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Who has 12 of 8?

  1. Mar 26, 2008 #1

    arivero

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    A 10 dimensional spinor has 8 degrees of freedom (it is Weyl-Majorana), and the Standard Model fermions sum 96 degrees of freedom.

    So 12 fermions of 8 degree of freedom each are the minimum to account for the standard model.

    Is there any theory, amateur or profesional, containing only this minimum for the fermion sector? (It should be OK if there is additional bosonic content, and/or of course gauge content beyond the SM).


    To note:
    -In the age of massless neutrinos, the sm only had 90 dof. So no match for it in groups of 8.
    -Thus, any old theory matching "12 of 8" is actually a prediction for right-handed neutrinos, and probably a prediction for massive neutrinos. But it is not very striking, because any GUT in the ways of SO(10) has it.
    -It is interesting, even without a theory to back it, to consider which fermions should fill a 8-tuple. First idea is to pair electron with neutrino and up with down.
     
    Last edited: Mar 26, 2008
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  3. Mar 27, 2008 #2

    arivero

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    Other consideration: an issue with the arrangement of two 4D particles in the 10D fermion is that each pair has two different 4D masses (in 10D, I guess it is massless). A way to ignore this issue is to use 24 fermions instead of 12, accounting the duplicate to some kind of "mirror matter". Not that I like it, but it makes explicit a point that is only implicit in the previous formulation: that in some sense we have 24 objects.
     
    Last edited: Mar 27, 2008
  4. Mar 27, 2008 #3
    Well...the fermions have to transform as representations of gauge groups, so you'd need something that contains the standard model, with the correct hypercharge normalization, as well as representations that contained things like a 3 + 3* of SU(3).

    I don't know THAT much group theory, but I'd suspect this probably means that you can't find what you're looking for.

    I do seem to recall something about N=8 SUGRA in 4 dimensions---I need to check some old class notes.
     
  5. Mar 27, 2008 #4

    arivero

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    I know the 4D fermions should transform as representations of gauge, but in this query I am not setting any constrain about the 10D ones; I guess that the theory should explain how to produce such transform but I am not asking for it just now; simply any theory having these "12 of 8" as basic pieces.
     
  6. Mar 27, 2008 #5
    Am I way off base if I suggest an improbable, but relevant, modified rishon theory in which two rishon flavors, P and N (P for "Positive" where the sign of isospin and hypercharge is the identical, and N for "Negative" where the sign of isospin and hypercharge is opposite) and their antirishons, each appearing in three different colors, give you twelve fermions? Not only that, but in theory these 12 rishons compose everything, including all guage bosons. And if you allow both chiral symmetry and isospin symmetry to be spontaneously broken at 45 degrees, you get maximal mixings between all quark-type and lepton-type composite states where P-rishons are massive and N-rishons are nearly massless, and between all right-handed and left-handed composite states where the right-handed states are pushed up into the heirarchy and the left-handed states are pushed toward masslesness. The quark-types appear as combinations of two rishons and an antirishon or two antirishons and a rishon (overall rishon number of 1 or -1), and are three times as abundant as the lepton-types. The lepton-types appear as combinations of three rishons or three antirishons (overall rishon number of 3 or -3). This setup describes only the first generation of composite fermions, but adding PPbar pairs allows for higher generations, with the limit of three generations set automatically when spontaneous fission by fall-apart is taken into account for generation-4 or higher composites. If each of these rishons has 8 degrees of freedom (and therefore are Weyl-Majorana), and there are 12 distinct varieties due to the combinations of isospin, hypercharge, and color charge, then you get your 96 degrees of freedom in the Standard Model and simultaneously explain most, if not all, mass mixings of observed SM fermions?

    Maybe I'm just way off base...
     
  7. Mar 28, 2008 #6

    jal

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  8. Mar 28, 2008 #7

    arivero

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    Hmm you have done the enumeration too fast, let me see.

    Your explanation is too fast, I do not quite follow it. For instance, electric charge is unclear.

    The quark-types appear as combinations of two rishons and an antirishon or two antirishons and a rishon (overall rishon number of 1 or -1), and are three times as abundant as the lepton-types.

    PP-N PN-N NN-N PP-P PN-P NN-P
    times colour times antiparticle. This is 6*3*2 = 36. So I guess you need a mechanism to forbid the combination of two different rishons with an antirishon, ie PN-P and PN-N. In this way PP-N, NN-N, PP-P and NN-P times colour times antiparticle are the 24 degrees of freedom of the quarks, right?

