All the lepton masses from G, pi, e

[CarlB]

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]

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:

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


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

Well within experimental precision

Regards, Hans

 

[Hans de Vries]

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]

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

 

[CarlB]

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

 

[Hans de Vries]

I loved the thumbnail, how did you do it?

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.

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.

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


Regards, Hans.

 

[Soshnikov_Serg]

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

 

[arivero]

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

Indeed even Polchinski book explains about this kind; one expects that Me*sqrt(G)*Exp(1/alpha or something so) to be of order unity, if G is the ultimate cutoff for electroamgnetism.

 

[Hans de Vries]

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.

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

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

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)

$$\begin{array}{llll} \mbox{equation:} & \mbox{\huge \frac{P^2_\mu}{P^2_e}}\ =\ 206.7682838 (54) & \mbox{\huge \frac{P^2_\tau}{P^2_e}}\ =\ 3477.441653 (83) & \mbox{\huge \frac{P^2_\tau}{P^2_\mu}}\ =\ 16.818061210 (38) \\ \\ \mbox{experim:} & \mbox{\huge \frac{m_\mu}{m_e}}\ =\ 206.7682838 (54) & \mbox{\huge \frac{m_\tau}{m_e}}\ =\ 3477.48 (57) & \mbox{\huge \frac{m_\tau}{m_\mu}}\ =\ 16.8183 (27) \end{array}$$
2) The projections P obey the Koide relation always
independent of the angle θ, while one of the projections
can become negative. ( to please Carl :^) )

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

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: outside3) 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 $(1+\sqrt{2})/\sqrt{6}$

$$P_{max}^2\ =\ 0.97140452079...$$

$$P_\tau \quad \ =\ 0.97140158810...$$

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

 

[CarlB]

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 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.

 

[arivero]

a power law n^5, reported on spp

http://groups.google.es/group/sci.physics.particle/browse_thread/thread/af72a41b7539e298?hl=es

 

[CarlB]

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

Let $$\lambda_{en}$$ 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

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

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:

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

Accuracy is very high.

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

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

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:
$$\int\;|\vec{B}|^2\;d^3r$$
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]

Carl, if you are going to review the pion thing, you could perhaps to include also Hans #349 ff and Taarik post #366.

 

[CarlB]

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:
$$\lambda^2_{\pi +}/\lambda^2_{\pi 0} = 1 + \lambda_\mu/\lambda_Z$$
Multiply by the square root of the mass of the Z:
$$\lambda_Z (\lambda^2_+/\lambda^2_0) = \lambda_Z + \lambda_\mu$$
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..

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

$$\lambda_Z(\lambda^2_{+}-\lambda^2_0) = \lambda_\mu \;\lambda^2_0$$

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

$$\lambda_Z\; (2 \lambda_Q) = \lambda_\mu \;\lambda_0$$

which is not quite linear in square root mass.

 

[CarlB]

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:
$$\left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right)$$

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:
$$\left(\begin{array}{ccc}I&J&K\\K&I&J\\J&K&I\end{array}\right)$$

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:
$$\left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right) = \left(\begin{array}{ccc}RR&RG&RB\\GR&GG&GB\\BR&BG&BB\end{array}\right)^2$$

or

$$\left(\begin{array}{ccc}I&J&K\\K&I&J\\JK&I\end{array}\right) = \left(\begin{array}{ccc}I&J&K\\K&I&J\\J&K&I\end{array}\right)^2$$

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

 

[CarlB]

After the pion, the next most basic meson is the rho. There are 5 rho resonances, the rho(770), rho(1450), rho(1700), rho(1900), and rho(2150). The last two are "omitted from the summary section" by the Particle Data Group. The lower three can be put into Koide form as follows:

$$\lambda_{\rho n} = 25.0544\sqrt{\textrm{MeV}}(10/7 -1/3 \cos(2/9 + 2n\pi/3)\;)$$

This is fairly similar to the formula for the pion:
$$\lambda_{\pi n} = 25.0544\sqrt{\textrm{MeV}}(6/5 -3/4 \cos(2/9 + 2n\pi/3)\;)$$

And of course the Koide formula uses the same angle 2/9.

I don't feel very comfortable about the rational values of the v = 10/7, 6/5, and s= 1/3 and 3/4 numbers. I feel better about the fact that v+s/2 tends to be constant when you are looking at two different resonance series. That is, 10/7 + (1/3)/2 is about equal to 6/5 + (3/4)/2 to 1%. There's a good example of this 2 to 1 ratio in some of the longer resonance series where you have more terms in the series.

