All the lepton masses from G, pi, e

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

 

"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

 

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

[tex]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[/tex]

where B and E are constants. The contribution to B is split into two parts, [tex]1 + \sqrt{2}\cos(2/9+2n\pi/3)[/tex], so we write [tex]B_v = B[/tex], [tex]B_s = \sqrt{2}\cos(2/9+2n\pi/3)[/tex], and [tex]E_s=\sqrt{2}\sin(2/9+2n\pi/3)[/tex]. 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 [tex]m = (B_v + B_s)^2 + E_s^2[/tex].

What's more interesting is that if you write down the vectors [tex](B_v,B_s,E_s)[/tex] 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 [tex](B_v,B_s,E_s)[/tex] 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 [tex](B_s,B_v,E_s)[/tex] 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):

[tex]\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}[/tex]

And the three vectors (which ignore the phase angle 2/9 because it is presumably cancelled in the neutrinos) are:
[tex]\begin{array}{ccc}
(1,& \sqrt{2},& 0)\\
(1,& -\sqrt{2}/2,& +\sqrt{3/2})\\
(1,& -\sqrt{2}/2,& -\sqrt{3/2})
\end{array}[/tex]

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:
[tex]\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)[/tex]


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}[/itex] with its top down and the origin in h/2,
thus:

[tex]\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)[/tex]

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:

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


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
https://www.physicsforums.com/showthread.php?t=196498

which is the same except for some coordinate switching:

[tex]\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)[/tex]

see (2.4) in the paper and also page 18

Regards, Hans

 

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:

[tex]\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)[/tex]

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

[tex]\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)[/tex]

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

https://www.physicsforums.com/showthread.php?t=46055&page=8


[tex]
\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}
[/tex]

Well within experimental precision

Regards, Hans

 

The value 0.222222047168 (465) does very nice indeed
as the Cabibbo angle as originally guessed.


[tex]
\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)
[/tex]

[tex]
\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)
[/tex]


Regards, Hans

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

 

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 [itex]e[/itex] and [itex]\tau[/itex]

[tex]\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)[/tex]

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 ?)

[tex]
\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}
[/tex]



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

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


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

 

I loved the thumbnail, how did you do it?

I'm still mulling over the concept of torque here. I spent the weekend making a java applet that graphs the discrete Fourier transform of the baryons. I got the graphical user interface (GUI) to run, but didn't see the patterns I expected, just noise. That could be defects in the program, defects in my understanding of how to use it, etc.

The reason for looking at discrete Fourier transforms with respect to masses was given by Marni Sheppeard here:
http://arcadianfunctor.files.wordpress.com/2007/11/fouriermass07.pdf

 

It's done with Povray, The projections are just shadows, they come
basically for free. One could also draw the 3D object multiple times with
each time one of the coordinates fixed. You would get "2.5 D" projections
with the balls and cylinders all aligned in the same 2D plane. Might give
a nice effect as well.

If the input is a number of mass spikes then the "noise" may well
be the correct result.


Regards, Hans.

 

Me = sqr((h/(2*Pi))*C/(4*Pi*G)) * Exp( - 16 * Pi)

Me = 9.08086 * 10 ^ -31 kg

it`s not for real electron, bat for "naked" electron
Soshnikov_Serg

 

Cleaned up + Neutrino masses preliminary.

The charged leptons are placed 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.

[tex]\left(\begin{array}{ccc}
e_z& e_y& e_x \\
\mu_z & \mu_y& \mu_x\\
\tau_z & \tau_y& \tau_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)[/tex]

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


Now:

1) The projections P on the vector (cos θ, sin θ, 1) lead
to the exact charged lepton masses if we use for the angle:

θ = 7/6 pi + 0.22222204717 (47)

[tex]
\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}
[/tex]



2) The projections P obey the Koide relation always
independent of the angle θ, while one of the projections
can become negative. ( to please Carl :^) )

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

Angle θ with pointer inside the cube: all projections positive.
Angle θ with pointer outside the cube: one projection negative.

The latter is a requirement for Koide's relation to describe the
neutrino mass-eigen-states. This setup does so naturally.
There are the following regions for θ:

075 - 105 degrees: inside
105 - 195 degrees: outside
195 - 225 degrees: inside
225 - 315 degrees: outside
315 - 345 degrees: inside
345 - 075 degrees: outside


3) What fixes the angle θ? For the charged leptons we have the
following numerical coincident involving the projection of tau:

The maximum value of a projection is 0.9855985... is [itex](1+\sqrt{2})/\sqrt{6}[/itex]

[tex]P_{max}^2\ =\ 0.97140452079...[/tex]

[tex]P_\tau \quad \ =\ 0.97140158810...[/tex]

It has to be said that the masses are still very sensitive to the very small
error (The muon mass uncertainty determines the uncertainty because of
Koide's relation)



4) The coordinates could even be interpreted as real coordinates
with the Pauli-Weisskopf interpretation of the wave function as a
continuous 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 (cos θ, sin θ, 1) with the z-axis is same as the
precessing angle of a spin 1 boson. In other words: A spin coupling
like adaption of the Yukawa coupling.


