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The significance of rational numbers Q+and the
coordinate system P(2π) for the energy
relationships in physics
Helmut Christian Schmidt
March 9, 2024
Student of
Ludwig-Maximilians-University Munich
Faculty of Physics
80539 Munich Germany
Helmut.Schmidt@campus.lmu.de
ORCID: 0000-0001-7072-204X
1
Abstract1
Philosophically it can be assumed that we can only recognize relationships.2
Thus, the cosmos can be understood as a set of rational numbers Q. Observa-3
tions from the past are the subset Q+for physics. For energy relations, a system4
of 3 objects, each with 3 spatial coordinates and common time, is sufficient. The5
quantum information (QI) from these 10 independent parameters results in the6
polynomial P(2). Each measurement results from coincidences of revolutions of7
objects relative to the firmament with the smallest unit π. The smallest ratios8
of the particles result from the 3 spatial dimensions in powers of πand can be9
expressed as 3 neutrinos ντ=π νµ= 1 νe=π−1. A transformation of P(2)10
into polynomials P(2π) provides the ratios of the energies from the number of11
ur-particles (C.F. von Weizs¨acker). P(2π) is compatible with quantum theory12
(QT) and general relativity (GR). With the minimum required 10 parameters,13
the neutron results in:14
15
Ep= (2π)4+ (2π)3+ (2π)2
16
Ee=−((2π)1+ (2π)0+ (2π)−1)17
18
The area between the objects with Epand Eeis the curved surface19
(Christoffel symbol) of the neutral, solid measuring device or the Earth.20
21
Emeasurement = 2(2π)−2+ 2(2π)−4−2(2π)−6
22
Etime = 6(2π)−8
23
mNeutr on/me=Ep+Ee+Emeasurement +Etime = 1838.683661124
measured : 1838.68366173(89)me.25
26
A photon is made up of neutrinos or can be viewed as two entangled electrons27
e−and e+. The charge is an energy ratio EC.28
EC=−π1+ 2π−1−π−3+ 2π−5−π−7+π−9−π−12
29
mproton =mneutron +ECme= 1836.15267363 me
30
1/(GNh) describes the curvature of the Earth from the number of ur-particles31
NEarth in relation to the speed of light c5and time.32
ρ/dt =√π4−π2−π−1−π−3/s33
hGNc5s9/m10 ρ/dt = 0,99999134
Further calculations such as mmuon/meand the planetary system show the35
advantages of P(2π). With an isotropic distribution of the ur-particles in a36
photon, H0 results in37
H0≈√π/c2s2/m2/ρ/s = 2.13 10−18 /s = 65.1km/s/Mpc38
and a photon energy with spatial curvature 1/ρ3m.39
λ≈√π/2/ρ3m= 1.0885mm40
2
1 Introduction41
The quantum theory of ur-objects or ur-alternatives proposed by C. F. von42
Weizs¨acker is interpreted as a quantum theory (QT) of information and is based43
on a two-dimensional Hilbert space with the universal symmetry group SU(2)44
or a bit [1,2]. In the following, an distinction is made between ur-particles,45
dimensions, objects, and energy (Figure 1). The number of ur-particles add up46
to s according to the objects i and the different dimensions (t, φ, r, θ).47
48
Figure 1: Classification into alternatives49
50
Since Newton, physics has been based on forces and the center of gravity.51
An alternative to the center of gravity is to consider a system of 3 objects, each52
with 3 spatial coordinates and the time of an observer. This corresponds to53
Einstein’s idea of explaining matter in terms of the metrics of spacetime. The54
10 independent equations in general relativity (GR) represent the parameters of55
a system [3,4]. Like the Planck time tP lanck =pGNh/c5, quantum theory can56
be combined with the metric via the number of ur-particles. With 1/(GNh), kg57
