View Poll Results: Multiple poll: Check all you agree. Logarithms of lepton mass quotients should be pursued. 24 27.91% Alpha calculation from serial expansion should be pursued 22 25.58% We should look for more empirical relationships 26 30.23% Pythagorean triples approach should be pursued. 21 24.42% Quotients from distance radiuses should be investigated 16 18.60% The estimate of magnetic anomalous moment should be investigated. 26 30.23% The estimate of Weinberg angle should be investigated. 21 24.42% Jay R. Yabon theory should be investigate. 16 18.60% I support the efforts in this thread. 47 54.65% I think the effort in this thread is not worthwhile. 30 34.88% Multiple Choice Poll. Voters: 86. You may not vote on this poll

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## All the lepton masses from G, pi, e

For numerical "coincidences", hep-ex/0412028 remembers that "the infamous Florida 2000 presidential election recount with the official result of 2,913,321 Republican vs. 2,913,144 Democratic votes, with the ratio equal to 1.000061".

 Recognitions: Science Advisor Once in a while one runs into a nice numerical coincident like this one relating alpha to a Yukawa potential: $$\mbox{\Huge \frac{\alpha}{2\pi}\ =\ \frac{e^{-2\pi r}}{2\pi r}}$$ Where $r$ is the quotient of two radii: $$r\ =\ \frac{r_c}{r_l}$$ in which rc is the Compton radius of the electron, muon or tau lepton and rl is the corresponding cut-off radius for which the electrostatic energy is equal to the magnetostatic energy of the classical electron, muon or tau lepton. It's the magnetic moment which is responsible for the magneto- static energy. One gets the following relations: $$\alpha =1/137.05268 -\ \quad \mbox{for the electron}$$ $$\alpha =1/137.04743 -\ \quad \mbox{for the muon}$$ $$\alpha =1/137.03796 -\ \quad \mbox{for the tau lepton}$$ $$\alpha =1/137.03599971\ \quad \mbox{experimental}$$ The differences stem from the differences of the magnetic moment analomies. Regards, Hans ================================================== PS: For the EM fields the following expressions were used: $$\textsf{E}_x\ = \frac{q}{4\pi\epsilon_o}\ \frac{x}{r^3}, \quad \textsf{E}_y\ = \frac{q}{4\pi\epsilon_o}\ \frac{y}{r^3}, \quad \textsf{E}_z\ = \frac{q}{4\pi\epsilon_o}\ \frac{z}{r^3}$$ $$\textsf{B}_x\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}x \right), \quad \textsf{B}_y\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}y \right), \quad \textsf{B}_z\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}z-\frac{1}{r^3} \right)$$ And for the total energy: $$\mbox{Energy:} \qquad E\ =\ \frac{1}{2}\left(\epsilon_o\textsf{E}^2 + \frac{1}{\mu_o}\textsf{B}^2\right)$$ $$E\ =\ \int \left\{ \frac{q^2}{32\pi^2\epsilon_o r^4}\ +\ \frac{\mu_o\mu_e^2}{8\pi^2}\left(\frac{3z^2}{r^8}+\frac{1}{r^6} \right) \right\} dx^3\ =\ \frac{q^2}{8\pi\epsilon_o r_o}\ + \ \frac{\mu_o\mu_e^2}{4\pi}\ \frac{3}{r_o^3}$$ For an electron: $$\frac{q^2}{8\pi\epsilon_o} = 1.1535385\ 10^{-28},\quad \frac{3\mu_o\mu_e^2}{4\pi}\ =\ 2.5862051\ 10^{-53}$$ and ro is the cut-off radius.
 Blog Entries: 6 Recognitions: Gold Member This last comment reminders me about Born calculation of the Lamb shift, in the train cominb back from Shelter Island. He had no serious argument to fix the cut-off in the renormalized equation, so he simply choose the compton lenght of the electron because it is the cutoff of particle creation.
 See: "Evidence for Detection of a Moving Magnetic Monopole", Price et al., Physical Review Letters, August 25, 1975, Volume 35, Number 8. This was the last of a series of balloon flights, launched in 1973, but not analyzed by myself until 1975, due to higher priority cosmic rays analysis then ongoing. The suggestion that the anomalous track could have been caused by a doubly fractionating normal nucleus is untenable. One would have expected to have seen billions of similar tracks, not quite as closely matched to the expected track of a magnetic monopole, first. No such similar events were ever detected. For further information, contact the administrator who can email me, as I do not regularly post at this forum. Or check www.sciforums.com where I do regularly post, and PM me. Whether the Large Hadron Collider [LHC] will create a magnetic monopole is highly debatable. It might also create miniature black holes, or strangelets.

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 Quote by taarik Please see my paper about mass quantization @ arxiv. hep-ph/0702140
http://www.slac.stanford.edu/spires/...fs?key=7074930
and the list of citations of McGregor
http://www.slac.stanford.edu/spires/...=NUCIA,A58,159
are interesting for the topics of this thread. A problem of quantisation of M instead of quantisation of M^2 is that it has some scent of classical group theory, thus one needs to see how many of the relationships are already explained in the quark model and check the extant cases.

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

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

the question, to me, is not why muon and pion have different mass, but why have they got almost the same mass. A conjecture is SUSY.

