What is new with Koide sum rules?

[arivero]

It is possible to use only the mass of the top, or the electoweak vacuum, and ask for a Koide waterfall chaining solutions until we arrive to a mass of the top equal to zero. There are five such chains, only three of them are actually "falls", and of those only one uses always the same solution of the Koide equation (see my paper, or this thread above). The waterfall is:

t:174.10 GeV--> b:3.64 GeV---> c:1.698 GeV --> s:121.95 MeV ---> u:0 ---> d:8.75 KeV

Note that the last triplet is even older than Koide, from Harari et al.

This descent uses only one input, Fermi scale, and the mases of c and s are even near of tau and muon that in the descent with two inputs. It supports then the idea of an unperturbed spectrum, where charged leptons are degenerated with some quarks, and then a perturbations that somehow commutes with the cause of Koide.

 

[MTd2]

The up quark is massless?

 

[arivero]

The up quark is massless?

Indeed

There are, with some variants, two main arguments here.

- You can consider it pragmatically, that we are just looking for solutions of the Waterfall that happen to produce a small mass for the quark up, say 3 KeV, starting from 175 GeV. That is six orders of magnitude, and it is a convenient device to count the zero mass solutions and then, barring catastrophes, do all the numerical search from it. It should work the same for the down quark in this case.

- Or, you can take seriously the requirement in order to avoid the theta problem of QCD, and claim that the mass of the up is really zero and its measured mass is of a secondary nature, so that really for s,d,u the masses are modified following
m'_x=m_x + {m_y m_z \over M} + ...
with M coming from QCD. Note that if the up is really massless at the point where you apply the formula, then s and d are not modifyed in first order.

 

[MTd2]

What do you mean by " its measured mass is of a secondary nature"? I don't understand how you hid its mass.

 

[mitchell porter]

I have been looking into "quark-hadron duality" for an approach to the Koide waterfall that I'm not yet ready to explain. But I have to point out something I just found in the literature - in "The origins of quark-hadron duality" by Close and Isgur: that one manifestation of this duality, is that the same formula can be expressed as the square of a sum or as a sum of squares - see page 4. Doesn't that sound like the Koide formula? - with the "square roots of the masses" as the basic quantities that you sum or that you square.

 

[arivero]

What do you mean by " its measured mass is of a secondary nature"? I don't understand how you hid its mass.

Where is the problem, exactly? I put the mass of the up equal to zero and then I use an expansion to produce a final mass, this is a very usual recipe. The problem is that M is an interaction scale which comes from the chiral scale of QCD, so it is not fundamental in the Koide waterfall, hence the name of "secondary"

BTW, I mean MeV, no KeV, of course.

 

[arivero]

I have been looking into "quark-hadron duality" for an approach to the Koide waterfall that I'm not yet ready to explain. But I have to point out something I just found in the literature - in "The origins of quark-hadron duality" by Close and Isgur: that one manifestation of this duality, is that the same formula can be expressed as the square of a sum or as a sum of squares - see page 4. Doesn't that sound like the Koide formula? - with the "square roots of the masses" as the basic quantities that you sum or that you square.

I am not sure. Koide seems about finite sequences, mostly triples, and duality is about sums over all the states.

 

[arivero]

t:174.10 GeV--> b:3.64 GeV---> c:1.698 GeV --> s:121.95 MeV ---> u:0 ---> d:8.75 MeV

Just for the record, this is the only waterfall with a sensible value of t/b for five steps to zero. If we aim for six steps, there is (only) other solution, rather more peculiar:

t:174.10 GeV-->c:1.859 GeV-->b:3.401 GeV-->s:132.23 MeV-->d:9.49 MeV --->u:0

Really it uses the same triples uds, scb and cbt, but instead of csu to land into the zero, it calls for bsd.

There is not monotonic solution of six steps compatible with t/b "desert" and crossing zero. And this two are really the only six step series with such compatibility. Of these, the former survives better when going to experimental values.

 

[MTd2]

Where is the problem, exactly?

Suppose the meson rho0 or omega0, where both are u anti u, where does it get its mass, from gluons only?

 

[arivero]

Suppose the meson rho0 or omega0, where both are u anti u, where does it get its mass, from gluons only?

Indeed, and this is true for most of the low mass pions. I am not conversant in QCD, but if you open a thread on the topic in the SM subforum, I will try to follow it.

The main use of a massless up quark is to solve the strong CP problem (again, a topic where someone in the SM subforum can be more conversant than me) Here you can see
http://arxiv.org/abs/hep-ph/9403203
to Banks, Nir, and Seiberg telling that they do not believe that the controversy has solved. Since then, data from lattice show up quark with a mass different of zero, but again it is not clear if they are already accounting for some QCD trick (such as the one I told above).

 

[arivero]

If someone wants to explore koide like equations, this tool seems useful.

from __future__ import division
from sympy import *
x, y, z, t = symbols('x y z t')
def koide(a,b,c):
    return 2*(a+b+c)**2-3*(a**2+b**2+c**2)
In [5]: koide (x,y,z)
Out[5]: 
     2      2      2                2
- 3⋅x  - 3⋅y  - 3⋅z  + 2⋅(x + y + z) 
In [7]: solve([koide(0,1,y)],x,y)
Out[7]: 
⎡⎛    ⎽⎽     ⎞  ⎛  ⎽⎽     ⎞⎤
⎣⎝- ╲╱ 3  + 2,⎠, ⎝╲╱ 3  + 2,⎠⎦

In [8]: solve([koide(0,1,y),koide(1,y,z)],x,y,z)
Out[8]: 
⎡⎛    ⎽⎽       ⎞  ⎛    ⎽⎽⎽           ⎽⎽⎽     ⎞  ⎛  ⎽⎽⎽       ⎞  ⎛  ⎽⎽⎽          ⎽⎽⎽     ⎞⎤
⎣⎝- ╲╱ 3  + 2, 0⎠, ⎝- ╲╱ 3  + 2, - 4⋅╲╱ 3  + 12⎠, ⎝╲╱ 3  + 2, 0⎠, ⎝╲╱ 3  + 2, 4⋅╲╱ 3  + 12⎠⎦

 

[arivero]

Sympy produces some errors and I am not sure how exhaustive it is, so I have turned to Mathematica for an exploration of the "Landscape" of koide coupled systems.

This attachment is a notebook to explore the set of solutions for u,s,d having a starting tbc triplet where t>b>c. So the input data is t and b, and of the two possible solutions for c we choose the one lightest than the b (and t) quark.