is eliminated and the Earth is described in terms of NEarth relative to a photon58
with c5, purely based on the dimensions with the units m and s.59
Bohr postulated the angular momentum quanta with L=nh/(2π) [5]. With the60
assumption of a finite number of ur-particles in an object, the energy ratios can61
also be formulated as polynomials P(2π) instead of the Schr¨odinger equation62
to establish a connection to epicycles in the planetary system. The following63
considerations arise exclusively from rational relationships Qbetween objects.64
Physics is knowledge from the past, i.e. half of the numbers without the zero65
Q+. A central point is the introduction of the most efficient coordinate system66
for metrics of the cosmos from polynomials P(2π) for neutral objects and P(π)67
for charged objects.68
3
2 Physics before the Standard Model69
2.1 Cosmos70
The cosmos consists of ratios and therefore rational numbers Q. At the center71
of every object there is a particle n= 1 from which all objects can be built.72
The quantum information can be formulated as a polynomial P(2).73
2.2 Physics: Q+
74
Only the ratios of spatial coordinates from the past can be measured.75
−t(n+ 1) <−t(n)<−t(0) = 0 t∈Q+n∈N+(2.1)76
Physics arise from the comparison between two objects and an observer. A77
system consists of at least 3 visible or invisible objects Oi, each with 3 spatial78
coordinates. Normalization with meters and seconds occurs on the Earth’s79
surface as object O0.80
Oii∈ {..., 0,1,2, ...}(2.2)81
The numbers of ur-particles with dimension d qd,i ∈Zrefers to the local82
spatial coordinates and time:83
qt,i qφ,i qr,i qθ,i
84
The number sof ur-particles begins at the center of the system.85
si=qt,i +qφ,i +qr,i +qθ,i s=Pd,i qd,i (2.3)86
At the end of qφ,i the object is complete and corresponds to its surface area87
and rest mass. The distributions of qφ,i,qr,i and qθ,i lead to different states88
(elementary particles, atoms, and molecules) with different densities and orbits89
(planets, lunar resonances, and eccentric orbits). For example, in the interior90
of the sun, due to nuclear fusion, the neutrino production follows the trend91
qθ> qr(see Chapter 2.9.) and indicates a change in the number of dimensions92
depending on R(t)/RSun .93
94
The smallest units of measurement are spatial coincidences in revolutions of95
πand measure time. In the following, the coordinate system for Orbiti(s) is96
referred to as epicyclic coordinates. The dimensions t= 2, φ= 1, r= 0 and97
θ=−1 are used in the exponent.98
The coordinates qk,i in epicyle πare the metric.99
Orbiti(s) = Ei=qt,i πt+4i+qφ,iπφ+4i+qr,i πr+4i+qθ,iπθ+4i(2.4)100
The orbit corresponds to the energy. Neutral ob jects are given by polynomials101
with base 2π.102
Orbiti(s) = Ei=qt,i (2π)t+4i+qφ,i(2π)φ+4i+qr,i(2π)r+4i+qθ,i(2π)θ+4i(2.5)103
4
The time is determined by full revolutions, so the calculated energy is also an104
action. The distinction between the moving and stationary location coordinates105
of objects is insignificant for the energy ratios Ei.Er,i =qr,iπr+4iis no different106
from the circular arcs Eφ,i and Eθ,i. Thus, the radius itself is curved, which107
enables a simple formulation of the GR. In a system, the location coordinates of108
objects move in a spiral in the direction of time analogous to geodetic lines. For109
the spatial location coordinates to be independent, there cannot be an algebraic110
function between the dimensions. The energy of neutrinos and electrons can only111
be specified in connection with a system of at least 3 objects. The minimum112
