All the lepton masses from G, pi, e

[arivero]

0.000 006 263 813 : Difference between electron and muon anomaly
0.000 000 084 639 : First vacuum polarization term of the electron
----------------------------------------------------------------------
0.000 006 348 452 : Sum of the vacuum polarization terms (the above)

0.000 006 353 732 : (+/-3 000) mass ratio of electron and the W boson.
0.000 000 002 998 : uncertainty from the Z mass

Hmm, I had tried the same with the data truncated at two loops for electron and three for the muon and I got a somehow weaker result. It seems one needs to add the three loop data for the electron, but it is a mess because then we have three new diagrams to exclude.

 

[Hans de Vries]

The other interesting observation is that the missing term
becomes equal in both cases to within experimental value.


0.00115877693 : mu/mZ + VP2
0.00115965218 : electron anomaly
0.00000087525 : missing

0.00116504602 : mu/mZ + me/mw (+ VP2 - VP2)
0.0011659208_ : muon anomaly
0.00000087478 : missing

0.00000000047 : missing1 - missing2
0.00000002668 : uncertainty due to Z (cancels if missings are subtracted)
0.00000000299 : uncertainty due to W mass


So there may be a single missing term.


Regards, Hans

PS: VP2 = 0.00000008464 : First vacuum polarization term of the electron
which is a second order term.

PPS: I'm not entirely sure if the term 0.00116504602 : mu/mZ + me/mw (+ VP2 - VP2)
should indeed not include VP2.

me/mW - VP2 = difference between muon and electron anomaly.
mu/mZ + VP2 = all self energy terms + first vacuum polarization term of the electron anomaly

 

[arivero]

Making it in a reverse way: as the missing term is already in the electron anomalous moment, assume it is sort of square of the first term. So it is mu^2/X^2. Solve for X:
X=sqrt(mu^2/.000000875015)=112.95 GeV

The quantity is interesting in two ways. We suspect of a neutral scalar H0 at 115 GeV, and it could have this role. But also mw*sqrt(2) is 113.87 GeV, so we can use the W particle, which was my first attempt for the missing term. Lacking of more theory, both are equally suitable: values up to 114.5 GeV are covered by the Z indeterminacy. The first has the advantage of not using an arbitrary 1/2 coefficient and it is neutral as the Z, but it has not been discovered (yet?), the second is an already discovered particle but we have used it for the "vacuum polarised graphs", and it is surprising to have it here too, even if squared.

 

[arivero]

References

For the sake of completeness, references. The industry of calculation of the anomalous moment seems to be based in Cornell, around a veteran named T. Kinoshi-ta. Other group does exist in North America around A. Czarnecki

http://arxiv.org/abs/hep-ph/9810512, from Czarnecki and Marciano, is the main entry point for the calculation up to order alpha^4. It is regretly a short preprint and it does not separate loop by loop, so one is referred to more detailed bibliography, which is not in the arxiv

The five diagrams for order alpha^2 appear well separated in Levine and Wright Phys. Rev. D 8, 3171-3179 (1973) http://prola.aps.org/abstract/PRD/v8/i9/p3171_1. I got from here the specific value we were using above.

Also some sums for 40 diagrams of order alpha^3 are presented there. Note that of these, 12 diagrams are vacuum polarisation loops, amounting perhaps to a contribution of 0.37 (alpha/pi)^3

The alpha^2 "polarisation loop", depending of the mass quotient of the external and internal lepton, is studied both analytic and numerically by Li Mendel and Samuel, Phys. Rev. D 47, 1723-1725 (1993) http://prola.aps.org/abstract/PRD/v47/i4/p1723_1

Also Samuel Li and Mendel provide a calculation of tau lepton, Phys. Rev. Lett. 67, 668-670 (1991) http://prola.aps.org/abstract/PRL/v67/i6/p668_1 up to order alpha^3. For this lepton, the contributions for quarks are already noticeable at this scale.

 

[Jay R. Yablon]

Do you habe any ideas about the tau lepton?

