- #51

arivero

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hep-ph/0606039 said:Notice that the 3 vertical columns are evocative of the mesons [tex]\eta_8, \eta_1, \pi^0[/tex] in their SU(3) decompositions.

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- #51

arivero

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hep-ph/0606039 said:Notice that the 3 vertical columns are evocative of the mesons [tex]\eta_8, \eta_1, \pi^0[/tex] in their SU(3) decompositions.

- #52

arivero

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CarlB said:I'm still mulling through the Koide paper, particularly page 6: http://www.arxiv.org/abs/hep-ph/0605074

By the way, I wonder if the possibility of setting the vector [tex](-\sqrt m_1 ,\sqrt m_2 ,\sqrt m_3)[/tex] both at 45 degrees of (1,1,1) and at 90 degrees of [tex](\sqrt{m_\tau},\sqrt{m_\mu},\sqrt{m_e})[/tex] (which is also at 45 degrees of (1,1,1) ) is ruled out by the phenomenology or on the contrary makes a good ansatz. This additional condition amounts to ask

[tex]\sqrt { m_1 m_\tau} =^? \sqrt{m_2 m_\mu}+\sqrt{m_3 m_e}[/tex]

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- #53

CarlB

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arivero said:Ernest Ma in hep-ph/0606039 does a toast to preons when reviewing tribimaximal mixing, he says:

Nice observation. And it sort of fits in with my belief that the leptons are SU(3) singlets.

By the way, I'm not at all sure that "tribimaximal" has become a part of the usual physics terminology. In response to a question about Koide's paper, I got a response from a fairly famous physicist as follows:

Carl

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- #54

arivero

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arivero said:By the way, I wonder if the possibility of setting the vector [tex](-\sqrt m_1 ,\sqrt m_2 ,\sqrt m_3)[/tex] both at 45 degrees of (1,1,1) and at 90 degrees of [tex](\sqrt{m_\tau},\sqrt{m_\mu},\sqrt{m_e})[/tex] (which is also at 45 degrees of (1,1,1) ) is ruled out by the phenomenology or on the contrary makes a good ansatz. This additional condition amounts to ask

Lets to work out this. The ortogonality condition can we rewritten as

[tex]\sqrt { m_1} = \sqrt{m_2} \sqrt{ m_\mu\over m_\tau}+\sqrt{m_3} \sqrt{m_e\over m_\tau}[/tex]

and then the 45 degrees condition

[tex]

{-\sqrt{m_1} + \sqrt{m_2} +\sqrt{m_3} \over

\sqrt{m_1+m_2+m_3} \sqrt{3}} = {1 \over \sqrt 2}

[/tex]

becomes

[tex]

\sqrt{m_2} (1-\sqrt{ m_\mu\over m_\tau})+\sqrt{m_3} (1-\sqrt{m_e\over m_\tau}) = \sqrt \frac 3 2 \sqrt{ m_2 (1+{ m_\mu\over m_\tau}) + m_3 (1+{m_e\over m_\tau}) + \sqrt{ 2 m_2 m_3 { m_\mu m_e \over m_\tau^2} }}

[/tex]

or

[tex]

\sqrt{m_2\over m_3} (1-\sqrt{ m_\mu\over m_\tau})+\ (1-\sqrt{m_e\over m_\tau}) = \sqrt \frac 3 2 \sqrt{ {m_2\over m_3} (1+{ m_\mu\over m_\tau}) + \ (1+{m_e\over m_\tau}) + \sqrt{ 2 {m_2 \over m_3} { m_\mu m_e \over m_\tau^2} }}

[/tex]

Hmm pretty unmanageable. Lets put numbers in...

[tex]

.75615 \sqrt{m_2\over m_3} + .983042

=\sqrt \frac 3 2 \sqrt{1.059459 {m_2\over m_3} + 1.00028756 + .0116954 \sqrt{ m_2 \over m_3} }[/tex]

and

[tex]\sqrt { m_1\over m_3} = .24384 \sqrt{m_2\over m_3} + .016957[/tex]

The question now, if the numbers are right, is to see if this ansatz is compatible with the measured oscillations

EDITED AGAIN: in principle it implies m2/m3=0.525,m1/m3=0.0375 (???) :sad: it does not seem to work

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selfAdjoint

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Kea said:

Are you sure that being on the cover of New Scientist counts as non-fringe evidence?