It takes about three to five hours to solve all the possible sets of coupled equations with all the possible sign combinations for the square roots.

Most of the solutions are either degenerate or with some mass value higher than the mass of the "charm". If we sort the non degenerated solutions according the mass of the answer, we find that the "natural" solution appears already in the second place:

{0.0701502, 0.0540148, 0.00354978} -->  (4.197,x,y) (1.35978,x, z)  (x, y, - z) 
{0.0922984, 0.00818075, 0.0000355252} --> Natural Waterfall (4.197,1.359,-x),(1.359,x,z),(x,z,y)
{0.0922984, 0.0381469, 0.0000355252}
{0.0922984, 0.0381469, 0.00272298}
{0.0922984, 0.0381469, 0.0082687} 
{0.0922984, 0.0530732, 0.00374946}
{0.103209, 0.0922984, 0.0000355252}
{0.12806, 0.0215823, 0.000876337}
{0.16781, 0.0536519, 0.00362577}
{0.191464, 0.0922984, 0.006624}    
{0.195452, 0.00606144, 0.000970879} -------->   (x,y,z) (x,y,4.197) (x,-y,1.359)
{0.195452, 0.0442732, 0.00606144}
{0.195452, 0.08461, 0.00606144}
{0.195452, 0.0922984, 0.00606144}
{0.195452, 0.195452, 0.00606144} 
{0.201489, 0.0922984, 0.00662588}
...

Note that the main evidence for the validity of the natural result is its orthogonality to leptons; any other similar b,c,s triple will work too, for this task. The "Natural" sequence is, as explained in the start of the thread, the one where we always happen to choose the same solution of the equation, in each step.

 

[arivero]

This second attachment is the landscape of all the combinations of Koide equations having boundary conditions at 0 and 174.10 GeV. So there is really one empirical input, the Fermi Scale (equal to top mass because the yukawa of the top is equal to one; you could also say that we are calculating all the possible paths between yukawa =1 and yukawa =0 with six quarks).

Mathematica finds 4140 different solutions, but most of them are degenerated, having at least two quarks with the same mass. Again, let me order them from lower to higher "bottom quark prediction". The table for "b,c,s d" (we assume the zero coupling is the up) nondegenerated starts:

2.56251 degenerated...
2.82403 degenerated...
3.20787, 1.69849 degenerated...
{3.40143, 1.85905, 0.132232, 0.00949385} {t,b,c}, {b,c,-s}, {b,s,d}, {s,0,d}
{3.41535, 1.8493, 0.132774, 0.00953271} {t,b,c}, {b,s,d}, {c,s,0}, {s,0,d}
{3.64088, 1.69849, 1.66928, 0.121946} {t,b,c}, {t,b,c'}, {b,c,-s}, {c',s,0}
{3.64088, 1.69849, 0.121946, 0} {t,b,c}, {b,c,-s}, {c,s,0},{c,s,-d}, 
{3.64088, 1.69849, 0.121946, 0.0151864} {t,b,c} {b,c,-s} {c,s,0} {s,0,d}
{3.64088, 1.69849, 0.261403, 0.121946} 
{3.64088, 1.69849, 1.69849, 0.121946}
{4.51619, 3.64088, 1.69849, 0.121946}
...

So Koide ansatz with the single input of 174.10 GeV predicts the right scale for the other quarks:
3.40..3.64 GeV for the bottom
1.85..1.69 GeV for charm
132...121 MeV for strange
0...9...15 MeV for down

 

[arivero]

Suppose the meson rho0 or omega0, where both are u anti u, where does it get its mass, from gluons only?

Ah, note that now there is a subtopic in the threasBootstrap thread about chiral symmetry breaking, it is somehow related to the question of QCD mass. https://www.physicsforums.com/showthread.php?t=485247&page=10

 

[arivero]

Could Brannen transformation matrix N = \left( \begin{array}{ccc} <br /> \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} \\ <br /> e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} \\ <br /> e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} \\ \end{array} \right)
be a symmetry of the fermion field, so that the yukawa coupling of the Higgs g\phi \bar\psi\psi somehow goes over to each field, say \phi (N \bar\psi) (N \psi) and the couplig g is really N square?

Really, I can not find any theory where the square root of masses is subjected to a fundamental symmetry. Nor to any symmetry whatever. Hmm.

 

[mitchell porter]

The Koide triplets overlap. So if each triplet derives from a set of more fundamental quantities - e.g. if a triplet is derived from the eigenvalues of a matrix - then the set of fundamental quantities associated with one triplet ought to overlap with the fundamental quantities behind an overlapping triplet. But then, if a symmetry mixes all the fundamental quantities for one triplet, from the perspective of the overlapping triplet, just a subset of its fundamental quantities is being transformed, and in a way that depends on quantities from "outside" (outside the overlap with the first triplet). This is not apriori impossible, but it's tricky to arrange.

Alternatively, the masses, or the sqrt-mass VEVs from which the masses derive, may be the symmetry-breaking solution of some potential, as in comment #40. But then one would like to know why the ordering of the masses from largest to smallest takes the very specific form "quark with +2/3 charge, followed by its -1/3 charge partner, repeated three times", when (given a potential completely symmetric in the sqrt-masses) any ordering was apriori possible.

 

[arivero]

Well, as for the ordering, I have tried (the mathematica notebooks above) all the possible orderings and most of them produce a next-to-top quark with a mass higher than the measured bottom quark. So the waterfall is one of the few orderings that creates a wide gap between top and bottom, and the best non-degenerated one.

My thinking on the symmetry was that just fixing top=1 then breaks in cascade all the symmetries. But I agree it is tricky.

Perhaps the quark side symmetries are only approximate and it just happens that the approximation in sbc cancels the one in bct and conspires to get an exact top quark value. Not rare thing in history of science, I can imagine a couple of famous examples.

 

[arivero]

Honestly, I am too old, or too tired, to understand how Koide produces naturally the mass from the square root mass here,
http://ccdb5fs.kek.jp/cgi-bin/img/allpdf?198105037
nor how he got the first mass formula from the spurion.

 

[arivero]

Ok I can find two kinds of works where the square root of yukawa coupling has some fundamental role. One is "flavons", as in http://arxiv.org/abs/1203.1489v3 Other is Composite Higgs, and particularly Contino (eg http://arxiv.org/abs/1005.4269v1) works like to make explicit this. My memory fails me, so perhaps Mitchell has already catalogued other cases.

Edith, also the higgs in NCG seems to be a product of two more elementary numbers, from bialgebras or elsewhere

 

[mitchell porter]

also the higgs in NCG seems to be a product of two more elementary numbers, from bialgebras or elsewhere

We had two NCG higgs papers this month, so perhaps we can discuss this.

