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Electric And Gravitational Fields Are Calibrated For Life

By!

Ian Beardsley!

of 2 39

Abstract………………………………………………………………..3

Electric Fields…………………………………………………………4

Here we talk about electric fields and explain just what the

natural constants are and what they mean.

1.0 The Mass of a Proton……………………………………………5

Here we find the mass of a proton in terms of quantum

mechanics’ h and classical gravity’s G, if we take the unit

of a second as a natural duration.

2.0 Proton-Seconds, Proton Radius, Hydrocarbons…….7

Here we formulate the idea of proton-seconds and find we

can do something without yet finding the second as a natural

duration of the atom, particularly in biochemistry’s

hydrocarbon skeleton’s.

3.0 Theory For The Atoms…………………………………………10

Here we introduce the idea of intermediary mass,

and that we can combine with Giordano’s Relationship

to produce the constant k, laying the groundwork for the

duration of a second as natural.

4.0 The Principle of Least Action………………………………..16

Here we use the equation for duration of the second

as natural so we can use it to formulate a prediction of the

radius of the proton. and develop a field equation of

gravity that uses classical gravity’s G and quantum

mechanics’ h.

5.0 Newton’s Universal Law of Gravitation………………….20

Here we develop a field equation of

gravity that uses classical gravity’s G and quantum

mechanics’ h.

6.0 Rigorous Formulation of Proton-Seconds………………22

Here we formulate our idea of proton-seconds more rigorously

so we can apply it to the periodic table as a mathematical

constructs.

7.0 Theory of Compounds…………………………………………..25

Here we show the elements and their ensuing

compounds are not just chemical constructs but

mathematical constructs.

8.0 The Cause of the Electric Fields……………………………..28

Here we propose a theory for the source of electric

Fields, and show they seem to suggest carbon to be

again at center stage.

9.0 Natural Units……………………………………………………….30

We show that strangely in terms of our theories here the

SI system of units comes out with its basis of kg, m, and s

as natural

10.0 Abstract Cosmology…………………………………………….37

We introduce the idea of abstract cosmology

𝔼

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Abstract

Here we look at the structure of Nature, what we think it is and why it is calibrated the way it is.

We look at the structure of matter, space, and time, and show through the natural constants,

which are a measure of the way matter space and time are calibrated, that this calibration is

such that it allows for life from the microcosmos calibrated with Planck’s constant h for the

subatomic particles such as protons, electrons, and photons, to the intermediary, like the

elements made from the subatomic particles, and their ensuing compounds, such as the

hydrocarbons the skeleton’s of life chemistry, to the Sun, the planets, and their moons on the

large scale of the macrocosmos. We show there is some indication the structure of the

microcosmos is formulated in terms of the intermediary realms such as the elements and their

compounds, and in terms of the larger scale, the planets, moons and stars. One might suggest

this structure has the purpose of allowing for life, such that carbon, the core element of

biological life chemistry, is at the core of the overall construct. One might suggest the reason

this scenario is, is so that the Universe can be conscious. We ﬁnd a recursive 6-fold symmetry

is behind all of this structure. A strange thing that falls out of this is that meters, second, and

kilograms are natural units."

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Electric Fields

We all know matter causes the gravitational ﬁeld and that it is an attractive force the acts only

between masses. It is only noticeable for large masses, not just larger than atoms or the

subatomic particles that make them up, but larger than say a boulder; we would be hard

pressed to measure the gravitational attraction between to grains of sand, which are even

millions of times more massive than the silicon and oxygen atoms from which they are made.!

But there are other kinds of force ﬁelds that allow matter to interact with one another, and the

one we are interested in here is the electric ﬁelds. We can measure these ﬁelds that are

associated with the subatomic particles the protons, and electrons. Unlike gravity, they are not

just attractive, but repulsive. Where gravity is a construct based on two, one mass attracts

another, electric ﬁelds are based on three. We notice that a proton and electron attract one

another, but that two protons repel one another and two electrons repel one another. The way

we have dealt with this is to say that the proton has positive charge, and the electron has

negative charge and established the convention that a positive and negative charge attract one

another and that two like charges repel one another. So while the positive proton attracts the

negative electron, the positive proton repels the positive proton, and the negative electron

repels the negative electron. What we mean by a proton having a positive charge physically we

just don’t know, nor of the electron having a negative charge either. We could have called the

proton negative and the electron positive, but we chose the convention to be the other way

around.!

