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Spacetime Operators, The Proton, And The Protoplanetary Disc!

By Ian Beardsley!

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Contents

Abstract………………………………………………………………………….3

Part 1: Earth/Moon/Sun System And The Proton………………………..4!

Important To Point Out………………………………………………………………………5

Introduction………………………………………………………………………6!

Earth/Moon/Sun System……………………………………………………….6!

Proton Radius……………………………………………………………………………………10

Part 2: The Solar Formulation……………………………………………..13

Equating The Lunar And Solar Formulations Yield

Our 1 Second Base Unit………………………………………………………………………17

Part 3: The Characteristic Length For Habitable Planets………..22

Part 4: The Ancient System of Measuring Time

and the Contemporary Second…………………………………………….25

Part 5: Biological Life Part of a Universal

Natural Process In The Universe………………………………………….35

Part 6: Integrating Analytic and Wave

Solutions of the Solar System………………………………………………42

Conclusion……………………………………………………………………..47

About Parts 7 and 8, The Spherical Cloud Collapse

And Cylindrical Protoplanetary Disc………………………………………50

Part 7: Radial Solution For Planetary Wave Equation…………………….. 51!

Part 8: The Earth Wave Equation Using Cylindrical Coordinates…………67!

Appendix 1…………………………………………………………………………79

Appendix 2: The Data For Verifying The Equations…………………..…81

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Abstract!

The formation of planetary systems from a protoplanetary disc of gas and dust has it solution

analytically in the pressure gradient balanced by the gravitational forces of the protostar and

centripetal forces of the disc’s rotation. It can be quite complex to solve planetary formation

like this. I have a solution of the Schrödinger wave equation for our planetary system and since

this is the wave equation you solve for electrons orbiting a proton in an atom in quantum

mechanics, the solution, like in the atom depends mostly on merely the orbital number of the

body. We can see in part 6 at the end of this paper the Earth solution in the wave equation

appears in the analytic modeling approach with the potential of making solutions simpler. This

might open a lot of doors. While more massive, luminous stars (blue stars) have some

advantages to habitable systems compared to medium (yellow stars like ours) and lighter stars

(red stars), and lighter stars have some advantages compared to medium and heavier stars, it

would seem medium, yellow stars like our Sun are best overall for life, if not intelligent life. This

approach with a wave solution integrated with the classical, analytical approach may open a lot

of doors and make it possible to show that perhaps life is part of a universal natural process in

the universe, where medium luminosity yellow stars like our Sun are ideal.!

By ﬁnding the solution to the Solar System of the Schrödinger Wave Equation used to solve the

atom in quantum mechanics, a characteristic length for habitable planets of star systems is

proposed. It gives the orbit of the Moon which we ﬁnd is proportional to the cube of the

planet’s mass to the cube of its Moon. The moon is included because our moon makes life

optimally possible because its orbit holds the Earth at its inclination to its orbit allowing for the

seasons, preventing cold and hot temperature extremes. Part of the solution involves the

property of our Earth/Moon/Sun system that it has a near perfect eclipse which is to say that

the the Moon as seen from the Earth near perfectly eclipses the Sun.!

The solution exists in the formation of spacetime operators that have a characteristic time of 1

second. This is the way it happens and is interesting because the second precedes the

development of modern physics and ultimately has its origins in the ancient Sumerians, who

were of the ﬁrst to settle down from wandering and gathering to invent agriculture, writing, and

mathematics — civilization in general. Our unit of a second stems from their base 60 counting

system. So that is looked at and it may be that the dynamics of base 60 are responsible for the

naturalness of the second. The spacetime operators also solve the proton that is the

fundamental unit of charge for atoms.!

It is being suggested that life might be part of a Universal Natural Process, not just because of

what we just said above, but because the same spacetime operators that solve the Solar

System and the atom also are based on a sixfold symmetry found in the hydrocarbons which

are are the skeletons of life chemistry.!

I think this theory presenting life as a natural universal process has value in that it gives us a

context with which to look at current times where we have sent men to the Moon and robotic

landers to the planet Mars. A time where we are planning a manned mission to Mars, as well as

making plans to colonize the red planet. In that we are now an interplanetary spacefaring

civilization, perhaps to soon become one that is interstellar.!

In parts 7 and 8 we ﬁnd to formulate the wave function for the Earth we need solve the

Schrödinger Wave Equation for both spherical and cylindrical coordinates."

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Part 1: Earth/Moon/Sun System And The Proton

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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.

The data in this paper is in Appendix 1 at its end.

I would like to clarify something here about the Moon:

I have shown the second in the paper is the base unit for both the Earth/Moon/Sun system as a

wave solution and the proton, making them connected to things like G, c, and h and that we

have this second when things are such that there is a perfect eclipse. The theory does not depend

on the eclipse, just the Radius of the Sun to the Radius of the Moon for the wave solution which

is the orbital radius of the Earth to the orbital radius of the Moon which is the condition of a

near perfect eclipse.

