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The Basis of the Universe as Six-Fold Symmetry!
By !
Ian Beardsley!
Copyright © 2022 by Ian Beardsley"
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It would seem at the basis of Nature is six-fold symmetry whether we are talking about biology,
like it’s hydrocarbons skeleton, which we do here, or the orbits of the planets. I show the earth
is the eigenvalue or characteristic root for the matrix that describes the solar system and to do
this I formulate the solar system as six-fold symmetry and suggest this could be taken as a
logos, a formulation of the language of Nature. I develop a constant k which is on larger scales
compared to Planck’s constant on the quantum scale, and I find it is is necessary to look at the
larger scales of chemistry using Avogadro’s number and the velocities of the planets, really to
predict the radius of the proton in terms of not just Planck’s constant, but in terms of gravity, G,
which takes us into the realm of Quantum Field Theory. Once we establish the idea of proton-
seconds we find we can formulate chemical compounds as mathematical, algebraic constructs
which is done here. The paper goes from fundamental particles like protons, to atoms, to
compounds, and to the planets suggesting perhaps it is all part of the same thing in making up
the Universe.!
of 3 49
Contents
1.0 The Constant k………………………………………….4
2.0 Orbital Velocity………………………………………….8
3.0 The Earth is the Basis or Eigenvalue…………….20
4.0 Hydrocarbons……………………………………………21
5.0 Six-fold The Orbits and Hydrocarbons…………25
6.0 Six-fold in the Electric Field………………………..27
7.0 Predicting The Proton Radius……………………..34
8.0 Rigorous Formulation of Proton-Seconds……..37
9.0 Theory of Compounds…………………………………41
of 4 49
1.0 The Constant k: Warren Giordano writes in his paper The Fine Structure Constant And
The Gravitational Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js
And it works, G is:
G=6.67408E-11 N(m2/kg2)
Let us reformulate this as by introducing Avogadro’s number because multiplying Planck’s
constant ‘ ’ by ‘ ’ is G if multiplied by and Avogadro’s number is
:
Equation 1.1
We just need the right units. We can get these:
Since grams and atom cancel we can work in grams even though our equations are in kilograms.
Let us not write H, since formally it is grams per mole of hydrogen but write
We have:
Or,…
Equation 1.2
Where
Equation 1.3
Let us say we were to consider Any Element say carbon . Then in general
h
1 + α
α
10
23
h
1 + α
10
23
N
A
= Avaga dro′ s − Nu mber = 6.02E 23atom s /gram
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
H = 6.02E 23
atom s
gr a m
⋅
1gr a m
atom
= 6.02E 23
ℍ = 1
gr a m
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
of 5 49
Equation 1.4
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
Equation 1.5
is not molar mass, and that is a variable determined by ; it is a mole of atoms multiplied
by the number of protons in . The reason we point this out, though it may already be clear, is
we wish to find the physical theory behind it. That is we need to find the physical explanation for
equation 1.4
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gr a m s
6proton s
N
A
=
6(6E 23pr oton s)
6gr a m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23protons
Z ⋅ gra m s
𝔼 =
Z ⋅ gra m s
Z ⋅ pr oton s
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
of 6 49
We know that
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the orbit
of earth as it relates to the sun. We have:
=
We can now write
Eq 1.6.
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
multiply both sides by and we have earth velocity on the left and the units stay the same on
the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put
in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Eq. 1.7
This brings up an interesting question: while we have masses characteristic of the microcosmos
like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to
become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)
which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where
we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find
that mass let us take the geometric mean between the mass of a proton and the mass of 1.44
solar masses. We could take the average, or the harmonic mean, but the geometric mean is the
squaring of the proportions, it is the side of a square with the area equal to the area of the
rectangle with these proportions as its sides. We have:
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E 30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
of 7 49
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Eq. 1.8
All we really need to do now is divide 1.7 by 1.8 and we get an even number that is the six of our
six-fold symmetry.
Eq. 1.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Equation 1.9
Where k is a constant, given
Equation 1.10
We can explicitly write the constant k:
Equation 1.11
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic
equation is:
Equation 1.12
Let us approximate 0.77 with 3/4. Since we have our constant
And
Then
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
800
s
m
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
of 8 49
Equation 1.13
Since our constant k in terms the Chandrasekhar limit is
Equation 1.14
2.0 Orbital Velocity Earth rotates through 15 degrees in one hour:
The distance then that the earth equatorial surface rotates through is where r is the earth
radius and theta is in radians.
