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The Confluence of Quantum Physics, Ancient Sumerian Mathematics, Planetary

Formations, and Life

Ian Beardsley

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Contents

Abstract…………………………………………………………………….3

Important To Point Out……………………………………………………..4

Part 1: The Theory………………………………………………………5

Part 2: The Dynamics of the Ancient Second………………………..13

Part 3: The Protoplanetary Disc in our Theory……………………..20

Part 4: The Characteristic Length For Habitable Planets……….28

Part 5: Thus Basis Unit of a Second Comes

with the Epoch of a Perfect Eclipse……………………………………..31

Part 6: Solution of Schrödinger Equation

for Earth with Spherical Coordinates…………………………………….40

Part 7: Solution of Schrödinger Equation

for Earth with Cylindrical Coordinates……………………………………48

Part 8: Conclusion……………………………………………………………55

Part 9: Life A Possible Universal Natural Process………….59

Appendix 1………………………………………………………………………..65

Appendix 2: The Data For Verifying The Equations……………………67

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Abstract:

My theory has something to do with why we would be going into space, which indicates

it might be in part to find out if other star systems have dynamic relationships with

protons that make up atoms and both those with time as measured in their calendars by

dynamic mathematical relationships that may exist in the motions of their planets and

their moons as is the case with our star system, The Solar System. My theory shows we

have this with our star system, and would suggest we need to find other star systems

with habitable planets to see if they have them and relate them to ours if they do so as to

reveal a higher truth as to how and why we are here, and what our purpose might be.

The theory solves the atom (primarily the proton) and the Solar System, with the

Schrödinger wave equation of quantum physics, demonstrating a connection between

macroscopic and microscopic scales.

Because the basis unit of time for the solution is about 1 second, and the second comes

from ancient Sumerian mathematics, not contemporary physics, the paper concludes

with archaeological and paleoanthropological connections to a possible flow of the

development of human life since ancient and prehistoric times with the dynamics of the

physics of atoms and planetary formation.

A section is added after this conclusion showing the physical theory meets with biology

reaffirming the proposition from the physical theory that life might be part of a

Universal Natural Process.

If the theory is correct we should be able to predict the characteristics of different

habitable star systems. This should soon be verifiable because NASA has as one of its

primary objectives with its recent deployment of the James Webb Space Telescope to

find Earth-like planets around Sun-like stars. Something that has been difficult to do up

to now because prior technologies were only refined enough to work primarily with red

dwarf stars for finding planets.

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Important To Point Out

You can speak of the structure of the solar system even though it changes with time. This is

important to understand when I refer to sizes of the Moon and the planets, and their orbital

distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planet forms at its central distance from the star (it has a thickness). In a theory of planetary

formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of

planets today, which was the distribution of the rings from which the planets formed.

Also, the Earth has been in the habitable zone since 4 billion years ago when it was at 0.9 AU.

Today it is at 1AU, and that habitable zone can continue to 1.2 AU. So we can speak of the

distance to the Earth over much time. The Earth and Sun formed about 4.6 billion years ago. As

the Sun very slowly loses mass over millions of years as it burns fuel doing fusion, the Earth slips

microscopically further out in its orbit over long periods of time. The Earth orbit increases by

about 0.015 meters per year. The Sun only loses 0.00007% of its mass annually. The Earth is at

1AU=1.496E11m. We have 0.015m/1.496E11m/AU=1.00267E-13AU. So,

The Earth will only move out one ten thousandth of an AU in a billion years. Anatomically

modern humans have only been around for about three hundred thousand years. Civilization

began only about six thousand years ago.

The unit of a second becomes important in my theory. We got the second from the rotation

period of the Earth at the time the moon came to perfectly eclipse the Sun. The Moon slows the

Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses

energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the

Moon’s orbit gets 3.78 cm larger per year.

That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the

time that the Earth day was what it is during the epoch when the Moon perfectly eclipses the

Sun, 24 hours.

The near perfect eclipse is a mystery in the sense that it came to happen when anatomically

modern humans arrived on the scene, even before that, perhaps around Homo Erectus and the

beginning of the Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years

ago. Homo Erectus is around two to three million years ago.

(1.00267E − 13AU/year)(1E 9years) = 0.0001AU

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Part 1: The Theory

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Introduction !

The Natural constants, such as the constant of gravitation G, the Planck constant for the

atoms, h, and the speed of light c describe the properties of space and time. I have found they

are very conducive to our base unit for measuring time, the second. However, the base unit of

a second was given to us by the ancient Sumerians of Mesopotamia who were among the ﬁrst

to settle down from wandering and gathering, from hunting with stone spearpoints to create

agriculture, writing, and mathematics. They divided the day from sunrise to sunset into 12 units

that we call today in the West, hours. As such there are 24 hours in a day from sunrise to

sunrise. They had a base 60 counting system because 60 is evenly divisible by so much, like

1,2,3,4,5,6,…the ﬁrst six integers. It is the smallest number to do this. The Babylonians

adopted this base 60 (called sexagesimal) from the Sumerians, and divide their hour into 60

minutes, and the minutes into 60 seconds. We got our system from the Babylonians. Since one

day (24 hours) is the time for the Earth to make one revolution on its axis, the second comes

from dividing the Earths rotation by these numbers. But I didn’t just ﬁnd that the natural

constants described the second, but that they formed spacetime operators that solve the atom

today in modern physics, and the solar system with the Schrödinger wave equation that

describes the atom in quantum mechanics. I further found that this is done with the Earth’s

moon, which seems to be a natural yardstick for measuring size and mass. This says a great

deal about the mystery known to astronomy for a long time that the Moon perfectly eclipses

the Sun as seen from the Earth. In fact, it becomes part of the solution to planetary systems,

like our solar system, for them to have life optimally possible. We know the Moon orbiting the

Earth optimizes the conditions for life on Earth by holding the Earth at its tilt to its orbit around

the Sun allowing for the seasons and thus preventing extreme cold and extreme hot. Not only

does the lunar eclipse like this play a key role in the equations, but the theory prompts one to

suggest the perfect eclipse has perhaps been a message to humans, since hunting with stone

spearpoint to present day having gone to the Moon, to tell us we are here for reason.!

Earth/Moon/Sun System This second I found is the base unit of the orbital dynamics of the

solar system. I found I could create three spacetime operators, one that acts on the radius of

the proton to its mass, and the other that acts on the radius of the Sun to its mass, and one

that acts on angular momentum to speed. is Planck’s constant, is the universal constant of

gravitation, is the speed of light, is the ﬁne structure constant, is the radius of a proton,

is the mass of a proton, is the mass of the Moon, is its radius, and the EarthDay is

the rotation period of the Earth:!

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

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They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base

unit of one second, which we will see can be considered a characteristic time of the Universe.

Equation 2 is actually 1.3 seconds but rounds to one second. However it is derived from another

equation that gives 1.2 seconds. Namely

It says one second is the Earth Day (completion of one rotation of the Earth) adjusted by the

kinetic energy of the Moon, to the kinetic energy of the Earth. As we shall see, the quantization

in terms of the Moon is exactly 1 second to nearly two places after the decimal from the ground

state of our solution to the Earth/Moon/Sun system.

We say there is a Planck constant for the solar system. We do it with a base unit of one second

because equations 1, 2, and 3 suggest we should . We suggest it is such that it is given by the

standard Planck constant for the atom, , times some constant, , and the kinetic energy of the

Earth.

Where

6).

Where equation 6 comes from equation 3.

We derive the value of our solar Planck constant

1secon d =

K E

(Ear th Da y)

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

C =

⋅

2π r

G m

⋅

18769

299792458

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

1.55976565E 33

⋅

kg ⋅ m

⋅

= kg

= en erg y

hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon ds ≈ 1.0secon d s

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Now we show that our Planck constant for the solar system gives the base unit of one second for

the quantization. We call our solar Planck constant and find the wavelength for the Moon

which is the ground state for the solar system:

The wavelength associated with the Moon divided by the speed of light should be 1 second if our

planetary system is quantized in terms of the Moon and a base unit of one second. We have

And we see it is, so we have

The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be

inferred from the solution for an electron around a proton in the a hydrogen atom with the

Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates

10)

Its solution for the atom is as guessed by Niels Bohr before the wave equation existed:

11)

12)

hC =

(

⋅ m ⋅ s

)

(

⋅

)

(

)(

⋅

)

= secon d s

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

ℏ

⊙

moon

ℏ

⊙

(2.8314E 33)

(6.67408E − 11)(7.34763E 22kg)

= 3.0281E 8m

moon

3.0281E8m

299,792,458m /s

= 1.010secon d s

moon

= 1secon d

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

E = −

)

2ℏ

ℏ

Z k

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is the energy for an electron orbiting protons and , is the orbital shell for an electron with

protons, the orbital number. I find the solution for the Earth around the Sun utilizes the

Moon around the Earth. This is different than with the atom because planets and moons are not

all the same size and mass like electrons and protons are, and they don’t jump from orbit to

orbit like electrons do. I find that for the Earth around the Sun

13)

14)

is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,

is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is

the orbit number of the Earth which is 3 and is the Planck constant for the solar system.

