# All the lepton masses from G, pi, e

• arivero

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Classical Distance Ratio Formula

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This one is the most “physical” yet. It’s also the simplest
and most accurate expression involving lepton mass ratios.
(fully within experimental range and exact to 0.0000073%)
.
.
.
$$\frac{d_{crad}}{d_{spin}} \ \ = \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \ a \ \frac{m_e}{m_\tau} \ ) , \ \ \ \ \ \ \ \ \ \ \ \ \ where \ \ a = \frac{1}{\frac {1}{2}(1+\frac{1}{2})} = \frac{4}{3}$$
.
.

It relates the ratio of two “classical distances” with the
lepton mass ratio’s. Since distances scale (inversely)
proportional to mass we may expect the distance ratio
to be directly proportional to the lepton masses as well.
So far so good.

The distance ratio is dependent only on the fine structure
constant and consequently can be calculated with a very
high precision:

1/206.68905011 (69) _ dist. ratio from fine structure const.
1/206.689035 (19) ___ dist. ratio calculated with mass ratios

1/206.7682838 (54) __ electron/muon mass ratio
1/3477.48 (57)_______ electron/tau mass ratio

The electron/muon ratio already equals the distance
ratio to within 0.038% A simple correction term using
the electron/tau ratio then brings it to 0.0000073%

The distance ratio is the same for all spin ½ particles.
The classical distances are:

=====================================
A lepton and anti-lepton spaced at this distance,
(which is inversely proportional to their mass),
have a (negative) potential energy which is equal
to their rest mass energy.

dspin:___Classical (Electron) Spin ½ Orbital
=====================================
A lepton and anti-lepton spaced at this distance,
(which is inversely proportional to their mass),
and orbiting at the frequency corresponding to their
rest mass, have an angular momentum of a spin ½
particle: $\sqrt{(\frac{1}{2}(1+\frac{1}{2}))} \ \hbar$

It turns out that the velocity at which the leptons
would have to orbit here is equal for all spin ½ particles.
It’s a dimension less constant when written as v/c and
the solution of the following equation:

$$\frac{v^2/c^2}{\sqrt{1-v^2/c^2}} \ \ = \ \ \sqrt{\frac{1}{2}(1+\frac{1}{2})}$$

Where the right hand term corresponds to the angular
momentum. When solved it gives:

$$\frac{v}{c} \ \ = \ \ v_{spin\frac{1}{2}} \ \ = \ \ 0.754141435281767 \ \ = \ \ \sqrt{\frac{1}{8}\sqrt{57} - \frac{3}{8}}$$

The relation between the distance ratio and the fine
structure constant is:

$$\frac{d_{crad}}{d_{spin}} \ \ = \ \ \frac{1}{2}\alpha / v_{spin\frac{1}{2}}$$

So that we can write exact to within 0.0000073%

$$\alpha \ \ = \ \ 2 \ v_{spin\frac{1}{2}} \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \frac{1}{\frac{1}{2}(1+\frac{1}{2})} \ \frac{m_e}{m_\tau} \ ) \ \$$

The following CoData 2002 values were used so you
can test it yourself. (Don’t forget to apply a factor of
$\frac{1}{\sqrt{1-v^2/c^2}}$ to the mass when calculating the angular
momentum)

fine structure const 0.07297352568 ____ 0.00000000024
Planck constant ____ 6.626 0693e-34 ___ 0.0000011e-34 J s
speed of light _____ 299792458 ________ (exact) m/s
electron mass ______ 9.1093826e-31 ____ 0.0000016e-31 kg
speed of light _____ 299792458 ________ (exact) m/s
electric constant __ 8.854187817e-12 __ (exact) F/m
electric charge ____ 1.60217653e-19 ___ 0.00000014e-19 C
clas_electron_rad __ 2.817940285e-15 __ 0.000000028e-15 m

These "classical distances" are loaded with theoretical problems.
The classical electron radius has been used to try to cut of
the electric field close to the electron in the hope to avoid
the infinity problem of the $1/r^2$ field: Doesn’t work.

The spin angular momentum is much to high to fit into any
guess of the electron size. The angular orbit here is much
larger then electron radius! The electric force is more then
40 times higher then the centrifugal force to allow such an
orbit, et-cetera.

However, as a side remark, I've been looking at not so very
different scenarios were it seems that it may be possible to
bring the spin back into classical EM field theory without the
use of any new physics.

Regards, Hans

Last edited:
Hans de Vries said:
However, as a side remark, I've been looking at not so very
different scenarios were it seems that it may be possible to
bring the spin back into classical EM field theory without the
use of any new physics.
Well, for sure spin is a non-classical parameter. It is measured as integer or half-integer multiples of the Planck Constant. So If you have a classical theory with spin, you have a classical theory predicting Planck Constant. And my, that's really new physics.

I am giving a glance to Gonzalez-Martin estimates, as well as Wyler's. At different decades, they postulated the use of quotient of volumes as a mean to determine mass ratios. It does not seem as accurate as this thread work.

a bit more bibliography

1) Hans, the thing that worries me about your last try is the dissappearance of the logarithmic expressions. It seems to be thus radically different.