In Estrada and Marcolli, we have an action with parameters "a ... e" that are functions of the Yukawa matrices, and some other "f" parameters that are more fundamental, and then we have various relations among a ... f that exist at unification energy.

This action is an expansion of a "spectral action functional" which includes variation of a Dirac operator. The Higgs field arises from those "inner fluctuations" of the Dirac operator that are associated with the finite noncommutative part F of the "almost commutative" space M x F that defines the model. (Inner fluctuations associated with M produce the gauge fields.) These fluctuations have the form u[D,u*], where [,] is a commutator and u,u* are I think unitary elements of the algebra A in the spectral triple <A,H,D>. Alejandro, are u and u* what you were talking about?

The new work this month has been about obtaining a 125 GeV Higgs in the noncommutative SM, either by imposing asymptotic safety and reproducing the Shaposhnikov-Wetterich argument (this is what Estrada and Marcolli did), or by including the scalar that gives Majorana mass to the RH neutrino, in the RG equations (this is what Chamseddine and Connes did). The AS argument gives the right value within a GeV; the other model just shifts the range of possible Higgs masses so that it includes the observed value.

Turning to the world of Koide relations, there have been a few studies of how the values of the expression in the Koide formula flow, for different triplets; and we also have the work of Sumino, which imposes boundary conditions on RG flow at the EWSB scale, in order to explain the exactness of the Koide relation for the pole masses.

So the obvious way to explain the Koide formula in the NCG context, would be to use a high-scale (unification-energy) version of the Sumino mechanism, that employs the "a...f constraints" to engineer the necessary low-energy relations. I'm not sure if this is possible, but if it will jumpstart discussion again, I'm willing to think about it...

 

[S.Daedalus]

This whole discussion is way above my level of expertise in high-energy physics, but I have a side question if I may: if I understood correctly, one of the main problems with the Koide approach is that it's a connection between the low-energy masses of the theory, which should have no fundamental significance due to renormalization group flow. But couldn't there be something like supersymmetry nonrenormalization at work, that is, the parameters m that turn up in the low energy theory are actually identical to the high energy masses?

 

[mitchell porter]

Actually those nonrenormalization properties disable the Sumino mechanism for preserving the Koide relation, because it relies on vertex corrections that no longer exist under susy! Koide and Yamagarbagea developed an alternative but it doesn't work as well. Still, perhaps one can hope that susy will simplify the RG equations in some other way.

Some resources.

 

[mitchell porter]

Another useful NCG paper is Kolodrubetz & Marcolli. Also see this lecture, especially slide 10. It seems that one wants to construct a cascade of effective field theories, with a Sumino model at the final stage.

Returning to comment #65... The original Koide triplet relates yukawas from a single mass matrix, but the new triplets for quarks all combine up-type yukawas with down-type yukawas, so the transformation looks unnatural. It's as if we need an extended Higgs mechanism that includes "up-down yukawas". We could suppose they are there and set them to zero... but what would they be? The Standard Model mass matrices tabulate coefficients of Yukawa terms in the Lagrangian. These new "up-down Yukawa terms" would require something new.

Nonetheless:

\left( \begin{array}{ccc} <br /> y^u_{11} &amp; 0 &amp; y^u_{12} &amp; 0 &amp; y^u_{13} &amp; 0 \\<br /> 0 &amp; y^d_{11} &amp; 0 &amp; y^d_{12} &amp; 0 &amp; y^d_{13} \\<br /> y^u_{21} &amp; 0 &amp; y^u_{22} &amp; 0 &amp; y^u_{23} &amp; 0 \\<br /> 0 &amp; y^d_{21} &amp; 0 &amp; y^d_{22} &amp; 0 &amp; y^d_{23} \\<br /> y^u_{31} &amp; 0 &amp; y^u_{32} &amp; 0 &amp; y^u_{33} &amp; 0 \\<br /> 0 &amp; y^d_{31} &amp; 0 &amp; y^d_{32} &amp; 0 &amp; y^d_{33} \\ \end{array} \right)

... if I may be permitted to introduce this interleaving of up and down Yukawa matrices, without exactly saying what it is; and if we suppose that the "up" and "down" parts are each diagonalized as much as possible, with diagonal entries ordered by size; then the Koide waterfall amounts to saying that there is a "Brannen symmetry" for each 3x3 block on the main diagonal.

edit: Whoops, I missed a stage. The Brannen symmetry relates the square roots of the masses. So we would be looking at blocks on the diagonal of a 6x6 matrix whose square is the matrix above.

edit #2: The Brannen transformation for a particular block could look like this:
\left( \begin{array}{ccc} <br /> 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \sqrt{2} &amp; e^{i\delta} &amp; e^{-i\delta} &amp; 0 &amp; 0 \\ <br /> 0 &amp; e^{-i\delta} &amp; \sqrt{2} &amp; e^{i\delta} &amp; 0 &amp; 0 \\ <br /> 0 &amp; e^{i\delta} &amp; e^{-i\delta} &amp; \sqrt{2} &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{array} \right)

 

[mitchell porter]

There's a new paper on Koide triplets today, in which the author experiments with a Brannen parametrization of up-type masses and down-type masses, and comes out with phases of 2/27 and 4/27. The usual Brannen phase for the charged leptons is 2/9, i.e. 6/27. These are numbers which I first saw on Marni Sheppeard's blog, and which I thought she discovered through discussion with Dave Look, so I'll be writing to the author to let him know - as well as to mention the tbcsud "waterfall" of triplets discussed in this thread. I think of the waterfall as real, and tend to dismiss those quark family triplets as spurious. Given the idea that there are unknown "Koide symmetries" responsible for the "authentic triplets", I suppose it's possible that the same symmetries could be present in uct and dsb too, but with a lot more noise.

 

[arivero]

There's a new paper on Koide triplets today,

...
I think of the waterfall as real, and tend to dismiss those quark family triplets as spurious.
...

Of course I am of the same opinion, but still I am sligthly amused that the waterfall uses delta_scb = 3 delta_L, and he gets delta_L= 3 delta_U.

 

[mitchell porter]

Or to put it another way, the scb angle is 2/3, the eμτ angle is 2/9, and the uct angle is 2/27.

We also have that the eμτ mass scale is 313 MeV (one-third the proton mass, i.e. constituent mass of a first-generation quark), and the scb mass scale is three times that.