Also Samuel Li and Mendel provide a calculation of tau lepton, Phys. Rev. Lett. 67, 668-670 (1991) http://prola.aps.org/abstract/PRL/v67/i6/p668_1 up to order alpha^3. For this lepton, the contributions for quarks are already noticeable at this scale.

Alejandro"

Do you and Hans have any close fits relating the tau magnetic moment with any lepton-to-electroweak-boson ratio?

Jay.

 

[arivero]

Alejandro"
Do you and Hans have any close fits relating the tau magnetic moment with any lepton-to-electroweak-boson ratio?

Hmm the answer past yesterday was yes, the answer today is more towards no. On one side the quantity m_e m_\tau/m_W^2 in of the right order of magnitude to do further corrections in our calculations, but we do not need it anymore, giving the uncertainty in the mass of Z. On other hand, and more concretely answering your question, the difference between anomalous moment of mu and tau can only be covered with a new quotient m_e/m_{X^+}, and the mass of the new X+ particle should be around 68-70 GeV. At these energy range, the LEP2 presented an slight "statistical" deviation, but no particle

 

[arivero]

mw=80525 (+-38)

80425

So I was getting a discrepance between calculations at home and calculations at work.

 

[Jay R. Yablon]

Mass Term Gordon-Like Magnetic Moment Decomposition

Dear Alejandro and Hans:

I just posted to my website http://home.nycap.rr.com/jry/FermionMass.htm , a Gordon-like decomposition of the Fermion mass into a term due to electromagnetic charge, and a term due to magnetic moment. This is still an early working draft.

I hope this can help you in your efforts by providing a covariant field theory context for your efforts to characterize the magnetic moments. I know that your efforts have helped me recognize that consideration of magnetic moments is likely to be a critical aspect of what I am attempting to do.



Jay.

 

[arivero]

I just posted to my website http://home.nycap.rr.com/jry/FermionMass.htm , a Gordon-like decomposition of the Fermion mass into a term due to electromagnetic charge, and a term due to magnetic moment. This is still an early working draft.

Probably Hans is also exploring this way, but I am not so optimistic about a direct connection; perhaps a semiclassical effect, could be. But even that is strange to manage. To me, it seems more as if the symmetry breaking mechanism of the electroweak group (and its vacuum value) were needing of the lepton masses is some misterious way.

From our quadratic formulae we can get rather intriguing equations. For instance this one:

{m_\tau\over m_Z} + {m_\mu\over m_W}=<br /> {m_\tau\over m_\mu} a_\mu^{s.e.} + {m_\mu\over m_e} a_\mu^{v.p.}<br />

Where a_\mu^{s.e.},a_\mu^{v.p.} are the self-energy and vacuum polarisation parts of the muon anomalous magnetic moment; note that the v.p. part depends internally of lepton mass quotients, while the s.e. is mass independent, in QED (in the full electroweak theory new dependences appear).

 

[arivero]

Some of the development of the thread uploaded at http://arxiv.org/abs/hep-ph/0503104

 

[Hans de Vries]

Hmm,

I was doing something else and ran just incidently into this one:

\sqrt{ \ 2 \ \frac{m_V}{m_Z} \ \frac{m_{\tau}}{m_e}} \ = \ 137.038 (12)

m_V is the vacuum expectation value of 246.22046 GeV (according to Jay)
The biggest uncertainty is from the tau mass.


Regards, Hans


\ \ \alpha \ \ \ = 1/137.03599911
m_V = 246220.46
m_Z = 91187.6 (+-2.1)
m_τ = 1776.99(+0.29-0.26)
m_e = 0.51099892 (+-0.00000004)

 

[arivero]

vacuum expectation value of 246.22046 GeV (according to Jay)
The biggest uncertainty is from the tau mass.

Also this vacuum should have a high uncertainness. I wonder where did Jay got so many digits from.

 

[Hans de Vries]

Also this vacuum should have a high uncertainness. I wonder where did Jay got so many digits from.