2) An history of numerological approaches to the fine structure constant is contained in
H. Kragh, "Magic Number: A Partial History of Fine-Structure Constant", Arch Hist Exact Sci 57 (2003) p 395-431
According copyright, private emailing of the paper, not massive, is allowed. So if someone wants to have a read of it, please tell me at my standard email address, arivero@... (where ...=unizar.es)

arivero said:
1) Hans, the thing that worries me about your last try is the dissappearance of the logarithmic expressions. It seems to be thus radically different.

We have another situation now. We relate the mass ratio's
with two "classical physical quantities" In fact these quantities
are about the two most elementary you can think of.

The 1st relates the rest mass to the electric field energy.

The 2nd relates the rest mass to the spin 1/2 ang. momentum.

It's remarkable that the ratio between these two equals the
electron/muon mass ratio to within 0.038%

It's even more remarkable that a simple correction term using
the other (electron/tau) mass ratio produces the right result
to within 0.0000073%! And all within experimental accuracy:

___________$$\frac{d_{crad}}{d_{spin}} \ \ = \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \ \frac{4}{3} \ \frac{m_e}{m_\tau} \ )$$

The only coefficient (4/3) is equal to the inverse square of
the spin 1/2 angular momentum value:

___________________$$\frac{1}{\frac{1}{2}(1+\frac{1}{2})} = \frac{4}{3}$$

Further, it's remarkable (and good news) that the ratio is
directly proportional to the mass ratio's. Both quantities are
distances and distances are supposed to behave that way.

For instance the classical electron radius is 206.7682838 times
larger than the classical muon radius, exactly the same ratio as
their masses have (but reversed).

These classical radii have another long forgotten property:
If something would rotate around this radius with a frequency
corresponding to the rest mass frequency then the speed
comes out to be c times alpha. This is true for the electron,
muon and tau.

The classical radii can be found in the Codata list. The other
value however has no public status. One has to solve a few
equations to get it. I called it a "classical value" because
probably nobody with a background in QFT or just QM in general
would ever bother to calculate it!

It simply looks which orbit a particle, with a certain rest mass
and a rotation frequency corresponding with that rest mass,
needs to have in order to give it the spin angular momentum
of a spin 1/2 particle.

Regards, Hans

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Renormalization group running of the coupling constant seems to inspire some of the formulae for alpha. While the real coupling is sensitive to the particle content and it should break down at the GUT scale, these theories prefer to work with some naive exponential coupling from the Planck scale, where they impose a simple value of alpha at these scale, say 2pi or 1 or infinity. So Nottale formula,
$$M_P/m_e = e^{\frac38 \alpha^{-1}}$$
has alpha going down from infinity to 1/137.41.

By aumenting the complexity of the formula, the "prediction" can be more precise. Recently an anonymous nicknamed "Quantoken" has suggested
$$(M_P/m_e)^2=2 \pi \alpha^{-1} e^{\frac23 \alpha^{-1}}$$
which gives alpha a value near 2 pi at the Planck scale, and a "prediction" of 1/137.07

I don't what to do of these formulae. My current opinion, as I said above, is that they work because they are kind of approximations to the running constant formulae. I had some curiosity because the factors 1/3 and 3/8 in the exponents differ a 1/24, so I wondered if another formula with such factor could be proposed. But to compare both formulae is not useful here, because the exponentials are nearly parallel.

EDITED: Probably http://members.lycos.co.uk/nigelbryancook/'s http://nigelcook0.tripod.com/ [Broken] falls also in the family here described. The key is to notice that any reference to Newton constant is, at the same time, a reference to the Planck mass scale.

Last edited by a moderator:
Magnetic Anomalies and Mass Ratios

I found this numerical relation intriguing as well:
Taken into account that there are only a handful
of mass-ratios to play with.

$$1 \ \ + \ \ \frac{m_\mu}{m_Z} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \ 1.001158692 (27)$$

$$1 \ \ + \ \ \frac{m_\mu}{m_Z} \ \ + \ \ \frac{m_e}{m_W} \ \ = \ \ 1.001165046 (30)$$

We recognize the magnetic moments of the leptons:

1.001159652187 ____ Electron’s Magnetic Moment
1.001158692(27) ___ 1 + mμ/mZ

1.0011659160 ______ Muon’s Magnetic Moment
1.001165046(30) ___ 1 + mμ/mZ + me/mW

The small term me/mW doesn’t do too bad either in
bridging the difference between the electron's and
muon’s (normalized) magnetic moments:

0.000006263813 ____ Difference of Magnetic Moments
0.000006353(3) ____ me/mW

The weak contributions are very small in the normal
anomaly calculations. For the electron they are
simply neglected. For the muon they are smaller than

At this scale one would expect to see nothing but QED.

Regards, Hans

PS: The CODATA mass values used:

W mass = 80.425___(38) GeV
Z mass = 91.1876__(21) GeV
e mass = 0.51099892(4) MeV
μ mass = 105.658369(9) MeV

This seems to be a case for "add 1 effect", as I.J. Good, above quoted, explains. The approximation becomes more impressive by counting from 1.00XXX... that from 0.00XXX...