energy components E are:113
Eντ=π114
Eνµ= 1115
Eνe=π−1
116
Ee−= 1 −π−1
117
Ee+=−1−π−1
118
The time qt,i reduced to a single measurement point in the system.119
2.3 Photon - Speed of Light120
A photon consists of entangled neutrinos Oiwith the world line relative to121
a larger object O0. The cohesion between the neutrinos and O0can also be122
understood as the entanglement of two electrons, e−and e+, with a minimum123
energy Eγ>0, and a binding energy Eθ,1,2=−2/π through the spin.124
Orbit1,2=Eγ= (gφ,1−gφ,2)(2π)φ+ (gr,1−gr,2)(2π)r−2πθ(2.8)125
c/f1,2/m =gφ,1−gφ,2Eφ,1,2= (gφ,1−gφ,2)(2π)φ>0126
n1,2λ1,2/m =gr,1−gr,2Er,1,2= (gr,1−gr,2)(2π)r= 0127
spin 1 = −2/π Eθ,1,2=−2/π128
c is the only required constant and describes the homogeneous density of129
the ur-particles per meter in the 4-dimensional time space n1,2λ1,2=c/f1,2.130
The energy component Er,1,2= 0 of the photon is independent of the length131
of the wave train n1,2λ1,2. The negative energy of spin 1 = −2/π indicates132
attraction in the gravitational field relative to a measuring device or earth. The133
interaction between two entangled and therefore immediately adjacent photons134
results from spin with an interaction energy Eθ,1,2=−2/π < 0. This applies to135
all entangled objects.136
5
2.4 Neutral Objects - Measuring Apparatus137
The surface of each object rotates at a frequency fi. Only a system of at138
least 3 objects provides conclusions about frequencies fiand radii qr,i.139
Orbiti(s) = Ei=qφ,i (2π)φ+4i+qr,i(2π)r+4i+qθ,i (2π)θ+4i(2.10)140
Eφ,i =c/m ti=c/m /fi=qφ,i(2π)φ+4i
141
1/f1,2= 1/f1−1/f2(2.11)142
Er,i =qr,i(2π)r+4i
143
The frequency 1/f1,2and the phase are unique relative to object O0. In a144
measuring device O0, the common time qt,0>0 is determined and can be145
represented in spatial coordinates. On the curved surface of O0, the number146
of ur-particles is standardized with meters and seconds. The number of ur-147
particles decreases gradually toward the center with each ur-particle, which can148
be interpreted as diffraction on the surface of O0. For each of the three spatial149
dimensions, there are three foci of O1and O2beneath the curved surface of O0:150
rf,1,2,φf,1,2,θf ,1,2with energies Ef,φ Ef,r Ef,θ (2.12)151
The energies of the space coordinates are summarized as Ef,space :152
Ef,space =Ef,φ +Ef ,r −Ef,θ (2.13)153
The total energy is calculated together with the combined time of the measure-154
ment.155
Ef=Ef,space +Ef,time (2.14)156
In detail, the dimensions φ, r, θ for objects O2and O1with the parameters157
λ∈ {4,3,2}and ν∈ {1,0,-1}are marked. According to the law of gravity158
F= (m1m2)r−2, F decreases with the second power of r. Similarly, Ef,space de-159
creases for neutral objects with (2π)−λ−ν. The algorithm is adopted to calculate160
stepwise Effrom high to low energies with 2 loops for O1and O2.161
for λ =φ2to θ2step −1 (2.15)
for ν =φ1to θ1step −1
if g2,λ >0then Ef,−λ−ν−1=−sgn(ν)g2,λ g1,ν (2π)−λ−ν/π attraction
if g2,λ <0then Ef,−λ−ν=−sgn(ν)g2,λ g1,ν (2π)−λ−ν/π repulsion
Ef,t =|g2,λg1,ν |(2π)−2φ2
next
next
Ef,−λ−ν−1=−sgn(ν)g2,λ g1,ν (2π)−λ−ν/π (2.16)162
replaces a Christoffel symbol from the GR, quantized with ∂d= 1/π163
Γµ
λν =gµρ(∂λgν ρ +∂νgλρ −∂ρgλν ) (2.17)164
A term g2,λ(2π)λ= 0 leads to 2 neutrinos with base π. (see Muon Chapter165
2.9). GR and the quantum field theory QFT are based on P(2π) and P(π),166
respectively.167
6
2.5. Neutron168
In the following, the objects Oiare marked with the indices p, e and N.169