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

eclipse the Sun.

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.

I wrote in the first paragraph at the bottom:

"...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."

I didn't say it wasn't a coincidence, but that it "prompts one to suggest" It has no bearing on the

theory.

Also, the Earth has been in the habitable zone since 4 billion years ago (0.9 AU) 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.

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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 condition 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

= energ y

hC = (6.62607E − 34)(1.55976565E33) = 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)

= 3.0281E8m

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)

=2.727E36J

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 36J

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

60mi n

)(

1min

60sec

)

86400

d a y

sec

2 ⋅ 3 ⋅ 4 = 24

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60 = 86400secon d s /day

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

of 11 82

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.

(

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 ds

= 0.833E − 15m

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Part 2: The Solar Formulation

of 14 82

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

So equation 5 becomes

This gives the mass of the Moon is

Putting this in equation 1 yields

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

of 15 82

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

; ; ; ;

is really . We can write 11 as

13.

We say. . That is

Let us see how accurate our equation is:

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

→

⊙

→ G

→ M

⊙

ℏ

→ L

ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

= ℏ

⊙

/2π

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

2ℏ

⊙

of 16 82

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!

14. !

Where is !

We have

15.

This has an accuracy of

Thus the solutions to the wave equation!

⊙

(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

⊙

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 %

of 17 82

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.

This gives:

16.

We remember that

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

2π r

G m

of 18 82

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

That is an accuracy of

Equation 18 can be written:

20.

From our equation:

6α

h 4π r

= 1.004996352secon ds

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)

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

1907

1932

= 98.7 %

1secon d = 2r

⊙

6α

h 4π r

= 1.004996352secon ds

of 19 82

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.

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

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 =

⊙

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 %

of 20 82

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 55:

16. !

And equating the orbital equations gives

26.

These last two yield

27.

The accuracy is

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 %

⊙

⋅ ℏ

⊙

2 3

⋅

⊙

⋅ ℏ

⊙

= 1

of 21 82

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 lunar, solar, and earth masses.

24.

(400.5986)

5.972E 24kg

1.9891E 30kg

= 1

0.98 = 1

1secon d = d

⊙

of 22 82

Part 3: The Characteristic Length For Habitable Planets

of 23 82

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 24 82

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)(1.989E30kg)

(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 25 82

Part 4: The Ancient System of Measuring Time and the Contemporary Second

of 26 82

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)(60mi n /h our) = 1440min /d a y

1440mi nutes/60 = 24minutes /h our

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

of 27 82

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

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 1

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

⊙

= 3.0281E8m

of 28 82

and,

From these it becomes clear that

Thus combining equation 2 with the following…

We write

From 3 we have

Which gives us

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 = 1.002 secon ds = 1.00secon d s

2π

of 29 82

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).

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.!

Conclusion

In conclusion we have that the unit of a second as derived from the Mesopotamians is a Natural

constant. As well we see the Hindu Vedic system describes Nature as well. It says

( )

1 =

(

)

2π

cos(45

∘

) = cos(π /4) =

⊙

= 2(6.957E8m) = 1.3914E 9m eters

⊙

= SolarRa diu s

= 1.496E11m

= Ear thOrbit

⊙

= 107.52 ≈ 108

moon

= 2(1.74E6m) = 3.48E6m

moon

= Lu n arRa diu s

moon

= 3.845E8m

moon

= Lu n ar Orbit

of 30 82

To see how this is connected to nature we would have to go into the significance of base 10

counting as we did with base 60 counting. But interesting here is that

because

13.

where

Which is to say radius of the Sun to the radius of the Moon in the planetary equation plays the

role of the number of protons (Z) in an element for the energy of the electron orbits in the

atomic equation and that the Moon perfectly eclipses the Sun as see from the earth (eq 15) might

be a condition for sophisticated habitable star systems, we know the Moon allows for the

seasons making life very successful on Earth by preventing temperature extremes. So in the

Hindu Vedic system the diameters of the Sun and Moon fitting into Earth orbit and Lunar orbit

the same amount might play a similar role in a theory. You would have to go into base 10 like we

did with base 60. We have…

moon

= 110 ≈ 108

(86,000sec /d a y)

43200

0.4sec /vispala

= 108,000vispala s /d a y

108(10)(10)(10) = 108,000

⊙

= 107.52 ≈ 108

moon

= 110 ≈ 108

⊙

6.957E8m

6,371,000m

= 109 ≈ 108

K E

= n

⊙

⋅

2ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

of 31 82

West

24 hours

60 minutes

60 seconds

India

60 hours

24 minutes

24 seconds

A sort of inversion of things. The incredible thing here is that the wonderful equations:

unify the Mesopotamian system with the Hindu system because

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

Equation 6 that says

1 hour=(60min/hour)(60sec/min)=3600sec=1 Western Hour

1 vispala=0.4sec

60 vispala=(60)(0.4sec)=24 sec=1pala

60 pala = (60)(24 sec)=1440 sec=24 min = 1 ghaticka = 1 HinduHour

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

1vi spal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

1 =

(

)

2π

1 =

(1440sec)

(3600sec)

2π

1 =

(1gh a t ick a)

(1h our)

2π

1 =

Hin du Hour

Wester n Hour

2π

of 32 82

where is the circumference of a unit circle, and we have unity on the left, so it could be if you

combine the Indian system of measuring time with the Western system you would have a system

of units for the solar system where the equations work out nicely.

