That Nefarious Word, “Somehow”

Too many of us are willing to abandon, without a fight, the analytical ability of the human brain, and to say that unrealistic effects are “somehow” possible. If your philosophical outlook is contaminated by capitulation to “somehow,” it is time to reconsider. There are many examples to illustrate this thesis; three of my favorites — and I trust that they are sufficient to make the point — are planetary motion, the kinetic theory of heat, and radioactive decay.

– – – – – – – – – –

In ancient times, life was a succession of miracles (and even today, sorry to say, a large proportion of people believe that we are frequently assaulted by hostile visitors from outer space via Unidentified Flying Objects). Looking to the heavens, one saw that the earth was obviously the center of the universe, with the sun, moon, and “fixed” stars rotating, with predictable regularity, about the earth. However, there was a group of stars, the “wanderers” or “planets,” that somehow had a kind of peculiar motion relative to the earth. 

To illustrate how the Ancients viewed the situation, consider the various parts of Fig. 1. In (a), an observer is at point P on the surface of the Earth. (To minimize distractions, the observer is near the Equator, and Mars is the only other planet.) Since the observer is on the side away from the Sun, the time (on the observer’s watch) is around 12 midnight. 

Fig.11-1

Fig. 1- How the Earth-centered Universe illusion is created. S = Sun; E = Earth; P = Observer on surface of Earth; M = Mars; B = a relatively distant star, such as Betelgeuse. (a)Relative to the Observer, it is around 12 midnight. (b)Twelve hours later, it is around 12 noon. (c)The Universe of (b), relative to the Observer’s original position in (a). (d)One year later than the Universe of (a). Mars has “wandered off.”

By coincidence, Mars is approximately overhead, as shown. Also, some relatively distant star, labeled B (perhaps Betelgeuse), is seen overhead. 

Now move ahead 12 hours, to Fig. 1(b). You and I know that the Earth rotates 180º in 12 hours, so observer P is now facing the sun, as shown. Mars and star B have not moved appreciably. 

But this is a very incomplete depiction of observer P’s thoughts. He (let’s assume that it is a male observer) is not aware that the Earth is rotating; in his judgment, the Earth is where it was in Fig. 1(a); in his judgment, the Universe is described in Fig. 1(c). He has seen Mars and star B all rotate around the Earth. 

Next, move ahead to midnight, one year later, as in Fig. 1(d). Star B is exactly where it was a year ago, but Mars has wandered off; careful observations reveal that Mars undergoes some kind of oscillation, with a period corresponding to 687 Earth days. 

It seems to be quite simple, and we wonder why the Ancients had it all wrong. After all, they were just as intelligent as we are today. Well, it is not simple at all. I omitted an important ingredient: the annual rotation of the Earth around the Sun. Relative to the observer at P, the motion of Mars becomes a mixture of the 365-day rotation of Earth and the 687-day rotation of Mars. This kind of peculiar complication is also true for the other planets, of course. What was the explanation? “This is the way the Deity set it up, so go about your business, and don’t ask too many questions.”

With humankind constantly embroiled in religious wars, it is easier to appreciate that belief in the Deity is not only distracting, but it can be deadly. 

The illusion that the earth was the center was eventually explained, of course, by the heliocentric model of Nicolaus Copernicus (1473-1543). This revealed, in a beautifully simple way, that the earth and planets were orbiting the sun.

– – – – – – – – – –

My second example deals with the mysterious fluid that somehow explained heat. It was thought that, when two rough surfaces slide against each other, friction squeezes out, or generates, this “heat fluid,” which manifests itself as a rise in temperature. 

The correct theory states that the particles of matter are in random motion, and when we rub rough surfaces together, we accelerate the motion (since acceleration equals force divided by mass). In other words, the temperature of each particle is correlated with its kinetic energy of motion. This explanation is difficult to swallow, but one can actually see the incessant random motion of minuscule particles — Brownian movement — with the aid of sufficient magnification using a microscope. This discovery was from a botanist looking at pollen grains [Robert Brown (1773-1858)]. Here, again, a beautifully simple explanation eventually wiped away centuries of “heat fluid.”