Now we look at how many degrees through which the earth rotates in 1 minute:
For seconds…
Now let’s look at the same for the distance the earth moves around the Sun in an hour, a minute,
and a second as well…
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
360
24
= 15
s = r θ
s = (6,378.14)(0.2618ra d ) = 1669.8k m /hr
24(60) = 1440min
360
1440
= 0.25
∘
= 0.0043633ra d
s = (6,378.14)(0.00436) = 27.83k m /min
24(60)(60) = 86,400s
360
∘
86,400
= 0.004167
∘
= 0.00007272ra d
s = (6,378,14)(0.00007272) = 0.464k m /s
(365.25)(24) = 8766hr /yr
360
8766
= 0.041
∘
= 0.00071678ra d
(149,598,000)(0.000716768) = 107,227k m /hr
(365.25)(24)(60) = 525960
of 9 49
And finally…
We have the following table:
As we can see I am in good
agreement with Martin
Zombek, Handbook of Space
Astronomy and Astrophysics
which provides data. Notice
27.83 km/min in earth
rotation is approximately the
29.786 km/sec in earth orbit.
That is a valuable clue. Now let
us consider
Earth: 1 AU=149,598,000km (average earth-sun separation).
The earth rotates through about one degree a day.
Earth Orbit:
Earth Rotation: r=6,378.14km
360
525960
= = 0.000684463
∘
= 0.000011946ra d
s = (149,598,000)(0.000011946) = 1,787.1k m /min
(365.25)(24)(60)(60) = 31557600
360
31557600
= 0.000011408
∘
= 0.000000199ra d
(149,598,000)(0.000000199) = 29.786k m /sec
360
365.25
= 0.9856
deg
d ay
≈ 1
deg
d ay
360
∘
= 2π r a dia n s = 6.283
ra d
year
0.9856
∘
= 0.017202424
ra d
d ay
(149,598,000)(0.017202424) = 2,573,448.201
k m
d ay
360
∘
1d ay
= 6.283
ra d
d ay
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Moon: sidereal month =27.83 days, r=405,400 km
The points to be made in this exploration
1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.
2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.786km/s
around the sun. These two are close to the same.
3. The synodic month is 29.53 days approximately equals the 29.786 km/s the Earth moves
around the sun. The sidereal month is 27.83 days is the 27.83 km/min through which the
Earth rotates at its surface.
With that information we have this mystery of sexagesimal in the earth-moon-sun orbital
parameters, solved. We see we can make the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree) is the distance the earth moves
around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are
360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).
We have:
1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always
close to that because its orbit is approximately circular. We have
s = r θ = (6,378.14)(6.283) = 40,075.0
k m
d ay
(27.83)(24) = 667.92
h ours
m onth
360
667.92
= 0.529
∘
= 0.0094
ra d
hr
(405,400k m)(0.0094r a d /hr) = 3,810.76k m /hr
(3,810.75k m /hr )(hr /60min) = 63.5k m /min = 1.0585k m /s
orbit rotat ion orbit m oon
29.786k m minute 1d ay k ilom eter
secon d 27.83k m degr ee secon d
=
mi n ⋅ d ay
deg
⋅
k m
s
2
(mi n)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
mi n ⋅ d ay
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
of 11 49
This is approximately the diameter of the Earth orbit. We define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
Moon orbits:
Earth completes a 360 degree orbit yields:
Where on the right it is in radians and is the radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
Where is the inverse of the golden ratio. .
We have:
Equation 2.1
Our base ten counting is defined
is defined
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
mi n
=
27.83k m
mi n
⋅
mi n
60sec
= 0.4638
k m
sec
θ
e
=
1d ay
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786)(14,400)(1k m /s)
(0.4638)(149,598,000)
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
of 12 49
, such that
which is given by
Thus since the diameter of the Earth orbit is
Then its radius is
Since we measure time with the Earth orbital period and that period is given by Kepler as
Then
Equation 2.2
Is approximately one year. In this section we set out two show the historical development of the
second by dividing up the motions of the Earth, Moon, and the apparent motion of the Sun into
units of 24, and 60 result in the solar system’s size as based around the Earth orbit giving us
Since we have established a connection between the microcosmos and the macrocosmos we
would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want
to know the universal gravitational constant in these units:
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T =
3
2
10 ⋅
ϕ
3
2
10
3
⋅ 360
∘
ϕ
100
360
∘
= 2AU
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
of 13 49
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
Equation 2.3
Which evaluates:
Converting this to meters per second:
Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s
is an average over a varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
5
3
GM
ϕ
v =
5
3
(40)(1)
0.618
= 6
AU
year
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
of 14 49
For the earth, we show now it is true for the moon as well, and indeed it would hold for any
system circular, or approximately circular. For the Sun and its planets
If you use astronomical units and Earth years. In general we use
Which has the constant of proportionality
If m is small compared with M, you can write
Since we want to work with Earth and the Moon m is not so small compared to M. So,…
=
=
And these do round neatly to one-exponent like this. The sidereal lunar month is
(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…
The tables give
Which is an average for the lunar orbit in its slightly elliptical orbit.