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. As well India has

1 =

(60pal a)

(60mi n)

2π =

(24min)

(60mi n)

2π

1 =

((60)(24sec))

((60)(60sec))

2π =

(60pal a)

(60sec)

2π =

(24sec)

(60sec)

2π

s = r θ

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

of 33 82

its mandalas, visual painted patterns that are said to represent the sounds of chants to use to

take them into deep meditational states. The patterns found in their mandalas match the

patterns created by sound putting sand grains on a plate with transducers to vibrate them with

tones in modern acoustics studies. Again, some have suggested 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…!

of 34 82

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 35 82

Part 5: Biological Life Part of a Universal Natural Process In The Universe

of 36 82

Introduction I have a theory for the star systems, the planets and their suns, wherein such

systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The

result is that star systems and atomic systems, systems on the macro scales and micro scales, are

governed by the same underlying principles. It came to pass that this theory indicated an

overlap with biological systems indicating that biological life could be part of the same idea, and

that is what I hope to pull out of that paper and develop here.

The theory for the planets, which also predicted the radius of a proton, which is one of the

fundamental particles out of which matter is made, suggested that star systems which support

life are a part of a Universal natural process. One of the conditions for star systems that support

life made use of the interesting fact that has mystified science for some time now, that the Moon

is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason

a moon would be necessary for life to be optimally successful on a habitable planet is that we

know our moon that orbits our Earth, makes life here successful because it holds the Earth at its

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

extreme cold.

Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in

terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a

natural yardstick. The strange thing is that the basis unit of time is one second and the second

was not developed with the planets and atoms in mind, so it is strange that it lands upon the

function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,

Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient

Sumerians, who invented civilization with settling down from following the herds and hunting

with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a

base 60 counting system, and found it convenient then to divide the Earth day (rotation period)

into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in

the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is

how we have the duration of a second we have today. They chose base 60 (sexagesimal) because

60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or

24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,

they gave us the 360 degree circle as well.

In this paper I show that the same unit of a second that describes planetary systems and atomic

system in common describes hydrocarbons, the skeletons of biological life chemistry. I further

show that it predicts the atomic radii of the hydrogen and carbon atoms from which such

skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,

carbon is the core element upon which life is made, and hydrogen is the simplest element,

element one in the periodic table of the elements, consisting of 1 proton, and is the most

abundant element in the universe by far, and plays the dominant role in all of chemistry. We will

see here that as much as the spacetime operators describe the solar system and the proton, they

describe the hydrocarbons as well, the skeleton of biological life chemistry. Further reason to

suggest life may be part of a Universal Natural Process.

of 37 82

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 ds = h ydr ogen(H )

of 38 82

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

of 39 82

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

of 40 82

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 − 27kg)

(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 − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

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

of 41 82

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 − 27kg)

(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 − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

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

of 42 82

Part 6: Integrating Analytic and Wave Solutions of the Solar System

of 43 82

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!

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:!

2. !

Since , we have that which gives from 2:!

⋆

= − ρ

(

⋆

−

)

P = c

⋆

= v

(1 − η)

ρ(r) = ρ

(

)

−s

= v

(1 − η)

≈

⋆

(1 − 2η)

= − ρ

(

⋆

⋅ 2η

)

P = c

dP/dr = c

dρ /dr

of 44 82

We integrate both sides:!

And, we have!

3. !

We take!

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

have!

4. !

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

5. and!

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

of 45 82

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!

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

Where , so!

7. !

So, essentially, by the chain rule!

to clarify things. The reason 7 is signiﬁcant is that equation 1, part 4 page 27!

Where

And Part 1 the spacetime operators are!

μm

= 1.38E − 23J/K

μ ≈ 2.3

= 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 R

⋅ R =

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 ds

(

6α

4πh

)

⋅

= 1secon d

of 46 82

So we have

8. !

We have!

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:!

10. or, !

Where!

from !

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

d R

= −

earth

ℏ

⊙

= −

earth

ℏ

⊙

d R

∫

= −

earth

ℏ

⊙

∫

d R

lnP = −

earth

ℏ

⊙

ln R + C

P(R) = P

(

)

−

ear th

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

earth

πM

ℏ

⊙

= GM

c(1secon d )

moon

ℏ

⊙

⋅

= 1.0secon ds

(

6α

4πh

)

⋅

= 1secon d

of 47 82

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

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:

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

1vi spal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

of 48 82

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 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.!