 Here is another surprise the t lepton mass can be obtained by taking 57 jumps of 29.318 MeV from the muon mass i.e t mass= 57x 29.318. Now this 57 number also helps us to include the electron mass as 57 times electron mass= 29.127 very close to 29.318. This leads us to thing that like in Nambu's and many other cases the basic unit appears from the electron mass. Now this also means the pion -muon= 57x electron , tau - muon =57x 29.318 =57x57x electron. which in turn leads to tau- pion =56x 29.318 =56 x 57 x electron. Hence the lightest hadron i.e pion and lightest unstable lepton i.e muon , two leptons muon and tau , lightest hadron and heaviest lepton i.e tau are all related through electron mass.

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 Quote by taarik Now this also means the pion -muon= 57x electron , tau - muon =57x 29.318 =57x57x electron. which in turn leads to tau- pion =56x 29.318 =56 x 57 x electron..
I prefer to write then

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

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

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

EDIT2. Perhaps Krolikowski has some argument for 58/2. Also, Ramanna (eg pg 16 of nucl-th/9706063)

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 Quote by arivero I prefer to write then...
Hmm, funnier:

$$( m_{\pi_0} - m_\mu )= \sqrt {m_e} \sqrt {m_\tau - m_\mu}$$

LHS and RHS still agree within a 0.3 %. No bad.

The above comments still apply. On other hand, if I recall correctly, the question about why the mass of the charged pion is higher, and not lower, than the neutral one was a touchy issue decades ago, and it required very high level theoretists to explain it.

 Refering to first version, it ia again amazing that the mass of the neutral pion is deteremined very precisely in terms of the three leptons. Now neutral pion has no relation with the three leptons. it does not decay into any of these particles. On the other hand the charged pion decays into electron and muon. Hence charged pion mass should have been related to lepton masses.

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 Quote by taarik Refering to first version, it ia again amazing that the mass of the neutral pion is deteremined very precisely in terms of the three leptons. Now neutral pion has no relation with the three leptons. it does not decay into any of these particles. On the other hand the charged pion decays into electron and muon. Hence charged pion mass should have been related to lepton masses.
I agree, it is misterious. Furthermore, forgetting the issue of integer multiples and the squaring of masses, the formula is very reminiscent of charged pion decay, you know, these $\prop m_\mu^2 ( m_\pi_+^2 - m_\mu^2)$ from textbooks.

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

EDITED: a purpose of the above formulas is to consider the limit $m_\mu \approx m_\tau$ where the former formula cancels and the two first ones in the above become the same. Also, the same cancellation and similarity happens in the other limit $m_e \to 0$. Simultaneous limit conflicts with Koide's.

 Blog Entries: 6 Recognitions: Gold Member My thinking during the last two years was: initially there is a symmetry where neutrinos have the same mass than neutral mesons and charged leptons have the same mass than charged mesons. Note the count of degrees of freedom. Of course one could also expect the dirac mass of neutrino and charged lepton to coincide. Then seesaw moves the mass of neutrinos out of reach and mixing, including CKM, and/or other unknown mechanism alter the mass eigenvalues of the mesons. The mechanism could be related to a mismatch between isospin in mesons and leptons. Namely, third generation mesons do not exist except bB.

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 Quote by arivero To put more intrigue, the mass difference between eta and the average of pion and muon (say, diff=427.2 MeV) also fits roughly in the obvious permuted formulae: $$\sqrt {m_\mu} \sqrt {m_\tau - m_e} \approx \sqrt {m_\tau} \sqrt {m_\mu - m_e} \approx \sqrt {m_\tau \pm m_e} \sqrt {m_\mu \pm m_e} \approx \sqrt {m_\mu} \sqrt {m_\tau} = 433.27 MeV$$
Hmm, Gell-Mann Okubo value for unmixed $\eta_8$ is
569.32 GeV, so $$\eta_8 - \pi^0 = 434.34 MeV$$

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 Quote by arivero Hmm, Gell-Mann Okubo value for unmixed $\eta_8$ is 569.32 GeV, so $$\eta_8 - \pi^0 = 434.34 MeV$$
For amateurs, it could be worthwhile to explain what the Gell-Mann Okubo is, or refer to a textbook. I like Donoghue Golowich Holstein, "Dynamics of the Standard Model". The formula appears in chapter VII, expression (1.6b). You get the formula from the following set of equations:
$$m^2_\pi=B_0 (m_u + m_d)$$
$$m^2_{K^0}=B_0 (m_s + m_d)$$
$$m^2_{K^\pm}=B_0 (m_s + m_u)$$
$$m^2_{\eta_8}=\frac 13 B_0 (4 m_s + m_u + m_d )$$
and so on.

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

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

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

$$m^2_\pi = (m_\mu + \sqrt { m_e (m_\tau-m_\mu)})^2 = B_0 \hat m$$
$$m^2_{\eta_8} = (m_\pi + \sqrt { m_\mu (m_\tau-m_e)})^2 = \frac 23 B_0 (2 m_s + \hat m)$$
Here you can see also one of the themes which were debatable in the sixties: the use of mass square instead of plain mass. For instance, it is because of it that our resulting equations
do not allow to cancel $B_0$ out.

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 Quote by taarik Here is another surprise the t lepton mass can be obtained by taking 57 jumps of 29.318 MeV from the muon mass i.e t mass= 57x 29.318. Now this 57 number...