Still, it could be pretty standard physics. The corrections to the magnetic moment are no very far away from formulae of this kind, with some powers of the quotient of masses etc. I'll check my QED.

EDITED: a recent review of muon magnetic moment calculations:
http://xxx.unizar.es/abs/hep-ph/0411168

Last edited:
Hans de Vries said:
The weak contributions are very small in the normal
anomaly calculations. For the electron they are
simply neglected. For the muon they are smaller than
At this scale one would expect to see nothing but QED.

Yep you are right here, we only expect some corrections depending on alpha. Perhaps it is coincidence, perhaps a fine tunning mechanism for a whole summation of diagrams.

By the way, while checking this I have got a clue of where Nottale could have got his 3/8 coefficient from. While he claims to relate it to angular prediction from GUT, a older apparition of 3/8 appears associated to the infrared-catastrophic term of the traditional calculation for the anomalous moment. For instance Sakurai' formula 4.462 contains
$${\alpha \over 3 \pi}{q^2 \over m_e^2} (\ln {m_e\over \lambda_{IR}} -\frac38)$$
So perhaps some argument about the control of the infrared catastrophe up to Planck mass scale could be retorted somehow to justify Nottale's formula.

Obviusly this 3/8 term was very apparent in all the old books when doing this calculation. I had seen it also in formulae 10.3 of Aitchison, who in turn refers to Bjorken Drell sect 8.6 and pg 172--6 for the infrared problem.

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Surprices!

These classical distance ratio's, $R_f$ for fermions and $R_b$
for bosons are really full of surprices! The whole list for
sofar:

1)

The ratio $R_b/R_f$ of the two is equal to the $m_W/m_Z$ ratio!
(to within 0.063% or sigma 1.2)

$$\cos \theta_W \ \ = \ \ \frac{m_W}{m_Z} \ \ = \ \ \frac{R_b}{R_f} \ \ = \ \ \frac{2 \beta_b/\alpha}{2 \beta_f/\alpha} \ \ = \ \ \frac{\beta_b}{\beta_f}$$

where $\beta_f, \beta_b$ are the classical spin 1/2 and spin 1 velocites.
This would mean that the electro-weak mixing angle would be a
numerical constant! with a physical geometric origin rather than
an arbitrary symmetry breaking parameter.

2)

Rf is equal to the muon-electron mass ratio to within 0.038%

$$R_f^{-1} \ \ =\ \ \frac{m_e}{m_{\mu}}$$

which improves to 0.0000073% if we add a correction term
involving the third lepton like this:

$$R_f^{-1} \ \ = \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \ \frac{4}{3} \ \frac{m_e}{m_\tau} \ )$$

1/206.6890501 first expression
1/206.7682987 second expression
1/207.7682838 mass ratio

3)

We can do something similar for the tau-muon mass ratio:

$$\beta_b R_f^{-1/2}}\ \ =\ \ \frac{m_{\mu}}{m_{\tau}}$$

which is accurate to within 0.090%. This becomes 0.000040%
when we add the following correction term with the other lepton:

$$\beta_b R_f^{-1/2}}\ \ =\ \ \frac{m_{\mu}}{m_{\tau}} \ (1 \ + \ \ \frac{3}{16} \ \frac{m_e}{m_\mu} \ )$$

1/16.80305 first expression
1/16.81829 second expression
1/16.8183 mass ratio

----
----

We did see that the clasical velocity can be defined as:

“The velocity of a mass with spin s rotating on an orbit
with a frequency corresponding to its rest mass and an
angular momentum $\sqrt{ s(s+1}\ \hbar$"

One gets a general solution for the 'classical velocity' of spin s:

$$\beta_s \ \ \ \ \ \ = \ \ \ \ \sqrt{\sqrt{\ \ s(s+1) \ \ + \ \ ( \frac{1}{2} \ s(s+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ s(s+1) }$$

Which solutions are dimensionless constants, independent
of the mass of the particle. The values of the common spins
are given below:

spin 0.0: __ 0.00000000000000000000000000000000
spin 0.5: __ 0.75414143528176709788873548859945
spin 1.0: __ 0.85559967716735219296923576621118
spin 1.5: __ 0.90580479773844104117525862119228
spin 2.0: __ 0.93433577808377694874713811004304
spin inf: __ 1.00000000000000000000000000000000

Weinberg’s Electro-Weak mixing angle becomes a dimension-
less constant as well and is given in the $\sin^2 \theta_W$ form as:

$$\sin^2 \theta_W \ \ \ \ = \ \ \ \ 1 \ - \ \frac{\beta^2_f }{\beta^2_b} \ \ \ \ = \ \ \ \ 0.22310132230086634541466926662604$$

$$\sin^2 \theta_W \ \ \ \ = \ \ \ \ 1 \ - \ \frac { \sqrt{\ \ \frac{1}{2}(\frac{1}{2}+1) \ \ + \ \ ( \frac{1}{2} \ \ \frac{1}{2}(\frac{1}{2}+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ \ \frac{1}{2}(\frac{1}{2}+1) } {\sqrt{\ \ 1(1+1) \ \ + \ \ ( \frac{1}{2} \ \ 1(1+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ \ 1(1+1) }$$