The reason I like this 2 to 1 relationship between v and s is because in the derivation of the previous post, it comes up with v + 2 s \cos(\delta + 2n\pi/3), so changing v and s by 2 to 1 means that the vector length is split between the constant part (the valence part) "v", and the variable or sea part "s".

I'm still thinking on Hans' equation. The kind of thing I'm looking at is geometric as in:
http://www.sparknotes.com/math/geometry2/theorems/section5.rhtml

 

[Kea]

After the pion, the next most basic meson is the rho...

Carl, of course it is nice to have a derivation of so many particle masses, but I am guessing that people will still complain that $v$ and $s$ constitute 2 parameters for 2 outcomes in each case (forgetting for the moment the nice 'coincidence' of choice of scale etc), so could you please elaborate a little on the field idea, and the choice of these variables.

 

[CarlB]

Well Kea, there are three masses so three degrees of freedom. The "v" and "s" only provide two degrees of freedom. The coincidence is in the 2/9. I'm working on a nice graphic that will show how the coincidence works, for inclusion in the paper, which I guess I ought to give the first cut on:
http://www.brannenworks.com/qbs.pdf

The idea is to plot the angle delta as a function of the power one chooses to take. That is to explore mass formulas that look like:

$$m^{k} = v + s\cos(\delta + 2n\pi/3)$$

where the three parameters are v, s, and delta, and to plot delta as a function of k. I believe that a nice coincidence will show up when k=1/2, that is the curves for the various mesons the approrpriate baryons, and the leptons will all converge to 2/9 at that value of k.

Carl

 

[arivero]

I think that one must consider some logical grouping for the mesons. For instance, when we recorded the sqrt(m1) sqrt(m2-m3) permutations, we related them to the level spliting of the flavour octet.

 

[CarlB]

Here's a mass formula for the pions that I hope the reader will find amusing. Begin with the mass formula for the charged leptons (since I want to use "n" for radial excitations of mesons, I will use "g" for the quantum number that gives the generation):

$$\sqrt{m_{e g}}/25.0544 = \sqrt{0.5} + \cos(2/9 + 2g\pi/3)$$

where $$25.0544 \sqrt{\textrm{eV}}$$ is a mass scaling constant that makes the leptons nice and we will use for the mesons as well.

The basic idea is to think of the mesons as having radial excitations "n" with quantum numbers like the spherical harmonics of the hydrogen atom, but also having color excitations following Koide's formula. This means that we get the hydrogen wave functions, but tripled.

Guess that the three lowest mass pions = $$(\pi, \pi(1300), \pi(1800))$$ with masses (139.57, 1300, 1812) are the n=1, l=0 states with three Koide generations distinguishing them. Let's use "g" for the generation number. Then the formula for these three pion masses turns out to be:

$$\sqrt{m_{\pi 10g}}/25.0544 = 1.196797 -0.743543\cos(2/9 + 2g\pi/3).$$

Make the same guess about the three lowest mass J=1 pions = $$(\pi_1(1400), \pi_1(1600), \pi_1(2015))$$. Their masses are (1376, 1653, 2013). Since J=1, these have n=2, l=1. Note the last of these, along with the last two in the next group of three, is hard to find in the PDG. It's listed on the page titled "further states":
http://pdg.lbl.gov/2007/listings/m300.pdf
The Koide formula for these j=1 pion masses is:

$$\sqrt{m_{\pi 21g}}/25.0544 = 1.6313 + 0.1792986\cos(2/9 + 2g\pi/3 + \pi/12).$$

The extra angle pi/12 means that these have a formula like the neutrinos.

And the three lowest J=2 pions are the $$(\pi_2(1670), \pi_2(1880), \pi_2(2005))$$ with masses (1672.4, 1880, 1990). These presumably have n=3, l=2. They also have a Koide formula:

$$\sqrt{m_{\pi 32g}}/25.0544 = 1.71477 + 0.0864566\cos(2/9 + 2g\pi/3 + 2\pi/12).$$

The above 3 equations give 9 masses and they work pretty well.

