# All the lepton masses from G, pi, e

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Evidence for a moving Magnetic Monopole in 1975

See:

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

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

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

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

Whether the Large Hadron Collider [LHC] will create a magnetic monopole is highly debatable. It might also create miniature black holes, or strangelets.

Mass quantization in terms of pion and muon mass difference

Gold Member

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

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Mass quantization

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

Gold Member
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.

Gold Member
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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Gold Member
I prefer to write then...
Hmm, funnier:

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

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

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

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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.

Gold Member
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.

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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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Gold Member
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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Gold Member
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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Gold Member
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...

Dirac gets 53 times the electron mass for the muon, in a paper that has been lately recalled by the guys working on strings and branes.

Gold Member
A pretty amazing coincident isn't it? The relation:

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

following from the previous post is as exact as:

1 : 1.0000067 (42)

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

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

Entering the octet, we are touching deep problems of the elders. There is a short work of witten http://www.slac.stanford.edu/spires/find/hep/www?j=PRLTA,51,2351 [Broken] about how the mass of the charged pion must always be higher than the neutral pion, even if only to avoid tachions in the limit of zero pion mass. Also, the mixing between $\eta_8[\itex] and [itex]\eta_0[\itex] to give [itex]\eta[\itex] and [itex]\eta'[\itex] was the U(1) headache, addressed by t'Hoft, http://www.slac.stanford.edu/spires/find/hep/www?j=NUPHA,B159,213 [Broken] and http://www.slac.stanford.edu/spires/find/hep/www?j=NUPHA,B156,269 [Broken] independently, and according Okubo still unclear. I have found even some recent work in the context of strings: http://www.slac.stanford.edu/spires/find/hep/wwwrefs?key=5864437 [Broken] Last edited by a moderator: Gold Member You could try to use the LLL algorithm to find formulae. It is an interesting idea. Actually, the team of D.H. Bailey have tried in the past to input the standard model masses etc in some of their algorithms; but I had no idea about their new work using LLL. Last edited: Actually, the LLL algorithm is a bit older than Bailey's PSLQ algorithm. The LLL is a little less efficient, but that's only a problem if you have thousands of significant digits of some number and try to find a formula in terms of known constants for it. For your work, the LLL may be better, because with the LLL you can look for simultaneous relations, the PSLQ can't do that (at least not in general, you can work with complex numbers, quaternions,... but you'll reach a limit beyond which you can't go). If you have 5 numbers that are known to ten digits then you have 50 digits of information. Ten digits may not be enough to detect a relation, but 50 may be enough provided, of course, that the five numbers are given by formulae of the same form that are specified by the same constants. See the Appendix of this article for details. Gold Member Last edited: I found that your link gave a good and understandable explanation. However, p. 9 ...They can be categorised as three coupling constants, and three mass ratios, and their empirically determined numerical values are approximately:12 gS .=4, e.=1/12, mN.=1/2× 10−10 The values of the coupling constants are rather more familiar in their squared forms: thus we have the gravitational fine structure constant … the ordinary (electromagnetic) fine structure constant, e2 .= 1/137, and … ---------- I need explanation with e.=1/12 and , e2 .= 1/137 I always thought that 12X12=144 not 137 jal Gold Member Well, I guess he takes the nearest simple fraction to sqrt(137), and he chooses 1/12. So it really compares 12 against 11.7, arguably not so bad. A peculiarity of this article is that a couple ways ago I have been told that the relationship between mass of electron and pion (!) had circulated as a conjecture in the sixties. I though it came from McGregor, but it is formula (6) in pg 11 of this paper. "So it really compares 12 against 11.7, arguably not so bad." If you find anything else that can help I'll put it in "How to build a universe" jal CarlB Science Advisor Homework Helper I've had some interesting results in rewriting the Koide equation as a sort of "field energy" equation. The idea is to treat the square in the mass as coming from the energy of a field. Field energies are quadratic, for instance, E&M field gives mass as m = (E^2 + B^2)/c^2, where I've left off some units. So begin with electromagnetism as a toy example. Then the thing to notice is that E and B end up quantized at different amplitudes. Magnetic monopoles are much heavier than electrons, so assume that when you quantize E and B, the contribution of B dominates, giving you m = B^2/c^2. From there, you assume that the angle I've called "delta" is 2/9exactly, and that the reason this doesn't exactly fit the electron, muon and tau masses is cause "E" does contribute slightly. That converts the Koide formula from being a two parameter fit, with mu and delta, (the mass scale and the angle), to being a two parameter fit with a B scale and an E scale. To write the masses we have $$m_n = |B|^2(1 + \sqrt{2}\cos(2/9+2n\pi/3))^2 + |E|^2(\sqrt{2}\sin(2/9+2n\pi/3))^2$$ where B and E are constants. The contribution to B is split into two parts, $$1 + \sqrt{2}\cos(2/9+2n\pi/3)$$, so we write $$B_v = B$$, $$B_s = \sqrt{2}\cos(2/9+2n\pi/3)$$, and $$E_s=\sqrt{2}\sin(2/9+2n\pi/3)$$. That is, the "v" field is a valence field that is shared between the electron, muon, and tau, and the "s" field is a sea field that distinguishes the three generations. Then the mass equation is $$m = (B_v + B_s)^2 + E_s^2$$. What's more interesting is that if you write down the vectors $$(B_v,B_s,E_s)$$ for the electron, muon, and tau, you get the tribimaximal mixing matrix (after scaling the B stuff and E stuff so that each vector has length 1). Another way of saying this is that the vectors $$(B_v,B_s,E_s)$$ are orthogonal. Making them orthonormal defines the tribimaximal neutrino mixing matrix. (Except that when you see it in the literature, it is usually has two columns reversed so you should put the three contributions in the order $$(B_s,B_v,E_s)$$ instead.) Using the best PDG numbers for the electron and muon masses to predict the tau mass, the equations for the charged lepton masses are (ignore the precision, I haven't had time to compute the ranges and fix everything up yet): $$\begin{array}{rcl} m_n &=& 313.8561002547\;\textrm{MeV}\;(1 + \sqrt{2}\cos(2/9 + 2n\pi/3))^2\\ &&+4.6929703\;\textrm{eV}\; (\sqrt{2}\sin(2/9+2n\pi/3))^2 \end{array}$$ And the three vectors (which ignore the phase angle 2/9 because it is presumably cancelled in the neutrinos) are: $$\begin{array}{ccc} (1,& \sqrt{2},& 0)\\ (1,& -\sqrt{2}/2,& +\sqrt{3/2})\\ (1,& -\sqrt{2}/2,& -\sqrt{3/2}) \end{array}$$ In the above, note that the angle 2/9 has been removed as it is presumably cancelled in the neutrinos, which also use a similar angle. And the scaling to B and E has been removed because in computing phases, one needs to normalize by particle number rather than energy. After dividing by the lengths of the vectors, sqrt(3), and turning the three vectors into a matrix, one has: $$\left(\begin{array}{ccc} \sqrt{1/3},& \sqrt{2/3},& 0\\ \sqrt{1/3},& -\sqrt{1/6},& +\sqrt{1/2}\\ \sqrt{1/3},& -\sqrt{1/6},& -\sqrt{1/2} \end{array}\right)$$ Carl Koide paper giving Tribimaximal mixing matrix, see eqn (3.2): http://arxiv.org/abs/hep-ph/0605074 Last edited: Hans de Vries Science Advisor Gold Member After dividing by the lengths of the vectors, sqrt(3), and turning the three vectors into a matrix, one has: $$\left(\begin{array}{ccc} \sqrt{1/3},& \sqrt{2/3},& 0\\ \sqrt{1/3},& -\sqrt{1/6},& +\sqrt{1/2}\\ \sqrt{1/3},& -\sqrt{1/6},& -\sqrt{1/2} \end{array}\right)$$ Carl Koide paper giving Tribimaximal mixing matrix, see eqn (3.2): http://arxiv.org/abs/hep-ph/0605074 Interesting, I noticed that one can write the above tribimaximal matrix for neutrino mixing as the x,y,z coordinates of a tetrahedron with sides of [itex]\sqrt{1/2}$ with its top down and the origin in h/2,
thus:

