Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

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




  • Prev by Date: Re: FindInstance returning incorrect results
  • Next by Date: Re: Re: (Mathematica wish list)
  • Previous by thread: Re: Non linear system solving
  • Next by thread: Wolfram Workbench 1.0 now available