

D. THE DENSITY OF MASS IN THE UNIVERSEA TUBE MODEL INSTEAD OF THE CUBE MODEL ABOVE
There are at lest two quite different ways to look at the basis volume. It could be thought of as a very long tube down which the photon moved. Since it involved one second the photon at C would sweep out a length of roughly 3 x 10^{8} meters or a small tube about 300,000 Km long and of such a diameter it would contain one photon.... in cross section roughly 3.81853 x 10 ^{ 11 }cm sq or 3.486 x 10^{6} cm radius i..e. VERY small, particularly relative to the length. The other way to look at this basis volume was to set up a cubical volume (roughly 1 cc) such that it would contain one photon ON THE AVERAGE in each second. This means that as one photon leaves, on the average another photon will enter this volume. There are many of the tubes of spaghetti from the first model entering and leaving this cube, and each has one photon so that each tube contributes roughly a microsecond of residency to the basis volume. Thus the cube model overlaps many of the "tubes" above, and roughly a million photons will enter and leave this basis volume in the one second.
The second "cubical" model was, for me, easier to manipulate conceptually and to obtain quantitative meaningful results. Thus that will be presented here as the primary model. Below is a tube model.
The volume, V_{d} , of the average photon in the tube is
V_{d} = p r ^{2} l’ = p d_{1}^{2} l’/4 eq 53
letting L_{1} (= 6 p^{5}) be the constant relating the interpole distance, d, to the wavelength, l, then
L_{1} d = l eq 4 or eq 53a
V_{d} =p d_{1}^{3} L_{1} /4 = L_{1} p d_{1}^{3} /4 eq 54
and converting the volume to a basis volume, V_{db} , and the diameter to a basis diameter, d_{b}, then
V_{db }= (L_{1} p d_{b}^{3} )/4 eq 55
But V_{db} is the disturbed volume in the unit cell the fraction of the spasons actually involved in the change.
V_{db}= N_{d} /N = 1/ D^{3} = T = x_{b} x_{b} D eq 56
This then is the cross section of the photon times the length traveled, and note that, because to the paired nature of the poles, it is this rather than perhaps two times this value that also might perhaps be expected. Because we know the value of D, that now allows us to compute the value of the energy of this "average" photon in the basis cell, it allows computation of the wavelength, and in fact all such properties including the frequency of this "average" photon and density of mass in the universe as a whole. V_{db} = 1/ D^{3} = 4.2486966942 x 10 ^{32} (pure number) eq 57 solving eq 64 (V_{db}= L_{1} p d_{b}^{3} /4) for d_{b} eq 58
d_{b }= (4/ p L_{1})^{1/3}(1/D) =2.10881486229 x10^{12 }(in b units) eq 59 =1/16.546978 D = [(2/3p)^{1/3 }(1/1000 p ^{17 }]=[(1/D^{3})(4/ p L_{1})]^{1/3}
l_{b} =6 p^{6 }d_{b }= 348.60775/D =1.21643511 x 10^{8 }(in b units) eq 60
f = D/l_{b }= 2.3559124 x 10 ^{18}(in cycles per second or Hz) eq 61
= D^{2} /348.60775 = (3 p/2)^{1/3 }(106 p^{26 }/6) (note f is independent of basis units)
E_{b} = hf = E_{r}/B = h D^{2}/ 348.60775 =1.5610774 x 10^{8}(erg/B) eq 62
r_{b }= r /B =E_{b} / C^{2 }= E_{b}/b^{2} D^{2}(per B) = h(3 p/2)^{1/3 }(6p^{4})(b^{2})=
=1.90075984 x 10 ^{ 29 }/b^{2 }(in gms / B) eq 63
= 1.7369288x 10 ^{ 29 }(in gms / B) eq 63a
r =1.90075984 x10^{29}/b^{5}(gms/cc)=1.517278x10^{29 }gm/cc eq 64
The density of mass in the universe as shown in eq 64 is a function of the photon model, of h, and of b, but the D term has dropped out and thus the density is independent of the speed of light. This implies that the density of mass in the universe is an independent and basic property of the universe. It will be linked to the expansion of the universe, the radius and the age of the universe, but it is independent of C or D. 1. COSMOLOGY – the size and age of our universea) HUBBLE'S LAWBased upon observation E. Hubble noted that far away objects appeared to be receding with a velocity, V, proportional to the distance, s, with the proportionality constant being the Hubble constant, H (note units in per second)
V = H s eq 65
The age of the universe, tu, and the radius of the universe, Ru, are related to H such that
tu = 1/H eq 66
Ru = C/H eq 67
Assuming that Hubble's constant is a true constant and differentiating equation 65 we obtain the gravitational form of Hubble’s equation :
ds^{2} /dt^{2} = dV/dt = H ds/dt = HV = H^{2} s eq 68
Taking the extremes ** of eq 68, the graviational form of Hubble’s equaion, this integrates to
s = so e^{+Ht} + B e^{Ht} eq 69
The e^{+Ht} term is the expanding universe term, and the second term represents a contracting universe term usually ignored in our universe. This is a linked  time reversal  contracting contraterrenne universe paired with our own universe.