The starting point for calculating the rest mass of the neutron with P(2π) is a170
comparison of two objects with a positive Epand a negative portion Ee. For171
a visible object, the energy is E=Ep−Ee>0. For a stationary object the172
derivatives are ˙qd,p = ˙qd,e = 0.173
Ep= (2π)4+ (2π)3+ (2π)2(2.18)174
A smaller object consists of electrons, and the energy175
Ee=−((2π)1+ (2π)0+ (2π)−1) (2.19)176
The axis of symmetry between the objects with energies Ep,e and ENis the177
surface of the measurement and results from 2.11, 2.14 and 2.15.178
EN= 2(2π)−2+ 2(2π)−4−2(2π)−6
179
Etime = 6(2π)−8(2.20)180
mNeutr on/me= (2π)4+ (2π)3+ (2π)2−(2π)1−(2π)0−(2π)−1+ 2(2π)−2+181
2(2π)−4−2(2π)−6+ 6(2π)−8= 1838.6836611 (2.21)182
theory : 1838.6836611memeasured : 1838.68366173(89)me[7]183
The decimal places of mneutron/mewas (2π)−8= 4 10−7, which is within184
1838.68366173(89).185
The calculation required only 10 terms, making it the most efficient186
method for determing the mN eutron/meratio. This result is unique187
because of the transcendental numbers πd.188
189
Figure 2: mN eutron /meas a polynomial P(2π)190
191
7
According to the algorithm, the two objects are separated by a parity oper-192
ator and arranged as a wave with φ,rand θ(Figure 1). The algorithm involves193
a Fourier transformation from the outer wave (6 terms) to the quantum infor-194
mation of the inner world (6 prefactors) .195
2.6 Proton196
As in QFTs, the kinetics between neutral objects are mediated by a number197
of other particles of electrons or photons with positive or negative energy. The198
electric charge should be calculated as a polynomial EC+=P(π) from the mass199
difference between the neutron and the proton. The approach for ECis part200
of Oeof the neutron and is itself a system of 3 objects (EC,1, EC,0, EC,−1). In201
EC,1there are three types of neutrinos πφ=π, πr= 1, πθ=pi−1.πrcan be202
viewed as the center of the charge, for a ”Coriolis force” of πφand πθand gives203
the minimum energy Ec,1:204
EC,1=−π1+ 2π−1<0 (2.22).205
From EC,1there is a transition to the object EC,0with a parity operator. It is206
assumed that after 4 space dimensions π−4the type of neutrinos remains the207
same νφ
−4and νθ
−4.208
Ec,0=−π−3+ 2π−5(2.23)209
The same applies in Ec,−1for two of the neutrinos νφ
−8and νθ
−8. The assump-210
tion is that the third neutrino πθabsorbs energy from the gravitational field.211
The phase changes to νµ
−12 due to a neutrino oscillation.212
Ec,−1=−pi−7+π−9−π−12 (2.24)213
Together with the neutron mass, the proton mass results:214
EC=−π1+ 2π−1−π−3+ 2π−5−π−7+π−9−π−12
215
mP roton =mNeutron +Ecme= 1836.15267363 me(2.25)216
In the neutron, Efhas several prefactors of 0. ECfills these positions ex-217
actly with powers of π(Figure 3).218
219
220
8
Figure 3: mP r oton/meas polynomial P(2π)221
222
mP roton/me= (2π)4+(2π)3+(2π)2−(2π)1−(2π)0−(2π)−1+2(2π)−2+2(2π)−4−223
2(2π)−6+6(2π)−8+(−π+2π−1−π−3+2π−5−π−7+π−9−π−12 ) = 1836.15267363224
(2.26)225
theory : 1836.15267363memeasured : 1836.15267343(11)me[7]226
The two terms π−10 and π−11 are the placeholders for valence electrons. The227
calculated proton mass corresponds to the measured value.228
The ±1/3eor ±2/3echarges of quarks are explained by the fact that the three229
objects always belong to a system.230
2.7 Muon231
For a charged particle, the energy ECis distributed between objects O1and O2
232
and measuring device O0:233
EC=−π1+ 2π−1−π−3+ 2π−5−π−7+π−9−π−12
234
The muon can be understood as consisting of 2 particles, each with a triple235
polynomial.236