It's basically "one over the square of" : The Fermi Coupling constant 1.16637 (1) times sqrt(2)

So it should be 246.2206 (11)

Regards, Hans.

 

[arivero]

That's just the value you want it to have! Do you have you any more data like that.

Funny, I have found a paper which, while aiming toward other goals, also uses a variant of QED discarding vacuum polarisation terms in order to get more explicit formulas. It happens in section 3 of
http://prola.aps.org/abstract/PR/v95/i5/p1300_1
which is titled "3. EXAMPLE: QUANTUM ELECTRODYNAMICS WITHOUT PHOTON SELF ENERGY PARTS".

The authors are a M.Gell-Mann and a F.E.Low, from Illinois.

 

[Hans de Vries]

Funny, I have found a paper which, while aiming toward other goals, also uses a variant of QED discarding vacuum polarisation terms in order to get more explicit formulas. It happens in section 3 of
http://prola.aps.org/abstract/PR/v95/i5/p1300_1
which is titled "3. EXAMPLE: QUANTUM ELECTRODYNAMICS WITHOUT PHOTON SELF ENERGY PARTS".

The authors are a M.Gell-Mann and a F.E.Low, from Illinois.

There are related follow-ups:

Quantum Electrodynamics at Small Distances,
Baker, Johnson, 1969.
http://prola.aps.org/abstract/PR/v183/i5/p1292_1


Quantum Electrodynamics Without Photon Self-Energy Parts
S. L. Adler and W. A. Bardeen 1971
http://prola.aps.org/abstract/PRD/v4/i10/p3045_1


Regards, Hans

 

[arivero]

Baker and Johnson have actually a whole forest of papers.

Following this review of classic bibliography, I also come across to the formula
<br /> {M_0^2 \over M_V^2}= {3 \over 2 \pi} \alpha<br />
which is world-famous, but I was unaware. Regretly it is about a single scalar charged particle, not a fermion, and the quotient against the vector boson gets this square dependence. The formula was found in
Radiative Corrections as the Origin of Spontaneous Symmetry Breaking by Sidney Coleman and Erick Weinberg, http://prola.aps.org/abstract/PRD/v7/i6/p1888_1 They even have a generalisation to SU(3)xU(1).

Incidentaly, one of these authors was contacted about our preprint 0503104, here is his statement: Given the current state of knowledge in the field, speculations concerning approximate numerical coincidences such as the ones you discuss do not constitute the degree of substantial new physics that is required for publication

 

[CarlB]

Hans, I am impressed. It's too many digits for the accuracy.

If the standard model is an effective field theory from a deeper level, then the fine structure constant should be calculated from a series in that deeper (unified) level.

One supposes that such a unified field theory would be extremely strongly coupled (otherwise it'd be visible), and that our usual perturbation methods would fail, and therefore that calculations would be impossible. However, this is a way out of this.

Our usual experience with bound states is that when two particles are bound together, we expect the bound state to have a higher mass than either of the particles contributing to it. Of course the total mass is a little less than the sum of the masses, E=mc^2 and all that. But if the particles are extremely strongly bound, then the mass of the bound state could be negligible compared to the mass of either free particle.

For example, the mass of a free up quark is unknown, but all indications are that it would require a lot of energy to make one, so its mass should be extremely large. Present experimental limits say it should be much larger than the mass of a proton.

Now doing quantum mechanics in such a nonperturbational region might seem impossible, but this is not necessarily the case. In fact, infinite potential wells make for simple quantum mechanics problems. Perturbation theory may not be needed or appropriate.

Looking at QFT from the position eigenstate representation point of view, the creation operators for elementary particles have to work in infinitesimal regions of space. Suppose we want to do physics in that tiny region. The natural thing we'll do is to use a Gaussian centered at the position.

One way of representing the potential energy between two objects bound by extreme energies is to suppose that they each stress space-time (in the general relativistic stress-energy manner), but in ways that are complementary. Thus the sum makes for less stress to space-time than either of the separate particles.

