All the lepton masses from G, pi, e

Hans de Vries

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

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
Phys. Rev. 95, 1300-1312 (1954)
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

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
[tex]
{M_0^2 \over M_V^2}= {3 \over 2 \pi} \alpha
[/tex]
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, Phys. Rev. D 7, 1888-1910 (1973) 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, [tex]E=mc^2[/tex] 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 [tex]\zeta(3) = \sum 1/n^3[/tex] 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

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)

Attached Thumbnails

 

Jay R. Yablon

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."

Best,

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, Phys. Rev. D 5, 3021-3047 (1972) (Spires)

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/ind...tle=Bakery:HdV

arivero

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
[tex]m_e+m_\mu+m_\tau=\frac 23 (\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})[/tex]

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:

[tex]M = \left( \begin{array}{ccc}
\sqrt{2} & e^{i\delta} & e^{-i\delta} \\
e^{-i\delta} & \sqrt{2} & e^{i\delta} \\
e^{i\delta} & e^{-i\delta} & \sqrt{2} \\ \end{array} \right) [/tex]

where [tex]\delta = .222 = 12.72[/tex] degrees, is the Cabibbo angle. Let

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

Then the eigenvectors and eigenvalues of M are approximately:

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

with the lepton masses in MeV.

Note that the sum of the eigenvalues of the matrix M are [tex]3 \sqrt{2}[/tex], and the square of this is 18. And [tex]M^2[/tex] 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 [tex]\sqrt{2}[/tex]

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

Carl, this is really very interesting.

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

[tex]\frac{m_\mu}{m_e}\ =\ 206.7682838 (54)[/tex]

[tex]\frac{m_\tau}{m_e}\ =\ 3477.441653 (83)[/tex]

[tex]\frac{m_\tau}{m_\mu}\ =\ 16.818061210 (38)[/tex]

Which is well within the experimental range:

[tex]\frac{m_\mu}{m_e}\ =\ 206.7682838 (54)[/tex]

[tex]\frac{m_\tau}{m_e}\ =\ 3477.48 (57)[/tex]

[tex]\frac{m_\tau}{m_\mu}\ =\ 16.8183 (27)[/tex]

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 [itex]\sqrt 2[/itex]'s on the diagonal of your matrix as you said:

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

[tex]1+206.7682838+3477.441653=\frac{2}{3} (\sqrt{1}+\sqrt{206.7682838}+\sqrt{3477.441653})^2[/tex]

Regards, Hans

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

arivero

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
[tex]
\tan \theta_c={\sqrt 3 (x_\mu - x_e) \over 2 x_\tau - x_\mu -x_e}
[/tex]
with [tex]x_l\equiv \sqrt{m_l}[/tex]- 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 [tex]\delta=0[/tex]

arivero

As a matter of notation, I'd call such matrix "[tex]A^\frac 12[/tex]" 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

I actually liked the close proximity to 2/9 of [itex]\delta\ = 0.2222220..[/itex] 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 [tex]\theta_C = 2/9[/tex]. 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 [tex]|e_r>[/tex], [tex]|e_g>[/tex], [tex]|e_b>[/tex], [tex]|\nu_r>[/tex], [tex]|\nu_g>[/tex], [tex]|\nu_b>[/tex], then the mass matrix shown is a result of a branching ratio for interactions of the form:

[tex]|e_r> -> |e_r>[/tex] 50%
[tex]|e_r> -> |e_g>[/tex] 25%
[tex]|e_r> -> |e_b>[/tex] 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 [tex]m_e[/tex]. 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 travelling in one direction with a right handed electron travelling 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

I did try this yesterday, :^)

[tex]M^\frac{1}{2} = \left( \begin{array}{ccc}
\sqrt{2} & e^{i\delta} & e^{-i\delta} \\
e^{-i\delta} & \sqrt{2} & e^{i\delta} \\
e^{i\delta} & e^{-i\delta} & \sqrt{2} \\ \end{array} \right) [/tex]

With [itex]\delta \ =\ \frac{2}{9}[/itex]

one gets the following eigenvalues:

3.3650337331519900946141875218449
0.82054356652318236464766296162669
0.057063387444112687143215689158409

Squaring the ratios gives the following mass ratios:

3477.4728371045985323130012254729 = [itex]m_\tau /m_e[/itex]
206.77031597272938861255033931149 = [itex]m_\mu /m_e[/itex]
16.818046733377517056533911325581 = [itex]m_\tau /m_\mu[/itex]

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 [itex]\delta[/itex] and which one may hope to be either a simple
mathematical constant (2/9) or another SM parameter. (Cabibbo)


Regards Hans