The usual electro-weak parameters g1 and g2 would become:

$$g_1^2 \ \ = \ \ e^2 \frac{\beta_b^2}{\beta_f^2} \ \ \ \ \ \ \ \ \ g_2^2 \ \ = \ \ \frac{e^2}{1-\frac{\beta_f^2}{\beta_b^2}}$$

where $e^2 = \alpha$

Regards, Hans

Last edited:
Hans de Vries said:
1)

The ratio $R_b/R_f$ of the two is equal to the $m_W/m_Z$ ratio!
(to within 0.063% or sigma 1.2)

$$\cos \theta_W \ \ = \ \ \frac{m_W}{m_Z} \ \ = \ \ \frac{R_b}{R_f} \ \ = \ \ \frac{2 \beta_b/\alpha}{2 \beta_f/\alpha} \ \ = \ \ \frac{\beta_b}{\beta_f}$$

where $\beta_f, \beta_b$ are the classical spin 1/2 and spin 1 velocites.
This would mean that the electro-weak mixing angle would be a
numerical constant! with a physical geometric origin rather than
an arbitrary symmetry breaking parameter.

Now there is an interesting story behind this:

I was looking at this peculiar coincidence:

0.00115869 = muon / Z mass ratio
0.00115965 = electron magnetic anomaly
0.00000635 = electron / W mass ratio
0.00000626 = difference between muon and electron magnetic anomaly

