All the lepton masses from G, pi, e

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    Lepton Pi

Multiple poll: Check all you agree.

  • Logarithms of lepton mass quotients should be pursued.

    Votes: 21 26.6%
  • Alpha calculation from serial expansion should be pursued

    Votes: 19 24.1%
  • We should look for more empirical relationships

    Votes: 24 30.4%
  • Pythagorean triples approach should be pursued.

    Votes: 21 26.6%
  • Quotients from distance radiuses should be investigated

    Votes: 16 20.3%
  • The estimate of magnetic anomalous moment should be investigated.

    Votes: 24 30.4%
  • The estimate of Weinberg angle should be investigated.

    Votes: 18 22.8%
  • Jay R. Yabon theory should be investigate.

    Votes: 15 19.0%
  • I support the efforts in this thread.

    Votes: 43 54.4%
  • I think the effort in this thread is not worthwhile.

    Votes: 28 35.4%

  • Total voters
    79
  • #351
Jay R. Yablon said:
Dear friends over at Physics Forums:

I have for more than two years been researching the possibility that
baryons may in fact be non-Abelian magnetic sources.

The result of this research are now formally and rigorously presented in
a paper at:

http://home.nycap.rr.com/jry/Papers/Baryon%20Paper.pdf

Thank you for presenting your work over here Jay. It's becoming more
interesting with each new paper.

I think we do share the viewpoint that much of the interesting and
important physics is the result of the special terms, like the spin term in
the Dirac equation for the fermions, and the extra non-Abelian term for
the Bosons. My feeling is that the universe would be just a dull homo-
geneous "soup" without these two terms.


Regards, Hans.
 
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  • #352
I never did expect numerical coincidences with hadron masses but
the pion shows a second one, again magnetic anomaly related.
Half the mass ratio of the (charged) two quark pion and the electron
is roughly equal to alpha. Now if we divide this value by 2pi then we
find a coincidence with the Muon's magnetic anomaly:

\mbox{\Huge $\frac{1}{\pi} \frac{m_e}{\ \ m_{\pi\pm}}\ =\ 0.001165407 (3) $}

We already had an Electron magnetic anomaly coincidence:

\mbox{\Huge $\left(\ \frac{m_{\pi^\pm}}{m_{\pi^0}} - 1\ \right)^2\ =\ 0.00115821 (26) $}

This adds to the list of Magnetic Anomaly related numerical coincidences:


0.001165920________ Muon Magnetic Analomy
0.001165407__(3)___ Electron / Pion mass ratio /pi.
0.001165892________ Electrom / Pion mass ratio +Vpi-Vmu

0.001159652________ Electron Magnetic Analomy
0.001159567________ Mass independent Magnetic Analomy
0.001158692__(27)__ Muon / Z boson mass ratio.
0.00115821___(26)__ Pion mass delta square.

0.0000063532_______ Muon Mag.Anomaly mass dependence
0.0000063558_(20)__ Electron / W mass ratio (2007)


The third value (+Vpi-Vmu) is closer. What we have done here
is to add the one-loop QED vacuum polarization difference for
a particle of muon mass and a particle of pion mass. A lepton
with the mass of a pion should have a higher magnetic anomaly
as the muon by this amount plus another 4-8% or so for higher
order VP loops which I can't calculate.

The "Mass independent Magnetic Anomaly" is the same for
all leptons. It is calculated here by subtracting the one loop VP
contribution from the electrons magnetic anomaly.

The third numerical coincidence was found earlier and relates
the electron / W mass ratio with the mass dependent part
of the Muon's magnetic anomaly. It was calculated by taking
the difference between the Muon and Electrons magnetic
anomalies and adding the Electrons largest (one-loop) VP term.Regards, Hans

Code:
 One loop vacuum polarization:

 m1/m2 = 273.132044975751,   one-loop vpol = 0.000006389950   (pion +/-)
 m1/m2 = 206.768283800000,   one-loop vpol = 0.000005904060   (muon)
 m1/m2 = 1.00000000000000,   one-loop vpol = 0.000000084641   (electron)
 
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  • #353
Hans de Vries said:
Thank you for presenting your work over here Jay. It's becoming more
interesting with each new paper.

I think we do share the viewpoint that much of the interesting and
important physics is the result of the special terms, like the spin term in
the Dirac equation for the fermions, and the extra non-Abelian term for
the Bosons. My feeling is that the universe would be just a dull homo-
geneous "soup" without these two terms.

Regards, Hans.

Hi Hans, thank you for your encouragement. :smile:

As you know I mentioned over on SPF, the most important results in the baryon paper have to do with confinment. I have extracted and consolidated these results and posted them at the link below.

http://home.nycap.rr.com/jry/Papers/Confinement%20Paper.pdf

This paper may provide an exact analytical solution to the problem of QCD confinement and the existence of short-range meson mediators of the nuclear interactions.

As you noted over on SPF, this paper evolves from the non-vanshing three form P = dF which would be the Yang-Mills (non-abelian) "magnetic current."

If I were to put the paper in one sentence it would be:

Baryons are non-Abelian (Yang-Mills) magnetic monopoles.

And if permitted a second sentence:

These Yang Mills monopoles exhibit all the properties of a baryon, including three fermions which via exclusion we connect with quarks, confinement of these quarks and their mediating gluons such that there is never any net flux of gluons or individual quarks across any closed two-dimensional surface of integration through the baryon current density, and emission of mesons which have short range such that these mesons, which account for interactions between baryons, are the only entities for which there is a net flux through the integration surface.

I appreciate any comments which the Physics Forums readers may have.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm
sci.physics.foundations co-moderator
 
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  • #354
Hans de Vries said:
0.001165407__(3)___ Electron / Pion mass ratio /pi.

Well, that is rare, as every you find :smile: ! I would expect the quotient between charged and neutral pion to be related somehow to the fine structure constant, and then this natural connection would propagate to the rest of findings in the thread. But again, too much exactitude.

In a related thems, did I tell that I was interested on the history of the calculation of g-2? It seems that a guy, who helped Kino****a to set up the software, is now in the dark side of accidental numeric coincidences.

I have been for the last four months working in a BOINC supercomputing platform, Zivis. I wonder how much time could take to repeat K. et al calculations; I guess that programming time should be higher than computing time.
 