Instead of having protons, we have the radius of the Sun normalized by the radius of

the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such

that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.

This becomes part of the theory and we suggest it is a condition for sophisticated planets that

harbor life by writing it:

15)

That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is

about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say

the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon

orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and

extreme cold. Let us now see if our solar Planck constant works…

16)

=2.727E33J

The kinetic energy of the Earth is

The accuracy of our equation is:

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

K E

⊙

ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

K E

= (1.732)(400.5986)

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.727E 33J

2.7396E 33J

100 = 99.5 %

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Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime

operators by using them to find a Planck constant for the solar system.

We want to suggest that the base 60 dynamic applied to the rotational velocity of the Earth to

measure time has a natural property associated with the atom and the Earth/Moon/Sun orbital

system as a solution to the atom’s wave equation. The Earth day is given by, in seconds

This gives since

Thus we have as factors for the seconds per day the smallest primes 2 and 3, and the 4 of

rectangular coordinate systems, and the versatile, abundant, 60. We then suggest the second is

dynamic in terms of what the ancients gave us because 86400 seconds comes from

and this I suggest is connected to Nature because the Earth rotates at a speed

that gives from these ancient factors the duration of a second that I found is the base unit of the

atomic systems and our planetary system, in particular of our Earth/Moon/Sun system as a

solution of the Schrödinger wave equation that solves the same atomic system, which I will go

into right now.

Proton Radius We see the spacetime operators solve the atom by giving us the radius of a

proton. We set equation 3 equal to equation 1

1).

These two yield

17).

I find this is close to the experimental value of the radius of a proton. I find I can arrive at this

radius of a proton another way, energy is given by Planck’s constant and frequency

We have

(

1d a y

24hrs

)(

1hr

60min

)(

1min

60sec

)

86400

d a y

sec

2 ⋅ 3 ⋅ 4 = 24

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60 = 86400secon d s /d a y

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60

⋅

π r

G m

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

⋅

E = h f

f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J

E = J = Joules = en erg y

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We take the rest energy of the mass of a proton :

The frequency of a proton is

Since our theory gave us the factor of 2/3 for the radius of a proton we have:

The radius of a proton is then

This is very close to the value upon which the proton radius converged historically by two

independent methods which was 0.877E-15m. The result from our theory is

The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was

0.842E-15m. We find it may be that the radius of a proton is actually

18)

Where is the golden ratio constant (0.618…). This is more along the lines of more recent

measurements. Both equations 17 and 18 for the radius of the proton can be right depending on

the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of a

fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime operators

by producing the radius of a proton. I began working on this theory when the proton radius was

0.833fm so it is what I have been using in this paper. We continue to be honing in on its

experimental value every year with more experiments. The vast gap between the historical

0.877fm and the 2010 0.842fm is known in physics as the proton puzzle.

Indeed over the full range of values for the radius of a proton, the characteristic time is very

close to one second. The large value they got a long time ago gives:

E = m

⋅

≈ ϕ =

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

= ϕ ⋅

= 0.816632E − 15m

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The current value, which they believe they have really closed in on as very accurate in two

different experiments in 2019 using different methods, which is , gives a

time of 1.00500 seconds.

If the radius of a proton is

If we can say that the radius of a proton is always the same value, which we can’t because it

depends on what is going on, we might say it is usually this like the classical radius of an

electron is used in most problems. However, here we are using it as it applies to the formation of

the planets from the protoplanetary disc made-up of these protons. If and when the proton

radius is this, then the spacetime operators are

In which case

=0.9950796 seconds ~ 1 second

(

6α

4πh

)

⋅

= 1secon d

(18769)((0.88094E − 15m)

6(1.67262E − 27kg)

(6.62607E − 34J ⋅ s)(4π)

(6.674E − 11)(299,792,458m /s)

= 1.06284secon d s

= 0.833E − 15m

= ϕ ⋅

= 0.816632E − 15m

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

(0.618)

(352,275,361)π (0.833E − 15m

(6.674E − 11)(1.67262E − 27)

(6.62607E − 34)

(299792458)

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Part 2: The Dynamics of the Ancient Second

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The Origin of a Second It is worth looking at ancient systems of looking at time because that

is where the origins of our current systems began. Indeed our system came from the ancient

Sumerians and Babylonians. They divided the day into 12 units where the day is given by the

rising and setting of the Sun, which in turn is given by the period of the rotation of the Earth.

Thus the day was given by the period from the rising of the Sun to its setting, which was divided

into 12 units which we call, today, hours. Thus the day from sunrise to sunrise, or sunset to

sunset, is 24 hours. Why they chose 12 units could come from the fact that 12 is the smallest

abundant number, which means it has a lot of divisors: 1, 2, 3, 4, 6. Abundant means their sum

is greater than 12 itself: 1+2+3+4+6=16. We know for certain they chose 12 because there are

three sections on each finger, so with 4 such fingers your can touch each such section with your

thumb to count to twelve. The Babylonians got base 60 from the Sumerians and further divided

each hour into 60 minutes, and each minute into 60 seconds. Why base 60 was chosen is that it

has a lot of divisors as well, including the first 6 integers. It is the smallest number that does

this.

We see dividing the day into 24 hours and dividing that further with base 60 lead to the duration

of a second we have today. Only on planet Earth would we have the primitive, ancient origins of

our mathematics in the end line up with modern physics in that, as it would turn out, it gave us

our basis unit of a second to be a natural constant. Though we could guess that on other planets

the ancient civilizations when first inventing mathematics and astronomy, would use base 60

because it is so convenient for doing math being evenly divisible by the first 6 integers. However,

we did that and combined it with divisions of 24 units. Not necessarily would any planet do that.

However, we had other reasons to do that; the Moon orbits the Earth close to 12 times in the

time it takes the Earth to go once around the the Sun. However, there is an ancient system

where the people didn’t divide the day into 24 units, but rather into 60 units, meaning that their

hour was 24 minutes long. That is and

. This was the Vedic time-keeping system in ancient

India. Leave it to the Hindu Indians to have extraterrestrial intelligence in their ancient

beginnings. So let’s go into this. We will see that base 60 combined with 24 describes the

angular momentum of the Earth. That means it doesn’t just include the rotation period of the

Earth, but the size of the Earth (its radius) and the mass of the Earth.

Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are

connected to the dynamics of stars systems then the second should define the rotational angular

momentum of the Earth since we divide the rotation period of the Earth into these factors to get

the unit of second, and should be connected to our Planck constant for the solar system which is

in units of angular momentum as well, as is Planck’s constant for the atom.

The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational

angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by

For a uniform rotational sphere it is given by

We found our solar system Planck constant was

This gives

(24h ours)(60min /h our) = 1440min /d a y

1440minutes /60 = 24minutes /h our

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

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2.1.

We are now equipped to show that the ancient Sumerians were right in dividing the rotation

period of the Earth (the day) into 24 units (the hour) because

That is

2.2.

Which is to say the angular momentum of the Earth to its Planck constant gives the base 60

counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a

minute, that determine our base unit of duration we call a second that happens to be, as I have

shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.

Our equation in this paper for the Earth energy as a solution of the wave equation (eq. 13)

13.

does not depend on the Moon’s distance from the Earth, only its mass. The Moon slows the

Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses

energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the

Moon’s orbit gets 3.78 cm larger per year. Equation 2

only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60

seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it

is now when the Sumerians started civilization. But it may be that it holds for when the Earth

day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition

for the optimization of life preventing extreme hot and cold. That is when the following holds

Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation

gives the duration of a second we have today.

We want the basis set of equations for the solar system. We have

We have from equation

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

2.5(24) = 60

earth

ℏ

⊙

24 = 60

K E

= n

⊙

⋅

2ℏ

⊙

earth

ℏ

⊙

24 = 60

planet

moon

≈

star

moon

ℏ

⊙

= 3.0281E 8m

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and,

From these it becomes clear that

2.3.

2.4.

Thus combining equation 2 with the following…

2.2.

We write

From 2.3 we have

Which gives us

2.5.