From Sheppeard's blog (1 2), I get that the mass scale for a uct triplet would be about 20 GeV.

edit: A few months ago I was thinking about what sort of model would produce just these "family phases" - what Zenczykowski calls δ_L, δ_D, δ_U - simply because that's easier to think about. I was interested in an Adler-type 3HDM (three-Higgs-doublet model) with circulant mass matrices. But you could take any model of the charged-lepton sector, that produces a Koide relation, and try to apply it separately to the up-type and down-type quarks - for example, Ernest Ma's supersymmetric model.

 

[arivero]

Or to put it another way, the scb angle is 2/3, the eμτ angle is 2/9, and the uct angle is 2/27.

Of course, it is very problematic to have angles which are not a submultiple of the circumference. Up to now, the main motivation for the factor of three was to consider the case where one of the masses is 0, fixing thus the angle, and then the orthogonality between the triples with 15 degrees and 45 degrees, this is pi/12 and pi/4. The angle of 2/3 I though of it as a perturbation from pi/4, the angle of 2/9 as a perturbation from pi/12.

 

[mitchell porter]

Since we don't know where these quantities come from, I don't think we can say that their form is problematic. Would their origin be easier to understand if they were simple fractions of π? Also, it's hard to think of e.g. 2/9 (the actual phase, for e-μ-τ) as a perturbation of π/12 (the phase for e-μ-τ, in the "modified waterfall" that lands on Harari-Haut-Weyers values for d-u-s masses), because normally a perturbation of a quantity x just gives you "x plus a small mess", it doesn't give you a simple rational number! I have noticed that 2/3 (possible phase for s-c-b) is obtained by the first two terms in the Leibniz formula for π/4, as if it were a truncation. One could start thinking about formulas with Grassmann variables, so all the higher terms vanish...

Another line of investigation would be to look for the significance of the "Brannen angle" in the other frameworks that manage to produce Koide triplets. Sumino, in his paper which tries to explain the exactness of the original Koide formula despite RG running, also presents an original derivation of the triplet itself (from the interactions in the scalar sector of his model). Then there's Ma, mentioned above; then there's Koide's original preon theory. Carl Brannen's formula plays no apparent role in any of these, but I wonder if they still look simple when expressed using his variables?

edit: Some comments on whether m_u=0 is still a live option. (For the general reader of this thread: Alejandro found a "waterfall" of interlocking Koide triplets which works well for the four heaviest quarks and which can be extended to the remaining quarks. The modified waterfall is a version adjusted so that the up quark has exactly zero mass. The heavy quark masses become less accurate but the Brannen angles assume interesting values, and the idea is that the real waterfall is a perturbed version of this modified waterfall, see his paper for details.)

Michael Dine gave a talk as recently as 2009 implying that it was still being considered by theorists like Seiberg and Kaplan. Dine's 1993 review "Topics in string phenomenology" points out two ways to get m_u=0 from string theory, one from anomalous discrete symmetry, the other from a horizontal symmetry as described in a series of papers (1 2 3).

From the other side, 1103.3304 gives in a few sentences (page 83) the reason why workers in lattice QCD might dismiss the m_u=0 option as an explanation for no strong CP violation. This argument needs to be confronted with the ideas in reference 3, listed above.

 

[arivero]

I am curious about how sensible the prediction of the top mass is to the factor 3 in the jump from leptons to quarks. So here is the "bc -l program"

define top(massfactor,anglefactor) {
me=0.000510998910
mmu=0.1056583668
mtau=((sqrt(me)+sqrt(mmu))*(2+sqrt(3)*sqrt(1+2*sqrt(me*mmu)/(sqrt(me)+sqrt(mmu))^2)))^2
m=(me+mmu+mtau)/6
pi=4*a(1); cos=(sqrt(me/m)-1)/sqrt(2); tan=sqrt(1-cos^2)/cos
delta=pi+a(tan)-2*pi/3
mc=massfactor*m*(1+sqrt(2)*c(anglefactor*delta+4*pi/3))^2
ms=massfactor*m*(1+sqrt(2)*c(anglefactor*delta+2*pi/3))^2
mb=massfactor*m*(1+sqrt(2)*c(anglefactor*delta))^2
mtop=((sqrt(mc)+sqrt(mb))*(2+sqrt(3)*sqrt(1+2*sqrt(mc*mb)/(sqrt(mc)+sqrt(mb))^2)))^2
return mtop
}

Newcomers can see that the same formula is used for mtop and mtau. And of course it is also the same formula, in angular form, for the rest of the calculation involving delta and m. Koide everywhere.

so with the factor 3 argued in my paper, we get

top(3,3)
173.2639415940

Which is in the center of the combination of Tevatron (173.18) and LHC (173.34). In fact, the weighed average of Tevatron and CMS (september) should be 173.265 ± .679 GeV, so the prediction is pretty in the center.

Which are the 1 sigma limits for the mass and angle factors, with this average? Well, pretty narrow, but still some place for perturbative corrections:

173.265+0.679
173.944
top(3,3.046)
173.9416301253
top(3.012,3)
173.9569973497
x=3.01;top(x,x)
173.9906886621
173.265-0.679
172.586
x=2.990;top(x,x)
172.5374243669

 

[arivero]

Probably it is a red herring, but some comments from mitchell have indirectly driven me to look at the mass formula for an stack of D-branes. I am not sure in how they are in the superstring case, but already in the bosonic string they look a lot as a generalisation of Koide mixing:

<br /> M^2 = \big((n + {\theta_i - \theta_j \over 2 \pi}) {R&#039; \over \alpha&#039;}\big)^2 + {N-1 \over \alpha&#039;} <br />

THis is f. 174 in arXiv:hep-th/0007170v3

For n=1, N=1, and i,j from 1 to 3 with i different of j, the stack of three D-branes looks Koide's formula. I am not sure of which is the mass formula in this case (nor in the fermionic/superstring case...)
It should be, if M^2 where instead a seesawed product of two masses, M^2=m_{ij} M_0


<br /> m_{ij} = {R&#039;^2 \over M_0 \alpha&#039;^2} (n + {\theta_i - \theta_j \over 2 \pi})^2 <br />

Note that the basic fact is that the sum of the three differences \theta_i - \theta_j is zero, as in the case of the sum of three cosines in Koide.