Now whatever, what we can say is that the magnetic
anomaly is totally dominated by photon (spin 1) interactions
coming from the first order $\alpha/2\pi[/tex] term while the difference of the muon and electron anomaly is almost entirely vacuum polarization interaction (spin 1/2), the result of virtual electrons and muons. This causes an asymmetry that results in a different anomaly for the muon and the electron which otherwise would be the same. when I saw this I immediately tried if Rf/Rb had any relation with mW/mZ. To my big surprise it turned out that they were equal! Regards, Hans Hans de Vries said: 0.00115869 = muon / Z mass ratio 0.00115965 = electron magnetic anomaly 0.00000635 = electron / W mass ratio 0.00000626 = difference between muon and electron magnetic anomaly Now whatever, what we can say is that the magnetic anomaly is totally dominated by photon (spin 1) interactions coming from the first order [itex]\alpha/2\pi[/tex] term while the difference of the muon and electron anomaly is almost entirely vacuum polarization interaction (spin 1/2), the result of virtual electrons and muons. I see. In the first case we are having neutral particles, and the Z ratio gets a role. In the second case we have a contribution from charged particles, and the W ratio gets a role. And Z is a neutral particle, and W is a charged particle. So at least we have a logical link. Some remarks from the historic literature: According H. Kragh (above quoted article), Sommerfeld definition of alpha is not very far from the quotient you are using for the rest of the electroweak thing. To be precise, the fine structure constant is defined as the quotient between wave K angular momentum, h/2pi and "limiting momentum", e^2/c. Or equivalently (H. Kragh, "the fine structure constant before quantum mechanics"), between the orbital velocity in wave K and the maximum velocity c. (Let me recall here that the eigenvalues of electron velocity are +-c only). Related to this we have the so-called "Sommerfeld puzzle". In the early time, before Dirac, he applied his quantization procedure to an orbiting electron plus relativistic mass correction, and well, he got the right formula for the fine structure of hidrogen. Without any reference to spin! When Dirac equation was known, time after, its only real news were to justify an arbitrary selection rule for level intensity. Hans, I can not read Sommerfeld german, but I guess you can, so I'd suggest you to try to find his relativistic calculation. It is around 1915-1916, a couple articles in Annalen der Physik and Akad. der Wiss Muenchen Sitzungsberichte. Last edited: arivero said: Related to this we have the so-called "Sommerfeld puzzle". In the early time, before Dirac, he applied his quantization procedure to an orbiting electron plus relativistic mass correction, and well, he got the right formula for the fine structure of hidrogen. Without any reference to spin! When Dirac equation was known, time after, its only real news were to justify an arbitrary selection rule for level intensity He may have got the right correction, but he did not get the total number of states right - he only got half because he didn't have spin. The Sommerfeld calculation was one of many false steps in the early days of QM, unlike the Dirac equation which actually made for progress. This is all fascinating stuff PFers! Are you any closer to making some specific, concrete predictions? Testable ones would be nice. zefram_c said: He may have got the right correction, but he did not get the total number of states right - he only got half because he didn't have spin. The Sommerfeld calculation was one of many false steps in the early days of QM, unlike the Dirac equation which actually made for progress. Actually, if my lecture notes are right, he got not half but double, and then he was forced to impose a selection rule to have it right. The selection rule being a change in one h unit of angular momentum, it is now clear, as you say, that it was related to a spin flip. But again, please count that I can not read German so I am looking at third hand -or more. lecture notes. Nereid said: Are you any closer to making some specific, concrete predictions? Testable ones would be nice. Nereid, I don't know. I plan to redo the perturbative calculation holding also the cubic v/c term -usually wiped out, but cointaining the infrared divergence- to see if some clue comes from there. I can not tell of Hans's plans -our main contact is via this thread-. Also I am not sure what kind of predictions should we be looking for; Weinberg angle is "predicted" by Hans, but this quantity very commonly predicted in theoretical works. Another scent to follow comes from L. De Broglie http://prola.aps.org/abstract/PR/v76/i6/p862_1. There, inspired by Feynman "Relativistic Cut-off" results, he suggests that electron charge self-energy can be regulated by asking for additional massive spin 1 bosons interacting with the electron. He argues that the total sum of charges for these bosons must counterweight the electron charge, but I do not quite follow his argument yet. Last edited: Point is, that there is an small nuisance in the calculation of the magnetic anomaly: it is calculated in the non-relativistic approximation v<<c of quantum fields, but the result contains the fine structure constant, which is a genuine relativistic quantity. So perhaps -only perhaps- HdV is seeing some conjure of the rest of the electroweak structure to grant a soft landing in the non-relativistic world. On other hand, there are other phenomena at the NR limit that could cover this role instead of the HdV equations. For an instance, the change in divergence rate of lepton self-energies. Last edited: arivero said: Actually, if my lecture notes are right, he got not half but double, and then he was forced to impose a selection rule to have it right. The selection rule being a change in one h unit of angular momentum, it is now clear, as you say, that it was related to a spin flip. But again, please count that I can not read German so I am looking at third hand -or more. lecture notes. Hi, Alejandro I found this subject very well covered in the book: "The story of spin" from Sin-Itiro Tomonaga (the QED one). Chapter one covers the pre- Dirac models of Sommerfield, Landé and Pauli with all the splitting levels. Landé basically improved Sommerfields "Fine Structure Formula" to make it suitable for atoms with a larger Z. It's covered in chapter 2. Landé 's "Erzats Model" made it to circa 1926. Tomonaga describes were it eventually failed in chapter 2. Pauli get's very close to the right solution but he doesn't accept the concept of spin until Thomas handles spin relativistically and gets the factor 2 that bothered Pauli so much. Chapter 3 then handles Pauli's Spin theory and then finally how Dirac derived his equation from the Klein Gordon one. How Dirac could make the orbital angular momentum a conserved quantity by adding the electron spin and how Dirac subsequently found the factor 2 in the magnetic moment of the electron when he applied an external field. Regards, Hans Nereid said: This is all fascinating stuff PFers! Are you any closer to making some specific, concrete predictions? Testable ones would be nice. I'd like to amend this ... since all measurements are inexact, and since there's ample history in physics of key constants being hard to pin down with accuracy (my favourite is http://www.npl.washington.edu/eotwash/gconst.html [Broken]), a good way to test the numerology work here is to state unambiguously what predictions of well-observed ratios (etc) are, and point out a) what the next few digits will be found to be, and b) which best measurements today are actually 'wrong'! A succinct, bold summary anyone? Last edited by a moderator: Nereid said: I'd like to amend this ... since all measurements are inexact, and since there's ample history in physics of key constants being hard to pin down with accuracy (my favourite is http://www.npl.washington.edu/eotwash/gconst.html [Broken]), a good way to test the numerology work here is to state unambiguously what predictions of well-observed ratios (etc) are, and point out a) what the next few digits will be found to be, and b) which best measurements today are actually 'wrong'! A succinct, bold summary anyone? First let me to tell that I have my doubts about the possibility of testing numerology. A senior bayesian, quoted somewhere above, has some ideas about measuring complexity and then how likely a numerological proposal is. But my point of view is that at most numerology can signal a subyacent theory. For a famous example, the fact of the three running constants being near equal at very high scale is traditionally interpreted a a signal of an unknown GUT theory. Said this, let me summarize the thread up to now. There are are least two or three different approaches condensed here; I have tryed to bring here all the interesting coincidences noted in the literature, and if I have failed to name any, I should be glad to head about it (here or privately). We have seen the idea of studying the logarithms of mass quotients, which seem very approachable with simple geometrical quantities. Then we have played a bit with a diversity of formulae, and then be have jumped to a very different area of coincidences with quotients of magnitudes in the electroweak group. About precision of each coincidence, I think HdV has done a very careful effort to remark it at each step. A beyond 0.1% has been more or less systematically requested in order to consider interesting a coincidence, as well as simplicity of the formula. Last edited by a moderator: $$\beta_b R_f^{-1/2}}\ \ =\ \ \frac{m_{\mu}}{m_{\tau}} \ (1 \ + \ \ \frac{3}{16} \ \frac{m_e}{m_\mu} \ )$$ I understood that this coefficient 3/16 is got from trial an error, and it is unrelated to angular momentum (while the 4/3 is). Am I in a mistake? Alejandro arivero said: $$\beta_b R_f^{-1/2}}\ \ =\ \ \frac{m_{\mu}}{m_{\tau}} \ (1 \ + \ \ \frac{3}{16} \ \frac{m_e}{m_\mu} \ )$$ I understood that this coefficient 3/16 is got from trial an error, and it is unrelated to angular momentum (while the 4/3 is). Am I in a mistake? Alejandro It's indeed trial and error. I did find a note a I made about: $$16/3 = \frac{(1(1+1))^2}{(\frac{1}{2}(\frac{1}{2}+1)) } \ \ \ because \ \ \ \beta_b R_f^{-1/2} \ \ \ contains \ \ \ \beta_b \beta_f^{-1/2}$$ but that might be too far fetched. Regards, Hans Heisenberg's Relation as a generic Mass Generator -- -- I came across this one in "Introduction to the Standard Model" from Cottingham and Greenwood, where it is used to explain the large meson and baryon masses compared to the quark masses. So a quark confined in a ~1 fm core results in momentum of [itex]p = \hbar/1 fm$
~ 200 MeV/c or equivalent to a mass of E = pc ~ 200 MeV, which is
much larger than the individual quark masses.