We need to find G in terms of Lunar Units (LU) as opposed to Astronomical Units (AU), and in
terms of the lunar month as one instead of the earth year. We see G has the same value we had
for solar-earth units:
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
1
k
= G
M + m
4π
2
1
k
= G
M
4π
2
1
k
= (6.674E − 11)
(
5.9722E 24kg + 7.347731E 22kg
4π
2
)
(6.674E − 11)
(
6E 24kg
4π
2
)
4E14
4π
2
=
1E14
π
2
= 1E13
a
3
= (1E13)(2,404,512)
2
= 5.78E 25
a = 3.8668E8m
a = 3.833E8m
of 15 49
=
We only need to convert to m, kg, s by using their relationships with LU, LM, and . That is,
for the orbital velocity is always six:
Where is the orbital velocity of the moon, and is the orbital velocity of the earth. The
difference comes when converting the system from lunar units in the case of the first to kg/m/s
and from astronomical units in the case of the second to kg/m/s. For example the first is:
Which is correct, the tables give that the lunar orbital velocity is 1000 m/s.
To find logos, is to find Nature not just as number, but as dimensionless whole
number, because fractions have inherent in them irrational numbers, which are
unending decimal expressions that at some point must be rounded to so many
figures to put them to use. Logos then is the language of Nature that transcends
conundrum because it would be utilized as relationship between point, plane, and
line as opposed to units. We have found that for the orbital velocity:
Equation 9.1
Where is the earth orbital radius. This results in the orbital velocity for a orbiting body is
Equation 9.2.
Which is logos because it is number, which is 6. The orbital velocity of a body is always 6
because:
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
M
⊕
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
v
m
= 6
3.8668E8m
2,404,512s
= 964.886m /s
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
e
v =
5
3
GM
ϕ
v
m
=
5
3
(40)(1)
ϕ
= 6
of 16 49
Where is the orbital velocity of the moon, and is the orbital velocity of the earth, That is G
is always 40, and M is always 1 providing the orbit is circular. Let us show this for Venus. Its
orbital distance VU (Venus units) is 1.082E11meters. It orbital period is 1.94E7s is the Venus
year (VY).
=
Thus we have
We can then express all orbital velocities as 6, but to find there values in a formal system of
units, we need to convert from these natural units to something like kg/m/s. Thus logos
translated into a language we understand can best be done in a square array, or as a matrix
transformation. We have
You will find this gives , ,
All of which are correct within the variations of these velocities in their deviations from a
perfectly circular orbit. Logos is:
Equations 9.3
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
of 17 49
, such that
which is given by
To say that logos is:
Is to say that for the solar system Kepler’s Law of Planetary Motion holds:
That in general
Because as such for Earth we have
For the Moon we have
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1VU
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
of 18 49
For Venus we have
Making
, ,
This is derived from
Which is to say
Is logos because
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
=
5
3
(40)(1)
ϕ
= 6
v
v
=
5
3
(40)(1)
ϕ
= 6
G
Mm
R
2
=
mv
2
R
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1VU
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
of 19 49
such that
which is given by
Which is the relationship between point, plane, and line.
From everything we have said G, the gravitational constant is about 40
We can do this for any planet and get G is approximately 40. We found the orbital velocity of any
planets is 6. This is true because as we have shown, the orbital distance of any planet is
This gives since
That orbital velocities are for the moon, earth, and venus
But what does this mean? It means since
And, v=6, that where the orbital velocities of the planets are 6, their distances from the sun are
all one.
This gives:
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
AU
3
(Ear th Dista nceMeters)
3
⋅
Sol arMa ssKilogra m s
M
⊙
⋅
(Ear thOrbit a l Per iod Secon d s)
2
year
2
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
GM
R
=
5
3
GM
ϕ
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
R =
GM
v
2
R =
40
36
≈ 1
of 20 49
But where six is
3.0 The Earth is the Basis or Eigenvalue
Since equation 1.9 is
Equation 1.9
And since we have said
Which we can write
Equation 3.1
Where r and T are orbital radii and orbital periods subscripted with their designations, then we
see Earth is the eigenvalue because
Equation 3.2.
Which means
Or,…
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
k v
e
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6 6 6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
(
6 k v
e
6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
k v
e
r
e
T
e
= v
e
of 21 49
Equation 3.6.
That is since Eigenvalues are characteristic roots, that is the value such that for the matrix A:
4.0 Hydrocarbons
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 4.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 4.2.
