Periodic Table of the Elements (Problem)
Posted: Tue Dec 17, 2013 11:02 am
Some of the most confusing paragraphs in Nothing But Motion are at the bottom of page 128 and the top of 129, where Larson describes how he derives the Periodic Table. I was trying to write a short paper to explain it, and ran into some problems--namely, it does not actually "add up." So let me explain, and hopefully somebody can spot what I am doing wrong. Here is my interpretation:
Larson then goes to build the Periodic Table, based in this 2E2 relationship... which my RS2 artificial reality did not like, since you end up with E2 = 2E2, or 1=2.
So, have I misinterpreted something here, or is there a math error, or did Larson make a mistake?
Larson defines two units here, D, the one-dimensional displacement units (aka speed displacements, direction reversals, "natural (single) units" or natural unit), and E, the 1D electric rotation. He then defines the relation between them: E = 2D. (I don't know why he uses so many terms for the same concept; it's very confusing.)The possible rotating combinations therefore constitute a series, successive members of which differ by two of the natural one-dimensional units of displacement. Since we will not encounter single units in these atomic structures, it will simplify our calculations if we work with double units rather than the single natural units. We will therefore define the unit of electric displacement in the atomic structures as the equivalent of two natural one-dimensional displacement units.
This just says that we measure atomic weight as "D" (net displacement) and the atomic number as "E" (net electric rotation). Fractional atomic weights are isotopic mass, addressed in Basic Properties of Matter. Since we are not dealing with weights yet, weights can be ignored.On this basis, the position of each element in the series of combinations, as determined by its net total equivalent electric displacement, is its atomic number. For reasons that will be brought out later, half of the unit of atomic number has been taken as the unit of atomic weight.
He now introduces a two-dimensional displacement, the magnetic displacement, as described earlier on the page. We'll call this M. So, M = E2, since the only one-dimensional rotation we have is the electric displacement. So far, so good.At the unit level dimensional differences have no numerical effect; that is, 13 = 12 = 1. But where the rotation extends to greater displacement values a two-dimensional displacement n is equivalent to n2 one-dimensional units.
This is just a simple substitution. Since E = 2D, then: M = E2 = (2D)2 = 4D2; this works fine.If we let n represent the number of units of electric displacement, as defined above, the corresponding number of natural (single) units is 2n, and the natural unit equivalent of a magnetic (two-dimensional) displacement n is 4n2,
Now here's where it falls apart... he correctly states that E = 2D, or D = E/2. He then takes M = 4D2, and attempts to solve for E by substitution: M = 4(E/2)2 = 2E2. Correct me if I'm wrong, but aren't you supposed to "square" first, with the result being M = 4(E/2)2 = 4(E2/4) = E2, NOT 2E2?Inasmuch as we have defined the electric displacement unit as two natural units, it then follows that a magnetic displacement n is equivalent to 2n2 electric displacement units.
Larson then goes to build the Periodic Table, based in this 2E2 relationship... which my RS2 artificial reality did not like, since you end up with E2 = 2E2, or 1=2.
So, have I misinterpreted something here, or is there a math error, or did Larson make a mistake?