  • #355
The so-called Yang- Mills "mass gap"

Dear Physics Forum friends:

I have posted an updated draft of my confinement paper to:

http://home.nycap.rr.com/jry/Papers/Confinement%20Paper.pdf

Last evening, I uncovered and added to the paper, a line of calculation
which explicitly and formally gives a non-zero rest mass to the meson
mediators of QCD without resort to the Higgs Mechanism, naturally
eliminates the propagator poles and therefore allows one to have "real"
rather than "virtual" particles without ad hoc tricks, formally renders
QCD a short-range interaction, and solves the Yang-Mills mass gap
problem. This is in addition to the confinement solution which has
already been posted for several days now.

Interested in feedback.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm
sci.physics.foundations co-moderator
 
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  • #356
Gerald Rosen

I believe we have not included G. Rosen in the references for the topic of this thread.

His http://prola.aps.org/abstract/PRD/v4/i2/p275_1 , for instance, gives one of these formulae with exponential relationship between Planck and electron masses. Other papers look for masses of electron and quark, value of the fine structure constant, etc. http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+rosen%2C+g&FORMAT=WWW&SEQUENCE=
 
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  • #357
arivero said:
Then we use http://wwwusr.obspm.fr/~nottale/ukmachar.htm to get the mass of the electron

<br /> \ln (M_P/m_e) = \alpha^{-1} \sin^2 \theta_W <br />

Note that this remark is mostly an elaborate rewritting of the argument about the self energy of the electron. For instance in a modern text such as Polchiski's string textbook, it appears as
<br /> \delta m \approx \alpha m \ln {1 \over m l}<br />
without any reference except "known since the 1930s", and then it is explained that l should to be expected to ultimate cutoff, thus the length of Planck, and that \delta m \over m should be expected to be of order O(1).

Thus the formula can be rewritten as
<br /> \alpha^{-1} {\delta m_e \over m_e} \approx \ln {m_P \over m_e}<br />

And Nottale's statement translates, in standard knowledge, as telling that the electromagnetic contribution to the mass of the electron is about 3/8.

While most of the "rediscoverers" of this relationship are outsiders, "out of the loop" people, in the case of Nottale it is sort of astonishing that he forgets to tell that he is recovering a "known" relationship and not a new one (he does not have, at all, new arguments for the 3/8 adjustment neither).

Another remark nobody does is that the argument can be reverted to predict the GUT-Planck-String scale from the low energy scale. Thus one could to try to predict muon or electron masses from the electroweak scale, then use the cutoff argument to predict the GUT scale, and the to use the seesaw to predict the neutrino scale.

EDITED: Nottale presentation can be found in http://luth2.obspm.fr/~luthier/nottale/ukmachar.htm Also, it is argued in http://luth2.obspm.fr/~luthier/nottale/arDNB.pdf that better results are met if one uses in a very peculiar way the fine structure constant running value instead of the infrared one. Indeed using alpha at m_e, one really "predicts" (except the justification of the 3/8) the value of Newton Constant. In any case, I am ashamed 1) about our inability to note that it was just the renormalisation of electron mass as usual, and 2) that Nottale hides or ignores this fact, after being in contact with QED theory during decades.
 
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  • #358
For numerical "coincidences", hep-ex/0412028 remembers that "the infamous Florida 2000 presidential election recount with the official result of 2,913,321 Republican vs. 2,913,144 Democratic votes, with the ratio equal to 1.000061".
 
  • #359
Once in a while one runs into a nice numerical coincident like this one
relating alpha to a Yukawa potential:

\mbox{\Huge $\frac{\alpha}{2\pi}\ =\ \frac{e^{-2\pi r}}{2\pi r}$}

Where r is the quotient of two radii:

r\ =\ \frac{r_c}{r_l}

in which rc is the Compton radius of the electron, muon or tau lepton and
rl is the corresponding cut-off radius for which the electrostatic energy
is equal to the magnetostatic energy of the classical electron, muon or
tau lepton. It's the magnetic moment which is responsible for the magneto-
static energy. One gets the following relations:

\alpha =1/137.05268 -\ \quad \mbox{for the electron}
\alpha =1/137.04743 -\ \quad \mbox{for the muon}
\alpha =1/137.03796 -\ \quad \mbox{for the tau lepton}
\alpha =1/137.03599971\ \quad \mbox{experimental}

The differences stem from the differences of the magnetic moment analomies. Regards, Hans==================================================

PS: For the EM fields the following expressions were used:

<br /> \textsf{E}_x\ = \frac{q}{4\pi\epsilon_o}\ \frac{x}{r^3}, \quad <br /> \textsf{E}_y\ = \frac{q}{4\pi\epsilon_o}\ \frac{y}{r^3}, \quad <br /> \textsf{E}_z\ = \frac{q}{4\pi\epsilon_o}\ \frac{z}{r^3}<br />

<br /> \textsf{B}_x\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}x <br /> \right), \quad \textsf{B}_y\ = <br /> \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}y \right), \quad <br /> \textsf{B}_z\ = <br /> \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}z-\frac{1}{r^3} \right) <br />

And for the total energy:<br /> \mbox{Energy:} \qquad E\ =\ \frac{1}{2}\left(\epsilon_o\textsf{E}^2 <br /> + \frac{1}{\mu_o}\textsf{B}^2\right) <br />

<br /> E\ =\ \int \left\{ \frac{q^2}{32\pi^2\epsilon_o r^4}\ +\ <br /> \frac{\mu_o\mu_e^2}{8\pi^2}\left(\frac{3z^2}{r^8}+\frac{1}{r^6} <br /> \right) \right\} dx^3\ =\ \frac{q^2}{8\pi\epsilon_o r_o}\ + \ <br /> \frac{\mu_o\mu_e^2}{4\pi}\ \frac{3}{r_o^3} <br />

For an electron:

<br /> \frac{q^2}{8\pi\epsilon_o} = 1.1535385\ <br /> 10^{-28},\quad \frac{3\mu_o\mu_e^2}{4\pi}\ =\ 2.5862051\ 10^{-53}<br />

and ro is the cut-off radius.
 
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  • #360
This last comment reminders me about Born calculation of the Lamb shift,
in the train cominb back from Shelter Island. He had no serious argument to fix the cut-off in the renormalized equation, so he simply choose the compton length of the electron because it is the cutoff of particle creation.
 
  • #361
Evidence for a moving Magnetic Monopole in 1975

See:

"Evidence for Detection of a Moving Magnetic Monopole", Price et al., Physical Review Letters, August 25, 1975, Volume 35, Number 8.

This was the last of a series of balloon flights, launched in 1973, but not analyzed by myself until 1975, due to higher priority cosmic rays analysis then ongoing.

