Chapter 2: The scaling exercise
Derivation
of the mediating mass
The scaling
laws of the equilibrium universe
The
equality of the equilibrium radius and Hubble’s distance of
acceleration
Understanding the enigma of the dimensionless
constant fc
The quantum
equilibrium of the galaxies
The
quantum drive for 3D-space with gravity as engineThe
grand correction: the idea of a vector composition of Hubble’s
distance
The
analysis of the vector composition of Hubble’s distance
DERIVATION OF THE MEDIATING MASS
Applying the uncertainty principle to the rest mass of the electron gives the Compton length. Extending this idea to the proton seems useless, for the diameter is much greater than its generalized Compton length. Still it is useful to use the idea of the generalized Compton length.
The uncertainty principle determines for electron and proton respectively:
mec λe = h with λe = h / me c and λp = h / mp c
The self energy of the particle, electron & proton, should be balanced by the same potential energy of the unit electric charge of both.
me c² = e² / 2εo re and mp c² = e² / 2εo rp .
Giving:
rp / λp = re /λe = α = e² / 2 εo hc
Where α is called the fine structure constant for electron or proton.
Consequently, above derivation can be repeated for force equilibrium between gravity and electrostatic force.
fe = e² / 2εo re²;
fge = 2G me Me / λge² = me ae = G mpl² / λge² =
h c /λge²
fp = e² / 2εo rp²;
fgp = 2G mp Mp / λgp² = mp ap = G mpl² / λgp² =
h c / λgp²
The relations; 2me Me = mpl² and 2mp Mp = mpl² can be derived as follows. In short λ is the generalized Compton length; am is the quantum acceleration in equilibrium to a gravitational mass of Mm with a similar λ (event horizon). With am λ² = G Mm and am 2λ = c² one determines
c² λ = 2 G Mm . Apply the quantum definitions and one proves:
2Mm mm = mpl² .In this way the force mm am can be compared to other forces. See also, the general validity for the scaling of event horizons.
So the balance of forces gives the dimensions of λge & λgp with respect to each specific Compton length.
re / λe = α and α λe² = λge²
rp / λp = α and α λp² = λgp²
1. In case λge , λgp is replaced by λe or λp respectively, fge or fgp are greater than fe or fp .
2. G drops out of the g-force relation. The mediating force depends on the characteristic dimensions of the particle and the quantum mechanic constants.
3. These relations are independent of the choice of any value of the factor (two), because this constant appears in the relation as a factor 2 (2m M = mpl²).
4. The accelerations ae and ap are not equal to aH.
Ad 2 &3 are important, for the mediating mass is not dependent of fc
See, Understanding the enigma of the dimensionless constant fc and the grand correction, the idea of the vector composition of Hubble’s distance.
Applying the conservation of angular momentum for the above relations of λge and λgp and taking into account the half spin properties, gives the Compton length of the mediating mass:
λm² = ¼ λge λgp
So: mm² = mp me / 4α
Because m is inverse proportional to the Compton length. Since the neutrons appear in the transmuted atoms, the relation seems only valid for the primordial species, Hydrogen.
The value of mm = 250.81* me with mp = 1836.153 * me and 1/α = 137.037. Putting the mediating mass just below the pi-meson group of (264 to 273)*me and above the rest mass of the muon, 206*me .
Trying to pinpoint the result of the mediating mass by including the atoms and taking the half-spin of the neutron also into account, was nearly pointless. For the spin of the neutron and proton are the same quantum wise and it seems clear from the onset that stable neutral atoms including the electrons, should have a mediating mass between that of the muon and that of the pi-mesons. Taking into consideration the slight instability of the neutron, always decaying into an electron and proton, it shows that the conditions to generate the mediating mass are difficult to achieve. Achieving the dimensions proper and the spins for both lepton (electron) and proton (quarks) simultaneously seems impossible to create in high energy experiments.
It explains actually, why the mediating mass is a one off feature in the law of physics. Only during the Big Bang, the creation ex nihilio (creation out of the Nothing), the equilibrium condition for proton and electrons could have been created in conjunction to gravity. After the broken equilibrium the situation seemed to be irrevocable. Apparently, the weak force dominates always in the decay of the neutron.
