up and down voltage of brain waves, the pendulum swings of our blood hormone levels, the cyclic procession of our days, months and years and at large scale, our
body’s journey from dust to dust.
The angular deviation theta, 8 from the initial reference direction of a single moving sphere, gets rotated to a new angle theta, 9 8 by a collision with another sphere. It has been shown that the new angle 0 is the previous angle, _ times twice the average distance traveled between collisions called the mean free path, here symbolized by delta, 6, divided by the diameter, D, of the sphere. Algebraically,
0-—0 the new angle is equal to twice the mean free path divided by the diameter
of the spheres times the original directional angle of the sphere’s motion. If we symbolize the time between collisions with tau, 1, after an elapsed time of
experimental observation, tf, we can say that the deviations from the initial direction
of the sphere changes like (a The exponent, t/, represents the time of the
experimental observation divided by the average time between collisions of the spheres, i.e. the time we’ve been watching, ft, is expressed as units of inter-collision interval, t. Of course, the circular deviation in the angle from the initial direction rotates repeatedly around a circle as the number of collisions increase. If a point on a circle marks the angular change resulting from each collision and the system runs long enough, it has been shown that the circle will eventually be completely covered
by points.
An estimate of the entropy, S, being generated by each sphere labeled with some index i, Si, is positive because the recurrent motion is deviating continuously
from the initial direction. It can be computed for each sphere as the /ogarithm of the intercollision time-averaged deviation from the initial direction, S; = “loe(=>) and the T
entropy of the whole n hard sphere system is the sum of the n entropies, which can be expressed as nxS;. If we keep books by registering the points when each sphere’s makes a stop on the top half of the sphere’s circle as 1 and the bottom
half as O (and we must arbitrarily decide between 0 and 1 if it falls exactly on the
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