The suggestion that the anomalous track could have been caused by a doubly fractionating normal nucleus is untenable. One would have expected to have seen billions of similar tracks, not quite as closely matched to the expected track of a magnetic monopole, first. No such similar events were ever detected.

For further information, contact the administrator who can email me, as I do not regularly post at this forum. Or check www.sciforums.com where I do regularly post, and PM me.

Whether the Large Hadron Collider [LHC] will create a magnetic monopole is highly debatable. It might also create miniature black holes, or strangelets.
 
  • #362
Mass quantization in terms of pion and muon mass difference

Please see my paper about mass quantization @ arxiv. hep-ph/0702140
 
  • #363
taarik said:
Please see my paper about mass quantization @ arxiv. hep-ph/0702140

Indeed your paper, your reference list
http://www.slac.stanford.edu/spires/find/hep/wwwrefs?key=7074930
and the list of citations of McGregor
http://www.slac.stanford.edu/spires/find/hep/www?c=NUCIA,A58,159
are interesting for the topics of this thread. A problem of quantisation of M instead of quantisation of M^2 is that it has some scent of classical group theory, thus one needs to see how many of the relationships are already explained in the quark model and check the extant cases.
 
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  • #364
Mass quantization

Thiis paper has been accepted for publication in Modern Physics Letters A.
The important thing is that while the charged pions and muon are related in sense via the decay of former into latter, the neutral pion and muon are not related in any sense. Yet there mass difference serves as a basic mass unit for both leptons and hadrons.
 
  • #365
taarik said:
Thiis paper has been accepted for publication in Modern Physics Letters A.
The important thing is that while the charged pions and muon are related in sense via the decay of former into latter, the neutral pion and muon are not related in any sense. Yet there mass difference serves as a basic mass unit for both leptons and hadrons.
very good news.

the question, to me, is not why muon and pion have different mass, but why have they got almost the same mass. A conjecture is SUSY.
 
  • #366
Here is another surprise
the t lepton mass can be obtained by taking 57 jumps of 29.318 MeV from the muon mass i.e t mass= 57x 29.318. Now this 57 number also helps us to include the electron mass as 57 times electron mass= 29.127 very close to 29.318. This leads us to thing that like in Nambu's and many other cases the basic unit appears from the electron mass.
Now this also means the pion -muon= 57x electron , tau - muon =57x 29.318 =57x57x electron. which in turn leads to tau- pion =56x 29.318 =56 x 57 x electron.
Hence the lightest hadron i.e pion and lightest unstable lepton i.e muon , two leptons muon and tau , lightest hadron and heaviest lepton i.e tau are all related through electron mass.
 
  • #367
taarik said:
Now this also means the pion -muon= 57x electron , tau - muon =57x 29.318 =57x57x electron. which in turn leads to tau- pion =56x 29.318 =56 x 57 x electron..

I prefer to write then

{ m_\pi - m_\mu \over m_e}= \sqrt {m_\tau - m_\mu \over m_e}

It should be nice to have a mathematical (group theoretical) argument for 57.

EDIT: It is a bit puzzling that if we fix the mass of tau, mu and electron to the experimental values, the above formula "predicts" 134.88 MeV, to be compared with the mass of the neutral pion (134.976 MeV). Naively one could expect the result to be more related with the mass of the charged pion, which is 4.6 MeV above.

EDIT2. Perhaps Krolikowski has some argument for 58/2. Also, Ramanna (eg pg 16 of nucl-th/9706063)
 
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  • #368
arivero said:
I prefer to write then...
Hmm, funnier:

( m_{\pi_0} - m_\mu )= \sqrt {m_e} \sqrt {m_\tau - m_\mu}LHS and RHS still agree within a 0.3 %. No bad.

The above comments still apply. On other hand, if I recall correctly, the question about why the mass of the charged pion is higher, and not lower, than the neutral one was a touchy issue decades ago, and it required very high level theoretists to explain it.
 
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  • #369
Refering to first version, it ia again amazing that the mass of the neutral pion is deteremined very precisely in terms of the three leptons. Now neutral pion has no relation with the three leptons. it does not decay into any of these particles. On the other hand the charged pion decays into electron and muon. Hence charged pion mass should have been related to lepton masses.
 
  • #370
taarik said:
Refering to first version, it ia again amazing that the mass of the neutral pion is deteremined very precisely in terms of the three leptons. Now neutral pion has no relation with the three leptons. it does not decay into any of these particles. On the other hand the charged pion decays into electron and muon. Hence charged pion mass should have been related to lepton masses.

I agree, it is misterious. Furthermore, forgetting the issue of integer multiples and the squaring of masses, the formula is very reminiscent of charged pion decay, you know, these \prop m_\mu^2 ( m_\pi_+^2 - m_\mu^2) from textbooks.

To put more intrigue, the mass difference between eta and the average of pion and muon (say, diff=427.2 MeV) also fits roughly in the obvious permuted formulae:
<br /> \sqrt {m_\mu} \sqrt {m_\tau - m_e} \approx \sqrt {m_\tau} \sqrt {m_\mu - m_e}<br /> \approx \sqrt {m_\tau \pm m_e} \sqrt {m_\mu \pm m_e}<br /> \approx \sqrt {m_\mu} \sqrt {m_\tau} = 433.27 MeV<br />

EDITED: a purpose of the above formulas is to consider the limit m_\mu \approx m_\tau where the former formula cancels and the two first ones in the above become the same. Also, the same cancellation and similarity happens in the other limit m_e \to 0. Simultaneous limit conflicts with Koide's.
 
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  • #371
My thinking during the last two years was:
initially there is a symmetry where neutrinos have the same mass than neutral mesons and charged leptons have the same mass than charged mesons. Note the count of degrees of freedom. Of course one could also expect the dirac mass of neutrino and charged lepton to coincide.
Then seesaw moves the mass of neutrinos out of reach
and mixing, including CKM, and/or other unknown mechanism alter the mass eigenvalues of the mesons.
The mechanism could be related to a mismatch between isospin in mesons and leptons. Namely, third generation mesons do not exist except bB.
 