For the hydrogen excitations, the total energy (or mass) of the hydrogen atom is approximately:
$$m_{H n} = 10^9 - 13.6 /n^2$$ eV.
The second term, the binding energy, is very small compared to the total energy of the hydrogen atom (about 1GeV) which is mostly due to the rest mass of the proton. Consequently, if I rewrite this as a formula for the square root of the mass of the hydrogen atom, the square root energy levels will still follow a 1/n^2 law:
$$\sqrt{m_{H n}} = 10^{9/2} - (10^{-9/2}\times 13.6/2) /n^2 \sqrt{eV}.$$

So based on the energies of the hydrogen atom, one might look for 1/n^2 dependency in the three Koide mass formulas given above, in order to unify the three equations. The three "v" terms need to be: (1.19680, 1.6313, 1.71477). These have n=(1,2,3). They can be fairly well approximated with the formula $$v_n = 16/9-\sqrt{1/3}/n^2.$$ The three "s" terms need to be (-0.743543, +0.1792986, +0.0864566 ). The sign change can be accounted for by making all three signs negative, but taking an extra phase of pi for the n=1 and n=2 case. This can be accomplished by adding a phase of $$nj\pi/2=n(n-1)\pi/2$$ to the phase angle. Then the numbers are fairly closely approximated by $$s_n = -0.75/n^2.$$ The resulting formula for nine pion masses is fairly compact:

$$\sqrt{m_{\pi nlg}}/25.0544 = 16/9 - \sqrt{3}/n^2 -(3/4)\cos(2/9 + 2n\pi/3 + jn\pi/2)/n^2$$

The above simplifications give an approximations of the v's as follows:
1.19680 -> 1.2004275
1.6313 -> 1.63344
1.71477 -> 1.71362
And the s's are approximated by:
-0.743543 -> -3/4
+0.1792986 -> 0.1875,
+0.0864566 -> 0.08333
which is fairly close.

The nine computed masses are as follows:
137.99 1271.2 1834.1
1365.5 1659.6 2032.4
1676.1 1883.0 1977.3

The nine actual masses are:
136.5(2.5) 1300(100) 1812(14)
1376(17) 1652(21) 2014(?)
1672.4(3.2) 1876-2003(?) 1974-2005(?)

Note: The error in the pi mass is due to the mass difference between the pi+ and pi0. And the masses labeled with (?) do not have mass spreads in the PDG. If more than one measurement is given, the range of measurements are given.

 

[CarlB]

Ooops. Correction on that mass formula. Managed to make three errors in it. (And what happened to the edit function within 24 hours?)

$$\sqrt{m_{\pi nlg}}/25.0544 = 16/9 - \sqrt{3}/n^2 -(3/4)\cos(2/9 + 2g\pi/3 + l\pi/12 + ln\pi/2)/n^2$$

In the above, for the pion states it applies to, l is assumed to be equal to n-1. The full set of quantum numbers is nlmg rather than nlg, but there is no energy dependency on m.

And I guess I should note that 16/9 = 2 - 2/9. So the valence part of the formula can be interpreted as the contribution from the seas of the two quarks (whose valence parts have canceled because one is a quark and the other is an anti-quark), and the 2/9 may be related to the 2/9 in the cosine.

I've got a lot more of these. One of the cool ones to analyze is the J/Psi. There are 6 states with masses fairly close together. Is there a way of dividing them up into two groups of three, where each group of three fits the above kind of formula? Let's see if I can get a solution edited into this post before it times me out.

 

[CarlB]

Remember Foot's gemoetric interpretation of Koide's formula in terms of the vector (1,1,1)? It turns out that the same may be said of the tribimaximal mixing matrix. This should be of interest to those interested in Koide's formula:
http://carlbrannen.wordpress.com/2008/06/29/the-mns-matrix-as-magic-square/

Maybe we will finally get a complete Koide type formulation for the elementary particles as a similar property applies to the CKM matrix as well. (That is, Kea is says that the CKM matrix can be written as a sum of a 1-circulant and 2-circulant matrix, and this is related to the Foot geometry, as shown above.)

Also, I should mention that I've got good results applying the Koide formula to the b-bbar and c-cbar mesons. It could be an interesting summer.

 

[arivero]

2008 review

CITATION: C. Amsler et al. (Particle Data Group), Phys. Lett. B667, 1 (2008) has some changes in W mass and width, as well as others.


The strange news: [/B]

with the move of W mass from 80.425 down to 80.398 ± 0.025 GeV the fit of me/mW to the muon anomalous difference, in post #41, should be worse. But the pdg now lists .62686, an increased difference. So both moves sort of cancel.

the positive review:

Alpha formula, in post #4, survives. But term 4 should be included in next revisions!
after term 3: 0.0072973525686533
f.s. constant 0.0072973525680.(+/-240) old
f.s. constant 0.0072973525692 .(+/-27) http://arxiv.org/abs/0712.2607

post #28, Pythagorean triples, would go better or similar, with actual tau mass.

