When you drop a ball, it falls under the influence of Earth's gravity. As you know, you can solve for the velocity and position of the ball as functions of time by integrating the acceleration of the ball with respect to time. Since the Earth is so massive compared to the ball, we can simply neglect its motion, although we know deep down that the momentum change of the ball as it falls is equal and opposite to the momentum change of the Earth.
What if the mass of the ball isn't so small? How would you analyze the motion of the Moon and the Earth, for example, or a binary star system in which the masses of the stars are comparable? Luckily, conservation laws allow us to simplify this problem and indeed reduce it to a single-particle problem. The particles move so that the momentum and angular momentum of the first particle with respect to the center of mass are equal and opposite to those of the second particle. Thus, we can solve the two-body problem by reducing it to the one-body problem.
This is looking almost like an inductive proof. Perhaps we can keep going and reduce the three-body problem to the two-body problem, and then reduce it to the one-body problem, which we can actually solve. Unfortunately, this procedure fails because there aren't other constants of the motion (like total momentum and total angular momentum) that we can use to reduce the three-body problem to the two-body problem. And indeed, despite three centuries of effort, during which many ways to approximate the solution have been developed, we still have discovered no "closed-form" solution to the three-body problem.
Now imagine "watching" the motion of 1023 particles in a small sample of gas. This number is so fantastically large that there is no way to imagine even writing down the initial state of all the particles in our lifetimes, never mind integrating them to find the state of the particles after even a few femtoseconds (10-15 s). It would seem that we must abandon all hope of applying mechanics to such systems, but through the marvelous insights of Clausius, Boltzmann, Maxwell, Gibbs, and Planck, we have learned that the enormous numbers of particles in such systems actually lead to a profound simplicity. In fact, even systems with only a few particles can illustrate this simplicity, and in this experiment you will explore the behavior of a model of a two-dimensional gas, such as can be found in layered semiconductor devices and at metal-semiconductor contacts.
In the second half of the nineteenth century a different way to describe mechanical systems was developed based on the assumption that an isolated system of particles is equally likely to be found in any possible configuration. This can be illustrated with the air in the room. Imagine releasing a special, tagged molecule of nitrogen into the air in the corner of the room. Let's call the molecule "Pat" for specificity. If you want to know where Pat is a moment after release, you would be well served to look very close to where Pat was released. Now seal up the room overnight and come back the next day. Where are you likely to find Pat (assuming he's still in the room)? Imagine dividing the room up into little cubes, each 1 cm on a side. It seems reasonable to assume that you are equally likely to find Pat in any cube in the room (or more generally, that the probability of finding Pat in a particular subvolume v in a room of volume V is simply the ratio p = v/V). This assumption is called the fundamental assumption of statistical mechanics, and we will explore in this experiment some consequences of this assumption and see how well they are born out by experiment. You will see that even if the number of particles isn't spectacularly large, the predictions we obtain from the fundamental assumption are quite good indeed.
You will conduct two experiments involving Squiggle balls placed inside an
arena. In the first experiment, balls will be placed on either
side of a moveable partition inside a rectangular arena, as illustrated in
the figure at the right. Using the fundamental assumption of
statistical mechanics, you will predict how likely the divider
will be found at different distances from the center position, and compare
your predictions with observations.
When a small system, such as a ping-pong ball, is placed in contact with a thermal reservoir, such as provided by a bunch of Squiggle balls bouncing around, the fundamental assumptions predicts that the probability of finding the ping-pong ball in a certain state is proportional to the Boltzmann factor, defined to be exp(-E/kT), where E is the energy of the ping-pong ball, T is the temperature of the thermal reservoir, and k is a constant called Boltzmann's constant. By dividing an area into two equal areas, and making one area higher than the other, you can investigate whether Boltzmann's factor describes the likelihood of finding the ping-pong ball in the high-energy state.
We will assume that the arena is flat and level, so that gravity does not favor one side over the other. Squiggle balls have a battery-powered motor and asymmetric rotor inside, so they spin and tumble "randomly". Unscrew a Squiggle ball to see how the insides work. Then make sure that the arena surface is level.