In general the conclusion can be that the conservation of mass in our cosmos, despite atomic transmutations is conserved unless baronic matter transforms into the matter or mass of a black hole. See, The speculation of the mediating mass for an atom with electric Z-charge.
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Chapter 2: The scaling exercise
THE SCALING LAWS OF THE EQUILIBRIUM UNIVERSE, OUR PRESENT- DAY UNIVERSE
The equilibrium condition of forces between our universe and the anti-universe is given by
aH RH² = G Mtot (1)
See, the equilibrium hypothesis; the relative acceleration constant.
And secondly, by the quantum transformation:
aH fc RH = c² with aH = mH c³ / fc h and
RH = h / mH c (2)
fc is a numeric or dimensionless constant, which is called the phase velocity factor or the conjugated factor in case of interference between normal and conjugated gravity. The problem is that fc works inverted with respect to c² making its interpretation confusing.
Taking fc = 2 and substituting it in rel. (2) into rel.(1) ;
c² RH = G Mtot or mH 2 Mtot = mpl² (3)
Here mpl² = hc / G ; the famous Planck constant.
Now divide Mtot by the number of mediating masses, electrical neutral, or the primordial ones generated during the Big Bang ( see derivation of mediating mass), the equilibrium relation (1) becomes:
aH λm² = G mm (4)
Where RH² / N = λm² and λm is the generalized Compton length of the mediating mass. RH, the equilibrium radius for Hubble’s distance fulfils the square root rule and in this way, the world at rest over all matter and the size of the universe are strictly related. The interpretation of the rel. (4) in terms of physics is that every neutral particle (proton and electron) is by the mediation of quantum mechanics in equilibrium with its counter part in the anti-universe. The g-spin of mm is delayed over a distance λm by an reaction force of mm aH where mm drops from both sides of equation (4).
aH (h / mm c)² = G mm ;
(mH c³ / 2h)(h / mm c)² = G mm ; mH mpl²=2mm³
Then aH = (2mm³ / mpl²)(c³ / 2h) and
RH = (mpl² / 2mm³)(h/c) .
The factor 2 stands for fc . It drops out of the relation for aH, and it makes aH independent of fc . Similarly it can be shown that mm is independent of fc . So only Mtot & RH show a dependence of fc.
Putting RH in rel. (3) gives:
Mtot = mpl (power 4) / 4mm³
If fc = 2, the factor 4 is valid, but if fc = 4, this factor is 16 in the relation for Mtot. Now calculate the table below, for:
mm = 250.81 * me ; me = 9.108 10exp(-31) kg ;
mpl = 5.435 10exp(-8) kg; h = 6.62 10exp(-34) mks dimensions;
G = 6.7 10exp(-11) mks dimensions; c = 2.99 10exp(+8) m/sec.
Mtot RH aH definitionfc
10exp(+53) kg 10exp(+26) m 10exp(-10) m/sec²
7.32 5.50 1.63 RH c²=GMtot
1.83 2.74 1.63 RH c² = 2G Mtot
0.814 1.83 1.63 RH c² = 3G Mtot
0.458 1.37 1.63 RH c² = 4G Mtot
1.83 2.74 none(GRT) RH c² = 2G Mtot
0.915 1.37 none(GRT) RH c² = 2G Mtot
3.66 2.74 none(Newton) RH c² = GMtot
The results for RH & Mtot based on GRT are also given in the table.
The observed data for RH & aH are 14 billion lyrs and 10exp(-10) m/sec². The theoretical value of aH is close enough to the one in the MOND-hypothesis. There are some reasons why these accelerations should not be in agreement due to the difference in their definitions. The main reason seems to be in the case of the MOND-hypothesis, the acceleration is a derived constant based on an effective value, while here the constant is the maximum value because it is valid at the equilibrium radius.
To rectify some previous error in the table, the derivation of the mass relation for GRT is given (a shortcut version). From the standard work on gravitation (1972) ,Milner, Thorne and Wheeler, the derivation of the critical mass density ρ in relation to Hubble’s constant H, based on GRT is:
c² / RH² = 8/3 π ρ where; H = c / RH .
Taking the view of the imbedded 3D-surface (volume) of radius RH in the 4D hyper surface of time and space, one may write:
Mtot = 4/3π ρ RH³; which gives
RH c² = 2G Mtot or
½ Mtot c² = G Mtot² / RH
One does not expect the term of ½ Mtot c² in this equality, for Mtot c² is the relativistic energy. But apparently for the critical energy of the universe, the rest mass is balanced against the rest mass of the gravitational energy.