Eµ,2= (2π)3−(2π)2−π1Eµ,1= (2π)0+π−1−(2π)−1(2.27)237
The powers of (2π) can be separated.238
Eµ,1,2+π0−π−1= (2π)3−(2π)2+ (2π)0−(2π)−1(2.28)239
A visible object with (2π)3>0 requires (2π)0>0 for a rest mass and gives the240
minimum energy Eµ,1,2. The decay is given by 0(2π)1= 0. E= 0(2π)4is the241
placeholder for the kinetic energies before decay. The surface of the object O0
242
marks the symmetry between E1,2=Eµ,1,2+π1−π−1and the energies E0,f
243
244
The terms E1from E2with different signs attract each other according to245
the following algorithm (2.15):246
E1,2= (2π)3−(2π)2+ (2π)0−(2π)−1(2.29)
attraction Eθ>> (2π)3(−(2π)−1)>> E0,f,1= (2π)−4/π =−2(2π)−5
E0,f,t =|2(2π)−8|
E0,1,2,−1=−(2π)2+ (2π)0(2.30)
repulsion −Er>> (−(2π)2)(2π)0>> E0,f,2= 2(2π)−2
E0,f,t =E0,f,t +|2(2π)−8|= 2(2π)−8
It is assumed that 0(2π)4leads to decay to form a neutrino or π0and 1/π247
interaction.248
E0,1,2,−2=−0(2π)4+π0
rejection 0(2π)4(−π0)>> Enu,f,3=−(2π)−4(2.31)
9
E0,1,2,−3= +0(2π)4−π−1
attraction −(2π)4(−π−1)>> Enu,f,3=−(2π)−3(2.32)
This means that there are 2 additional terms in O0with −(2π)−3and249
−(2π)−4.250
In summary the remaining mass of the muon is:251
mµ/me= (2π)3−(2π)2+ (2π)0−(2π)−1
252
+2(2π)−2−(2π)−3−(2π)−4−2(2π)−5+ 4(2π)−8
253
+EC= 206.7682833 (2.33)254
theory : 206.7682833memeasured : 206.7682830(46)me[7]255
256
257
Figure 4: mM uon/meas polynomial P(2π)258
259
In the muon, the frequency of the placeholder is 0(2π)1. The decay results260
from dividing the energies mµ/meinto 3 new objects with the kinetic energy261
Eφ.262
Eφ= (2π)3−(2π)2=Eφ,νµ+Eφ,νe+Eφ,e (2.34)263
The probabilities of the impulses and energies result from the QT.264
Eνµ=Eφ,νµ+π0+? (2.35)265
Eνe=Eφ,νe+? −π−1(2.36)266
The complete values of Eνµand Eνeresult from absorption only to form a267
three-polynomial. The following applies to the energy balance of the electron:268
Ee=Eφ,e + (2π)0−π0−(2π)−1+π−1+EC( 2.37)269
2.9. Tau270
The first particle with the factor (2π)4is the proton. The tau should there-271
fore have the factor 2(2π)4and thus indicates a particle that is composed of 4272
objects.273
10
274
Figure 5: mT au /meas polynomial P(2π)275
276
Along with EC, the first estimate is:277
mτ/me= 2(2π)4+ 2(2π)3−3(2π)2−2(2π)1−2−(2π)−1+EC=278
3477.34memeasured : 3477.23me[22] (2.38)279
2.10 Fine-strucure constant280
For a free electron, at least the placeholders are available, starting with π4. Even281
if the background for the fine structure constant is not yet known, it should be282
calculable from ratios of energies and thus with a series expansion from π.283
284
Figure 6: Fine-strucure constant as polynomial P(π)285
286
1/α =π4+π3+π2−1−π−1+π−2−π−3+π−7−π−9−2π−10 −2π−11 −2π−12 =287
137.035999107 measured : 137.035999084(21) (2.39)288
2.11 Gravitational constant – Planck constant289
The unit kg is not required in this theory. GNhis considered a common constant.290
The Planck time tp=pGNℏ/c5(2.38) describes the smallest possible time291
interval using the known laws of physics. The ratio of a photon relative to a292
neutral measuring device is c5. One interpretation of this could be that the293
11
density of the ur-particles in a 5-dimensional space around a body is constant.294
The minimum energy in the gravitational field should be in relation to a neutrino295
pi4= (πφ)3, together −pi2= (πθ)3for the explanation of the Centrifugal force.296
EG,2,1=π4+ 0π3−π2
297
As a counterpart EG,0are in 5 dimensions298
EG,0=−π4−5−pi2−5=−π−1−pi−3
299
with the sum EG
300
EG=π4−π2−π−1−π−3(2.40)301
With the common constant of h, GNand c5follows:302