    The lepton-types appear as combinations of three rishons or three antirishons (overall rishon number of 3 or -3).

    PPP PPN PNN NNN
    times antiparticle. This is 4*2 = 8, and it is colour neutral.


    but adding PPbar
    pairs allows for higher generations, with the limit of three generations set automatically when spontaneous fission by fall-apart is taken into account for generation-4 or higher composites.


    Ok, thus it disqualifies for the count of 96, if we need to set automatically by hand a mechanism; we need really two extra mechanism, one to purge the extra quarks and another to produce generations. I have no problem about extras, my own sBootstrap theory needs to cut out three chiral +4/3 quarks. And even superstrings neeed GSO truncation. But in this case it is not easy to claim that we have naturally a "12 times 8" structure.
     
  9. Mar 28, 2008 #8

    arivero

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    Hmm let me try again. I suppose your leptons are PPP and NNN, while quarks are PNN and PPN. Ah, ok, then this is the famous fermion cube, well. That works if only P, or only N, is coloured. In this case it is true that there are three times more PPN and PNN than NNN and PPP. Total we get 8 combinations here, the 8 corners of the cube. To get the rest of the generation, we use antiparticles. And then we are in a 16; we still need to argue that all of them are Dirac fermions, to add the 32 degrees of freedom.

    But you are basically right here, that another way to produce 8 degrees of freedom is to consider the fermion cube. This is really the Standard Way, half of a representation of SO(10), isnt it?

    but it any case, this thing does not fit your definition of rishons and antirishons. Hmm have I forgot to consider the order of the rishons? I never got how different orderings TTV TVT VTT are supposed to give different charges.
     
    Last edited: Mar 28, 2008
  10. Mar 29, 2008 #9
    Yeah, I just spewed all that out there without explaining a whole lot. Sorry. I'll try to explain these things a little better.

    Yes. In fact, this can be done by invoking the symmetry rules and Pauli Exclusion Principle. Now I will warn you that my model requires the isospin of the individual rishons to be + or -1/6, rather than the traditional + or -1/2 when using the traditional baryon number basis; it did not work unless I required this, and I will show you why;

    P(I=+1/6, Y=+1/3), N(I=-1/6, Y=+1/3), -P(I=-1/6, Y=-1/3), -N(I=+1/6, Y=-1/3)

    Q = I + Y/2

    PPP = positron, NNN = electron-neutrino
    PP-N = up quark, NN-P = down quark
    -P-PN = up antiquark, -N-NP = down antiquark
    -P-P-P = electron, -N-N-N = electron-antineutrino

    These states all share one interesting feature; the isospin of all members of the composite are of the same sign, summing to + or -1/2. Thus, composite fermions are required to share the common isospin grouping of 1/2, while any composite fermion that has any isospin different is apparently unallowed by this mechanism.

    So, states that have two cancelling rishons and leave one by itself are forbidden (that excludes PN-N, PP-P, NP-P, and NN-N plus conjugates). This eliminates 8 states from the entire grouping, regardless of color rules.

    Well, the same mechanism as above will exclude PPN, PNN, and conjugates. Composites which have three rishons that do not all share identical flavors are forbidden, removing 4 additional states from the complete line-up, once again regardless to color rules. I think one of the secrets here is that you treat the color charge matrix of the rishons to be 1/3 of the gluon color charge matrix. Also, the opposite flavors are assigned opposite color charges; this adds additional reasoning for excluding the PPN, PNN, and conjugates from the line-up, as well as the 8 states removed from the candidate quarks, since one can require states that have a lone rishon-color charge to be unallowed.

    If we take this in common vernacular of QCD, we could say that quarks are either Red (R), Green (G), or Blue (B). That makes our common colored SU(3)_c gluons R-G, G-B, B-R, R-B, B-G, and G-R. Dividing G-B or B-G by 3 gives us the first rishon color charge set, which I will refer to as "a", where "a" has SU(3)_c color charge g_3 = +1/6, g_8 = +sqrt(3)/6 for P and -N, and g_3 = -1/6, g_8 = -sqrt(3)/6 for -P and N. Similarly, (B-R)/3 and (R-B)/3 give us the second rishon color "b", with SU(3)_c charges of g_3 = +1/6, g_8 = -sqrt(3)/6 for P and -N, and g_3 = -1/6, g_8 = +sqrt(3)/6 for -P and N. Lastly, (R-G)/3 and (G-R)/3 give us "c", with SU(3)_c charges of g_3 = -1/3, g_8 = 0 for P and -N, and g_3 = +1/3, g_8 = 0 for -P and N. So we essentially have a setup where only colors of the same sign are combined in the basic composites, and all others appear to be unallowed for three-rishon composites.