Contrary the classic derivation of Newton gives;
RH c² = G Mtot .
The disappearance of the factor 2 denotes that the centre of mass of the universe is in absolute rest.
The new exiting result is that the quantum accelerated universe (but in an equilibrium of forces) has the same mass relation as in GRT for fc = 2. The overall mass and the Hubble distance are equal. However the distance travelled by light in the same time interval is twice that of the Hubble distance in the equation for GRT, for in the accelerated model Hubble’s distance is an accelerated distance. Similarly, if fc = 4 , then Hubble’s distances are both equal for the two cases and equivalent to the observed Hubble distance, but the overall mass is a factor two different. The factor
fc = 4 can not be explained on theoretical consideration and is discussed in Understanding the enigma of the dimensionless constant fc. No conclusion can be drawn just as yet from the table, because of this factor.
Note that the expressions fc mH Mtot = mpl² and
mH mpl²= fc mm³ relate the rest mass of matter to the quantum mass of vacuum mH and the Planck mass. This most important fact was overlooked for a long time, but it finally confirmed, the statement that rest mass of a particle is entirely a feature carried by the carrier field. Thus empty space! (According to Denaerde).
Note: Hubble’s distance in GRT is the criterion for an open, closed or eternally expanding universe. In the pyramid theory the universe is eternally expanding. Secondly, Hubble’s acceleration is constant during the evolution of the universe, so Hubble’s constant H cannot change.
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Chapter2: The scaling exercise
THE EQUALITY OF THE EQUILIBRIUM RADIUS AND HUBBLE’S DISTANCE OF ACCELERATION
As by after thought, it is not difficult to prove the square root rule of Hubble’s distance divided by the square root of the number of all mediating masses to be equal to the generalized Compton length. Secondly, one also has to prove that the accelerated Hubble distance is equal but not necessary coincides with the equilibrium radius of the universe. So far only a supposition was made due to the square root rule that this was valid.
So let us list all the relevant equations: (fc = 2)
aH 2s = c² aH RH² = G Mtot giving
c² (RH²/ s) = 2G Mtot
aH λ² = G mm c² λ = 2G Mm giving
2mm Mm = mpl²
Mm = mpl² / 2mm
λ = h / (mm c)
aH = (mm³ c³) / (mpl² h)
s = (mpl² h) / (2mm³ c)
Elimination of λ and s, gives:
RH² / Nb = λ s = (mpl² h²) / (2mm (power 4) c²)
Mtot = mpl (power 4) / 4mm³ Mtot / Mm = Nb = mpl² / 2mm²
Here the relation: 2mH Mtot = mpl² should be valid and it is a generalisation of the mirror mass relation used every where in this scaling exercise. Without this supposition the proof is not valid. See, also The validity for the scaling of the event horizons. (Again it is valid for the quantum mechanic objects).
Substitution of Nb, gives:
RH² = (mpl² h)² / (2mm³ c)² = s²
Which is the same result if one had assumed that s = RH !
Note: Since N or Nm = ¼ (mpl / mm) (power4) and Nb are already mentioned, the ratio between Mtot and mH, the Hubble mass, can be given:
NH = ? (mpl / mm ) (power 6)) = Mtot / mH
Later, at the end of chapter 3 it is possible to understand that mH could also be the equivalent graviton mass of the smallest quantum of accelerated vacuum.
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Chapter 2: The scaling exercise
UNDERSTANDING THE ENIGMA OF THE DIMENSIONLESS CONSTANT fc
The only supposition one can make about the constant acceleration of quanta, is the pull of an external universe, the anti-universe, probably consisting of anti-matter(for reasons of symmetry of the laws of nature) and the equality of mass generating gravity in both universes, which in summation allows the balance of force, given:
aH RH² = G Mtot which converts to
c² RH = 4G Mtot
With fc = 4 to let Hubble’s distance corresponds to the observed value.
Straightaway one factor 2 of fc is explained by the kinetic property of the integrated path under constant acceleration given in general s = ½ a t² . The second factor 2 is certainly an integer and follows from the comparison overall mass relations, one based on GRT and other based on force equilibrium. Because both RH are equal in both relations, the factor 2 has not any other value, meaning twice the overall mass in the one model with respect to the one overall mass in the other model. Still it is not explained where the factor 2 comes from.
