All the lepton masses from G, pi, e

[CarlB]

from post #342:
it allows 4 generations if you count m=0

If you'll look carefully at the formula, you will find that putting 4 into the equation gives you the same masses as if you'd put 1. This is because trig functions repeat every 2 pi.

I'm preparing a submission for Phys Math Central on the subject of Koide's mass formulas for the hadrons. I'm supposed to have something ready by January 10th. The latest cut is here:
http://www.brannenworks.com/koidehadrons.pdf

Carl

 

[granpa]

I was using this formula:

a mass of zero can be added to numerator and denominator without changing anything.

 

[arivero]

I
I'm preparing a submission for Phys Math Central on the subject of Koide's mass formulas for the hadrons. I'm supposed to have something ready by January 10th. The latest cut is here:
http://www.brannenworks.com/koidehadrons.pdf

Carl

Hi Carl,

I would use some more group notation. Instead of your 6x6=36, I'd vote by 6 x bar6 = sum of irreps.

Second, besides to the "extraordinary" case you put, I'd include more of the "Ordinary" case, meaning the original argument of Koide about his formula for composite particles, along the lines of the original PhysRev.

 

[granpa]

there are obviously an infinite number of formulas relating the masses of the electron muon and tauon. the beauty of the koide formula is its simplicity and elegance. especially the fact that Q=2/3 which is exactly halfway between the upper and lower limits of 1 and 1/3.

as wikipedia puts it:

Not only is this result odd in that three apparently random numbers should give a simple fraction, but also that Q is exactly halfway between the two extremes of 1/3 and 1.

the formula involving cos given above seems to me to lack this elegance. it should be obvious that anyone can always find a formula involving cos that will pass through any 3 points quite easily.

 

[CarlB]

Hi Carl,

I would use some more group notation. Instead of your 6x6=36, I'd vote by 6 x bar6 = sum of irreps.

Hmmm. The only place I've got 36 in the thing right now is the calculation for the square of a set of MUBs. In terms of color, this would be
(3+\bar{3})\times (3+\bar{3})
i.e. (R+G+B+/R+/G+/B)^2,
and I guess the symmetry would be, (hems and haws whilst trying to recall Georgi), uh, let's see, maybe
(3+\bar{3})\times (3+\bar{3})<br /> = 8+\bar{8}+6+\bar{6}+3+\bar{3}+1+\bar{1}?

Yes, as usual, your comment is brilliant and I will incorporate it.

In a certain sense, what I'm doing is taking the
3\times \bar{3} = 8+1
definition of gluons and claiming that I can compute in qubits with it using 3x3 matrices. [If you can read this, then PhysicsForums let me edit out a silly comment about 6+3-bar.]

Because both can be faithfully represented by 3x3 complex matrix addition and multiplication. From the symmetry point of view, it's a matter of selecting a certain ratio of the coupling constants.

Second, besides to the "extraordinary" case you put, I'd include more of the "Ordinary" case, meaning the original argument of Koide about his formula for composite particles, along the lines of the original PhysRev.

You mean the 1982 article I suppose. I guess I'll have to hike over to the university and read the article. Somehow I don't think I've ever downloaded it. One of the things I'm trying to do is to avoid talking about the preon model. There are two reasons for this. First, it doesn't apply to hadrons, and second, it gets into hot water with Coleman Mandula or (worse) spin statistics.

I think I should be ready to release a first cut in a few days. It still doesn't have "results" and the abstract is obsolete. But I've redone the results section to be make an argument that hangs together better.

 

[CarlB]

Okay, I've got this thing past the first cut. It is time to ask for advice and for people to point out obvious problems etc.

http://www.brannenworks.com/koidehadrons.pdf

 

[CarlB]

The most recent arXiv included a hep-th article that I thought was pretty cute.

Symmetries of Nonrelativistic Phase Space and the Structure of Quark-Lepton Generation

Piotr Zenczykowski.

According to the Hamiltonian formalism, nonrelativistic phase space may be considered as an arena of physics, with momentum and position treated as independent variables. Invariance of x^2+p^2 constitutes then a natural generalization of ordinary rotational invariance. We consider Dirac-like linearization of this form, with position and momentum satisfying standard commutation relations. This leads to the identification of a quantum-level structure from which some phase space properties might emerge. Genuine rotations and reflections in phase space are tied to the existence of new quantum numbers, unrelated to ordinary 3D space. Their properties allow their identification with the internal quantum numbers characterising the structure of a single quark-lepton generation in the Standard Model. In particular, the algebraic structure of the Harari-Shupe preon model of fundamental particles is reproduced exactly and without invoking any subparticles. Analysis of the Clifford algebra of nonrelativistic phase space singles out an element which might be associated with the concept of lepton mass. This element is transformed into a corresponding element for a single coloured quark, leading to a generalization of the concept of mass and a different starting point for the discussion of quark unobservability.
http://arxiv.org/abs/0901.2896

 

[arivero]

Let Me be the mass of the electron, Mp be the mass of the proton, and Mn be the mass of the neutron. Then observe that

Mn/Me - Mp/Me is approximately equal to ln(4*pi) = 2.5310...

Of course, it is really the quotient between the electromagnetic isospin breaking, (Mn-Mp), and the mass of the electron.

ellipsoid of volume

4pi(4pi-1/pi)(4pi-2/pi) = 1836.15...

Which is approximately equal to Mp/Me (a very old result)

Hmm not sure how old is this "ellipsoid". A really old approximation to Mp/Me as 6 pi^5 has been discussed in the middle of this thread.

I hesitate to put in measured values since Mp/Me has changed several times. At one time the ellipsoid volume was within one standard deviation of the accepted CODATA value, but the accepted value changed. I simply rest my case with the mathematical formulae.