    Four-rishon composites and higher, though, are allowed in the event that they have a rishon-number of zero. The gluons turn out to be four-rishon colored composites; just use the rishons listed to build your various colors of quarks, then see what would pass between them in order to exchange colors. The W particles are interesting, though. They have isospin of 1, and as such have to be constructed as six-rishon states with PPP-N-N-N (W+) and -P-P-PNNN (W-) in color-neutral configuration.

    Taking the U(1) part of the symmetry to be exact, and hence conserving hypercharge, one may require isospin symmetry to be spontaneously broken, and hence mix the P and N states at nearly 45 degrees (identically for the -P and -N). In the assumption that P and N start out with equal mass before the symmetry breaking, we can cause the P to become massive, and the N to become nearly massless. In this limit, we have the ability to treat states with N-N attachments as being nearly the same mass as those without an attachment, while states with P-P attachments have significantly greater mass. If one tunes the P-P repulsion so that it is very large, and also so that it is based on the P mass (and hence you can say that N-N is also repulsive, but negligibly small due to negligible N mass), we can establish the massive generations. In the process, PN and -P-N become repulsive but small, and PP and NN are attractive (with PP on the order of the P mass, while NN is negligible since it is on the order of the N mass; this allows up and down quarks to be near in mass). However, we can also show that a fourth-generation fermion would automatically fall apart into three less-massive first generation fermions because of the availability of a lower-energy state;

    4-gen neutrino = NNNP-PP-PP-PP-P --> NNN + PPP + -P-P-P or NN-P + N-P-P + PPP
    4-gen down quark = NN-PP-PP-PP-P --> NN-P + PPP + -P-P-P or N-P-P + N-P-P + PPP
    4-gen up quark = N-P-PP-PP-PP-P --> N-P-P + PPP + -P-P-P
    4-gen electron = -P-P-PP-PP-PP-P --> PPP + -P-P-P + PPP

    The mixing at 45 degrees would suggest that an individual rishon could be P or N about half of the time, hence the probability of being in any one flavor state is nearly 1/2. Since any composite fermion would fail to exist if it entered an unallowed state, the rishons are constrained to reverse isospin in groups of three, so that;

    (1/2)x(1/2)x(1/2) = 1/8 probability over the sum of all possible states

    PPP --> NNN mixes at 1/8 probability
    PP-N --> NN-P mixes at 1/8 probability
    P-N-N --> N-P-P mixes at 1/8 probability
    -N-N-N --> -P-P-P mixes at 1/8 probability

    Throwing in the unusual permutations that result from mixings of second and third generation objects gets even more interesting. Since an up quark is not just PP-N, but may also encompass the states of near-equal mass PP-NN-N and PP-NN-NN-N, we have to account for mixings with higher-mass generations like charm (PP-NP-P and PP-NP-PN-N) or top (PP-NP-PP-P). Similarly for the down-type quarks, leptons and neutrinos. Mixings of the Z-type (weak neutral current) proceed via isospin-flip by even numbers of rishons, while mixings of the W-type (weak charged current) proceed via isospin-flip by odd numbers of rishons. Then we must also take into account the mixing of boson states, which can come in so many varieties as the number of rishons goes up. I have already been through this excersize, and found that the mixing between B and W_3 that produces Z and the photon is amazingly reconstructed in this model given that the B and W_3 start with appropriate masses (which probably falls from a Higgs or Higgs-like mechanism). I have a hunch that if the probabilities alone are calculated out, then the mass mixings will fall into line closely with experimental results. The last piece then is the mass-creating Higgs or Higgs-like mechanism that must first exist, and then we can calculate the masses of fermions, bosons, and all else to very high accuracy when compared to experiment.

    This does require the assumption that isospin-symmetry breaking is spontaneous after a fashion... but the mechanism that limits to three generations is a result of the fact that the P-P repulsion is very large. Otherwise, if the sum of masses of three first-generation fermions was larger than the mass of a fourth-generation fermion, then we would see fourth, fifth, and sixth-generation fermions formed in nature.
     
