# New Paper on Hadrons and Koide’s mass formula

I’ve got a paper on the hadrons ready to submit to Phys Math Central. This is a fairly new peer reviewed open access journal for which I have a “pass” that allows me to avoid having to pay the \$1500 submission fee, so long as I submit before January 31. This is a big deal and I want to do it right, so I’m looking for advice from readers.

The paper as it stands is here:
Koide mass formulas for the hadrons, 49 pages, LaTeX.

The subject is the extension of Koide’s lepton mass formula to the neutrinos and then to the hadrons. I’ve written the background section so it should be accessible to typical grad students in physics.

I’ve put this together as an example of applying quantum information theory to the practical problem of the hadron masses. This all is fairly simple stuff and it uses very basic ideas in quantum mechanics.

Quantum information treats the information contained in quantum states. For a colored particle, that information is “red”, “green” or “blue”. But the usual method of modeling the color force is instead to approximate the color force as a modification to the Coulomb force.

On looking at the problem as a color force problem, we find that the excited states of a color bound state, in the quantum information approximation, must come in triplets and these triplets are related by the discrete Fourier transform on 3 variables.

So we apply the discrete Fourier transform to triplets of particles and what do we find. Well we get a simple equation relating the charged lepton masses which implies Koide’s mass equation. And we get an implication for a neutrino version. These two equations tell us how leptons look when they are excited, as if the leptons were composite particles.

The mesons and hadrons are composite particles and also have excitations. So, naturally, we plug those excitation mass numbers into the discrete Fourier transform and what do we get. Yes, we get more copies of the lepton mass equations.

The paper includes 33 mass equations that fit triplets of mesons, and 6 more that fit triplets of baryons.

Filed under particle physics, physics

### 59 responses to “New Paper on Hadrons and Koide’s mass formula”

1. Kea

Excellent news. Hopefully the reviewers will be reasonable people who don’t think the physics content of one’s paper depends on the hardness of one’s hats.

2. kneemo

Quite a tour de force, Carl. Have you sent this to Tommaso? I’m sure he’ll have some great suggestions concerning the overall organization, as the experimental particle physics community is likely your primary audience. I, for one, think some images to help with the Berry-Pancharatnam topological phase discussion would be helpful.

A few small typos I found:

“The pure denstiy matrix…” pg. 12 first paragraph

“FOne would then write…” pg. 15 third paragraph

“problem of findint” pg. 22 first paragraph

Overall, excellent work though. Your paper adds much support for the importance of circulants in particle physics. Any nonperturbative unified theory of particle physics should have something to say about such circulants, especially if such a theory ends up being a matrix theory. 😉

3. carlbrannen

Okay, I put the thing through a spell chequer, and now it’s got about two dozen improvements. I also like the idea about a picture showing BP phase and I’ve got those from other stuff I’ve written, in particular my book on density matrices.

By the way, I’m not sure whether to add it, but there is a subtlety in the comment where it notes that two degrees of freedom could be either spin-1/2 and get BP phase, or spin-0 and not. And that is that it isn’t that easy to get these kinds of things to come out. If you take two spin-1/2 objects with BP phase charge n=1 and combine them aligning, the result is a spin-1 object and it gets a BP phase “charge” of n=2.

So it messes things up to say that neutrinos and electrons are made from similar things. To fix it, my instinct is that you have to go to braid statistics. Along that line, note equation (11) of this:

New Kinds of Quantum Statistics
Frank Wilczek
http://arxiv.org/abs/0812.5097

Basically, another way of describing the BP phase is as one of these new kinds of quantum statistics. I’m thinking I need to add this as a reference.

4. kneemo

Yup, sounds like you need to write down some BP charge “fusion rules” for various particle species (spin 0, spin 1/2, spin 1, etc.), as fusion rules specify the possible value of the charge when you combine two charged particles in anyon theory.

Some good references are:

Preskill’s KITP talk

and especially page 5 of
Non-Abelian Anyons and Topological Quantum Computation – Freedman et al.

It appears you have a non-abelian anyon model with multiple fusion channels.

5. carlbrannen

Obviously y’all are going to have to write the papers connecting this up as non abelian anyons.

The difference between what I’m doing and what is going on in with anyons or braids is that I’m primarily concerned with what happens to one quantum object. So to me, the various RG, RB, BR, etc., objects are components to a single path. The path loops back on itself, but it is the path of a single particle. With the braid theory, they’re keeping track of three (or more) particles simultaneously. So these are inherently multiparticle quantum mechanics.

