Meeting a Terrific Challenge

Discussion of Larson Research Center work.

Moderator: dbundy

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Joined: Mon Dec 17, 2012 9:14 pm

Re: Meeting a Terrific Challenge

Post by dbundy » Thu Feb 01, 2018 3:06 pm

At the end of the previous entry, I wrote that we would introduce a new direction in the mathematical approach of our RSt. I've already written about a new scalar math and introduced the idea of three units of numbers or three number systems based on sqrt of 1, 2 and 3. The genesis of this idea and others that follow stems from the 3d stack of 8, 1-unit cubes, or the 2x2x2 stack of cubes we call Larson's cube (LC).

The fact that the 3d geometric figure of the LC perfectly corresponds to the numbers of the binomial expansion of the tetraktys reveals a deep and fundamental connection, or union, of numbers and geometry. It's hard to overestimate the significance of this discovery, as the pursuit of such a union has driven the historical development of traditional mathematics, from ancient times. I've written much about this on my website at

Here, though, I want to show how the math of the LC, in the form of the binomial expansion of the Tetraktys, should permit us to calculate the values of the LRC's RSt and thus the 1d, 2d and 3d properties of the S|T combos, which should, if valid, yield the magnitudes of the observed properties of the physical entities, which correspond to these various combos.

The key is understanding the expansion, or progression of the scale, or size, of the oscillating LC, since such an expansion corresponds to the sum of all its multi-dimensional scalar motion. To understand this, we must recognize that each physical dimension has two "directions." In the case of the unit LC, we can see that multiplying each dimension by the corresponding coefficient of that dimension, yields the correct number of dimensional units for a given dimension:

Recall that there is 1 0d point, 3 1d lines, 3 2d planes and 1 3d volume in the LC, and that these numbers are identical to the 1 3 3 1 sequence of numbers of the 4th line of the binomial expansion of the Tetraktys (BET). So, we can write this as:

10 x 1 point, 11 x 3 lines, 12 x 3 planes and 13 x 1 volume, or

10 (1mp) + 11 (3dp) + 12 (3qp) + 13 (1op) = 1(1) + 1(3x2) + 1(3x4) + 1(1x8) = 1+6+12+8 = 27 poles, where mp = monopole, dp = dipole, qp = quadrupole and op = octopole.

This is the multi-pole sum of the unit LC. So, to sum multiple units of the LC, we simply sum the coefficients of each dimension times the number of poles in that dimension:
n0(1mp) + n1(3dp) + n2(3qp) + n3(1op), or

n0(1) + n1(6) + n2(12) + n3(8).

On this basis, the progression of the LC, in terms of poles is, 27, 125, 343, 729, or

33, 53, 73, …(2n+1)3, n = 1, 2, 3, …∞

With this much understood, we can proceed to calculate the multi-dimensional magnitudes of the S|T combos. We have long since shown that summing the 1d numbers of the fundamental S|T combo, 1/2 + 1/1 + 2/1 = 4|4 num into the numbers of the triple combo, calculates the neutrino number, enabling us to calculate all the combos corresponding to the quarks and leptons of the first family of the LST's standard model of particle physics. For example, in the material sector:

Electron = 3(6|6) = 18|18;
Anti-Up = (2(6|6) + 4|4 = 12|12 + 4|4 = 16|16;
Down = ((2(4|4) +6|6 = 8|8 + 6|6 = 14|14;
Neutrino = 3(4|4) = 12|12;
Anti-Down = ((2(4|4) +6|6 = 8|8 + 6|6 = 14|14;
Up = (2(6|6) + 4|4 = 12|12 + 4|4 = 16|16;
Positron = 3(6|6) = 18|18;

with the inverse of these corresponding to the cosmic sector of the RST universe.

Combining these into the proton, neutron and helium atom entities, all the 1d charges work out perfectly, corresponding to the electrical properties of the observed entities, even in the beta decay process. However, these 1d numbers are not suitable to calculate the 2d and 3d magnitudes of the combos. We need to be able to calculate with 2d and 3d numbers to do this.

Of course, the vector mathematics of the LST are totally unsuited for this purpose. We must have a new scalar mathematics to calculate the multi-dimensional scalar magnitudes of the S|T combos, and I have shown why this is so on the LRC website. It's quite a story, but it's necessary to understand it, if one is to comprehend what we mean, when we refer to multi-dimensional scalar numbers and magnitudes.

Once it's understood how the scalar motion of an oscillating volume can form the basis for a multi-dimensional scalar mathematics, capable of describing the multi-dimensional magnitudes of scalar motion in various combinations, just as an oscillating vector motion can be used as the basis for multi-dimensional vector mathematics, describing magnitudes of vector motion, a whole new world of possibilities beckons us forward.