With regard to the true nature of “heat,” there were plenty of hints, such as: The spread of a noxious odor from one end of a room to the other requires that molecules rapidly move through the air. But Brownian movement finally revealed that molecules are in constant motion. The motion is perfectly elastic; no outside agent has to supply energy to keep a can full of real particles bouncing around at constant temperature. As the molecules collide, individual speeds momentarily increase and decrease, but the average kinetic energy of each particle (that is, mv2/2) remains constant. 

Figure 2 is a plot of average velocities versus mass if the above can contains a mixture of electrons, protons, and water molecules. These are freely-moving “particles” in the sense that they can travel a relatively large distance before striking another particle. Interactions due to electric charge are ignored. At “room” temperature, 293ºK = 20ºC = 68ºF, the average electron velocity is 115,000 m/s, while that of the heavier photons is “only” 2694 m/s, and that of water molecules is 635 m/s. Momentum is conserved: mass X velocitybefore a collision is equal to mass X velocity after, with the equation satisfied for each of the three directions: left-right, front-back, and up-down.

Fig.11-2

Fig. 2- Average velocity, meters/second, of a freely-moving particle at room temperature, 293ºK = 20ºC = 68ºF, versus the mass of the particle, kilograms. It is assumed that each particle can travel an appreciable distance before striking another particle; interactions due to electric charges are ignored. The plot is a straight line on this log-log paper. The values for electron, proton, and water-molecule “particles” are indicated by dots.

– – – – – – – – – –

Finally, let us consider radioactive decay. For example, a uranium U238 nucleus contains 92 protons and 146 neutrons, for a total of 238 mass particles. All of the residents of the nucleus, whether they are protons or neutrons, are called nucleons. The “cross-section” through a U238 nucleus is schematically shown in Fig. 3. The element has a half-life of 4.51 billion years; that is, starting with a pure sample of U238, half of it will undergo spontaneous fission (it will violently fall apart) in 4.51 X 109 years (which can also be expressed as 109.654 years.)

The diameter of a nucleus is, of course, much smaller than that of an atom. The uranium nucleus has a diameter of 0.000136 angstrom = 13.6 femtometers = 13.6 fm. (The diameter of a typical atom is 1 angstrom; the number of fm in 1 meter is 1 followed by 15 zeros.) The nucleus is bound together by the strong (nuclear interaction) force. This is different from gravitational and electromagnetic forces, and it overwhelms the repulsion between like charges (positive protons) provided the distance is less than 1.4 fm. This latter value is the range over which the strong force operates; it rapidly decreases to zero beyond this radius. (Needless to say, we have no idea as to what the strong force is, any more than we know what gravitational, magnetic, or electrostatic forces are, but the strong force solved the mystery of what holds the nucleus together if all of those protons are repelling each other.) 

Despite the difficulties we have in observing these approximate spheres of femtometer diameters, the picture that emerges is the following: The protons and neutrons are in rapid motion, interacting little with each other. The movements are approximately independent because of the short range of the strong force. Electrostatic repulsion is a relatively weak contribution, and temperature is not a factor in radioactive decay. (The goings-on in a nucleus are shielded from the outside jostling between atoms, which is the basis of “temperature” to us.) 

For the U238 nucleus, Fig. 3 suggests the following: Its volume is given by
1300 cubic femtometers, while the volume of each strong force’s sphere of influence, or SOI, is 11.49 cubic femtometers.

According to the volume ratio, we can model the nucleus as a sphere divided into 115.4 SOIs. In other words, visualize that the uranium nucleus consists of 115.4 different regions, each having a diameter of around 2.8 femtometers (or volume of 11.49 cubic femtometers). Since there is a total of 238 nucleons, but only 115.4 SOIs, many of these SOIs will contain two or even three nucleons at a given time.