The non Abelian stuff Wilczek talks about in
http://arxiv.org/abs/0812.5097
is algebraically related to the non Hermitian pure density matrices in a fairly simple way.

We’re considering braids of 3 objects, use x, y, and z for the Pauli spin matrices (or Clifford algebra generators). The braid is small enough that no two braids commute. Then the remaining braid rules are: $\sigma_{xy}\sigma_{yz}\sigma_{xy} = \sigma_{yz}\sigma_{xy}\sigma_{yz}$ $\sigma_{yz}\sigma_{zx}\sigma_{yz} = \sigma_{zx}\sigma_{yz}\sigma_{zx}$ $\sigma_{zx}\sigma_{xy}\sigma_{zx} = \sigma_{xy}\sigma_{zx}\sigma_{xy}$
and Wilczek’s solution is $\sigma_{xy} = (1+i)(1+xy)/2$
which defines a swap between the x and y states.

Now I’m using R, G, and B which are related to x, y, and z by
R = (1+x)/2 or x = 2R-1,
G = (1+y)/2 or y = 2G-1,
B = (1+z)/2 or z = 2B-1.
and you might want to rewrite i using i=xyz, but either way you can write a braid operator in terms of the projection operators for MUB spin.

6. Dave

Hi Carl,
On pg. 11, I think there is a typo somewhere in here…
exp(i tan−1(3 −√8)) ≈ exp(i 0.28599378)

7. carlbrannen

Thanks Dave,

It’s fixed. I screwed up the final calculation. The number 0.2859… should be 0.16991845…. This is the angle whose cosine and sine are $\sqrt{1/3} \pm \sqrt{1/6}.$

Carl

8. Dave

You probably knew this, but it’s also (arcsin(1/3))/2.
Or, half the angle (19.47deg) where the lower points of an inscibed tetrahedron fall on the unit sphere.

9. Dave

…and (arccos(1-2/9))/4=.16991845

10. Dave

even simpler… (cos(19.47))/2=sqrt(2/9)

11. carlbrannen

Dave,

No, I’m stunned. That’s very much appreciated news to me. I’m going to add something into the paper to that effect.

The reason it’s interesting to me (and also no doubt to Marni Sheppeard) is because the number 2/9 appears as the mysterious charge or amplitude later on, see the text just below equation (67). Also known as “that damned number”.

If you give me a real name and I’ll add a note to the effect that you pointed this out. Otherwise, I’ll attribute it to “Dave”.

12. Kea

Cool, Dave! I’m sure you realise just how cool.

13. David Lackey

Hi Carl,
I would be very honored to be mentioned in your amazing paper, Thank You!

Looking at pg.14, 1st paragraph, I noticed tan(19.47)=sqrt(1/8).

maybe relevant?…
tan(90-19.47)=sqrt(8), and tan((90-19.47)/2)=sqrt(1/2)

14. Daniel de França MTd2

I saw that you sugested at physics forums that in order to find the 3 generations of Garrett Lisi’s model, one should use your method. But you haven’t yet applied your method to find the havier quarks. So, how will you find them?

Carl: I suggested one way to expand E8 to 3 generations was to look at the discrete Fourier transform as described in the paper. The first applications are with heavy quarkonium. These are mesons made with heavy quarks. Of course quarks themselves are not observable so I can’t find them.

15. Mike

Pretty good Dave, you probably also recognized this as part of the geometry of the star hexagon in a two dimensional drawing.
And Carl, guess this would connect with one of your previous papers on the Star of David.
Are you trying to “keep it simple” for this paper,
or have you revised the more “precise value”?

Carl: Mike, the value 2/9 is not as exact enough for measurement. But there always seem to be corrections to qm formulas and 2/9 is pretty close. So I decided it’s better to stick with the 2/9. Part of the reason for this is that it’s at least possible that the error could appear not as a correction to delta, but instead as a correction to the tau mass in particular.

16. Daniel de França MTd2

I got it now. The Top quark is an excited state in E8.

17. Mike

Hi Carl, Then for future reference you might also be interested in the approximation cos (77.16) = 2/9.

77.16 is close to the apex angle of the large triangle in the heptagon or ~ sin (90/7) = 2/9.

Both the heptagon and the hexagon fit in the
Cosmological Circle from which alpha,
the fine structure constant, is derived; and a very close approximation to your latest value for delta is on the fqxi essay page…

We noted alpha^3, pi, and 1.8 in our essay. 1.8 = 9/5.