  11. Mar 29, 2008 #10
    Well, yes I think you are getting what I tried to say.

    That is the idea, for sure.

    This is why I immediately identified what I was saying as a modified rishon theory. I kept some of the conventions, but changed what I thought needed changing. First of all, I decided that having T and V have opposite hypercharge (as it is according to Harari) was entirely insufficient and confusing, so I equated -T to P, T to -P, V to N, and -V to -N. This allowed me to identify P and N as having the same baryon/lepton number (and hence the same hypercharge) so that definitions of particle vs. antiparticle are drawn along the lines of hypercharge exactly. Since the U(1) portion of the symmetry is conserved, this makes it real easy to decide what is a rishon and what is an antirishon. Furthermore, the naming convention follows from the product of hypercharge and isospin, so the rishons with identical sign for isospin and hypercharge take the symbol "P" (for "Positive", the sign of the resulting product), and the rishons with opposite sign for isospin and hypercharge take the symbol "N" (for "Negative", the sign of the resulting product). Furthermore, I wanted to preserve the SU(3)_c structure, rather than try to pass it off as resulting from a magnetic quantum number set (1, 2, and -3) on a linear scale (as I have seen done in some more recent rishon theory papers, and found to be insufficient in my mind as well). I also discarded the idea that different charges arise from different orderings, but held to the idea that different orderings change the handedness of the composite fermions, which I can explain briefly if need be. And I also assumed that spin-3/2 states could arise, but figured that the breaking of chiral-symmetry may cause all spin-1/2 fermions to be left-handed, and all spin-3/2 candidates to be right-handed (thus requiring right-handed matter to be unable to interact with weak vector bosons, but still able to interact with scalar Higgs bosons).

    One more thing... note that since P-P is a strongly-repulsive grouping, this may explain why matter and antimatter, on the whole, appear to seperate on cosmological scales. But it still seems to need another mechanism to allow opposite electric charges to attract? Or perhaps that is built in as well, and becomes non-negligible or even dominant for composite state interactions?
     
    Last edited: Mar 29, 2008
  12. Mar 29, 2008 #11

    arivero

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    Generically, what I do not like of the model is that the pairs P-P are the carpet for a lot of nuissances to be to swap under, and then the model becomes very adhoc... it does not even predict the number of generations. I like to argue on this, because my sBootstrap predicts it :rolleyes:, but also because the main question of this thread was to produce the 96.

    Hmm? This is automagical, exchange of odd spin does the trick.
     
  13. Mar 29, 2008 #12
    Well, I would like to think that the isospin-breaking is, in essense, responsible for the three-generation structure; the three generations emerge because there is no way to break down a three-generation structure into a discrete number of first-generation objects unless you go to a fourth-generation state in the transition, and the fourth-generation structure is so massive that the availability of a state in which this object becomes less energetic by decaying spontaneously into three first-generation objects requires it to never be "seen" in nature. So, to say that the model does not adequately explain the three-generation structure is, I think, a slightly naive view. In the universe we live in, the only structure we currently observe limits us to three-generations; this does not, in my mind, prevent the universe from behaving differently if symmetry breaking was not spontaneous.

    I do like one idea of mine alot; that is, not only is the broken electroweak symmetry spontaneously broken, but why not all symmetries that are broken? Why not consider all symmetry-breaking to be spontaneous? After all, what other reason would matter, unless the cause was spontaneous?

    Oh, yeah... true. That happens on its own, without any further explanation needed. Thanks!
     
  14. Mar 29, 2008 #13
    I should elaborate on the three-generation thingy...

    okay, so I am going to do the exercise here that I did on my own earlier in greater detail and effort.

    If you have an object that is composed of five rishons, the only way it can decay is to separate into a three-rishon fermion and a rishon-antirishon boson. Such bosons are quite massive, and so it is unlikely that a five-rishon state will readily decay into a three-rishon state without being highly energetic (i.e. the process must proceed via virtual boson, and hence has a very small probability of occuring). Similarly for the seven-rishon state, which can decay into either a three- or five-rishon state by emission of a four- or two-rishon boson, respectively. The bosons are automatically very massive, and the fermions nearly massless, but symmetry breaking supplies us with fermion masses to work with, and also with a set of massless bosons (i.e. the photon and gluons) among the other naturally massive ones. A nine-rishon state, however, has the necessary ingredients (6 rishons and 3 antirishons, or 6 antirishons and 3 rishons) to "fall apart" into three three-rishon fermions, without any mediation by massive bosons. Hence, the moment a seven-rishon state gains an N-N pair, it will be inclined to instantly decay into three three-rishon states, probably even long before the N-N pair can flip into a P-P pair. And hence, the structure of the model results in a three-generation picture of our universe, as long as the generations fill a mass-heirarchy with large gaps.

