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S|T Combos as Preons

Posted on Monday, March 26, 2007 at 05:56AM by Registered CommenterDoug | CommentsPost a Comment

In the previous post, I introduced the S|T combos as preons, the LST community’s historical name for theoretical entities, composing fermions of the standard model (SM).  Sundance Bilson-Thompson has recently extended the concept to include the bosons of the SM, as well, publishing a paper on it last Summer. Lee Smolin happened to read the paper and realized that it was just what he and his colleague, Fotini Markopoulou, were seeking in order to explain the braid-like spacetime structures in loop quantum gravity (see story in New Scientist).

Smolin was so excited with these developments that he’s quoted as saying, “I’ve been jumping up and down about these ideas,” which is also an apt description of my reaction when I read about it, because, just as Smolin had been looking for something to explain his spacetime braids, I had been looking for some way to combine S|T combos.  For months, the only way I could see that they could be combined did not produce anything interesting.  Combining four S|T units into a square, or a cube, did not get us very far, because a square has as many vertices as it does sides, and the equation for the net inward or outward motion didn’t change as a result.  I just couldn’t see how to get started.

However, when I saw the braided triplets of Bilson-Thompson’s preon-inspired concept of helons, I realized I could use groups of three S|T units, instead of groups of four, and follow his lead, as I explained in the previous post below, in modeling the SM with S|T triplets.  Nevertheless, I knew that this would only give us a start, because the braided helons can’t explain the origin, or nature, of spin, mass, or Cabibbo mixing, whatever that is, and the S|T combos have to do this, or else we have nothing worth pursuing. 

Fortunately, it appears that the S|T combos (obviously, all this is very tentative, of course) should go far beyond the braids of Smolin and company, in explaining the origin and nature of spin, mass, and, who knows, maybe even Cabibbo mixing, whatever that is!  Here’s how it might work out for mass. If we denote each of the three vertices of the S|T units forming the triplet, as A, B, C, and the red (SUDR ) end of a given S|T, as “r,” the  blue (TUDR) end as “b,” and the middle term as “i” for “inward,” then we easily see that
  1. A = ar + cb
  2. B = ab + br
  3. C = bb + cr

where a, b, and c denote the three S|Ts in a given triplet.  Figure 1 below illustrates the scheme:

LabeledTriplet.png

Figure 1.  Labelling the S|T Triplet 

Recall that the minimum value for these magnitudes is one unit.  For the blue units this is one TUDR, or ds|dt = 2|1, and, for the red units, it is one SUDR, or ds|dt = 1|2.  Each S|T unit constitutes an arbitrary combination of SUDR and TUDR units, possibly corresponding to different frequencies of bosons.  If the number of SUDRs and TUDRs in the S|T combo are equal, then the middle term is balanced and colored green, but, if they are unequal, then the middle term is colored red or blue, depending upon the direction of the imbalance.  Hence, we see that the possible values for the properties of a given triplet are determined by its quantum degrees of freedom, which depend on the inherent displacement values of its nodes, rather than on any spatial orientation.  In other words, we are attempting to use the fundamental degrees of freedom, used to describe space and time magnitudes in our RST theory, to give rise to matter, with properties, as described in the LST theory of the standard model (SM).

We can list a few of the initial displacement values in a table, as shown below.

STValueTable.png

Table 1. First Five Possible SUDR|TUDR Combinations

Using table 1 above, we can select values corresponding to the red and blue nodes of the S|T units in a given triplet, and then calculate its color.  For example, for the triplet in figure 1 above, we can pick the following values, which yield an all red combination:

  1. A = ar + cb = 4|8 + 6|3 = 10|11
  2. B = ab + br = 4|2 + 5|10 = 9|12
  3. C = bb + cr = 2|1 + 2|4 = 4|5

Since the summed values of the nodes, constituting the A, B, and C vertices of the triplet, are all less than unity, we must color them red, as shown in figure 2 below:

 

LabeledRedTriplet.png 

Figure 2.  A Red S|T Triplet

Then, the space|time value of the triplet is the sum of the values of its vertices:

T = A + B + C = [(10|11) + (9|12) + (4|5)] = 23|28,

which is the same displacement value we get if we calculate the middle value of each S|T and then sum them

  1. ai = ar + ab =  4|8 + 4|2 = 8|10
  2. bi = br + bb = 5|10 + 2|1 = 7|11
  3. ci = cr + cb = 2|4 + 6|3 = 8|7
  4. T = ai + bi + ci = [(8|10) + (7|11) + (8|7)] = 23|28,

where, again, the subscript “i” indicates the inward magnitude of the middle term of each S|T in the triplet. In other words, this all-red triplet has a net inward space|time displacement, ds|dt = 23|28, which we are want to tentatively identify with five, natural, units of mass, but who knows.  The problem then becomes, what is charge? One possibility is that unit charge is due to the phase difference between the nodes of opposite color; that is, an electron’s charge differs from a positron’s charge, because, when its expanding, the positron is contracting, but this is going to take some work, given that there are three possible red and three possible blue nodes in each triplet.  It seems to require that all the electron’s red nodes have to be in sync, while all the positron’s blue nodes also have to be in sync, but, at the same time, opposite in phase relative to the electron. 

The question is, what happens when one of the nodes of a triplet is “the odd man out” color wise, but either in or out of phase with either its companion nodes, or the anti nodes of an antiparticle? Things really get complicated in a hurry, with just a few degrees of freedom.

Of course, we realize that we are very fortunate indeed to even have candidates for all these properties.  I think the LST community would be envious, if they knew what we have. I’m sure Lee Smolin would be jumping for joy.

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