EDIT: Michael Rios suggested, last year, to use three coincident branes to emulate Koide.
EDIT2: Today is the birthday of Lubos Motl, this is my birthday gift: string theory becomes predictive

 

[arivero]

Unrelated to the previous comment, except for the fact that strings dof come in groups of 8, it could be worthwhile to rethink again the 12x8 ideas in the light of Koide. The "Koide waterfall" in #58 above, with mass of up quark exactly zero in three Koide steps, provides, if we also use the orthogonality condition, some intriguing pairing of leptons and quarks:

t:174.10 GeV
b:3.64 GeV
tau, c:1.698 GeV
mu, s:121.95 MeV
e, u:0
d:8.75 MeV

On other hand, the most naive way of building a multiplet with 8 degrees of freedom is to use an electroweak pair of Dirac fermions: neutrino, electron for instance, or any up, down combination. This is still possible here, and even it could be convenient if we consider that we are going to broke this pairing of leptons and quarks. But looking at this table, we could take it serioustly and consider that one lepton and the three colours of a quark should be the components of a multiplet. Then the unpaired quarks would correspond to see-sawed neutrinos and the whole table is

\nu_1,t: 174.10 GeV
\nu_2,b: 3.64 GeV
\tau,c: 1.698 GeV
\mu,s: 121.95 MeV
\nu_3,d: 8.75 MeV
e,u: 0

Of course we have sixteen degrees of freedom in each line and it is still to see how they should be managed in groups of 8, either by chirality or by particle/antiparticle.

What is intriguing in any case is the mu,s pairing: a charged lepton with a down type quark. It could point to the need of using a SU(2)xSU(2) L-R symmetry.

 

[mitchell porter]

Pati-Salam as we know it, doesn't allow such a scheme. The orthodox way to embed the waterfall in Pati-Salam would be to use the conventional generation structure (three sets of two "four-color quarks", one of which divides into an up-type quark and a neutrino, the other of which divides into a down-type quark and a charged lepton), a selection of Higgses (there must at least be one to break SU(4)_c to SU(3)_c and another to break U(1)_B-L x SU(2)_R to U(1)_Y), then work out the 3x3 Yukawa matrices for the "four-color quarks", and finally the effective Yukawas for the SM quarks and leptons. And since the waterfall has that intricate structure, probably the best way is via flavons: the Yukawas are VEVs of "flavon" fields. (Koide himself uses flavons in his yukawaon models of recent years.) We can then try to obtain a Pati-Salam waterfall from flavon symmetries.

This doesn't have the simplicity of just directly associating e-mu-tau with u-s-c (or with s-c-b, as might have been suggested by Georgi-Jarlskog), but at least it is a type of theory which it is known can be constructed. If you do it this way, the orthodox way, you do get to preserve the direct association of muon with strange quark. So you might suppose that the second generation is a sort of pivot, where there is approximate equality of masses, connecting waterfall Koide triplets on the quark side, and the usual family Koide triplets on the lepton side.

Intriguingly, if you imagine interleaving the Yukawas for charged and uncharged leptons in the fashion of #73, then the Brannen transformation matrix for family Koide triplets (rather than sequential, waterfall triplets) looks like this:
\left( \begin{array}{ccc} <br /> 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \sqrt{2} &amp; 0 &amp; e^{i\delta} &amp; 0 &amp; e^{-i\delta} \\ <br /> 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; e^{-i\delta} &amp; 0 &amp; \sqrt{2} &amp; 0 &amp; e^{i\delta} \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\ <br /> 0 &amp; e^{i\delta} &amp; 0 &amp; e^{-i\delta} &amp; 0 &amp; \sqrt{2} \\ \end{array} \right)
(or the obvious counterpart where the two interleaved blocks change places).

As I wrote in #73, an "interleaving of Yukawa matrices" has no physical meaning that I can identify. But what I like about this perspective is that the "family Brannen transformation" and the "sequential Brannen transformation" could both plausibly be part of some larger algebraic structure. In both cases they're based on a 3x3 block within the 6x6 matrix, it's just that the spacing is different. So the idea is that a Pati-Salam embedding of waterfall + original Koide could result from a flavor symmetry containing that "larger algebraic structure", with family symmetries dominating on the lepton side and sequential symmetries dominating on the quark side, and with the second generation providing the bridge.

 

[arivero]

I have scanned a couple of collections which show the history of Koide before Koide. The first set pivotes on Harari-Haut-Weyers and its refutations, the second set is some extra articles of the same age, found while I explored the first selection.

While they will be useful mostly to Carl Brannen, perhaps Mitchell and other crowd can enjoy them too.

https://docs.google.com/open?id=1UflBQIr-r3RsEigmr1ty2kTMpSU_5gM2Z-DFtWfKW7rewRqkMlReePUCCGDj

https://docs.google.com/open?id=1vRfLIl-wvb7BRNlS4z9uukYDFwDLb71OC2x5iB9Z9O4vB4PnIwW634ZOhATP

And yes, I use a monitor which can pivote 90 degrees. But nowadays you can always cancel the gravitational sensor of your iPad, can you?

 

[arivero]

Wow, Zenczykowski paper was accepted for PhysRev D last Thursday (Dec 13, 2012).

There's a new paper on Koide triplets today,

If it is still v2, I am a bit sorry that he has not changed the references (for instance, to include the waterfall) but still a good thing.

 

[arivero]

Just for the record, it is interesting to look to the solutions in the lepton side ascending from the e-mu-tau triple. Remember we are conjecturing a descent where some leptonic object partners with every quark. It could be reasonable to think of Dirac mass terms for neutrinos, for instance.

-, t:174.10 GeV
-, b:3.64 GeV
tau, c:1.698 GeV
mu, s:121.95 MeV
e, u:0
-, d:8.75 MeV

Now, once we have broken the pairing, we can use Koide separately in each sector... just to see if it has some sense.

To ascent from mu, tau to the next two levels, the equation with the above values has discriminant cero, we should look with some care with branch of the answer is it really taken, but anyway here you have both branches. For both of them, the second step is unique, due to negative roots forbidding other solutions.

mtau
1776.96888139816566506171
((sqrt(mtau)-sqrt(mmu))*(2+sqrt(3)*sqrt(1-2*sqrt(mtau*mmu)/(sqrt(mtau)-sqrt(mmu))^2)))^2
7211.73510126774064895083
((sqrt(mtau)+sqrt(m1))*(2+sqrt(3)*sqrt(1+2*sqrt(mtau*m1)/(sqrt(mtau)+sqrt(m1))^2)))^2
268928.53716239525673236427

((sqrt(mtau)-sqrt(mmu))*(2-sqrt(3)*sqrt(1-2*sqrt(mtau*mmu)/(sqrt(mtau)-sqrt(mmu))^2)))^2
1812.91990902666662582893
((sqrt(mtau)+sqrt(m1))*(2+sqrt(3)*sqrt(1+2*sqrt(mtau*m1)/(sqrt(mtau)+sqrt(m1))^2)))^2
121946.96009306194199844666

I like this second branch: the (tau, nu1, nu2) triplet equal to (1.777, 1.813, 121.95). It could be saying that the lepton sequence moves to increase the gap between the two final states. And while we left the 174.1 GeV endpoint, we still are in a nice mass range.