Fig.11-3

Fig. 3- “Cross-section” through a uranium U238 nucleus. There are 92 protons and 146 neutrons for a total of 238 nucleons. The diameter is 13.6 X 10-15 m = 13.6 femtometers = 13.6 fm. Also shown is the “sphere of influence,” or SOI, of the strong force, which rapidly decreases to zero beyond a radius of 1.4 fm.

Although only half of the nuclei do so, it is convenient in what follows to refer to the nucleus that fissions in exactly 4.51 billion years. 

What is the cause-and-effect that leads a U238 nucleus to disintegrate in 4.51 billion years? This question is worthy of a very strong somehow. However, since this essay is not designed to be a mystery story, I hereby reveal the approximate answer: There are 1 followed by 52 zeros ways in which the uranium nucleons can distribute themselves, but in 4.51 billion years they can only occupy 1 followed by 40 zeros of these positions. Dividing (that is, subtracting the number of zeros), this implies that there are 1 followed by 12 zeros (or one trillion) unstable configurations. When the nucleons momentarily fall into one of these unstable modes, the nucleus explodes. And, to repeat, this occurs once in 4.51 billion years!

The calculations are, of necessity, greatly simplified. Because of symmetries, the effective number of possible orientations is somewhat less than indicated. Also, current nuclear theory says that the protons and neutrons tend to form concentric “shells”, similar to the electron shells that surround the nucleus, so there is not a completely chaotic mixture of protons and neutrons. In what follows, however, I will ignore the effects of symmetries and shells. The final conclusion, that there is a huge number of combinations, remains valid. (Here it is a pleasure to be in the politically-correct arena, where a few trillion more or less make very little difference.) 

The reasoning behind the somehow explanation follows: Because of their movements, the 92 protons and 146 neutrons occupy a different distribution of SOIs from moment to moment. Many of these distributions are unstable: most notably, if approximately 46 protons and 73 neutrons accumulate near one “side” of the nucleus, while the remaining half accumulate near the opposite side, this encourages a dumbbell-like shape. The nucleus then splits apart near the middle of the dumbbell, and the two daughter nuclei fly apart. The daughters may be unstable, and not every unstable nucleus falls apart via the dumbbell route, but the general explanation is the same: for radioactive nuclei, an unstable configuration of nucleons eventually occurs.

A few simple calculations show how the U238 nucleus can fall apart after 4.5 billion years of togetherness. We start out with deceptively small numbers: Recall that the U238 nucleus contains 92 protons, 146 neutrons, and 115.4 SOIs. First, consider the 92 protons. In how many ways can they distribute themselves over the 115.4 SOIs? Because of the mutual repulsion between protons, it is reasonable to assume that there will be a maximum of one per SOI. Therefore, if we line up the SOIs from left to right, the distributions can extend from

Fig.11-4

where P = proton, 0 = no proton. The top row represents an extreme condition in which the first 92 SOIs contain a proton, while the remaining 23.4 SOIs do not contain a proton. The bottom row represents the extreme in which the last 92 SOIs contain a proton, and so forth. Of course, any in-between distribution is also valid. 

How many combinations are possible? Any elementary algebra book should give the number of different combinations of n elements taken r at a time. Here we have n = 115.4 SOIs and r = 92 protons. We get a huge number of combinations: 1 followed by 24 zeros.

Next, consider the 146 neutrons. In how many ways can they distribute themselves over the 115.4 SOIs? Despite the strong force, neutrons (and protons) are kept apart by their incessant and rapid motion. It is reasonable to assume that, at a given instant, there will be one neutron per SOI, plus 30.6 wandering leftovers. Now we have n = 115.4 SOIs and r = 30.6. The number of combinations that the 30.6 neutrons can form is 1 followed by 28 zeros. 

Finally, we end up with a huge monster number of zeros: Nucleus “togetherness” is one in which the family members largely ignore each other. We can assume that the proton and neutron distributions are independent. If that is the case, each of the proton distributions can be combined with all of the neutron distributions, and vice versa. The grand total number of proton and neutron combinations is the product of the individual number of combinations, or 1 followed by 24 + 28 = 52 zeros. 