In that case, if we represent the stress of each particle with a Gaussian, we end up deriving a potential energy that, for very low energies, works out as proportional to the square of distance. This is the classic harmonic oscillator problem, and the solution in QM is well known without any need for perturbation theory.

Now your series for the fine structure constant used a Gaussian form. Coincidence? I doubt it.

My suspicion is that this is a clue. My guess is that there is a unified field theory with equal coupling constants for everything, and that the strong force is strong because it has fewer coupling constants multiplied together in it. That would have to do with the factor in the exponential. This all has to do with my bizarre belief that even the leptons are composite particles.

Carl Brannen

As an aside, I once decided to see if the sum of inverses of cubes \zeta(3) = \sum 1/n^3 could be summed similarly to how the sum of inverses of squares or fourth powers could be summed. I wrote a C++ program that computed the sum out to 5000 decimal places (which requires a lot of elementary mathematics as the series converges very very slowly), and then did various manipulations on it to search for a pattern.

The most useful thing to know, when trying to determine if a high precision number is rational, is the series obtained by taking the fractional part of an approximation and inverting it. It's been over a decade, but I seem to recall that the name for this is the "partial fraction expansion".

 

[arivero]

particle spectrum

Attached (!) I have drawn the whole elementary particle spectrum at logarithmic scale. Honoring Yablon, I have put a 1/137 line also between the tau and the electroweak vacuum.

There are four clearly distinguished zones, usually called the electromagnetic breaking, the chiral breaking, the hadronic scale (or SU(3) gap) and the electroweak breaking scale. SO I have encircled them with green rectangles.

(EDITED: If you are using the Microsoft Explorer, you will need to expand the jpg to full screen or almost)

 

[Jay R. Yablon]

Checking in

Hi Alejandro:

Thanks for the recognition.

While I have stayed off the board for awhile I have not been inactive. I am working on a paper with a well-known nuclear theorist in Europe. I won't get into details yet, but I think you all will find it interesting once we are ready to "go public."



Jay.

 

[arivero]

Still more references, if only for a future observer/reader of the thread...

Stephen L. Adler (of the anomaly fame) pursued during the seventies an eigenvalue condition in order to pinpoint the value of the fine structure constant:

Short-Distance Behavior of Quantum Electrodynamics and an Eigenvalue Condition for alpha, http://prola.aps.org/abstract/PRD/v5/i12/p3021_1 (http://www.slac.stanford.edu/spires/find/hep/www?j=PHRVA,D5,3021 )

PS: specially for new readers, please remember we are trying an at-a-glance view of this thread (and other results) in the wiki bakery http://www.physcomments.org/wiki/index.php?title=Bakery:HdV

 

[arivero]

Koide formula

We missed this one. It is a published formula, its exactitude is one of the better ones in the thread, and a very short review can be checked online as hep-ph/9603369
m_e+m_\mu+m_\tau=\frac 23 (\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})

Or, if you prefer: The vector having by components the square roots of leptonic masses helds and angle of 45 degrees with the vector (1,1,1). If you use experimental input, the angle is 45.0003 plus/minus 0.0012 degrees, according Esposito and Santorelli.

Jay can be specially interested on this property, because it derives almost trivially from asking trace preservation both of the mass matrix M and its square root (the 45 degrees being the maximum possible aperture without negative eigenvalues in square root of M).

EDITED: http://ccdb3fs.kek.jp/cgi-bin/img_index?198912199 is the first paper from Koide. (even before the improved values of tau?) From other references, it seems that the research was framed in the general context of "democratic family simmetry" ( a degenerated mass matrix filled with ones, so that when rotating to eigenvectors only the third generation is naturally massive).

 

[CarlB]

Here's another lepton mass formula, one that gives all the mass ratios, and involves the Cabibbo angle:

Consider the matrix:

M = \left( \begin{array}{ccc}<br /> \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} \\<br /> e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} \\<br /> e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} \\ \end{array} \right)

where \delta = .222 = 12.72 degrees, is the Cabibbo angle. Let

r = e^{2 i \pi/3}, s = 1/r.