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

⊙

planet

moon

≈

star

moon

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

of 49 82

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!

8. !

10. or, !

From page 46."

d R

= −

earth

ℏ

⊙

P(R) = P

(

)

−

ear th

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

of 50 82

About Parts 7 and 8, The Spherical Cloud Collapse And Cylindrical Protoplanetary Disc

We solve the Schrödinger wave equation in spherical coordinates and choose quantum

numbers so that the spherical cloud collapses under its angular momentum to form a ﬂat disc.

This approach, however, has the probability density peak along the axis around which the

cloud rotates and we want it in plane of rotation. However, it does account for a sphere at the

center which is needed for the protostar or star, which becomes The Sun. This method gives

the energy states for the orbits:!

which agree with the Earth. However, we need to solve the Schrödinger wave equation in

cylindrical coordinates to have a high probability density in the plane of a ﬂat disc, the

protoplanetary disc, to have a high probability of the Earth in n=3 orbit and for a peak of the

ground state at . We propose is quantized in terms of the Earth and the Moon."

= −

ℏ

⊙

2μa

= n

⊙

2ℏ

⊙

ρ = a

of 51 82

Part 7: The Earth Wave Equation Using Spherical Coordinates!

of 52 82

The radial solution to 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 !

1. !

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

(

−

ℏ

⊙

∇

+ V

)

ψ = E ψ

of 53 82

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!

9. Or,…!

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

R(r)

(

)

R(r) −

2μr

ℏ

⊙

(

−G

− E

)

− l(l + 1) = s

of 54 82

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!

18. , !

And, our equation becomes!

(

)

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

x = rϵ

y(r)

= ϵ

y(x)

d x

of 55 82

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.!

21. !

From , we have!

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

2μGM

ℏ

⊙

2j + k + 1

k = 2l + 1

of 56 82

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

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

Earth orbit is!

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

⊙

6.96E8m

1737400m

= 400.5986

of 57 82

=2.727E36J

The kinetic energy of the Earth is

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

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):

= (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 36J

2.7396E 33J

100 = 99.5 %

E =

2ℏ

K E

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

= 1.7865E 35J

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 )

of 58 82

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.

The Radial Wave Function

We have the solutions of the radial wave equation is equation 12

12. !

We know that!

, , , !

So, if we express the indices in terms of the quantum numbers and we have!

31. !

Z = Z

2ℏ

n = 5

K E

(5.683E 26kg)(10140m /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 = 2proton s = Heliu m(He)

2ℏ

= e

−x/2

(k+1)/2

(x)

k = 2l + 1

(k + 1)/2 = l + 1

j + l + 1 = n

j = n − l − 1

= e

−x/2

(l+1)

2l+1

n−l−1

(x)

of 59 82

From equations 16 and 26 we have!

16. !

26. !

33. !

And, since !

34. !

31 becomes!

35. !

We recall equation 13:!

13. !

So equation 35 becomes!

36. !

Where is the normalization factor that has already absorbed a factor of . Equation 36

must be normalized so that a probability of 100% is 1. This comes from the condition that!

It is!

37. !

So our ﬁnal equation for the radial wave function is!

(

)

= −

2μE

ℏ

⊙

= −

ℏ

⊙

2μa

ϵ =

x = ϵr

x =

= e

−r/na

(

)

l+1

2l+1

n−l−1

(

)

y(r) = rR(r)

n,l

= Ae

−r/na

(

)

2l+1

n−l−1

(

)

2/na

< ψ (r)

ψ (r) > = 1 =

∫

∞

n,l

(r))*R

n,l

(r)r

A =

(

)

(n − l − 1)!

2n[(n + l)!]

of 60 82

38. !

Where is equation 23:!

23. !

The thing is planetary systems are not atomic systems, in atoms all the electrons are the same,

as well as the protons. And, planets don’t jump from orbit to orbit like electrons do. 23 is a kind

of ground state, but we can also consider it:!

23b. !

by looking at:!

28. !

26. !

21. !

16. !

33. !

25. !

This gives!

So, !

If we are at Earth orbit (n=3) with angular momentum number , Then 38 is!

n,l

(

)

(n − l − 1)!

2n[(n + l)!]

−r/na

(

)

2l+1

n−l−1

(

)

ℏ

⊙

ℏ

⊙

= n

⊙

2ℏ

⊙

= −

ℏ

⊙

2μa

2μGM

ℏ

⊙

2j + k + 1

(

)

= −

2μE

ℏ

⊙

ϵ =

n = j + l + 1

E =

2μℏ

⊙

μ = M

l = 0

of 61 82

39. !

for the Earth. Clearly, this is going to have a probability of 1 at for a shell of some

thickness that is going to decrease both closer in and further away from . We know

that!