It turns out that the U238 nucleus cannot even come close to this many combinations in 4.51 billion years: How many seconds are there in 4.5 billion years? 1 followed by 17 zeros. How many different combinations can occur in the nucleus each second? Let us go from the sublime to the ridiculous: Assume, although we know that it is unrealistic, that a nucleon can move from one SOI to an adjacent SOI, a center-to-center distance of 2.8 fm, at the speed of light. The time taken to traverse this distance is the reciprocal of 1 followed by 23 zeros. In 4.51 billion years, therefore, 1 followed by 17 + 23 = 40 zeros is the number of movements — not even close to the total number of possible combinations, 1 followed by 52 zeros. 

The above implies that there are many SOI combinations that probably result in spontaneous fission. The main point of the above exercise is this: It also implies that the fields, positions, and momentums of the nucleons today determine how and when the nucleus will fission 4.51 billion years from now! Perhaps 1 followed by 20 zeros computer hackers, working for 1 followed by 10 zeros years, can come up with the numerical answers that will convert these probablies into certainties.

The message from the above examples is that one should seek simple conjectures before surrendering tosomehow philosophy.

Radioactive decay also addresses an aspect of the “uncertainty principle” that some physicists call forth to justify a weird  conclusion: That it is fundamentally impossible to elucidate all of the present motions of a system of subatomic particles in order to predict future behavior. But if photons, electrons, protons, and neutrons all look like diffuse wiggles, precise position and momentum become uncertain. The viewpoint espoused in the present essay is that all of this uncertainty is merely a reflection of our ignorance. “Particles” such as electrons are there, all right, but they are tiny wave packets, not compact baseballs flying around other (nuclear) baseballs.

A classical physicist, given all of the fields, positions, and momentums at the present time, can calculate (in principle) an electron’s position and momentum at any future time. But because the electron is a diffuse field of charge associated with a wave packet, in calculating its “future whereabouts,” an approximate answer, rather than no answer at all, is to be expected.

 

Appendix

The average velocity of a freely-moving particle, according to the kinetic theory of heat, is
v = (3kT/m)1/2,
where v = velocity, meters/second,
k = Boltzmann’s constant = 1.381 X 10-23 J/K,
T = absolute temperature, ºK,
m = mass of the particle, kilograms.
If the “particle” is an electron at room temperature, for example, we have
v = (3 X 1.381 X 10-23 X 293/9.109 X 10-31)1/2 = 115,400 m/s.

With regard to radioactive decay:

The number of different combinations C(n,r) of n elements taken r at a time is
C(n,r) = n!/[r!(n – r)!].

For the uranium nucleus, we have n = 115.4 SOIs and r = 92 protons. Therefore, the protons in the U238nucleus can form the following number of combinations:
CP = 115.4!/(92!23.4!) = 10189.4/(10142.11023.0) = 1024.3. [In evaluating factorials, one can use Stirling’s formula:
x! = xxexp(-x)(2
πx)1/2, or log(x!) = (x + ½)log x – 0.43429x + 0.3991.] 

Next, consider the 146 neutrons. Now we have n = 115.4 SOIs and r = 30.6. The number of combinations that the 30.6 neutrons can form is
CN = 115.4!/(30.6!84.8!) = 10189.4/(1033.210128.2) = 1027.9. The grand total number of proton and neutron combinations is 1024.3 X 1027.9 = 1052.2.
The number of seconds in 4.51 billion years is
60 X 60 X 24 X 365 X 4.51 X 109 = 1017.15. The time taken for a nucleon to move from one SOI to an adjacent SOI, a center-to-center distance of 2.8 fm, at the speed of light, is
2.8 X 10-15/(3 X 108) = 10-23.03 second. In 4.51 billion years, therefore, there would be
1017.15/10-23.03 = 1040.2 movements.

Advertisements

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s