Which describes mass per meter over time, which is:
Equation 4.3:
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
k
r
2
e
T
2
e
= v
e
λ
A
x = λ
x
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
α
1
α
2
of 22 49
Equation 4.4.
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 4.5.
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 4.6
We take the square root to get meters:
Equation 4.7
We multiply that with the value we have in equation 1.4:
Equation 4.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 4.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 4.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 4.11
Equation 4.12
If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of
the elements in the periodic table occur for dividing by something less that 1-second. It seems
the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that (See appendix 1 and appendix 2):
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m)
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
S
p
= 2.953E − 15m
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
p
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6pr oton s = 6m
p
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
of 23 49
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it.
1
4.13
I then developed the concept of proton-seconds where time is associated with atoms through the
number of protons they have.
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a
measure of energy over time where space and time are concerned it must play a role. Of course
the radius of a proton plays a role since squared and multiplied by it is the surface area of our
proton embedded in space. The gravitational constant is force produced per kilogram over a
distance, thus it is a measure of how the surrounding space has an effect on the proton giving it
inertia. The speed of light c has to play a role because it is the velocity at which events are
separated through time. The mass of a proton has to play a role because it is a measurement of
inertia itself. And alas the fine structure constant describes the degree to which these factors
have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional
analysis.
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
m
p
=
3r
p
18α
2
4πh
Gc
4π
See Appendix 2: Sexagesimal And The Cosmic Calendar, page 39
1
Fig. 1
of 24 49
Equation 4.14
We find six seconds gives 1 proton is hydrogen:
Equation 4.15
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements.
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon ds = hydr ogen(H )
1
α
2
m
p
h 4π r
2
p
Gc
of 25 49
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We see the radius of a proton is given by carbon (1 second):
Equation 4.16
The experimental radius of a proton is:
Equation 4.17
5.0 Six-fold The Orbits and Hydrocarbons
We now look at the orbits in six-fold symmetry with earth the eigenvalue equation 3.19 from
section 3.0 and and the hydrocarbons, the skeletons of life chemistry which we could say have
an eigenvalue of six as well:
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
= 0.833
±
0.014f m
of 26 49
Eq 3.19.
Eq. 4.14.
Eq 4.15.
We see that
Where
We can take
Where is the eigenvalue and is given by
We have for the velocity of the earth:
Eq 5.1.
Where the value of 1/k is exactly 773.5 m\s. Earlier we estimated it as 800 m\s , but we used the
orbital velocity of earth equal to exactly 30,000 m/s and it is actually 29790 m/s, so the result is
(
6 k v
e
6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
1
α
2
m
p
h4πr
2
p
Gc
= 6proton ⋅ seconds = carbon(C )
1
α
2
m
p
h4πr
2
p
Gc
= 1proton ⋅ 6seconds = hydrogen(H )
k v
e
=
1
α
2
m
p
h 4π r
2
p
Gc
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
1 + α ≈ 1
k v
e
= 6 = λ
1
α
2
m
p
h4πr
2
p
Gc
= 6 = λ
λ
A
x = λ
x
v
e
=
4πh
k α
4
⋅
r
2
p
m
2
p
⋅
1
Gc
of 27 49
about the same. is sort of like a Planck’s constant for a larger physical scale and is based on G
the universal gravitational constant, which we have said describes the effect of matter on space
over distance and time, and h is Planck’s constant describing energy over time on the quantum
scale. Putting our value for k in equation 5.1 for the velocity of the earth, we have:
Eq 5.2.
Where is Avogadro’s number without units, as it turns out it is a natural constant. It is
At this point we could go in another direction and consider a protoplanetary cloud the
collapses into a rotating disc around that evolves into the planets orbiting a star which
could involve h and because it is the is the mass of a proton which would be the
primary mass unit that makes up a predominantly hydrogen cloud. But in this paper we
will continue along another line. Since we have the velocity of the earth in terms of
gravity G and Planck’s constant h, h always coupled with c for obvious reasons (looking
at computations in quantum mechanics, E= where here lambda is wavelength not
the eigenvalue) we want to now find this six-fold basis not in the gravitational fields, but
in the electric fields. We turn to that now…
6.0 Six-fold in the Electric Field
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 6.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
k
v
e
=
3
2
⋅
1
N
A
𝔼
⋅
r
2
p
m
3
p
ch
3
2π G
3
N
A
E
N
A
H = 6.02 E 23
atoms
gram
⋅
1gram
atom
= 6.02E 23
m
p
m
p
hc /λ
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
of 28 49
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from equation 3.16
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
Fig. 2
of 29 49
And we have
Equation 6.2
We get
Equation 6.3
Is carbon. We see the six-fold symmetry that is the carbon is due to the cube having 6
faces. I think I can be more accurate using space expanding in a sphere as opposed to a cube. We
have that
Equation 6.4
Equation 6.5
Equation 6.6.
