It is a well known fact---especially among divers---that the gravitational pressure under the water increases at the rate of about 15 pounds per square inch (i.e. one atmosphere) for every 34 feet we descend into a body of water (tank, reservoir, lake, river, ocean, etc.). A pound is a measure of weight, mg, where g is the local acceleration due to gravity. Pressure, therefore, is force or weight (mg per area r2) that is to say: p = mg/r2.
But in the case of our gravity equations,
F = mg = mGM/r2
It's the pressure potential (M/r2 or mass per area) that constitutes the variable in each separate application, for the acceleration due to gravity must be supplied from the constant G. This constant G is measured as 6.6732 x 10-8 centimeters cubed per gram per second squared: cm3/gm-sec2. But if we analyze this structure, we find that it separates readily into two parts:
(1) The units (cm2/gm) which are to be divided into the variable M/r2 so as to measure them in the c.g.s. System;
(2) The constant acceleration due to the gravity of the universe as a whole---6.6732 x 10-8 cm/sec---a measure that should be constant during any particular century throughout the universe's surface as a whole, provided that the hypersphere is nowhere oblate and hence that all points along its surface are at an equal distance Ru from its origin-center.
For example, By factoring out the small mass m, the equation for the local acceleration due to gravity here on the earth becomes:
g = GMearth / r2earth
g = 6.6732 x 10-8 x 5.977 x 1027
(6.37747 x 108)2
Whence, g = 980.6646 cm/sec2, which agrees very well with the measure given in my dictionary of 980.665 cm/sec2.
But just what did we do to derive this answer? We had to supply the two
Mearth = 5.977 x 1027 grams and rearth = 6.37747 x 108 centimeters to obtain our local terrestrial pressure potential M/r2. But the constant universal acceleration, distributed through this pressure potential, 6.6732 x 10-8 cm/sec, had to be supplied from the universal constant of gravity G.
Now this is a very important point. In the universal case of any given point at the surface of the universe in relation to no other masses than the mass of the universe Mu itself regarded as being centered at the Origin-Center exactly one radius of the universe Ru away, we should not have to find the measures of these universal factors, for at the surface of the universe with its constant pressure these also should always be constant and therefore already figured into G.
Consequently, as long as we use the water-density centimeter-gram-second measuring system in which one cubic centimeter of water always weights one gram, we can affirm that the universal constant of gravity G itself should give us the acceleration due to the gravity of the universe as a whole as 6.6732 x 10-8 centimeters per second squared (the cm2/gm now being reduced to unity).
But we also know that the velocity of a satellite revolving around another body can always be found by equating the outward thrust of its centrifugal force with the inward pull of gravity towards the body:
Fc = mv2/r = Fg = mg
Hence, v2 = gr.
But suppose that we now wish to find the velocity of a photon traveling at the velocity of light c around the surface of the universe at the distance of one radius of the universe Ru from its origin center.
v -----------limit-----------> c
g -----------limit-----------> G
r ------------limit----------> Ru
Therefore, c2 = GRu.
But we have just discovered the formula for the radius of _the universe Ru !!!
Ru = c2/G
But if G = 6.6732 x 10-8 cm/sec2 and if c = 2.997,925 x 1010 cm/sec, then Ru = 1.346,81 x 1028 centimeters (or since 1 light-year = 9.46053 x 1017 cm.), then Ru = 14.2361 billion light-years !!!
[This does not imply that this is the total age of the universe, but
only the age of the universe after which optical light had been created. The Mayan
Calendar reveals that the age of the universe is 16.40082 billion years old which accuracy
is determined by the Mayan Teacher, Itsamna, a "skyperson".]
So the radius of the universe is some 14 billion light-years, which means, of course, that it must have been created some 14 billion years ago, for Ru = cTu. But just how do you know for sure that this is really true? We know it to be true heuristically because, as the sequel will show, it leads to one discovery right after another; and we know it pragmatically because all these discoveries in turn prove to be true. Let us now move on to formulate our first mass-space inverse particle formula.
