In the following sections the physics of the Einstein gravitons in relation to N-law of gravity and the section about the all around rotation of empty space, one makes use of the vector properties of the Einstein gravitons. The gravitons as a group are mostly not relativistic under normal conditions around a star. So the two components of the graviton, the linear and the precession component can be treated as separated, because of the right angle between them. In Newton’s law of gravity this separation for the components of the gravitons can be retraced.
Firstly, in a straight fall of a mass point to the centre of gravity, then the precession component is excluded. Secondly, for the circular path of a mass point around this centre, in which the linear component of acceleration component is constant along the circumference and compensated by the centripetal force. The mass for the graviton follows from m(R) = h / (Rlin(R) c ).
In the sections The infinity transformation for…, and the relaxation condition for the gravitons the equality between the mass (rest mass) of a matter body and the product of the number of the gravitons and their local energy, the mass at a certain radius, is used without questioning the ratio for the atomic inertia mass and mediating mass of the gravitons. Apparently, these not careful ad hoc suppositions are allowed because very often one deals in ratios for the dimensions, but it is always possible to make systematic errors.
Depending on the subject discussed in a section one of the components is not considered. In case of the section, The all around rotation of empty space the part for the linear component is not considered and implicitly used in the calculation.
Discussion
First it was thought that the physics behind the sun wheel drive or the drive of the gravitons with respect to self rotation of the sun was complicated, but it turned out to be reasonable simple once the proper relation of the gravitons was understood.
The basic relation behind the mechanism of the sun wheel is the angular momentum of the gravitons.
∫ b(R) dR = h ∫ {λ/R}power 3/2 dR = - 2h{λ(pow 3/2) / √Ro} (1 )
However, let us first begin with some discussion about the physics of the drive. In fact it is very Newtonian, because the gravitons have two components in different directions with their interaction to matter one expects the mechanism of action to reaction, which means in some way the gravitons are involved in the generation of inertia spin, meaning the rotation of the sun matter around an axis.
The first remark is that the rotation axis is always perpendicular to the plane of the eclipse of the planets for reasons understood better at the end of this section.
The second remark is that the inertia spin can only be generated as a consequence of the loss of energy or the original rest mass of the atoms in the sun. It is assumed that all conversion at fusion of the atomic processes is only releasing heat dissipating energy and not gravitons. In the reaction to the dissipation also inertia losses are compensated by gravitons due to the conservation of angular momentum of the gravitons and in the reaction to this the angular momentum or the self rotation of the sun matter increases. The statement is that without fusion burning no additional angular momentum for inertia spin of the sun matter can be generated.
At first instance the conservation law due to fusion seems to be violated. However one should realize that overall together with the gravitons, the increase of the inertia spin, one conserves angular momentum.
The proof of above remarks is relative simple in terms of mathematics, because of the simple relation of (1), but it is not easy to understand why without fusion no inertia spin is generated, because still the gravitons interact with a macroscopic matter surface at an angle of 60° . The main reason is explained in the all around rotation of empty space. An additional point is that if the mass of the sun does not change then the force balance of equilibrium including the precession component of the gravitons to the infinitely faraway, does not change. Everything stays as it is, stationary, a perpetual lossless process of cycling gravitons.
If the mass of the universe does not change then the stored angular momentum of the gravitons does not change. Empty space rotates and expands accelerated, certainly in the before discussed sense of quantum mechanics, but it does not arise to a rotation of matter in a star and in grander scale the same is valid for the galaxies provided no matter is converted or generated. It seems one has to make a distinction between the primordial angular momentum stored in empty space from the onset onwards and the angular momentum attained later on due to whatever valid process of matter creation or dissipation. It seems that the driving mechanism for the rotation of the galaxies is the generation of matter via the super massive black mirrors in their centers. See the onset of the relativistic membrane theory.
Note, an omission in this discussion is made, because the product of the number of gravitons and the angular momentum was not considered. Here is of not much consequence and remedied in the following section.