** Taking the original Hubble equation V= Hs = ds/dt, this integrates directly to s = so e^{+Ht} (eq 69a) but lacks the second contracting universe term B e^{Ht} (eq 69b) which is needed in the following derivation and should not be ignored so the extra step is really required to obtain the more general result.
Gravity is nothing more, or less, than the expansion of the universe. The expansion; however, is related directly to the presence of mass, or energy in the universe.
F_{2} = m_{2} a = m_{2} d^{2}r/dt^{2} = G m_{1} m_{2} / r^{2} eq. 70
d^{2}r/dt^{2} = G m_{1} / r^{2} eq. 71
r^{2 }d^{2}r/dt^{2} = G m_{1 } eq. 72 On the left are spacetime and only spacetime terms, on the right are mass terms. Thus mass is related ONLY to spacetime.
substituting m_{1} C^{2} = E_{1 eq 6a}
C^{2} r^{2} d^{2}r/dt^{2} = G E_{1}
On the left are spacetime terms and only spacetime terms, on the right is energy. Thus if energy is quantized, then spacetime also must be quantized, and the gravity term is directly related to space time, and also is quantized. The gravity thus occurs only where mass/ energy is present.
but d^{2}r/dt^{2} = H^{2} r eq. 73
thus H^{2} = G m_{1} /r^{2} =(4/3) p r^{3} r G/r^{2} =4/3 pr G eq. 74
And finally the expansion of the universe must happen ONLY at the same point where gravity occurs, and all are really the same phenomena, just looked at in different ways.
There also is an energy of time flow. As time passes the photon poles are further and further apart relative to the distance at any given time. Thus the energy is less and less.
DE_{t} = t m Dt eq 75
where DE_{t} is the change in energy of mass, m, in a time interval Dt, and is the proportionality constant, but it also is:
t = H C^{2} ( = 1850.908 ergs/ gm sec) eq 76
and eq 75 could also be rewritten as:
DE_{t} = t m Dt = H C^{2}m Dt = mC^{2 } H Dt eq 77
The evaluation of t is 1850.908 ergs per gm second or 0.1850908 Joules per Kg sec. Starting with eq 78 the expansion term s_{o} e^{+Ht} taken for a "short time" relative to the 15 billion years total may be reduced to eq 75. For very long times, or near the start of the universe the exponential should really be used.
2. A NEW LOOK AT GRAVITY
There is a new way of looking at gravity that may be “peculiar” but it does give instructive results. The fundamental Newtonian equation for gravity is:
F = m a = G m_{1} m_{2} / r^{2} eq 76
where r is the radius to the center of the mass m_{2 } and G is the gravitational constant 6.6732 x 10^{8} dyne cm^{2} / gm^{2} Then taking a man for example mass, m_{1} , standing on a large planet such as earth which is mass, m_{2}, the acceleration on the man will be
F = m_{1} a = m_{1 }ds 2 /dt^{2} = G m_{1} m_{2} / r^{2} eq 77
and dropping the m_{1 }from the equation, and letting the mass of the planet with radius s be its volume (4/3 p s^{3}) times its density, r , the equation now becomes:
ds ^{ 2} /dt^{2} = G r 4/3 p s^{3} / s^{2}= 4/3 p r G s eq 78
At this point equation 74 is identical to equation 78 if we realize that the "constant" terms are equal, and thus:
H^{2} = 4/3 p r G eq 79
and the expansion / contraction of the universe equation 69 can be rewritten with H^{2} being substituted for 4/3 p r G. Thus gravity is nothing more or less than the expansion of the universe, gravity is just an acceleration, no different from any other acceleration, and not a "special force" at all. The need for a gravitron etc. disappears.