And, for Earth, r=1.496E11m. One of the diﬀerences in the solar system solution with the

atomic solution is that while the electron orbiting the proton is constant, the planet orbiting the

Sun is variable from planet to planet. So far in our solutions the moon of the Earth is a constant

that has appeared not just with the Earth, but with Jupiter and Saturn.!

Angular Wave Solutions We have the wave equations are!

4. !

5. !

And, we solved 4, the radial component. But we need to solve 5, and this must be broken up

into two equations, the polar ( ) and the azimuthal ( ), to get the Hydrogen wave functions.

The product of the azimuthal and polar solutions are the spherical harmonics. The product of

all three are the hydrogen wave functions. In the atom electrons exist in spherical shells, but for

the solar system, it formed from a spherical cloud that collapsed into a disc, where we can take

the disc as only having a radial component. However, this radial component started from a

rotating sphere of gas and dust, so our wave equation will be diﬀerent that with the atom. Let’s

tackle this now.!

We can separate the variables in 5 by writing!

40. !

5 becomes:!

41. !

Mulitplying by !

42. !

3,0

(r) =

−3/2

(

1 −

27a

)

−r/3a

r = a

Δr

r = a

ℏ

⊙

(2.8314E33J ⋅ s)

(6.674E − 11)(5.972E 24kg)(7.34763E 22kg)

= 3.7256E6m

R(r)

∂

∂r

(

∂

∂r

)

R(r) −

2mr

ℏ

⊙

(

V(r) − E

)

Y(θ, ϕ)sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

Y(θ, ϕ) +

Y(θ, ϕsin

θ )

∂

∂θ

Y(θ, ϕ) = 0

Y(θ, ϕ) = f (θ )g(ϕ)

f (θ )g(ϕ)sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) +

g(ϕ)sin

∂

∂θ

g(ϕ) = − l(l + 1)

sin

sinθ

f (θ )

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) + l(l + 1)sin

θ +

g(ϕ)

∂

∂ϕ

g(ϕ) = 0

of 62 82

They ﬁnd that since the two therms in plus the term in , that those in equal and those

in equal where which is the magnetic moment quantum number. They know this by

guessing and ﬁnding that it works, which is the way things are often done in quantum

mechanics. We have!

43. !

44. !

The solution to 44, the azimuthal angle equation is obviously!

45. , or. !

Where the subscript m indicates there are m solutions. Just plug it in and you will see it works,

the exponential function is clearly the solution to an equation of that form from any course in

diﬀerential equations. We now want to solve 43, the polar equation…!

43. !

It can be written!

46. !

Evaluate the ﬁrst term:!

47. !

Equation 46 becomes!

48. !

We change variables to , which means we have the derivatives in terms of this new

variable:!

−m

sinθ

f (θ )

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) + l(l + 1)sin

θ = m

g(ϕ)

∂

∂ϕ

g(ϕ) = − m

g(ϕ) = e

imϕ

(ϕ) = e

imϕ

sinθ

f (θ )

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) + l(l + 1)sin

θ f (θ ) = m

sinθ f (θ )

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) + l(l + 1)sin

θ f (θ ) − m

f (θ ) = 0

sinθ f (θ )

∂

∂θ

(

sinθ

∂

∂θ

)

f (θ ) = sin

f (θ )

dθ

+ sinθcosθ

d f (θ )

dθ

sin

f (θ )

dθ

+ sinθcosθ

d f (θ )

dθ

+ l(l + 1)sin

θ f (θ ) − m

f (θ ) = 0

x = cosθ

d f (θ )

dθ

d f (x)

d x

dθ

= − sinθ

d f (x)

d x

f (θ )

dθ

= − cosθ

d f (x)

d x

+ sin

f (x)

d x

of 63 82

Substituting this into 48 gives!

49. !

Dividing through by !

50. !

Using , we have!

51. !

We can write this in primes for compactness!

52. !

This is the associated Legendre equation and it reduces to the Legendre equation when m=0.

That is when m=0…!

53. !

The solutions to the associated Legendre equation are the associated Legendre polynomials.

Associated Legendre polynomials are generated from Legendre polynomials with!

54. !

where are Legendre polynomials and the Legendre polynomials themselves are

generated from!

55. !