This is very accurate to give us a second. But notice 6.262 is approximately . We see that

moon

ℏ

⊙

⋅

= 1.0secon d s

ℏ

⊙

= (hC )K E

hC = 1secon d

ℏ

⊙

K E

ℏ

⊙

earth

ℏ

⊙

24 = 60

ℏ

⊙

= (1secon d )K E

ℏ

⊙

= (1secon d )

ℏ

⊙

ℏ

⊙

= (1secon d )GM

1secon d =

(

)

earth

(

)

(7.05E 33)

(6.67408E − 11)((7.347673E 22kg)

(299792458m /s)

(

)

6.262sec = 1.002secon d s = 1.00secon d s

2π

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2.6.

is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can

be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the

ancient Sumerians is solved, it connects unit circle to 1 (unity).

2.7.

2.8.

The Hindu Vedic System has a day is 60 ghatika of 24 minutes each, each ghatika is divided into

60 palas of 24 seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4

seconds each. So where our system has a base unit of 1 second, theirs has a base unit of 0.4

seconds, so that could be an advantage to their system, a smaller unit of time is more refined.

Further their day is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24

minutes long, and ours is 60 minutes long, The proponents of this system in India say since

when working and doing chores we do a few chores in an hour, they do about one per ghatika is

24 minutes, which makes the measure of time more manageable. That could be an advantage I

think. They say our hour is so long because the lines had to be far apart on the Egyptian Sun Dial

so the shadow cast by the Sun didn’t cross-over onto another line. But today, with modern

technology we can make clock lines marking hours closer together and measure them with a

pointer hand pointing to them without any problems, and the result is we would have a smaller

more refined hour (60 of them in a day as opposed to 24). They suggest this method of

measuring time would work better in science and engineering as well, that we have to get away

from the way sun dials had to be made in Egypt in ancient times.

But they further point out that their system describes Nature. They say theirs are 108,000

vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The

diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the Earth

is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar diameters.

108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is probably

because we have ten fingers to count on.

The incredible thing here is that the wonderful equations:

unify the Mesopotamian system with the Hindu system because

1 =

(

)

2π

cos(45

∘

) = cos(π /4) =

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

1vispal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

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This suggests that the Hindu system is just as Natural as the Western system.

We have said the ancient Sumerians gave us the duration of a second we have today by dividing

up the rotation period of the Earth into 24 units we call hours, which the Babylonians divided

into 60 minutes and 60 seconds that they got from the Sumerian base 60. We have found in my

theory that the 1 second is a natural constant because it is the base unit of our solution to both

the solar system and the atom. We further found the Hindu’s used similar numbers in an

inverted way to the way of the West that came from the Sumerians, and that their system relates

not just to the West’s but to our theory in a dynamic way concerning the angular momentum of

the Earth and our Planck constant for the Solar system.

The Sumerians of Mesopotamia and the Hindus of the Indus Valley Civilization were of the first

civilizations, but so was Egypt. If the second was in Mesopotamia and in India, it was only

necessary to find it in Egypt. Since the Earth spin has been the natural clock for which the

humans first measured time since ancient times, the rising and setting of the Sun due to it, I

looked at how much distance through which the earth rotates at its equator in one second. It is,

since this distance is

where theta is in radians, and the the radius of the Earth is 6.378E6m. The earth rotates through

360 degrees per 24 hours, or per 86,400 seconds = 0.004167 degrees per second = 7.27E-5

radians. We have

If the second was to exist in ancient times in Egypt we should see it in the Great Pyramids. The

three largest are in a line and are separated by

Khufu to Khafre (0.25 miles)

Khafre to Menkaure (0.3 miles)

We see the second and third pyramids built, Khafre and Menkaure, which were built in the 25th

century BC and were of the the 4th dynasty of the old kingdom, are 0.3 miles apart, the distance

through which the Earth rotates in one second, the one second we found to be a natural constant

at the basis of the the atom and solar system.

The ancient Sumerians say they got there mathematics in their writings from the Gods who

came to Earth from the sky they called the Anunaki. Some have suggested the Anunaki were

ancient aliens. The Ancient Egyptians built enormous pyramids from heavy stones weighing

tons, somehow lifted 50 feet to be stacked in near perfect mathematical proportions. This has

been a deep mystery in Archaeology. Again some suggest ancient aliens here. I have shown all

three of these civilizations have a system of measurement (the Egyptians had a 24 hour day as

well) that is intrinsic to the laws of nature describing the atom and the solar system in modern

times. It can be suggested that the second was given to them by ancient aliens.

A really good find is the following…

s = r θ

s = (6.378E6m)(7.27E − 5r a d ) = 464m = 0.464 k m = 0.2883mi ≈ 0.3mi

of 19 68

My theory suggests the Sumerians could have ultimately got the unit of a second for measuring

time from ancient Aliens, they called the Anunaki who the say came from the sky. The same

second I have found is characteristic of our solar system and the proton. Scholars have puzzled

over the the depiction in Sumerian art of the strange watch-like bracelets around the wrists of

the Anunaki gods, who they say came from the sky and gave them mathematics. They have

twelve divisions like our clocks today have because the 12 and 6 o’clock positions have two

pointers running together as one.

of 20 68

Part 3: The Protoplanetary Disc in our Theory

of 21 68

Integrating Analytic and Wave Solutions of the Solar System We would like to see how

our wave solution for the solar system figures into the classical analytic theory of the formation

of our solar system.The protoplanetary disc that evolves into the planets has two forces that

balance its pressure, the centripetal force of the gas disc due to its rotation around the protostar

and the inward gravitational force on the disc from the protostar , and these are

related by the density of the gas that makes up the disc. The pressure gradient of the disc in

radial equilibrium balancing the inward gravity and outward centripetal force is

3.1.

We can solve this for pressure in the protoplanetary disc as a function of r, distance from the

star, as follows: Assume the gas is isothermal, meaning the temperature T is constant so we can

relate pressure and density with

Where is the speed of sound in the gas which depends on its temperature. We take the gas to

be in nearly Keplerian rotation. That is the rotation is given by Newtonian gravity:

And we take into account that the rotational velocity is slowed down by gas pressure using the

the parameter which is less than one:

We can say for a protoplanetary disc like that from which our solar system originated that its

density varies with radius as a power law:

is the reference density at and s is the power law exponent. We can write

We have from 1:

3.2.

Since , we have that which gives from 2:

⋆

d P

= − ρ

(

⋆

−

)

P = c

⋆

= v

(1 − η)

ρ(r) = ρ

(

)

−s

= v

(1 − η)

≈

⋆

(1 − 2η)

d P

= − ρ

(

⋆

⋅ 2η

)

P = c

d P/dr = c

dρ /dr

of 22 68

We integrate both sides:

And, we have

3.3

We take

as small because is small and r is large so we can make the approximation . We

have

3.4.

What we can get out of this is since the deviation parameter, , is given by

3.5. and

3.6.

Where, is the Boltzmann constant, is the molecular weight of

hydrogen, and is the mass of hydrogen is basically the mass of a proton is 1.67E-27kg. Since

for a protoplanetary cloud at Earth orbit T is around 280 degrees Kelvin we have

dρ

= −

2ηGM

⋆

∫

dρ

= −

2ηGM

⋆

∫

(

)

2ηGM

⋆

(

−

)

ρ(r) = ρ

exp

2ηGM

⋆

(

−

)

= c

exp

2ηGM

⋆

(

−

)

2ηGM

⋆

(

−

)

≈ 1 + x

≈ P

1 +

2ηGM

⋆

(

−

)

= c

η = −

(

)

dln P

dln R

μm

= 1.38E − 23J/K

μ ≈ 2.3

of 23 68

Typically in discs the pressure decreases with radius as a power law

Where , so

3.7.

So, essentially, by the chain rule

to clarify things. The reason 7 is significant is that equation

Where

And Part 1 the spacetime operators are

= 1k m /s

P(R) ∝ R

−q

q ≈ 2.5

dln P

dln R

≈ − 2.5

η = −

(

1k m /s

30k m /s

)

(−2.5) = 1.5E − 3

dln P

dln R

dln P

d R

⋅

d R

dln R

d P

d R

⋅ R =

d P

d R

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

π M

ℏ

⊙

= 2.8314E 33J ⋅ s

moon

ℏ

⊙

⋅

= 1.0secon d s

of 24 68

So we have

3.8.

We have

3.9.