Hopefully, it's a valid world that can be shown to correspond to magnitudes of observed properties of matter, but, of course, there's no guarantee. It may turn out to be an exercise in futility, but I think it's worthwhile to explore it and see where it leads.

More next time.

Posts: 84
Joined: Mon Dec 17, 2012 9:14 pm

Re: Meeting a Terrific Challenge

Post by dbundy » Wed Feb 14, 2018 11:25 am

The idea that multi-dimensional scalars exist isn't in question. They do, even though the idea may seem preposterous to LST mathematicians and physicists. However, that's only because they've yet to recognize scalar motion and the reciprocity of space and time, as shown by Dewey Larson.

Multi-dimensional scalar motion is clearly evident, when we consider an oscillating volume. Such a volume has 0, 1 and 2 dimensional components, as well as a three-dimensional volume and it's clear that they can all be identified mathematically as well as geometrically. The mathematics of the oscillating volume is defined by the multi-dimensional units of Larson's cube, the 2x2x2 stack of 1-unit cubes, in which the radius of the inner volume, which is just contained by the stack, is an integer magnitude and the radius of the outer volume, which just contains the stack, is a non-integer magnitude.

While this may not appear to be very useful at first, I think it will turn out to be very useful eventually, since the ratios between the radii, areas and volumes of the two balls are integral magnitudes. We will explore those magnitudes later, but first we need to understand that the way they arise is fundamental, and this bids us to ponder the theoretical and philosophical implications.

Recall that in the previous post, I showed the startling correspondence between the tetraktys and the LC. We saw how, when the binomial expansion of the tetraktys (BET) is regarded as the expansion of two "directions" in each of three physical dimensions, the result is the mathematical equivalence of the unit LC's geometry. That is, the numerical expansion of the fourth line of the tetraktys,


is equivalent to the geometric expansion of the LC, the 2x2x2 stack of unit cubes, since the LC can be considered to consists of

1 0d point, 3 1d lines, 3 2d planes and 1 3d volume.

As explained in my previous post, when we expand the LC in the two "directions" of each of these dimensions, we can write the expansion of the multiple dimensions in terms of a sum of multi-poles:

10(1mp) + 11(3dp) + 12(3qp) + 13(1op) = 1(1) + 1(3x2) + 1(3x4) + 1(1x8) = 1+6+12+8 = 27 poles,

or, more generally,

n0(1mp) + n1(3dp) + n2(3qp) + n3(1op) = (2n +1)3 poles, n = 1, 2, 3, ...∞. [1]

where mp = monopole, dp = dipole, qp = quadrupole and op = octopole.

Of course, nature doesn't expand by these integer numbers of cubes with their right angles, but by the non-integer geometric magnitudes of balls, with their respective radii, and, as my earlier analysis has shown, there are three of these found in the LC:

√1, √2, and √3.

Consequently, the expansion of the LC, in terms of the numbers of the tetraktys, can be thought of as the geometric expansion of the 1d, 2d and 3d non-integer magnitudes, as well as the 1d, 2d and 3d integer magnitudes of [1], recalling that numbers have, as magnitudes do, three properties: magnitude, dimension and "direction."

Since each dimension has two "directions" in which magnitude can exist, the binomial expansion of the tetraktys enables us to identify it with the LC, up to its three, non-zero, dimensions. Beyond that, however, the powers of two that we find in Pascal's triangle, while exhibiting some of the properties of the LC expansion, loses its dimensional correlation with the LC. It is the expansion of the 3d level of the triangle that constitutes what we might call a 3d scalar number, and we can expand its integer magnitudes as shown in [1] above, or its non-integer magnitudes, as I will show now.

To do this, we simply multiply our 3d number, which we will designate with an uppercase R, by n, where n = 1, 2, 3, ...∞:

1 x R = 1mp+3dp+3qp+1op = (1x20)+(3x21)+(3x22)+(1x23) = 1+6+12+8 = 27;
2 x R = 2(1+6+12+8) = 2+12+24+16 = 54;
3 x R = 3(1+6+12+8) = 3+18+36+24 = 81;
n x R = n(1+6+12+8) = n+6n+12n+8n = 27n.

The dimensions of the terms of the expanded number ascend left to right, from 0 to 3, so there are n 0-dimensional units, followed by 6n 1-dimensional, 12n 2-dimensional, and 8n 3-dimensional units in the number. The magnitude and dimension of these units, ud, are:

u1 = √1,
u2 = √2,
u3 = √3,

as appropriate for each non-zero dimension:

n x Rud = 1n+6nu1+12nu2+8nu3 = 27nud

It's important to understand that the subscript number of the units, while identical to the exponential number of respective poles in each term, is an actual magnitude of the corresponding unit, so that the 1d unit (√1) is multiplied by the 1d coefficient (6n), the 2d unit (√2), by the 2d coefficient (12n), and the 3d unit (√3), by the 3d coefficient (8n).