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selfAdjoint said:Are you sure that being on the cover of New Scientist counts as non-fringe evidence?

No, it doesn't really. But when more String theorists realise that one can describe moduli with ribbon diagrams,

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- #58

arivero

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Some days ago, CarlB did some comment on meson masses, joking that he could try about searching Koide's formulae there.

Well, just in case, here attached is a C program do to it with the six basic charged mesons, plus the dump of the output for the general case of plus and minus square roots. Koide quotient is in the first column of data, the other thing are "angles against the diagonal" in six, n, and 3 dimensions, with N being the number of mesons involved in a particular calculation.

Input data is

Code:

```
float mass[6] = {139.57018, /*pm0.00035*/
493.677, /*pm 0.016*/
1869.3, /*pm 0,4*/
1968.2, /*pm 0.5 */
5279.0, /*pm 0.5 */
6286 /*pm 5 */
};
```

Corresponding to pion, kaon, D, Dstrange, B, Bcharmed. In the listing I have used uppercase D and B for the stranged and charmed versions of d and b.

Program compiles as usual: gcc -lm koide.c -o koide

If you run it, you could want to sort -n -k 4, and perhaps to grep output.txt -v -

Most of the coincidences are artifacts coming from having pairs of particles with nearby mass, d D and b B. You could want to run some "null tests" with values having the same pattern, say {0.001,0.002,300,300,1000,1000}. Exact integer multiples (nor fractions), or near exact, are most probably artifacts.

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- #59

arivero

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Why am I posting this in the thread of preons? Well, because, as veteran readers of PF know, I got the idea last year of having the preons directly from quarks: a pair of quarks would compose a boson, and then this boson is supersymmetrically transfomed to the fermion we are aiming for. Sort of bootstrap.

I described the idea one year ago in hep-ph/0512065. It is not bad, it produces only a extra degree of freedom in the neutrino sector and six horrible 4/3 coloured degrees of freedom in a quark/antiquark sector, but I would hope they can be eliminated on the grounds of representation theory.

It has the adventage that we know the masses of the, er, subquarks, and also we know how they bind: with SU(3) colour. The binding via SU(3) colour gives some substance to infrarred mass relationships, as Koide's, that are troublesome to be planted in the GUT scale (albeit some people do). Ideally the IR limit of QCD explodes the coupling constant, and then justifies the trick.

But the problem is that we already have the "susyleptonic" sector of this theory, and they are the charged mesons of the previous post. It is not a badly broken thing, because the pion has more of less the same mass than muon and the D particles are about the same mass than the tau. But the electron has no partner near, neither the B particles.

So here is why I think it is unlikely to find Koide relationships for mesons: because I think that in the limit of unbroken supersymmetry, leptons would derive Koide relationship from the fact they are partners of sleptons, which happen to be mesons, thus composites. In this limit, then, mesons also meet Koide relationships, and furthermore they are degenerated in pairs. But I find unlikely that Koide can survive in both sectors after symmetry breaking, and if it survives in leptons, my guess is that it will break in the mesonic part.

Telling this, and against myself, I can not but notice that besides the obvious triplets [tex](0, m_\pi, m_{D_s})[/tex] and [tex](0, m_\pi, m_{D})[/tex], also the triplet [tex](0, m_k, m_{B_c}[/tex]) is, unexpectedly,very Koide-like. Moreover, when the 0 is substituted by another of the particles, D or B, it counterweights nicely, climbing Koide's relation from 2:3 to 2:5.

I described the idea one year ago in hep-ph/0512065. It is not bad, it produces only a extra degree of freedom in the neutrino sector and six horrible 4/3 coloured degrees of freedom in a quark/antiquark sector, but I would hope they can be eliminated on the grounds of representation theory.

It has the adventage that we know the masses of the, er, subquarks, and also we know how they bind: with SU(3) colour. The binding via SU(3) colour gives some substance to infrarred mass relationships, as Koide's, that are troublesome to be planted in the GUT scale (albeit some people do). Ideally the IR limit of QCD explodes the coupling constant, and then justifies the trick.

But the problem is that we already have the "susyleptonic" sector of this theory, and they are the charged mesons of the previous post. It is not a badly broken thing, because the pion has more of less the same mass than muon and the D particles are about the same mass than the tau. But the electron has no partner near, neither the B particles.