 

[mitchell porter]

Suppose we have six flavors of quark in an SU(4) gauge theory. For the moment, suppose there are no other quantum numbers... Then we will have a 6x6 yukawa matrix.

Next, suppose that these yukawas are flavon vevs, and that the flavon potential has a discrete symmetry generated by the four "sequential" (#73) and two "family" (#82) Brannen transformations, for particular values of δ.

And now, let us augment this "theory", so that the usual electric charges for the quarks arise or are introduced, and so that the usual Pati-Salam higgsing of SU(4) to SU(3) occurs. It seems that the first step should introduce a "checkerboard" texture to the 6x6 yukawa matrix, and then the second step should "double" the yukawa matrix, so there's one 6x6 yukawa checkerboard for three-color quarks and another 6x6 yukawa checkerboard for leptons.

Finally, let us suppose that the sequential symmetries dominate the quark yukawas, and the family symmetries dominate the lepton yukawas (though the residual family symmetries in the quark yukawas may be strong enough to produce recognizable Koide phases of 2/27 for up quarks and 4/27 for down quarks). This can give us the waterfall for the quarks, and the original Koide relation for the charged leptons.

edit: I think the first thing to do, would be to create the theory of the second paragraph. That would be practice at constructing a theory in which a Koide waterfall of masses arose from a 6x6 yukawa matrix.

 

[arivero]

Finally, let us suppose that the sequential symmetries dominate the quark yukawas, and the family symmetries dominate the lepton yukawas (though the residual family symmetries in the quark yukawas may be strong enough to produce recognizable Koide phases of 2/27 for up quarks and 4/27 for down quarks). .

I ack that the publication of Żenczykowski paper has biased me to consider again the family symmetries. It could be that we are seeing different aspects of a larger discrete symmetry group... but how? The real problem is that it does not seem a permutation group because for any pair of masses we exchange, we find that one Koide equation keeps invariant but obviously others, containing only one of the two masses, are not preserved. by the way, the publication of Phys. Rev. D 86, 117303 (2012) officially raises the number of cites of Brannen's and of myself on this topic!. I get a citation to hep-ph/0505220 so that the author can refer indirectly to internet forums with a "Brannen, as cited in...". And Carl gets a second citation, directly to http://brannenworks.com/MASSES2.pdf

 

[mitchell porter]

I think it would be instructive to express all the fermion pole masses as multiples of the Brannen mass parameter for the original Koide triple, M_L, and then use the relationship between the top mass and the Higgs VEV to express the latter in the same units. M_L is presumably the fundamental quantity in the waterfall (because it apparently comes from QCD or SQCD), but I don't think we've thought about how to get the Fermi scale from it. Yet surely this should be playing a role in our thinking about the Higgs.

 

[mitchell porter]

Today I was looking at two new and two old papers. The new papers are "Neutrino Mass and Mixing with Discrete Symmetry" by King and Luhn, and "Top-quark and neutrino composite Higgs bosons" by Adam Smetana. The old papers - well, one was a thesis, Francois Goffinet's thesis, http://cp3.irmp.ucl.ac.be/upload/theses/phd/goffinet.pdf, and the other was the co-authored paper resulting from it, "A New Look at an Old Mass Relation".

Together, they should have something to say about how to extend the waterfall to the neutrinos, to the mixing angles, and to the properties of the Higgs sector. Goffinet's concept of "pseudo-mass" was invented precisely to link the Koide relation to mixing angles. King and Luhn review flavon models with discrete family symmetries, for the neutrino sector. And Smetana tries to get the Fermi scale by having both a top condensate and a neutrino condensate, in a broad class of models featuring a gauged flavor symmetry. To get the numbers right he ends up needing a large number of right-handed neutrinos, so probably he is still missing something essential, but it begins to make the connection I called for in the previous comment.

 

[arivero]

Today I was looking at two new and two old papers. The new papers are "Neutrino Mass and Mixing with Discrete Symmetry" by King and Luhn,

S_4 is intriguing. It is equal to Z_2 \times Z_2 \ltimes S_3, and I wonder if these, say, four copies of S_3 could be the four copies acting in the waterfall. Also, where does it come from? Speculatively, could be a subgroup of SU(5)_flavour -the sBootstrap group-. And more speculatively, what about the permutations of the 4 components of a spinor?

 

[arivero]

Given that a lot of Koide stuff seems related (hat tip to de Vries and Brannen here) to this matrix

\begin{pmatrix}-\frac{2\,\mathrm{sin}\left( t\right) }{\sqrt{6}} &amp; \frac{\mathrm{sin}\left( t\right) }{\sqrt{6}}+\frac{\mathrm{cos}\left( t\right) }{\sqrt{2}} &amp; \frac{\mathrm{sin}\left( t\right) }{\sqrt{6}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{2}}\cr \frac{2\,\mathrm{cos}\left( t\right) }{\sqrt{6}} &amp; \frac{\mathrm{sin}\left( t\right) }{\sqrt{2}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{6}} &amp; -\frac{\mathrm{sin}\left( t\right) }{\sqrt{2}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{6}}\cr \frac{1}{\sqrt{3}} &amp; \frac{1}{\sqrt{3}} &amp; \frac{1}{\sqrt{3}}\end{pmatrix}

I have setup a wxMaxima notebook to play with it. Not that I like Maxima, I used it in a VAX and it was already superseded by REDUCE when Mathematica come. But it comes with Ubuntu and has a graphical interface, which Reduce has not.

 

[mitchell porter]

Koide has a model of leptons with an S4 symmetry embedded in a flavor SU(3).

 

[mitchell porter]

Since this is a Koide thread we have to mention Zenczykowski's latest, though it is about "family triples", and not what I call the "sequential triples" of the waterfall... He's still building on the generalization of the e-mu-tau 2/9 parameter to u-c-t and d-s-b; he proposes that another parameter, which is just "1" for e-mu-tau, is also "1" for the quarks if you use Goffinet's concept of "pseudo-mass". If that's true it's a breakthrough, as well as a headache for the waterfall, because aren't we getting overloaded with too many relationships at once?

