Triparitie K(8)*K(8)*K(8) model for proton decay in the Standard
- To: mathgroup at smc.vnet.net
- Subject: [mg71378] Triparitie K(8)*K(8)*K(8) model for proton decay in the Standard
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Thu, 16 Nov 2006 00:53:12 -0500 (EST)
- References: <f5%5h.24836$TV3.13526@newssvr21.news.prodigy.com>
http://en.wikipedia.org/wiki/Proton_decay p \longrightarrow e^{+}+\pi^{0} \longrightarrow e^{+}+2\gamma (Picture:p->e(+)+Pi(0)->e(+)+2*gamma) Some thought on this K(n)3 type model gives the proton decay as to the second secular energy level ( usually a forbidden transition): p(+)-> 7*Pi(0)+e(+) I give seven Pi(0)'s as that's how the energy seems to divide up! The K(8)^3 model also agrees with the proton magnetic moment as: mu_p=N[14/Sqrt[8*Pi]]=2.7926 It seems to make more sense to use K(8) as octonion as the quark model in the proton than a lepton/ quaternon. That model give very different results that the 12by12 did! K(8)*K(8)*K(8) as three quarks that are U(1)*SU(3) like: M24= {{0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0}, { 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0}} Polynomial: Factor[CharacteristicPolynomial[An[24], x]] (-9 + x)(-6+ x)2(-1 + x)7(2 + x)14 So that is 14 at -2 7 at 1 ( detla 3) 2 at 6(delta 8) 1 at 9(delta 11) Scale as paired spin states for a proton: 1836.1089/(28*2)=32.7877 me s = 1836.1089/(28*2) 32.787658928571425` 3*s 98.36297678571427` 8*s 262.3012714285714` 11*s 360.6642482142857` Pi zero mass in me's is 264.137 me by my very old data. That state would be the only proton decay state that gives a particle. p(+)->Pi(0)+?(+) I got an estimate of the lifetime of a proton to the Pi(0) decay using my graph model. What I assumed was that the largest root 9 was a Markov -Binet: f[n_]=(9^n-9^(-n))/(9-1/9) and the state with is 8 units up from the ground state is: 8=f[n] I got the solution that n=1.94083 That no good without a time scale: ( using my old numbers from 1970's book) as a Planck black body vibration time scale h*v=mp/56*c2 v=2*Pi*mp*c2/(56*hbar) Gamma_lifetime=v^n=1.083757655067437*10^45 sec Since the universe is only 1030 sec old or there abouts, we aren't likely to see a proton decay today, ha, ha... That's just a crude estimate to see if I could actually do it. There is also another even less likely decay into a Lamda zero particle at 2183.2me from the top energy state at 2196.7731 me: p(+)=Lambda(0)+e(+) 2.4946053955461785*10^48 sec The middle state +1 has no physical particle state near it. It has a lifetime near 10^34 seconds as well. That's why the proton is so stable. Further work on K(n) matrices suggests the very high energy decay of: p(+)-> Hyperion(-)+e(+)+e(-)+neutrino Of a Delta (-) hyperion ( spin 3/2) near 2401 me as a K(15) graph. This last decay is just idle speculation from the hypergeometery, graph spectra and relative energy levels. Roger Bagula