So here is why I think it is unlikely to find Koide relationships for mesons: because I think that in the limit of unbroken supersymmetry, leptons would derive Koide relationship from the fact they are partners of sleptons, which happen to be mesons, thus composites. In this limit, then, mesons also meet Koide relationships, and furthermore they are degenerated in pairs. But I find unlikely that Koide can survive in both sectors after symmetry breaking, and if it survives in leptons, my guess is that it will break in the mesonic part.

Telling this, and against myself, I can not but notice that besides the obvious triplets [tex](0, m_\pi, m_{D_s})[/tex] and [tex](0, m_\pi, m_{D})[/tex], also the triplet [tex](0, m_k, m_{B_c}[/tex]) is, unexpectedly,very Koide-like. Moreover, when the 0 is substituted by another of the particles, D or B, it counterweights nicely, climbing Koide's relation from 2:3 to 2:5.

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arivero

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arivero

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It is not rare to see the mass of the pion as proportional to the square root of the mass of the constituyent quark. I have just seen it in a old work by Olivier Martin. Hmm.

Ah, another funny reference for the preon thread. Peskin 1981 is scanned in KEK:

http://ccdb4fs.kek.jp/cgi-bin/img_index?8202032 [Broken] Enjoy!

Ah, another funny reference for the preon thread. Peskin 1981 is scanned in KEK:

http://ccdb4fs.kek.jp/cgi-bin/img_index?8202032 [Broken] Enjoy!

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CarlB

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Nice link. "For theorists, my requests are more serious: ... The third, and most pressing, is to learn how to compute the mass spectrum of quarks and leptons in composite models, even in models too simple to be realistic. It is, after all, only through the computation of the quark and lepton masses that the idea of compositeness can really fulfill its promise."

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arivero

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http://es.arxiv.org/abs/hep-ph/0508278

http://es.arxiv.org/abs/hep-ph/0604169

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CarlB

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arivero said:

http://es.arxiv.org/abs/hep-ph/0508278

http://es.arxiv.org/abs/hep-ph/0604169

Hmm. Tetrahedral group? I wonder where that would come from. See figure 6.4 on page 93(107):

http://www.brannenworks.com/dmaa.pdf

- #65

arivero

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It seems a different context, but I have not traveled across your whole work :uhh:CarlB said:See figure 6.4 on page 93(107):

In any case, it is sort of consolation to see that people in the academy as Zee and Ma are keeping these topics live.

On my own side, I do not see how to fit A4 in my susypreonic model.

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arivero

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http://www.sns.ias.edu/~adler/Html/preons.html [Broken]

I have just noticed Adler's preons in the final chapter of his book on quaternionic quantum mechanics. He generalises Harari model by allowing two different permutations; one for T/V and another for spin +/-. In his generalisation, the three generations come from the three differents ways of ordering ++- +-+ -++ the spins, in the same way that the three colours come from TTV TVT TTV.

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At what threshold energies are the quark "preons" expected to be detected in a proton collider experiments?I do not like composite models of quarks and leptons; they seem to me just as decomposition of phonemes: it can be done, but it is not linguistic. Still, I could be wrong. Another PF inhabiatant, Carl Brannen, likes them enough to have developed his own model. And I'd guess he is not the only one around here.

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arivero

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I refer to "Families from Spinors",

http://prola.aps.org/abstract/PRD/v25/i2/p553_1

As far as I can find, the earliest use five clifford algebra generators to build the fermions appears in the body of this article (and some extra hints in the appendix). If you know of earlier work, please tell me!

Actually I think I understand how it happens, and that really we need to take six generators to build both the fermions and their susy partners, in this way it is a 6-dimensional algebra and it explains why the "Newton Institute Spectral Triple" (as I hve christened it, to differ from "Les Houches Spectral Triple") is 6-dimensional.

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arivero

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I have rescued Peskin's note. It is actually very interesting to check page 3. His description of models of the second kind hints why such models are forcefully related to spin and then to Clifford algebras. He does not point towards Wilczek and Zee papers, probably because they are contemporary, but the explanation in W & Z follows similar trends.Ah, another funny reference for the preon thread. Peskin 1981 is scanned in KEK:

http://ccdb4fs.kek.jp/cgi-bin/img_index?8202032 [Broken] Enjoy!