Now, let us consider the force balance according to the square root rule, which is dividing Hubble’s distance by the square root of the number of mediating masses.
aH (RH² / N) = G mm ; with N mm = Mtot
aH λ² = G mm ; with λ = RH / √ N
and λ is the generalised Compton length.
Having reached this point in the reasoning, one should realize that the Compton length in the relation, like Hubble’s distance, is the distance under constant acceleration, but the distance travelled by light over the same interval of time is twice that. So consequently one should rectify our assumption.
So instead of λ substitute ½ λ in the above relation.
aH (½λ) ² = G mm with the definition of aH one finds:
mH = 16 mm³ / mpl² ; where aH = (4mm³ c³) / (mpl² h). Then aH is a factor 4 higher than 1.6 10exp(-10) m/sec² and RH is a factor 4 smaller
than 27.4 billion lyrs.
The calculation with 2λ instead of λ is left to the reader, but again the result is not what we want.
Thus what does the half wavelength of the Compton length in physics mean? It means for a Fermi particle that the half wavelength is the spin of the particle. To us it seems inconceivable that the rectilinear wavelength of acceleration coincides with the spin dimensions destroying its particle properties or the particle collapses. So for all cases, that the accelerated length coincides with the generalized Compton length, then the light path over the same time interval is twice that. It turns out that the Compton length is always equal to the accelerated length, because even if the phase velocity is 2c, it does not affect the atomic calculations of the scaling. See The grand correction: the idea of a vector composition of Hubble´s distance.
Since fc returns in both relations:
fc mH Mtot = mpl² and fc mm³ = mH mpl² ; one should try to find the independent parts of fc not scaling with it.
1.From the derivation of the mediating mass mm , we know that its value is independent of fc .
2.With N mm = Mtot , both relation give N = mm² / mH² , which is independent of fc .
3.The other combination of the two relations gives;
fc² N = (mpl / mm) power 4.
Splitting in dependent and independent parts on either side of the equality sign, gives;
N mm (power 4) = ( fc / mpl² )²
The right hand part in the equality is determined by the constants of nature.
The separation in part shows the result of
RH= h / mH c = mpl² h / (fc mm³ c) = h² / (fc G mm³) ;
Where mm is independent of fc .
In preliminary conclusion, the observed Hubble distance in the acceleration model points directly to the presence of another universe, because this universe affects the fundamental constants of nature, otherwise Hubble’s distance should have been 27.4 lyrs , the one for fc = 2 .
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Chapter2: The scaling exercise
THE QUANTUM EQUILIBRIUM OF THE GALAXIES
On average the galaxies have about 10 billion stars of an average mass of the Sun, which is 2.0 10exp(+30) kg. Already the square-root rule between the number of mediating masses N and the overall mass of the universe is known. The ratio of the number of galaxies with respect to Mtot is Ng, while the ratio of the number of stars to Mtot is Ns.
Under the assumption that aH for each mass system is the same, one can calculate the equilibrium radii to find out if these confirm the square-root rule.
RH / λm = √ N RH / Rg = √ Ng RH / Rs = √ Ns
Rg & Rs are respectively the equilibrium radii of galaxy and Sun:
aH Rg² = G Mg and aH Rs² = G Ms
Naturally, the radii fit above rule, for the mass of each system is proportional with the number of mediating particles per system.
Galaxy; Mg = 10exp(+40) kg Rg = 6000 lyrs
Sun ; Ms = 2. 10exp(+30) kg Rs = 10exp(+15) m ≈ 1 lyr
Earth ; Me = 5. 10exp(+24) kg Re = 10exp(+12) m ≈ 0.001 lyrs
For the moment, Rs & Re seem without meaning but show how incredibly small aH is. Rg as an equilibrium radius, tells us when stars escape from the galaxy, if the radius is greater than Rg .
By using: aH = (4mm³/mpl²)(c³/4h) and λm = h / mm c (fc = 4)
One can derive: Mg / Rg² = mm / λm² = 2.43 kg/m² or 955 kg/m². The first is based on the mediating mass and the other on the proton mass. The last makes equilibrium radius a factor 19.8 smaller than the one based on the mediating mass, which should be the correct one for this calculation. Still, because spherical symmetry is used, Rg is too small at the plane of eclipse of the galaxy. It should be corrected for the elongation of the mass distribution in the eclipse.