Equation 6.7.
Equation 6.8.
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71protons ≈ 6pr oton s
1/6α
2
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
m
p
=
3r
p
18α
2
4πh
Gc
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon ds = hydr ogen(H )
of 30 49
Equation 6.9
Equation 6.10
Periodic table as cycles of six (3x6=18)
Hydrocarbons: The Skeltons of Life Chemistry
We said essentially at the outset if we consider it more explicitly that the periodic table is
structured the way it is because its cycle of 18 groups is founded in six-fold symmetry because 18
is 6 X 3=2 X 3 X 3 and 6 is 2 X 3 and 2 and 3 are the smallest prime numbers meaning no
number can be reduced further than factors of these, so these are the basis set at the crux of
Nature.
We said that the hydrocarbons are the skeleton of life and we now further point out that their
components hydrogen and carbon are in inverse relationship with one another, as seen in the
figure to the left which interprets the relationship between carbon and seconds and hydrogen
and seconds and how they define the periodicity of the Periodic Table of the Elements by
inscribing proton cross-sections in regular hexagons that tessellate with one another and are
made up of tessellations of equilateral triangles. The inverse nature is in that for carbon six
protons pack in one hexagon around a center to make a ring and for the hydrogen the proton
inscribes in one hexagon touching all sides its radius the apothems.
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71protons ≈ 6pr oton s
of 31 49
We said that the
hydrocarbons are the
skeleton of life and we
now further point out
that their components
hydrogen and carbon
are in inverse
relationship with one
another, as seen in the
figure to the left which
interprets the
relationship between
carbon and seconds and
hydrogen and seconds
and how they define the
periodicity of the
Periodic Table of the
Elements by inscribing
proton cross-sections in
regular hexagons that
tessellate with one
another and are made
up of tessellations of
equilateral triangles.
The inverse nature is in
that for carbon six
protons pack in one
hexagon around a center
to make a ring and for
the hydrogen the proton
inscribes in one hexagon
touching all sides its
radius the apothems.
of 32 49
We find it useful to write out the constants in terms of their units we have used to describe
space, time, and matter and their properties in the previous sections…
, ,
We have that
Which is the units of the gravitational constant G is
Which is the units of force in Newtons is
The Bohr radius, or mean orbital radius of an electron in hydrogen for its ground state is
Is units of energy in joules.
We now express the inverse relationship between carbon ( ) and hydrogen ( ) by writing
G = N
m
2
kg
2
= kg
m
s
2
⋅
m
2
kg
2
=
m
3
kg ⋅ s
2
h = J ⋅ s = kg
m
s
2
⋅ m ⋅ s = kg ⋅
m
2
s
c =
m
s
r
p
= m
m
p
= kg
hc
m
2
p
=
m
3
kg ⋅ s
2
hc
m
2
p
=
(6.626E − 34)
1.6726E − 27E − 27)
2
⋅
299,792459
1
= 7.10048E 28N
m
2
kg
2
hc
r
2
p
= kg
m
s
2
(6.626E − 34)
(0.833E − 15m)
2
⋅
299,792459
1
= 286,274N
a
0
=
4πϵ
0
ℏ
2
m
e
e
2
= 5.29177E − 11m
hc
a
0
= 7.1E − 5J
ℂ
ℍ
t
6
ℍ = m
p
t
6
= kg ⋅ s
t
1
ℂ
=
t
1
6m
p
=
s
kg
hc
m
2
p
⋅ t
6
ℍ =
hc
m
2
p
⋅ 6sec ⋅ m
p
=
m
3
s
hc
r
2
p
⋅
t
1
ℂ
=
hc
r
2
p
⋅
1sec
6m
p
=
m
s
of 33 49
We multiply the former with the inverse of the latter:
And we see that the second maps the first into a factor of 6 which is what they should do.
But what we really want to do is is form a geometric mean:
Because it is energy, we will say the energy of our hydrocarbon in hexagonal tension.The
ionization energy of carbon is 1.804051E-18 Joules and that of hydrogen is 2.18E-18 Joules., the
energies to remove an electron (to move it from ground state to infinity). These are very close to
exactly the squares of our 1.4142E-9 Joules for the energy of our hexagonal tension for the
hydrocarbon. Why would we have the square of the ionization energies? Because the energy is
not just over a line, but two lines that form an area, and area is the square of a line. The
ionization energy of hydrogen is given by the Rydberg equation:
where and and A=2.18E-18J
The Electron affinity of carbon is 153.9kJ/mol
Electronegativity of carbon is 2.55
First ionization of carbon is 11.2603eV
IE=11.2603eV=1.8041E-18J.