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  • #372
arivero said:
To put more intrigue, the mass difference between eta and the average of pion and muon (say, diff=427.2 MeV) also fits roughly in the obvious permuted formulae:
<br /> \sqrt {m_\mu} \sqrt {m_\tau - m_e} \approx \sqrt {m_\tau} \sqrt {m_\mu - m_e}<br /> \approx \sqrt {m_\tau \pm m_e} \sqrt {m_\mu \pm m_e}<br /> \approx \sqrt {m_\mu} \sqrt {m_\tau} = 433.27 MeV<br />
Hmm, Gell-Mann Okubo value for unmixed \eta_8 is
569.32 GeV, so \eta_8 - \pi^0 = 434.34 MeV
 
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  • #373
arivero said:
Hmm, Gell-Mann Okubo value for unmixed \eta_8 is
569.32 GeV, so \eta_8 - \pi^0 = 434.34 MeV

For amateurs, it could be worthwhile to explain what the Gell-Mann Okubo is, or refer to a textbook. I like Donoghue Golowich Holstein, "Dynamics of the Standard Model". The formula appears in chapter VII, expression (1.6b). You get the formula from the following set of equations:
m^2_\pi=B_0 (m_u + m_d)
m^2_{K^0}=B_0 (m_s + m_d)
m^2_{K^\pm}=B_0 (m_s + m_u)
m^2_{\eta_8}=\frac 13 B_0 (4 m_s + m_u + m_d )
and so on.

Asuming isospin, up and down have the same mass, and thus you can get a combination of neutral kaon, pion and eta8.

If works well with the neutral particles; it is not only that it does not account for isospin; the idea does not account for EM interactions neither. Old timers extract an extra EM relation via "Dashen's theorem", but I think to remember there was some work of Witten or some other genious about this kind of corrections.

EDITED: Indeed we could use the above expressions to reformulate our equations in terms of the mass m_s and \hat m \equiv m_u = m_d, with SU(3) flavour breaking to global SU(2) isospin x U(1) as it happened in the papers of 1960s on global symmetries.

m^2_\pi = (m_\mu + \sqrt { m_e (m_\tau-m_\mu)})^2 = B_0 \hat m
m^2_{\eta_8} = (m_\pi + \sqrt { m_\mu (m_\tau-m_e)})^2 = \frac 23 B_0 (2 m_s + \hat m)
Here you can see also one of the themes which were debatable in the sixties: the use of mass square instead of plain mass. For instance, it is because of it that our resulting equations
do not allow to cancel B_0 out.
 
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  • #374
taarik said:
Here is another surprise
the t lepton mass can be obtained by taking 57 jumps of 29.318 MeV from the muon mass i.e t mass= 57x 29.318. Now this 57 number...

http://adsabs.harvard.edu/abs/1962RSPSA.268...57D

Dirac gets 53 times the electron mass for the muon, in a paper that has been lately recalled by the guys working on strings and branes.
 
  • #375
Hans de Vries said:
A pretty amazing coincident isn't it? The relation:

\mbox{\Huge $\frac{m_{\pi^\pm}}{m_{\pi^0}}\ =\ 1+ \left(\frac{m_\mu}{m_Z}\right)^\frac{1}{2}\ =\ 1.0340344(55)$}

following from the previous post is as exact as:

1 : 1.0000067 (42)


About this one, it can have interesting implications: if the mass of the muon goes to zero, assume so it happens with the masses of up and down, then (global) isospin symmetry is restored. On the contrary, if Z goes to cero (and W) but the quark masses are different, the restored symmetry is only gauge G-W-S isospin and the charged pion decays into the neutral one. What is amazing in this argument is that if Z goes to infinity the two pions can not tunnel one into another, but from the point of view of the electroweak scale the masses of quarks are negligible, thus global isospin is restored again.

The traditional current algebra formula for the pion mass(^2) difference puts it in terms of the fine structure constant and the pion decay constant, e^2 / F_\pi^2 times some other factors.

Entering the octet, we are touching deep problems of the elders. There is a short work of witten http://www.slac.stanford.edu/spires/find/hep/www?j=PRLTA,51,2351 about how the mass of the charged pion must always be higher than the neutral pion, even if only to avoid tachions in the limit of zero pion mass. Also, the mixing between \eta_8[\itex] and \eta_0[\itex] to give \eta[\itex] and \eta&amp;amp;amp;#039;[\itex] was the U(1) headache, addressed by t&amp;amp;amp;#039;Hoft, http://www.slac.stanford.edu/spires/find/hep/www?j=NUPHA,B159,213 and http://www.slac.stanford.edu/spires/find/hep/www?j=NUPHA,B156,269 independently, and according Okubo still unclear. I have found even some recent work in the context of strings: http://www.slac.stanford.edu/spires/find/hep/wwwrefs?key=5864437
 
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  • #377
  • #378
Count Iblis said:
You could try to use the LLL algorithm to find formulae.
It is an interesting idea. Actually, the team of http://crd.lbl.gov/~dhbailey/dhbpapers/ have tried in the past to input the standard model masses etc in some of their algorithms; but I had no idea about their new work using LLL.
 
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  • #379
Actually, the LLL algorithm is a bit older than Bailey's PSLQ algorithm. The LLL is a little less efficient, but that's only a problem if you have thousands of significant digits of some number and try to find a formula in terms of known constants for it.

For your work, the LLL may be better, because with the LLL you can look for simultaneous relations, the PSLQ can't do that (at least not in general, you can work with complex numbers, quaternions,... but you'll reach a limit beyond which you can't go). If you have 5 numbers that are known to ten digits then you have 50 digits of information. Ten digits may not be enough to detect a relation, but 50 may be enough provided, of course, that the five numbers are given by formulae of the same form that are specified by the same constants.

See the Appendix of this article for details.
 
  • #380
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  • #381
I found that your link gave a good and understandable explanation.
However,
p. 9
...They can be categorised as three coupling constants,
and three mass ratios, and their empirically determined numerical
values are approximately:12
gS .=4, e.=1/12, mN.=1/2× 10−10
The values of the coupling constants are rather more familiar in their
squared forms: thus we have the gravitational fine structure constant
… the ordinary (electromagnetic) fine structure constant, e2 .= 1/137, and …
----------
I need explanation with e.=1/12 and , e2 .= 1/137
I always thought that 12X12=144 not 137
jal
 
  • #382
Well, I guess he takes the nearest simple fraction to sqrt(137), and he chooses 1/12. So it really compares 12 against 11.7, arguably not so bad.

A peculiarity of this article is that a couple ways ago I have been told that the relationship between mass of electron and pion (!) had circulated as a conjecture in the sixties. I though it came from McGregor, but it is formula (6) in pg 11 of this paper.
 
  • #383
"So it really compares 12 against 11.7, arguably not so bad."
If you find anything else that can help I'll put it in "How to build a universe"
jal
 
  • #384
I've had some interesting results in rewriting the Koide equation as a sort of "field energy" equation. The idea is to treat the square in the mass as coming from the energy of a field.