The actual calculation
The frame work for the calculation is already discussed in the section of the all-around rotation etc. and is given by rel.(1) for the gravitons. The expression for the angular velocity under the condition of gravity is:
ω²(R) = λ c²/ R³ (2)
The ratio between the actual material ω , inertia spin, of the sun’s body and that of the gravitons of rel.(2) , determines how close the outer surface of the sun matter follows the ω of the gravitons. It is the first step to be calculated. Further determine the angular momentum of the sun as a solid body. It is:
b = M Rg² ω (3)
With Rg² = 0.4 Ro² . Ro is the radius of the surface and Rg is the gyration radius of the inertia momentum of all the mass concentrated at a point of the distance Rg from the rotation axis.
M = 2.0 10exp(+30) kg , Ro = 7.0 10exp(+8) m , ωav = 25.4 days and
λ = 1.5 10exp(+3) m .
ωav = 2π * 25.4 / (24*3600) = 1.85 10exp(-3) rad
ωg = 6.274 10exp(-4) rad , acc. to rel. (2).
The ratio ωav / ωg = 2.94 . If Rg = 4.43 10exp(+8) m due to the factor √ 0.4 of Ro then ωg = 1.25 exp(-3) rad . The ratio becomes ωav / ωg = 1.99
The remarkable thing is that ωav , the solar spin is far greater than the graviton spin at Ro. From the classic point of view this can be explained, because the initial angular momentum due to conservation of this momentum increases the rotation frequency due to reduction in the solar radius, which in turn is caused by loss of energy by fusion, not inertia loss. In terms of the graviton drive with loss of angular momentum it is not that strange. The solar spin increases because the gravitons carry over angular momentum. So the solar spin can be higher than the graviton frequency.
Making the supposition that the factor 1/0.4 =2.5 to explain the discrepancy for the ratios 2.94 and 1.99 seems too easy. One has to dig deeper and ask the question, what happens to the change of the matter spin due to loss of rest mass in the sun?
The total differential of the momentum for the rel. (3) becomes:
db = M R² ω {dM/M + 2dR/R} (4)
Secondly from rel.(2) : ω = c √λ/ R(pow..3/2) and dλ / λ = dM / M
For per definition: λc² = G M
and with dω = 0 (constant ω), one finds similarly as for rel.(4):
dR / R = 1/3 dM/M (5)
Substitution of (5) into (4):
db = ω M R² * 5/3 * (dM / M)
The increase of energy in spin is : (dω =0):
dW = ω db = ω² R² 5/3 dM
So the factor 5/3 is not related to the square power of ω but to a single ω. The ωav of 1.85 10exp(-3) rad converts to the ωg of 6.274 10exp(-4) rad by:
ωg = 1.85 10exp(-3) * √0 .4 / (5/3) = 7.0 10exp(-4) rad
The error is 12% may be caused by the choice of the model that the solar matter behaves like a solid instead of a liquid. Although the dR is not zero because the condition of dω =0 was used.
The constrained of dR = 0 points to a solid body with no change in shape, but the matter density changes due to dM-loss. It is not realistic. One needs to know the time dependence of ω(t) and over a period of billion of years it is unknown, because previous theories never considered graviton loss in relation to inertia. The present-day algorithms for computers consider solely the nuclear synthesis in relation to electromagnetic loss and particle loss.
Similarly with the constrained of db = 0 , the conservation of angular momentum it is not realistic to assess, because of the time dependence of ω(t) is unknown.
In conclusion, it is amazing that for constant angular velocity the result is so close to reality. It describes the drive reasonable well, otherwise it was impossible to find the same ωav of 25.4 days. The next section treats a number of case studies in a systematic manner for this spectacular mechanism of star drive by the gravitons.
THE CASE STUDIES OF THE SUN WHEEL DRIVE
Introduction
The mechanism of the graviton drive is complex, much more than was anticipated. Only when the end result of the analysis in the section of the infinity transformation and the paradox for the gravitons was understood the relaxation model became clear.
The law of the conservation of the number of gravitons with respect to the angular momentum of the single graviton group split the model for the graviton drive in two parts.
1.The situation for the equation in which the number of gravitons is constant.
2.The situation of short circuit or relaxation, because the condition of quantum relaxation is reached.
In the case that the number of gravitons is constant, the time constants could be determined for the cases which at first sight gave the most important results. The analyses were not exhausted and much more thorough checks are needed.
In the short circuit situation the model equations are different and are given in the last paragraph.
The model equations for a constant number of gravitons.
Here, such things as how the equations behave under conservation of energy or conservation of angular momentum are considered. All these questions can be resolved in Newtonian sense using the insight given by Einstein’s gravitons.