We have (2.18E-18+1.8E-18)/2=1.99E-18 and
We can predict the radius of a proton if we can derive an expression for 1 second. We approach
that next…
(
hc
m
2
p
⋅ 6sec ⋅ m
p
)
(
r
2
p
hc
⋅
6m
p
1sec
)
= 6r
2
p
= 4.163E − 30m
2
hc
r
2
p
↦
hc
6r
2
p
hc
r
2
p
= 286,274N
hc
6r
2
p
= 47,712N
(
hc
m
2
p
⋅ t
6
ℍ
)(
hc
r
2
p
⋅
ℂ
t
1
)
= 1.4142E − 9kg
m
2
s
2
= 1.4142E − 9J = 2E − 9J
ΔE = A
(
1
n
2
f
−
1
n
2
i
)
= 2.18E − 18J
n
f
= ∞
n
i
= 1
C + IE → C
+
+ e
−
(1.414E − 9)
2
= 1.99E − 18
of 34 49
7.0 Predicting The Proton Radius We consider two paths, one with velocity c in one
medium, the other with velocity v in another. In order to go from one point two another over
two paths, the refraction is such that the sine of the angle of incidence equals the sine of the
angle of refraction.
We have the two paths are travelled in a time t:
This is the mysterious Nature of reality, the path of least time is taken. This falls under the
general heading The Principle of Least Action, attributed to firstly the French Natural
Philosopher and mathematician Louis Maupertuis of the early eighteenth century. I say
mysterious, because as it is said in physics, for something to know the path of least action and
take it, it is as if it has explored first all paths between A and B to know which one would be the
path of least action. Everything in physics comes to this principle. It is called a principle, not a
theory, law, or rule. Yet it seems to be the way Nature behaves, and it is mysterious. Richard
Feynman applied it to quantum mechanics, probably because the mysterious Planck’s constant
that governs quantum mechanics is in Joule-seconds, energy over time, and this is the terms in
which action is formulated mathematically.
In our scenario here we regard matter, the proton in particular, as the cross-section of a
hypersphere. Our two mediums are hyperspace and space and the least action principle applies
in the same mathematical form. This is abstract cosmology, that really is the underlying
mathematics is common to all systems, that in effect they are manifestations of one another.
We have
We make the approximation
So that
We are saying t=6 seconds is the proton, and r is the radius of a proton r=0.833E-15m. Thus
The radius of a hydrogen atom is .
sin α
sin β
=
v
c
t = r
p
c + v
c v
t
r
c v − v = c
t
r
c v = c
v =
r
t
v =
0.833E − 15m
6s
= 1.389E − 16m /s
R
h
= 1.2E − 10m
of 35 49
And we have our one second as a natural constant with respect to the atom. We see it occurs at
t/2 which is at half the radius of a hydrogen atom. We want to deal with that size or around it;
our velocity comes from it and it can be thought of as a proton drift velocity akin to the electron
drift velocity in a wire that gives rise to an electrical current. We find if we derive the equation
that represents these computations, we have
Where,…
, ,
Radius of hydrogen atom
Remember our constant k equation 4.16 (Don’t forget to divide by two somewhere):
Since we have the equation of the radius of a proton is given by, by evaluating it at one second
which is carbon:
t =
1.2E − 10m
1.389E − 16m /s
= 863,930 . 8855s
t
2
= 431,965 . 4428s
t
k
= (773.5m /s)(431,965 . 4428s) = 334,125, 270m
334,125, 270m
c
=
334,125, 270m
299,792, 459m /s
= 1.11452193secon d s
v =
r
t
t = 6s =
1
α
2
m
p
h 4π r
2
p
Gc
1
t
= α
2
m
p
Gc
h 4π r
2
p
v =
r
p
t
= α
2
m
p
Gc
h 4π
R
H
= 1.2E − 10m
t =
R
h
v
= R
h
⋅
1
α
2
m
p
h 4π
Gc
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
of 36 49
And 1 second in terms of the atom is given by
Then the equation for the radius of a proton is:
Equation 7.1.
And since
Then,…
Equation 7.2
Let’s verify equation 7.1:
Remember that in equation 2.1, which is
That, must remain coupled. is determined by the number of protons in .
Making the approximation 9/8~1 we can write equation 2.3 as (We have picked up the fraction
9/8 by making several approximations)
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6 ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Da y = (6)1.2secon d s
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
=
K Eof Moon
K Eof Ear th
⋅ Ear th Da y
r
p
=
9
8
2
1
1.67E − 27
(6.626E − 34)(299,792, 459)
(6.674E − 11)4π
3
⋅
1.2E − 10
6.02E 23
= 0.93f m
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
N
A
𝔼
N
A
𝔼
of 37 49
Equation 7.3
Which gives
We form constants:
Equation 7.4
Equation 7.5
And we have the Equation:
Equation 7.6.