Then the eigenvectors and eigenvalues of M are approximately:

(1,r,s), \sqrt{m_e / 157},
(1,s,r), \sqrt{m_\mu / 157},
(1,1,1),\sqrt{m_\tau / 157},

with the lepton masses in MeV.

Note that the sum of the eigenvalues of the matrix M are 3 \sqrt{2}, and the square of this is 18. And M^2 has diagonal entries of 4, so a trace of 12. Since 12/18 = 2/3, the relationship between the lepton masses and their square roots already discussed on this thread is automatically provided by the choice of the diagonal value as \sqrt{2}

The reason for using the Cabibbo angle is that the Cabibbo angle gives the difference between the quarks as mass eigenstates and the quarks as weak force eigenstates. If you believe, as I do, that the quarks and electrons are made from the same subparticles (which I call binons and which correspond to the idempotents of a Clifford algebra), then it is natural that the Cabibbo angle enters into the mass matrix for the leptons. Note that the Cabibbo angle is small enough that its sine is close to the angle.

Carl

 

[Hans de Vries]

Here's another lepton mass formula, one that gives all the mass ratios, and involves the Cabibbo angle:

Consider the matrix:

M = \left( \begin{array}{ccc} <br /> \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} \\ <br /> e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} \\ <br /> e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} \\ \end{array} \right)

where \delta = .222 = 12.72 degrees, is the Cabibbo angle. Let

r = e^{2 i \pi/3}, s = 1/r.

Then the eigenvectors and eigenvalues of M are approximately:

(1,r,s), \sqrt{m_e / 157},
(1,s,r), \sqrt{m_\mu / 157},
(1,1,1),\sqrt{m_\tau / 157},

with the lepton masses in MeV.

Carl

Carl, this is really very interesting.

If one takes 0.222222047168 (465) for the Cabibbo angle then you
get the following lepton mass ratios:

\frac{m_\mu}{m_e}\ =\ 206.7682838 (54)

\frac{m_\tau}{m_e}\ =\ 3477.441653 (83)

\frac{m_\tau}{m_\mu}\ =\ 16.818061210 (38)

Which is well within the experimental range:

\frac{m_\mu}{m_e}\ =\ 206.7682838 (54)

\frac{m_\tau}{m_e}\ =\ 3477.48 (57)

\frac{m_\tau}{m_\mu}\ =\ 16.8183 (27)

The first one is no surprise because we solved the Cabibbo angle
to get this result, but the second surely is! This means that your
formula is predictive to 5+ digits!

Furthermore, the values given also are exact for Koide's formula
which is due to the \sqrt 2's on the diagonal of your matrix as you said:

m_e+m_\mu+m_\tau=\frac{2}{3} (\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2

1+206.7682838+3477.441653=\frac{2}{3} (\sqrt{1}+\sqrt{206.7682838}+\sqrt{3477.441653})^2

Regards, Hans

P.S. The scale factor would be 156.9281952 (123)

 

[arivero]

The first one is no surprise because we solved the Cabibbo angle
to get this result, but the second surely is! This means that your
formula is predictive to 5+ digits!

Hmm? Ah, you mean that you have used Carl's matrix to find a value for the Cabibbo angle, do you? It is actually very surprising to be able to find this angle from leptons; it is the mixing parameter ... of quarks!

Incidentally, the first model that originated Koide's formula also had a prediction for this angle, it was
<br /> \tan \theta_c={\sqrt 3 (x_\mu - x_e) \over 2 x_\tau - x_\mu -x_e}<br />
with x_l\equiv \sqrt{m_l}- Ref http://prola.aps.org/abstract/PRL/v47/i18/p1241_1 Note that the context was similar to Carl's argument, ie a composite model for quarks and leptons.

Surely this result can be combined into Carl's matrix to get Koide's democratic mass matrix (http://prola.aps.org/abstract/PRD/v28/i1/p252_1,http://prola.aps.org/abstract/PRD/v39/i5/p1391_1)
which is more of less similar to Carl's but having \delta=0

 

[arivero]

M = \left( \begin{array}{ccc}<br /> \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} \\<br /> e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} \\<br /> e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} \\ \end{array} \right)

As a matter of notation, I'd call such matrix "A^\frac 12" or something similar, given that its eigenvalues relate to square roots of masses.