#44, Weinberg angle, enters the 1-sigma! In 2004 it was to within 0.063% or sigma 1.2). Now with the 2008 pdg, it predicts W mass within 0.029% and sigma 0.94

The prediction is more spectacular if you see post #66, column "calculated in MeV"

 

[arivero]

Some works are borderline between standard "texture" research and our thread.
In http://arxiv.org/pdf/hep-ph/9703217v1, they suggest
[tex]
\sqrt {\m_t\over\m_c} = {m_\tau\over m_\mu}
[\tex]
In http://arxiv.org/pdf/hep-ph/0106286v2, a lot of such sqrt textures are invoked as popularly known.
In http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198812215 and http://dx.doi.org/10.1016/0370-2693(87)91621-2 Christof Wetterich leads other attempts.

 

[arivero]

Alpha formula, in post #4, survives. But term 4 should be included in next revisions!
after term 3: 0.0072973525686533
f.s. constant 0.0072973525680.(+/-240) old
f.s. constant 0.0072973525692 .(+/-27) http://arxiv.org/abs/0712.2607

Correction, the full calculation is in http://www.physics-quest.org/, so we can compare
0.00729735256865385342269 theoretical
0.0072973525692 .(+/-27) measured http://arxiv.org/abs/0712.2607
quotient:

$$1.000000000075 \pm 0.000000000369$$

 

[Hans de Vries]

Correction, the full calculation is in http://www.physics-quest.org/, so we can compare
0.00729735256865385342269 theoretical
0.0072973525692 .(+/-27) measured http://arxiv.org/abs/0712.2607
quotient:

$$1.000000000075 \pm 0.000000000369$$

Wow!

That's a nice surprice! New measurements plus a revised value of the eighth-order
QED contribution to the anomalous magnetic moment of the electron from Kinogarbagea
brings our series in line with experiment to a fraction of the (improved) error.

$\alpha^{-1}~=~~$ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
$\alpha^{-1}~=~~$ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)
$\alpha^{-1}~=~~$ 137.035 999 095 829 Theoretical value from alpha series.

New measurement from Gabrielse's group: http://arxiv.org/abs/arXiv:0801.1134v1
Revision of the eight-order QED term: http://arxiv.org/abs/0712.2607
The alpha series: http://physics-quest.org/fine_structure_constant.pdf

This happened a while ago already. somehow we missed it. Thank you for posting!



Regards, Hans.

 

[arivero]

To complete the record:

$\alpha^{-1}~=~~$ 137.035 999 108 (450) Original conject. from Codata 2004.
$\alpha^{-1}~=~~$ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
$\alpha^{-1}~=~~$ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)

$\alpha^{-1}~=~~$ 137.035 999 095 829 Theoretical value from alpha series.

Wow!
This happened a while ago already. somehow we missed it. Thank you for posting!

I somehow fused it with the already reported, somewhere in the middle of the thread, 2006 update.

As for the calculation, the first order formula
$$\alpha^{-1/2}+ \alpha^{1/2}=e^{\pi^2 \over 4}$$

is crying a word: duality. Problems are:
1) It is not so clear how the succesive corrections are applied.
2) the formula for dyon energy (particle with electric+monopole charge) is
$$M^2 \approx e^2 + e^{-2} = \alpha^{1}+ \alpha^{-1}$$
so this formula is a kind of rare square root of the usual duality formula.

3)Still we haven't got a clue for the precise choosing of e(pi^2/4) except that it works.

 

[Hans de Vries]

To complete the record:

$\alpha^{-1}~=~~$ 137.035 999 110 (450) Original conject. from Codata 2004.
$\alpha^{-1}~=~~$ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
$\alpha^{-1}~=~~$ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)

$\alpha^{-1}~=~~$ 137.035 999 095 829 Theoretical value from alpha series.

Both the direct measurement from the Quantum Hall effect (CODATA 2004) and
the indirect one from the Harvard g/2 measurements are now in agreement.

As for the calculation, the first order formula
$$\alpha^{-1/2}+ \alpha^{1/2}=e^{\pi^2 \over 4}$$

is crying a word: duality. Problems are:
1) It is not so clear how the succesive corrections are applied.
2) the formula for dyon energy (particle with electric+monopole charge) is
$$M^2 \approx e^2 + e^{-2} = \alpha^{1}+ \alpha^{-1}$$
so this formula is a kind of rare square root of the usual duality formula.

3)Still we haven't got a clue for the precise choosing of e(pi^2/4) except that it works.

To explain it is the hard thing... Do you have a link for the this dyon
mass formula?Regards, Hans

 

[arivero]

To explain it is the hard thing... Do you have a link for the this dyon
mass formula?

In fact, the point of not having recognised it immediately is proof of the amateurish character of the internet forums

Remember the lore: if a theory goes with a coupling constant a, the dual theory goes usually with a coupling constant 1/a.