A pseudoconvention in this thread is to put both the exact quotient between left and right sides of the equality AND the one-sigma error of this quotient. In this way it is clear how much we can expect the fit to change.
A tradition in this thread is to put both the exact quotient between left and right sides of the equality AND the one-sigma error of this quotient. In this way it is clear how much we can expect the fit to change.

In any case, according current pdg data:

mn − mp = 1.2933317 ± 0.0000005 MeV
Me= 0.510998910 ± 0.000000013 MeV

thus
(mn-mp)/me=2.530987199 ± .000001 MeV
and
(mn-mp)/me :: ln(4*pi) = 0.999 985 36 ± 0.000 000 4

As Hans says, the important point is that the agreement is of 99.998 5 %. The one-sigma adjustment is not important because the explerimental precision is a lot better. But this thread stresses that reaching the empirical bar of 99.99% in a simple formula in not easy.

 

[CarlB]

Walter Smilga has rewritten his justification of Wyler's approximation of the fine structure constant in a nice clean article:

Probing the mathematical nature of the photon field
Walter Smilga

The mathematical content of the interaction term of quantum electrodynamics is examined under the following assumption: It is presumed that the apparent degrees-of-freedom of the photon field reflect the kinematical degrees-of-freedom of the two-particle state space of massive fermions, rather than independent degrees-of-freedom of the photon field. This assumption is verified by reproducing the numerical value of the fine-structure constant.
http://arxiv.org/abs/0901.4917

Of course I see everything through the lens of what I'm doing. So I see the above as evidence that in understanding the color force, one should do something similar with gluons. My recent paper analyzes the mesons in terms of what happens to the quarks, pretty much ignoring the gluons. And it also finds that things turn out simpler than expected.

 

[legna777]

http://foros.astroseti.org/viewtopic.php?t=4529&postdays=0&postorder=asc&start=180

Entonces Legna, (perdona mi ignorancia) ¿qué significa la cifra resultante de 522?



saludos

Existen, como sabes, 5 grupos excepcionales ( a veces los llamo especiales ) de Lie y son:

\[<br /> E8,E7,E6,F4,G2\]


\[<br /> dim(E8)=248\]<br />

\[<br /> dim(E7)=133\]<br />

\[<br /> dim(E6)=78\]<br />

\[<br /> dim(F4)=52\]<br />

\[<br /> dim(G2)=14\]<br />

En el grupo excepcional \[<br /> E8\] están contenidos los otros 4 grupos excepcionales de Lie.
Este grupo \[<br /> E8\] tiene 240 raíces no nulas, número que es el producto
de los 5 primeros términos de la sucesión de Fibonacci.

\[<br /> 240=1\times2\times3\times5\times8\]

\[<br /> dim(E8)+dim(E7)+dim(E6)+dim(F4)+dim(G2)=\Omega\]



\[<br /> \Omega+(8-5-3-2-1)=522\]

Saludos

La tabla de caracteres del grupo \[<br /> E8\] se compone de una matriz de \[<br /> 453060\times453060\]

http://gaussianos.com/category/noticias/page/3/

\[<br /> \frac{\pi^{5}}{120}\] es el factor de volumen de una hiperesfera de 10 dimensiones

http://es.wikipedia.org/wiki/3-esfera

Momento magnético anómalo del electrón

Límite del error experimental máximo:





<br /> \[<br /> 1+\frac{\Omega+(\frac{\pi^{5}}{120})^{-1}}{453060}=1+\frac{\alpha}{2\pi}+\frac{2\alpha^{2}}{3}(\frac{\alpha}{2\pi})-\frac{4}{3}(\frac{\alpha}{2\pi})^{2}\]

http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment

Saludos

 

[legna777]

http://foros.astroseti.org/viewtopic.php?t=4529&postdays=0&postorder=asc&start=210

 

[legna777]

Hola

\[<br /> (\frac{m_{e}}{m_{e}}+\frac{m_{\mu}}{m_{e}}+\frac{m_{\tau}}{m_{e}})\approxeq(2^{7}+\frac{1}{\sqrt{2^{7}}})^{-1}(\frac{m_{W-}}{m_{e}}+\frac{m_{W-}}{m_{e}}+\frac{m_{W-}}{m_{e}})\]

\[<br /> (\frac{m_{e}}{m_{e}}+\frac{m_{\mu}}{m_{e}}+\frac{m_{\tau}}{m_{e}})\approxeq(\alpha^{-1}(M_{Z})-\sin(\frac{2\pi}{6}))^{-1}(\frac{m_{W-}}{m_{e}}+\frac{m_{W-}}{m_{e}}+\frac{m_{W-}}{m_{e}})\]


\[<br /> m_{e}=\] masa del electrón

\[<br /> m_{\mu}=\] masa del muón

\[<br /> m_{\tau}=\] masa del tauón

\[<br /> m_{W-}=\] masa del bosón W con carga negativa -1= 80,398 Gev

<br /> <br /> \[<br /> \alpha^{-1}(M_{Z})=128,95\] o valor de la "constante" estructura fina a la escala de unificación electrodebil o masa del bosón Z

muon-electron mass ratio

http://physics.nist.gov/cgi-bin/cuu/Value?mmusme|search_for=atomnuc!

tau-electron mass ratio

http://physics.nist.gov/cgi-bin/cuu/Value?mtausme|search_for=atomnuc!