 

[ohwilleke]

Keep up the good work! This is really facinating reading.

 

[Hans de Vries]

Hmm? Ah, you mean that you have used Carl's matrix to find a value for the Cabibbo angle, do you? It is actually very surprising to be able to find this angle from leptons; it is the mixing parameter ... of quarks!

I actually liked the close proximity to 2/9 of \delta\ = 0.2222220.. without
the necessity of the latter actually being the angle used to systemize
the quark eigenstate mixing observed in weak hadronic decay.

After all the goal is to reduce the number of arbitrary parameters :^)

Regards, Hans

 

[CarlB]

Hans, thanks for supplying more accuracy. The fit to the lepton masses can't be any better or worse than the 2/3 formula fit. From there, my guess is that the Cabibbo angle fit is at least partly based on chance. That is, if you look at the CKM matrix, the other numbers aren't showing up. Maybe that's due to suppression from the high quark masses, and the Cabibbo angle being close is due to the SU(3) associated with the up, down and strange quarks having roughly similar masses.

By the way, I wonder what happens to the fit if you use \theta_C = 2/9. One wonders if there is a relationship between the masses of the up, down and strange, and the deviation of the formula from 2/9. If there were, then maybe the same relationship would clear up the rest of the CKM matrix.

Arivero, thanks for the references. I'm not associated with a university, so if it doesn't show up on a google search I have to drive over to the University of Washington to look it up. I'll go by there this afternoon, I can hardly wait. And you're right that the formula isn't really any more predictive than the 2/3 formula, except for the Cabibbo angle coincidence.

If one assumes that the leptons and quarks, along with their various families and colors, are made from various combinations of just two subparticles each of which come in three equivalent colors, which I've been calling the |e_r&gt;, |e_g&gt;, |e_b&gt;, |\nu_r&gt;, |\nu_g&gt;, |\nu_b&gt;, then the mass matrix shown is a result of a branching ratio for interactions of the form:

|e_r&gt; -&gt; |e_r&gt; 50%
|e_r&gt; -&gt; |e_g&gt; 25%
|e_r&gt; -&gt; |e_b&gt; 25% (and cyclic in r, g, b)

with the Cabibbo angle providing the phases for the last two interactions. That is, a particle of a given color has a 50% chance of staying that same color, and 25% chances of switching to one of the other two colors. This has a direct interpretation in terms of angles, if you break it up into left and right handed parts.

I should mention what all this has to do with Higgs-free lepton masses.

Consider the Feynman diagrams (in the momentum representation) where each vertex has only two propagators, a massless electron propagator coming in, and a massless electron propagator coming out, and a vertex value of m_e. When you add up this set of diagrams, the result is just the usual propagator for the electron with mass. Feynman's comment on this, (a footnote in his book, "QED: The strange theory of matter and light"), is that "nobody knows what this means". Well the reason that no one knows what it means is because these vertices can't be derived from a Lorentz symmetric Lagrangian.

But what the above comment does show is that it is possible to remove the Higgs from the standard model (along with all those parameters that go with it), if you are willing to assume Feynman diagrams that don't come from energy conservation principles.

You can take the same idea further by breaking the electron into left and right handed parts, and then assuming that the Feynman diagrams always swap a left handed electron traveling in one direction with a right handed electron traveling in the opposite direction. This preserves spin and is similar to the old "zitterbewegung" model of the electron.

If you take the same idea still further, and assume the electron, muon and tau are linear combinations of subparticles, as described above, then the lepton mass formula is natural to associate with the Cabibbo angle. By the way, the standard model includes a Higgs boson to take away the momentum from the left handed electron reversing direction, but its otherwise the same thing.

Mass is weird because it is only the mass interaction that allows left and right handed particles to interact. That's why the Higgs bosons are supposed to be spin-0, to allow the coupling of left and right handed fermions.