The graviton equation for Newton’s law of gravity has a total differential.
R² = λ Rlin(R)
Giving the total differential:
2 dR /R = dM / M + dRlin / Rlin (1)
Where dλ / λ = dM / M from λ c² = G M .
If dRlin = λ , as is the condition considered in the previous sections then by substitution:
dR /R = ½ dM / M + ½ λ² / R²
λ² / R² is of higher order and always smaller than dR / R ; so considering this outcome, it is determined that the total differential of rel. (1) does not play a role. A reminder seems to be necessary. It is confusing but if λ is considered in relation to R then it is as a low energy graviton, while dλ / λ is always related to the mass M .
The graviton equation for the integrated angular momentum is:
b1 = ∫ b(R) dR = -2 h λ(pow 3/2) / √R
The total differential is:
db1 / b1 = 3/2 dM / M - ½ dR / R (2)
The minus sign in the equation for b1 does not return in rel. (2). The polarity sign in a total differential means b1 is proportional to M to the power of 3/2 or (negative) inverse proportional to the square root of R. It is useful in reading the differential after manipulating a whole set of differentials.
The second graviton equation corresponding to the direct orbital momentum. It is:
b2 = h (λ / R)(pow 3/2) (3a)
Giving the total differential:
(db2 / b2)g = 3/2 dM / M – 3/2 dR / R (3b)
This equation should correspond to the angular momentum of a mass point in inertia is:
bs = M Rg² ωs (4a)
This expression is the same for the self rotation of the sun, where R can be the gyration radius if the inertia momentum of the solid body. The proof that the product of the number of gravitons NR and b2 for rel. (3a) is the same as in rel. (4a), is given in the section of the infinity transformation and the paradox for the gravitons. By not taking into account the number of gravitons in the total differential of rel.(3b), then NR is constant.
If the gyration radius coincides with the surface radius of the sun then
bs = M R² ω
with the total differential:
(db2 / b2)s = dM / M + 2dR / R + dω / ω (4b)
The more elaborate expression with Rg and ωs accounted for in the total differential, is given in the last paragraph.
The energy of the angular momentum for matter spin, not used for the gravitons is:
W = ½ M R² ω²
having the total differential:
dW / W = (db2 / b2)s + dω / ω (5)
Only the energy conservation of the material spin is understood. The conservation of energy for rel. (2) and (3a) is excluded and it is not that important compared to the conservation of angular momentum for these relations.
The expression for the angular velocity of a mass point in circular orbit around a gravity centre is:
ω² = λ c² / R³
Having a total differential of:
dω / ω = ½ dM /M - 3/2 dR / R (6)
Note ω is implicitly used in the rel. (2) and (3).
The analyses in the following paragraph consider a number of conditions for this coupled system of first order differential equations.
Applications.
Case 1.
The classic case for a orbiting mass or the self rotation of a mass in a gravity field is given by:
(db2 / b2)s = dM / M + 2dRs / Rs + dωs / ωs (7)
dωg / ωg = ½ dM /M - 3/2 dRg / Rg (8)
In fact the spin radius is independent of the gyration radius. For a planet or orbiting mass Rs = Rg and ωs = ωg , making the rel. (7) and (8) :
(db2 / b2)s = 3/2 dM / M + ½ dRg / Rg (9a)
There is no coupled system. It is a single differential equation.
The expression (9a) is used in the following more simple cases, representing the self rotation of rel.(4a) .
For the orbital momentum of rel.(7) then dM = 0, no change of matter in orbit:
(db2 / b2)s = ½ dM / M + ½ dRg / Rg (9b)
Case 2.
It shows the three coupled equations for the angular momentum. One is for the inertia and the other two are for the gravitons. Although rel. (11) is the differentiated form of rel. (12) .
(db2 / b2)s = 3/2 dM / M + ½ dRg / Rg (10)
(db2 / b2)g = 3/2 dM / M – 3/2 dR / R (11)
db1/b1 = 3/2 dM/M - ½ dR /R (12)
Momentum conservation might be determined by :
db1 + (db2)s + (db2)g = 0
This is not allowed because of equation (12).
Case 3
The next severe step is db1 = 0 or (db2)s + (db2)g = 0. For these the condition dR/ R = 3dM / M is valid in the rel. (10) , (11) and (12).