Saludos

\[<br /> [\frac{M_{Z}}{m_{e}}-(\frac{m_{e}}{m_{e}}+\frac{m_{\mu}}{m_{e}}+\frac{m_{\tau}}{m_{e}})]=174764=\sum_{p=2}^{137}p^{2}\]

Donde los corchetes en
\[<br /> [\frac{M_{Z}}{m_{e}}-(\frac{m_{e}}{m_{e}}+\frac{m_{\mu}}{m_{e}}+\frac{m_{\tau}}{m_{e}})]

significa la función parte entera de lo que contiene

Y <br /> \[<br /> \sum_{p=2}^{137}p^{2}\] es el sumatorio del cuadrado de todos los números primos menores o iguales a 137

\[<br /> M_{Z}=91,1876\; Gev\]




\[<br /> [(\frac{M_{W}}{m_{e}}){\cos\theta_{W}}]=\sum_{p=2}^{127}p^{2}=138834\]

Recuérdese que para la inversa de la "constante" de acoplamiento electromagnética en la escala de unificación elctrodébil ( masa bosón Z ), 128,95; el número primo inmediatamente inferior a su parte entera [128,95] es el primo de Mersenne 127

\[<br /> 127=2^{7}-1\]

Saludos

 

[legna777]

Hola

\[<br /> N_{C}(E8)+(1+\cos\theta_{W})(1^{3}+2^{3}+3^{3}+5^{3}+8^{3})=(137\cdot\ln137)^{2}\]

\[<br /> 1\times2\times3\times5\times8=240\] Cantidad de raíces no nulas del grupo E8. Los 5 primeros números de la sucesión de Fibonacci.



\[<br /> N_{C}(E8)=453060=\] Número de la matriz generadora de los caracteres del grupo E8

\[<br /> (1^{3}+2^{3}+3^{3}+5^{3}+8^{3})=[\sqrt{N_{C}(E8)}]=[\sqrt{453060}]=673\]

 

[legna777]

Hola

\frac{1}{\ln(\frac{\alpha^{-1}-21\varphi}{2})}=\frac{\pi^{4}}{384}

Donde Pi^4/384 es la densidad de enpaquetamiento de hiperesferas en dimensión 8. Grupo E8.

http://mathworld.wolfram.com/HyperspherePacking.html


Alpha= constante estructura fina= (137,035999084...)^-1
Phi= 1+SQR(5) /2 = número aureo= 1,618033989...

 

[legna777]

Hola

\[<br /> (\frac{m_{PK}}{m_{e}})^{2}\cdot\varphi\cdot\cos(2\pi/10)\approxeq\alpha^{-21}\]

\[<br /> m_{PK}=\] masa de Planck= 2.176 44 x 10E-8 kg

http://physics.nist.gov/cgi-bin/cuu/Value?plkm|search_for=universal_in!



<br /> \[<br /> m_{e}=\] masa del electrón= 9.109 382 15 x 10E-31 kg

http://physics.nist.gov/cgi-bin/cuu/Value?me|search_for=atomnuc!

\[<br /> \alpha^{-1}=137,035999084\]

\[<br /> \varphi=\frac{1+\sqrt{5}}{2}\]= límite del cociente entre

2 términos consecutivos de la sucesión de Fibonacci.

\[<br /> 21=1\times3\times7\]= septimo término de la sucesión de Fibonacci

1, 2, 3, 5, 8, 13, 21,...

\[<br /> [\alpha^{-1}]=137\]

\[<br /> [\alpha^{-1}]=137=2^{2}-1+2^{3}-1+2^{7}-1\]

Saludos

 

[legna777]

Hola

\[<br /> \ln((\frac{\ln\ln\varphi}{\sum_{p}^{137}1/P}+137)/21)\approxeq\sum_{p}^{137}1/p\]


Donde: \[<br /> \sum_{p}^{137}1/p=1,872603466...\]<br /> es el sumatorio del inverso de los números primos menores o iguales a 137

<br /> \[<br /> \varphi=\frac{1+\sqrt{5}}{2}=1,618033989...\]

 

[legna777]

Hola

<br /> \[<br /> (\frac{M_{Z}}{m_{e}})(\frac{\alpha}{2\pi})=\frac{m_{\mu}}{m_{e}}a_{\tau}a_{e}\]

<br /> \[<br /> m_{Z}=91,1876\]<br /> $\: Gev$= 1.625566473 E-25 Kg

<br /> \[<br /> m_{e}=9.10938215E-31\: Kg\]

<br /> \[<br /> \alpha=1/137,035999084\]

\[<br /> a_{\tau}\]= momento magnético anómalo del tau=\[<br /> \frac{g-2}{g}\]=<br /> \[<br /> 1.00117721\]

The tau lepton anomalous magnetic moment

http://arxiv.org/abs/hep-ph/0702026




\[<br /> a_{e}=\frac{g_{e}-2}{g_{e}}=1.00115965218\]= momento magnético anómalo de electrón.

http://physics.nist.gov/cgi-bin/cuu/Value?ae|search_for=atomnuc!


<br /> \[<br /> m_{\mu}=1.8835313E-28\: Kg\]=masa del muón

 

[legna777]

Hola

http://img522.imageshack.us/img522/8949/27nx3mz9.jpg

[/URL]

Espacio de 10 dimensiones ( 9 espaciales y una temporal )
6 dimensiones enrolladas-compactadas.

http://img224.imageshack.us/img224/9769/29pm8.jpg

[/URL]

http://foros.astroseti.org/viewtopic.php?t=4529&postdays=0&postorder=asc&start=165

 

[legna777]

Hola

http://img385.imageshack.us/img385/4326/30ya8.jpg

Masa de la Tierra= 5,9736E 24 Kg

 

[legna777]

Hola

exp(137 + 2Phi) X Tiempo de Planck= Edad del Universo=

13700 millones años= 13700 x 1E6 x 365 x 24 x 60 x 60= 4,320432E 17 segundos

Phi= (1 + SQR(5) )/2= numero aureo= 1,618033989...

exp( 137 + 2Phi ) = 8.012176286E 60

Tiempo de Planck= 5,3912 E -44 segundos

http://es.wikipedia.org/wiki/Tiempo_de_Planck

Good is the architect of Universe

 

[legna777]

http://foros.astroseti.org/viewtopic.php?t=4529&postdays=0&postorder=asc&start=165

