Relative charges of the Standard Model fermions modeled on a cube, the Koide formula, and the Golden Ratio.

[pascal2]

Hi folks. Been a long time since I posted here. 

Some while back I came up with a geometrical model able to deal with the relative charges of ALL the Standard Model fermions with a single figure- a cube embedded in a reference plane along a body diagonal (a diagonal that runs through the cube's center and two diametrically opposed vertices). If instead of thinking of the charges in terms of thirds (the usual way), we think of them as whole numbers, then the neutrinos have charges of +0 or -0, the lighter quarks +1 or -1, the heavier quarks +2 or -2, and the electrons +3 or -3, in sets of alternating polarity, so +0, -1, +2, -3 for normal matter, and -0, +1, -2, +3 for antimatter. Now we use the vertices of the cube to mark each of these relative charges. If we use the body diagonal as a rotational axis, we can rotate the cube until the sines OR cosines (for any particular angle) fall into these relative absolute values. The default angle is arctan(sqrt27), or 79.10660535....  Two of the vertices are IN the reference plane along the body diagonal, and these represent the two neutrinos. Then the three other vertices closest to these neutrino vertices represent triplets that will have the lengths 1, 2, and 3x the minimum nonzero value. 

But one can go beyond the default angle. You can add or subtract ANY multiple of 30 degrees and you still get the correct charge magnitude sequences. And since there are 12 such angles around a circle you get exactly 12 variants on this theme- interesting because there are 12 Standard Model fermions for each of matter and antimatter. A coincidence? Somehow I don't think so. For the default angle, you have to use the SINES to get the triplets- but for that angle PLUS (or MINUS) 30 degrees, you switch to the COSINES. The next increment, plus or minus, of 30 degrees brings you back to sines, and so on.

When you sum the charge triplets (which occupy the vertices of an equilateral triangle that is normal to the reference plane) that sum is always ZERO.

And, when you SQUARE the sines or cosines of the various utilized angles, you always get multiples of 1/28. This is no accident either- I worked out that the denominator of this fraction is always ONE MORE than the value under the square root sign in the default angle (arctan(sqrt27).  Anyway, for any set of sines or cosines for any particular angle in this system, the sum of the sines and cosines squared is always exactly 1. Curiously the sums of the numerators of these fractions across triplets of squared sines or cosines is always 42.

So now, having dealt with the relative charges, I'll now move on to the MASSES. The Koide formula relates the different masses along the electron, muon, and tauon sets perpendicular to the charges in tables of these properties. See https://en.wikipedia.org/wiki/Koide_formula.  While it has been challenged many times over the years, it still has some interesting aspects to it.

I mentioned above the triplets of squares of sines and cosines that sum to 42 re the charge model. The particular ones that come from the sines (or cosines) relevant to the charges are always 27, 3, and 12.  Those that are NOT used for the charges always have numerators of the squares that are 1, 25, and 16. Notice that 1, 25, and 16 are squares of 1, 5 and 4. If you look at the WIki page on the Koide formula you'll see a square figure in the upper right that is supposed to be a representation of the square roots of the masses of the electron, muon, and tauon. I measured these and found that the edges of these component squares are (relative to the smallest, representing the square root of the electron mass), 1, 5, and 4, just as the square roots of the numerators of the squares of the sines or cosines not used for the charge series in the cube-in-plane model.

I divided the mass of the tauon by 16 and it comes out to a bit larger than 111. If 1776 MeV/C^2 exactly, then you get 111 exactly. I also took the value of the mass of the muon and divided that by 25, and if you take this as 105 exactly, the result is 4.2 exactly. Then, if you multiply 4.2 by 10, and divide 111 by 42 you get almost exactly the square of the Golden Ratio, for some reason.

I'd done work some years back relating to the decimalized expansions of terms in straight-line relations in the Pascal Triangle (that is, the edge-parallel diagonals, the columns, the shallow diagonals, and the rows). Ratios between these decimalized expansions are, respectively, 1/9 (or, 1/(5+sqrt16)), 1/(5+sqrt15), and 1/(5+sqrt35) and 1/11. (or 1/(5+sqrt36). Note that there is a difference of exactly 1 between the terms under the square roots in the denominators between the pairs of ratios for the edge-parallel diagonals and the columns, on the one hand, and those for the shallow diagonals and the rows. A pair of related pairs.

Now, we know that you can get the Golden ratio by summing terms along shallow diagonals, which generates the Fibonacci numbers, and then taking the limit ratio for pairs of contiguous Fib numbers. If you take the reciprocal of the Golden Ratio, you get that number minus one, exactly- that is 1.618033989... versus 0.618033989...  When you square the Golden Ratio, you instead ADD one, so 2.618033989.

I found that for the values of the the ratios of consecutive columns and shallow diagonals in the Pascal Triangle, if you take THEIR reciprocals, you don't subtract or add 1 to the former. Instead you have to add or subtract 10x the original nonreciprocal value. 

I've wondered whether this has any relation, in what I wrote above about the masses of the Standard Model fermions, and the ratios of their reduced values (that is, divided by 16 for the Tauon, and 25 for the Muon), in having to multiply the latter by 10 when taking the ratio that is so close to the square of the Golden Ratio. 

Anyway, that's about it.

Jess Tauber