hGNc5s8/m10√π4−π2−π−1−π−3= 0.999991 (2.41)303
The units meter and second must appear in this formula. The value of GN
304
is only known up to the fifth digit [7]. In this respect, result 1 can be assumed.305
h and c are already precisely defined. The only parameter that remains to306
be determined by measurement is GN. The only force that holds the world307
together is the number of ur-particles. According to Gauss’s integral theorem,308
the distribution of energy within an object is irrelevant. Only the size and309
rotation of a body defines the total energy.310
3 Planetary system311
3.1. Sun – Earth – Moon312
The Sun, Earth and Moon system are unique in the planetary system. Rel-313
ative to the Earth, the bound Moon has largely the minimum energy in angular314
velocities ( ˙φ,˙
θ). In the photon the radial component is Er,1,2= 0 (2.3). Seen315
from the Earth, the visual angle is on average the same and results in an ex-316
tremum for the energy. The system is in a basic state. For Earth, c is therefore317
easy to calculate. The Earth is described by the equatorial and polar diameter318
as well as the star rotation time. As a substitute for the number of unknown319
particles NEarth , c is the only required parameter. A photon has a longitudi-320
nal dimension lP hoton with a beginning n1and an end n1+ 1 and corresponds321
to lP hoton =lEarth on the Earth’s surface. Orthograde to the photon is the322
maximum ratio one to DEarth2(DEarth =equatorial Earth diameter)323
c m Day/DE arth2= 1/(2π) (3.1)324
This formula gives the Earth’s equatorial radius an accuracy of 489 m. Together325
with the contour line above sea level, c could also be exact (e.g., the 1000 km326
wide Congo Basin is almost 500 m opposite the Pacific).327
Sun, Earth REarth = 6356.75 km, and the bound Moon has a stable ratio of328
radii and orbits. For the three spatial dimensions, 23= 8 is the ratio between329
the rotations/orbital periods of the celestial bodies.330
RMond /(REr de +RMond )=8/(2π)=4/π RErde (4/π −1) = 1736,9km (3.2)331
12
Based on the pole diameter, the deviation is only 0.00011.332
The distance between the Moon and the Earth DMoon−Earth increases by 38.2333
mm per year:334
d/dtDMoon−E arth = 38.2mm/384400 km/year = 3.1510−18/s (3.3)335
H0 [11]: H0 = (67.8±0.9)kms−1M pc−1= 2.2181018 /s336
d/dtDMoon−E arth (1 −1/pi) = 2.14710−18/s ≈H0 (3.4)337
For now, it is speculated that the distances between celestial bodies are also the338
result of the expansion of the entire universe.339
3.2. Calculations of orbits in the planetary system340
The advantage of the solar system is that apoapsis and periapsis are directly341
observable, while in the atom some energy levels degenerate. At the center of342
the system is the sun with the focus of time tF ocus . With the assumption of343
Rsun ≈tF ocuscand normalization to Rsun = 696342 km, the orbits result in344
resonances:345
r2
apoapsis +r2
periasis =R2
sunEnrapo/periasis =Rsun √En(3.5)346
After 3 dimensions for the surface of the sun, 3 invisible dimensions follow up347
to Mercury. The first three terms for φ,r, and θresult in apoasis and periasis348
with an accuracy of approximately 1
:349
Mercury
rperiapsis = 696342kmp(2π)5−0(2π)4+ (2π)3= 69775692km
Measurement : 69.81 106km rel.Deviation = 0.0005
With 2 half-integer quantum numbers the apoapsis is
rapoapsis = 696342kmp1/2(2π)5−1/2(2π)4+ (2π)3= 46006512km
Measurement : 46,002 106km rel.Deviation = 0.0001
Like with ladder operators, orbits can be constructed iteratively.