This is very different from say the electromagnetic mass model which
leads to the "classical radius of the electron" for instance. The use of
Heisenberg's uncertainty relation is of course a perfect way to obtain a
cut-off for any $1/r^2$ field in order to avoid infinities.

Just like the $1/r^2$ field energy models we get the relation that the
radius is inversely proportional to the mass. If we now define a
"Heisenberg radius" for the electron or any other arbitrary particle
in the above way then we see the following:

$$r_h = 137.03599911 \ r_0 \ \ \ \ or \ \ \ \ 3.861592678. 10^{-13} m$$

where $r_0$ is the classical (electron) radius. It relates to it with
the value of alpha. We could thus assign a "coupling constant"
of 1.0000000 to the "Heisenberg field".

Such an Heisenberg radius fits much better to the other classically
derived radii. We have these two classical velocities for spin 1/2 and
spin 1 particles defined by:

“The velocity of a mass with spin s rotating on an orbit
with a frequency corresponding to its rest mass and an
angular momentum $\sqrt{ s(s+1}\ \hbar$"

And one can now express our radii for fermions and bosons as:

$$r_f = \beta_f \ r_h$$
$$r_b = \beta_b \ r_h$$

with the classical velocities:

$$\beta_f$$ = 0.75414143528176709788873548859945
$$\beta_b$$ = 0.85559967716735219296923576621118

So these fit much better here than with the so much smaller
classical electron radius. It's also independent of the charge.
The "Heisenberg radius" itself corresponds to an orbital speed
of c when you require that the frequency equals the rest mass
frequency.

If we do the same for the classical calculation of the magnetic
moment then we get.

$$r_m = 1.0011596521859 \ r_h$$

So it's the small anomaly that tips it over the edge (the speed
would have to pass c by the same amount if not explained
otherwise)

Regards, Hans

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Hans de Vries said:
--
--

I came across this one in "Introduction to the Standard Model" from
Cottingham and Greenwood, where it is used to explain the large
meson and baryon masses compared to the quark masses.

So a quark confined in a ~1 fm core results in momentum of $p = \hbar/1 fm$
~ 200 MeV/c or equivalent to a mass of E = pc ~ 200 MeV, which is
much larger than the individual quark masses.
Yep, I remember this argument. I believe it evolved from the other use of the distance-momentum undeterminacy, to relate the range of a virtual particle and its mass. Thus, considering that nuclear force is a short interaction about 1 fm, it was suggested to look for a massive particle around 200 MeV, which we know now to be the pion (and its family).

The argument for quarks seems more touchy: glueballs, asymptotic freedom, partons, mass gap...

The use of
Heisenberg's uncertainty relation is of course a perfect way to obtain a
cut-off for any $1/r^2$ field in order to avoid infinities.
And jointly with the relativistic cut-off (which Bethe uses for the Lamb shift), they explain why QFT is less divergent than classical field theory.

If we do the same for the classical calculation of the magnetic
moment then we get.

$$r_m = 1.0011596521859 \ r_h$$

Hmm, this needs a little more elaboration; the other distances are related to forces, but what about this one?

arivero said:
The argument for quarks seems more touchy: glueballs, asymptotic freedom, partons, mass gap...
The idea of kinetic glueball energy as the main mass generator
for hadrons is not unlike this kind of Heisenberg-like mass generation
via the momentum for particles confined to a small space.

Heisenberg being behind the fermionic degeneracy force preventing
White Dwarfs (electrons) and Neutron stars (neutrons) from collapsing
further:

http://scienceworld.wolfram.com/physics/ElectronDegeneracyPressure.html

With Pauli's exclusion principle mentioned just as often for the same...

http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c3

So it's a fermionic thing then with glueballs being bosons...

Hadronic mass... It's basically what attracts us to earth.

In the case of kinetic glueball energy we are then attracted to
earth mainly by massless particles not coupled to Higgs field...

For glueballs one would expect some form of field energy, as the
result of the self interaction, and something similar to potential
energy...

Well, questions enough and LHC still 3 years away!

arivero said:
Hmm, this needs a little more elaboration; the other distances are related to forces, but what about this one?
Just using the classical equation $\mu = \frac{1}{2}evr$ for the magnetic moment
to obtain a radius if we require the rest mass frequency.

Hans.

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quarks!

hep-ph/0502200 does a regression fit of ln(m/M) for quark masses. The three points of each generation are in line, after they choose the right "\bar M. S. masses" (hmm).

Intriguingly, a fourth generation b-quark predicted with this scheme should have exactly the same mass that the Z boson.