The ﬁrst six Legendre polynomials are!

sin

(

sin

f (x)

d x

− cosθ

d f (x)

d x

)

+ sinθcosθ

(

−sinθ

d f (x)

d x

)

+l(l + 1)sin

θ f (x) − m

f (x) = 0

sin

f (x)

d x

− cosθ

d f (x)

d x

− cosθ

d f (x)

d x

+ l(l + 1)f (x) −

sin

f (x) = 0

cosθ = x

sin

θ = 1 − cos

θ = 1 − x

(1 − x

)

f (x)

d x

− 2x

d f (x)

d x

+ l(l + 1)f (x) −

1 − x

f (x) = 0

(1 − x

)f′ ′ (x) − 2x f′ (x) + l(l + 1)f (x) −

1 − x

f (x) = 0

(1 − x

)f′ ′ (x) − 2x f′ (x) + l(l + 1)f (x) = 0

l,m

(x) = (−1)

(1 − x

)

d x

(x)

(x) =

−1)

d x

(1 − x

)

of 64 82

And the ﬁrst few associated Legendre Polynomials are!

, !

I am thinking if a spherical cloud collapses into the protoplanetary disc, then m=0 in the

azimuthal wave equation. The entire wave equation is the radial times the polar times the

azimuthal. The product of the azimuthal and polar waves are the spherical harmonics and

ﬂatten out the spherical cloud into a protoplanetary disc according to which ones you use.

Working with ChatGpt we come up with….!

The Earth Wave Equation

The compete wave function for the Earth is!

56. !

The cloud collapsing under its rotation has m=0, so the azimuthal part is!

57. !

As such, the spherical harmonics reduce to their polar components!

58. !

Where are the Legendre polynomials. To have a protoplanetary disc we need spherical

harmonics that break the spherical symmetry. The lowest-order spherical harmonic that

introduces ﬂattening is the quadrupole term which has l=2. We have!

59. !

The total wave equation is!

60. !

(x) = 1, P

(x) = x, P

(x) =

(3x

− 1), P

(x) =

(5x

− 3x)

(x) =

(35x

− 30x

+ 3),

(62x

− 70x

+ 15x)

0,0

(x) = 1, P

1,1

(x) = − 1 − x

, P

1,0

(x) = x, P

2,2

(x) = 3(1 − x

)

2,1

(x) = − 3x 1 − x

, P

2,0

(x) =

(3x

− 1), P

3,3

(x) = − 15( 1 − x

)

3,2

(x) = 15x(1 − x

), P

3,1

(x) = −

(5x

− 1) 1 − x

3,0

(x) =

(5x

− 3x)

Ψ(r, θ, ϕ) = R

n,l

⋅ Y

l,m(θ,ϕ)

g(ϕ) = e

i⋅m⋅ϕ

= 1

l,0

(θ, ϕ) =

2l + 1

4π

(cosθ )

2,0

(θ, ϕ) =

16π

(3cos

θ − 1)

Ψ(r, θ, ϕ) = R

n,0

(r)Y

0,0

(θ, ϕ) + R

n,2

(r)Y

2,0

(θ, ϕ)

of 65 82

is the spherically symmetric component and and are the

radial wave functions for and . The term will dominate, so we have!

61. !

Thus we have!

62. !

Putting in our value for :!

Which becomes because the Laguerre polynomial for a collapsing cloud is because

n=3 and l=2, not . And must be squared because .!

63. !

Where!

This is an interesting thing because for the spherical rotating cloud to collapse into a disc, we

have to choose at least because in the case for ﬂattening we need and

, which means it is ﬂat for n=3 or greater. I say interesting because is Earth

orbit and this is the optimally habitable planet in the Solar System. Further interesting because

for n=1 we have , which means at the center of the disc it is spherically symmetrical. This

is good because that is where the Sun is, which is a sphere. This comes from the spherical

cloud that collapses into the protoplanetary disc to have a bulge at its center, the protostar that

when it ignites, starts fusion, becomes a star. This introduces radiation pressure which blows

the lighter elements further out which form the gas giants Jupiter and Saturn, and allows

heavier elements to remain close in forming the terrestrial planets.!

While Chatgpt suggested , as the proper spherical harmonic for the sphere to collapse

into a disc, it also says that the term peaks at and , along the

rotation axis and has a minimum at , the equatorial plane, which is not what we want

0,0

(θ, ϕ) =

4π

n,0

(r)

n,2

(r)

l = 0

l = 2

n,2

(r)

Ψ(r, θ, ϕ) ≈ R

n,2

(r)Y

2,0

(θ, ϕ)

Earth

(r, θ, ϕ) = R

n,2

(r) ⋅

16π

(3cos

θ − 1)

n,2

(r)

earth

(r, θ, ϕ) =

(

)

(3 − 0 − 1)!

2 ⋅ 3[(3 + 0)!]

−r/3a

(

)

(

)

⋅

π(3cos

θ − 1)

(x) = 1

(x)

(

)

l = 2

earth

(r, θ, ϕ) =

72 3πa

−

⋅

(

)

⋅ (3cos

θ − 1)

ℏ

⊙

(2.8314E33J ⋅ s)

(6.674E − 11)(5.972E 24kg)(7.34763E 22kg)

= 3.7256E6m

n = 3

l = 2

l ≤ n − 1

n = 3

l = 0

l = 2

(3cos

θ − 1)

θ = 0

∘

θ = 180

∘

θ = 90

∘

of 66 82

because it suggests higher probability density along the axis around which the disc spins. But

at this time, I have suggested doing the Schrödinger wave equation in cylindrical coordinates,

so as to start with the protoplanetary disc, already ﬂat. While that is useful, so has been the

spherical coordinates collapsing into a disc after solved because it gave the right Energy

solution for Earth orbit that I arrived at in the original theory. Namely:!