Venus
rapoapsis = 696342kmp2(2π)5+ 3(2π)4−(2π)3= 107905705km
Measurement : 107.4128 106km rel.Deviation = 0.004
rperiapsis = 696342kmp2(2π)5+ 3(2π)4+ (2π)3= 109014662km
Measurement : 108.9088 106km rel.Deviation = 0.001
350
Table 1: Apoapsis and periapsis of Mercury and Venus (3.6)351
rV enus/rM ercury = 6123.80/2448.57 = 2.50094 (3.7)352
The ratios of the radii of Mercury and Venus can be viewed as quantum numbers.353
354
13
3.3. Orbital periods in the planetary system355
The orbital periods of the planets result iteratively from the sun, Mercury356
and their focus. These calculations are always performed without π, but also357
involve polynomials, especially P(8). The factor
½
leads to the relative speed358
(Table 2). These orbital periods complement the observations of the Titius-Bode359
law [8] and resonances [9,10], but are currently still speculative.360
Mercury orbital period relative to a solar rotation of 25.38 days361
25.38 day 1/2(8 −1−1/2/8) = 88.04 day measured: 87.969 day
Orbital period of Venus without moon:
1/2(83−82+ 0 ∗8 + 1) day = 224.5day measured: 224.70 day
Earth’s orbital period:
1/2(83+ 3(82+ 8 + 1)) day = 365.5day measured: 365.25 day
Orbital period of the moon:
1/2(82−81−1) day = 27.5day measured: 27.322 day
Orbital period of Mars with two moons in resonance:
1/2∗3(83−82+ 8 + 2) day = 687 day measured: 686.98 day
362
Table 2: Orbital period in the planetary system in P(8) (3.8)363
364
3.4. H0 and cosmic microwave background radiation365
(CMBR)366
With the assumption of Q+, the expansion of the universe is already given. As367
shown with the neutron, the wave or photons on the surface of a solid, neutral368
object are Fourier transformed into the quantum information tfin the interior369
(see 2.39).370
hGNs3/m5c5s5/m5√π4−π2−π−1−π−3= 0.999991 (2.39)371
NEarth = 1/(hGN)s3/m5is distributed isotropically in the 5 dimensional372
space of c5with curvature ρ/dt =√π4−π2−π−1−π−3/s = 9.337501/s in373
time. π4and π2describe the contrasts using φand θaround a center r.374
NEarth = 1/(hGN)m5/s3/s = 2.2611962 1043 /s =c5s5/m5ρ/dt (2.40)375
For a photon in D5 the energy is: c5s5/m5=ρ/dt. For observation, the expan-376
sion of the universe results in H0 at the boundary of D3 and D2 with the ratio377 √π.378
H0theory =√π/c2s2/m2/ρ/s = 2.13 10−18 /s = 65.1km/s/Mpc (2.61)379
Measurement: H0 = 67.8±0.9km/s/Mpc [11]380
One possible interpretation is that visible and invisible particles are added381
to every object over time. With an isotropic distribution of the ur-particles in382
a visible photon, H0 results in a photon energy factor of 1/2 and a spatial cur-383
vature of 1/ρ3m. Isotropic photons indicate statistically uniformly distributed384
ur-particles in a vacuum, and the wavelength λ385
14
λtheory =√π/2/ρ3m= 1.0885 mm (2.61)386
Measurement: λ= 1.063 mm (peak of the spectral radiance dEλ/dλ)387
The theoretical values support at least a significant proportion of the CMBR388
monopoly term [11].389
Summary and Conclusions390
Exact predictions for the masses of elementary particles arise solely from391
the assumption of rational numbers Qin the universe and form physics built392
from Q+. Dimensions (φ,r,θ) and ur-particles are the required categories for the393
quantum information P(2). What is essential is the introduction of the most394
efficient coordinate system for describing the cosmos from polynomials P(2π) for395
neutral objects and with P(π) for charged objects. For 3 dimensions there are396
3 different centers of gravity with π, 1 and 1/π corresponding to the neutrinos397
ντ,νµand νe, respectively. At least 3 objects, each with 3 location coordinates,398
are required to determine the location at a specific time. This corresponds399