In the case of T-duality the coupling have dimensions. Or said otherwise, Tduality is possible because the theory has a dimensional scale. But for ElectroMagnetic duality, the coupling is adimensional. The (square of the?) magnetic energy of a monopole goes as hbar/alpha.

Actually, a lot of the initial formulae you suggested in 2004 were of the form x+1/x for some constant x; it is hard of believe that you were not inspired, at least inconsciently, by the usual popular remarks on duality. Again, I failed to notice it, so perhaps both did.

A dyon is a particle having both elemental and monopole charges. So its mass square has two contributions.

I saw the formula for the mass in a remark about http://www.slac.stanford.edu/spires/find/hep/www?irn=251658"

The dyon mass formula appears in the last page of Montonen-Olive preprint, and they quote towards to older articles.

To put some stuff in handwritten from Witten itself, see slide 25 of http://math.berkeley.edu/index.php?...tManager_op=downloadFile&JAS_Document_id=2101 in http://math.berkeley.edu/index.php?...ntManager_op=viewDocument&JAS_Document_id=116
and then slide 28

 

[arivero]

Perhaps I have cited very heavy artillery. I apologize; I have never put a lot of interest on these topological objects; it is a long history.

So another source is Zee's book on QFT, and it redirects to http://www.maths.ed.ac.uk/~jmf/Teaching/EDC.html" , from Figueroa-o'Farrill.

In section 1.3.1 we are told that the mass of a monopole always meets the Bogomol’nyi bound. In section 1.3.2 the bound is "saturated" in the Prasad–Sommerfield limit. So we have the initials BPS. The saturated bound is the formula for the mass of a monopole with both electric and magnetic charge I told above, depending unfortunately of $$\alpha^1$$ instead of $$\sqrt{\alpha} \; (=e)$$.
Alert: Section 1.4.2 explains another earlier calculation of Witten, and it builds the "Noether" charge of a dyon as funtion of charges q g plus a theta-vacuum, in the shape
$$N={q \over e} + (e g) {\theta \over 8 \pi^2}$$
and then argues that N must be an integer. But note that q and g are not fully adimensional; they are defined as, say q=(n e) and (g=4 pi m/e), so the factors of e there in the formula are intended only to cancel the internal ones. The argument in 1.4.2 amounts to prove that when m=1, n is also quantized.

 

[arivero]

Could it be better to write the first order approximation squared, ie

$$e^{\pi^2/2} -2 \approx \alpha + \alpha^{-1}$$

In such case, how could the corrective terms be applied?

 

[CarlB]

Or is it useful to begin with a first order equation:
$$\sqrt{\alpha} = i^{\ln(i)} = i^{i\pi/2}$$
$$= (i^i)^{\pi/2} = (e^{-\pi/2})^{\pi/2} = e^{-\pi^2/4}$$

Or did I make an algebra mistake here?

 

[Kea]

Assuming that the first order relation is indicative of duality, this would be Langlands, or electric magnetic duality, most simply understood via relations coming from the modular group, or its braid group cover. This suggests interpreting the Gaussian term as a trace factor (usually called $d$ in knotty algebra maths), or a normalisation arising from geometry of non integral dimension, not unlike Deligne's Gaussian for the discrete Fourier transform, except that the factor of $\pi$ presumably results from an infinite sequence of terms.

 

[arivero]

Or is it useful to begin with a first order equation:
$$\sqrt{\alpha} = i^{\ln(i)} = i^{i\pi/2}$$
$$= (i^i)^{\pi/2} = (e^{-\pi/2})^{\pi/2} = e^{-\pi^2/4}$$

Or did I make an algebra mistake here?

:rofl::rofl: No, I think it is fine. Of course $i=e^{i\pi /2}$ is a nice trick to get this power of pi. Actually it shows that the seed chosen by Hans is not a random number.

Most important perhaps, you can -as Hans hinted in 2004, calling this a "gaussian"- also write in a formal way

${i \over \sqrt \pi} \int_{-\infty}^\infty e^{ (x-0) (x\pm \pi)} dx = e^{- {\pi^2 / 4}}$ or if you prefer ${1 \over \sqrt \pi} \int_{-\infty}^\infty e^{ - (x-0) (x\pm i \pi)} dx$

it is because of this second detail that I have wasted, poorly, the weekend looking into soliton, instanton, dyons etc...ons. Not only the x-(1/x) part of the equality is typical of this kind of theories; also the integral of gaussians is the part to get exact values of energy, barrier penetration, mass etc. So the "1-st order" (Hans call the one you wrote "0-th order") formula has common points with non-perturbative QM/QFT/Qthing in both sides of the expression, which is intriguing.

 

[Count Iblis]

You only got ten digits, that's not enough to guess the correct formula.