He mentions the usual problem, that these relations work best for low-energy masses. We've previously discussed Sumino's efforts to have family gauge bosons cancel out certain QED corrections, so that Koide's relation may be exact; but I was always curious about whether there might be some dual description of physics, in which, rather than thinking of the UV as fundamental, you thought of the physics as "IR + new degrees of freedom at a series of higher energies" - the idea being that the cause of Koide relations might be more transparent in this hypothetical "infrared first" formulation.

Well, I wonder if this paper by Davide Gaiotto (from January 2012) might be relevant: "Domain Walls for Two-Dimensional Renormalization Group Flows". "Renormalization Group domain walls are natural conformal interfaces between two CFTs related by an RG flow. The RG domain wall gives an exact relation between the operators in the UV and IR CFTs." It seems a tiny step towards what I had in mind.

 

[arivero]

If that's true it's a breakthrough, as well as a headache for the waterfall,

The waterfall could happily miss the last triple, d-u-s, in exchange by one of the "standard" ones, but d-u-s does a better prediction of the down mass that d-s-b.

A motivation to follow this track could be, put all the quarks in the faces of a cube, such that all the equations of the waterfall are the faces that meet in some vertex. You will notice that his cube has a property, that opposite faces have opposite weak isospin. You can also notice that we only need three equations to fix the faces.

One of the vertexes of this cube is DSB, and of course is opposite vertex has the faces of the up-type quarks. This is the only axis that does not correspond to a waterfall symmetry, and on other hand the DUS vertex is the only axis which is used in both extremes.

Going to discrete groups, S4 is the group of permutations of the four "Z3 axis" in a cube, while the subgroup S3 is contained in four not-very-different ways, each of them being the permutations that keep one of such axis invariant (you can exchange fully the vertex by the opposite, to implement the Z2 subgroup of S3).

 

[arivero]

The real problem of S4 is to know the physics content, the objects we are permuting. The suggestion of putting quarks (or leptons) in the faces of a cube is rarely seen in the literature.

About PZ and pseudomasses, I think it is not very different of the initial objections to Harari-Haut-Weyers, they also do a similar trick, or a trick that can be interpreted as taking only the diagonal of the undiagonalised mass matrix.

By the way, I note that the abbreviations 2/27 and 4/27 are first used by Sheppeard in her note 342

 

[mitchell porter]

The pseudomass adds up the contributions of all the mass eigenstates to one of the weak eigenstates. So it looks like "sequential" triples, like in the waterfall, apply to mass eigenstates, family triples (as in the original Koide formula) apply to weak eigenstates, and this wasn't noticed until recently because, for charged leptons, the weak eigenstates are the same as the mass eigenstates.

For the quarks, we can then think of the waterfall as the dominant chain of relationships, and then the mixing parameters encode the rotation away from waterfall mass values, required to produce family triples with 2n/27 phases.

For the leptons, perhaps the family triples dominate, and a waterfall is weak or nonexistent. (I'm still not clear on whether right-handed neutrinos could have masses of the order of the quarks, as in #81, and then give rise to the observed small masses via seesaw.)

p.s. Chris Quigg had a paper yesterday - "Beyond Confinement" - in which 2/27 shows up as the exponent in a relation between the top mass and the nucleon mass, in a unified theory!

 

[arivero]

I hated macsyma and I am transferring my hate to its free twin, maxima, but still this is interesting. I got again the waterfall while I was trying to solve for the full S4 symmetric set of Koide equations.

This is, I was trying to find mass values in the six faces of a cube such of for each vertex we have a Koide equation. The group of rotations of the cube is S4; imposing Koide explictly breaks the symmetry as it gives different values to different faces. Not a very convincing motivation, but ok to play a little bit.

Now, in maxima. You define
K(x,y,z):=x^2+y^2+z^2-4*(x*y+y*z+z*x);
so that
expand(K(x,y,z)*K(x,-y,z)*K(x,y,-z)*K(x,-y,-z));
is a degree 8 even polynomial on three variables;
we put all of it

Q(x,y,z):=z^8-28*y^2*z^6-28*x^2*z^6+198*y^4*z^4-1172*x^2*y^2*z^4+198*x^4*z^4-28*y^6*z^2-1172*x^2*y^4*z^2-1172*x^4*y^2*
z^2-28*x^6*z^2+y^8-28*x^2*y^6+198*x^4*y^4-28*x^6*y^2+x^8;

and now we can use maxima "eliminate" to do the equivalent, I guess, of Sylvester matrix.
factor(eliminate([Q(1,x1,x2),Q(1,x2,x3)],[x2])) shows two terms that just validate x^2=x3^2. So in my first step I also canceled these factors:

step1:factor(eliminate([Q(1,x1,x2),Q(1,x2,x3)],[x2]))/(x3-x1)^8/(x3+x1)^8;

In the next step, factor(eliminate([paso1[1],Q(1,x3,x4)],[x3])) happens to have as a factor the polynomial for Q(1,x1,x4) and then it trivially tell us that the whole system of vertexes (1,x1,x2),(1,x2,x3),(1,x3,x4),(1,x4,x1) has solutions. As we want to exhibe actually some solution, I cancel the factor first

step2:factor(eliminate([step1[1],Q(1,x3,x4)],[x3]))/factor(Q(1,x1,x4)^12);

and now I cross each of the factors against the extant equation Q(1,x1,x4)

for i:1 thru 16 do (
pol:part(part(step2[1],i),1),
sl:solve([pol,Q(1,x1,x4)],[x1,x4]),
for k:1 thru length(sl) do (
  s:ev([x1,x4],sl[k]),
  if featurep(s[1],real) and featurep(s[2],real) then 
     (s:abs(s),
     if s[1]>s[2] then s:[s[2],s[1]],
     print(s,float(s)), 
     )
  )
);

The process solves to:
four complete x1,x2,x3,x4,x1 cycles
two sequences x1,x2,x3,x4
two sequences x1,x2,x3

The solutions for the waterfall triplets cbt and bcs appear in the list, numerically as sqrt(t)=10.12, sqrt(b)=1.464, sqrt(s)=0.267, with sqrt(c)=1, in one of the not-closing sequences. I guess I need an expert on discrete groups in order to understand what is going on.

PS. wow, now I notice that Q surely is the trick that Goffinet uses to avoid the square roots somewhere in his thesis.

 

[mitchell porter]

Exciting progress!

wow, now I notice that Q surely is the trick that Goffinet uses to avoid the square roots somewhere in his thesis.

http://cp3.irmp.ucl.ac.be/upload/theses/phd/goffinet.pdf (section 3.2.1).

I guess I need an expert on discrete groups in order to understand what is going on.