Hypothetical Neutrino mass-states eV from here: (page 48)
http://arxiv.org/abs/hep-ph/0603118
are used in the drawing below.


Regards, Hans

 

On the subject of applying Koide's mass formula to the baryon resonances, I've added a <a href="http://carlbrannen.wordpress.com/2007/12/16/regge-trajectories-and-koides-formula/">blog [Broken] post</a> on how this fits in with Regge trajectories.

The short version is that Regge trajectories look like M = sqrt(L), where M and L are mass and angular momentum. This comes from an assumption of flux tubes that have energy (and therefore mass) proportional to their length R, but angular momentum proportional to R^2. These flux tubes have different energies per unit length. The formula applies to baryons with the same quantum numbers except for angular momentum.

Koide's mass formula looks like M = E^2 where E is a field strength, and it applies to baryons with identical quantum numbers. That is, it applies to groups of three resonances that share identical angular momentum.

If you combine the two mass formulas, you end up concluding that the flux tubes that give the Regge trajectories have diameters that are proportional to the Koide field strength.

 

Soon I am going to release a paper with some new coincidences based on square roots of masses. Here's an example.

Let [tex]\lambda_{en}[/tex] be the (positive) square root of the mass of the nth electron, that is the square roots of the masses of the electron, muon, and tau. The Koide equation can be written as

[tex]\lambda_{en} = 25.0544\sqrt{\textrm{MeV}}(\sqrt{1/2} + \cos(2/9 + 2n\pi/3)\;)[/tex]

The lightest meson is the pion. It comes in three varieties (with the same quantum numbers), the pion, pi(1300), and pi(1800). The square roots of their masses are given by an equation similar to the above:

[tex]\lambda_{\pi n} = 25.0544\sqrt{\textrm{MeV}}(6/5 -3/4 \cos(2/9 + 2n\pi/3)\;)[/tex]

Accuracy is very high.

To express the relationship as a linear one in terms of square roots of masses, we have:

[tex]4\lambda_{\pi_1} + 3\lambda_{e_3} =
4\lambda_{\pi_2} + 3\lambda_{e_2} =
4\lambda_{\pi_3} + 3\lambda_{e_1} = \lambda_{\pi_1} + \lambda_{e_3}
=\sqrt{138}+\sqrt{1777}\;\sqrt{\textrm{MeV}}.[/tex]

Linear relationships on mass are suggestive that the elementary particles are collections of objects that don't interact enough to completely change their character. An example is the masses of the atomic nuclei. The number of nucleons is approximately proportional to the mass of the nuclei (and atom). And nuclei mass can be thought of as mostly having to do with neutrons and protons and only a little to do with the force that keeps them together.

Linear relationships on square root of mass are suggestive that the mass (or energy) comes from an object which is linear but is proportional to the square root of energy. For example, the energy in a magnetic field is:
[tex]\int\;|\vec{B}|^2\;d^3r[/tex]
The magnetic field is a linear object. That is, if two objects each have magnetic fields and they are superimposed, then by linear superposition, the total magnetic field is the sum of the magnetic fields. The energy in the field is proportional to the square of the field and so if we wish to do a linear operation on the object creating the energy / mass, we need to first take the square root of the energy / mass.

The first thing a physicist does to a linear data stream is to take the Fourier transform of it. For a discrete set of 3 masses this would be the discrete Fourier transform. Marni Sheppeard wrote a short paper showing that the Koide formulas are related to discrete Fourier transforms:

http://arcadianfunctor.files.wordpress.com/2007/11/fouriermass07.pdf

I'll eventually get back with more.

 

Arivero,

I'm looking at things that relate the generation structure of the fermions with the excitation structure of the mesons and baryons through square roots. The use of square roots is supposed here to get the situation to one where there is something linear going on.

Hans #349 is possibly appropriate because it has a square root in it, but it's written as ratios. Maybe I could rewrite it in square root form. Hmmm, let's see. Hans has, writing things in square root mass terms:
[tex]\lambda^2_{\pi +}/\lambda^2_{\pi 0} = 1 + \lambda_\mu/\lambda_Z[/tex]
Multiply by the square root of the mass of the Z:
[tex]\lambda_Z (\lambda^2_+/\lambda^2_0) = \lambda_Z + \lambda_\mu[/tex]
The right hand side is nice because it is linear in square roots but the left hand side is not. So rewrite the ratio of the pion masses..