 

[legna777]

Esto se pone interesante con la colaboración de todos :lol:

Cabibbo–Kobayashi–Maskawa matrix

In the Standard Model of particle physics, the Cabibbo–Kobayashi–Maskawa matrix (CKM matrix, quark mixing matrix, sometimes also called KM matrix) is a unitary matrix which contains information on the strength of flavour-changing weak decays. Technically, it specifies the mismatch of quantum states of quarks when they propagate freely and when they take part in the weak interactions. It is important in the understanding of CP violations. A precise mathematical definition of this matrix is given in the article on the formulation of the standard model. This matrix was introduced for three generations of quarks by Makoto Kobayashi and Toshihide Maskawa, adding one generation to the matrix previously introduced by Nicola Cabibbo. This matrix is also an extension of the GIM mechanism, which only includes 2 of the 3 current families of quarks

Perdón, señores, se me olvido el link de la cita

http://en.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix

\[<br /> \theta_{C}=\] Ángulo de Cabibbo

\[<br /> \theta_{W}=\] ángulo mezcla electrodebil

\[<br /> \theta_{W}=\frac{4\varphi\theta_{C}}{3}\]

\[<br /> \varphi=\frac{1+\sqrt{5}}{2}\]

\[<br /> \alpha^{-1}=137,035999084\]

<br /> \[<br /> \frac{20\pi\alpha}{3^{2}}\approxeq\sin^{2}\theta_{C}=(0,2257)^{2}\]

\[<br /> 1+\sin(3\theta_{C})+2\cdot3^{2}\approxeq\frac{1}{\sin^{2}\theta_{C}}\]

Ángulo áureo= \[<br /> \frac{2\pi}{\varphi}\]

\[<br /> \frac{\tan(2\pi/\varphi)}{2\cdot3^{2}}\approxeq\mid V_{cd}\mid^{2}=(0,2256)^{2}\]




<br /> \[<br /> \left(\frac{1}{\mid V_{ud}\mid\cdot\mid V_{us}\mid\cdot\mid V_{ub}\mid}+\frac{1}{\mid V_{cd}\mid\cdot\mid V_{cs}\mid\cdot\mid V_{cb}\mid}+\frac{1}{\mid V_{td}\mid\cdot\mid V_{ts}\mid\cdot\mid V_{tb}\mid}\right)^{-1}=\pi^{4}+\frac{2}{5^{2}}\]<br />

Saludos

 

[legna777]

\left(\sum^{137}_{p=2}p^{2}\right)\frac{\sqrt{8}}{\pi}=\frac{M_{W}}{m_{e}}

Sum over all prime numbers <= 137

M_{W}=80,398 Gev

 

[legna777]

Gravitational radiation from binary systems

http://en.wikipedia.org/wiki/Gravitational_waves

P=\frac{32\cdot G_{N}^{4}\cdot M_{\odot}^{2}\cdot M_{\oplus}^{2}\cdot(M_{\odot}+M_{\oplus})}{5c^{5}d}\cdot\frac{1}{(1-e^{2})^{\frac{7}{2}}}\cdot(1+\frac{73e^{2}}{24}+\frac{37e^{4}}{96})$

<br /> P=\frac{32\cdot G_{N}^{4}\cdot M_{\odot}^{2}\cdot M_{\oplus}^{2}\cdot(M_{\odot}+M_{\oplus})\cdot t_{year}}{5c^{5}d}\cdot\frac{1}{(1-e^{2})^{\frac{7}{2}}}(1+\frac{73e^{2}}{24}+\frac{37e^{4}}{96})=M_{pk}\cdot c^{2}\cdot\sqrt{10}\cdot\left[\frac{1}{(1-e^{2})^{\frac{7}{2}}}\cdot(1+\frac{73e^{2}}{24}+\frac{37e^{4}}{96})\right]^{2}$

M_{\oplus}=5,9722\cdot10^{24}Kg$\medskip{}

M_{\odot}=1,98842\cdot10^{30}Kg$ \medskip{}

Where :
A.U, d=1,49597870700\cdot10^{11}m$\medskip{}

t_{year}=315581449,76\, s$\medskip{}

<br /> M_{pk}= Planck'mass = 2,17644 E-8 Kg

<br /> e=1,6710218\cdot10^{-2}$\medskip{} ( Earth's Orbit eccentricity )

Particle Physics Booklet (rpp-2008-booklet ), pág 6 :http://pdg.lbl.gov/2008/download/rpp-2008-booklet.pdf

 

[legna777]

Efecto casimir: energía del vacio entre 2 placas cuadradas conductoras de area L y a una distancia d

<br /> \[<br /> E_{vacio\; Casimir}=-\frac{\pi^{2}\hbar cL^{2}}{720d^{3}}\]

Fuerza de Casimir:


\[<br /> F_{vacio\; Casimir}=\frac{\pi^{2}\hbar cL^{2}}{240d^{4}}\]

Presión de Casimir:

<br /> \[<br /> P_{vacio\; Casimir}=\frac{\pi^{2}\hbar c}{240d^{4}}\]

http://en.wikipedia.org/wiki/Casimir_effect

Investigation into Compactified Dimensions: Casimir Energies and Phenomenological Aspects

http://arxiv.org/abs/0901.3640v1

Página 31

==========================================================


<br /> \[<br /> E_{vacio\; Casimir}=-\frac{\hbar cL^{2}}{[\frac{240}{\alpha^{-1}\pi^{3}}+\ln(\frac{M_{PK}}{m_{e}})]\sqrt{2}d^{3}}=-\frac{\pi^{2}\hbar cL^{2}}{720d^{3}}\]

Donde: \[<br /> \alpha^{-1}=137,035999084\]

\[<br /> M_{PK}=2,17644E-8Kg\]

\[<br /> m_{e}=9,10938215E-31Kg\]

http://foros.astroseti.org/viewtopic.php?p=77266#77266

 

[arivero]

I noticed a classical argument from Nigel
http://nige.wordpress.com/carl-brannen-and-the-koide-formula-how-to-derive-the-koide-formula/
about Koide-ish square roots, which could be of some value: Simply consider the total kinetic energy of the mass center of a system of three particles and compare it with the sum of the energies of the components.