Imagine a very small electronic particle (we find out later that it is the ergoelectron) whole energy E actually takes the form of a spherical surface (i.e. 4pr2) gravity field 4pG. The lines of electric force surrounding an electric charge q take the form of a spherical surface 4pq. So, similarly the lines of gravity force surrounding the universal acceleration due to gravity G also form a spherical surface gravity field 4pG. The size of the field cannot be determined unless we know the radius r, but whether the radius r is large and the field energy sparse or the radius r is small and the field energy much more concentrated, the energy remains the same, 4pG. But let's suppose that the energy of this 4pG field is the particle's rest energy E = mc2. Then the mass of this mysterious particle has to be 4pG/c2. But we now know that the radius of the universe Ru = c2/G. And so we arrive at our first inverse mass-space particle formula:
massergoelectron = 4p/Ru= 9.330,4681 x 10-28
gram = (9.109,558,6 x 10-28 g) (electron mass units), since the (+ mass of
electron) = 1.024,250,3 emu.
(Above paragraph referenced from Chapter IV, mark "FF")
Now, if we place this ergoelectron on the hypotenuse of a triangle such that we can place the magneton e/Ru on one leg and the electron on the other leg at right angles to it, we obtain our first space-time-energy
Rule of Squares: 4p/Ru =
(((16p2-e2)1/2/Ru)2 + (e/Ru)2)1/2
But now for the first time in history, we have a formula for the electron itself: (16p2 -e2)1/2/Ru. Now, let's see if this is really so. The constant "p" (which we all know) is equal to 3.1415926 and the constant "e" (which we all know) is equal to 2.7182818. So, (16p2-e2)1/2 must be equal to 12.268,847. Dividing this by the radius of the universe Ru = c2/G = 1.34681 x 1028 cm, we obtain the quantity of the mass of the electron to five figures 9.10956 x10-28 gram.
But how can you take something so vague as the radius of the universe and arrive at something so exact as the mass of the electron? Because we now have an exact measure for the radius of the universe Ru! In fact, we can now back-calculate from the more exact measured quantity of the electron mass 9.109,558,6 x 10-28 gram and establish a still more exact measure for Ru = 1.346,810,3 x 1028 cm = 14.236,098,000 light-years.
Indeed, we can progress one step further: we can extend to maybe seven decimal places the very poorly known value of the universal constant of gravity G = 6.673,214,7 x 10-8 (cm2/g)cm/sec2! This is a stratagem absolutely necessary if we are ever to obtain any exactitude in our new intercategorical physics.
Let's try another angle! Many physicists today claim that the photon can have a momentum p but never a rest mass m = E/c2, because photons are always traveling in space at the velocity of light c and so never have a chance to come to a stop where they can have a rest mass. But we can now correct this erroneous idea. Photons always travel at the velocity of light c in space --- true? True! --- but our new intercategorical physics teaches us that space is not the only category in which things can really exist. We have at least seven categories in all --- plus some five composite categories. So we find that the photon, while traveling at the velocity of light c in space, can also be completely at rest in the flow of time --- like a boat drifting down a river...
It's like our friends speeding in their familiar blue car down a six-lane divided expressway. At first we're parked beside the road at a pre-arranged rendezvous point quietly awaiting them. Then all at once we spot them as they come racing by us at goodness only knows what velocity. But we take off after them and soon catch up with them and come up right beside them in the fast lane. Now we're speeding down the highway side by side. So close are our velocities to each other that, as we look across to them, we feel as if they were parked there right beside us, for relative to each other we're neither of us moving at all. In fact, a boy from our back seat has no problem in throwing a ball of twine through his open window to a friend of his in the back seat of the other car, and the twine stretches out perpendicular to the trajectory of both cars as if we were parked beside each other in a parking lot.
Well, so it is with the photon. It speeds by us in such a flash that we couldn't possibly measure its mass even if it were large enough for us to detect. But in the category of time the situation is altogether different. The moment we step into time in our minds, the photon and we are traveling side by side, for we're now both traveling at the velocity of light together.
Fortunately, however, we don't have to actually measure the mass of the photon. We can calculate it very quickly from our mass-wave-radius equation we gave you back on page 2 of Chapter 2:
mrw = h/2pc = 3.517,739,4 x 10-38 gram-centimeter.
For, if the proton is constantly traveling around the spherical surface of the universe, its wave-radius can be nothing else than the radius of the universe Ru itself. Hence the formula for the mass of the photon:
mphoton = h/2pcRu = 2.611,904 x 10-66 gram.
But what happens to the velocity c of the photon when we encounter it in time traveling with us together at the velocity of light c of the universe's expanding time flow? It's velocity is now absorbed into its mass to form a new particle, the kineton, whose mass is exactly equal to the momentum of the photon:
mphoton = h/2pcRu = 2.611,904 x 10-66 gram.
