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The evidence for an approximate "lepton as fourth colour" symmetry is so overhelming in the particle spectrum that Hanlon's razor does not seem to apply. Still, my own incompetence fails to recognise an adequate model.

Point is, once we have massive see-saw-able neutrinos, we have enough leptons to organise approximate multiplets with three colours of a quark plus one colour neutral lepton:

  • $(ν_1,t_r,t_g,t_b)$ at about 174.10 GeV
  • $(ν_2,b_r,b_g,b_b)$ around 3.64 GeV
  • $(τ,c_r,c_g,c_b)$ around 1.698 GeV
  • $(μ,s_r,s_g,s_b)$ around 121.95 MeV
  • $(e,u_r,u_g,u_b)$ with null mass.
  • $(ν_3,d_r,d_g,d_b)$ about 8.75 MeV

But you can see the problem: there are two charged leptons with the two second generation quarks, and then two neutral leptons with the third generation! So something must be done with the L-R SU(2)xSU(2) charges of the model, or with the quark assignment. In fact I believe to remember that the paper of Harari-Haut-Weyers, from which the mass assignments for s.u.d are taken, had a model where the right and left quarks were permuted between generations, so I should expect more work to exist in the literature.

My question is, do you know of some sort of "twisted Pati-Salam L-R model" where the above multiplets are valid?

EDIT 1: the representations,

To start to play, they should not be taken from the standard model, but from Left-Right symmetric models with some Pati-Salam symmetry.

This means that both leptons and quarks are in representations of $SU(2)_R$ and $SU(2)_L$ with Right and Left isospins of $\pm 1/2$ where they are in the doublet and 0 where they are in the singlet. So the objection of Lubos Motl, below, is that for instance the muon above is in +1/2 of the left doublet while the strange is the -1/2 of the left doublet. And of course the same problem for the conjugate multiplet, in the $SU(2)_R$ side.

But this was my original question! Is it possible to twist the symmetries and generation-wise charge assignments to allow for such a mix?

EDIT 1.1: To clarify, my question is on the second and third generations. The first generation multiplet $(e^L,u^L_{rgb})$ as well as its conjugate $(e^R,u^R_{rgb})$ are usual multiplets of $SU(4) \times SU(2)_L \times SU(2)_R$, the former being doublet in the L and singlet in R, the later being singlet in L and doublet in R. In this case there should be nothing surprising about a Higgs mechanism preserving SU(4), no more that the preservation of SU(3) in the Standard Model. Nobody is surprised that the three up quarks have equal mass, the three down quarks another mass but equal for all of them, and still $u^R$ and $u^L$ have different electroweak properties.

EDIT 1.2: Note for instance Gabriele Honecker version of supersymmetric Pati-Salam, http://inspirehep.net/record/614377?ln=es, http://inspirehep.net/record/1185446?ln=es, where one generation has different representation that the other two.

EDIT 2: the masses (just at motivation, not the real question!)

Lubos points out that the values are numerological, but how they are? Well, this is irrelevant to the question, but it can be of marginal interest: the series is choosen so that all of the values fit with Koide formula: (174.10,3.64,1.698), (3.64,1.698,0.12195), (1.698,0.12195,0),(121.95,0,8.75). So the only input is 0 for up and 174.10 for top. Besides, the last triplet has the proportions of Harari-Haut-Wylers model: up equal to zero and $m_d/m_s$ is $\tan^2 15$.

If some of you check the triplets agains Koide, remember take the negative sign for $\sqrt {m_s}$ in the second one. In this way, it is orthogonal to the charged lepton triplet. The link between the scb triplet and the charged lepton triplet is exploited to predict the masses after the breaking of the multiplets, by assuming that all the Koide equations are still happening.

EDIT 2.1: when Koide is written as $m_k=M(1+\sqrt 2 \cos ({2 \pi \over 3} k + \delta))^2$, then it can be seen by inspecting above that $M_{scb}=3M_l$ and $\delta_{scb}=3\delta_l$. Asuming that this relation also survives to the breaking of "SU(4)", then it is possible to use as input the mass of electron and muon to predict all the other masses. And it works: the predictions are

$173.26, 4.197, 1.77696, 1.359, 92.275, 5.32, .03564$;

and the experiments (pdg2012) give, respectively,

$173.5 \pm 1.0, 4.18 \pm 0.03, 1.77682 (16), 1.275 \pm 0.025, 95 \pm 5, \sim4.8, \sim2.3$

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I edited the question title, "some sort of" instead of "well-known". I was tempted to stress "valid", but really, as you see, I feel open about this. Just "published" should be enough. –  arivero Dec 9 '12 at 18:33
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2 Answers

I am going to try to elaborate an answer by myself to show what I am thinking about, but please feel free to add your own.