This condition for dR is not allowed as will be shown in the paragraph for the second order total differential.
Case 4.
(db2 / b2)s +(db2 / b2)g ≤ db1 / b1 or (db2 / b2)g ≤ (db1 / b1)
The combination has always the same sense of polarity as db1 . The sum should always the smallest due to the polarity sign (db2 / b2)s or
(db2 / b2)g .
If (db2)s = 0 then (db2)g ≤ db1 and if (db2)g = 0 then (db2)s can be greater than db1. The condition that (db2)s or the sum of both can be greater than db1 is not a stable situation. Only for some time this condition can be maintained until the quantum relaxation condition is reached. Case 4 shows the validity for the relaxation mechanism in a dissipating mass system.
Case 5.
(db2)s = 0 gives: Then dR / R = - 3dM / M
(db2 / b2)g = 6 dM / M = -2dR / R (14a)
db1 / b1 = 3dM / M = - dR/ R (14b)
It looks highly unstable because rel. (14a) is always greater than (14b) .
Case 6.
The energy conservation of inertia.
dW / W = (db2 / b2)s + dω / ω = 2dM / M - dR / R (15)
This by substitution of rel. (6) in rel. (9a).
With dW = 0 follows dR / R = 2dM / M
(db2 / b2)s = 5/2 dM / M = 5/4 dR / R ;
(db2 / b2)g = 3/2 dM / M = 3/4 dR / R ;
db1 / b1 = ½ dM / M = 1/4 dR / R .
Again (db2) is greater than db1 and energy conservation cannot be a very stable condition.
Case 7
(db2 / b2)g = 0 then follows; dR / R = dM / M .
(db2 / b2)s = 2dM / M = 2dR / R
db1 / b1 = dM / M = dR / R
Apparently, (db2)s > db1 . (b2)s seems possible because of the condition in the total differential of the second order, but the condition cannot be maintained. In the next case one find the perpetual condition for the planets.
Case 8.
Take equation (9b) in combination with (11) and (12). Apply (db2)g = 0
Again (dM / M = dR / R) is:
(db2 / b2)s = dM / M = dR / R
db1 / b1 = dM / M = dR / R= (db2 / b2)s
Both db’s are equal. It shows that this condition can be continuous in time.
Case (9).
(dω = 0) then dM / M = 3dR / R or dR / R = 1/3 dM / M (see eq. 8).
(db2 / b2)s = 5/3 dM / M = 5dR / R
(db2 / b2)g = dM / M = 3 dR / R
db1 / b1 = 4/3 dM / M = 4 dR / R
Take the smallest combination for the sum of :
(db2 / b2)s - (db2 / b2)g = 2/3 dM / M = ½ db1 / b1
It means that one of the b2’s is inverse proportional to the other, but the sum must have the same polarity as b1 , it is still not a stable condition.
The cases for the gyration radius Rg and the angular velocity ωs .
The equation for a solid body with inertia momentum of Rs = (√0.4) Rg .
(db2 / b2)s = dM / M + 2dR / R + dωs / ωs
(db2 / b2)s = dM / M + f dR / R + dωs / ωs
With f = 1.265 and Rg = R
(db2 / b2)g = 3/2 dM / M - 2/3 dR / R
db1 / b1 = 3/2 dM / M - ½ dR /R
Applying (db2)g = 0 gives : dR / R = dM / M
Or (dω = 0) gives : dR / R = 1/3 dM / M
Elimination gives expression in which f and (dωs / ωs) do not disappear. In general case (7) and case (9) are repeated. It looks like the growth rate of the instable condition will be less severe.
Conclusion.
The general tendency for the behaviour (b2)s with respect to (db1) is understood from these exercises.
If (db2 / b2)s +(db2 / b2)g ≥ db1 / b1 or (db2 / b2)g ≥ (db1 / b1) then no
stable condition exists and a relaxation mechanism is the consequence.
The quantum condition for (dR / R) is determined in the section for the relaxation condition for the gravitons and because the equality for (dR / R = dM / M or whatever factor in front of dM / M) it leads to an assessment of the time constants for some of the possible modes of relaxation.
The scrutiny of the total second order differential of b1 .