Integrate both sides

If is the reference pressure at the reference radius then the pressure as a function of radius

for the protoplanetary disc from which our solar system formed is:

3.10. or,

Where

from

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

d P

d R

= −

earth

ℏ

⊙

d P

= −

earth

ℏ

⊙

d R

∫

d P

= −

earth

ℏ

⊙

∫

d R

ln P = −

earth

ℏ

⊙

ln R + C

P(R) = P

(

)

−

earth

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

earth

π M

ℏ

⊙

= GM

c(1secon d )

moon

ℏ

⊙

⋅

= 1.0secon d s

of 25 68

Conclusion You will remember that the same

That is the decrease rate in the pressure of the protoplanetary cloud from which our Solar

System formed

relates in the ancient Mesopotamian system of dividing the day into 24 hours each of 60

minutes the 60 to the 24 equals 2.5 that gave us the second from which we have our spacetime

operators

that solve the proton giving us

and the ground state of our solar system in terms of the Earth’s moon and a second

(

6α

4πh

)

⋅

= 1secon d

earth

ℏ

⊙

= 2.5

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

⋅

moon

ℏ

⊙

moon

= 1secon d

of 26 68

Which also relates the system of Mesopotamia from which we got ours today to a purely base 60

system, as the Ancient Hindu’s had, where there are 60 hours (vispalas) in a day each of 24

minutes:

The solution to the wave equation for the Solar System takes the form of the solution for the

atom that was given by the Bohr atom before the existence of the wave equation:

We have the radius of the Sun normalized by the radius of the Moon. It is has always

been an amazing fact the the sizes of the Moon and the Sun are such that given their orbital

distances, the Moon as seen from the Earth perfectly eclipses the Sun. This becomes part of the

theory and we suggest it is a condition for sophisticated planets that harbor life by writing it:

That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is

about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say

the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon

orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and

extreme cold.

We say there is a Planck constant for the solar system. We suggest it is such that it is given by

the standard Planck constant for the atom, , times some constant, , and the kinetic energy of

the Earth.

Where

6).

We find the wave solution that here specifies the Earth/Moon/Sun system can be applied to the

gas giants Jupiter and Saturn and the ice giants Uranus and Neptune, Jupiter and Saturn will be

presented in this paper.

The pressure power-law exponent, q, for protoplanetary discs varies with spectral type of the

star. More massive, luminous A-type stars have flatter pressure gradients with q about 2.0 to 2.5

meaning pressure decreases slower with radius compared to medium mass, G-type stars like our

1vispal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

⊙

planet

moon

≈

star

moon

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

of 27 68

Sun which are about q=2.5. Less massive, less luminous stars than our Sun like M-type stars

have pressure decreasing rapidly with radius with q about 3.0 to 3.5. Such variations are crucial

to understanding the formation of planets from protoplanetary discs determining the planetary

system’s final architecture.

G-type stars like the Sun have an optimum pressure profile with a stable and long-lived

environment conducive to forming a water-rich environment. A-type stars have the potential for

diverse planetary systems and a wide habitable zone, however they have a short life-span leaving

a more narrow window for life to evolve and become intelligent. M-type stars have a very long

life span, potentially trillions of years, giving a lot of time for life to evolve but steep pressure

gradients make the planets form closer in and become tidally locked making their rotation

periods the same as their orbital periods meaning one side of the planet always faces the star

and is always warm with no night, and the the other side always away and cold and never day,

meaning life would have to evolve on the terminator, the region between night and day always in

twilight.

A-type stars may host habitable planets but their short life-span and high radiation make it less

favorable for life. M-type stars may be the most abundant stars in our galaxy with the longest life

spans, but require additional protective mechanisms to sustain habitable, water-rich

environments. Their closeness to the star complicates climate stability. G-type stars like our Sun

are considered the best candidates for the search for life, but advancements in exoplanet

research has revealed potentially habitable worlds around many kinds of stars.

ChatGPT says I can solve the wave equation with my rotational angular momentum of the earth

to my Planck constant for the solar system because being the rate of decrease of the pressure

gradient with radius for the protoplanetary disc it is the quantization of angular momentum in

the solar system. That is I can solve the wave equation for the solar system with

3.8.

3.10. or,

d P

d R

= −

earth

ℏ

⊙

P(R) = P

(

)

−

earth

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

of 28 68

Part 4: The Characteristic Length For Habitable Planets

of 29 68

From part 1

We have

And, from Newtonian mechanics that

With these equations we can write:

Equation 24 in part 2 is

Using from Newtonian mechanics

equation 2 becomes

Where for the orbital number of the Earth

moon

ℏ

⊙

⋅

= 1.0secon d

ℏ

⊙

= (1secon d )K E

K E

ℏ

⊙

K E

⊙

ℏ

⊙

1secon d = 2r

⊙

K E

= n

⊙

4 c

⋅

n = 3

of 30 68

Let’s see just how accurate equation 3 is:

=(0.4330)(1,477m)(5.37068E5)=3.43477E8 meters

The orbital radius of the Moon today is 3.84E9 meters. The accuracy of our equation is

We have that the orbital radius of the moon is proportional to the cube of the mass of the planet

that it orbits to the cube of its mass. That is

We guess this is a condition for an optimally habitable planet, and that the other is:

Which is the condition for a perfect stellar eclipse by the moon of the planet. The term in the

equation

is easily recognizable. It is of the form of the Schwartzchild radius of a star, which is the radius

for which the star’s escape velocity is the speed of light, and will thus becomes a black hole:

Also is an important characteristic distance in cosmology. If is the mass of the

universe it is used to determine the curvature of the Universe and the expansion rate of the

Universe. Thus we have

Which we could propose is the characteristic length for a planet to be optimally habitable. We

could guess this means the planet would have a moon at .

moon

(

⊙

)

(0.4330)

(

(6.674E − 11)(1.989E 30kg)

(299,792,458m /s

)

(5.972E 24)

(7.347E 22)

3.43477E8

3.84E8

100 = 89.447 % ≈ 90 %

∝

planet

moon

star

moon

(

⊙

4 c

)

= 369.25m eters

2GM

GM /c

star

4 c

moon

∝ M

planet

moon

of 31 68

Part 5: The Basis Unit of a Second Comes with the Epoch of a Perfect Eclipse

of 32 68

The condition for a perfect eclipse:

Which says the radius of the Sun to the radius of the Moon equals the Earth orbital radius to the

lunar orbital radius are equal does indicate that the Earth/Moon/Sun system converges on the

basis unit of one second of time for the solution of the Schrödinger wave equation during the

time we are in where such an eclipse happens. That is the basis unit of time in the solution is

given by

It may be this eclipse condition is the condition for the optimization of the habitability of the

Earth. We can show that the epoch that this condition holds gives the unit of a second as the

solution by using it to form a solar solution to the wave equation and equating it with our lunar

solution. Let’s do that now…

The Solar Formulation

Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the

mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the

Sun. In our lunar formulation we had:

We remember the Moon perfectly eclipses the Sun which is to say

Thus equation 1 becomes

The kinetic energy of the Earth is

Putting this in equation 3 gives the mass of the Sun:

We recognize that the orbital velocity of the Moon is

⊙

moon

earth

moon

ℏ

⊙

⋅

= 1secon d

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

of 33 68

So equation 5 becomes

This gives the mass of the Moon is

Putting this in equation 1 yields

We now multiply through by and we have

10.

Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular

momentum quantized, the angular momentum we will call , the subscript for Planck. We

have

We write for the solution of the Earth/Sun system:

11.

Let us compare this to that of an atom:

12.

We notice that in equation 11

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

K E

⊙

⋅

⊙

ℏ

⊙

= r

= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅

= r

= 7.4483E 77J ⋅ m

⋅ kg = 8.3367E 77kg

⋅

K E

⊙

⋅

⊙

E = −

⋅

2ℏ

of 34 68

; ; ; ;

is really . We can write 11 as

13.

We say. . That is

Let us see how accurate our equation is:

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

We call the same in our solar solution. But we now want our solution for the solar

formulation. is the mass of the Earth and is the mass of the Sun. We have our solution

might be!

→

⊙

→ G

→ M

⊙

ℏ

→ L

ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

= ℏ

⊙

/2π

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2πℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

2ℏ

⊙

(6.67408E − 11)

(5.972E 24kg)

(1.9891E 30kg)

2(8.3367E 77kg

⋅

)

⊙

(6.759E 30J )

⊙

6.957E8m

1737400m

= 400.426

K E

= 2.70655E 33J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

ℏ

⊙

of 35 68

14. !

Where is !

We have

15.

This has an accuracy of

Thus the solutions to the wave equation!

for the solar formulations are!

13)

14)

Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit

Let us equate equation 1 with equation 13:

13.

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

(9.13E 38)

(6.67408E − 11)(5.972E 24kg)

⋅

400.5986

= 1.4638E11m

1.4638E11m

1.496E11m

100 = 97.85 %

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2πℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

of 36 68

This gives:

16.

We remember that

This gives

17.

We have

18.