The key to thinking about gravity is to realize that the surface of the earth is NOT and inertial reference system at all, in fact I can feel the acceleration as I sit here, and thus is not a proper frame of reference. There is a zero G reference in the center of the earth, and it perhaps could be used, but the surface of the earth is definitely an accelerated system. Taking the central point the surface is actually accelerating very slightly away from that point. That implies that the whole earth is very like a big balloon being blown up slowly, it is expanding.
Then why do we not measure this expansion? The very yard stick which we propose to use also is expanding and since it is made of the exact same materials as the earth it is expanding at the same rate. Thus the acceleration we feel is the only major evidence of the expansion of the universe locally. There is another analogy to this "gravitational" term. The so called Corriolis forces which had to be invented to account for the actual rotation of the earth under a projectile fired either north or south. This virtual force was required because we persisted in using an improper assumption that the earth was a stationary reference plane. When we account for the rotation then the need for the "Corriolis force" disappears.
Similarly when we account for the real acceleration in the expansion of all mass in the entire universe the need for a separate "gravitational" force also disappears, and this explains why no "gravitron" has so far been found. There is no need for a gravitron to carry a fictious force. The whole universe is expanding, and thus the "real" motions of planets includes their own more or less spherical expansion combined with what would appear to be an increasing spiral orbit if we had an absolute meter stick to measure the distances. Since we and our measures are part of the universe, we can not do this, and must account for the changing distance by inventing an apparent force. If we account for the expansion of the universe we no longer need the force. Thus gravitation is no more or no less than the expansion of the universe which also is no more than time passing as new spasons are created to provide the change needed to allow time to pass.
3. STRONG and WEAK NUCLEAR FORCESThe strong and weak nuclear forces are in nature different from the two body forces, which now include Gravity linked into electromagnetic forces as they all are simply two body photon interactions. The strong nuclear force will be assumed as an induction to be a THREE body interaction and the weak nuclear force will be assumed to be a FOUR body interaction. All the two body interactions will be found to be proportional to 1/r^{2} while all of the three body will be proportional to 1/r^{6} and four body interactions proportional to 1/r^{12} . To complete this list even though they are not being developed at this time, five body interactions would be proportional to 1/r^{20}. The two body forces are interactions of two things taken one at a time, i.e. 2 factorial divided by 0 factorial 2!/0!=2. The three body forces are three things taken two at a time, but exclude two things one at a time already counted, which is 3 factorial divided by 1 factorial = 3!/1!=2x3= 6. The four body forces are four things taken three at a time, excluding the lower interactions and they are 3x4 = 12 = 4 factorial divided by 2 factorial 4!/2!){five body would be 5!/3! = 5 x 4 = 20}. F_{g or em} = k/r^{2}
F_{s} = B_{1}/r^{6} eq 80
F_{w} = B_{2}/r^{12} eq 81 and since by definition E º dF/dr, differentiating, and letting the constant be subsumed into the proportionality constants: E_{g or em} = k’/r
E_{s} = B_{1}’/r^{5} eq 82
E_{w} = B_{2}’/r^{11} eq 83 These force equations should describe the electrical charge density, Ec, of nuclear isotopes. R. Hofstader has reported this data for numerous isotopes, and using that it is possible to evaluate B_{1} and B_{2}, taken in the form:
E_{c} = B_{1}/r^{6} + B_{2}/r^{12 } eq 84
The charge density is related only to the proton atomic number, A, and the neutrons do not contribute to charge. However, the strong and weak forces operate on both neutrons and protons. Thus we need to relate neutron count to proton count or both to the total atomic weight of the nucleus. This then requires one other relationship. Accounting for neutron to proton ratios the stable isotopes^{136} are related to proton count such that:
E_{c} = 1.53  0.7667 A^{1/3} eq 85 Equation 84 was placed into a least square error fit program which varied B_{1 }and B_{2 }to evaluate B_{1} and B_{2} and then the values found placed back into the same program but with those values to find the curve that best fit the data. Also remember that the strong nuclear force term is positive inside the nucleus and repulsive outside the nucleus, so this requires two equations to account for this. This program is shown below:
FORTRAN Computer program for Fs and Fw DIMENSION R(20), E(20), NAME(20) 1 READ (5,100,END=900)Z,A,NAME 100 FORMAT (2F10.0,20A1) RHALF=1.05*A**(1./3.) RC=RHALF3.0287 BONE=920. BTWO=328460. EHALF=0.6407 EC=1.530.7667*(A**0.33333) DO 10 J=1,20 R(J)=0.5*J IF(R(J).LT.RHALF GO TO 11 150 EA=BONE/((R(J)RHALF+3.00287)**6BTWO/((R(J)RHALF+3.0287)**12) E(J)=0.5*EC*EA/EHALF 500 GO TO 12 11 S=2.*RHALFR(J)RC 151 E(J)=EC+EC*((BONE/S**6)+BTWO/(S**12))/EHALF)*0.5 12 CONTINUE 10 CONTINUE WRITE(6,200) Z,A,NAME,EC,RHALF WRITE(6,300) R WRITE(6,300) E 200 FORMAT(20(1X,F5,3)) WRITE(6,444) 444 FORMAT GO TO 1 900 STOP END
The actual measured data and the calculated curves are shown below. The values shown in the program are the best fit with B_{1 }being 920 and B_{2 }being 328460 when Ec is given in units of 10^{19} coulombs per cubic centimeter and the radius is given in Fermis i.e. Fermis to the 6th power or Fermis to the 12th power for the respective constants. There is less than 1% total deviation from the reported data for the 12 isotopes calculated. (He4, C12, O16, Mg24, Ca40, V51, Co59, Sr88, In115, Sb122, Au197, and Bi209). Because of use of equation 85, the isotopes below about C12 should be run on an individual basis, not with the general stability zone equation. The fact that a simplistic relationship like equation 84 would fit the whole family of isotopes strongly supports the hypothesis of strong and weak nuclear forces.
Fig 4 Charge Density of selected isotopes from computer program
4. THE AGE AND SIZE OF THE UNIVERSE
From H and eq 66 and eq 67 it is possible to calculate the age and radius of the universe. Substituting the density of matter (1.5172778 x 10 ^{ 8}) in the universe into eq 74 (H^{2} = 4/3 p r G), with the relatively poorly known^{147} constant G = 6.6732 x 10^{8 }dyne cm^{2}/gm^{2} (error ±0.0031 or 230 ppm) H may be found:
H = 2.05941 x 10^{18 } per sec eq 86
(with expected error of ±0.00047, this compares to the pure number derivation eq 113 where H was found to be 2.059429x 10^{18 } per sec.) The limiting factor in the accuracy in H above is the value of the gravitational constant with 230 ppm error. Still this value of H is far more accurate any other value to date. The age of the Universe:
tu= 1/H = 4.85575 x 10^{17 } seconds or 15.387 billion years eq 87
The radius of the Universe:
Ru = C/H = 1.4557 x 10^{28 } cm (or x 10^{26 } meters) or 15.387 billion light years. eq 88
This compares to guesses near 15 billion Ly with at least ± 2 billion Ly error in observed values.
From later indexing theory the radius would appear to require 2^{164 } spasons which is a value of 1.489 x 10^{28 }cm and a value of H = 2.0131x 10^{}^{18 }per sec thus the values all agree well with the probable error stated. But this derivation merely sets an upper limit for H and is not exact, i.e. the prior value is the more correct one. 
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