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

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

$$\sqrt{m_n}\ =\ \sqrt{1/3} + \sqrt{2/3}\cos(2/9 + 2n\pi/3)$$

I see that this "A4-symmetry" was already found here by Ma:
http://arxiv.org/PS_cache/hep-ph/pdf/0606/0606024v1.pdf

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

On the other hand, Garrett Lisi uses a 3d quark matrix here:
http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.0770v1.pdf

which is the same except for some coordinate switching:

$$\left(\begin{array}{ccc} -\sqrt{1/3},& -\sqrt{1/3}, & -\sqrt{1/3} \\ -\sqrt{1/2},& +\sqrt{1/2}, & 0 \\ -\sqrt{1/6},&-\sqrt{1/6},& \sqrt{2/3} \end{array}\right)$$

see (2.4) in the paper and also page 18

Regards, Hans

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CarlB
Homework Helper
Hans,

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

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

I need to go back and look at the baryon masses and see if this sort of formula fits them better.

Hans de Vries
Gold Member
Hans,

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

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

I need to go back and look at the baryon masses and see if this sort of formula fits them better.

One can see the matrix elements as vertices of the unit cube as well,
with one vertex at 0.0 and the three nearest given by:

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

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

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

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

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

Well within experimental precision

Regards, Hans

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Hans de Vries
Gold Member
The value 0.222222047168 (465) does very nice indeed
as the Cabibbo angle as originally guessed.

$$\mbox{Cabibbo-Kobayashi-Maskawa:}\ \ \left( \begin{array}{lll} 0.9753 & 0.221 & 0.003 \\ 0.221 & 0.9747 & 0.040 \\ 0.009 & 0.039 & 0.9991 \end{array} \right)$$

$$\left( \begin{array}{lll} \cos(2/9) & \sin(2/9) & 0 \\ \sin(2/9) & \cos(2/9) & 0 \\ 0 & 0 & 1 \end{array} \right)\ \ \ =\ \ \ \left( \begin{array}{lll} 0.9754 & 0.2204 & 0.000\ \\ 0.2204 & 0.9754 & 0.000 \\ 0.00 & 0.00 & 1.000 \end{array} \right)$$

Regards, Hans

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

Hans de Vries
Gold Member
We can put this all in a picture (see below) like this:

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

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

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

Now:

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

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

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

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

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

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

Regards, Hans

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CarlB