rapidly as $v_b$ exceeds the average velocity of gas particles $\langle v\rangle$. For instance, collision frequency per unit area decreases from its initial value by a factor of 1000 when $v_b=2\langle v\rangle$.

This effect is also accompanied by “weaker” collisions at the conductive wall when considering particle velocities relative to the box, ultimately causing the box to slow down. But howmuch does the box’s speed decrease by? It would slow down until $v=v_{\mathrm{critical}}$ and the pressure on the conductive wall overcomes than that on the insulated wall again. Then, the box will accelerate towards the conductive wall and the cycle repeats. However, this won’t continue indefinitely. There are several ways to justify this as well. For instance, as the gas in the vicinity of the insulated wall heats up, the maximum pressure difference (so when the velocity of the box is at minimum) of the conductive and insulated walls $P_c-P_i$ will decrease. This means two things:

1. The maximum speed of the box between two consecutive “slow-downs” will get smaller
2. The $v_{critical}$ necessary for $P_c>P_i$ will also get smaller

Eventually, when the entire gas reaches $T_b$, $v_{critical}$ will just be zero, so the box will eventually come to a full stop, as needed. From the two observations above, we can conclude that the box will eventually return to rest after its velocity-time curve undergoes something that resembles damped harmonic oscillations.