Presumably, on this basis, we can calculate the units of n-dimensional motion in a given reciprocal number (RN) = n(1/2+1/1+2/1) = n(4|4). The procedure we will use requires us to take the ratio of various geometric quantities, based on these 3d numbers, as I will discuss next time.

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Joined: Mon Dec 17, 2012 9:14 pm

Re: Meeting a Terrific Challenge

Post by dbundy » Wed Feb 28, 2018 11:57 am

The most enduring challenge that Western science has faced down through the centuries, since the days of the brilliant Greeks, like Pythagoras and Euclid, is the fact that there are geometric magnitudes for which there are no corresponding numbers. It is the same challenge facing the modern LST community today: The integer numbers of quantum physics are irreconcilable with the geometric magnitudes of general relativity, creating the same kind of impasse the ancients faced. Believe it or not, digital approximation to analog magnitudes is still incomplete, in spite of the spectacular sight of 4k video and other technological achievements of the 21st century.

Perhaps Kronecker was right, after all, even though the world has followed Dedekind cuts and Cantor sets into the realm of real numbers, notwithstanding the objections of philosophers. Larson assumes Kronecker was right: "God made the integers, all the rest is the work of man," as he has taken a revolutionary approach to answering the question of how our analog world emerges from God's integers, even though that was not his intention, per se.

As it turns out, though, the 3d scalar expansion of numbers can be in-separately connected to the 3d scalar expansion of geometry, by recognizing the identity of the tetraktys and LC, uniting the magnitudes of numbers, with those of geometry, if I'm not mistaken.

What makes this possible is the fact that the ratio of the irrational numbers in the expanding LC, such as π, cancel out in these ratios, leaving only integers, or the square roots of integers. Consequently, the expanding cubes of the LC, though not corresponding to the 2d areas and 3d volumes of physical expansion, or progression, directly, nevertheless determine the boundaries of these associated 2d areas and 3d volumes, and the ratio of these turns out to be integer magnitudes, or their square roots, in terms of a given 2d or 3d unit.

For example, the ratio of the unit 3d volume with the radius square root of 3, to the surface area of the 2d sphere, with the radius square root of 2, is 1, 2, 3 ...n, in units of √3. The ratio of the same volume to the circle area of the 2d volume (its cross section) is 2, 4, 6, ...2n, again in units of √3.

Given that the LC defines three volumes, areas and radii, there are lots of combinations to consider, and I will try to deal with these systematically, in due course, but first the issue of how these magnitudes combine has to be addressed.

Recall that in the previous post, it was shown how the numerical expansion of the LC stack of 8 one-unit cubes can be calculated using the numbers of the fourth line (dimension) of the tetraktys, as shown in the table below:


This expansion of the geometric stack via its correspondence to the tetraktys may seem non-intuitive at first glance, because, when we think of a stack of 8 cubes added to another stack of 8 cubes, we intuitively think we should have 8+8 = 16 cubes.

However, when we are adding 23 + n3, the sum is equal to (2+n)3, not 8+n3. In the case of 1 stack +1 stack, for instance, we have:

(2x2x2) + (2x2x2) = 23 + 23 = 43 = 64 cubes, not 8 + 8 = 16 cubes.

With this much understood, we can calculate the magnitudes of the volumes defined by the magnitude of a given stack and the respective radii determined by that stack. We just use the multi-dimensional number line shown in the table below:


The chart omits the 2d number line, but it shows the progression of all three radii, which we can use to calculate the magnitudes of the volumes and areas of each of the three balls of any given LC combo and their ratios, and this capability should enable us to calculate the magnitude of the 2d and 3d scalar motion associated with a given RN, n(1/2+1/1+2/1).

If this is the correct approach, it ought to enable us to calculate the 2d and 3d magnitudes of a given S|T combo, such as those found in the LRC's version of the LST's standard model of particle physics, and thereby hopefully match their observed magnetic and mass properties.

Posts: 84
Joined: Mon Dec 17, 2012 9:14 pm

Re: Meeting a Terrific Challenge

Post by dbundy » Thu Mar 15, 2018 6:57 pm

Ok, the next step in our challenge is to relate the expansion of our 3d number system (i.e. √1, √2, √3), based on the radii of the the three balls defined by the LC, to our reciprocal numbers that constitute the S|T unit combos of the LRC's version of the standard model (sm).