Problem is, that $\nu_1$ and $\mu$ are not in the same SU(2) state that $t$ and $s$.

Both $\mu_L$ and $s_L$ are $SU(2)_R$ singlets, but in $SU(2)_L$ the former is $+1/2$, the later is $-1/2$

And of course the opposite problem happens with $\mu_R$ and $s_R$; they are $SU(2)_L$ singlets but they differ in $SU(2)_R$.

On other hand, $\bar \mu_R$ is a $SU(2)_R$ singlet and $-1/2$ in $SU(2)_L$, and similarly with $\bar \mu_L$. Of course lepton number becomes -1 instead of +1, but perhaps we can cope with this. It could then be sensible to build the multiplets as

$(\bar ν_1,t_r,t_g,t_b)$



$(\bar μ,s_r,s_g,s_b)$



There is an extra point scored here: that it makes Koide equation of leptons have the same look that Koide for quarks, with two pieces of the same kind and a different one.

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There can't be any multiplets of a viable gauge group - which would include the electroweak $SU(2) $ - that look like that because it's known that the left-handed parts of the quark in your first two "multiplets" form an electroweak doublet, for example, while the remaining components of these "multiplets" - two flavors of neutrinos - surely don't. I can find dozens of similar inconsistencies in your list. Equivalently, the trace of the electroweak $T_3$ in most of your would-be multiplets is not zero but it must be zero because the trace of every generator of a non-Abelian group must vanish in every representation.

You just organized the particles into random multiplets because of numerological motivations (proximity of masses?) whose basic physics is obviously indefensible. So of course that such random sets of particle species can't form multiplets.

Moreover, you seem to misunderstand the nature of the true lepton and quark fields because you clump the left-handed ones and right-handed ones. They come from different representations of the gauge group which must be discussed separately. The left-handed ones are doublets, the right-handed ones are singlets, and so on. It's because the electroweak interactions are chiral. You clumped completely different representations into "unified" groups regardless of their actual transformation group, by linking them to the observed particles. But the observed particles and their masses aren't coming from fields that transform uniformly under the gauge group, as the doublet-vs-singlet chiral character mentioned above indicates.

Why don't you try to learn how the fields actually transform in the electroweak theory, and then study its possible generalizations? It doesn't make any sense to try to find would-be grand unified theories if you don't understand the group theory in the more modest and more well-established electroweak theory, and you obviously don't.

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I guess that my question is if I can use different L-R charge assignments in the L-R SU(2) groups for each generation, so that all the particles in the same multiplet go to the same state in the same representation. I have expanded a bit about it, it was clear from the context I was not thinking of electroweak $T_3$ but L-R $T_3$ –  arivero Dec 9 '12 at 15:19
Dear Alejandro, it just doesn't matter. These multiplets are impossible as representations of any non-Abelian group because they don't pass basic tests I have mentioned. A representation of a group isn't an arbitrary collection of objects that you clumped because of some crazy numerological reason concerning masses - that have nothing to do with the multiplets. A representation is made of basis vectors that must transform into each other so their quantum numbers must reflect the actual group - and the quantum numbers are the charges under forces, not masses! Your collections can't be reps. –  Luboš Motl Dec 9 '12 at 19:54
Ok, so your answer to my question is "No, such arrangement can not exist" and not "such arrangement should need a complicated Higgs sector to put masses in". It is a bit puzzling because the only exotics here are the ($\mu$-s) and the ($\nu$-t) multiplets. –  arivero Dec 9 '12 at 21:01
Or, are you claiming that $(u_r,u_g,u_b)$ should not be considered to have a symmetry under SU(3) colour, because $u^R$ and $u^L$ have different electroweak assignments? It makes sense to tell it, but it is mostly a point of view, not a point of principle. –  arivero Dec 9 '12 at 21:26
After some extra reading, I feel your answer is not helpful because it fails to mention relevant cases, well known by you, where different generations have different representation and charge assignments. For instance Cvetic-Shiu-Uranga or Gabriele Honecker models. –  arivero Dec 17 '12 at 16:12
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