((9/5 + (alpha^3 x pi^1/3))^-1 – 3/9 = 0.2222220466 ~ 2/9

Also as an historical aside and “confirmation”
an even simpler formulation from harmonic proportion was found from the Foundation Stone of ancient geometry. This was the Greek template for the formation of matter. The star tetrahedron is related to Metatron’s Cube which is an aspect of the Foundation Stone. Basically the proportions involved include 144/648 = 162/729 = 2/9.

18. Dave

Hmm, interesting….
If you have a sphere of radius 1 with an inscribed tetrahedron, then a circle placed within a face will have a radius of sqrt(2/9), exact.

19. Dave

I think we’re barking up the wrong tree focusing on 2/9 rad. If you take the cube root of 2/3 rad, you get 2/9 rad, (and the other 2 roots, of course.)

20. Dave

And that makes the extra neutrino term(pi/12), actually pi/4, before taking the cube root.

21. carlbrannen

Dave,

You’re quite right that before taking the cubed root the value is 2/3 instead of 2/9, and the pi/12 becomes pi/4. And pi/4 is the correct Berry-Pancharatnam or topological phase. But it turns out that I don’t have any better explanation of 2/3 than 2/9.

In the paper, the reason that the pi/12 gets a cubed root while the 2/9 does not is that the pi/12 only appears when you go around the color space and return back where you started. It’s a topological phase. On the other hand, the 2/9 is supposed to come from a QFT correction to the vertex. So it applies to the vertices individually; you don’t have to string together a sequence that gets you back where you started.

As far as what this has to do with various geometric objects, I have no idea. The basic problem is that there are a lot of ways of getting 2 and 3 out of mathematics.

22. Dave

The binding energy of the quarks is commonly listed as 313MeV, (proton 938.27MeV), and is “1” the natural lepton mass scale (where tau is 5.6617).
Or, where the angle is pi/2.
Since most of the nucleon mass is binding energy, this might be the “ground state” energy of a single up-up bond.

23. Dave

I have a very simple formula for the top mass, which involves a recognition that in natural units, the top quark and electron are nearly equidistant from “unity”…

(cos(2(arctan(3-sqrt8))))^2=8/9 exact
..or just take (2/9)*4, or 2/3+2/9

electron in “natural” units=.00162812978=e_n

1/(e_n/(8/9))=545.957

545.957 * 313.8563Mev=171.352GeV

(313Mev converts natural units to MeV)

PDG top mass=171.2 (/171.352=.999113)

Don’t know if it’s a coincidence, but that flip around unity might illustrate the essential difference between leptons and quarks.

24. Mike

Interesting Dave, more connections… first, thanks for posting this,
another curious connection.

Dave
January 22, 2009 at 2:24 pm

“Hmm, interesting….
If you have a sphere of radius 1 with an inscribed tetrahedron, then a circle placed within a face will have a radius of sqrt(2/9), exact.”

***

As you probably know, 8/9 is the whole tone ratio in classical harmonics and is also represented by the ancients as a squaring of the circle. A circle with circumference 8 is “squared” (nearly the same area) by a square with a perimeter of 9. This is the “main” construction in the Cosmological Circle. The perfect fifth 3/2, or here 2/3 was represented by a triangle in a circle. 2/3 = 6/9 = 486/729 (a tri-circle with total circumference of 729 is circumscribed by a circle of 486 circumference). 8/9 – 6/9 = 2/9. And from the Foundation Stone 8/9 = 648/729. 144 = 12^2 360/12 = 30. Ratios common to both the Foundation Stone and the Cosmological Circle.

Carl’s now “blessed number” delta again,

((144 – (alpha^3 / 30)^1/2))/648 = 0.2222220466 ~ 2/9

Same value as above post:

January 22, 2009 at 8:05 am

((9/5 + (alpha^3 x pi^1/3))^-1 – 3/9 = 0.2222220466 ~ 2/9

25. Dave L.

Here is a very convincing description of the arrangement of quarks in a baryon…

http://www.unclear2nuclear.com/shape.php

26. David Lackey

I found a simple formula for the neutron-proton mass difference.

2006 CODATA neutron=1.00866491597
proton=1.00727646677

/ .33693868676 = 2.9936156209 = Nc
/ .33693868676 = 2.9894948439 = Pc

(.3369.. converts amu to natural units based on the Carl’s marvelous lepton mass formula)

(1/N^.5)+15=15.5779656=Nt

1/(Nt^2)=.00412077137

Nc – .00412077137 = 2.9894948495 = Pt

Pt / Pc = 1.00000000189187

It might be a coincidence, but it popped up as soon as I put the N-P difference into my natural mass values spreadsheet this morning.

