**Dumbbell Wall Collisions**

The pressure exerted by a gas is due to molecular collisions against a surface. The pressure is equal to the change in momentum caused by the collisions, per unit time and per unit area. Here we are concerned with differences between monatomic and diatomic molecule collisions. Therefore we need to compare the change in momentum caused by a collision of a single billiard-ball shaped mass, and the change in momentum caused by a dumbbell-shaped mass.

Analyzing dumbbell collisions is a good example of how the physics of a seemingly simple problem gets rather complicated, as soon as the geometry gets even a little bit complicated. And we are only going to treat the case of collisions with an idealized, perfectly flat, frictionless, wall. To quote Richard Feynman: "Mathematical analysis is not the grand thing it is said to be; it solves only the simplest possible equations. As soon as the equations get a little more complicated, just a shade - they cannot be solved analytically. But the numerical method...can take care of any equation..." And the numerical method is what we will ultimately resort to.

**Problem Geometry**

Consider the spinning dumbbell shown below. The mass
at each end is m/2 kg. It is colliding with an immovable wall parallel to the
y-z plane. Prior to the collision the rotation velocity at each mass is v_{r1},
and the center of gravity is moving with a translation velocity v_{x1}.
The translation velocities in the y- and z-directions, and spin velocity perpendicular
to the plane of the figure, are irrelevant.

Effect of a Collision

Assuming a perfectly elastic and frictionless collision,
after the collision the rotation velocity is v_{r2}, and the center
of gravity is moving with velocity v_{x2}, where

The total kinetic energy for the velocities of interest is

Equation (1) conserves total kinetic energy, as it
must. For an impact angle θ=0, the rotation velocity is unchanged, and
the sign of the translation velocity is reversed. The change in momentum is
the same as for a single mass m traveling with the same translation velocity
v_{x1.}

For θ =π/2,
the rotation velocity and translation velocity switch places. The change in
momentum is m(v_{x1}-v_{r1}). However the mass at the other
end of the dumbbell also collides with the wall, experiencing a change in momentum
of m(v_{x1}+v_{r1}). Therefore the total change is again the
same as for a single mass m traveling with the same translation velocity. For
any other impact angle θ the change in momentum is different for the dumbbell
and the single mass.

Monte-Carlo Pressure Calculation

For pressure, what we are really interested in is the average change in momentum, for dumbbells arriving with random velocities and angles. One immediate complication is that for a completely random set of incoming rotation angles, the angle at which the impact occurs is not completely random. It can be shown that the expected value of sinθ at impact, for a uniform distribution for θ prior to impact, is

When v_{r1}>v_{x1}the calculation
becomes messier than I care to bother with. Another serious complication is
that sometimes both masses collide with the wall, and more often not. Determining
the conditions for a double collision also involves a messy calculation (there
are actually two classes of double collisions, one where the center of gravity
still moves towards the wall following the 1^{st} collision, and another
where it bounces away from the wall after the 1^{st} collision).

**Summary**

The major difference between dumbbell collisions and billiard-ball collisions is the interchange of energy between the translation and rotation velocities caused by a dumbbell collision. Despite this difference, the Monte-Carlo runs indicate that equal pressure is exerted by dumbbells and billiard balls. This result is confirmed by an alternate argument in Section 1. The average energy before and after the dumbbell collisions was also distributed equally among all degrees of freedom.

To the list of Physics of Sound subsections

Note added October 1, 2000: the original simulation
modeled the actual flight of the dumbbell. (Viewing a movie of the simulation
is a good reality check on the equations and code). For 100,000 trials the pressure
results were equal to within the Monte-Carlo noise, but when I ran 10^{7}
trials it was clear that there was a difference on the order of 0.15%. Turns
out that the rotation angle changed by 2 degrees or so every time increment.
At the final increment, one end is slightly past the boundary, and I used linear
interpolation with the prior position to find the impact angle at the boundary.
This created a small bias in the results. I subsequently wrote a new simulation.
The initial velocities and angle are still selected randomly. But given the
initial conditions, each path is deterministic, and the path equations can be
solved very accurately numerically. The results are now typically equal to within
.001% for 10^{7} trials.