As well unlike gravity, we don’t know that electric charge is a function of mass, and indeed the

proton, which is about 1836 times more massive than the electron carries a charge equivalent

to that of an electron. This is what we want to explore, which is why nature is calibrated as it is.

We hope to show that this calibration allows for life. For instance, if the charges in the protons

are equivalent to one another, but the masses are diﬀerent, then the acceleration of a proton

attracted to the electron is less than the acceleration of the same electron attracted to the

same proton, and this in turn determines the structure of the periodic table of the elements,

and their properties, that allows for life, among other things.!

Our formulation of electric ﬁelds is geometric. We explain the electric ﬁeld by placing an

arbitrary surface around a point charge and say the ﬁeld exiting the surface, covers the surface

such as it equals the volume of charge contained in it. We say the electric ﬁeld, the ﬁeld, has

a ﬂux, which is itself dotted over the area of the surface , is the charge in coulombs divided

by a constant epsilon determined by experiment contained in that volume:!

Thus Q and epsilon are measures for the degree to which this holds, and represent the

calibration of Nature. If they were other than they are, things would work diﬀerently, or not at

all, not just in terms of the electric ﬁeld, but in terms of other aspects of Nature, which seem to

be properly calibrated to allow for something as complex as life, or other aspects of Nature,

like the balance in stars that allows them to give warmth and light to the planets. We measure

in the laboratory if we say charge q in coulombs is q= 1.602E-19C, then ,

where . Thus the electric ﬁeld is force per coulomb at a point in space!



∮



E ⋅ d



a =

= 8.988E 9

4πϵ

of 5 39

In my work A Theory of Space, Time, and Matter (Beardsley 2022) I developed a theory for the

mass of a proton in terms of G, h, and c which are the gravitational constant, Planck’s

constant, and the speed of light. Thus with G of classical gravitational theory and h of quantum

mechanics we are in the realm of quantum ﬁeld theory which attempts to unify quantum

mechanics with gravitation theory. I wrote section 1…!

1.0 The Mass of a Proton Gravity is a property of space measured by the universal constant

of gravity, G:

Equation 1.1

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

E =

⋅

4πr

4πϵ

⋅

= k

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

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Equation 1.2.

Which describes mass per meter over time, which is:

Equation 1.3:

It must be adjusted by the fine structure constant . It is my guess the factor should be which

is 18,769.:

Equation 1.4.

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 1.5.

We take the square root to get meters:

Equation 1.6.

We multiply that with the value we have in equation 1.4:

Equation 1.7.

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 1.8.

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 1.9.

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

Equation 1.10.

Equation 1.11.

If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of

the elements in the periodic table occur for dividing by something less that 1-second. It seems

kg ⋅ s

1.82E − 16

kg ⋅ s

(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m)

= 4π r

= 8.72E − 30m

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon d s

= 1.67E − 27kg

1E − 26kg ⋅ s

1second

= 1E − 26kg = 6protons = 6m

1E − 26kg ⋅ s

6secon d s

1E − 26kg ⋅ s

1second

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the duration of a second is natural. If it is, since it was formed by a calendar based on

reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon

orbital mechanics. I find it is, that:

That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic

energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by

very small amounts over millions of years. The solar system has evolved towards this since the

explosion of life called the Cambrian, and will slowly decay away from it.

1.12

Where

1.13.

Is the fine structure constant squared is the ratio of the potential energy of an electron in the

first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared.

I then developed the concept of proton-seconds in section 2.0 where time is associated with

atoms through the number of protons they have.