Field energies are quadratic, for instance, E&M field gives mass as m = (E^2 + B^2)/c^2, where I've left off some units. So begin with electromagnetism as a toy example.

Then the thing to notice is that E and B end up quantized at different amplitudes. Magnetic monopoles are much heavier than electrons, so assume that when you quantize E and B, the contribution of B dominates, giving you m = B^2/c^2.

From there, you assume that the angle I've called "delta" is 2/9exactly, and that the reason this doesn't exactly fit the electron, muon and tau masses is cause "E" does contribute slightly.

That converts the Koide formula from being a two parameter fit, with mu and delta, (the mass scale and the angle), to being a two parameter fit with a B scale and an E scale. To write the masses we have

m_n = |B|^2(1 + \sqrt{2}\cos(2/9+2n\pi/3))^2 + |E|^2(\sqrt{2}\sin(2/9+2n\pi/3))^2

where B and E are constants. The contribution to B is split into two parts, 1 + \sqrt{2}\cos(2/9+2n\pi/3), so we write B_v = B, B_s = \sqrt{2}\cos(2/9+2n\pi/3), and E_s=\sqrt{2}\sin(2/9+2n\pi/3). That is, the "v" field is a valence field that is shared between the electron, muon, and tau, and the "s" field is a sea field that distinguishes the three generations.

Then the mass equation is m = (B_v + B_s)^2 + E_s^2.

What's more interesting is that if you write down the vectors (B_v,B_s,E_s) for the electron, muon, and tau, you get the tribimaximal mixing matrix (after scaling the B stuff and E stuff so that each vector has length 1).

Another way of saying this is that the vectors (B_v,B_s,E_s) are orthogonal. Making them orthonormal defines the tribimaximal neutrino mixing matrix. (Except that when you see it in the literature, it is usually has two columns reversed so you should put the three contributions in the order (B_s,B_v,E_s) instead.)

Using the best PDG numbers for the electron and muon masses to predict the tau mass, the equations for the charged lepton masses are (ignore the precision, I haven't had time to compute the ranges and fix everything up yet):

\begin{array}{rcl}<br /> m_n &amp;=&amp; 313.8561002547\;\textrm{MeV}\;(1 + \sqrt{2}\cos(2/9 + 2n\pi/3))^2\\<br /> &amp;&amp;+4.6929703\;\textrm{eV}\; (\sqrt{2}\sin(2/9+2n\pi/3))^2<br /> \end{array}

And the three vectors (which ignore the phase angle 2/9 because it is presumably canceled in the neutrinos) are:
\begin{array}{ccc}<br /> (1,&amp; \sqrt{2},&amp; 0)\\<br /> (1,&amp; -\sqrt{2}/2,&amp; +\sqrt{3/2})\\<br /> (1,&amp; -\sqrt{2}/2,&amp; -\sqrt{3/2})<br /> \end{array}

In the above, note that the angle 2/9 has been removed as it is presumably canceled in the neutrinos, which also use a similar angle. And the scaling to B and E has been removed because in computing phases, one needs to normalize by particle number rather than energy.

After dividing by the lengths of the vectors, sqrt(3), and turning the three vectors into a matrix, one has:
\left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; \sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)


Carl

Koide paper giving Tribimaximal mixing matrix, see eqn (3.2):
http://arxiv.org/abs/hep-ph/0605074
 
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  • #385
CarlB said:
After dividing by the lengths of the vectors, sqrt(3), and turning the three vectors into a matrix, one has:
\left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; \sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)

Carl

Koide paper giving Tribimaximal mixing matrix, see eqn (3.2):
http://arxiv.org/abs/hep-ph/0605074

Interesting, I noticed that one can write the above tribimaximal
matrix for neutrino mixing as the x,y,z coordinates of a tetrahedron
with sides of \sqrt{1/2} with its top down and the origin in h/2,
thus:

\left(\begin{array}{ccc}<br /> z_1,&amp; y_1,&amp; x_1\\<br /> z_2,&amp; y_2,&amp; x_2\\<br /> z_3,&amp; y_3,&amp; x_3<br /> \end{array}\right)\ =\ <br /> \left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; \sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)

The angle of 2/9 radians is then a simple rotation around the z-axis
to get your form of Koide's lepton mass formula:

\sqrt{m_n}\ =\ \sqrt{1/3} + \sqrt{2/3}\cos(2/9 + 2n\pi/3)I see that this "A4-symmetry" was already found here by Ma:
http://arxiv.org/PS_cache/hep-ph/pdf/0606/0606024v1.pdf

and that there is another group X24 which is larger which could
incorporate quarks here:
http://aps.arxiv.org/PS_cache/hep-ph/pdf/0701/0701034v3.pdf

On the other hand, Garrett Lisi uses a 3d quark matrix here:
http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.0770v1.pdf
https://www.physicsforums.com/showthread.php?t=196498

which is the same except for some coordinate switching:

\left(\begin{array}{ccc}<br /> -\sqrt{1/3},&amp; -\sqrt{1/3}, &amp; -\sqrt{1/3} \\<br /> -\sqrt{1/2},&amp; +\sqrt{1/2}, &amp; 0 \\<br /> -\sqrt{1/6},&amp;-\sqrt{1/6},&amp; \sqrt{2/3}<br /> \end{array}\right)

see (2.4) in the paper and also page 18

Regards, Hans
 
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  • #386
Hans,

I ran into the Garrett matrix at Bee's blog:
http://backreaction.blogspot.com/2007/11/theoretically-simple-exception-of.html

Of all the places where it comes up, the physically most convincing one to me is the neutrino mixing angle. If I had known in advance when I was writing the above post that I would get that matrix, I might have started from that end.

I need to go back and look at the baryon masses and see if this sort of formula fits them better.
 
  • #387
CarlB said:
Hans,

I ran into the Garrett matrix at Bee's blog:
http://backreaction.blogspot.com/2007/11/theoretically-simple-exception-of.html

Of all the places where it comes up, the physically most convincing one to me is the neutrino mixing angle. If I had known in advance when I was writing the above post that I would get that matrix, I might have started from that end.

