Koide Waterfall

http://www.vixra.org/abs/1111.0062

So why is, the name of this prezi, "Koide Waterfall"?

arXiv:1101.5525 Rodejohann and Zhang

• Notice that the triplet of charm, bottom and top quark also fulfill Koide formula

4 sqrt s sqrt d = s + d

http://www.arxiv.org/abs/1111.7232

1978 Harari Haut Weyers

• Try to predict Cabibbo angle
• Find a particular fix of mass values for down and strange quark when up is zero

vixra:1111.0062 A.R

arXiv:1111.7232 A.R.

• Notices that with the minus sign in a square root, also strange, charm and bottom are a Koide compliant triplet

Phys.Lett.B698:152-156,2011

Simple: set the yukawa coupling of the top quark to one, set the coupling of the up quark to zero, look for solutions (with any sign of the roots) doing the sequence

http://www.arxiv.org/abs/1101.5525

Phys.Lett. B78 (1978) 459 http://www.sciencedirect.com/science/article/pii/0370269378904859

top->bottom->charm->strange->up->down

1981 Y Koide

• Uses composite quarks and leptons to try to predict the Cabibbo angle
• Finds a very predictive formula for the mass values... of charged leptons!

Yeah, strange, up, down is the original solution of Harari et al.

• Furthermore, this sign allows to see the quark triplet almost orthogonal to the lepton triplet, so one can be produced from the other.

Phys.Rev.Lett. 47 (1981) 1241-1243 http://dx.doi.org/10.1103/PhysRevLett.47.1241

There are only five solutions

• of these four come in pairs from the equation
• In each pair, one of them has the bottom mass is higher than the top.
• Actually, the unpaired solution is the interesting one

One more thing...

Does it sound useful? Or, at least, amusing?

Please share this presentation!

http://prezi.com/e2hba7tkygvj/koide-waterfall/

105.6583668(38) MeV

0.510998910(13) MeV

Inputs

4197.57589(15) MeV

174.10 GeV

All the basic scales of the Standard Model fermions (well, ask Carl Brannen for neutrinos) are already here!

1359.56428(5) MeV

3.64 GeV

92.274758(3) MeV

The last combination of the

Tevatron, arXiv:1207.1069, is

1776.96894(7) MeV

1.698 GeV

t

b

c

s

u

d

121.95 MeV

173.18 ± 0.94 GeV

173.263947(6) GeV

0 MeV

4.18 ± 0.03 GeV

1.275 ± 0.025 GeV

8.75 MeV

1776.82 ± 0.16 MeV

95 ± 5 MeV

173.07± 0.52 ± 0.72

We find a "tree level" approximation to the final values

Compare with the experimental measures from PDG 2013: agreement within one sigma.

THIS IS THE SCRATCHPAD... NOT PART OF THE PRESENTATION

x > 0 AND ...

( 1 + 9 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2

1) THE LADDER

( 1- 8 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2

x= 8 sqrt(3)

+

( -1 +8 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2

(2 sqrt(3) + x)^2 / (8+x) = 3/2

0

( -1 -9 sqrt(3)+x)^2/(196+2 sqrt(3)+x) = 3/2

From this tree-level:

• The electron seems to get mass with some mechanism that preserves Koide, and produces the ladder we have seen in the first part.
• The up quark does not seem to preserve Koide when getting its (lattice-calculated?) mass.

x=19+5 sqrt(3)

( 5+sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2

-

(-2 +x) ^2 /(8+x) = 3/ 2

-

0, (-1+ sqrt(3))^2, (1+sqrt(3))^2

im

( -3 +sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2

+

(+2 +x ) ^2 / (8+x) = 3/2

x=4

im

( 3- sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2

0

(-2 sqrt(3) +x)^2 / (8+x) = 3/2

( -5-sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2

x=-1+sqrt(3)

-

-

2) The WATERFALL

#!/usr/bin/python

from math import sqrt

class Masa(object):

def __init__(self, valor, origen, origenpos):

self.valor = valor

self.origen = origen

self.origenpos= origenpos

#bueno, menos esto, lo demas estilo fortran :-DDD

series=[[Masa(0,(0,0),[0,0]),Masa(1,(0,0),[0,0])]]

def soluciones(m1,m2):

#return [1]

respuestas=[]

B=m1+m2

for x in (+sqrt(m1),-sqrt(m1)):

for y in (+sqrt(m2),-sqrt(m2)):

A=x+y

D=3*(2*A**2-B)

if - 0.00001 < D < 0.00001 :

D=0

if A > -0.00000001:

respuestas.append((2*A)**2)

if D > 0 :

if 2*A+sqrt(D) > 0:

respuestas.append((2*A+sqrt(D))**2)

if 2*A-sqrt(D) > 0:

respuestas.append((2*A-sqrt(D))**2)

respuestas.sort()

anterior=respuestas[0]+234234354;

validas=[]

for x in respuestas:

if abs(x-anterior)>0.000001: validas.append(x)

anterior=x

if (m1,m2) == (1,0): validas=[validas[0]] #caso especial, para quitar degeneraciom x -- 1/x

return validas

#para ser mas precisos, hay y reducirlas si en algun caso

#difienren en menos de 1E8

for rango in range(2,6):

nuevas=[];

for serie in series:

for i in range(0,len(serie)):

for j in range(0,i):

if [i,j] > serie[len(serie)-1].origenpos: # control degeneracy

for m in soluciones(serie[i].valor,serie[j].valor):

nuevaserie=serie + [Masa(m,(serie[i].valor,serie[j].valor),[i,j])]

nuevas.append(nuevaserie)

series=nuevas

print len(series), len(series[0])

result=[]

for serie in series:

espectro= [ masa.valor for masa in serie]

normespectro=[ x/max(espectro)*174.10 for x in espectro]

normespectro.sort()

normespectro.reverse()

anterior=normespectro[0]+10000

degen=0

for x in normespectro:

if abs(x-anterior)<0.0000001: degen+=1

anterior=x

if degen==0:

result.append([normespectro, [masa.origenpos for masa in serie], [masa.origen for masa in serie] ])

result.sort()

print "no degeneradas", len(result)

for x in result: print x

174.10

765.08965

16

3.64

see also http://www.physicsforums.com/showthread.php?p=3989113

7.4641

1.698

top 19^2+3*25+190 sqrt(3)

bottom 16

charm 4+2 sqrt(3)

strange 4-2 sqrt(3)

up 0

d (4-2 sqrt(3))^2 / 4+2 sqrt(3)

.5359

.12195

• Some perturbation m_s m_d / M could be invoked.

0

0.03647

0.00875