To reiterate the Koide situation for quarks ...

Recall that there are $6$ basic leptons and $18$ quarks on the hexagon in the tetractys. This factor of $3$, due to color, now appears in the Koide scale for the new quark triplet. So if the charged leptons use a natural scale $\mu_{l} = 1 \equiv 313.8$ MeV, then the $(b,c,s)$ triplet uses a scale $\mu_{q} = 3$ (ie. the proton mass). The phases are $\theta_{l} = 2/9$ and $\theta_{q} = 6/9$. Both triplets, and the neutrino triplets, use the same central $r$ parameter of $\sqrt{2}$. The neutrinos require the phase adjustment of $\pm \pi /12$ and a much smaller scale, which is $\mu_{\nu} = 3.14 \times 10^{-10}$. This determines all quark, lepton and mirror masses, via the Koide formula

$m_k = \mu (1 + \sqrt{2} \cos (\theta + 2 \pi k/3))^2$

for $k = 1,2,3$, and using a simple extension to six dimensions in order to include the other quarks. This extension replaces the $\pi /4$ mixing phase with a $\pi /3$ phase, as expected for a higher dimensional information space.

You are all free to check this yourselves.

8 years ago

And whereas $\sqrt{2}$ comes from $\cos \pi/4$, with the phase $\pi /3$ we get a mixing parameter of $2$. This is close to the $r$ parameter required for the missing quarks, $(t,u,d)$, as noted yesterday.

ReplyDeleteIf this site had voting, I'd vote this post up. This is a milestone synthesis on which further progress will be built.

ReplyDeleteLOL, Mitchell! You could put a link to it over there, just for the laugh of seeing how many minutes it took for it to be deleted.

ReplyDeleteNote also that $2$ times the $1/2$ from $\cos \pi/3$ is the coefficient in the secondary Koide formula for the last three quarks, namely $1$. This is another mixing parameter, notably associated also to $\pi/4$, by tangent.

ReplyDeleteAnd the bosons are Fourier dual to the fermions ...

ReplyDeleteMy doubt with r=2 is that it is the saturation parameter, isn't it? I mean, the one for the extreme n-tuple (0,0,...,0,1).

ReplyDeleteSo?

ReplyDeleteI made a link at the cosmology thread.

ReplyDeleteAh, I see.

ReplyDelete