    1st gen base:

    PPP --> e+, -P-P-P --> e-
    PP-N --> u, -P-PN --> -u
    P-N-N --> -d, -PNN --> d
    -N-N-N --> nu_e, NNN --> -nu_e

    Now the first- and second-generations, when you treat N-N as negligible in mass due to P/N mixing, also includes states with extra N-N pairs;

    1st gen total:

    PPP + PPPN-N + PPPN-NN-N --> e+, -P-P-P + -P-P-PN-N + -P-P-PN-NN-N --> e-
    PP-N + PP-NN-N + PP-NN-NN-N --> u, -P-PN + -P-PNN-N + -P-PNN-NN-N --> -u
    P-N-N + P-N-NN-N + P-N-NN-NN-N --> -d, -PNN + -PNNN-N + -PNNN-NN-N --> d
    -N-N-N + -N-N-NN-N + -N-N-NN-NN-N --> nu_e, NNN + NNNN-N + NNNN-NN-N --> -nu_e

    2nd gen total:

    PPPP-P + PPPP-PN-N --> mu+, -P-P-PP-P + -P-P-PP-PN-N --> mu-
    PP-NP-P + PP-NP-PN-N --> c, -P-PNP-P + -P-PNP-PN-N --> -c
    P-N-NP-P + P-N-NP-PN-N --> -s, -PNNP-P + -PNNP-PN-N --> s
    -N-N-NP-P + -N-N-NP-PN-N --> nu_mu, NNNP-P + NNNP-PN-N --> -nu_mu

    3rd gen total:

    PPPP-PP-P --> tau+, -P-P-PP-PP-P --> tau-
    PP-NP-PP-P --> t, -P-PNP-PP-P --> -t
    P-N-NP-PP-P --> -b, -PNNP-PP-P --> b
    -N-N-NP-PP-P --> nu_tau, NNNP-PP-P --> -nu_tau

    sample fourth-generation-like combos that could possibly "instantly" fall apart:

    PPPN-NN-NN-N --> PPP + NNN + -N-N-N = e+ -nu_e nu_e (e+ --> e+ -nu_e nu_e)
    PPPP-PN-NN-N --> PPP + NN-P + P-N-N = e+ d -d (mu+ --> e+ d -d)
    PPPP-PP-PN-N --> PPP + N-P-P + PP-N = e+ -u u (tau+ --> e+ -u u)
    PP-NN-NN-NN-N --> PP-N + NNN + -N-N-N = u -nu_e nu_e (u --> u -nu_e nu_e)
    PP-NP-PN-NN-N --> PP-N + NN-P + P-N-N = u d -d (c --> u d -d)
    OR --> PPP + NN-P + -N-N-N = e+ d nu_e (c --> e+ d nu_e)
    PP-NP-PP-PN-N --> PP-N + N-P-P + PP-N = u -u u (t --> u -u u)
    OR --> PPP + N-P-P + P-N-N = e+ -u -d (t --> e+ -u -d, not entirely likely)
    P-N-NN-NN-NN-N --> P-N-N + NNN + -N-N-N = -d -nu_e nu_e (-d --> -d -nu_e nu_e)
    P-N-NP-PN-NN-N --> P-N-N + NN-P + P-N-N = -d d -d (-s --> -d d -d)
    OR --> PP-N + NN-P + -N-N-N = u -d nu_e (-s --> u d nu_e, not entirely likely)
    P-N-NP-PP-PN-N --> P-N-N + N-P-P + PP-N = -d -u u (-b --> -d -u u)
    OR --> PPP + N-P-P + -N-N-N = e+ -u nu_e (-b --> e+ -u nu_e)
    -N-N-NN-NN-NN-N --> -N-N-N + NNN + -N-N-N = nu_e -nu_e nu_e (nu_e --> nu_e -nu_e nu_e)

    Does this help clarify the idea?
     