2.0 Proton-Seconds, Proton Radius, Hydrocarbons We show carbon, the core element

of life is six-fold symmetric with hydrogen in terms of the natural constants that characterize

space, time, and matter:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

We find one second gives six protons which is carbon:

K E

moon

K E

earth

(EarthDay) ≈ 1second

18α

4πh

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G : 6.67408 × 10

−11

c : 299,792,459m /s

α : 1/137

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Equation 2.1

We find six seconds gives 1 proton is hydrogen:

Equation 2.2

For time t greater than 6 seconds we have fractional protons. For t<6 we the have other

elements.

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. Here is the code for the program:

#include <stdio.h>

#include <math.h>

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6second s = hydrogen(H )

h 4π r

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int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

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 need an interpretation of equation 2.1, which was

Matter is that which has inertia. This means it resists change in position with a force applied to

it. The more of it, the more it resists a force. We understand this from experience, but what is

matter that it has inertia? In this analogy we are suggesting a proton is a three dimensional

bubble embedded in a two dimensional plane. As such there has to be a normal vector holding

the higher dimensional sphere in a lower dimensional space. Thus if we apply a force to to the

cross-section of the sphere in the plane there should be a force countering it proportional to the

normal holding it in a lower dimensional universe. It is actually a 4-dimensional hypersphere

whose cross-section is a sphere. This counter force would be experienced as inertia. (Fig. 2)

Since plank’s constant h is a measure of energy over time where space and time are concerned it

must play a role. Of course the radius of a proton plays a role since squared and multiplied by

it is the surface area of our proton embedded in space. The gravitational constant is force

produced per kilogram over a distance, thus it is a measure of how the surrounding space has an

effect on the proton giving it inertia. The speed of light c has to play a role because it is the

velocity at which events are separated through time. The mass of a proton has to play a role

because it is a measurement of inertia itself. And alas the fine structure constant describes the

degree to which these factors have an effect. We see the inertia then in equation 6 is six protons

over 1 second, by dimensional analysis.

h 4π r

= 6proton ⋅ secon d s = carbon(C )

4π

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We see the radius of a proton is given by carbon (1 second):

Equation 2.3.

The experimental radius of a proton is:

Equation 2.4.

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared:

Equation 2.5.

But we need to find the unit of a second as a natural duration in terms of the atom, which I did

with an innovation applied to Giordano’s Relationship in section 4.0. Here we will call it section

3.0

3.0 Theory For The Atoms We have determined the mass and radius of the proton, but still

need to derive the second to open it up to other realms and I find this can be done by deriving a

constant k by theorizing the atom, something made up of protons, neutrons and electrons. But I

have derived such a constant from Giordano’s relationship.

We now formulate what I call Giordano’s Relationship: Warren Giordano writes in his paper

The Fine Structure Constant And The Gravitational Constant: Keys To The Substance Of The

Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’

is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant

numerically, but neglecting any units.

Let’s do that

(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js

4πh

= 8.288587 × 10

−16

m = 0.829f m

= 0.833

0.014f m

1 + α

Fig. 2

of 11 39

And it works, G is:

G=6.67408E-11 N(m2/kg2)

Let us reformulate this as:

Equation 3.1

Where and H=1 gram/atom

Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so on

for 6E23 atoms per gram. Thus,…

Since grams and atom cancel we can work in grams even though our equations are in kilograms.

Let us not write H, since formally it is grams per mole of hydrogen but write

We have:

Or,…

Equation 3.2

Where

Equation 3.3

Let us say we were to consider Any Element say carbon . Then in general

Equation 3.4

We have

and

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

(1 + α)

H = 6.0003

⋅ s

= Avaga dro′ s − Num ber = 6.02E 23a tom s /gr a m

H = 6.02E 23

atoms

gra m

⋅

1gra m

atom

= 6.02E 23

ℍ = 1

gra m

atom

(1 + α)

⋅ N

ℍ = 6.0003kg

⋅

h(1 + α)N

ℍ = 6Gx

x = 1.00kg

𝔼

ℂ

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gra m s

6protons

6(6E 23protons)

6gra m s

of 12 39

12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And,

Therefore we always have:

Equation 3.5

This works nicely because we formulated molar mass nicely; we said element one (hydrogen)

which is one proton and one electron has one gram for a mole of atoms. Historically this was

done because we chose carbon (element six) to have 12 grams per mole, and determined what

the mole was such that it would hold. The reason this works is that hydrogen is one proton and

has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any neutrons,

and the neutron has the same mass as the proton, and our theory makes use only of protons (in

this instance of its formulation) equation 4.3

Comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because

it is important, that in equation 4.5

is not molar mass, and that is a variable determined by ; it is the number of a mole of

atoms multiplied by the number of protons in . The reason we point this out, though it may

already be clear, is we wish to find the physical theory behind it. That is we need to find the

physical explanation for equation 4.4

It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in

kilograms, seconds, and meters, it may mean these units of measurement have some kind of a

meaning. We can in fact write it:

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23protons

Z ⋅ gra m s

𝔼 =

Z ⋅ gra m s

Z ⋅ protons

⋅ 𝔼 = 6E 23

x = 1.00kg

⋅ 𝔼 = 6E 23

𝔼

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

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We know that

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of

1.00 to astronomical units, years, and solar masses, as these are connected to the orbit

of earth as it relates to the sun. We have:

We can now write

Eq 3.6.

This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital

velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum

mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We

multiply both sides by and we have earth velocity on the left and the units stay the same on

the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put

in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:

Eq. 3.7

This brings up an interesting question: while we have masses characteristic of the microcosmos

like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to

become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)

which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where

we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find

that mass let us take the geometric mean between the mass of a proton and the mass of 1.44

solar masses. We could take the average, or the harmonic mean, but the geometric mean is the

squaring of the proportions, it is the side of a square with the area equal to the area of the

rectangle with these proportions as its sides. We have:

(1 + α)

⋅ N

𝔼 = 6.000kg

⋅

(1.98847E 30)

⊙

1.4959787E11m

⋅

year

3.154E 7s

1.8754341E64

⊙

⋅ year

(1 + α)

⋅ N

𝔼

year

= 8.2172E 32M

⊙

2π AU/year

4π

(1 + α)

⋅ N

𝔼 ⋅ v

= 422.787kg

of 14 39

We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .

We have the intermediary mass is:

Eq. 3.8

All we really need to do now is divide 4.7 by 4.8 and we get an even number that is the six of our

six-fold symmetry.

Eq. 3.9

The six of our six-fold symmetry.

We have something very interesting here. We have

This is:

Equation 3.10

⊙

= 1.98847E 30kg

= 1.67262E − 27kg

= (2.8634E 30)(1.67262E − 27) = 69.205kg

(1 + α)

⋅ N

𝔼 ⋅ v

= 6.1092 ≈ 6

69.205kg

6kg

= 6

k v

= 6

of 15 39

Where k is a constant, given

Equation 3.11

We can take the velocity of earth as being 30,000 m/s by rounding it. We have

Using , we write

Eq. 3.12

The primordial element from which all others were made.

We can explicitly write the constant k:

k =

800

30,000

800

= 37

37.5 = 6.123734357

⋅

h 4π r

= 1.004996352secon d s

k v

= 6

h 4π r

Gck v

= 1proton ⋅ secon d

(K . E . Moon)(Ear th Da y)

(K . E . Ear th)

≈ 1secon d

h 4π r

Gck v

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= 1proton

h 4π r

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= 6proton s

h 4π r

Gck v

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= ℍ

ℍ = hydrogen

of 16 39

Equation 3.13

It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf

will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic

equation is:

Equation 3.14

Let us approximate 0.77 with 3/4. Since we have our constant

And

Then

Equation 3.15

Since our constant k in terms the Chandrasekhar limit is

Equation 3.16

The innovations here were the idea of an intermediary mass, and where is a variable,

Avogadro’s number multiplied by the number of protons in . We can now develop the second

as a natural duration in the atom and we will do this here in section 4.0 which was section 5.0.

4.0 The Principle of Least Action We consider two paths, one with velocity c in one

medium, the other with velocity v in another. In order to go from one point two another over

two paths, the refraction is such that the sine of the angle of incidence equals the sine of the

angle of refraction.