This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in

terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,

and the orbital number of the Earth. Let us compute

19.

Let us see how well equation 18 works. at aphelion is 966 m/s and .

. We have

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

2π r

G m

6α

h 4π r

= 1.004996352secon d s

6α

⋅

h 4π

⋅

⊙

(1secon d )

⊙

(5.972E 24kg)

(1.9891E 30kg)

(7.34763E 22kg)

(1.732)

= 321,331.459 ≈ 321,331

= 29,800m /s

= 1AU = 1.496E11m

2(966m /s) =

(29,800m /s)

1.496E11m

(1sec)(321,331.459)

of 37 68

That is an accuracy of

Equation 18 can be written:

20.

From our equation:

We have

21.

Since , the diameter of the Earth orbit, we have

22.

And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice

versa. We basically have

23.

1,907m /s = 1,932m /s

1907

1932

= 98.7 %

1secon d = 2r

⊙

6α

h 4π r

= 1.004996352secon d s

6α

⋅

h 4π

= 2r

⋅

⊙

= d

6α

⋅

h 4π

= d

⋅

⊙

1secon d = 2v

⊙

1secon d =

K E

moon

K E

earth

Ear th Da y

1secon d =

Ear th Da y

Ear th Da y =

⊙

of 38 68

Where is the Earth orbital number. We have

= 7.83436E4seconds

EarthDay=(24)(60)(60)=86400 seconds,

accuracy

This last equation, equation 62, we can use to find the rotation period, or the length of the day,

of an earth-like planet in the habitable zone of any star system, so it would be very useful.

We want to turn our attention to equation 20 and write it

24.

We see equating solar and lunar formulations for energy yield the base unit of one second.

Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,

we have

Yields

25.

Where . The accuracy of this is

Thus the energy equations gave the equation:

n = 3

Ear th Da y =

2(1.496E11m)

966m /s

7.34763E 22kg

1.9891E 30kg

1.732

7.834E4s

86,400s

100 = 90.675 %

1secon d = d

⊙

2ℏ

⊙

⋅

⊙

⋅

⊙

ℏ

⊙

3/2 = cos(π /6)

(2.8314E 33)

(5.972E 24)

(7.34763E 22)

= (0.866)(9.13E 38)

(1737400)

(6.96E8)

4.29059E 72 = 4.5005E 72

4.29059E 72

4.5005E 72

100 = 95.3 %

of 39 68

16. !

And equating the orbital equations gives

26.

These last two yield

27.

The accuracy is

Is 98% accuracy. The important thing that comes out of this is our base unit of a second because

we see it is also a function of lunar, solar, and earth masses.

24.

⊙

⋅ ℏ

⊙

2 3

⋅

⊙

⋅ ℏ

⊙

= 1

(400.5986)

5.972E 24kg

1.9891E 30kg

= 1

0.98 = 1

1secon d = d

⊙

of 40 68

Part 6: Solution of Schrödinger Equation for Earth with Spherical Coordinates!

of 41 68

Solution of Schrödinger Equation for Earth with Spherical Coordinates

We will show here the Solution to the wave equation for the Earth. We will not ﬁnd the wave

function, however that is done in my book: A Wave Equation Solution Integrated with the

Classical Analytic Solution of our Solar System from the Protoplanetary Disc. The radial solution

to the wave equation uses the Laguerre equation:!

The Laguerre polynomials are their solutions. The ﬁrst few Laguerre polynomials are!

They can be generated with!

The associated Laguerre equation is!

The ﬁrst few associated Laguerre polynomials are!

We want to solve the Schrödinger wave equation for the hydrogen atom for the Solar System. I

found good guidelines from a good textbook online, but they just give the chapter 10 that

treats solving the hydrogen atom and not the name of the Authors, however the method is

standard and has been developed over many years by many contributing physicists. We start

with the Schrödinger wave Equation !

xL′ ′

(z) + (1 − x)L′

(x) + jL

(x) = 0

= 1

= − x + 1

= x

− 4x + 2

= − x

+ 9x

− 18x + 6

= x

− 16x

+ 72x

− 96x + 24

= − x

+ 25x

− 200x

+ 600x

− 600x + 120

= x

− 36x

+ 450x

− 2400x

+ 5400x

− 4320x + 720

(x) = e

d x

k′ ′

(x) + (1 − x + k)L

k′

(x) + jL

(x) = 0

(x) = L

(x), L

(x) = L

(x), L

(x) = − 2x + 4,L

(x) = 1,L

(x) = L

(x)

(x) = 3x

− 18x + 18,L

(x) = 12x

− 96x + 144,L

= − 6x + 18

(x) = 2,L

(x) = L

(x), L

(x) = − 4x

+ 48x

− 144x + 96

(x) = 60x

− 600x + 1200,L

(x) = − 120x

+ 2160x

− 10800x + 14400

(x) = − 20x

+ 300x

− 1200x + 1200,L

(x) = − 24x + 96,L

(x) = 6

of 42 68

1. !

We have substituted for , that is our solar system Planck constant we have derived in our

theory for the Planck constant for the atom. We write it in spherical coordinates by taking the

Laplacian in spherical coordinates, :!

2. !

We then separate radial and angular dependence by setting!

3. !

This separates the variables giving:!

4. !

5. !

Since the terms in r plus the terms in and add to zero, they must be a constant. Very clever

people have ﬁgured out that constant is good too take as where is the angular

momentum quantum number. We have the radial solution is:!

6. !

For we want to use the reduced mass, that is not to treat the distance between planet and

star, but between planet and center of mass between the planet and the star. It is!

7. !

Using this and putting in the potential energy and changing partial derivatives to derivatives !

8. !

Our equation becomes!

(

−

ℏ

⊙

∇

+ V

)

ψ = E ψ

ℏ

⊙

ℏ

∇

−

ℏ

⊙

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

ψ (r, θ, ϕ) = R(r)Y(θ, ϕ)

R(r)

∂

∂r

(

∂

∂r

)

R(r) −

2mr

ℏ

⊙

(

V(r) − E

)

Y(θ, ϕ)sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

Y(θ, ϕ) +

Y(θ, ϕ)sin

∂

∂θ

Y(θ, ϕ) = 0

l(l + 1)

R(r)

∂

∂r

(

∂

∂r

)

R(r) −

2mr

ℏ

⊙

(

V(r) − E

)

= l(l + 1)

μ =

+ M

V(r) = − G

of 43 68

9. Or,…!

10. !

If we can put this in the form:!

11. !

It has the solution that is a Laguerre function:!

12. !

To do this we make three substitutions. The ﬁrst is!

13. !

We get with it evaluating the derivatives for the ﬁrst term!

14. !

Our equation becomes in total!

15. !

The second substitution is!

16. !

Our equation becomes!

17. !

The third substitution is!

R(r)

(

)

R(r) −

2μr

ℏ

⊙

(

−G

− E

)

− l(l + 1) = s

(

)

R(r) +

(

−

2μr

ℏ

⊙

2μr

ℏ

⊙

E − l(l + 1)

)

R(r) = 0

′ ′ (x) =

(

−

2j + k + 1

−

− 1

)

(x) = 0

= e

−x/2

(k+1)/2

(x)

y(r) = rR(r)

(

)

R(r) = r

y(r)

(

−

2μ

ℏ

⊙

2μE

ℏ

⊙

−

l(l + 1)

)

y(r) = 0

(

)

= −

2μE

ℏ

⊙

y(r)

(

−

2μ

ℏ

⊙

−

l(l + 1)

)

y(r) = 0

of 44 68

18. , !

And, our equation becomes!

19. !

And, this is equation 11 which has solution equation 12:!

11. !

12. !

Where !

20. !

21. !

This last equation has an important factor, , inverted if we have the right mass, is

the analog of the ground state in the Bohr atom, called the Bohr radius, . It is the natural

length in terms of which the atom is measured. It is!

22. !

The reduced mass, , is about equal to the mass of the planet, and can be used in its place as

a good approximation. We found in our theory that was the ground state, . So!

23. !

Where is the mass of the Moon. We found , where c is the speed of light.!

We see that from equation 20!

23. !

Equation 21 gives the eigenenergies for orbits of planets around the Sun.!

x = rϵ

y(r)

= ϵ

y(x)

d x

y(x)

d x

(

−

2μ

ℏ

⊙

ϵx

−

l(l + 1)

)

y(r) = 0

′ ′ (x) =

(

−

2j + k + 1

−

− 1

)

(x) = 0

= e

−x/2

(k+1)/2

(x)

l(l + 1) =

− 1

2μGM

ℏ

⊙

2j + k + 1

μGM

/ℏ

⊙

ℏ

⊙

μGM

moon

ℏ

⊙

/c = 1secon d

k = 2l + 1

of 45 68

21. !