The second order is careful differentiation of the relation
b1 = -2h λ(pow 3/2) / √R
The result is not too bad for some analyses.
d²b1 / b1 = 3/2 d²M / M - ½ d²R / R + ¾ (dR / R – dM / M)² (16)
If (dM / M = dR / R) then (db1 / b1 ≠ 0)
d²b1 / b1 = 3/2 d²M / M - ½ d²R / R
which by differentiation of (dM / M = dR / R) has the outcome:
d²b1 / b1 = d²M / M = d²R / R
If (R / R = 3 dM / M) then eq,(16) is :
d²b1 / b1 = 3/2 d²M / M - ½ d²R / R + 3(dM)² / M²
With d(M dR) = d(3R dM) giving: d²R / R = 3d²M / M + 6(dM)² / M²
It follows (d²b1 / b1 = 0) . the second order of b1 is zero.
The condition (dR / R = 3dM / M) makes as well
(db1 / b1 = 0) as (d²b1 / b1 = 0) .
The short circuit condition.
For completeness sake the two conditions determining the short circuit equations are given, because these conditions are already discussed in other sections.
Equation (1):
The balance between angular momentum of the inertia and the gravitons, considered for the most simple situation without details as ωs and Rg . Giving the equality:
(b2)s = NR (b2)g
M R² ω = M √(λ R) = NR h (λ / R)(pow 3/2)
See eq. (3a) and (4a). Here (NR = M / mR) and the number is not constant any longer.
So M drops out of the equality and is replaced by mR . It leads to the section of The relaxation condition for the gravitons.
A super visual inspection from the previous paragraph might give the impression that higher excitations than dR = λ, such as dR = 3λ or less dR = ?λ are possible. None of it is valid, because of the discussion in the mentioned section. The confusion consists that terms of (3dR / R), etc. determine the time constants and not the bandwidth of the base line decay.
Equation (2):
The other condition is given by:
NR b1 = constant
For the conservation of angular momentum, here the integrated momentum, the product of the number of gravitons and that of the single graviton group is important. It was an oversight at the beginning of this exercise that product was not understood and so one finds some traces of it in the previous paragraph, without diverting the essence of that paragraph too much. The proof is:
dNR / NR = - db1 / b1 = -(3/2 dM / M - ½ dR / R)
The inverse proportionality between NR & b1 is determined by the differentiation of the above product. The outcome, the negative polarity sign, results in the explanation why the relaxation oscillation continues under loss of mass in which the bandwidth is determined by ± ½λ. See the figure. Without the conservation of the gravitons driving the inertia by means of b1 the oscillation is not possible.
Notes
The classic conservation law of angular momentum of Newton is apparently valid for the gravitons. Rotation of a material body cannot begin without an initial presence of angular momentum. A angular momentum can not be created from nothing. So far it has not been proven that the loss of inertia in energy generates an angular momentum of the gravitons determining a gravitational field. The other way around has been proven with this theory of the sun wheel.
So without self rotation initially, it should not be possible to start the turning of the sun around an axis, which means any optional axis, a choice from infinite possibilities. In previous comments it was suggested as a possibility. In some sections of the book Physics 2, remarks like these were made and here is the proof of the opposite and the rectification.
One has to realize that the set of the three equations for the angular momentum are homogeneous to time. In other words no time delays between infinity and the radius R are allowed for the gravitons (Or the group velocities are that small that the effects of propagation are neglectible. The inhomogeneous solution of the equations might be of significance in the relativistic sun wheel drive of the super massive black hole.
THE INFINITY TRANSFORMATION AND THE PARADOX OF THE GRAVITONS
For a mass point orbiting in a central potential well fig. 1 shows the two regions that are valid for the angular momentum.
b = M R²ω (1) with ω² = λ c² / R³ (2)
giving: b = M c √ λ / √R (3)
and integrated over the R-interval to infinity for the gravitons:
b = -2NR h λ(pow3/2) / √R (4)
M is the mass of the orbiting planet and NR is the number of the gravitons needed to maintain the orbit for the mass M. Here the centripetal force is not considers explicitly, because it is implicitly used in the expression (4) for the gravitons and given in the force balance of (2).