[tex]\lambda_Z(1 + (\lambda^2_{+}-\lambda^2_0) / \lambda^2_0) = \lambda_Z + \lambda_\mu[/tex]

[tex]\lambda_Z(\lambda^2_{+}-\lambda^2_0) = \lambda_\mu \;\lambda^2_0[/tex]

Write [tex]\lambda_+ = \lambda_0 + \lambda_Q[/tex] where Q is the contribution to the mass field that comes from charge and is small compared to [tex]\lambda_0[/tex]. Keeping first order in Q we have:

[tex] \lambda_Z\; (2 \lambda_Q) = \lambda_\mu \;\lambda_0[/tex]

which is not quite linear in square root mass.

 

I should write down my derivation of the pi meson mass formula.

The pi+ is made from a up quark and an anti-down quark. They have to have opposite colors, but other than that, if you know the quantum numbers of one, you know the quantum numbers of the other.

Arbitrarily, consider the QM problem of an up quark moving in the field of an anti down quark. Make this a qubit kind of problem by ignoring position and momentum information. Then the only thing the up quark can do is change its color. The anti down quark will also change color, but this is defined by conservation laws and so we don't need to worry about it; if the up quark is blue, then the anti down quark is anti blue.

Over the long term, just looking at the up quark, there are nine transitions going on, from {R, G, B} to {R, G, B}. This amounts to a scattering matrix. We have nine amplitudes, write them as a matrix:
[tex]\left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right)[/tex]

SU(3) is an unbroken symmetry so we can assume RR = GG = BB, as well as RG = GB = BR and GR = BG = RB. This requires the matrix to be circulant. Let I = RR, J = RG, K = GB, so the matrix has to be of the form:
[tex]\left(\begin{array}{ccc}I&J&K\\K&I&J\\J&K&I\end{array}\right)[/tex]

The RG and GR amplitudes are the time reversals of each other. It follows that they must be complex conjugates, that is, K = J*. That makes the above circulant matrix Hermitian.

Suppose that J and K are nonzero, but I is zero. Is this possible?

The action of J on an arbitrary state is to increment its color (that is, if you think of the three scattering matrix terms that all must be equal to J, their action can apply to any state and increments the color). The action of K is to decrement its color. Since both these processes are possible, it is also possible to have one followed by the other. Such a process would leave the colors unchanged. Therefore I cannot be zero.

This is an argument similar to the one Feynman made when he contemplated what happens when you insert an intermediate state between the initial and final states; the path integral formulation says that you have to sum over the intermediate states.

Suppose my initial state is G and my final state is B. Following Feynman, we insert an intermediate state, which can be R, G, or B. Then we sum over intermediate states. We find that:

GB = GR RB + GG GB + GB BG

We can insert intermediate states between all the other 8 amplitudes. We end up with a full set of nine equations:

RR = RR RR + RG GR + RB BR,
RG = RR RG + RG GG + RB BG,
RB = RR RB + RG GB + RB BB,
GR = GR RR + GG GR + GB BR,
GG = GR RG + GG GG + GB BG,
GB = GR RB + GG GB + GB BB,
BR = BR RR + BG GR + BB BR,
BG = BR RG + BG GG + BB BG,
BB = BR RB + BG GB + BB BB.

We can conveniently rewrite the above equations in matrix form:
[tex]\left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right)
= \left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right)^2[/tex]

or

[tex]\left(\begin{array}{ccc}I&J&K\\K&I&J\\JK&I\end{array}\right)
= \left(\begin{array}{ccc}I&J&K\\K&I&J\\J&K&I\end{array}\right)^2[/tex]

The above has three solutions:
[tex]\begin{array}{rcl}
I&=&1/3,\\
J&=&\exp(+2n\pi/3)/3,\\
J&=&\exp(-2n\pi/3)/3\end{array}[/tex]
for n = 1,2,3.

This is good because there are exactly three pi mesons. Let's see if we can get Koide's formula for them.

We have nine processes describing the meson, that is, 3 copies of I, 3 copies of J, and 3 copies of K. The J and K processes are related by time reversal, but we really don't know how these relate to the I process. To convert these nine processes into a description of the meson, let's suppose that two constants convert the above amplitudes into field amplitudes, say "v" for the I processes and "s" for the J and K processes. The J and K processes are complex conjugates and are complex numbers. We also don't know how to convert these complex numbers into field strengths. Let's suppose that to convert them into a field strength requires a complex phase (maybe due to non commutativity like Berry phase), so they need to be multiplied by [tex]exp(i \delta)[/tex] for the J and the complex conjugate for the K.

Then the total field strength for the three particles is given by

v + 2 s \cos(\delta + 2n\pi/3)

With delta = 2/9, v = 6/5 and s = -3/4, this is the Koide formula for the lowest energy mesons, the pi mesons. With delta = 2/9, v = sqrt{1/2}, and s = +1, this is the Koide formula for the leptons. More to come later.