In order to meet the terrific challenge of determining why the mass of the proton (~1836 MeV/c2) is so much more than the mass of the electron (~ .511 MeV/c2), we anticipate having to understand the 3d properties of the S|T combos.

However, it will also behoove us to understand the other two families of particles, the second and third families of quarks and leptons found in the LST's standard model of particle physics, especially their leptons, since the mass of these particles varies from that of the electron, but their charges do not.

The mass of the muon lepton is ~ 106 MeV/c2, and the mass of the tau lepton is ~ 1776.82 MeV/c2, and an investigation of these mass differences from the electron mass avoids the complication of multiple quarks with different masses in combination.

I don't know why taking this course in the development didn't occur to me before, except I guess it seemed too daunting to contemplate, so I just didn't, but, as it turns out, the solution might just follow as a consequence of the fundamental postulates, like all the rest! And why should we be surprised?

Now, the reason I think this might be so is I suspect the mysterious limit to three families of quarks and leptons has to do with the limit of the universe to three dimensions, and if that is the case, then the question becomes one of how this connection is realized. The only way I've been able to connect them is through our 3d number system. That is to say, by changing the unit measure for each family to the radius of the corresponding ball of the LC.

On this basis, the unit of the first family would be the √1, that of the second family √2, and that of the third √3. The unit of the first family has worked out very well, as I have shown now for years, but only in terms of integers not square roots. Yet, we can express the RN's of our sm combos in terms of square roots by simply substituting the square root unit for that of the integer:

S|T = 1(√1)/2(√1) + 1(√1)/1(√1) + 2(√1)/1(√1) = 4(√1)|4(√1),

which gives us the neutrino of the first family (called the electron family), when it is multiplied by the three S|T units constituting a neutrino, or 3 x 4(√1)|4(√1) = 12(√1)|12(√1) num. From there, we can calculate the num value of each quark and lepton and their anti-particles, as shown below:


Where each balanced S|T unit (green) in a combo has a value of 4(√1)|4(√1) num, and each unbalanced S|T unit (red or blue) has a value of 6(√1)|6(√1) num. This works out to 14, 16 and 18 num in both "directions" from the neutrino's 12 num. Again, we have been amazed at how well these preon combos have worked out, in terms of 1d, or electrical properties, especially in terms of plus and minus beta decay, and the formation of protons and neutrons and the elements of the periodic table.

Yet, we have not been able to find the 2d (magnetic) or the 3d (mass) properties of these combinations, using the 1d RN, so the idea of identifying the three families with RNs employing the three different radii of the LC might be a good approach.

Since 12 is the square root of 144, we can use this value for the num of the electron neutrino, and this gives us √324|√324 (or 182) num for the electron & positron.

Accordingly, the neutrino of the second family's RN would be based on the √2 unit, giving it a value of 12(√2)|12(√2), or √288|√288 num. Thus, for the second family (called the muon family), the three green S|T units, constituting its neutrino, would have a value of √288/3, or √32:

υμ = √32 + √32 + √32 = √288

and unbalancing each subsequent combo by adding an additional 2d space or time unit to each of these in turn, gives us the sequence:

√72 + √32 + √32 = √392; √72 + √72 + √32 = √512; √72 + √72 + √72 = √648, or

-√648, -√512, -√392, √288, √392, √512, √648

Therefore, the 2d muons have a num of √648. Now, a μ- decays into a muon neutrino, √288 num, and a W- boson, so next we need the num of the 2d W- boson, which would be,

W- = 30(√2)|30(√2) num, or √1800|√1800 num,

given the 2d unit substitution in its calculation. Consequently, the difference between the num of the W bosons and the muons has to equal the num of the muon neutrinos:

√648 - √1800 = √288.

Like the 1d boson of beta decay, the 2d boson quickly decays into an electron and its anti-neutrino, but with a difference: Instead of a 1d electron and its 1d anti-neutrino, these would both have to be 2d combos, with 18(√2) num and 12(√2) num, respectively. I don't know what to make of that, yet.

As might be expected, the 3d case works out as well. The num of the neutrino of the third family (called the tau family), based on the 3d unit, √3, is 4(√3)|4(√3), or √432. Based on this value, the two tauons have a num of √972|√972 and the num of the 3d W bosons works out to √2700|√2700.

Hence the difference between the 3d tauons and the 3d bosons is √972 - √2700 = √432, which is the magnitude of the 3d tauon neutrino.

The decay of the tauons is almost identical to the decay of the muons, with the exception that the much heavier mass of the tauons allows their W bosons to decay into hadrons as well as leptons.

This must be yet another clue to the elusive 3d mass property of these combos. Hopefully, it'll help get us to the bottom of this mystery sooner than later.

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