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.
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 vr1, and the center of gravity is moving with a translation velocity vx1. 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 vr2, and the center of gravity is moving with velocity vx2, 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 vx1.
For θ =π/2, the rotation velocity and translation velocity switch places. The change in momentum is m(vx1-vr1). However the mass at the other end of the dumbbell also collides with the wall, experiencing a change in momentum of m(vx1+vr1). 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.Excluding θ =0, when vr1=0 energy is always transferred from translation velocity to rotation velocity. For small values of vx1, energy is transferred from rotational velocity to translation velocity for most collision angles. So a unique feature of the dumbbell, as compared to the single mass, is this interchange of energy between rotation and translation velocities following most collisions.
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 vr1>vx1the 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 1st collision, and another where it bounces away from the wall after the 1st collision).Fortunately, for people like me who don't like messy calculations, there is an alternative means of arriving at an answer. I wrote a Monte-Carlo dumbbell simulation in Matlab, and let it fling 100,000 dumbbells against a wall. A random value of θ prior to impact was drawn from a uniform distribution over 0-2π, and vx1 and vr1 were drawn from two zero-mean Gaussian distributions with equal variances. Only molecules with positive values of vx1 will collide with the wall, and the probability that a molecule will collide with the wall in a given time interval is proportional to vx1. Thus the probability distribution of molecules colliding with the wall is different than the probability distribution of molecules in general. This is accounted for by using only the positive velocities and by weighting the number of collisions by vx1. The result of the Monte-Carlo run was that the pressure exerted by the dumbbells was the same as the pressure exerted by the single masses. Also the average kinetic energies in the rotation and translation velocities were equal before and after the collision. The probability that both masses collide with the wall turned out to be 22% (value corrected Sept 13, 2000. Originally I neglected to weight the 2nd collision count by the velocity vx1, which resulted in an incorrect value of 16%).
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 107 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 107 trials.