Note; the above result is independent of the presence of a super massive black hole in the centre of the galaxy.
The next step in the discussion of the quantum equilibrium is the discovery of a conceptional mistake. The mistake is that if aH is constant in empty space, it acts at the centre of a mass system and consequently, it is the onset to accelerate the system. However, it is not what the derivation of the scaling laws tells us, especially the generalized square-root rule. So in case aH is acting on every neutral atom individually, it is in equilibrium with each neutral atom individually. Adding all atoms together each under force equilibrium, summing zeros actually, the overall system should also be in absolute equilibrium. See also, The equality of the equilibrium radius and Hubble’s distance. The square-root rule will lead to the discovery of the accelerated expansion of vacuum or empty 3D-space, discussed in the next section.
Taking the consequences of the conceptional mistake and returning to the equilibrium hypothesis, it is clear that the idea about the equilibrium condition of the super massive black holes is obsolete, which is absolutely true if only one kind of gravity should have existed in nature. However the equilibrium condition for black holes is immediately understood in relation to conjugated gravity. Namely, the formation of the super massive black hole is a consequence of the disappearance of conjugated matter in the early universe. See Quasars, super massive black holes and scaling of these enigmas.
Note: At the end of this study, it can be concluded that the equilibrium hypothesis was absolute correct. The universes are in force equilibrium with gravity. The result is that empty space expands accelerated linearly and rotationally, but in fact the mass centre of the galaxy drifts along, for every atom, mediating mass is a balance of opposing forces. See also, The assessment of the average number of black holes in the universe.
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Chapter2: The scaling exercise
THE QUANTUM DRIVE FOR 3D-SPACE WITH GRAVITY AS ENGINE; HUBBLE’S DISTANCE
The definition of Hubble’s distance is the time that light travels since the beginning of our cosmos. The reciprocal value of the age of the universe is Hubble’s constant, which can be expressed in the red shift, Doppler shift, of the light escaping the galaxies of certain radial velocity with respect to us as observers. The local velocity of a galaxy to its neighbour further away has a slightly increased red shift and in this manner one can correlate the distance of a galaxy with its velocity, which may even give e.m. shifts from infra red to radio frequencies. The general expression of Hubble’s constant H is: 1/R (dR/dt) = H. Where R is the distance and dR/dt is the red shift translated to radial velocity.
In the previous section about the equilibrium of the galaxies one can translate the square-root rule into that of the ratio of spherical surfaces of the overall universe and that of the galaxies which should be proportional to Ng , the mass ratio between the two. However and this is crucial, the ratios of the volumes between the two, overall universe and galaxy, is in violation to the one of the mass ratio.
Rg = Mtot / Mg = 4π RH² / 4π Rg² but Nv = (4/3)π RH³ / (4/3)π Rg³ .
The ratio between the volumes is bigger than the number of galaxies Ng. This can only be explained by the expansion of vacuum or empty space per unit of time.
For fc = 2 aH 2RH = c² with aH tg = c & RH = Rg√ Ng follows:
Rg / tg = vg = c / 2√ Ng
Where vg seems to be the expansion velocity of empty space at the surface for the equilibrium. It makes every galaxy a generator of empty space (three dimensional quanta), with their centre as a point of reference independent of the overall ‘centre’ of the universe. Note: the expansion rate of empty space per mediating mass follows by replacing Ng by
N = Mtot / mm = ¼(mpl / mm)(power 4) giving;
vm = c mm² / mpl² (independent of fc)
For Ng = 10exp(+10) galaxies vg = 1.5 km/sec at an equilibrium radius of 6.0*10exp(+19) m or 6000 lyrs ( Mg = 10exp(+40) kg ). The expansion of empty space equivalent with a mass leaving the equilibrium radius is about a factor of hundred too small compared to the radial velocities at the outer rim of the galaxies, where the stars escape with an average speed of 200 km/sec. Here the diameter of the average galaxy is hundred thousand lyrs, a factor of hundred different in dimension, which should explain every thing. However, an equilibrium radius of hundred thousand light years corresponds to a very big mass of the galaxy. Here again the above discrepancy between the ratios the surfaces and the volumes is confirmed. Another way of explaining it, is that because the expression; aH RH = ½ c² is valid for any velocity at a certain distance less than RH , the derived velocity is the one corresponding to a distance at a constant acceleration aH. So empty space moves accordingly, but it reaches after 14 billion years the half the c-velocity. Or still in other words, every centre of mass generating gravity has the velocity for empty space of 100 or 200 km /sec at a radius of hundred thousand lyrs.