All together, the paper is a hint about why Clifford modelers in the internet are not considered in the mainstream: the mainstream tried these tricks in the late seventies and failed.

http://es.arxiv.org/abs/hep-ph/0508278

http://es.arxiv.org/abs/hep-ph/0604169

Amusingly, the papers quote as main paper http://www.slac.stanford.edu/spires/find/hep/www?irn=310417 [Broken]

I have just noticed Adler's preons in the final chapter of his book on quaternionic quantum mechanics. He generalises Harari model by allowing two different permutations; one for T/V and another for spin +/-. In his generalisation, the three generations come from the three differents ways of ordering ++- +-+ -++ the spins, in the same way that the three colours come from TTV TVT TTV.

http://arxiv.org/abs/hep-th/9610190 and http://arxiv.org/abs/hep-ph/0201009 are claimed to be follow-ups.

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arivero

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Still, the class b models group up to five SU(2) labeled particles because they need to produce a fermion. I like more my approach of getting composite bosons and elementary fermions. This is because my model predicts both the number of generations and the number of top quarks.

Consider N families with q interacting up quarks and p interacting down quarks. We have the obvious constraing 0 =<p=<N, 0 =<p=<N. If we ask for a matching of degrees of freedom, consider that for a given charged fermion we have 2 N degrees of freedom. Supersymmetry between composites and elementary particles requires the following equations:

p*q = 2N (for u+d) and p*(p+1)/2 = 2N for (d+d). From the second one we have

that the number of generations must be such that (1 + 16 N) is a perfect square and N>=p >= (-1+sqrt (1+16N))/2, so that N>=3. The lowest possibility is N=3. Then p=3 and q=2: we have 3 light downĺike quarks and two light uplike ones.

(ok, what about N=5 (fails because p=4, but q=2.5) or N=14 (ok with p=7, q=4) and so on? The point is that the union of up+antiup and down+antidown should also sum the number of degrees of freedom of the neutrinos, the providing a further constraint, still to be researched)

Consider N families with q interacting up quarks and p interacting down quarks. We have the obvious constraing 0 =<p=<N, 0 =<p=<N. If we ask for a matching of degrees of freedom, consider that for a given charged fermion we have 2 N degrees of freedom. Supersymmetry between composites and elementary particles requires the following equations:

p*q = 2N (for u+d) and p*(p+1)/2 = 2N for (d+d). From the second one we have

that the number of generations must be such that (1 + 16 N) is a perfect square and N>=p >= (-1+sqrt (1+16N))/2, so that N>=3. The lowest possibility is N=3. Then p=3 and q=2: we have 3 light downĺike quarks and two light uplike ones.

(ok, what about N=5 (fails because p=4, but q=2.5) or N=14 (ok with p=7, q=4) and so on? The point is that the union of up+antiup and down+antidown should also sum the number of degrees of freedom of the neutrinos, the providing a further constraint, still to be researched)

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CarlB

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Recently, I've been looking at generalizations of the idempotency relation [tex]\rho^2 = \rho[/tex], for more general density matrices. Among the spinors, these define the available quantum numbers. An easy example is the Dirac algebra. A complete set of primitive idempotents are the diagonal matrices with only a single 1 on the diagonal and the rest zero. These are the density matrices made from the four orthogonal spinors which each has a single one and the rest zero entries.

The above definition relies on the choice of a representation. To make it rely less, one could write down the detailed equations for the matrix square: [tex]\rho = \rho^2.[/tex] This would be sixteen quadratic equations in sixteen unknowns, starting with:

[tex]\begin{array}{rcl}

\rho_{11} &=& \rho_{11}\rho_{11} + \rho_{12}\rho_{21} + \rho_{13}\rho_{31} + \rho_{14}\rho_{41},\\

\rho_{12} &=& \rho_{11}\rho_{12} + \rho_{12}\rho_{22} + \rho_{13}\rho_{32} + \rho_{14}\rho_{42},\\

&&...\\

\end{array}[/tex]

and going through the other 14 values. The above equations are fairly simple. There are sixteen variables that appear on the left. These variables appear in pairs on the right. A given pair either contributes in exactly one row or it does not contribute to any. With sixteen elements, there are 256 possible pairs but only sixteen of these contribute to a row.

We can associate an ugly finite group structure with these 16 idempotency equations. The structure is on a group with 17 elements. The elements are the above [tex]\rho_{mn}[/tex] plus zero, but since we're treating them as a group instead of complex numbers, we can call the group elements [tex]h_{mn}[/tex] and 0. The group multiplication rule is that [tex]h_{ij}h_{kl} = \delta_j^k h_{il}[/tex], and zero times anything is zero.