<br /> E_{CM}=\frac 12 {{ (\sqrt{2E_1m_1}+ \sqrt{2E_2m_2}+\sqrt{2E_3m_3})^2} \over m_1+m_2+m_3}<br />

and compare it with the total Energy E_T=E_1+E_2+E_3[/tex]. <br /> <br /> In the simplest &quot;dirac delta&quot; partition where E_1=E_2=E_3<br /> we have that<br /> <br /> &lt;br /&gt; { E_{CM} \over E_T} = \frac 13 {{ (\sqrt{m_1}+ \sqrt{m_2}+\sqrt{m_3})^2} \over m_1+m_2+m_3}&lt;br /&gt;<br /> <br /> and then Koide &quot;3/2&quot; is the condition that the CM energy is one half of the total, free, kinetic energy. Funny.Disclaimer: I am not sure if I am being careful or standard when calling E_{CM} to this energy.<br /> Edit: A guess is that some related but quantum conditions could be found by playing with the available phase space for scattering, using golden rule and all that.

 

[arivero]

and then Koide "3/2" is the condition that the CM energy is one half of the total, free, kinetic energy. Funny.
.

Of course, the total energy in an arbitray reference frame is the energy of the CM in this frame plus the ("rotational?") kinetic energy in the CM reference frame. No jokes about Lubos website, please

So Koide is the relationship for masses in a frame where
- the kinetic energy of all the particles is the same, and
- The "traslational" energy of the CM in this frame is equal to the "rotational" energy of the particles in the CM frame.
- all the particles travel in the same line and orientation. (this is, to take the positive square roots and one dimensional sum)

 

[CarlB]

Hi Alejandro,

This is interesting. I'm going to have to play around with it as I'm not very intuitive about relativistic mechanics.

Marni and I just finished up the FFP10 conference in Perth, Australia. It went very well. I had a couple grad students ask for copies of my latest paper, which puts the Koide relation in the context of spin path integrals:
http://www.brannenworks.com/Gravity/EmergSpin.pdf

One of the talks was about quasars. It seem that there's evidence that they have intrinsic redshifts. They're born with very high redshifts, and then the redshifts decrease, in quantum jumps. The jumps correspond to changes in z of about 0.229, pretty close to the 2/9 in the Koide formula.

So I typed up a few extra PPT slides and wrote up an outline of a theory of gravity where gravity becomes repulsive inside of a black hole so that instead of having a single event horizon, they end up with a series of shells where their gravity is neutralized. The basic idea is to describe gravity as a vector field (see The River Model of Black Holes), and describe the vector length as a Wick rotated boost of something proportional to the gravitational flux. I'll post something at my blog when I get the time.

 

[arivero]

This is interesting. I'm going to have to play around with it as I'm not very intuitive about relativistic mechanics.

Actually, it is classical. It is not easy to read across Nigel's blog, it is not very concrete and it deviates from the path or misses the target very easily. But it is really straightforward, what he does in the deep, even without knowing it, is to ask for classical quantities where the angle with the vector \vec \sqrt m_i is relevant. And he finds one.

Let me rewrite it in one line:
<br /> {E_{CM} \over E_{CM} + E_{Rot}}={E_{CM} \over E_{Tot}}=<br /> { \frac 12 (\sum m_i) ({\sum m_i \sqrt {2 E_i/m_i}\over \sum m_i})^2 \over \sum E_i}<br /> = { ({\sum \sqrt m_i \sqrt {E_i} })^2 \over \sum m_i \sum E_i}<br /> = \Big({ \vec m^{1/2} . \vec E^{1/2} \over |\vec m^{1/2}| |\vec E^{1/2}| }\Big)^2<br />

This is a general formula for freshman physics, and then if we put all the three energies equal it simplifies to
{E_{CM} \over E_{Tot}}=<br /> { E_{1/3} ({\sum \sqrt m_i })^2 \over 3 E_{1/3} \sum m_i } = \frac 13 {({\sum \sqrt m_i })^2 \over \sum m_i }<br />

And thus Koide's happens to be the mass restriction that allows necessary conditions to find a frame where simultaneusly E_{CM}=E_{Rot} and E_1=E_2=E_3. Note that E_{Rot} in classical kinematics is frame-invariant.

With a little more of work, using the approximation v << c, it should be possible to do the same arguments in the relativistic kinematics of 3 body decay. Also, It could be seen not as an explanation of the origin of the relationship but as a consequence with kinematic repercusions.

Marni and I just finished up the FFP10 conference in Perth, Australia. It went very well. I had a couple grad students ask for copies of my latest paper, which puts the Koide relation in the context of spin path integrals:
http://www.brannenworks.com/Gravity/EmergSpin.pdf

Good to hear of it.

 

[arivero]

Of course, if all the three masses are equal, then it is the kinetic energy triple the one which holds Koide's relationship. Amusing.

The only thing I am worried here is about taking the positive sign of the square root for all the three velocities; this is an extra requirement to the reference frame, to see all the three particles in the same direction. And also that they are all the three in the same line. On other hand, the "different sign" case is the one used by Carl in the neutrinos.

 

[enotstrebor]

Plus c and h, of course.

The idea is to collect here in only a thread all the approximations voiced out during the summer. .

For a single 4\pi relationship for the electron generations you might want to see Apeiron 16(4) 475-484 (2009).