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12.
We now want to add to this section the equation of the mass of a proton.
Equation 7.7.
8.0 Rigorous Formulation of Proton-Seconds
We can actually formulate this differently than we have. We had
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are
lower and upper limits in an integral, then we have:
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
1
t
1
1
α
2
m
p
h 4π r
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton
of 38 49
Equation 8.1
This Equation is the generalized equation we can use for solving problems. Essentially we can
rigorously formulate the notion of proton-seconds by considering
Equation 8.2
Is protons-seconds squared where current density is and ( can also be
). We say
Equation 8.3
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is
Equation 8.4
Dividing Equation 8.2 through by t:
Equation 8.5
Which is proton-seconds. Dividing through by t again:
Equation 8.6
We see that if where and then J is I/m2 (current per square meter)
is analogous to amperes per per square meter which are coulombs per second through a surface.
Thus we are looking at a number of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qd t = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= protons
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
of 39 49
Equation 8.7
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.
!
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h 4π r
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y
of 40 49
We are now equipt to do computations in proton-seconds. We use equation 3.6 from two to
three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal
proton that we found described its radius (My feeling is we introduce the factor of 2 because
carbon is 6 protons +6 neutrons and 2 times 6 is twelve):
Now we integrate from phosphorus is 15 protons=0.396 seconds to aluminum is 13 protons =
0.462 seconds which is to integrate across silicon (divide your answer 10 by 2 to get protons):
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 8.8
Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus
enabling the construction of logic circuits that you can use to make computing machines. But
this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the
same group as aluminum, just above it. This results in a theory for AI elements as mathematical
constructs, that we will go into now.
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= protons
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.376 − 4.685) = 5pr oton s /secon d = boron
of 41 49
9.0 Theory of Compounds
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are
carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to
show not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 42 49
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them. But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 9.1
A = (Al, Si, P )
B = (G a, Ge, As)
A ×
B =
B
C
N
Al Si P
Ga Ge A s
= (Si ⋅ As − P ⋅ Ge)
B + (P ⋅ G a − Al ⋅ As)
C + ( Al ⋅ Ge − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsin θ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g/m ol
Si(A s − G a) + Ge(P − Al )
SiG e
=
2B
Ge + Si
of 43 49
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 9.2
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − G a) +
Ge
B
(P − Al ) =
2SiGe
Si + Ge
of 44 49
Appendix 1
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E 8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
of 45 49
"##$%&'()*+)Sexagesimal And The Cosmic Calendar
My book Abstract Cosmology could have easily been called The Book of Six, it was
Buckminster Fuller who said “Nature employs 60 degree coordination”. Sixty degrees are the
degrees of an equilateral triangle. The triangle is the structure that encloses an area with the
least amount of sides. Buckminster Fuller said, systems of triangles are the the only inherently
stable patterns. The regular hexagon, a six-sides polygon tessellates as equilateral, 60 degree,
inherently stable equilateral triangles. It was the scientist Shubnikov who said among the living
organisms the pattern with which we most frequently meet is five-fold symmetry. It is well
known that the physical, like snowflakes, are six-fold symmetry. Thus we should not be
surprised that our Abstract Cosmology is founded on six-fold symmetry. Two and three are the
smallest prime numbers and their product is six. Two times six is twelve, the number most
evenly divisible by whole numbers for its size (it is a so-called abundant number: divisible
evenly by 1,2,3,4,6, their sum is 16 which is greater than twelve itself and, three times six is
eighteen, the cyclical Nature of the periodic table (18 groups). !
At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and
hunting, to invent agriculture and and build ceramic homes, developed the first mathematics,
and it was this sexagesimal (base 60) that passed on to the Babylonians, then ended up with
the Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into
60 seconds of time, but who divided the sky into hours, minutes, and seconds of arc which
was measured in time counted by the rotation of the Earth, and its orbital period around the
Sun, and the orbital period of the Moon around the Earth.!
We are suggesting nature is founded on six-fold symmetry, six which is constructed by
multiplying together the first two prime numbers, 2 and 3. Two factorial is two, three factorial is
6. Two times six is twelve, the number of months in a year, approximated by our moon’s
approximately 12 orbits around the earth in the time it takes the earth to go around the sun
once. There are four weeks in a lunar month, and!
!
Is the number of seconds in the 24 hour Earth day. The duration of one second comes from
dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,
and Ancient Greeks and it is believed they did this because 60 is evenly divisible by!
1,2,3,4,5,6,…12,15,20,30,60,…!
The moon orbits the Earth in approximately 30 rotations of the Earth, hence the 30 day month.