  15. Mar 30, 2008 #14

    arivero

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    Indeed! Now I follow; you are telling that extra generations are build from pairs preon/antipreon and then any state having three or more such pairs can decay into three "first generation" preons.

    You need a truncation or some dinamics forbidding extra content. Your current proposal is that only pairs (P,-P) change the generation number, while pairs (N,-N) and (P,-N) are irregular in some way.

    What I like is that there is some smell of 2^5=32 components in the whole setup, for instance if you only look to "P" preons.

    Have we discused this already, in the preon thread? After all, it is a variant of the Fermion Cube, preached recently by Carl B, time ago by Mark William Hopkins, early 1980 by Adler, and late 1979 by Glashow, perhaps inspired or related to some drawings on SU(3)xSU(2)xU(1) for articles in the Scientific American in the seventies.
     
    Last edited: Mar 30, 2008
  16. Mar 31, 2008 #15

    arivero

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    To summarize:

    we have two nice arrangements the standard model into exactly 8 x 4 x 3:

    1) to fill the 8 with electroweak symmetry and to use colour a la Pati to create four of these octets: r,g,b and l(eptonic).

    2) to fill the 8 with the fermion cube (or half SO(10)) and to use the CPT discrete symmetries to produce the four copies (left/right, particle/antiparticle)

    more ideas?
     
  17. Mar 31, 2008 #16
    Actually, the beauty of it is that there does not need to be any extra interaction beyond a strong (P, -P) repulsion, since the wide difference in energy between a fourth-generation-like object and three much lighter first-generation objects is clearly overwhelmingly favoring the three-body fall-apart without needing to be mediated by massive vector bosons (especially if you follow the existing spectroscopy, since a fourth-gen b' quark would be yet heavier than a t quark).

    Yes, and thats the idea of the isospin breaking! The 45 degree mixing of P and N breaks the pure SO(10) into the 2^5=32 components since P gains all of the mass from N, and N falls to near masslesness.

    And yes, I believe I did leave some partial tidbits in the preon thread before...
     
  18. Mar 31, 2008 #17
    And hence the prediction of various mixings...

    You see, in this model, the up and down quarks, as well as the elctron and its neutrino (basically, anything first-generation), have three-rishon, five-rishon, and seven-rishon components. The Second generation objects have a five-rishon and a seven-rishon component. The third-generation objects have only the seven-rishon component. This is because the extra N-N pairs contribute almost no mass, and so would fall to the same mass as the structure without the extra N-N. If these states can be treated as acting together to form each generation, then we get some very interesting predictions for how the states mix based on 3-rishon and 2-rishon transitions (1-rishon transitions are not allowed because they would form an unallowed state). For example;

    PPPN-N (five-rishon positron) --> PPPP-P (five-rishon muon) is a 2-rishon transition, and would best be attributed to the weak neutral current (which will inhibit the process due to the large mass of a virtual Z boson).

    -PNNN-NN-N (seven-rishon down-quark) --> PP-NP-PN-N (seven-rishon charm-quark) is a 5-rishon transition, and would best be attributed to the weak charged current (once again inhibited by the large mass of a virtual W boson).

    These examples illustrate the variety of interactions that must be accounted for to produce the theoretical mixings that can occur in this model. If one takes all of the transitions that occur into account, then a matrix of standard-model mixings can be tested to see if the model works based on the developing experimental results. Interestingly, this approach predicts the smallest mixings will be between the charm and strange quarks on the quark side, and for the leptons, the muon and muon-neutrino. This would seem to agree well with the smaller CP violation in D-decays, whereas the K and B-decays show much more significant CP-violation in general.

    Since the probability of a rishon being either P or N at any given time is then 50%, or 1/2, we can calculate probabilities for certain transitions by taking 1/2 to a power equal to the number of rishons transitioning in the interaction, and these probabilities can be summed as you would in a Feynman diagram. The contribution of any one component can be guaged as the calculated probability of that component divided by the some of all components in the same interaction. Hence, an example;

    transition u --> s W+

    PP-N is not included
    PP-NN-N --> NN-PP-P (5-rishon, 1/2^5=1/32)
    PP-NN-NN-N --> NN-PP-PN-N (5-rishon, 1/2^5=1/32)

    1/32+1/32 = 1/16, now we want to know how much of the reaction falls from five-rishon states only; (1/32)/(1/16) = 1/2. The mediating W+ boson is accounted for in the phase-space, and its large mass will make this reaction more rare than it would be otherwise.

