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Koide Waterfall

Some advances in the use of Koide formula.
by Alejandro Rivero on 19 August 2013

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Transcript of Koide Waterfall

0.510998910(13) MeV
105.6583668(38) MeV
1776.96894(7) MeV
92.274758(3) MeV
1359.56428(5) MeV
4197.57589(15) MeV
173.263947(6) GeV
1776.82 ± 0.16 MeV
95 ± 5 MeV
1.275 ± 0.025 GeV
4.18 ± 0.03 GeV
173.07± 0.52 ± 0.72
Compare with the experimental measures from PDG 2013: agreement within one sigma.
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
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!
arXiv:1101.5525 Rodejohann and Zhang
Notice that the triplet of charm, bottom and top quark also fulfill Koide formula
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
Furthermore, this sign allows to see the quark triplet almost orthogonal to the lepton triplet, so one can be produced from the other.
The last combination of the
Tevatron, arXiv:1207.1069, is
173.18 ± 0.94 GeV
Does it sound useful? Or, at least, amusing?

Please share this presentation!

http://prezi.com/e2hba7tkygvj/koide-waterfall/
THIS IS THE SCRATCHPAD... NOT PART OF THE PRESENTATION
0, (-1+ sqrt(3))^2, (1+sqrt(3))^2
(2 sqrt(3) + x)^2 / (8+x) = 3/2
(-2 +x) ^2 /(8+x) = 3/ 2
(+2 +x ) ^2 / (8+x) = 3/2
(-2 sqrt(3) +x)^2 / (8+x) = 3/2
0
-
+
0
-
+
x=4
x > 0 AND ...
x= 8 sqrt(3)
( 5+sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2
( -3 +sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2
( 3- sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2
( -5-sqrt(3)+x)^2 / (20+2 sqrt(3)+x) =3/2
1) THE LADDER
2) The WATERFALL
im
im
-
x=19+5 sqrt(3)
x=-1+sqrt(3)
-
( 1 + 9 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2
( 1- 8 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2
( -1 +8 sqrt(3) +x)^2/(196+2 sqrt(3)+x) = 3/2
( -1 -9 sqrt(3)+x)^2/(196+2 sqrt(3)+x) = 3/2
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)
765.08965
174.10
3.64
7.4641
1.698
.5359
.12195
0
16
0
0.03647
0.00875
One more thing...
So why is, the name of this prezi, "Koide Waterfall"?
top->bottom->charm->strange->up->down
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
Yeah, strange, up, down is the original solution of Harari et al.
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
174.10 GeV
3.64 GeV
1.698 GeV
121.95 MeV
0 MeV
8.75 MeV
Inputs
t
b
c
s
u
d

mu, pi
tau
electron
We find a "tree level" approximation to the final values
All the basic scales of the Standard Model fermions (well, ask Carl Brannen for neutrinos) are already here!
Some perturbation m_s m_d / M could be invoked.
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.

http://www.arxiv.org/abs/1101.5525
http://www.arxiv.org/abs/1111.7232
http://www.vixra.org/abs/1111.0062
Phys.Lett. B78 (1978) 459
http://www.sciencedirect.com/science/article/pii/0370269378904859
Phys.Rev.Lett. 47 (1981) 1241-1243
http://dx.doi.org/10.1103/PhysRevLett.47.1241
Phys.Lett.B698:152-156,2011
see also http://www.physicsforums.com/showthread.php?p=3989113
#!/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
4 sqrt s sqrt d = s + d
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