I need to go back and look at the baryon masses and see if this sort of formula fits them better.
One can see the matrix elements as vertices of the unit cube as well,
with one vertex at 0.0 and the three nearest given by:

\left(\begin{array}{ccc}<br /> z_1,&amp; y_1,&amp; x_1\\<br /> z_2,&amp; y_2,&amp; x_2\\<br /> z_3,&amp; y_3,&amp; x_3<br /> \end{array}\right)\ =\ <br /> \left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; +\sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)

Connected to the Lepton mass ratios by a rotation around the z-axis:

\left(\begin{array}{c}<br /> \sqrt{m_\tau} \\<br /> \sqrt{m_\mu} \\<br /> \sqrt{m_e}<br /> \end{array}\right)\ =\ C<br /> \left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; +\sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)\left(\begin{array}{c}<br /> 1 \\<br /> \cos(2/9) \\<br /> \sin(2/9)<br /> \end{array}\right)

with "2/9" replaced with 0.222222047168 (465) one gets
the precise lepton mass ratios, as we know since your post here:

https://www.physicsforums.com/showthread.php?t=46055&page=8


<br /> \begin{array}{llll}<br /> \mbox{equation:} &amp;<br /> \mbox{\huge $\frac{m_\mu}{m_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_e}$}\ =\ 3477.441653 (83) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_\mu}$}\ =\ 16.818061210 (38) \\ <br /> \mbox{experim:} &amp;<br /> \mbox{\huge $\frac{m_\mu}{m_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_e}$}\ =\ 3477.48 (57) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_\mu}$}\ =\ 16.8183 (27) <br /> \end{array}<br />

Well within experimental precision

Regards, Hans
 
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  • #388
The value 0.222222047168 (465) does very nice indeed
as the Cabibbo angle as originally guessed.<br /> \mbox{Cabibbo-Kobayashi-Maskawa:}\ \ \left(<br /> \begin{array}{lll} 0.9753 &amp; 0.221 &amp; 0.003 \\ 0.221 &amp; 0.9747 &amp; 0.040 \\ 0.009 &amp; 0.039 &amp; 0.9991 <br /> \end{array}<br /> \right)<br />

<br /> \left(<br /> \begin{array}{lll} \cos(2/9) &amp; \sin(2/9) &amp; 0 \\ \sin(2/9) &amp; \cos(2/9) &amp; 0 \\ 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)\ \ \ =\ \ \<br /> \left(<br /> \begin{array}{lll} 0.9754 &amp; 0.2204 &amp; 0.000\ \\ 0.2204 &amp; 0.9754 &amp; 0.000 \\ 0.00 &amp; 0.00 &amp; 1.000<br /> \end{array}<br /> \right)<br />Regards, Hans

http://en.wikipedia.org/wiki/CKM_matrix
 
  • #389
We can put this all in a picture (see below) like this:

Place the charged leptons in a 3d coordinate space using the
tribimaximal neutrino mixing matrix: The coordinates determine
the percentage of neutrino mass eigen-states each neutrino
flavor has, with the matrix mirrored, swapping e and \tau

\left(\begin{array}{ccc}<br /> \tau_z &amp; \tau_y&amp; \tau_x\\<br /> \mu_z &amp; \mu_y&amp; \mu_x\\<br /> e_z&amp; e_y&amp; e_x<br /> \end{array}\right)\ =\ <br /> \left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; +\sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)

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


Now:

1) The projections P on the vector (sin θ, cos θ, 1) lead to the
exact charged lepton masses if we use 0.22222204717 (47) for θ,
(the Cabibbo angle for Quark mixing ?)

<br /> \begin{array}{llll}<br /> \mbox{equation:} &amp;<br /> \mbox{\huge $\frac{P^2_\mu}{P^2_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{P^2_\tau}{P^2_e}$}\ =\ 3477.441653 (83) &amp;<br /> \mbox{\huge $\frac{P^2_\tau}{P^2_\mu}$}\ =\ 16.818061210 (38) \\ \\ <br /> \mbox{experim:} &amp;<br /> \mbox{\huge $\frac{m_\mu}{m_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_e}$}\ =\ 3477.48 (57) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_\mu}$}\ =\ 16.8183 (27) <br /> \end{array}<br />



2) The projections P obey the Koide relation (exact for any θ):

\left(\ P_e+P_\mu+P_\tau\ \right)^2\ =\ \frac{3}{2}\left(\ P^2_e+P^2_\mu+P^2_\tau\ \right)


3) The coordinates could even be real coordinates using the
Pauli-Weisskopf interpretation of the wave function as a continous
charge-spin density distribution:

The angles with the z-axis of the charged leptons are equal to the
precession angle of spin 1/2 particles and the precessing speed would
be equal to phase frequency of the charged leptons if the torque is 2.
Also, the angle of (sin θ, cos θ, 1) with the z-axis is same as the
precessing angle of a spin 1 vector boson.


Regards, Hans
 

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  • #390
I loved the thumbnail, how did you do it?

I'm still mulling over the concept of torque here. I spent the weekend making a java applet that graphs the discrete Fourier transform of the baryons. I got the graphical user interface (GUI) to run, but didn't see the patterns I expected, just noise. That could be defects in the program, defects in my understanding of how to use it, etc.

The reason for looking at discrete Fourier transforms with respect to masses was given by Marni Sheppeard here:
http://arcadianfunctor.files.wordpress.com/2007/11/fouriermass07.pdf
 
  • #391
CarlB said:
I loved the thumbnail, how did you do it?

It's done with Povray, The projections are just shadows, they come
basically for free. One could also draw the 3D object multiple times with
each time one of the coordinates fixed. You would get "2.5 D" projections
with the balls and cylinders all aligned in the same 2D plane. Might give
a nice effect as well.

CarlB said:
I'm still mulling over the concept of torque here. I spent the weekend making a java applet that graphs the discrete Fourier transform of the baryons. I got the graphical user interface (GUI) to run, but didn't see the patterns I expected, just noise. That could be defects in the program, defects in my understanding of how to use it, etc.


If the input is a number of mass spikes then the "noise" may well
be the correct result.


Regards, Hans.
 
  • #392
Me = sqr((h/(2*Pi))*C/(4*Pi*G)) * Exp( - 16 * Pi)

Me = 9.08086 * 10 ^ -31 kg

it`s not for real electron, bat for "naked" electron
Soshnikov_Serg
 
  • #393
Soshnikov_Serg said:
Me = sqr((h/(2*Pi))*C/(4*Pi*G)) * Exp( - 16 * Pi)
Indeed even Polchinski book explains about this kind; one expects that Me*sqrt(G)*Exp(1/alpha or something so) to be of order unity, if G is the ultimate cutoff for electroamgnetism.
 
  • #394
Cleaned up + Neutrino masses preliminary.