A further step would be to look for an S4-symmetric potential where these solutions are the local minima.

 

[arivero]

http://cp3.irmp.ucl.ac.be/upload/theses/phd/goffinet.pdf (section 3.2.1).[/QUOTE]

Indeed it is the same polynomial.

z^8-28*y^2*z^6-28*x^2*z^6+198*y^4*z^4-1172*x^2*y^2*z^4+198*x^4*z^4-28*y^6*z^2-1172*x^2*y^4*z^2-1172*x^4*y^2*
z^2-28*x^6*z^2+y^8-28*x^2*y^6+198*x^4*y^4-28*x^6*y^2+x^8;

  4         3         3        2  2             2        2  2       3
(%o3) z  - 28 y z  - 28 x z  + 198 y  z  - 1172 x y z  + 198 x  z  - 28 y  z
           2           2           3      4         3        2  2       3
 - 1172 x y  z - 1172 x  y z - 28 x  z + y  - 28 x y  + 198 x  y  - 28 x  y
    4
 + x

The minor improvement is that here we are sure that it is an ·"if and only if" relationship; Goffinet, in the text, was worried that the squaring could be introducing spureous solutions. As we have shown that this poly decomposes exactly in the product of the four possible sign combinations of Koide equation, now we are in position to grant that every solution of Goffinet's matrix version (3.30) of the equation is really a Koide solution.

Let me copy here this equation 3.30, setting the determinant of M as a function of the traces in M and M^2:

|M| = {2 \over 3 * 32^2} {(7 (Tr M)^2 - 8 Tr M^2)^2 \over Tr M}

 

[arivero]

The following Mathematica code will help to find the solutions to the S4 symmetric Koide system.

If this post is the only one you are going to read, remember that we are organising three generations in opposite faces of a cube, and each corner must agree with Koide equation. A way to solve this is to fix one face, say to unity, and then check to four corners of this face.

Please use the code line-by-line; it is listed here without EOF separators! Also, please verify that you select the non common factor in each step, the order could change between versions of Mathematica (this is done with version 9.0 in the free trial period)

K[u_, v_, t_] := u u + v v + t t - 4 (u v + v t + t u)
G[m1_, m2_, m3_] =FullSimplify[K[Sqrt[m1], Sqrt[m2], Sqrt[m3]] K[-Sqrt[m1], Sqrt[m2], Sqrt[m3]] K[ Sqrt[m1], -Sqrt[m2], Sqrt[m3]] K[Sqrt[m1], Sqrt[m2], -Sqrt[m3]]]
Expand[G[1, a, b]]
step1 = FactorList[Resultant[G[1, x, y2], G[1, y2, x2], y2]]
Resultant[step1[[2, 1]], G[1, x2, y], x2]
step2a = FactorList[Resultant[step1[[3, 1]], G[1, x2, y], x2]]
step2b = FactorList[Resultant[step1[[4, 1]], G[1, x2, y], x2]]
step2 = {step2a[[2, 1]], step2a[[4, 1]], step2b[[3, 1]], 
   step2b[[4, 1]]};
step2[[1]]
s1 = N[Solve[{step2[[1]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s2 = N[Solve[{step2[[2]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s3 = N[Solve[{step2[[3]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s4 = N[Solve[{step2[[4]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
sol = Join[s1, s2, s3, s4];
{1/x*174.1, y/x*174.1} /. sol

You can be intrigued that the solutions are more detailed than a simultaneus Solve[] of the system of four equations G[1, x, y2]==0, G[1, y2, x2]==0, G[1, x2, y]==0, G[1, y, x]==0. I am intrigued too. It seems that some of the particular solutions found by the Resultant method are embedded inside a continuous spectrum of solutions, and then the Solve method avoids listing them twice. I wished to know more on the relationship between resultants and continuous solutions.

It is amusing that the first triplet of the list is Rodejohann-Zhang triplet,
{1, {x -> 102.50258, y -> 2.1435935}}. Scale it times 174.1/102.50 and you get 1.69849, 174.1, 3.64088

My own triplet appears later, as it is generated by the last polynomial... it is
1, {x -> 2.1435935, y -> 0.071796770}. Use the same scale factor than before, and you get 1.69849, 3.64088, 0.12195

Both triplets are of the kind that becomes hidden in the continuous under Solve.

I am not sure about why this resolvent method does not find solutions with a zero, for instance 1.69849, 0.12195, 0. They can be searched by starting from G[0, x, y2]==0, G[0, y2, x2]==0, G[0, x2, y]==0, G[0, y, x]==0

IN:  Solve[{G[0, x, y2] == 0, G[0, y2, x2] == 0, G[0, x2, y] == 0, 
  G[0, y, x] == 0, x == 0.12195, G[r, x, y2] == 0, G[r, y2, x2] == 0, 
  G[r, x2, y] == 0, G[r, y, x] == 0, y >= y2}, {x, y, y2, x2, r}]

OUT: 
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562, x2 -> 0.000628625,   r -> 0},
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 0.00875562,  y2 -> 0.00875562, x2 -> 0.12195, r -> 0.261411},
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562, x2 -> 0.12195,  r -> 3.13693}, 
{x -> 0.12195, y -> 1.69854, y2 -> 0.00875562,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 1.69854, y2 -> 1.69854,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 1.69854, y2 -> 1.69854,  x2 -> 0.12195, r -> 3.64099}, 
{x -> 0.12195, y -> 1.69854,  y2 -> 1.69854, x2 -> 0.12195, r -> 43.6919}, 
{x -> 0.12195,  y -> 1.69854, y2 -> 1.69854, x2 -> 23.6577, r -> 0}}

EDIT: ok, a faster recipe could be

pols:factor(eliminate([G(1,x,a),G(1,a,y)],[a]))/(y-x)^4$
f1:ev(part(pols[1],1),[y=x])$
float(sol1:solve([f1,G(1,a,x)],[x,a]));

the output has the following positive solutions:
[x = 29.85640584694755, a = 0.12453316162267], 
[x = 29.85640584694755, a = 650.4292237442922], 
[x = 29.85640646055102, a = 199.4974226119286], 
[x = 29.85640646055102, a = 13.92820323027551]
[x = 2.143593539448983, a = 102.5025773880714], 
[x = 2.143593539448983, a = 0.071796769724491], 

and thus
(%i58) mc
;
(%o58)                              1.69854
(%i59) mc * 2.143593;
(%o59)                           3.64097845422
(%i60) mc *  0.07179;
(%o60)                           0.1219381866
(%i61) mc * 102.50257;
(%o61)                          174.1047152478