You will conduct an experiment in which Squiggle balls are placed on either side of a moveable partition inside a rectangular arena, as illustrated in the figure at the right. We will assume that the arena is flat and level, so that gravity does not favor one side over the other. Squiggle balls have a battery-powered motor and asymmetric rotor inside, so they spin and tumble "randomly". Unscrew a Squiggle ball to see how the insides work.
We will imagine dividing up the rectangular area of the arena into squares of side equal to the diameter of the squiggle balls, which we will take to have unit dimension. For simplicity, we will start by considering an arena 2 units high by 6 units wide, and we will put two Squiggle balls on each side. Given the symmetry of this situation, you would expect that the average position of the dividing bar will be in the middle of the arena. You might even expect that this would be the most probable position for the bar, but you would also expect that the bar could be found closer to one end or the other. How can we predict the probability of finding the bar at a given distance from the center?
We will use the fundamental assumption of statistical mechanics to analyze the probabilities. With the bar in the middle, we have two balls on the left that can be positioned in six squares. How many ways can they be arranged? There are 6 choices for where to put the first ball (the blue one, for example), and then 5 remaining choices for the second (red) one, for a total of 30. [If the blue and red balls were fundamentally indistinguishable, we would have to divide this number by 2, because there would be no difference between having the "blue" ball in the first cell and the "red" ball in the second, versus having the "red" ball in the first cell and the "blue" ball in the second cell.] By the same reasoning, there are 30 distinct ways to arrange the two balls on the right. Now, for each arrangement of the two balls on the left, there are 30 ways to arrange the ones on the right; hence, there are 900 distinct arrangements of the balls when the bar is in the middle.
What about when the bar slides to the right one ball diameter. Now the number of possible arrangements on the left has increased to 8 x 7 = 56, and the number on the right has been reduced to 4 x 3 = 12, so that the total number is 672, which is less than the 900 possible when the bar is in the middle position. You can check that the following table describes all the possibilities:
| Position | Arrangements on left | Arrangements on right | Total | Probability |
|---|---|---|---|---|
| 1 | 2 | 90 | 180 | 7% |
| 2 | 12 | 56 | 672 | 26% |
| 3 | 30 | 30 | 900 | 35% |
| 4 | 56 | 12 | 672 | 26% |
| 5 | 90 | 2 | 180 | 7% |
The probability column has been calculated by dividing the number of arrangements for each position of the bar by the total number of arrangements, obtained by summing the total column to get 2604. The distribution is represented in the following figure, which shows that the most probable position is in the middle and that the probability tapers off the farther from the center the bar strays.
There is a useful way to express the number of ways that r flips out of N come up heads, or that r squares out of N are occupied by Squiggle balls. As you saw above, there are N choices for where to put the first ball, N-1 choices for the second ball, and so forth down to N - r + 1 choices for the last ball. A convenient way to express the product of these numbers is using factorials: N! / (N-r)!. This is called the permutations of N objects taken r at a time. If the objects are all the same — if switching any pair of balls does not produce a different arrangement — then this number should be divided by the number of ways you can switch around the r balls, which is r!. This is the usual situation, and so we will use the notation
for the number of combinations of n objects taken r at a time. If you are using Excel to compute this quantity, you can use the COMBIN(n,r) function, which is built in. In Kaleidagraph, you can define this function yourself. Open the library by Macros|Library.... Add the definition
comb(n,r) = n! / ((n-r)! * r!);
and close the library. Now you can use this function in your calculations in Kaleidagraph.
You can confirm that the probabilities shown in the table above are unchanged by using the combinations rather than the permutations. We will stick with the combinations of N objects taken r at a time for the remainder, since this will touch base with Pascal's triangle and any combinatorics that you may have studied before.
Experiment 2: Boltzmann Factor
You will use a second arena and two stop watches for this experiment. The arena has a single divider, which you should place in the middle to make two equal halves. The divider is high enough to allow ping-pong balls and other small spheres to pass underneath, but low enough to block Squiggle balls. By measuring the fraction of time the ping-pong ball spends in the upper half as a function of the height difference, you can test the prediction of the Boltzmann factor.
Updated 10/18/00 by Peter N. Saeta .