Phi From Trigonometry (Not The Usual Way).

[tetrahedron]

So I've found a somewhat convoluted relationship between the relative partial charges of Standard Model Fermions and the Golden Ratio, as preposterous as that might sound at first hearing. A couple of years back I had discovered that the relative partial charges in the Standard Model Fermions (plus or minus 0/3, 1/3, 2/3, and 3/3) could be represented as normals to a reference plane from the vertices of a cube embedded in the plane through a body diagonal through the cube. A body diagonal goes between diametrically opposed vertices as well as the center of the cube- it is the longest straight-line length within the cube volume. So embedded, two of the vertices are already on the body diagonal line, and so their distance to the plane is 0.

 

These then represent the matter and and antimatter neutrinos, which have a charge (ideally) of + or - 0/3. If we take any of the OTHER six vertices and rotate one to an angle of arctan(squareroot 27), then the sines of the angles subtended by these vertices against the reference plane will fall into a relative length of 1, 2, and 3, thus giving the relative charges of the down, antidown, up, antiup quarks as well as the electron and positron. The SQUARES of the sines are all multiples of 1/28, which relates to the original choice of 27 under the square root sign at the top of the post. The denominator is always 1 more than the original number (though this has nothing to do with the charge relations using the trigonometry). Arctan(squareroot27) isn't the only angle that will work. Arctan(squareroot27) PLUS OR MINUS ANY MULTIPLE OF 30 degrees will work as well, though half will use the sines and half the cosines, alternatively.

 

I've been playing with other initial numbers but using the same procedure looking for additional patterns now for years without success. Today I finally found one, though it relates primarily to pure mathematics and not to physics. 27 is 3 cubed. I wondered this evening what would happen if I used 3 to the 4th power (81) instead. So taking the square root we get 9, obviously. Then we take the arctan, and from that angle the sine, then the sine squared. The inverse of this value is 1.0123456789... We then subtract 1, giving .0123456789...and take the square root, which is .1111111111..., or 1/9. The tan(arcsin  1/9) is .1118033989...... We multiply by 10, then add 0.5. This gives the Golden Ratio, 1.618033989.... The relationship between these systems is a bit of a stretch, but it IS there. It makes me wonder what would happen if we ran the original system through this latest procedure that gave the Golden Ratio. I'll work on that tonight.

 

Jess Tauber

Edited April 13, 2019 by pascal

[tetrahedron]

It turns out that .1118033989 is exactly half one tenth the square root of 5, so it relates to one of the standard formulae for generating the Golden ratio, that is (squareroot5+1)/2 or (squareroot5-1)/2 which is the inverse, given that the inverse of 1.618033989 is 0.618033989.

 

Jess Tauber

Edited April 12, 2019 by pascal