The zitterbewegung model was based on noticing that the only eigenvalues of the electron under the operator that measures electron velocity are +c and -c. So the assumption was that the electron moved always at speed c. Of course having the electron suddenly reverse direction in order for the zitterbewegung to work is a violation of conservation of momentum, but the fact is that you do get the usual electron propagator out of all this.

By the way, to do these calculations, it helps to choose a representation of the gamma matrices that diagaonalizes particle/antiparticle and spin-z. That is, the four entries on the spinors will correspond to left-handed electron, right handed electron, left handed positron and right handed positron. When you do this, the mass matrix will no longer be diagonal.

There are a lot of other things that come out of this, and most of them are quite noxious to physicists. For example, in order to have the left handed electron be composed of three subparticles, the subparticles have to carry fractional spin. Like fractional charge, the idea is that fractional spin is hidden from observation by the color force. Another example is that one tends to conclude that the speed of light is only an "effective" speed, and that the binons have to travel faster. Also, the bare binon interaction violates isospin but is also a very strong force.

Convincing the physics world to simultaneously accept so many hard things to swallow is essentially impossible.

Carl

 

[Hans de Vries]

By the way, I wonder what happens to the fit if you use \theta_C = 2/9.

I did try this yesterday, :^)

M^\frac{1}{2} = \left( \begin{array}{ccc}<br /> \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} \\<br /> e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} \\<br /> e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} \\ \end{array} \right)

With \delta \ =\ \frac{2}{9}

one gets the following eigenvalues:

3.3650337331519900946141875218449
0.82054356652318236464766296162669
0.057063387444112687143215689158409

Squaring the ratios gives the following mass ratios:

3477.4728371045985323130012254729 = m_\tau /m_e
206.77031597272938861255033931149 = m_\mu /m_e
16.818046733377517056533911325581 = m_\tau /m_\mu

Which are exact to circa one part per million.


Effectively two parameters are predicted, The first comes from
Koide's formula which was brought to our attention by Alejandro
and which you reworked into your matrix. The second comes from
the parameter \delta and which one may hope to be either a simple
mathematical constant (2/9) or another SM parameter. (Cabibbo)


Regards Hans

 

[CarlB]

Hans, I just realized that the branching ratios I gave, i.e. |e_r&gt; -&gt; |e_g&gt; 50%, 25%, 25%, were probably incorrect. The reason is that when you compute a probability in QM you do it by taking
P = |&lt;a|b&gt;|^2,
and that means that the mass matrix has to be included four times, not twice. That means that the branching ratios actually are:

|e_r&gt; -&gt; |e_r&gt; 66.67%
|e_r&gt; -&gt; |e_g&gt; 16.67%
|e_r&gt; -&gt; |e_b&gt; 16.67%

I realized that the numbers had to be wrong when I was working out the branching ratio from another point of view. This is somewhat speculative, so bear with me, please. Remember that the unusual thing about SU(3) color is that it appears to be a perfect symmetry...

Suppose that nature has a particle that travels at some fixed speed near the speed of light and exerts an extreme stress on space-time. The stress being very high says that the energy contained in the particle is very high. Nature wishes to reduces this stress by cancelling it.

Suppose that there is a hidden dimension, and the stress has sinusoidal dependence on that hidden dimension. That is, when you average the stress over the hidden dimension, you get zero, but when you integrate the square of the stress over the hidden dimension you get a number that corresponds to a very high energy per unit volume.

One way that nature could arrange to minimize the total energy of the particle is by ganging it up with another particle of the same sort, but arranging for the phase of the other particle, in the hidden dimension, to cancel the first. Thus you would compute the potential energy of the combined particles by first summing their stresses, and then integrating over all space.

Because the cancellation would depend on how far apart the particles were, this would result in a force. It turns out that the force that results from this sort of thing is, to lowest order, compatible with the usual assumptions about the color force. That is, the force is proportional to the distance separated.