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Electro Weak masses as Coupling Generations

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.
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It's interesting to organize the mass-ratio coincidences given in this thread as
generations of a coupling constant. Instead of $e=\sqrt{\alpha}$ we will use the slightly
adapted value of $\varrho = \sqrt{\alpha/2\beta_f}$ where $\beta_f$ is our "classical fermion velocity" and
$\alpha/2\beta_f$ is our "classical distance ratio"

$$\begin{array}{llclccll} \mbox{Coupling Generation 4: } & m_e/m_Z & = & \beta_f/\pi & \mbox{times} & \varrho^4 & \mbox{ (Z boson) } & 1.0027295 \\ \mbox{Coupling Generation 4: } & m_e/m_W & = & \beta_b/\pi & \mbox{times} & \varrho^4 & \mbox{ (W boson} & 1.0033602 \\ \mbox{Coupling Generation 3: } & m_e/m_{\tau} & = & \beta_b & \mbox{times} & \varrho^3 & \mbox{ (tau lepton) } & 1.0012880 \\ \mbox{Coupling Generation 2: } & m_e/m_{\mu} & = & 1 & \mbox{times} & \varrho^2 & \mbox{ (mu lepton) } & 1.0003833 \\ \mbox{Coupling Generation 0: } & m_e/m_e & = & 1 & \mbox{times} & \varrho^0 & \mbox{ (electron) } & 1.0000000 \\ \end{array}$$

So we get very simple expressions with reasonable precision (last column)
for all known ElectroWeak masses based on the fine structure constant and
our two classical velocities $\beta_f$ and $\beta_b$ for spin 1/2 and spin 1 particles which
define the ElectroWeak mixing angle.

The classical velocities are defined by:

“The velocity of a mass with spin s rotating on an orbit
with a frequency corresponding to its rest mass and an
angular momentum $\sqrt{ s(s+1}\ \hbar$"

These are exact dimensionless constant ratio's with c given by:

$$\beta_f$$ = 0.75414143528176709788873548859945 for fermions
$$\beta_b$$ = 0.85559967716735219296923576621118 for bosons

From the general formula for arbitrary spin s:

$$\beta_s \ \ \ \ \ \ = \ \ \ \ \sqrt{\sqrt{\ \ s(s+1) \ \ + \ \ ( \frac{1}{2} \ s(s+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ s(s+1) }$$

We see that the ratio of the two $\beta_f/\beta_b$ is equal to the $m_W/m_Z$ ratio (to within
0.063% or sigma 1.2) and thus defines (as another numerical coincidence) the
electro weak mixing angle.

$$\cos \theta_W \ \ = \ \ \frac{m_W}{m_Z} \ \ = \ \ \frac{\beta_f}{\beta_b} \ \ = \ \ 0.8814185598789792074327801204579$$

All the mass ratio's are now defined here by the coupling constant alpha
and the two components that define the electroweak mixing angle.

Regards, Hans

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Hans de Vries said:
Magnetic Anomalies and Mass Ratios
I found this numerical relation intriguing as well:
Taken into account that there are only a handful
of mass-ratios to play with.
$$1 \ \ + \ \ \frac{m_\mu}{m_Z} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \ 1.001158692 (27)$$
$$1 \ \ + \ \ \frac{m_\mu}{m_Z} \ \ + \ \ \frac{m_e}{m_W} \ \ = \ \ 1.001165046 (30)$$

Let me add that a possible next term is 1/2 (mu/W)^2

$$\ \ \frac{m_\mu}{m_Z} \ \ + \ \ \frac{m_e}{m_W} + \ \ \frac12 ({m_\mu \over m_W})^2 \ \ = \ \ .00116590899$$

to be compared with current experimental value .0011659208(6)

arivero said:
.00116590899 to be compared with current experimental value .0011659208(6)
... or to the theoretical QED-only value .0011658471 or to the theoretical electroweak value .0011658487

This is sort of a problem when aiming to fit more terms: should the expansion go towards the experimental value (which is thought to include hadronic contributions) or towards the pure electroweak value (then revealing some hidden property of the electroweak theory)?

If we opt towards the second possibility, we have the advantage of being not constrained by experimental data anymore. But the theoretical value, I read, was calculated using a Z0=91.1875 and W=80.373, then giving worse agreement. On the other hand, higgs (and quarks?) is involved in the theoretical calculation, correcting the mass of W.

Incidentally, state-of-art theoretical calculation, Phys Rev D 67, 073006 (2003), is very readable.

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Hans de Vries said:
generations of a coupling constant. Instead of $e=\sqrt{\alpha}$ we will use the slightly
adapted value of $\varrho = \sqrt{\alpha/2\beta_f}$ where $\beta_f$ is our "classical fermion velocity" and
$\alpha/2\beta_f$ is our "classical distance ratio"

I supposse this adjustment is done so that the $$m_\mu/m_Z$$ correction of the anomalous magnetic moment becomes exactly the first order perturbative correction $$\alpha/2\pi$$. Another possibility is to define $\varrho$ (or an $$\alpha'$$) so that the correction to the anomalus magnetic moment of electron becomes the whole experimentally known correction.

