- #281
arivero
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I think that the drawing is at least the second version from De Rujula, but the only one he found in his archive as it is used in some publication. I saw another version during a talk in my hometown, I was in the first course of the undergraduate physics studies, and someone did a series of talks addressed to secondary school teachers. Around 1985-1986 then.
Yes, all the masses are in MeV. What we are doing here is just apply the mass formula [itex]M(a,b)= k (z_a+z_b)^2[/itex]. When one of the charges is constant, say z_q, and the other three sum zero [itex]z_1+z_2+z_3=0[/itex], then we have [itex]k(z_q+ z_i)^2[/itex] is a koide formula if and only if [itex]3 z_q^2 = z_1^2+z^2+ z^3[/itex]. The zero sum rule is the tradicional cosine of the most common version here in the thread. I have used alpha instead of delta for the phase because I am doing more jumps than usual. Koide phase is not only periodic 2 pi /3; it also allows reflections on pi/4 and pi/2. So I am not using the usual phase for the leptons but a reflected one.
Besides checking that I am recovering the usual koide formula, I was interested on the singular points and what kind of particles I get. I went with MeV masses because I was tired of exact roots of two, roots of three etc.
The particles (or pairs of preons) are organised according representations of SU(3)xSU(2). When the SU(2) is a singlet, it means the two particles are from the SU(3), ie they are pairs from the ones I call d,s,b. When the SU(2) is a doublet, one of the particles is a "c" or a "u". For these particles, the charge "z_c" does not come from the cosine but from the rule of the average of squares. I mean, the are the "z_0" of usual koide formula.
In the case pi/4 the charge of "s" coincides also with z_0, so what happens is that we get a massless lepton and that the s particle can also act to produce a pretty exotic koide tuple in the quark sector ds,ss,bs
(More properly, the "antiquark" sector, as a pair such as ds is an antiquark)
I am impressed by the "sneutrino" sector, as I had not anticipated the obvious thing that half of the
scalar neutrinos are massless, independently of the koide phase.
Yes, all the masses are in MeV. What we are doing here is just apply the mass formula [itex]M(a,b)= k (z_a+z_b)^2[/itex]. When one of the charges is constant, say z_q, and the other three sum zero [itex]z_1+z_2+z_3=0[/itex], then we have [itex]k(z_q+ z_i)^2[/itex] is a koide formula if and only if [itex]3 z_q^2 = z_1^2+z^2+ z^3[/itex]. The zero sum rule is the tradicional cosine of the most common version here in the thread. I have used alpha instead of delta for the phase because I am doing more jumps than usual. Koide phase is not only periodic 2 pi /3; it also allows reflections on pi/4 and pi/2. So I am not using the usual phase for the leptons but a reflected one.
Besides checking that I am recovering the usual koide formula, I was interested on the singular points and what kind of particles I get. I went with MeV masses because I was tired of exact roots of two, roots of three etc.
The particles (or pairs of preons) are organised according representations of SU(3)xSU(2). When the SU(2) is a singlet, it means the two particles are from the SU(3), ie they are pairs from the ones I call d,s,b. When the SU(2) is a doublet, one of the particles is a "c" or a "u". For these particles, the charge "z_c" does not come from the cosine but from the rule of the average of squares. I mean, the are the "z_0" of usual koide formula.
In the case pi/4 the charge of "s" coincides also with z_0, so what happens is that we get a massless lepton and that the s particle can also act to produce a pretty exotic koide tuple in the quark sector ds,ss,bs
(More properly, the "antiquark" sector, as a pair such as ds is an antiquark)
I am impressed by the "sneutrino" sector, as I had not anticipated the obvious thing that half of the
scalar neutrinos are massless, independently of the koide phase.
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