The reason the calculation works out this way is quite generic. That is, any force based on minimization of a potential, with the potential having a nice rounded bottom, will be approximately harmonic. So this coincidence really doesn't mean much in and of itself.

If it weren't for the Pauli exclusion principle, nature could cancel the first with another particle traveling in the same direction. So nature instead reduces the stress by combining several particles traveling in somewhat different directions.

It turns out that if you analyze this problem from the point of view of Clifford algebra (that is, you assume that the stress take the form of a Clifford algebra), there is no way to get a low energy bound state out of two particles. Instead, you have to go to three. Details are beyond the scope of this post. So let's assume that nature combines three particles, with their phases (as determined by an angular offset in the hidden dimension) different by 120 degrees. The fact that 360 degrees divides equally into three multiples of 120 degrees gives the explanation for why SU(3) is a perfect symmetry, but the details are beyond the scope of this post.

Let's assume that the "center of mass" of the three particles travels in the +z direction. Let the red particle be offset in the +x direction, with the green and blue offset appropriately around the z-axis. The three particles travel on a cone centered around the z-axis.

Let the opening angle of the cone be \theta_b, where b stands for "binding angle". We expect that b will be as small as nature can get away with, but that it will be balanced by Fermi pressure. That is, if \theta_b is too small, the probability of the three particles being near each other goes down, and this raises the total energy.

Then the unit velocity vectors for the three particles are:
V_R = (s_b , 0,c_b)
V_G = (-s_b/2, s_b\sqrt{3}/2,c_b)
V_B = (-s_b/2,-s_b\sqrt{3}/2,c_b),
where s_b, c_b = sin, cos(\theta_b),

If the speeds of the individual particle are c', then the speed of the bound particle is c&#039;\cos(\theta_b). If we assume that the bound particle is a handed electron, then this says that c' is faster than the speed of light by a factor of \sec(\theta_b).

In other words, the subparticles would have to be tachyons that travel at some fixed speed faster than the speed of light. There is some experimental evidence for the existence of this sort of thing. It consists of observations of gamma ray bursts. EGRET observed a double gamma ray burst with a delay of about an hour between bursts. It's called 940217 in the literature and there are plenty of theoretical attempts (failures) to explain it.

If the gamma ray burst were caused by a collection of tachyonic particles all traveling in the same direction, then a single burst of tachyons would generate time separated bursts of gamma rays as the tachyons traveled through regular matter if the regular matter was distributed into two lumps.

I made the argument that binons might be an explanation for high energy cosmic rays at the PHENO2005 meeting. There are about a half dozen good reasons for expecting exactly the odd sort of behavior seen in the Centauro events from a binon. The reasoning is beyond the scope of this post. I will soon get around to putting up a copy of the argument on the PHENO2005 website.

Anyway, the above description of a tachyon bound state would also apply to the left handed electron. Assume that the red for the left handed electron is also oriented in the +x direction, but the bound particle is traveling in the -z direction.

Then the unit vectors for the left-handed electron are:
Then the unit velocity vectors for the three particles are:
V_R&#039; = (s_b , 0,-c_b)
V_G&#039; = (-s_b/2, s_b\sqrt{3}/2,-c_b)
V_B&#039; = (-s_b/2,-s_b\sqrt{3}/2,-c_b).

This means we can now compute angles between <R'|R>, and <G'|R>:

&lt;R&#039;|R&gt; = s_b^2 - c_b^2,
&lt;G&#039;|R&gt; = -s_b^2/2 - c_b^2,
&lt;B&#039;|R&gt; = -s_b^2/2 - c_b^2,

The above give the cosines of the angles between the unit vectors. It's well known that probabilities in QM for things with an angle theta between them follow a law proportional to 1+cos(theta). After normalizing to unit probability, it turns out that the branching ratios do not depend on \theta_b. (This is also what you'd expect from relativistic length contraction in the z direction.) Instead, you get exactly the branching ratios that just happen to fit the fermion mass matrix:

P_{R&#039;R} = 2/3,
P_{G&#039;R} = 1/6,
P_{B&#039;R} = 1/6,

Carl