And, further the spherical cloud collapsing into a disc gives a sphere at the center, which is the

Sun. Cylindrical coordinates, beginning with a disc, cannot do that. So we have to combine

both methods. This is the next chapter in the paper…!

= −

ℏ

⊙

2μa

= n

⊙

2ℏ

⊙

of 67 82

Part 8: The Earth Wave Equation Using Cylindrical Coordinates!

of 68 82

Cylindrical Coordinates Wave Solution!

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 69 82

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 70 82

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 71 82

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 think 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)

⋅ ρdρdϕd z = 1

∫

∞

∫

2π

R(ρ)

⋅

Z(0)

ρdρd ϕ ⋅

∫

δz

−δz

dz ≈ 1

δz

R(ρ) = Ae

−βρ

of 72 82

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)

⋅ ρdρdϕd z = 1

⋅

2π

⋅

πσ

⋅

∫

∞

−2βρ

⋅

∫

2π

dϕ ⋅

∫

∞

−∞

−

2σ

= 1

∫

∞

−2βρ

ρdρ =

(2β )

4β

∫

2π

dϕ = 2π

of 73 82

This gives!

or. !

The ﬁnal normalized wave equation is:!

30. !

The wave function for n=3, m=0 (Earth orbit) will use a Laguerre polynomial for the radial

solution to account for the higher energy and additional nodes. The form would be!

31. !

are the associated Laguerre polynomials, is the normalization constant, and is a

parameter related to the potential and system properties. We have!

32. !

The particular Laguerre polynomial for is!

33. !

The normalized wave function for n=3, m=0 is!

34. !

Because the angular and axial z part are unchanged from the n=1, m=0 solution. We already

have the normalization constants for the angular and z-axis components. Let’s do the radial

component:!

This can be integrated by expanding the integrand, and integrating it term by terms with the

gamma function, for which there are ﬁve terms. The gamma function is!

∫

∞

−∞

−

2σ

= σ 2 π

⋅

4β

= 1

A = 2β

Ψ(ρ, ϕ, z) = 2βe

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/2σ

n,m

(ρ) = N

n,m

−βρ

n−

−1

(2βρ)

(x)

n,m

3,0

(ρ) = N

3,0

−βρ

(2βρ)

(x)

(x) =

− 3x + 3

3,0

(ρ, ϕ, z) = N

3,0

(2β

− 6βρ + 3)e

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/(2σ

)

∫

∞

3,0

(ρ)

ρdρ = 1

3,0

∫

∞

(2β

− 6βρ + 3)e

−βρ

ρdρ = 1

of 74 82

Where . So, for instance the ﬁrst term is k=5 gives!

…and so forth for four more terms. The end result is!

35. !

The ﬁnal wave function for the Earth orbit (n=3,m=0) is:!

36. !

We need to determine the parameter . ChatGpt tried this by trying to make the probability

density sharply peaked around and chose the as the Earth orbital radius of 1 AU. In

order to determine , it had to solve a cubic. It tried to get some estimates to do this using the

rational root theorem. It got a probability for of .!

I then told ChatBpt that would not be at n=3 Earth orbit, that the Earth would not be the

ground state, and that I thought that would occur at n=1. But that further I had reason to

believe that is given quantized by the Moon, so is!

37. !

So it redid the calculation estimating roots of the cubic and got a probability that at , the

probability density is approximately 0.9%, suggesting the peak was present but not

exceedingly sharp.!

I then told ChatGpt that there was no need to guess at roots to the cubic because cubics have

a generalized method of solution similar to the way quadratics do and to check its resources

for the method. It said: “You are absolutely right!” and proceeded to solve the cubic. The result

is that the ground state does peak strongly at my value for in terms of the Moon. So here is

what it did…!

The radial part of the wave function is from equation 36:!

∫

∞

−αρ

dρ =

Γ(k + 1)

k+1

Γ(n) = (n − 1)!

∫

∞

−2βρ

dρ =

Γ(6)

(2β )

120

64β

8β

3,0

2β

3,0

(ρ, ϕ, z) =

2β

(2β

− 6βρ + 3)e

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/(2σ

)

ρ = a

= 1AU

0.855a

ℏ

⊙

ρ = a

of 75 82

38. !

The radial probability density is!

39. !

Then we set the derivative of this to zero at . To do this, take the natural log of both

sides and diﬀerentiate. This gives!

40. !

