- #31
mormonator_rm
- 184
- 1
arivero said:
Okay. This all looks great here, and the p=2, q=3 model does appear to reproduce the SM, but how do you actually create a physical limit that will stop it at p=2 and q=3? To show an example, I am going to try to build the fermions from your model here, with the extra constraint that you cannot have more than one heavy quark type involved in a combination; I am also going to use terminology where the highest generation involved is the "p" or the "q". So we can build anything from anything that fits the pattern...
(p=1, q=0) : (uuS) = exotic, (u-uS) = nu_e ?
(p=1, q=1) : (duS) = -d, (d-uS) = e-, (d-dS) = nu_e ?, (-d-dS) = u
(p=1, q=2) : (sdS) = -c, (s-dS) = nu_mu, (s-uS) = mu-, (-s-uS) = s
(p=2, q=1) : (cdS) = -b, (c-dS) = tau+, (c-uS) = nu_tau, (-c-uS) = exotic
(p=1, q=3) : (bdS) = -t, (b-dS) = nu_tau', (b-uS) = tau'-, (-b-uS) = b'
(p=1, q=4) : (b'uS) = -b'', (b'-uS) = tau''-, (b'-dS) = nu_tau'', (-b'-dS) = t'
(p=3, q=1) : (tuS) = exotic, (t-uS) = nu_tau''', (t-dS) = tau'''+, (-t-dS) = b'''
(p=4, q=1) : (t'uS) = exotic, (t'-uS) = nu_tau'''', (t'-dS) = tau''''+, (-t'-dS) = b''''
(p=1, q=5) : (b''uS) = -b''''', (b''-uS) = tau'''''-, (b''-dS) = nu_tau''''', (-b''-dS) = t''
Need I continue further? At first glance, from what mass order seems to emerge so far, it seems it might follow a well known pattern;
p, q
1, 0
1, 1
1, 2
2, 1
1, 3
1, 4
3, 1
4, 1
1, 5
1, 6
1, 7
5, 1
6, 1
7, 1
1, 8
1, 9
1, 10
1, 11
8, 1
9, 1
10, 1
11, 1
12, 1
...
where the "p" types are predecessors of the next type-step in the familiar groupings 1,1,2,3,5,8,13,... and so on (Fibinacci Sequence (sp?)) with "q" types as predecessors in groupings of (Fibinacci + 1, i.e. 2,2,3,4,6,9,14,...) between every "p" type grouping. Thus, if allowed to go into infinite generations, the "p" type quarks will at first become much more massive than "q" type quarks of the same generation, and then suddenly drop below the "q" type quark masses of similar generation as the generations go higher. However, the pattern above suggests that quarks would be arranged, by ascending mass, in this kind of ordering;
u, d, s, c, b, b', t, t', b'', b''', b'''', t'', b''''', b'''''', t''', b''''''', t'''', b(8), b(9), t''''', b(10), t'''''', b(11), t''''''', b(12), b(13), t(8), b(14), b(15), t(9), b(16), t(10), b(17), t(11), b(18), b(19), t(12), b(20), b(21), b...
So, the whole Fibinacci idea kind of breaks down, and then I have very little hope of finding another pattern worth looking for. But to avoid this whole connundrum, it would be very wise to find a way to physically limit the acceptable fermions primarily to the second, third and fourth lines, which could give the three generations. If you impose the limits, somehow, of q > 0 and p + q < 4. This would produce just rows two through four minus the nu_tau entry, and adding the t-quark and nu_tau' from row five; it also eliminates both of the plausible exotics by default, and gives some structure;
0,1 1,1 2,1
0,2 1,2
0,3
The remaining nu_tau' could fill the vacancy left by the elimination of nu_tau in the row above, and would then, essentially, be nu_tau. The new limited list becomes;
(p=0, q=1) : (d-dS) = nu_e, (-d-dS) = u
(p=1, q=1) : (udS) = -d, (u-dS) = e+
(p=0, q=2) : (u-sS) = mu+, (-u-sS) = s
(p=1, q=2) : (dsS) = -c, (d-sS) = nu_mu
(p=2, q=1) : (dcS) = -b, (d-cS) = tau-
(p=0, q=3) : (dbS) = -t, (d-bS) = nu_tau
This gives us all of the SM fermions, with superstrings incorporated. The arrangement requires "q" to not be equal to zero at any time, so that would actually seem to suggest that the down quark is the logical choice for a fundamental particle when compared with the up quark. Hence, you have now reduced everything to composites of d and S, with d being the only justifiable fundamental quark. Both d and S could be thought of as fermions, though, but S has no electric or color charges...
I didn't think I would get such an interesting answer when I started writing this post, it is a little surprising to me, to be honest... I thought the idea would show a serious flaw of some sort if I worked it out. But it still leaves you with the question of how to physically justify those quirky constraints q > 0 and p + q < 4...
You know, I just thought of something, and you're going to hate me for this...
Doesn't this almost equate to my P/N modified rishon model? I mean, with S being completely neutral, it is close to being like the N rishon, and the down quark is complimentary to S the same way as the P rishon is to the N rishon... they could be treated as two slightly-different approaches to obtaining the physical electric charges for composite fermions. Where P and N have certain different features in hypercharge and isospin, it seems that d and S are the fundamental quark and electron-neutrino (from (d-dS) --> S), having hypercharge and isospin units that are normal to the current SM fermions! In this context, (uS) is nothing but a CPT operator for d, turning d to -d, with the idea that the electron-neutrino is a purely supersymmetric particle! Yet, it seems like our two different models are nothing more than two different languages for expressing hypercharge and isospin, with some differences in motivation, to get the same properties that we know the SM fermions to already have. I will give you this; my modified rishon model has rishons with unusual units of isospin, which leave them with sqrt(5)/sqrt(13) of the normal sqrt(I_z^2 + Y^2/2) vector as compared to the natural quarks and leptons in your sBootstrap. But my modified rishon model has 1) a physical motivation for limiting to three generations, as well as 2) an explanation of the structure of vector bosons, while this attempt I made to clarify sBootstrap to myself seems to have none of these features...
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