Yesterday, Mitchell mentioned a Higgs mass formula by Dharwadker et al, based on the four colour theorem. This formula correctly predicted the Higgs mass. It is

$m_{H} = \frac{1}{2} (m_{W^{+}} + m_{W^{-}} + m_{Z})$

I'm not yet sure what to make of this, but since it must come from the tetractys diagonal, it suggests something like a degenerate boson Koide triplet $(W^{+}, W^{-}, Z)$. On taking the twisted Fourier dual, this is a $(e^{+}, e^{-}, \nu_{R})$ triplet. Note that it includes no generation multiplicity, but it seems to be about quark lepton mixing. If we include mirror states, we could interpret this triplet as $(e^{+}, e^{-}, \gamma)$, a basic annihilation triplet.

Recall that there are three ways to obtain each of these bosons via composition of $Z$ color states. Thus the boson triplet comes from three copies of the $Z$ color triplet.

6 years ago

Using the mass squares, there was a calculation of the Weinberg angle by Hans de Vries that I retorted to produce a mass of 123 GeV.

ReplyDeleteBut it is even simpler for a mass of about 125 GeV, then the sum of mass squares of W, Z and H -lacking a better letter- is equal to the square of 174 GeV. Reading it from the coupling constants in the Lagrangian, and depending of what normalisation you use for the Higgs quartic coupling lambda, it could be as telling as

g^2+(1/2)g'^2+lambda =1

Where g and g' are as usual the SU(2) and U(1) couplings.

Oh my, yes, Alejandro! The top quark ...

ReplyDeleteI am beginning to be quite convinced by all these coincidences ... and not particularly by the $3$ sigma evidence itself, lol.