    OR

    transition t --> u Z

    PP-N is not included
    PP-NN-N is not included
    PP-NP-PP-P --> PP-NN-NN-N (4-rishon, 1/2^4=1/16)

    The entire reaction is mediated by seven-rishon objects only. The mediating Z boson is produced as a real particle, since the t-quark is massive enough to produce it, so the process occurs easily in the available phase-space.

    OR

    transition b --> anything

    NN-P is not included
    NN-PN-N is not included
    NN-PP-PP-P --> NN-PP-PN-N (2-rishon, 1/2^2=1/4; b --> s Z suppressed by Z mass)
    NN-PP-PP-P --> NN-PN-NN-N (4-rishon, 1/2^4=1/16; b --> d Z suppressed by Z mass)
    NN-PP-PP-P --> PP-NP-PP-P (3-rishon, 1/2^3=1/8; b --> t W- suppressed by W mass)
    NN-PP-PP-P --> PP-NP-PN-N (5-rishon, 1/2^5=1/32; b --> c W- suppressed by W mass)
    NN-PP-PP-P --> PP-NN-NN-N (7-rishon, 1/2^7=1/128; b --> u W- suppressed by W mass)
    NN-PP-PP-P --> NN-PP-PP-P (7-rishon, 1/2^7=1/128; b --> self, unsuppressed)

    1/4 + 1/16 + 1/8 + 1/32 + 1/128 + 1/128 = 31/64, selecting b --> c Z only we find that, without suppression, the transition would go to a charm quark with probability (1/32)/(31/64) = 2/31, or about 7% of the time. Since the W- mediates, the actual figure will be much less than 7% due to suppression.

    I tend to think that the actual mass-squared mixing matrix elements could be calculated as 2*P_12*P_21*M_1*M_2, where P_12 is the probability of fermion 1 transitioning to fermion 2, P_21 is the probability of fermion 2 transitioning into fermion 1, M_1 is the mass of fermion 1, and M_2 is the mass of fermion 2. This falls from the 2x2 matrix where M_1^2 is at the upper left, M_2^2 at the lower right, and the mixing elements in the upper right and lower left. From there, the commonly-named "mixing angle" can be calculated through invoking unitarity and Gell-Mann's usual relations.
     
  19. Mar 31, 2008 #18
    Well, I should note this is common practice for calculating the mixings of mesons and baryons... I guess it would be much closer related to the Cabbibo-Kobayashi-Maskawa matrix, since that after all does deal with quarks as individual entities. I suppose one could use the above idea to obtain theoretical values for V_ud, V_cs, V_tb, and so on.
     
  20. Mar 31, 2008 #19
    One more thought... if the bag model fails so badly because chiral symmetry has to be broken arbitrarily at the edge of the bag, then perhaps the modified rishon theory could solve this problem by the very constituents breaking the chiral symmetry, and thus the bag has nothing to do with the symmetry-breaking? Just a thought...

    I like the fermion cube filling the 8 degrees of freedom, because it just seems to be so convenient. Also, the symmetries are very helpful, whether exact or broken. I find that I tend to like solutions where I get to use spontaneous symmetry breaking and guage principles (such as with Weinberg-Salam and Glashow in the 70's), rather than resorting to reductionism. Whether this is ultimately the solution or not, only time will tell I suppose.
     
  21. Apr 1, 2008 #20

    arivero

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    Indeed it is... Really it seems there are only these two possibilities: either SU(3) (or Pati Salam?) acts in each 8-plet and SU(2) jumps across the CPT copies, or SU(2) and CPT act in the 8-plet and SU(3)PatiSalam acts across the copies. But while the former allows for a nice build of the 8-plet, the later is ugly.

    This fact, that SU(3) and SU(2) get to be "orthogonal" between them, has been noticed here and there, and an unifiyed SU(3)xSU(2) needs to use CPT symmetries to separate.

    About preons in general: it is true they can be used to generate all of these objects and predictions, but they do not offer a way to fix a given model except via experiment. In some sense, all D-brane stuff is preonic, the branes being Chan-Paton labels. Furthermore, a preon theory does not offer a way to avoid going even deeper; because of it I devised the sBootstrap, to get the good things from preons but still offer a clausure.
     
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