27. Mike

Ok Dave, Thanks for the link on “triangular” quarks,
and where is Carl?

15?

[audio src="http://www.earthtech.org/michael/bio/puzzling.wav" /]

28. Dave

Hi Mike,
Since the Neutron & proton are composed of three quark-bonds, the real number is probably 15/3.

29. Dave

Wow!
When 1/15 , as input to Carl’s lepton formula is 48.7026950deg, and it’s cos^2=0.43555555556!

30. Kea

Dave, you should write these cool things down carefully in a small paper, and try to publish it.

31. carlbrannen

The way to distinguish numerology from physics is that physics uses multiple examples of a formula working while numerology typically gives one only a single coincidence. That said, I just noticed that the 15.577xxx in David Lackey’s comment of February 7, is that the .577 part is the first 3 digits of the Euler-Mascheroni constant. I suspect that having a good memory for this sort of numerical coincidences is not a good thing.

32. Mike

Well anyway, numero neato Dave… 15 again!

(7/5)^2 x 2/9 = (2 x 7^2)/15^2 = 0.435555555…

33. Dave

The neutron has 3 electric charge bonds:
2/3*-1/3=-2/3, of which there are 2 = -4/9
and one of -1/3 * -1/3 = +1/9 (Total is -3/9)

The proton also has 3: +2/3*-1/3=-2/9 of which there are 2 = -4/9
and one +2/3*+2/3=+4/9 (Total is 0/9)
Since the difference between the is -1/3, and 15 is that 3rd, the total would be 45 in reciprocal natural units, 1/45=.022222…

34. Dave L

OK, I figured out a possible solution to where 15 comes from…

down -> up + W-
neutron -> proton + e + v

The W boson is 256.162 (sqrt=16.005) in natural units, and since the Carl’s lepton equation always adds a unit just before taking the square to obtain the “real” mass value, let’s do the same here… 15+1.005=16.005.
Since we must conserve mass, perhaps that extra “unit” is used up in the creation of the 2 leptons.
Perhaps it IS that “+1” in the lepton formula.

When the quite massive W decays into electron & anti-neutrino, this is where the mass might go.
But the weird thing is, wouldn’t this be negative mass, since we are subtracting it from the neutron to create the proton? Is “negative mass” positive charge”?

35. Dave

Thought I’d quickly mention something really cool, see http://arxiv.org/abs/hep-ph/0609131

In it, Pestieau & Castro describe the standard relationship between the W, Z, and B bosons (Mw^2 + MB^2 = Mz^2), but the B (carrier of weak hypercharge) turns out to be 137.08 in Brannen units! (Carl, hope you don’t mind if I call them that)
BTW, the weak mixing angle is 28.1547 deg, or acos(W/Z), using 2006 PDG for W & Z , and
(atan 1.5)/2 = 28.154966!

36. Dave

e=.302775638, 1/e=3+e

(1-e)/(1+e)=.535183758 = tan(28.154966)

e*h-bar/2Mp=.050639933136=nuclear magneton in natural units=un (h-bar=1, Mp is in “natural” lepton mass units)

Proton mag moment/un=2.792847356 (2006 CODATA)

Actual Proton mag. moment in natural units=
0.1414296033=up

up/sqrt(2)=0.1000058316

37. Dave

Using the same method as above, the muon magnetic moment (including muon g-factor) is
0.45021923962

Proton mag moment / muon mag moment=
.1414296033 / .45021923962 = .31413496078

.31413496078 / pi = 0.099992263612

38. Dave

at the end of

they give a charm mass of 986MeV

986MeV/313.8564MeV=3.141564
(where 313MeV is the same “lepton unity” value as the Feb 2 post above)

3.141564/pi=0.99999091232(!)

39. carlbrannen

Dave, that’s a nice coincidence, but when they only give three digits accuracy for the 986 MeV, we only end up with 3 digit accuracy in ratios involving it.

40. Dave

You’re absolutely right, I’ve wished there was a way to edit posts more than once!