The whole thing can be reversed. Given a group, we can define an idempotency relation by examining each possible product, and putting a quadratic term into the row that is given by the group product.

Since I'm playing with a model where the elementary fermions are composed of three preons, the natural group to apply this to is the permutation group on three elements.

Before doing that, an easier problem is the even subgroup of the permutation group on three elements. This is an easy problem. Call the three elements [tex]I, J, K[/tex]. The group definitions and the resulting quadratic equations are:

[tex]\begin{array}{rcccl}

I&=& (123) &=& I^2 + 2JK,\\

J&=& (231) &=& K^2 + 2IJ,\\

K&=& (312) &=& J^2 + 2IK.\end{array}[/tex]

The above quadratic equations are what one gets when one enforces the idempotency relation on a circulant matrix:

[tex]\left(\begin{array}{ccc}

I&J&K\\K&I&J\\J&K&I\end{array}\right)^2 =

\left(\begin{array}{ccc}

I&J&K\\K&I&J\\J&K&I\end{array}\right)[/tex]

And as we've discussed before, the circulant matrices are a natural road to Koide's formula. This makes the permutation group of three elements worth suffering through. In the permutation group on three objects, there are six elements. You get six quadratic equations in six unknowns. The solution is arduous (at least for me), but the results are interesting.

The complex number that replaces the identity group element [tex]I[/tex] takes on the absolute values of weak hypercharge among the elementary fermions (and nothing else). That is, it takes on the values 0, 1/6, 1/3, 1/2, 2/3, 1. This seems natural because weak hypercharge is associated with a U(1) group.

The fact that it doesn't give the negative values doesn't bother me too much because the antiparticles take negative quantum numbers anyway. I suspect that if one changes the group slightly, one can get the negatives and positives. In particular, the permutation group on three elements does not include the Poincare symmetry. So the permutation group is a simplified problem.

There are three odd permutation elements. In the group, these are the elements that square to unity, which reminds one of the generators of SU(2). In solving the 6 quadratic equations, you will find that your task is eased considerably if you assume that the three permutation elements take the same complex values. In any case, the sum of those three complex values (i.e. triple one of them if they are all the same), turns out to be the weak isospin numbers that one associates with the weak hypercharge numbers of the fermions (i.e. consistent with the assignment of weak hypercharge to [tex]I[/tex]).

Having done this, there are still two group elements to deal with, [tex]J[/tex] and [tex]K[/tex]. These are the permutations that one thinks of as defining the twist of the corner of a cube as either a positive twist or a negative twist. Another way of saying this is that if we have a group consisting of the identity I, a left L, and a right R, we want to have IR = LL = R, IL = RR = L, LR = RL = I.

The values for these complex numbers can be assigned consistently to the elementary fermions so that the difference between the left and right handed particles of a pair always has the same value. That is, the difference between the neutrino left and right is the same as the difference between the down left and right, etc. This suggests that these two values can be interpreted as the quantum numbers of the Higgs.

I typed the algebra calculations into my paper

http://www.brannenworks.com/dmfound.pdf

which unfortunately includes a lot of denstiy matrix theory that no one wants to read before the algebra calculations above, which appears right now in section (5.1) page 33. To get anyone interested in reading it, I'll have to strip all that off.

Carl

- #72

arivero

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Arivero, I didn't see your link to the Wilczek and Zee article until just now. I'll drop by the University and down load it this week. It seems like it's right down my line and could be very interesting.

Get also Casalbuoni and Gatto Phys Lett B vol 90B p 81 and vol 88B p 306. They are the first ones using the word "Clifford" in the context of preons.

- #73

arivero

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Haim Harari and Nathan Seiberg http://dx.doi.org/10.1016/0370-2693(81)90871-6 [Broken] 1981

Adler, Stephen L http://www.slac.stanford.edu/spires/find/hep/www?r=IASSNS-HEP-87/33 [Broken] 1987

The later is simply spin permutations. The former I have not checked, but it is interesting to note the coauthor.

Adler, Stephen L http://www.slac.stanford.edu/spires/find/hep/www?r=IASSNS-HEP-87/33 [Broken] 1987

The later is simply spin permutations. The former I have not checked, but it is interesting to note the coauthor.

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