Introduce a dimensionales variable!

And we ﬁnd we have the cubic to solve:!

41. !

We then proceed to solve this by a general method for cubics. We put it in standard form!

Eliminate term by making substitution!

This gives!

Which becomes!

This is the depressed cubic!

3,0

(ρ, ϕ, z) =

2β

(2β

− 6βρ + 3)e

−βρ

P(ρ) =

3,0

(ρ)

ρ =

4β

(2β

− 6βρ + 3)

−2βρ

ρ = a

4β

− 16β

+ 12β −

= 0

β =

− 16k

+ 12k − 3 = 0

+ 4k

+ 3k − 0.75 = 0

+ ak

+ bk + c = 0

a = − 4,b = 3,c = − 0.75

k = y −

= y +

(

y +

)

− 4

(

y +

)

+ 3

(

y +

)

− 0.75 = 0

y −

161

108

= 0

of 76 82

Where p=-7/3, and q=-161/108. The discriminant of the depressed cubic determines the

nature of the roots.!

We then apply Cardano’s formula:!

Now we can determine .!

Put in to reconstruct the wave equation…!

+ py + q = 0

Δ =

(

)

(

)

Δ = 0.085 > 0

y =

−

+ Δ +

−

− Δ

−

161

216

≈ 0.745

Δ ≈ 0.086 ≈ 0.29

y ≈ 1.779

k ≈ 3.112

β =

≈

3.112

ℏ

⊙

= 3.707E6m

β ≈ 8.4E − 7m

−1

3,0

(ρ, ϕ, z) =

2β

(2β

− 6βρ + 3)e

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/(2σ

)

2 × 8.4E − 7

. . . ≈ 9.7E − 7...

of 77 82

With our value of and evaluating :!

Which suggests a signiﬁcant concentration of probability around . So our value for

works great for quantized in terms of the Moon.!

Conclusion

We have said the solution for the energy level for the wave function for the Earth is for a

gravitational potential.!

41. !

However, atoms are diﬀerent than planetary systems. My theory suggests the energy for Earth

is in the wave solution from spherical coordinates in the previous section (the kinetic energy of

the Earth):!

42. !

Which we have shown works very well. Where is the radius of the Sun, is the radius of

the Moon, is the mass of the Earth, is the mass of the moon, and

is a Planck constant for the solar system. We have said the wave equation is!

43. !

But we must alter it to rather accommodate the energy level of my theory in cylindrical

coordinates. To do this we need to represent diﬀerently than with the atom. Since the energy

level in terms of is!

44. !

We need . I found the protoplanetary disc was quantized in terms of the the rotational angular

momentum of the Earth according to equation 10 page 49 (part 6):!

P(a

)

P(a

) =

3,0

(ρ)

ρ =

4β

(2β

− 6βρ + 3)

−2βρ

ρ = 0.92a

ρ = a

= G

⊙

= n

⊙

2ℏ

⊙

ℏ

⊙

= 2.8314E33Js

3,0

(ρ, ϕ, z) =

2β

(2β

− 6βρ + 3)e

−βρ

⋅

2π

⋅

(πσ

)

1/4

−z

/(2σ

)

ℏ

⊙

of 78 82

10. or, !

Thus if we want and your solution in terms of the rotational angular momentum of the Earth

, the solar system Planck constant , the radius of the Sun , the radius of the Moon

, the mass of the Moon , the mass of the Earth , and the speed of light c, then for to

be on the order of 1.00E-6 and in units of meters to the minus one that we got from solving the

cubic, that is!

45. !

Where where we have pointed out is related to the second from the 24 hour

day and 60 second minute in 2.5(24)=60 (Part 4). Thus the energy E_n is!

46. !

We need to ﬁgure in , the orbital number. It would seem for we have to multiply this result

by where n=3 for Earth orbit. 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. So the solution to the wave

function for the Schrödinger wave equation in cylindrical coordinates modeling a ﬂat disc, is!

47. !

= !

=2.4997E32J!

This suggests the equation for in atomic systems (equation 44)!

44. !

is in planetary systems!

48. "

P(R) = P

(

)

−

ear th

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

earth

ℏ

⊙

β = 2

ℏ

⊙

earth

⋅

⊙

⋅

4ℏ

⊙

earth

/ℏ

⊙

= 2.5

ℏ

⊙

earth

⋅

⊙

⋅

(n + 1)

= 2.4997E32J

= (n + 1) ⋅

ℏ

⊙

earth

⋅

⊙

⋅

(4)

7.348E 22kg

(0.4)

1.737E6m

6.957E8m

(7.348E 22kg)

(5.972E 24kg)

(3E8)

ℏ

⊙

= (n + 1)

ℏ

⊙

of 79 82

Appendix 1

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 80 82

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)(3.844E 8m)

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

ℏ

⊙

of 81 82

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

of 82 82

The Author!