The charged leptons are placed in a 3d coordinate space using the
tribimaximal neutrino mixing matrix: The coordinates determine the
percentage of neutrino mass eigen-states each neutrino flavor has.

\left(\begin{array}{ccc}<br /> e_z&amp; e_y&amp; e_x \\<br /> \mu_z &amp; \mu_y&amp; \mu_x\\<br /> \tau_z &amp; \tau_y&amp; \tau_x<br /> \end{array}\right)\ =\ <br /> \left(\begin{array}{ccc}<br /> \sqrt{1/3},&amp; +\sqrt{2/3},&amp; 0\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; +\sqrt{1/2}\\<br /> \sqrt{1/3},&amp; -\sqrt{1/6},&amp; -\sqrt{1/2}<br /> \end{array}\right)

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

1) The projections P on the vector (cos θ, sin θ, 1) lead
to the exact charged lepton masses if we use for the angle:

θ = 7/6 pi + 0.22222204717 (47)

<br /> \begin{array}{llll}<br /> \mbox{equation:} &amp;<br /> \mbox{\huge $\frac{P^2_\mu}{P^2_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{P^2_\tau}{P^2_e}$}\ =\ 3477.441653 (83) &amp;<br /> \mbox{\huge $\frac{P^2_\tau}{P^2_\mu}$}\ =\ 16.818061210 (38) \\ \\ <br /> \mbox{experim:} &amp;<br /> \mbox{\huge $\frac{m_\mu}{m_e}$}\ =\ 206.7682838 (54) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_e}$}\ =\ 3477.48 (57) &amp;<br /> \mbox{\huge $\frac{m_\tau}{m_\mu}$}\ =\ 16.8183 (27) <br /> \end{array}<br />
2) The projections P obey the Koide relation always
independent of the angle θ, while one of the projections
can become negative. ( to please Carl :^) )

\left(\ P_e+P_\mu+P_\tau\ \right)^2\ =\ \frac{3}{2}\left(\ P^2_e+P^2_\mu+P^2_\tau\ \right)

Angle θ with pointer inside the cube: all projections positive.
Angle θ with pointer outside the cube: one projection negative.

The latter is a requirement for Koide's relation to describe the
neutrino mass-eigen-states. This setup does so naturally.
There are the following regions for θ:

075 - 105 degrees: inside
105 - 195 degrees: outside
195 - 225 degrees: inside
225 - 315 degrees: outside
315 - 345 degrees: inside
345 - 075 degrees: outside3) What fixes the angle θ? For the charged leptons we have the
following numerical coincident involving the projection of tau:

The maximum value of a projection is 0.9855985... is (1+\sqrt{2})/\sqrt{6}

P_{max}^2\ =\ 0.97140452079...

P_\tau \quad \ =\ 0.97140158810...

It has to be said that the masses are still very sensitive to the very small
error (The muon mass uncertainty determines the uncertainty because of
Koide's relation)
4) The coordinates could even be interpreted as real coordinates
with the Pauli-Weisskopf interpretation of the wave function as a
continuous charge-spin density distribution:

The angles with the z-axis of the charged leptons are equal to the
precession angle of spin 1/2 particles and the precessing speed would
be equal to phase frequency of the charged leptons if the torque is 2.
Also, the angle of (cos θ, sin θ, 1) with the z-axis is same as the
precessing angle of a spin 1 boson. In other words: A spin coupling
like adaption of the Yukawa coupling.Hypothetical Neutrino mass-states eV from here: (page 48)
http://arxiv.org/abs/hep-ph/0603118
are used in the drawing below.Regards, Hans
 

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  • #395
On the subject of applying Koide's mass formula to the baryon resonances, I've added a <a href="http://carlbrannen.wordpress.com/2007/12/16/regge-trajectories-and-koides-formula/">blog post</a> on how this fits in with Regge trajectories.

The short version is that Regge trajectories look like M = sqrt(L), where M and L are mass and angular momentum. This comes from an assumption of flux tubes that have energy (and therefore mass) proportional to their length R, but angular momentum proportional to R^2. These flux tubes have different energies per unit length. The formula applies to baryons with the same quantum numbers except for angular momentum.

Koide's mass formula looks like M = E^2 where E is a field strength, and it applies to baryons with identical quantum numbers. That is, it applies to groups of three resonances that share identical angular momentum.

If you combine the two mass formulas, you end up concluding that the flux tubes that give the Regge trajectories have diameters that are proportional to the Koide field strength.
 
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  • #397
Soon I am going to release a paper with some new coincidences based on square roots of masses. Here's an example.

Let \lambda_{en} be the (positive) square root of the mass of the nth electron, that is the square roots of the masses of the electron, muon, and tau. The Koide equation can be written as

\lambda_{en} = 25.0544\sqrt{\textrm{MeV}}(\sqrt{1/2} + \cos(2/9 + 2n\pi/3)\;)

The lightest meson is the pion. It comes in three varieties (with the same quantum numbers), the pion, pi(1300), and pi(1800). The square roots of their masses are given by an equation similar to the above:

\lambda_{\pi n} = 25.0544\sqrt{\textrm{MeV}}(6/5 -3/4 \cos(2/9 + 2n\pi/3)\;)

Accuracy is very high.

To express the relationship as a linear one in terms of square roots of masses, we have:

4\lambda_{\pi_1} + 3\lambda_{e_3} = <br /> 4\lambda_{\pi_2} + 3\lambda_{e_2} = <br /> 4\lambda_{\pi_3} + 3\lambda_{e_1} = \lambda_{\pi_1} + \lambda_{e_3}<br /> =\sqrt{138}+\sqrt{1777}\;\sqrt{\textrm{MeV}}.

Linear relationships on mass are suggestive that the elementary particles are collections of objects that don't interact enough to completely change their character. An example is the masses of the atomic nuclei. The number of nucleons is approximately proportional to the mass of the nuclei (and atom). And nuclei mass can be thought of as mostly having to do with neutrons and protons and only a little to do with the force that keeps them together.

Linear relationships on square root of mass are suggestive that the mass (or energy) comes from an object which is linear but is proportional to the square root of energy. For example, the energy in a magnetic field is:
\int\;|\vec{B}|^2\;d^3r
The magnetic field is a linear object. That is, if two objects each have magnetic fields and they are superimposed, then by linear superposition, the total magnetic field is the sum of the magnetic fields. The energy in the field is proportional to the square of the field and so if we wish to do a linear operation on the object creating the energy / mass, we need to first take the square root of the energy / mass.