The \mu (n=1)to e (n=0) mass ratio m_{\mathrm{e_1}}/m_{\mathrm{e_0}} is \sqrt{2}(4\pi\varrho_1)^{(3-1)} where \varrho_1=.96220481
while the \tau (n=2) to \mu (n=1) mass ratio m_{\mathrm{e_2}}/m_{\mathrm{e_1}} is \sqrt{2}(4\pi\varrho_2)^{(3-2)} where \varrho_2=.94635968.

Thus the first and second (n=1,2) generation mass ratio (m_{\mathrm{e}(n)}/m_{\mathrm{e}(n-1)}) form is \sqrt{2}(4\pi\varrho_n)^{3-n}.

As the W, proton and electron masses are a function of (4\pi\varrho)^{3}, i.e

m_x = M_{\mathrm{sp}} ~(2S ~(4\pi\varrho)^3/\varsigma )^{(S ~C ~M)}

whereM_{\mathrm{sp}}=\sqrt{m_p \cdot m_e}, \varrho=0.9599737853, S is the spin quantum number (1/2,1), C is the charge quantum number (+1,-1) and M is the matter quantum number (matter= +1 antimatter= -1),

You can get the masses of the W, proton, electron and its generations using the single formula for particle x (x=W,p,e,mu,tau)

m_x = M_{\mathrm{sp}(n)} ~(2S~(4\pi\varrho)^3/\varsigma)^{(S ~C ~M)},

where M_{\mathrm{sp}(n)} = M_{\mathrm{sp}} ~S^{-n/2}(4\pi\varrho_n)^{(6S n - S n(n+1)))} and \varrho_n = 1 - log(1 + 64.7564~n/S)/(112S) are used, and generation n is \{0,1,2\}.

 

[legna777]

m_{e}=\frac{V_{H}}{\sqrt{2}\cdot\Bigl(e^{12}+\frac{m_{Z}}{m_{e}}\Bigr)}


<br /> $V_{H}=\; vacuum\; Higgs\; \; &quot;mass&quot;=\frac{246,2205691\cdot E^{9}\cdot1,602176487\cdot E^{-19}\: Gev}{c^{2}}$\medskip{}<br />


<br /> $m_{Z}=mass\; Z\; boson=91,1876\; Gev$<br />

<br /> $e^{\Bigl(12+\frac{\cos(2\pi/5)}{10}\Bigr)=\frac{m_{Z}+m_{W}}{2m_{e}}\;\; m_{e}=\: electr\acute{o}n\; mass\; m_{W}=bos\acute{o}n\: W=80,398\: Gev}$\medskip{}<br />


<br /> \[<br /> \frac{V_{H}}{\sqrt{2}}+\frac{V_{H}}{\pi^{4}}=\sum_{q=1}^{6}m_{q}-\sum_{l=1}^{6}m_{l}\]<br /> \medskip{}<br />

<br /> \[<br /> \sum_{q=1}^{6}m_{q}=sum\; over\; all\; six\; quarks\;\;\]<br /> \medskip{}<br />

<br /> \[<br /> \sum_{l=1}^{6}m_{l}=m_{e}+m_{\mu}+m_{\tau}+\nu_{e}+\nu_{\mu}+\nu_{\tau}=sum\; leptons\; masses\]<br /> <br />


<br /> \[<br /> \frac{V_{H}}{\sqrt{2}}-\frac{V_{H}}{\pi^{4}}=m_{Z}+m_{w}\]<br /> \medskip{}<br />

<br /> \[<br /> bos\acute{o}n\; Higgs\; mass=\frac{\sqrt{2}\cdot V_{H}}{e}=m_{H}\backsimeq128\; Gev\]<br /> \medskip{}<br />

<br /> $e^{\bigl(12+(\sin^{2}eff^{lep}(\phi_{W}))^{-1}/10\bigr)}=\frac{m_{H}}{m_{e}}$<br />

<br /> <br /> $\sin eff^{lep}(\phi_{W})=\sqrt{0,2315...}=\ln(\varphi)$\medskip{}<br /> <br />

<br /> $\varphi=\frac{1+\sqrt{5}}{2}$<br />

<br /> \[<br /> \frac{\sin\theta_{C}}{\sqrt{4\pi\alpha}}\Bigl(\frac{V_{H}}{\pi^{4}}\Bigr)=\sum_{l=1}^{6}m_{l}=m_{e}+m_{\mu}+m_{\tau}+\nu_{e}+\nu_{\mu}+\nu_{\tau}\]<br /> \medskip{}<br />

<br /> $\theta_{C}=Cabibbo\; angle=13,04^{\circ}$\medskip{}<br />

<br /> $\alpha=fine\; structure\; constant=(137,035999084...)^{-1}$<br />

"En este punto aun tenemos 4 bosones gauge (Wi\textgreek{m}(x) y B\textgreek{m}(x)) y 4 escalares $\xi\overrightarrow{(x)}$ y h(x)), todos ellos sin masa, lo que equivale a 12 grados de libertad[/color] (Conviene notar que un bosón vectorial de masa nula posee dos grados de libertad, mientras que un bosón vectorial masivo adquiere un nuevo grado de libertad debido a la posibilidad de tener polarización longitudinal: 12 = 4[bosones vectoriales sin masa] × 2 + 4[escalares sin masa]). P. W. Higgs fue el primero en darse cuenta de que el teorema de Goldstone no es aplicable a teorías gauge, o al menos puede ser soslayado mediante una conveniente selección de la representación. Así, basta con escoger una transformación:"

http://es.wikipedia.org/wiki/Mecanismo_de_Higgs


http://arxiv.org/PS_cache/hep-ph/pdf/0001/0001283v1.pdf

A Finely-Predicted Higgs Boson Mass from A Finely-Tuned Weak Scale

Lawrence J. Hall, Yasunori Nomura
(Submitted on 13 Oct 2009 (v1), last revised 19 Oct 2009 (this version, v2))

