Ok, in the last entry, I explained how, like Hamilton’s quaternion solution, which enabled him to go from 2D geometric rotation to 3D geometric rotation, by exploiting four mathematical dimensions, rather than the three he thought it would take, our mathematical solution must be based on the number four, rather than the number 3, in order to go from the triplets of the preons to the quads of the periodic table of elements (4(12 )= 22= 4 is the initial magnitude of the Wheel of Motion version of the periodic table).
The reason this appears to be the case is that we must use the radius of the balls in Larson’s cube, rather than the cubes (i.e. the continuous numbers rather than the discontinuous). The preons use the cubes, since they are simply counting the relative number of inverse oscillations that are combined together in the preons.
However, just as our numbers on the new number line are not equal, given the quotient interpretation, even though they are equal, given the difference interpretation, so too our 3D oscillations (SUDRs & TUDRs) can represent equal, but opposite, space/time and time/space displacements from unity (i.e. equal up and down displacements from no displacements), even though they are not equal in magnitude (speed). The magnitude of the upper one (blue) is four times the magnitude of the lower one (red).
This means we must count the total number of oscillations, red and blue, in a given preon, and then apply the quotient interpretation of the numbers to get the relative magnitudes, For instance, the up quark has two blue and a green in the midpoint of its three S|T units. Since the color green of the midpoint indicates a balance of red and blue, and the color blue of the midpoint indicates an imbalance of one more blue than red, we have a total five blues and three reds in the up quark (we have to count the two, opposing, units that balance the initial S|T unit before it is unbalanced in one or the other “direction.”)
On the other hand, the down quark has three green and one red, so it has a total of three blues and four reds. Now the proton consists of two up quarks and one down quark, while the neutron consists of one up quark and two down quarks. Hence, the red:blue ratio for the proton is 10:13, while for the neutron it’s 11:11. But when we add the 6 red and 3 blue of the electron to the total of the proton, its ratio is changed to 16/16.
Adding up the total blue:red magnitudes for the proton, we get (32/1):(16/2) = 64/16 = 4, while for the neutron we get (22/1):(11/2) = 44/11 = 4. This is a surprising, but welcome result, since it again conforms with the LST’s standard model of particle physics: The proton and neutron only differ by the positive charge of the proton and a slight difference in mass. The trouble is, the difference in mass is not in accord with the standard model. Whatever mass might turn out to be in our RST theory, it seems unlikely that the higher numbers of the proton would make it slightly lighter than the neutron, rather than the other way around.
Nevertheless, the good news is that, when we get to the atomic level, our triplets of S|T preon units turn out to be quads of S|T atomic units, which are ready to be compounded into the periods of the Wheel of Motion.
Now, the next step is to treat the 4 of the proton and the 4 of the neutron as inverse numbers, in the sense that the patterns of their constituent preons are inverses. I’ll explain how that works in the next post.