With the energy ratio NR , coming from M / mR , also the number of gravitons at radius R , the equality of both the rel. (3) and rel. (4) can be made. mR is the mass energy of the graviton at R.
mR c R = -2h λ (5)
mR and R are constant for a circular orbit. Dimension wise the rel. (5) is not correct, for it should have been realized that only the derivative of rel. (4) , the gradient db/ dR can be matched to rel.(3). By plugging in the transformation:
T(R) = CR / R
Rel.(5) can be corrected:
½ (mR c / h) R = - CR λ / R (6)
With the definition for the gravitons: Rlin = h / (mR c)
rel. (6) becomes:
½ R² / Rlin = - CR / R (7)
where R² = λ Rlin(R) satisfies the rel. (7)
with CR = - ½
Because the transformation
T(R) = (1/ b) db/dR = - ½ (1//R) (8)
becomes after integration:
ln(b1/b2) = ln (√(R2/R1)
or b1 √R1 = b2 √R2
If R2 >> R1 then b2 << b1 .It is a well behaved transformation.
For R ? ∞ then T ? 0 or b ? 0 . As fig.1 shows, T(R) is the infinity transformation between the classic- and the overall graviton angular momentum.
For an elliptic orbit or parabolic trajectory the expression of T(R) = -½ (1/ R)
becomes more complex T = T(R, φ) having a angle φ in polar coordinates and mR (R) and NR (R) are not constants any longer.
It must be possible to determine the transformation of each orbiting mass in a many body problem and analyze the behavior for the relaxation after many revolutions of all the different masses by using Lagrange principle. One might show that the lowest energy state for the many body problem is a system of nested circular orbits of the planets.
A more wishful application of rel.(8) is the relaxation procedure due to fusion burning in the central star. All these complexities are left to the interested bright young students of astronomy or physics.
Conclusion with this derivation, it is proven that the whole procedure for the theory of the gravitons according to its multiple definition, is a valid method to express the quantum mechanics of empty space in gravitons.
Comments
The infinity transformation is tricky, because from the relativistic point of view the gravitons reach the c-velocity at R ? ∞ . Now the reason to short circuit the discussion, is to state that the gravitons all relaxate to Hubble’s radius, which is the quantum parameter fulfilling the condition 2mH Mtot = mpl ²
The second remark is that due to the 60° angle between the components of the graviton , the relativistic graviton increases the angle (going 90°) in favour of the tangential component, while the linear component can hardly increase due to the hyperbolic motion. It is discussed a bit more in the section of the onset of the membrane theory of relativistic gravitons.
The paradox between the angular momentum classically and gravitationally for the same mass.
Nothing gives so much pleasure as finally to write down a paradox which was not understood for some time.
The classic ratio of the angular momentum between two radii for the same mass is:
B1 / B2 = (M R1² ω1) / (M R2² ω2) = √ (R1 / R2) (1)
So B is proportional to the square root of R.
It is never possible to make the ratio equal to one if R1 ≠ R2 .
Similarly the ratio for the graviton momentum becomes:
B1 / B2 = (NR1 b1) / (NR2 b2) = √ (R1 / R2) (2)
Using b = h (λ / R)(pow 3/2) ; NR = M / mR ; mR = 1 / Rlin and
R² = λ Rlin then (1 / mR) = R² / λ .
The amazing thing for the paradox is that in the relation (2) the B-ratio might be have the unit of one, then showing the conservation of momentum.
B1 / B2 = (NR1 b1) / (NR2 b2) = 1
b1 / b2 = NR2 / NR1 (3)
then b is inverse proportional to NR .
How to prove this? Take the expression for the angular momentum
B = NR b
Take the total differential.
dB = dNR b + db NR
or
dB / B = dNR / NR + db / b (4)
For dB = 0
dNR / NR = - db / b
In other words for the exchange of gravitons between two states b1 and b2 under de condition of conservation of momentum, then the number of gravitons is inverse proportional to the single momentum of a graviton group. However the initial value of NR cannot be the reciproke of b, because NR is dimensionless.
In the situation of an elliptical orbit under the condition of conservation of the classic momentum then the number of gravitons between the different positions of the radii the gravitons are instantaneous exchanged and adapted according rel. (4). The relation (4) is the law of the conservation of the number of gravitons: The product of b * NR = constant .
This law is used in the quantum shift for the precession of the polar axis of Earth. Without the exchange mechanism for the number of gravitons the coupling between the Earth rotation and the orbital momentum could not be proven.