But what happens to the velocity of c of the photon when we encounter it in time traveling with us together at the velocity of light c of the universe's expanding time flow? Its velocity is now absorbed into its mass to form a new particle, the kineton, whose mass is exactly equal to the momentum of the photon:
mkineton = h/2pRu = 7.830,293 x 10-56 gram
The kineton is a macrowave microparticle of antimatter in time, which we shall encounter again further down the road.
But what happens when we catch up with the kineton traveling at the velocity of light? We pass into the category of energy and find a new particle called the energion, the basic energy particle in the universe---at least as far as we human beings are concerned. Now the energion has a mass that exactly corresponds to the energy of the photon. So to find its mass we need to multiply the mass of the photon in space by c2 or the mass of the kineton in time by c:
menergion = hc/2pRu = 2.347,463 x 10-45 gram
Its formula can also be expressed as U/Ru, for U turns out to be our unified field constant hc/2p = 3.161,587,4 x 10-17erg-cm. (Above paragraph referenced from Chapter IV, mark "AA")
Moreover, the frequency of the energion is most astonishing:
The frequency of the energion is written as f = E/h = mc2/h.
So, the frequency of the energion is calculated to be Gc/2pp = 318.402,15 cycles per second. It just so happens to be the frequency of our top brain wave the (phi) wave of 318.4 cycles per second, some five octaves (25 = 32x) faster than the alpha wave in its mean of about 10 cycles per second. But even more amazing is the fact that this is also the frequency at which the graviton 1/Mu = 1/R2u = 5.512,989 x 10-57 gram traveling at the light-substance velocity of c3 takes to travel its figure-eight course (pRu + pRu = 2pRu) through the origin-center of the universe and back again, for 2pRu/c3 = 2p/Gc. Moreover, since the multiples of this particle cover the spectrum of radio waves, we call it appropriately also the radio-wave energion! It belongs as the microparticle to that form of energy matter we call ergomatter.
What happens, then, when we multiply the mass of the energion by the velocity of light c? We then pass into the mesomatter (i.e. mesonic matter) category of mass and find still another new particle called the thermeton:
mthermeton = hc2/2pRu (But Ru = c2/G), hence = Gh/2p = 7.037,518 x 10-35 gram.
We call it the infrared thermeton because its frequency and wavelength
bring it within the infrared range of the spectrum. It is the basic heat particle of the
(Above paragraph referenced from Chapter IV, mark "BB")
We must now multiply the mass of the thermeton in turn by the velocity
of light c. And when we do so, we make a discovery so momentous that it completely changes
our whole conception of the structure of matter! For the mass of the thermeton x c gives
us the photoproton!
Ghc/2p = GU !!! (for the definition of U, the Unified Field Constant, cf. above )
mphotoproton = 2.109,795,2 x 10-24 gram and
(dividing by the mass of the electron 9.109,558,6x10-28 gram, we get mphotoproton
= 2316.0235 emu.
(Above paragraph referenced from Chapter IV, mark "DD")
Then we make our great discovery! The photoproton is actually a
muopioproton almost exactly one muon and one pion greater in mass than the proton of
(Above paragraph referenced from Chapter IV, mark "CC")
Let's just see how this works:
The Photoproton GU = 2316.0235 emu in mass, minus one Muon (-206.7684
emu), minus one charged Pion (-273.12658 emu)
is a value of 1836.1285. Now, if we subtract 34 P-gyrinos (which is 34 x 0.000,578,667 = -0.01967) from this value, we have a value for the Photoproton that is equal to 1836.10885. This leaves us with a difference of 0.00005 emu from the mass of a proton which is 1836.1089 emu.
The 34 P-gyrinos, as we shall see later, link the photoproton with the microatom whose charges can induce the production of an atom of selenium #34! What could be more appropriate than that muopioprotonic photomatter should be linked to photovoltaic and photoconductive selenium #34!
Moreover, we're now getting a picture of a more complex ether of microparticles (photons of light, kinetons of flow, energions of energy, and thermetons of heat). This complex ether with its four evident aspects is constantly flowing down through innumerable vortexes inside particles, masses, and even galaxies to give rise to the ever down-flowing force of gravity ---like water flowing down the drain in a bathtub.
Go to Chapter IV
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