41. Dave

I was wondering about the geometric quantity which the 3 leptons are the cube roots of:
(1+sqrt2*cos(3*12.73239))*313.85637 MeV =662.6804 Mev

When the 3 down-type quarks are converted using 662.6804 MeV as the new conversion constant the latest PDG values yield sqrt (geometric) masses of:
d=.0872094 s=.396155 b=2.517518

Amazingly, not only do they sum to 3.00088 (as the leptons do), but after taking away the “1+” term, (d+s)/b = -0.999418854, just like (elec+muon)/tau = -1.

42. Dave

Sorry, if (d-1+s-1)/(b-1)=-1, then d+s+b=3
~ so both can’t be significant.

43. Dave

Hi Carl,
Since the electron mass is the magic threshold for creating stable matter from light, I was looking at the relationship between it’s gravitational and charge force numerators. Check this out…

“From their data, the researchers obtained a value of the fine structure constant, a number that characterizes the inherent strength of the electromagnetic force. As expected theoretically, the newly obtained value of 1/128.5 is significantly larger than the 1/137 observed for a fully screened electron.”

44. Dave

Sorry, the latex html I put in after “check this out” didn’t appear above. It is

ke^2 / (Gme^2)^1/3 = 1/128.5^2

45. Dave

In geometric units, the e, u, t masses sum to “6”, or 3.3569994E-027 kg = M_leptons

M_leptons * c / 2pi = 1.6017399 E-019 = e_lep

e_lep /elem. chg = .999728

If this is more than a coincidence, that means charge has dimensions of momentum.

46. Rhys

Do you really want to throw SU(3) colour symmetry out the window? If not, your matrix in equation 16 must have three identical eigenvalues i.e. s=0, and it is just a multiple of the identity matrix.

47. carlbrannen

Rhys, if the three eigenvectors corresponded to red, green, and blue, then sure enough they would have identical eigenvalues. However, colored particles don’t exist in nature as independent particles so that wouldn’t be very physical.

Instead, the matrix is for the interaction between three colored particles. Its eigenvectors correspond to different ways of mixing colored quarks. The three solutions all have equal amplitudes and so the three solutions each correspond to one red quark, one green, and one blue. This is neutral in color, which is okay.

48. Rhys

I think I was being a bit silly. The condition following from your equation 15 is that V|abc> transforms the same way as |abc> under SU(3) rotations (I don’t like your notation, but I’ll use it here). This means that V must transform as V -> U V U^(-1) when |abc> -> U|abc>.

But these transformations, for U in SU(3), don’t preserve the “one-circulant” property of V which you claim is required.

49. carlbrannen

Rhys, yes, the restriction to SU(3) would give you a V which would have to be a constant times the identity matrix. [edit: not quite true, see next comments] The constant would be called the “strong coupling constant”.

The “color matrix” V is an abbreviation for a lot of complicated stuff. For an example of this, see the first equation on page (4) of Statistical Mechanics of semi-classical colored Objects Phys.Lett. B478 (2000) 161-171.

In their paper, they’re interested in the interaction energy between quarks and so they pay attention only to the alpha and beta indices. My model is keeping track of the beta and beta’ indices only, that is, just one of the quarks. This is in analogy to how you model a 2-body interaction by keeping track of the difference between positions of the two bodies, rather than both sets of individual positions. If you know that the color of one quark is “red”, then by color neutrality you know that the color of the other quark, plus free gluons, is “anti-red”.

Their paper assumes a simple interaction with a single gluon, so they don’t have as many arbitrary parameters and can evaluate the color matrix from perturbation theory. See their table on top of page 4. But I’m interested in the exact non perturbational theory (which cannot be calculated) so I have to assume a more arbitrary potential.

But even an arbitrary interaction needs to be assembled from stuff that treats the three possible quark colors equally. That’s not a general symmetry of SU(3) perturbation theory; it’s a symmetry of the three states in the “3” irrep of SU(3). It’s a statement about how very complicated non perturbational amplitudes must depend on color.

50. Rhys

Colour SU(3) is not just a symmetry of perturbation theory. If you’re trying to approximate QCD, you can’t just ignore it as you have.

51. carlbrannen

Rhys, I’m not trying to “approximate QCD” only. It’s necessary to consider other forces because of asymptotic freedom. I’m trying to approximate all the forces between the quarks. This is limited only by Hermitian conjugacy. The non color forces do not change color and consequently end up on the diagonal of V, which destroys its perfect SU(3) symmetry.

52. Rhys

Nothing can violate colour SU(3) unless you want to throw QCD away altogether, and you would need a damn good reason for doing that. And I’m not sure why you’re mentioning asymptotic freedom; quantities like hadron masses depend on the (strongly coupled) IR behaviour of the theory.