Abstract: If supersymmetry is broken directly to the Standard Model at energies not very far from the unified scale, the Higgs boson mass lies in the range 128-141 GeV. The end points of this range are tightly determined. Theories with the Higgs boson dominantly in a single supermultiplet predict a mass at the upper edge, (141 \pm 2) GeV, with the uncertainty dominated by the experimental errors on the top quark mass and the QCD coupling. This edge prediction is remarkably insensitive to the supersymmetry breaking scale and to supersymmetric threshold corrections so that, in a wide class of theories, the theoretical uncertainties are at the level of \pm 0.4 GeV. A reduction in the uncertainties from the top quark mass and QCD coupling to the level of \pm 0.3 GeV may be possible at future colliders, increasing the accuracy of the confrontation with theory from 1.4% to 0.4%. Verification of this prediction would provide strong evidence for supersymmetry, broken at a very high scale of ~ 10^{14 \pm 2} GeV, and also for a Higgs boson that is elementary up to this high scale, implying fine-tuning of the Higgs mass parameter by ~ 20-28 orders of magnitude. Currently, the only known explanation for such fine-tuning is the multiverse.

http://arxiv.org/abs/0910.2235

 

[legna777]

<br /> \[<br /> \theta_{C}=Cabibbo\; angle\]<br /> \medskip{}<br />

<br /> <br /> \[<br /> \tan\theta_{c}=\sin^{2}eff^{lep}(\phi_{W})=0,2315...=\ln^{2}(\varphi)\]<br /> <br /> <br /> <br /> <br /> <br />


http://pdg.lbl.gov/2009/reviews/rpp2009-rev-vud-vus.pdf

 

[legna777]

"Standard" parameters


A "standard" parameterization of the CKM matrix uses three Euler angles ($\theta_{12}\,,\,\theta_{13}\,,\,\theta_{23}$)and one CP-violating phase $(\delta_{13})$
Couplings between quark generation i and j vanish if $\theta_{ij}=0$ .Cosines and sines of the angles are denoted $c_{ij}$ and $c_{ij}$ $s_{ij}$ respectively. $\theta_{12}$ is de Cabibbo angle.

The currently best known values for the standard parameters are:


$\theta_{12}=13.04\pm0.05^{\circ}$
$\theta_{13}=0.201\pm0.011^{\circ}$
$\theta_{23}=2.38\pm0.06^{\circ}$
$\delta_{13}=1.20\pm0.08^{\circ}$<br /> <br />


<br /> \[<br /> \begin{bmatrix}c_{12}c_{13} &amp; s_{12}c_{13} &amp; s_{13}e^{-i\delta_{13}}\\<br /> -s_{12}c_{23}-c_{12}s_{23}s_{13}e^{i\delta_{13}} &amp; c_{12}c_{23}-s_{12}s_{23}s_{13}e^{i\delta_{13}} &amp; s_{23}c_{13}\\<br /> s_{12}s_{23}-c_{12}c_{23}s_{13}e^{i\delta_{13}} &amp; -c_{12}s_{23}-s_{12}c_{23}s_{13}e^{i\delta_{13}} &amp; c_{23}c_{13}\end{bmatrix}\]<br /> <br />

http://en.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix

<br /> \[<br /> \frac{\alpha}{\sin(\phi_{W})}=\frac{\theta_{13}}{\theta_{12}}\]<br />


<br /> \[<br /> \theta_{12}+\theta_{13}+\theta_{23}-\delta/2=\frac{2\pi}{24}\; rad.\]<br /> <br />

<br /> \[<br /> 2\theta_{12}+\theta_{13}+\theta_{23}\backsimeq\phi_{W}\]<br />

 

[legna777]

Alexis Monnerot-Dumaine: The Fibonacci Fractal

http://alexis.monnerot-dumaine.neuf.fr/articles/fibonacci fractal.pdf

<br /> \[<br /> \frac{\sin eff^{lep}(\phi_{W})}{\frac{m_{W}}{m_{Z}}}=\frac{D_{H}}{3}=\frac{\ln\varphi}{\ln(1+\sqrt{2})}\]<br />

Hausdor Dimension=3=Dh

FRACTAL GEOMETRY IN QUANTUM MECHANICS, FIELD THEORY AND
SPIN SYSTEMS

H. Kröger

Physics Reports 322 (2000) 81-181

[Link Deleted]

It´s free, not my work. Great work

In geometry, an icosahedron (Greek: εικοσάεδρον, from eikosi twenty + hedron seat; pronounced /ˌaɪkɵsəˈhiːdrən/ or /aɪˌkɒsəˈhiːdrən/; plural: -drons, -dra /-drə/) is a regular polyhedron with 20 identical equilateral triangular faces, 30 edges and 12 vertices. It is one of five Platonic solids.

It has five trianglular faces meeting at each vertex. It can be represented by its vertex figure as 3.3.3.3.3 or 35, and also by Schläfli symbol {3,5}. It is the dual of the dodecahedron, [/color]which is represented by {5,3}, having three pentagonal faces around each vertex.

http://en.wikipedia.org/wiki/Icosahedron

<br /> <br /> \[<br /> -\frac{\sin\theta_{C}}{\sqrt{4\pi\alpha}}\backsimeq\cos(138,186965^{\circ})=\cos(\widehat{\Omega})_{d}icoSHD=\]<br /> <br />cosine icosaedral dihedral angle

Symmetry :dodecahedral manifold

<br /> \[<br /> \frac{2\pi}{12+\varphi^{-1/2}}=\widehat{\phi_{W}}\; rad\]<br /> <br />

<br /> \[<br /> \frac{2\cdot m_{Z}}{m_{e}}+\alpha_{s}^{-1}(m_{Z})\cdot\varphi^{2}=\exp(12+\varphi^{-1/2})\]<br /> \medskip{}<br />

<br /> \[<br /> \exp(\varphi^{5}+1)+\alpha_{s}^{-1}(m_{Z})\cdot\varphi^{2}=\frac{m_{Z}}{m_{e}}\]<br /> \medskip{}<br />

<br /> \[<br /> -\sin\Bigl(\frac{2\pi}{\varphi^{2}}\Bigr)-\cos\Bigl(\frac{2\pi}{\varphi^{2}}\Bigr)=\frac{\alpha(q=0)}{\alpha_{s}(m_{Z})}=\frac{(137,035999084)^{-1}}{0,1178}\]<br /> <br />

 

[Vanadium 50]

First, the paper link you posted to is not "free". It is "stolen". I have since deleted it.