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arivero said:
I supposse this adjustment is done so that the $$m_\mu/m_Z$$ correction of the anomalous magnetic moment becomes exactly the first order perturbative correction $$\alpha/2\pi$$. Another possibility is to define $\varrho$ (or an $$\alpha'$$) so that the correction to the anomalus magnetic moment of electron becomes the whole experimentally known correction.

If you take the magnetic moment anomaly of the electron times 2pi as an adjusted "coupling value" instead
of alpha then $\varrho$ becomes 1/√207.0023039 and:

$$\begin{array}{llclccllll} \mbox{Coupling Generation 4: } & m_e/m_Z & = & \beta_f/\pi & \mbox{times} & \varrho^4 & \mbox{ (Z boson) } & 91187.6 & 91215.2 & 1.0003031 \\ \mbox{Coupling Generation 4: } & m_e/m_W & = & \beta_b/\pi & \mbox{times} & \varrho^4 & \mbox{ (W boson} & 80425 & 80398.8 & 1.0003257 \\ \mbox{Coupling Generation 3: } & m_e/m_{\tau} & = & \beta_b & \mbox{times} & \varrho^3 & \mbox{ (tau lepton) } & 1776.99 & 1778.73 & 1.0009848 \\ \mbox{Coupling Generation 2: } & m_e/m_{\mu} & = & 1 & \mbox{times} & \varrho^2 & \mbox{ (mu lepton) } & 105.658369 & 105.777953 & 1.0011318 \\ \mbox{Coupling Generation 0: } & m_e/m_e & = & 1 & \mbox{times} & \varrho^0 & \mbox{ (electron) } & 0.51099892 & 0.51099892 & 1.0000000 \\ \end{array}$$

Which gives a significant improvement for the electroweak bosons. (Last 3 columns: measured in MeV,
calculated in MeV, precision)

Regards, Hans

PS: mathematical symbols directly as characters instead of latex:

√ α β γ ψ Ψ μ τ φ π Σ Π ∫ ∂ ∞ ≈ ≠ ≤ ≥ ≡ ½ ⅓ ¼ ⅔ ¾ ⅛ ⅜ ⅝ ⅞

The're nicer in Times New Roman though:

√ α β γ ψ Ψ μ τ φ π Σ Π ∫ ∂ ∞ ≈ ≠ ≤ ≥ ≡ ½ ⅓ ¼ ⅔ ¾ ⅛ ⅜ ⅝ ⅞

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Hans de Vries said:
PS: mathematical symbols directly as characters instead of latex:

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ½ ? ¼ ? ¾ ? ? ? ?

Hans I don't know how to make these characters.
If you have a moment could you say how to type a few of them?

I have tried [ alp*ha ] and & alp*ha ;

removing * and spaces but don't get anything that can survive the next edit.

Speaking of precision, our result for the muon anomalous moment quotients to 0.99998987, 1.0000530 or 1.0000517 depending if we compare to experimental measure, QED standard prediction or electroweak standard prediction. I think our estimate relates to some leptonic-only calculation in the electroweak setup. I wonder if there is some way to cancel Adler' kind of triangles across generations.

And I am using the "honest" version of the quotient, referring only to the additive correction. Putting the 1s inside, as if it were a multiplicative correction (to the nonanomalous g=2), the quotient is 0,9999999882 and so on.

I am really furious about this being an unpublishable unscientific claim. When I think Nobel, I am not thinking prize anymore... but dynamite.

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marcus said:
I have tried [ alp*ha ] and & alp*ha ;
removing * and spaces but don't get anything that can survive the next edit.

Haha! Marcus, I am also asking myself what happens with alpha. More exactly, why our mass-based formulae have lost any reference to the coupling constant alpha, while Schwinger first estimate was just alpha/2pi.

To me, some interplay of the SU(2)xU(1) triangle anomaly, generations, couplings and constants, can be suspected. And it could even be consistent with Hans's considerations on angular momentum and velocities.

But even if coincidental or unrelated (or perhaps related!), it could be time to name the recent numerology (well, halfbaked theory) of Jay R. Yablon, who has suggested that Higgs electroweak vacuum, <v>=246 GeV, should be invited to the game. For instance mass of tau= alpha times <v>. Or some other things involving powers of alpha and the sin of weinberg angle. Hmm.

marcus said:
Hans I don't know how to make these characters.
If you have a moment could you say how to type a few of them?

I have tried [ alp*ha ] and & alp*ha ;

removing * and spaces but don't get anything that can survive the next edit.

Hi, Marcus.

It seems you've got only the characters in the basic Ascii set. The first 256
characters of the Western. ISO 8859-1 characterset. You might need to set
the character encoding of your Browser to this character set under the

I got the other via Word (Insert, Symbol) You should be able to copy and
paste them from here.

α β γ δ ε ζ η θ ι κ λ μ ν ο π ρ ς σ τ υ φ
Γ Δ Θ Λ Ξ Π Σ Φ Ω ∏ ∑ ∩ ∆ ∂ ∫ ∂ ∞
√ ≈ ≠ ≤ ≥ ≡ ½ ⅓ ¼ ⅔ ¾ ⅛ ⅜ ⅝ ⅞

Regards, Hans