Taking the consequence, could mean that every galaxy is completely isolated from its neighbours after 14 billion years, but many galaxies are embracing super structures, for the influence of gravity even at these large distances can not be neglected. So at Hubble’s distance the quantum equilibrium radius is 14 billion years, which is the distance to expand space around all its matter in the universe, continuing perpetually.
Not deliberately but a error comes to light: Because, if one repeats above calculation for fc = 4 then aH RH = ¼ c² , which makes velocity of the quanta or the material system at the end of the acceleration ½√2 c instead of c. This is the topic in the section of the grand correction: the idea of a vector composition of Hubble’s distance.
Note, that no correction of the escape velocity at the equilibrium radius is needed, even if one considers the impact of the vector composition of Hubble’s distance, because the escape velocity for either the material or the immaterial equilibrium radius is also composed velocity vector with an angle of 60 degrees between the components.
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Chapter2: The scaling exercise
THE GRAND CORRECTION: THE IDEA OF A VECTOR COMPOSITION OF HUBBLE’S DISTANCE.
In the previous section it was discovered that for fc = 4 the relation aH RH = ¼ c² the end velocity of the accelerated quanta of empty space ½√2 c instead of c, while all other quantum relations as the definitions of
aH = mH c³ / fc h and RH = h / mH c
are based on the light velocity c.
In the section Understanding the enigma of the constant fc, it was determined that fc only depends on the constants G, h & c contracted to the Planck mass. So one needs to leave all scaling relations unchanged and the fundamental transformation equation, as 2s = c² cannot be determined by fc.
Left for correction is the change of the balance of force.
aH RH² = 2G Mtot
The essence is the addition of the factor 2: Meaning two gravitational sources or twice the force is needed to balance the constant force of acceleration to make RH in agreement to the observed value of 14 billion light years. Then again the relation aH RH = ½ c² is restored. However, although the overall mass of the universe is the same as in the case of
fc = 4 , aH and RH are twice as high. See the table.
Similarly, nearly as an act of desperation, one can try the possibility of
as s = 2c², using a phase velocity of 2c. For in a Newtonian universe of the accelerated quanta of space there exits no restriction on any velocity. Again the result is shown in the table, but Hubble’s distance and the overall mass are increased enormously. In taking stock of the predicament one is into, there were things very clear. The acceleration aH is determined by the Compton wave length of the mediating mass, which cannot scale differently and the end velocity should be the c-velocity at Hubble’s distance at 13.7 billion lyrs. Then the realization dawned that the acceleration distance in comparison to the standard meaning of Hubble’s distance, could be subjected to vector summation. So what if next to the rectilinear acceleration an angular momentum was driving these quanta? It should resolve our predicament in one stroke.
Mtot RH aH
10exp(+53) kg 10exp(+26) m 10exp(-10) m/sec²
7.32 11.0 1.63 aHRH=2c² RH c² = ½ GMtot
7.32 5.50 1.63 aHRH=½c² RH c² = G Mtot
1.83 2.74 1.63 aHRH=½c² RH c² = 2G Mtot
0.458 2.74 3.27 aHRH=½ c² RH c² = 4G Mtot
0.458 1.37______ _ _1.63 aHRH=¼c²_ RH c² = 4G Mtot
1.83 2.74 none(GRT) RH c² = 2G Mtot
0.915 1.37 none(GRT) RH c² = 2G Mtot
1.83 1.37 none(Newton) RH c² = G Mtot
See, The analysis of the vector composition of Hubble´s distance.
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Chapter2: The scaling exercise
THE ANALYSIS OF THE VECTOR COMPOSITION OF HUBBLE’S DISTANCE
The vector summation for the angular momentum of the actual acceleration of empty space is not that difficult. Since at Hubble’s distance 13.7 billion lyrs the end velocity is ½√2 c, is suggests a angle of 45 degrees somehow. But RH becomes √2 * 13.7 billion = 19.4 billion lyrs at the end velocity of c. So wrong guess!