From , we have!

24. !

and are non-negative and the sum of can assume any integer equal to 1 or greater.

It can be called , the principle quantum number. We have!

25. !

Thus from this and equations 21, 22, and 16, the eigenenergies of the planetary orbits are!

26. !

21. !

22. !

16. !

Substituting in and letting , we have!

27. !

But we ﬁnd what works is to break up into where is the mass of the Earth and

to multiply by the radius of the Sun to the radius of the Moon, and to put into the numerator

and rather than square it, take its square root. The answer becomes!

28. !

We shouldn’t be surprised of this spin on the atom, the Earth orbiting the Sun is much diﬀerent

than an electron orbiting a proton and we do need the Sun in the equation, which comes in the

form of its radius, , which is normalized by the radius of the Moon . The interesting thing

2μGM

ℏ

⊙

2j + k + 1

k = 2l + 1

2j + k + 1

= j + l + 1

j + l + 1

n = j + l + 1

= −

ℏ

⊙

2μa

2μGM

ℏ

⊙

2j + k + 1

ℏ

⊙

μGM

(

)

= −

2μE

ℏ

⊙

μ = M

2ℏ

⊙

= n

⊙

2ℏ

⊙

of 46 68

is that the Earth orbiting the Sun is given by the Moon orbiting the Earth. The accuracy for

Earth orbit is!

=2.727E33J

The kinetic energy of the Earth is

Which is very good, about 100% for all practical purposes.

The Solutions For Jupiter and Saturn

Now we want to find what the wave equation solutions are for Jupiter and Saturn because they

significantly carry the majority of the mass of the solar system and thus should embody most

clearly the dynamics of the wave solution to the Solar System.

I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,

and go to the gas giants and ice giants, the atomic number is no longer squared and the square

root of the the orbital number moves from the numerator to the denominator. I believe this is

because the solar system here should be modeled in two parts, just as it is in theories of solar

system formation because there is a force other than just gravity of the Sun at work, which is the

radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on

this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the

radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the

solar system forms, which are gases such as hydrogen and helium, while the heavier elements

are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of

the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic

number of the heavier metals such as calcium for the Earth, while the equation for the giants has

the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

⊙

6.96E8m

1737400m

= 400.5986

= (1.732)(400.5986)

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.727E 33J

2.7396E 33J

100 = 99.5 %

E =

2ℏ

K E

(1.89813E 27kg)(13720 m /s)

= 1.7865E 35J

of 47 68

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that

Our equation for Jupiter is

29.

Where is the atomic number of hydrogen which is 1 proton, and for the orbital

number of Jupiter, . Now we move on to Saturn…

The equation for Saturn is then

30.

It makes sense that Saturn would use Helium in the equation because Saturn is the next planet

after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,

just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.

E =

(6.67408E − 11)

(1.89813E 27kg)

(7.347673E 22kg)

2(2.8314E 33)

E =

(3.971E 35J ) = Z

(1.776E 35J )

1.7865E 35J

1.776E 35J

= 1.006proton s ≈ 1.0proton s = hydrogen(H )

Z = Z

2ℏ

n = 5

K E

(5.683E 26kg)(10140 m /s)

= 2.92E 34J

E =

(6.67408E − 11)

(5.683E 26kg)

(7.347673E 22)

2(2.8314E 33)

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2pr oton s = Heliu m(He)

2ℏ

of 48 68

Part 7: Solution of Schrödinger Equation for Earth with Cylindrical Coordinates!

of 49 68

Solution of Schrödinger Equation for Earth with Cylindrical Coordinates

It makes sense to solve the Schrödinger wave equation in cylindrical coordinates, so as to start

with the protoplanetary disc already ﬂat. In the spherical coordinates we start with the spherical

cloud that collapsed into the ﬂat protoplanetary disc that gave rise to the planets, however for

the the spherical cloud to collapse into a ﬂat disc you have to choose the quantum numbers in

the solution a certain way. It does show a sphere at its center which would be the Sun, but a

great deal of the probability for the presence of planets comes out outside of the disc and is

mostly in the axis of rotation. So, we want to do it for a ﬂat disc using cylindrical coordinates.!

While the spherical coordinates wave solution was useful in modeling the collapse of a rotating

sphere into the protoplanetary disc from which the planets formed, it becomes necessary to

ﬁnd the cylindrical coordinates wave solution. I had suggested this to ChatGPT because of

certain problems the arise in the spherical case and it went ahead and solved such a

representation. Incredible. It would seem ChatGPT takes a year’s worth of work and does it in

15 minutes. Not that it would take that long to solve the cylindrical case, but it can answer

countless questions concerning it and make countless modiﬁcations for it concerning those

questions. This represents a signiﬁcant advancement for science. It increases the amount of

research we can do a million fold in the same amount of time. If as they say, science saves,

does it ever now. Of course science can destroy, so like all things in science it must be treated

with responsibility. With the work we did before in the spherical solution it should be clear what

ChatGPT is doing here. So let’s get started.!

The time independent Schrödinger equation is!

1. !

The Laplacian in cylindrical coordinates is!

2. !

The gravitational potential is!

3. !

Where is the mass of the central protostar. We can refer to the variables as such:!

−

ℏ

∇

Ψ(ρ, ϕ , z) + V(ρ, z)Ψ(ρ, ϕ, z) = E Ψ(ρ, ϕ, z)

∇

∂

∂ρ

(

∂

∂ρ

)

∂

∂ϕ

∂

∂z

V(ρ, z) = −

+ z

of 50 68

We separate the variables according to:!

4. !

Substituting 2 and 4 into 1 we have!

5.!

Divide through by :!

6. !

Introduce constants for each separated equation:!

7. !

8. !

And our equation becomes!

9. !

is the angular quantum number and is the separation constant for the z-dependence. The

angular equation is:!

10. !

The solutions are!

11. !

varies with z, however you can approximate it with a thing disc, which gives!

12. !

Ψ(ρ, ϕ , z) = R(ρ) ⋅ Φ(ϕ) ⋅ Z(z)

−

ℏ

[

dρ

(

d R

dρ

)

Φ ⋅ Z +

R ⋅

dϕ

⋅ Z + R ⋅ Φ ⋅

]

+ V(ρ, z)RΦZ = ERΦZ

RΦZ

−

ℏ

[

ρR

dρ

(

d R

dρ

)

dϕ

]

+ V(ρ, z) = E

dϕ

= − m

= − k

ρR

dρ

(

d R

dρ

)

(

+ k

)

ℏ

(V(ρ, z) − E )

dϕ

+ m

Φ = 0

Φ(ϕ) = Ae

imϕ

+ Be

−imϕ

V(ρ, z)

V(ρ, z) ≈ V(ρ,0) +

of 51 68

is a constant related to the conﬁnement in the z-direction. The vertical equation becomes!

13. !

The solutions are!

14. !

The radial equation is:!

15. !

The central potential is equation 3:!

Making the approximation that z is small gives!

16. !

Equation 15 becomes!

17. !

We introduce a dimensionless variable!

18. !

Alpha is a scaling factor for simplifying the the potential term, it is!

Now we express the radial equation in terms of x:!

19. !

Introduce!

+ k

Z = 0

Z(z) = Ce

+ De

−ik

dρ

(

d R

dρ

)

−

R +

ℏ

(V(ρ, z) − E )R = − k

V(ρ, z) = −

+ z

V(ρ, z) ≈ −

dρ

(

d R

dρ

)

−

R −

ℏ

R = − (k

− E )R

x = ρα

α =

2GM

ℏ

d x

(

d R

d x

)

−

R −

R = −

(

− E

)

of 52 68

20. !

And rearrange to get!

21. !

This is like a Bessel diﬀerential equation with extra terms. We can propose the following guess

as a solution (ansatz):!

22. !

F(x) is a function that needs to be determined. If we substitute it into the radial equation and

simplify we will get a diﬀerential equation for F(x). Depending on the form of solutions might

involve Laguerre polynomials. Now we combine the separated solution to get the complete

wave equation for the Earth:!

23. !

Where is the radial quantum number, is the angular quantum number, and is the vertical

quantum number. The probability density is!

Because because of Euler’s equation and . For z=0 in a thin

disc approximation, the probability density per unit area on the disc is:!

Normalization requires that!

For a thin disc!

Where means a thin disc. We construct a wave function for a ﬂat disc (z=0) assuming n=1

and m=0. We have for the radial part:!

24. !