It is very important to understand the inverse proportionality between the number of gravitons in a position and that of the belonging angular momentum, for this property is used between left and right hand side of the equality in b-momentum of the precession shift.
See The precession shift of the polar axis of Earth.
Note: also the integrated momentum b can be subjected to conservation;
(NR *∫ b dR) = constant. This is the driving mechanism for the oscillation of the solar spin relaxation under the condition of fusion.
THE RELAXATION CONDITION FOR THE GRAVITONS
Following the derivation about the section of the infinity transformation and the paradox, one can take the classic expression of the angular momentum in which the ω, the circular velocity, is already substituted.
Then b becomes: b = M c √λ √R (1)
Instead of using the integrated form of angular momentum of the graviton, the normal or differentiated relation is used, because it is dimension wise correct.
b = NR h (λ / R)pow. 3/2 (2)
NR and M are respectively the number of the gravitons and the mass, in which again mR = M / NR , the graviton energy at radius R. Substitute in rel. (1) and (2) and take the equality.
(mR c / h) √λ√R = (λ / R)pow. 3/2
(mR c / h) R² = λ (3)
mR and R are the only variables with respect to the central mass λ and h is the uncertainty constant. The rel. (3) is the basic tool to understand the physics behind relaxation mechanism of the graviton drive.
In first order it is allowed to write mR = dM = M1 – M2 and R² converts to 2R dR, giving:
(mR c / h) 2R dR = λ (4)
If time is considered and fusion included giving rise to a decrease in λ (Mstar) then the relation (1) and (2) divert from each other. Of course if
mR = dM = M1 – M2 = 0 then the dimensions in rel. (4) are not correct any longer. dM = M1 – M2 is the difference in inertia while one looks for the energy difference for the gravitons. For now as a tool, it has to do. In the section, the case studies of the sun wheel drive the rel. (4) is applied.
If time and fusion are considered the diversion of relations (1)and (2) is expressed in rel. (4). The gravitons stored outside the radius R up to infinity contribute to the material orbit at R which can vary freely in radial direction. So empty space stores the gravitons which cannot disappear instantaneously with a change of dM , then causing a relaxation phenomena. The relaxation criterion is expressed in rel. (4) and it should be dimensionally correct.
2 (dM) c dR R = λ h (5)
in a slightly different notation.
(dM) c/ h has the dimension of the reciproke length of a graviton by definition.
Two situations can be considered:
Case 1. If dM = mλ = h / (λc) then rel. (5) becomes;
2 dR R = λ²
or dR = ½λ (λ/R) or dR / R = λ² / (2R²)
With the fundamental graviton λ , dR is always of a higher order than dM , because λ/R is always small. This criterion seems not correct.
Case 2. Other wise in reference to the section A non linear harmonic
string theory etc. , dR could be equal to λ or ½λ . The rel. (4) or (5) becomes:
(dM c/ h) R = ½
dM c/ h is the mass of the graviton with a length Rlin = ½R, but reciprocal to its mass.
Here something similar as the Bohr’s condition for the electron of the H-atom. In other words the stored energy of the gravitons can relaxate if
Rlin = ½R for the orbit of radius R. So dR = λ and Rlin = ½R are related.
In conclusion, if dR / R is known then dM / M can be determined, for it is allowed to write for dR , λ or ½λ .This is the band width for the relaxation, while λ in first instance gives the onset for the calculation of the time constants. For the band width no excited states of 2λ or 3λ are allowed. See discussion in a nonlinear harmonic string theory for Einstein gravitons.
For Earth the orbit radius is 1.5 10exp(+11) m , the radius of the Sun surface is 7.0 10exp(+8) m , and the fundamental graviton mass is
1.5 10exp(+3) m. The orbital ratio is 1.0 10exp(-8) and the surface ratio of λ/R is 1.5 10exp(+3) / 7.0 10exp(+8) = 2.1 10exp(-6). Knowing the star mass, dM, the mass per second can be determined giving the time constants for the relaxation. These constants might be of significance in the correlation of the geological periods for Earth.