January has 31 days, February 29, March 31, April 30 May 31, June 30, July 31, August 31,
September 30, October 31, November 30, December 31. This averages out to!
!
We divide a circle into 360 units called degrees, which is six squared times ten. As such the
equilateral triangle has each angle equal to 60 degrees. There are approximately 360 earth
rotations in the time it orbits the sun once, Hence in one day the Earth moves through
approximately one degree in one day in its 365 day journey around the Sun. The sidereal
month, the time it takes the moon to return to the same position against the background of the
stars is 27 days 7 hours 43 minutes. We see the power of sexagesimal (base 60) in computing
time with the month divided into days, hours, and seconds where there are 60 minutes in an
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60
2
= 86,400
31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31
12
=
367
12
= 30.58333
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hour and 60 seconds in a minute because we can convert this easily into seconds with such a
system as here, the duration of this sidereal month in seconds:!
!
The length of the synodic month, the completion of the phases of the moon, or the time it takes
the moon to return to the same position with respect to the Sun, is 29 days 12 hours 44
minutes 3 seconds. It is a little longer than the sidereal month because the earth has moved
with respect to the sun by the time the moon has come back around to a full orbit. We have:!
!
We looked at the arithmetic mean of the days in the month over a year, Let us look at the
harmonic mean:!
=!
!
Gathering like terms in the denominator:!
=0.225806+0.034482759+0.133333=0.39362!
!
Which is the most frequent value for the days in a month. Now let’s look at the geometric mean
which tempers the various values with one another.!
=30.493!
But perhaps it is more telling to take the mean between the individual days of the month that
we use:!
!
!
0.032258+0.034482759+0.033333=0.10007!
!
!
(24 ⋅ 27 + 7) ⋅ 60
2
+ 43 ⋅ 60
1
+ 0 ⋅ 60
0
= 2,358,000 + 2,580 + 0 = 2360580sec/m onth
(24 ⋅ 29 + 12) ⋅ 60
2
+ 44 ⋅ 60
1
+ 3 ⋅ 60
0
= 2,548,800 + 2,640 + 3 = 2551443sec /month
12
1
31
+
1
29
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
12
0.032258 + 0.034482759 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258
12
0.39362
= 30.486
12
31 ⋅ 29 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31
31 + 29 + 30
3
=
90
3
= 30
3
1
31
+
1
29
+
1
30
3
0.10007
= 29.799 ≈ 30
3
31 ⋅ 29 ⋅ 30 = 29.98888 ≈ 30
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We see that the arithmetic mean gives us exactly a 30 day month, hence we always speak
loosely saying a month is 30 days. !
Indeed when Buckminster Fuller speaks of Nature employing 60 degree coordination he is
speaking of the equilateral triangle as the fundamental unit of one for area. That is the six-sided
regular hexagon can be divided into six equilateral triangles, giving it an area of six. Indeed he
takes the equilateral triangle as one because not only does it enclose an area with the fewest
sides, it has the smallest area for its perimeter; a regular hexagon with a perimeter the same
size encloses more area. I would like to say the Ancient Sumerians, Babylonians, and Greeks
made this connection, because in their sexagesimal (base 60) system of notation, one is a
straight line with a triangle on top. See figure below. Ancient Mesopotamian Cuneiform for
writing. It is not sexagesimal notation in the strict sense, as they have a unique symbol for a
tens column, but it was this sexagesimal structure that lead to the 60 minute hour and 60
second minute, as well as the angular division projected onto the sky in 60 minutes of arc in a
degree and 60 seconds of arc in a minute as a way of mathematically defining time dynamically
in terms of the motions of the Earth and moon.!
For equation our equation Earth day needs to be shorter. A long time ago it
was; the Earth loses energy to the moon. The days become longer by 0.0067
hours per million years. Our Equation
Is actually 1.2 seconds.
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
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We have
24-20=0.0067t
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). The dinosaurs went extinct 65 million years ago giving small mammals a
chance to evolve paving the way for humans.
24-x=0.0067t
x=23.5645 hours
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic
calendar:
What is the next term?
20+3+0.57735+0.4714=24hours
Which bring us to today.
It would seem we could have divided that day, not into 24 hours but 20 hours and get our 1.2
seconds as a duration for measuring time:
Both work because
24=2x2x6
20=>20+20+20=60
Both have an intimate structural relationships with sexagesimal.
24h ours
1.2
= 20hours
3cos(0
∘
) +
2
3
cos(30
∘
) = dinosaur − ex t inct ion =
20hrs + 3hrs +
3
3
hrs +
2
3
=
(365.25d ays)(24hours)(60sec)(60sec)
(365.25d ays)(20hours)(60sec)(60sec)
= 1.2
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The Author