Did you end up submitting the paper? What did the referees’ reports say?

53. carlbrannen

Asymptotic freedom means that as far as the color force goes, the quarks are essentially free particles at very short distances. Read the wiki article. “Free particle” means that there is no color force. Consequently, electric forces cannot be ignored. Hence exact SU(3) cannot realistically be used for modeling exact quark potentials. Go look in the phenomenological literature you will see that they use a combination of an electric coulomb potential (1/r) and a linear quark potential (proportional to r).

Just because the SU(3) force is a perfect symmetry doesn’t mean that the particles that suffer it are perfect SU(3) particles. They have other quantum numbers and forces that are not SU(3) symmetric and those other forces contribute to bound states.

I submitted the paper to Phys Math Central. The referees did not complain about the potential. They complained that the paper wasn’t connected to the other results known on quarks and consequently wouldn’t be an advance in the field. So I decided to write another paper that would more directly tie in with what was known. That is Spin Path Integrals and Generations, which was submitted to Foundations of Physics earlier this month and is now in review.

In particle physics, one often finds that there are multiple ways of looking at something that seem to be entirely unrelated but actually have deep connections. What I am doing is of this nature. Instead of examining things from the point of view of position and momentum I’m looking at the particles as if the interaction occurred at only a single point. This gives an approximation that is dual to the usual one. They’re related approximations, but not as one being an improvement to the accuracy of the other.

The usual approx. is most accurate for heavy quarks at low energies. Mine is most accurate for energetic light quarks. But from the referee reports I didn’t see how I could argue this with them. All they wanted to talk about was how quarks are modeled using the usual ways. I decided that I needed to put a more fundamental paper in the literature showing that this was a natural thing to do, hence the FoP paper.

Difficulty in publishing is natural for ground breaking papers. There are plenty of examples. I’ve barely started to try to get this published.

54. Pingback: The Proton Spin Puzzle « Mass

55. Dave

In many books on control theory, they use either the force-current or force-voltage analogy to find a common mapping between electrical and mechanical quantities, but it’s the force-current analogy which preserves the topology of the schematic representation.

Since I=Q/t and F=p/t, this indicates that charge might be (potential?) momentum.

Now, e/mc=.955189, and exp(-2pi*a)=.9551846 (e=elem charge, m is the “lepton unity” mass, a=fsc).

It might be a coincidence, but I thought I’d mention it since it links such fundamental quantities in such a simple way.

56. Dave

In thinking about the inter-quark distance in a proton, I had the thought that perhaps it’s related to the lepton unity wavelength, thinking that there might be some kind of resonance binding the quarks into a triangle.

It turns out that if you draw a triangle with sides of wavelength 3.95031E-015 m (from h/mc=wavelength, where m is of course 5.594998 E-28 kg) – then the inscribed circle has radius 1.1404E-015 m, (i.e. 95% of the current rough measurements of the proton radius)

57. Dave

Here’s a diagram of 3 quarks, with each pair separated by the “lepton unity” wavelength. You can access it at… 1. Each wavelength (drawn as a perfect sinewave with amplitude of “1”) is 3.95031E-015 m.

2. The outer circle is the typical nuclear radius (1.2E-015 m )

3 The inner circle is the proton rms charge radius (.8768E-015 m)

58. Dave

It’s ironic that under the “Summary Table” on Wiki’s “natural units” page, it says that Heaviside-Lorentz (H-L) units do not commit to a unit of mass.
According to that table, e^2=4pi*a.

exp(-4pi*a) = 0.9123777, and
e^2/(mc)^2 = 0.9123879.

Maybe this means we can treat m as a H-L unit value on a par with c, h-bar, vac. permeability, etc.

Of course, this only works if you’re willing to consider that charge is dimensionally equivalent to momentum, which it is in the most common control theory analogy, as mentioned above.
For example, see the table on pg. 113 of “Fundamentals of modeling and analyzing engineering systems”, which can be previewed on Google books.
After all, charge is nothing more than particles tossing virtual photon momentum back and forth.

59. Alejandro Rivero

I noticed now in http://www.valdostamuseum.org/hamsmith/SnGdnPion.html a old calculation of Tony Smith, retorting the quark composition in Gell-Mann-Okubo formula, and getting then a pion mass of 126 MeV. Very close to the mass from Koide if you set the angle for a massless electron, albeit I do not think it is a common origin.