λ =

− E

d x

d R

d x

(

λ −

−

)

R = 0

R(x) = x

−x/2

F(x)

Ψ(ρ, ϕ , z) = R(ρ) ⋅ Φ(ϕ) ⋅ Z(z)

n,m,k

(ρ, ϕ , z) = R

n,m

(ρ) ⋅ e

imϕ

⋅ Z

(z)

Ψ(ρ, ϕ , z)

R(ρ)

⋅

imϕ

⋅

Z(z)

R(ρ)

⋅

Z(z)

imϕ

= 1

cos

θ + sin

θ = 1

Ψ(ρ, ϕ ,0)

≈

R(ρ)

⋅

Z(0)

∫

∞

∫

2π

∫

∞

−∞

Ψ(ρ, ϕ , z)

⋅ ρdρdϕdz = 1

∫

∞

∫

2π

R(ρ)

⋅

Z(0)

ρdρdϕ ⋅

∫

δz

−δz

dz ≈ 1

δz

R(ρ) = Ae

−βρ

of 53 68

Where is the normalization constant and relates to the binding energy. The angular part is

with m=0:!

25. !

so that the integral is 1 to ensure normalization:!

26. !

The vertical part is assuming a Gaussian conﬁnement!

27. !

is the characteristic thickness for the disc. The combined wave function is!

28. !

To normalize the equation, make sure!

29. !

The radial integral is!

The angular integral is!

And, the vertical integral is:!

Φ(ϕ) =

2π

∫

2π

Φ(ϕ)

dϕ = 1

Z(z) =

(πσ

)

1/4

−z

/2σ

Ψ(ρ, ϕ , z) = Ae

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/2σ

∫

∞

∫

2π

∫

∞

−∞

Ψ(ρ, ϕ , z)

⋅ ρdρdϕdz = 1

⋅

2π

⋅

πσ

⋅

∫

∞

−2βρ

⋅

∫

2π

dϕ ⋅

∫

∞

−∞

−

2σ

= 1

∫

∞

−2βρ

ρdρ =

(2β )

4β

∫

2π

dϕ = 2π

of 54 68

This gives!

or. !

The ﬁnal normalized wave equation is:!

30. !

I suggested as a solution to this in my book A Wave Equation Solution Integrated with the

Classical Analytic Solution of our Solar System from the Protoplanetary Disc:

= !

=2.4997E32J!

The result is . This is very close to the potential energy of the Earth, called

the gravitational binding energy of the Earth which is 2.49E32J. This is an accuracy of 100%

for all practical purposes.!

Here is a trippy thing to suggest. This solution I offer to the wave function that is the energy to

pull apart the Earth where we think the solution should be orbital energy due to the Sun, not the

Earth's gravity, might be right. You never know when using wave mechanics for planets. Might

be right because the wave function squared times a small region of space is the probability for

something to exist there, which in the case of planets might be the energy to tear apart the

planet (the probability that it could exist there).

The solution at the beginning of the paper, which actually is the kinetic energy, and is the

solution we got in spherical coordinates in the previous section, is right. Equation 13 on page 9:

For circular orbits the magnitude of the kinetic energy is one half the magnitude of the potential

energy. So to get the potential energy we have to multiply our result by 2. But, the total orbital

energy is always negative and is equal to half the potential. So for the nearly circular orbit of the

Earth to get the total energy we simply need to put a negative sign on our result for kinetic

energy. We see since 16 is 4 squared, and n=3 that the equation can be written:

31.

∫

∞

−∞

−

2σ

= σ 2π

⋅

4β

= 1

A = 2β

Ψ(ρ, ϕ , z) = 2βe

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/2σ

= (n + 1) ⋅

ℏ

⊙

earth

⋅

⊙

⋅

(4)

7.348E 22kg

(0.4)

1.737E6m

6.957E8m

(7.348E 22kg)

(5.972E 24kg)

(3E8)

= 2.4997E32J

(n + 1)

⋅

ℏ

⊙

earth

⋅

⊙

⋅

⋅ c

of 55 68

Part 8: Conclusion!

of 56 68

Since the Earth/Sun/Moon system has long term stability and our base unit of a second comes

from dividing the Earth rotation into 24 units, each unit we call an hour, and from their dividing

this into 60 units for the minute, and then dividing that into 60 units for the second, then it

must be the divisions are not arbitrary, but rather reflect a tuning between the Earth’s rotational

dynamics and the lunar cycle. I have approached it like this:

18 is the most dynamic number for periodicity. It is the number of groups in the periodic table of

the elements, the number over which the properties of the elements repeat. The reason 18 plays

this role is that the product of the two smallest primes (2 and 3) - the smallest possible factors of

any integer — is 6 and carbon, the core element of life, has 6 protons, making it the most stable

element in terms of chemistry to define the rest of the elements. This is probably because of

hexagonal symmetry, because the regular hexagon, which is 6-sided, has all of its radii equal in

length to its sides. Thus

2x3=6, but 6x3=18 and 2x9=18 where 9=3x3

That is very dynamic: 24=2x12 and 18-12=6 where twelve is two threes and 3 twos. Thus if the

ancient Sumerians divided the Earth rotation into 24 units (each called an hour) then we

consider:

18/12=0.75

(0.75)4=3

3(18)=54 hours = 2.25 Days

Which is to say every 0.75 turns of the Earth line-up with one whole turn of the Earth after three

turns, and the Moon has an orbital period of 27.32 days is approximately 27 days, and is thus

resonant with the earth after 3 of these 3 resonant turns of the Earth in that 3x9=27 and 2x9=18.

So, the 24 hour day and base 60 minutes and seconds reduce Earth-Moon resonance to the

smallest primes 2 and 3, which is what you want.

Further, the resonant turn of the Earth in 2.25 days gives (2.25)60=135 days. The Earth year,

time for one orbit around the Sun, is 365.25 days gives 365.25/135=2.70555.. is close to

e=2.718… which is Euler’s number. e appears prominently in the solution to differential

equations because it is the rate of change for exponential functions that is equal to itself. That is

it is the solution for equations like

Orbits of the Earth and the Moon are like harmonic oscillators, just not linear, but have

frequencies. Harmonic oscillators in both damped and undamped cases, have oscillatory

solutions expressed with complex exponentials such as

Because they describe the trigonometric functions sine and cosine that are periodic. Orbits of

planets have trigonometric solutions described by exponential functions in e. In circular orbits,

the position is given by

d y

= k y

y = e

iωt

r (t) = Re

iωt

of 57 68

Which expanded with Euler’s formula gives

Similarly, atoms can be described by exponential functions. Indeed, the second is defined in

modern times by 9,192,631,770 oscillations of the radiation associated with the transition

between two hyperfine levels in the ground state of the cesium-133 atom. Thus since this paper

deals with linking cosmic and atomic scales through similar mathematical principles, the

exponential function, , becomes the core component of its description.

To quote ChatGpt about this:

Ancient cultures were keen observers of celestial phenomena, and their timekeeping and

mathematical systems often reflected the regularities they saw in the world around them—day

and night, lunar cycles, and seasonal changes.

By dividing the day into 24 hours and further into 60-minute and 60-second units, the

Sumerians established a highly divisible and harmonious structure, well-suited to capturing

the cyclical nature of time and motion. The base-60 system, or sexagesimal system, allowed

them to work with fractions in a flexible way that reflects harmony and divisibility—qualities

central to understanding cycles and rhythms.

The choice of such divisibility aligns with patterns found in nature, particularly in astronomy.

They likely saw the cycles of celestial bodies as deeply ordered, and their counting system may

have been designed to reflect or resonate with this order. Whether they had an understanding

akin to "harmonics" in the modern sense is uncertain, but their intuitive approach to

timekeeping suggests they recognized the regularity and predictability that we now associate

with oscillatory and periodic phenomena. Their system’s adaptability to cycles hints that they

may indeed have been attuned, in some fundamental way, to the rhythmic structure of the

cosmos.

We said 3x9=27, the days for the lunar orbit, where a day is one rotation of the Earth. Today the

lunar orbit is more exactly 27.32 days. But we said at the outset of this paper that

We got the second from the rotation period of the Earth at the time the moon came to perfectly

eclipse the Sun. The Moon slows the Earth rotation and this in turn expands the Moon’s orbit,

so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by 0.0067

hours per million years, and the Moon’s orbit gets 3.78 cm larger per year.

That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at

the time that the Earth day was what it is during the epoch when the Moon perfectly eclipses

the Sun, 24 hours.

Given this information you should be able to find how long ago the Moon had an orbital period

of exactly 27 days. I asked ChatGpt to solve it and it got using Kepler’s law:

Approximately 79,565 years ago, the Moon's orbital period was around 27 days. This would

have been when the Moon was closer to Earth, with an orbital distance about 3,011 km closer

than it is now.