The first thing a physicist does to a linear data stream is to take the Fourier transform of it. For a discrete set of 3 masses this would be the discrete Fourier transform. Marni Sheppeard wrote a short paper showing that the Koide formulas are related to discrete Fourier transforms:

http://arcadianfunctor.files.wordpress.com/2007/11/fouriermass07.pdf

I'll eventually get back with more.
 
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  • #398
Carl, if you are going to review the pion thing, you could perhaps to include also Hans #349 ff and Taarik post #366.
 
  • #399
Arivero,

I'm looking at things that relate the generation structure of the fermions with the excitation structure of the mesons and baryons through square roots. The use of square roots is supposed here to get the situation to one where there is something linear going on.

Hans #349 is possibly appropriate because it has a square root in it, but it's written as ratios. Maybe I could rewrite it in square root form. Hmmm, let's see. Hans has, writing things in square root mass terms:
\lambda^2_{\pi +}/\lambda^2_{\pi 0} = 1 + \lambda_\mu/\lambda_Z
Multiply by the square root of the mass of the Z:
\lambda_Z (\lambda^2_+/\lambda^2_0) = \lambda_Z + \lambda_\mu
The right hand side is nice because it is linear in square roots but the left hand side is not. So rewrite the ratio of the pion masses..

\lambda_Z(1 + (\lambda^2_{+}-\lambda^2_0) / \lambda^2_0) = \lambda_Z + \lambda_\mu

\lambda_Z(\lambda^2_{+}-\lambda^2_0) = \lambda_\mu \;\lambda^2_0

Write \lambda_+ = \lambda_0 + \lambda_Q where Q is the contribution to the mass field that comes from charge and is small compared to \lambda_0. Keeping first order in Q we have:

\lambda_Z\; (2 \lambda_Q) = \lambda_\mu \;\lambda_0

which is not quite linear in square root mass.
 
  • #400
I should write down my derivation of the pi meson mass formula.

The pi+ is made from a up quark and an anti-down quark. They have to have opposite colors, but other than that, if you know the quantum numbers of one, you know the quantum numbers of the other.

Arbitrarily, consider the QM problem of an up quark moving in the field of an anti down quark. Make this a qubit kind of problem by ignoring position and momentum information. Then the only thing the up quark can do is change its color. The anti down quark will also change color, but this is defined by conservation laws and so we don't need to worry about it; if the up quark is blue, then the anti down quark is anti blue.

Over the long term, just looking at the up quark, there are nine transitions going on, from {R, G, B} to {R, G, B}. This amounts to a scattering matrix. We have nine amplitudes, write them as a matrix:
\left(\begin{array}{ccc}RR&amp;RG&amp;RB\\GR&amp;GG&amp;GB\\BR&amp;BG&amp;BB\end{array}\right)

SU(3) is an unbroken symmetry so we can assume RR = GG = BB, as well as RG = GB = BR and GR = BG = RB. This requires the matrix to be circulant. Let I = RR, J = RG, K = GB, so the matrix has to be of the form:
\left(\begin{array}{ccc}I&amp;J&amp;K\\K&amp;I&amp;J\\J&amp;K&amp;I\end{array}\right)

The RG and GR amplitudes are the time reversals of each other. It follows that they must be complex conjugates, that is, K = J*. That makes the above circulant matrix Hermitian.

Suppose that J and K are nonzero, but I is zero. Is this possible?

The action of J on an arbitrary state is to increment its color (that is, if you think of the three scattering matrix terms that all must be equal to J, their action can apply to any state and increments the color). The action of K is to decrement its color. Since both these processes are possible, it is also possible to have one followed by the other. Such a process would leave the colors unchanged. Therefore I cannot be zero.

This is an argument similar to the one Feynman made when he contemplated what happens when you insert an intermediate state between the initial and final states; the path integral formulation says that you have to sum over the intermediate states.

Suppose my initial state is G and my final state is B. Following Feynman, we insert an intermediate state, which can be R, G, or B. Then we sum over intermediate states. We find that:

GB = GR RB + GG GB + GB BG

We can insert intermediate states between all the other 8 amplitudes. We end up with a full set of nine equations:

RR = RR RR + RG GR + RB BR,
RG = RR RG + RG GG + RB BG,
RB = RR RB + RG GB + RB BB,
GR = GR RR + GG GR + GB BR,
GG = GR RG + GG GG + GB BG,
GB = GR RB + GG GB + GB BB,
BR = BR RR + BG GR + BB BR,
BG = BR RG + BG GG + BB BG,
BB = BR RB + BG GB + BB BB.

We can conveniently rewrite the above equations in matrix form:
\left(\begin{array}{ccc}RR&amp;RG&amp;RB\\GR&amp;GG&amp;GB\\BR&amp;BG&amp;BB\end{array}\right)<br /> = \left(\begin{array}{ccc}RR&amp;RG&amp;RB\\GR&amp;GG&amp;GB\\BR&amp;BG&amp;BB\end{array}\right)^2

or

\left(\begin{array}{ccc}I&amp;J&amp;K\\K&amp;I&amp;J\\JK&amp;I\end{array}\right)<br /> = \left(\begin{array}{ccc}I&amp;J&amp;K\\K&amp;I&amp;J\\J&amp;K&amp;I\end{array}\right)^2

The above has three solutions:
\begin{array}{rcl}<br /> I&amp;=&amp;1/3,\\<br /> J&amp;=&amp;\exp(+2n\pi/3)/3,\\<br /> J&amp;=&amp;\exp(-2n\pi/3)/3\end{array}
for n = 1,2,3.

This is good because there are exactly three pi mesons. Let's see if we can get Koide's formula for them.

We have nine processes describing the meson, that is, 3 copies of I, 3 copies of J, and 3 copies of K. The J and K processes are related by time reversal, but we really don't know how these relate to the I process. To convert these nine processes into a description of the meson, let's suppose that two constants convert the above amplitudes into field amplitudes, say "v" for the I processes and "s" for the J and K processes. The J and K processes are complex conjugates and are complex numbers. We also don't know how to convert these complex numbers into field strengths. Let's suppose that to convert them into a field strength requires a complex phase (maybe due to non commutativity like Berry phase), so they need to be multiplied by exp(i \delta) for the J and the complex conjugate for the K.

Then the total field strength for the three particles is given by

v + 2 s \cos(\delta + 2n\pi/3)

With delta = 2/9, v = 6/5 and s = -3/4, this is the Koide formula for the lowest energy mesons, the pi mesons. With delta = 2/9, v = sqrt{1/2}, and s = +1, this is the Koide formula for the leptons. More to come later.
 
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