We have the two paths are travelled in a time t:

This is the mysterious Nature of reality, the path of least time is taken. This falls under the

general heading The Principle of Least Action, attributed to firstly the French Natural

Philosopher and mathematician Louis Maupertuis of the early eighteenth century. I say

k =

(1 + α)

⋅ N

𝔼

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

1 + α

⋅ N

𝔼

= Mm

ℏ

1/2

ℏ = h /2π

k =

8π

(1 + α) ⋅ N

𝔼

sin α

sinβ

t = r

c + v

c v

of 17 39

mysterious, because as it is said in physics, for something to know the path of least action and

take it, it is as if it has explored first all paths between A and B to know which one would be the

path of least action. Everything in physics comes to this principle. It is called a principle, not a

theory, law, or rule. Yet it seems to be the way Nature behaves, and it is mysterious. Richard

Feynman applied it to quantum mechanics, probably because the mysterious Planck’s constant

that governs quantum mechanics is in Joule-seconds, energy over time, and this is the terms in

which action is formulated mathematically.

In our scenario here we regard matter, the proton in particular, as the cross-section of a

hypersphere. Our two mediums are hyperspace and space and the least action principle applies

in the same mathematical form. This is abstract cosmology, that really is the underlying

mathematics is common to all systems, that in effect they are manifestations of one another.

We have

We make the approximation

So that

We are saying t=6 seconds is the proton, and r is the radius of a proton r=0.833E-15m. Thus

The radius of a hydrogen atom is .

And we have our one second as a natural constant with respect to the atom. We see it occurs at

t/2 which is at half the radius of a hydrogen atom. We want to deal with that size or around it;

our velocity comes from it and it can be thought of as a proton drift velocity akin to the electron

drift velocity in a wire that gives rise to an electrical current. We find if we derive the equation

that represents these computations, we have

c v − v = c

c v = c

v =

0.833E − 15m

= 1.389E − 16m /s

= 1.2E − 10m

t =

1.2E − 10m

1.389E − 16m /s

= 863,930.8855s

= 431,965.4428s

= (773.5m /s)(431,965.4428s) = 334,125,270m

334,125,270m

299,792,459m /s

= 1.11452193secon d s

of 18 39

Where,…

, ,

Radius of hydrogen atom

Remember our constant k equation 4.16 (Don’t forget to divide by two somewhere):

Since we have the equation of the radius of a proton is given by, by evaluating it at one second

which is carbon:

And 1 second in terms of the atom is given by

Then the equation for the radius of a proton is:

Equation 4.1.

And since

v =

t = 6s =

h 4π r

= α

h 4π r

v =

= α

h 4π

= 1.2E − 10m

t =

= R

⋅

h 4π

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

Gπ

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

4πh

= 8.288587 × 10

−16

m = 0.829f m

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

2 ⋅

4π

⋅

𝔼

of 19 39

Then,…

Equation 4.2.

Let’s verify equation 4.1:

Remember that in equation 4.1, which is

That, must remain coupled. is determined by the number of protons in .

Making the approximation 9/8~1 we can write equation 5.1 as

Equation 4.3

Which gives

We form constants:

Equation 4.4.

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof Ear th

⋅ Ear th Da y = (6)1.2secon d s

3 2

⋅

Gπ

⋅

𝔼

K Eof Moon

K Eof Ear th

⋅ Ear th Da y

1.67E − 27

(6.626E − 34)(299,792,459)

(6.674E − 11) 4π

⋅

1.2E − 10

6.02E 23

= 0.93f m

2 ⋅

4π

⋅

𝔼

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

k =

2π

= 6.93E − 9kg

of 20 39

Equation 4.5

And we have the Equation:

Equation 4.6

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12.

We now want to add to this section the equation of the mass of a proton.

Equation 4.7.

We now want to write a new section for this paper that uses this in Newton’s Universal Law of

Gravitation in section 5.0…

5.0 Newton’s Universal Law of Gravitation Newton’s equation is

And, the mass of a proton is equation 4.7

Letting we have the gravitational attraction between two protons or neutrons is

Equation 5.1.