THE TIME CONSTANTS OF THE SUN WHEEL DRIVE COMPARED TO THE GEOLOGICAL TIMESCALE
The overall loss of energy of the Sun one can look up in the Encyclopedia Britannica or via Internet . The solar mass is M = 2.0 10exp(+30) kg and the loss of energy dW = 3.86 10exp(+33) erg /sec, which converts to
3.86 10exp(+26) joule /sec (watt). For comparison the loss of e.m. radiation is 0.966 10exp(+25) watt. To find the loss of mass per second divide by c² , making dW = 4.29 10exp(+9) kg /sec.
From the previous section, the relaxation condition etc. , determine the quantum shift in dR =λ or ±½ λ . It is the band width for departure of the angular momentum db = 0 or also (db / b = λ / R) . Higher excitations as dR = 3λ are not allowed. As determined in that section, db = 0 has the condition (dM / M = λ / R). Many more possibilities of condition were found in the section The case studies of the sun wheel drive etc. , which are not evaluated here.
With the radius for the sun surface R = 7.0 10exp(+8) m and the orbit radius of Earth R = 1.5 10exp(+11) m and the fundamental
λ = 1.5 10exp(+3) m, the following table is calculated.
If the Sun had a diameter of the Earth orbit then the relaxation constant τ should have been 148 000 years. For the present–day diameter of the Sun the time constants are between 10 or 30 million years. One should realize the band width of relaxation for the angular momentum of the base line has not changed for the conditions dω = 0 or db = 0.
For instance the time constant of the base line is {2.0 10exp(+30) / 4.29 exp(+9)} in seconds and then divided by 3.15 10exp(+7) sec to find the time constant in years. It makes the base time constant much greater than the age of our cosmos.
Further it is assumed that the overall loss of energy does not contain conversion energy into gravitons and also that the fusion rate is constant over time or varies only slightly over the period of 500 million years. As it turns out the ratio determining (dM /M) only contains the mass energy and it does not contain a factor in which the ratio between mediating- inertia mass plays a role. In the beginning of this exercise it was thought this factor might to be included.
Again turn to the Encyclopedia Britannica and look up the paragraph about the geological time scale for the major geologic periods in the evolution of Earth. Apparently a period of 4.5 billion years, the overall life of the solar system, is divided in two relevant periods, the pre-.Cambrian of 4.03 billion years and the Cambrian of 505 million years. The Cambrian consists of 10 distinguished periods of an average of 50.5 million years. These are not regular and vary between 10 and 71 million years.
The time constants in the table for the to days- radius of the Sun fit reasonable good for a logarithmic behavior of the relaxation of the periods in the geological timescale of 20 to 65 million years.
The most recent period, named the Cenozoic, of 65 million years is subdivided in 7 periods with an average of 9.3 million years. Some of these last 19 million and the most recent lasts 2.5 million years.
However it is clear from the table that a proper fit to the geologic events here on Earth requires a detailed computer algorithm for fusion burning and matter convection. A period of 500 million years is a long time for the sun wheel physics.
First it was thought that the period of 150 thousand years was suppressed if the geological time scale and it should have played a role for the precession shifts of the polar axis of Earth. Until it was realized that the relaxation time constant gave the Sun’s diameter as the Earth orbit. Nonetheless 17 periods of relaxations for the Sun drive might be deduced from the geologic events here on Earth. Although premature it is not necessary concluded that every geologic period corresponded to a relaxation shift of the Sun’s drive. What can be concluded without too much doubt is that every relaxation of the Sun’s drive correlates with a polar shift of Earth, especially for the Cenozoic period. Such polar shift should have caused major climatic catastrophe on Earth.
It is very ironic to realize that three and a half centuries Newton’s physics including Einstein’s, forbade a positional shift of the rotation axis of Earth and only the quantum physics of the Einstein gravitons put the spanner in the works.
Comments
The author is aware the geologic scientific studies of the geologist Maloof in his article for the geologic soc. of am. Bulletin; sept. / okt. 2006; the true polar wander. Although this study is based on a fixed polar axis for the solid core of Earth, it might be certainly possible to adapt his observations to conform the theory of the precession shift for the polar axis. At least it appears to be a first step in the right direction.
Another recent article is from P.Hoffman, The snow ball Earth, which has the hypothesis that once the present-day equator region were located in cold areas, but not necessary those of the present-day poles. It might suggest that the polar axis was tilted to the plane of the eclipse. The last is a jump of conclusion by this author.