to the 10 required equations in the GR. The calculation of mneutron/melectron
400
parameters with an accuracy of 10 digits is the key to determing the ratio of401
the rest masses of all the elementary particles [12]. All this knowledge requires402
a distinction between two objects by an observer, generally a neutral object.403
Due to the spacial curvature, the information is combined to form 3 different404
focal points and ultimately to form a common point at time tf. The algorithm405
for this purpose a Christoffel symbol (GR) as the opposite pole, quantized with406
∂d= 1/π. It is a Fourier transform and returns the quantum information P(2).407
The outside world is reflected on the surface of the measuring device into the408
inside. The particles themselves are weightless. With hGN, kg is eliminated.409
QT and GR are united by the common constant hGN. With the curvature of410
space in time dρ/dt, all constants in 5 dimensions can be explained by ratios.411
ρ/dt =√π4−π2−π−1−π−3/s412
hGNc5s9/m10 ρ/dt = 0.999991413
The three-dimensional nature of space was already discussed by Ehrenfest in414
1920 [13] and with higher-dimensional branes in the context of string theory415
[14]. According to the theory of polynomials P(2π), we live in a universe that416
switches the parity operator between two objects each with three spatial di-417
mensions (φ, r, θ). This means that the view of the world is the surface of a418
three-dimensional space, on the edge of vacuum with the highest anisotropy or419
complexity. The Big Bang model is assumed to be the standard model of the420
universe and is based on the conservation of energy over time. An alternative421
interpretation is that visible and invisible particles are added to every object422
over time. The gradient of the gravitational field of each object is therefore the423
cause of cosmic microwave background radiation. With an isotropic distribution424
of the ur-particles in a visible photon, H0 results and the photon energy has a425
spatial curvature of 1/ρ3m.426
H0≈√π/c2s2/m2/ρ/s = 2.13 10−18 /s = 65.1km/s/Mpc427
λ≈√π/2/ρ3m= 1.0885 mm428
Measurement: λ= 1.063 mm (peak of the spectral radiance dEλ/dλ)429
15
The theoretical values suggest at least a significant proportion of the CMBR430
monopoly term [11].431
The number of NEarth in the Earth begins in the center with one neutrino.432
Away from the surface of Earth, D2 begins with photons with an area (m2) and433
then linear structures D1 for the filaments of the universe until D0 with the434
background radiation.435
In a rational space, a photon has a beginning and an end through imme-436
diately adjacent neutrinos, or two entangled electrons e+and e−, respectively.437
There entangled neutrinos result in an energy for the electric charge ECand438
the proton mass:439
Ec=−π1+ 2π−1+π−3−2π−5+π−7−π−9+π−12
440
mP roton =mN eutron +Ecme= 1836.15267363 me(2.32)441
Polynomials yield P(2π) that contains more information than the time-symmetric,442
squared equations from QT, QFT and GR with constants. This finding is con-443
sistent with the basic principles of GR and the Noether theorem for the meaning444
of energy, which were newly discussed in 2022 [15]. The calculations with P(π)445
of P(2π) surpass the results of various variants of axiomatic quantum field the-446
ory [16-26] in terms of accuracy and brevity according to Ockham’s razor. New447
approaches to emergence [27-34] through chance, selection and self-organization448
are emerging.449
450
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The author declares that no monies, grants, or other assistance were received577
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