Second, Kroeger doesn't mention icosahedra - or polyhedra at all - in his paper.

Third, if this thread has taught us anything, it's that one can always toss a simple equation together to get some "interesting" value. By itself, this means nothing. This is why physicists call this "numerology".

Fourth, the ratio of two fundamental constants at two different scales cannot possibly be physical - as they are at two different scales.

Fifth, even if one takes everything at face value, the agreement is not impressive. Plugging in alpha_s at the Z, one gets a value for alpha at q=0 of 1/(137.4 ± 2.3). While that covers the correct value, the agreement is not so impressive when the error bars are added.

Finally, I'd like to remind everyone in this thread about the PF rules for overly speculative posts.

 

[arivero]

Yes, I am sorry that while this thread was active people was more aware of how touchy the issue is. We could close it perhaps.

 

[legna777]

First, the paper link you posted to is not "free". It is "stolen". I have since deleted it.

Second, Kroeger doesn't mention icosahedra - or polyhedra at all - in his paper.

Third, if this thread has taught us anything, it's that one can always toss a simple equation together to get some "interesting" value. By itself, this means nothing. This is why physicists call this "numerology".

Fourth, the ratio of two fundamental constants at two different scales cannot possibly be physical - as they are at two different scales.

Fifth, even if one takes everything at face value, the agreement is not impressive. Plugging in alpha_s at the Z, one gets a value for alpha at q=0 of 1/(137.4 ± 2.3). While that covers the correct value, the agreement is not so impressive when the error bars are added.

Finally, I'd like to remind everyone in this thread about the PF rules for overly speculative posts.

First, the paper is not "stolen": is public

Second: I do not say that the paper mentions nothing around icosahedra. I´ts is my opinion

Third: The fine structure constant QED at transfer momentun 0 ( or mass =0 )is:

(137,035999084... )-1

The
effective coupling equals the fine-structure constant a at
the Thomson limit(q2=
* 0)and is expected to increase
logarithmically as jq2
j increases at large jq2

PHYSICAL REVIEW LETTERS Volume 81 number 12 21 September 1998

"Measurement of the Running of Effective QED Coupling at Large Momentum Transfer
in the Space like Region"

The link is: PUBLIC, not stolen

http://dspace.lib.niigata-u.ac.jp:8080/dspace/bitstream/10191/1710/1/p2428_1.pdf





Four: Alpha_s is the coupling strong

Fifth: the ratio of two fundamental dimensionless constants at two different scales can possibly be physical

 

[Vanadium 50]

First, the paper is not "stolen": is public

No, it's not. It's in Physics Reports, which means that Elsevier owns the copyright. The fact that you have to download it from a file-sharing site rather than a scientific site should have been a big hint.

Second: I do not say that the paper mentions nothing around icosahedra. I´ts is my opinion

I'd like to remind everyone in this thread about the PF rules for overly speculative posts. .

 

[legna777]

No, it's not. It's in Physics Reports, which means that Elsevier owns the copyright. The fact that you have to download it from a file-sharing site rather than a scientific site should have been a big hint.



I'd like to remind everyone in this thread about the PF rules for overly speculative posts. .

Your are in true:

The fact that you have to download it from a file-sharing site rather than a scientific site should have been a big hint.

correctely deleted the link, I am sorry around this fact

 

[CarlB]

With journals, you used to get a pile of paper "reprints" to mail around to people. Now some journals instead give you an electronic copy of the paper. You're allowed to put this on your own website but it still isn't quite a public copy in that no one else is allowed to do this. I've got a paper to be published this month in IJMPD and here's what they say:

You may post the postprint on your personal website or your institution's repository, provided it is accompanied by the following acknowledgment:
Electronic version of an article published as [Journal, Volume, Issue, Year, Pages] [Article DOI] © [copyright World Scientific Publishing Company] [Journal URL]
http://www.worldscinet.com/authors/authorrights.shtml

So when you see a copy of a scientific paper on the web it's not necessarily stolen. If it's on a file sharing website that's a bad sign, but there are also papers that have been released by the journals.

 

[Hans de Vries]

I'd say that this last remark reveals all. Either legna is mocking of the rest of the thread (which is reasonable, but perhaps and argument more to close the thread after all these years) or a part of legna is mocking of other part of self. On other hand, it is true that the relationships are, if not under 1%, at least within 3%. But it is clear that they fail on the side of simplicity, when compared to the rest of the thread.

To make people realize that one can come up with hundreds of thousands
of formulas with an equal or lower error you could introduce the following
posting requirements:

1) Every numerical coincidence posted should include the error percentage.
2) The posts should contain the "prediction power" of the "formulas"

The "prediction power" can for instance be defined by the simple formula:

\mbox{PP} ~~=~~ \frac{10^{-\frac43 S}}{\mbox{error}}

where S is the number of symbols like 1,2,3,+,-,*/,sin,log...

The value of 1/PP is then the amount of similar complex formulas which give
an equal or better prediction, most likely to be a very big number...


Regards, Hans

 

[jtbell]

This thread is closed pending a moderation decision.

Update: it has been decided to leave this thread closed.