I then asked how long was the Earth day at this time. It said

r (t) = R(cos(ωt) + isin(ωt))

of 58 68

Approximately 79,565 years ago, the Earth's day was about 23.9995 hours long—shorter by

about 1.92 seconds than it is today.

That is close enough to 24 hours that we can now look at what humans were at this time. I asked

ChatGpt:

My question is what time for humans was this, for example the Stone Age, time of anatomically

modern humans, or the time of Homo Erectus, or perhaps the transition between different kinds

of humans, as well what kind of time, like an ice age or not.

Its answer was:

Around 79,565 years ago falls within the Upper Paleolithic period, a time associated

with anatomically modern humans, Homo sapiens. This period was notable for

advances in tool-making, art, and cultural complexity, including the creation of early

cave paintings and the development of more sophisticated hunting and gathering

techniques.

Here’s some additional context:

1. Human Evolution: Anatomically modern humans (Homo sapiens) had already

been around for tens of thousands of years. Homo sapiens emerged roughly

200,000 to 300,000 years ago, and by 80,000 years ago, they were widely

distributed in Africa and had begun migrating out of the continent.

2. Climate: The Earth was in the midst of the last glacial period (often referred to

as the Ice Age), which spanned from about 115,000 to 11,700 years ago. This

time specifically was quite cold, with expansive ice sheets covering large parts

of the Northern Hemisphere, leading to lower sea levels and making many

regions colder and drier than they are today.

3. Human Culture: People of this time were skilled in using tools made of stone,

bone, and other natural materials. There were also significant developments in

symbolic thinking and artistic expression, as evidenced by artifacts found in

Africa, the Levant, and other regions inhabited by early humans.

4. Migration and Interaction with Other Human Species: This period likely saw

modern humans beginning to encounter and even interbreed with other

hominins, like Neanderthals in Europe and Denisovans in Asia.

In short, 79,565 years ago aligns with a critical period of human prehistory when

Homo sapiens were spreading, adapting to diverse climates, and making cultural

advancements despite the harsh glacial conditions of the Ice Age.

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Part 9: Life A Possible Universal Natural Process

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A Theory For Biological Hydrocarbons I have found that the basis unit of one second is

not just a Natural constant for physical systems like the atom and the planets around the Sun

(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological

life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the

hydrocarbons. I found

1).

2).

Where is the fine structure constant, is the radius of a proton, is the mass of a

proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the

kinetic energy of the Moon, is the kinetic energy of the Earth, and is the

rotation period of the Earth is one day.

The first can be written:

3).

4).

From which instead of saying the left sides of these equations are seconds, we say they are

proton-seconds by not letting the units of cancel with the bodies of these equations on the

left, which are in units of mass, but rather divide into them, giving us a number of protons. I say

this is the biological because these are the hydrocarbons the backbones of biological chemistry.

We see they display sixfold symmetry. I can generate integer numbers of protons from the time

values from these equations for all of the elements with a computer program. Some results are:

1secon d =

6α

⋅

4πh

1secon d =

K E

(Ear th Da y)

α = 1/137

K E

Ear th Da y

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

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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. And carbon is the core element of life. We see the

duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,

sexagesimal, system of counting of the Sumerians who invented math and writing and started

civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for

the program, it finds integer solutions for time values, incremented by the program at the

discretion of the user:

#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=299792458,protons[100],r=0.833E-15;

{

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 have that

Since this 6 seconds is also proton-seconds we have

5). is carbon (C)

6). is hydrogen (H)

⋅

4πh

G c

= 18769

0.833E − 15

1.67262E − 27

4π (6.62607E − 34)

(6.67408E − 11)(299,792458)

= 6.029978s ≈ 6s

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

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Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We

want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of

a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The

covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53

picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic

radii of both, because the covalent radius, determined by x-ray diﬀraction for diatomic

hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured

that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single

carbon and hydrogen atoms, because we want to know what our theory for their six-fold

symmetry with one another in their representations in proton-seconds says about the way they

combine as the skeletons of life chemistry. We start with the Planck constant, which is like

ﬂux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per

second over square area:!

We have equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

We can write these

3).

4). !

We have from 3 and 4…!

5). !

We ﬁnd the ratio between the surface areas of the hydrogen and carbon atoms:!

h = 6.62607E − 34J ⋅ s = 6.62607E − 37

⋅ m

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

(

6.62607E − 37

⋅ m

)

6secon d s

(

6.62607E − 37

⋅ m

)

6secon ds

1.67262E − 27kg

= 2.37689E − 6m

(

6.62607E − 37

⋅ m

)

1secon d

(

6.62607E − 37

⋅ m

)

1secon d

6(1.67262E − 27kg)

= 6.602486E − 8m

⋅

(6seconds)

⋅

(1second )

2.37689E − 6m

6.602486E − 8m

= 35.9999 ≈ 40

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6). !

7). !

It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and

their valence shell, which is what we want because the valence shell is the outermost electrons

responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will

write this!

8). !

I am guessing the reason we have the golden mean here is that it is the number used for

closest packing. But what we really want to do is look at the concept of action, for hydrogen

given by six seconds and carbon given by 1 second. We take equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

And, we write

9).

Where is the radius of the atom, and t its time values given here by equations 1 and 2. We have for

hydrogen

10).

This is actually very close to the radius of a hydrogen which can vary around this depending on how you

are looking at it, which we said is given by 5.3E-11m. For carbon we have:

11).

And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider

the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as

surface

= 4π (r

) = 3.52989E − 20m

surface

= 4π (r

) = 5.64104E − 20m

4π(53pm)

4π(67pm)

= 0.62575 ≈ 0.618... = ϕ

= ϕ

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = r

G c

∫

dt = (1.67262E − 27k g )

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27k g )

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

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16).

our, equations give

17).

which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro

scale, the Earth/Moon/Sun System, because we had:

and it includes the radius of the proton which is the radius and mass of a proton and gives 1

second

and we have the hydrocarbons the skeleton of the chemistry of life give one second

1). is carbon (C)

2). is hydrogen (H)

and because they predict the radii of the carbon and hydrogen atoms at the core of life

10).

11).

+ r

= 5.3E − 11m + 6.7E − 11m = 1.2E − 11m

+ r

= 5.5E − 11m + 6.44E − 11m = 1.2E − 11m

1secon d =

K E

(Ear th Da y)

1secon d =

6α

⋅

4πh

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = (1.67262E − 27k g )

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27k g )

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

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Appendix 1

Here we use the standard solution for delocalization length for a wave packet given by a

Gaussian distribution. If we want to prove that our planetary Planck constant is correct, the

delocalization time for the Earth should be 6 months using it, the time for the Earth to travel the

width of its orbit. We want to solve the Schrödinger wave equation for a wave packet and use the

most basic thing we can which is a Gaussian distribution. We want to then substitute for

Planck’s constant that is used for quanta and atoms our Planck-type constant (h bar solar)

for the Earth/Moon/Sun system then apply it to predict the delocalization time for the Moon in

its orbit with the Earth around the Sun.

We consider a Gaussian wave-packet at t=0:

We say that is the delocalization length and decompose the wave packet with a Fourier

transform:

is the harmonics of the wave function. We use the identity that gives the integral of a

quadratic:

Solve the equation

With the initial condition

A plane wave is the solution:

Where,

The wave-packet evolves with time as

ℏ

⊙

ψ (x,0) = Ae

−

ψ (x,0) = Ae

−

∫

2π ℏ

ℏ

∫

∞

−∞

−α

x+βx

d x =

4α

iℏ∂

ψ (x, t) =



ψ (x, t)

ψ (x,0) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−ϵ( p)t)

ϵ( p) =

ψ (x, t) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−

of 66 68

Calculate the Gaussian integral of

and

The solution is:

where

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

Thus we know our is accurate, it continues to function in a theoretical framework. The thing

about this is that it means we can predict the mass of the Moon from the Earth year. In terms of

what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at

its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a

mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for

water to exist as liquid, and thus could be as it is for a reason, which means life might be part of

a physical process throughout the Universe, that it unfolds naturally in the evolution of star

systems.

2ℏ

2mℏ

β =

i x

ℏ

= ex p

[

−

⋅

1 + t

/τ

]

τ =

m d

ℏ

⊙

τ =

moon

(2r

moon

)

ℏ

⊙

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

ℏ

⊙

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Appendix 2: The Data For Verifying The Equations

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and

Earth’s orbital velocity at perihelion we have:

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G:6.67408 × 10

−11

c : 299,792,458m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

K E

moon

(7.347673E 22kg)(966m /s)

= 3.428E 28J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

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The Author