The only other quick check, which gives immediately the desired effect is to insert an angle of 60 degrees. A half distance of rectilinear acceleration has half the c velocity (aH Rhalf = 1/8 c²), while the light speed is c at twice the distance taking the angle of 60 degrees into account. The angular acceleration correspond to ½√3 RH (RH = 13.7 billion lyrs), which corresponds to an energy of 3 / 8 c² . So at the corresponding Hubble distance along the line of sight the end velocity is c and not, as previously thought ½√2 c. It is an absolute incredible result!
Now turn the result around and start with the part for the angular momentum of 3/8 c² (or half of it, to 3/16 c² ,in desperation), then the component of the angular momentum RH is 20.7 billion lyrs and the linear accelerated component is 6.85 billion lyrs, making 27.4 billion lyrs as a vector summation and it is impossible to construct an other resultant of 13.7 billion lyrs. Of course one can try any other possibility.
Note, the rotational part means that there is angular momentum in equilibrium with the gravitational pull next to the linear acceleration. And secondly, the quantum mass is a rest mass and can not change reaching the light speed. Thirdly, the dimension of the angular momentum, as vector, is the same as the energy equality as s = c². Mostly the factor 2 is missing, because here one has a couple of force (force times length) for a mass centre at rest.(see note below)
Having found a specific solution, one has to go back to base. The vector composition of the generalized Compton length could be the fundamental cause for the composed Hubble distance. So the equilibrium of force
aH λ² = G mm ; λ could be a composted vector. The mediating mass is a Fermi- kind of creature, which means the magnitude of the spin vector is ½λ . In a vector diagram under an angle of 60 degrees one has a decomposition of ½λ and ½√3 λ for the resultant of λ. It is possible that the angular momentum corresponds to half spin and the part of ½√3 is unknown to us in particle physics, but this distinction is not coming out of the scaling procedure which uses λ entirely.
To analyse above twist of nature, one has to make a brilliant deduction, for there is no play in the scaling exercise between the equivalence of the accelerated Compton length and the normal definition of the Compton length. Going a step further, the half spin definition according to the special relativity theory and likewise for the Compton length is valid for non accelerated space, while here it is assumed that every where space is accelerated. So the conclusion is revealing, that in one theory for equal length the definition can be completely different with regard to the other.
This makes that the angular momentum of ½√3 λ in the theory of acceleration and it makes the half spin in the general relativity theory and the linear part of ½λ of the theory of acceleration is conform the length of λ in GRT or the special relativity theory. The resultant of the generalized accelerated Compton length is composed of ½λ- linear and ½√3λ- angular based on an angel of 60 degrees. It is an extraordinary and completely unexpected result, which proves the above deduction of the composed Hubble distance.
The final conclusion and confirmation of this exercise in scaling of
chapter 2 is exceptional. The cosmology of the accelerated expanding universe of two opposing universes in force equilibrium dragging against gravity generated by both, is absolute equivalent to a cosmology of one universe based on the general relativity theory with critical density accordingly. However the world physics for both models is completely different.
Note that the dimension of energy for the angular momentum by a couple of force, has a factor 2 which is not explained. It will not be proven but the driving couple of the spinning Compton vector spends half its energy to time and the other part in our spatial space. See, The indirect proof of the hidden energy to time
In summary the complete vector relations for Hubble’s distance.
a1 s1 = ½ v1² , a2 s2 = ½ v2² , v1² + v2² = c² ,
aH = √ (a1² + a2²) , RH = √ (s1² + s2²)
a1 a2 = s1 s2 = v1 v2 = 0 ;
the criterion of the right angles of the vector components.
The Lorenz factor of mH is not excessive: γ = 1 / √ (1 - v²/ c²) with
v1 / c = ¼ , v2 / c = ¾ giving γmax = √ (7 / 16) = 1.512 .
Although the Lorenz factor is not excessive, a theoretical treatment of the relative acceleration (see relative acceleration) from the relativistic point of view based on mH, is a necessity, but it is beyond the competence of the author, for it appeals to the highly mathematical procedures in theoretical physics. See for further remarks on mH, the indirect proof of the hidden energy to time.
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