a reply to: micpsi
Hm, how does Salam's model account for electrons with charge -9/9 e if it only uses a preon with charge +5/9 and a preon of -4/9? The only answer
possible is that in addition to a X preon and a Y preon, Salam's model also has an anti-X preon (charge -5/9) and anti-Y preon (charge +4/9), in which
case the electron could be said to be composed of e = Y - anti-X, but now Salam's model would be employing four
species of preons (two preons +
And actually there is a mathematical proof which proves how the Singular Primordial Preon Theory is the most simplest model possible. It reduces the
Standard Model to a single preon (and its antipreon).
There are seven species of charge observed in the Standard Model: +e, +2e/3, +e/3, 0, -e/3, -2e/3, and -e.
Let N be the amount of preons in a set of items, and S be the number of item (preon) species. In a set where S=2 (1 preon + 1 antipreon), the number
of non-repeating permutations "P" can be computed by,
P = N+1
Note that in all instances where N is inferior to 6, the resulting amount of permutations will be inferior to the seven observed.
(N < 6)+1 = P < 7
Furthermore, the individual charge (e/N) of the species of preons will form permutation sums which are incompatible with observed particle charges.
N=1: [+e, -e] (incompatible)
N=2: [+e, 0, -e] (incompatible)
N=3: [+e, +e/3, -e/3, -e] (incompatible)
N=4: [+e, +e/2, 0, -e/2, -e] (incompatible)
N=5: [+e, +3e/5, +e/5, -e/5, -3e/5, -e] (incompatible)
N=6 is the smallet set which will yield seven permutations, and in whose permutations total charges match exactly observation:
N=6: [+e, +2e/3, +e/3, 0, -e/3, -2e/3, -e] (perfect match)
It can also be proven that more than two species of preons is useless. Permutations for sets with S=3 number of item species can be found by:
P = (N+1(N+2))/2
Note that as N is explored, the resulting number of permutations fails to reach seven.
N=1: (1+1(1+2))/2 = 3
N=2: (2+1(2+2))/2 = 6
N=3: (3+1(3+1))/2 = 10
Similarly, permutations for sets with S=4 is found by:
P = (N+1(N+2(N+3)))/6
Once again, as we browse N, we find that the system fails to yield exactly seven permutations:
N=1: (1+1(1+2(1+3)))/6 = 4
N=2: (2+1(2+2(2+3)))/6 = 10
This goes on until S itself reaches 7 - but then, seven species of preons would defeat the whole point of reductionism in the first place.