And we see we lost the constant G and have a quantum mechanical equation for gravity. We

want to write the equation of acceleration and we will do it for the moon first:

Equation 5.2.

Where is the radius of the moon (1,737,400 meters) and is the number of proton making

up the moon. The mass of the moon is 7.342E22 kg and the mass of a proton is 1.6726E-27 kg

meaning it has 4.39E49 protons. This gives

Equation 5.3

𝔼

= 1.99E − 34m

= k

𝔼

2π

⋅

𝔼

F = G

2π

⋅

𝔼

= m

F =

2π

𝔼

Gh c

2π

𝔼

⋅

(

)

Gh c

2π

𝔼

⋅

(

)

= 1.61m /s

of 21 39

Which is correct. We used G=6.674E-11, h=6.626E-34, c=299,792,459, and the radius of a

proton is 0.83rt3E-15. and . We consider the factors

and. to get

Equation 5.4.

If we do this for the Earth we get since the mass of the Earth is 5.97212E24 kg, and again the

mass of a proton is 1.6726E-27 kg meaning it has 3.57E51 protons. Radius of the earth 6,371,000

meters. This gives

Equation 5.5.

Which is correct (9.81 m/s2)

and. to get

Equation 5.6.

We took this little digression to see what is there and we find for the moon we evenly get 5

billion, and the earth ten hundred billion. What we really want to do is look at equation 5.1

We said it is a quantum mechanical equation for gravity between two protons. To see this as a

ﬁeld equation simply substitute for m, and we have:!

Equation 5.7. !

Where N is the number of protons in the body from which the ﬁeld emanates. The innovations

laid out so far in this paper lead to a more rigorous approach to our idea of proton-seconds,

which in turn leads to a mathematical structuring of the elements in the periodic table, beyond

the chemical structuring we have today. We address this now in sections 6.0 and 7.0…"

= 1.2E − 10m

𝔼 = 6.02E 23

1.2E − 10m

1,737,400m

= 6.9E − 17

𝔼

= 7.3E 25

𝔼

⋅

= 5E 9

Gh c

2π

𝔼

⋅

(

)

= 9.73m /s

1.2E − 10m

6,371,000m

= 1.88E − 17

𝔼

= 5.93E 27

𝔼

⋅

= 1E11

F =

2π

𝔼

a =

⋅

2π

⋅

𝔼

⋅

⋅ N

of 22 39

6.0 Rigorous Formulation of Proton-Seconds

We can actually formulate this differently than we have. We had

But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are

lower and upper limits in an integral, then we have:

Equation 6.1

This Equation is the generalized equation we can use for solving problems.

Essentially we can rigorously formulate the notion of proton-seconds by considering

Equation 6.2

Is protons-seconds squared where current density is and ( can also be

). We say

Equation 6.3

Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a

number of protons. It is

Equation 6.4

Dividing Equation 6.2 through by t:

Equation 6.5

Which is proton-seconds. Dividing through by t again:

h 4π r

= 6proton s

h 4π r

= 1proton

h 4π r

∫

dt = ℕ

∫

qdt = t

∬

ρ(x, y, z)d x d y



J = ρ



ρ = Q /m

Q /m

Q =

∫

ρd V

ℕ =

h 4π r

∫

= t

∬

ρ(x, y, z)d x d y

of 23 39

Equation 6.6

We see that if where and then J is I/m2 (current per square meter)

is analogous to amperes per per square meter which are coulombs per second through a surface.

Thus we are looking at a number of protons per second through a surface. Thus we write:

Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to

protons in the output from our program and 1.0 seconds is carbon (C). We see we have the

following theorem:

Equation 6.7

So as an example,…

Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second

which is carbon and we have protons per square meter.

h 4π r

∫

= proton s



J = ρ



ρ = Q /m

v = m /s

h 4π r

∫

= 6

∫

1.0

0.5

−2

dt = − 6(1 − 2) = 6

h 4π r

∫

∬



J ⋅ d



h 4π r

∫

1.0

0.5

∬



J ⋅ d



S = − 3

(

1 −

0.25

)

= 9

proton s

secon d



J(x, y, z) = (0,0, J ) = − J



S = d x d y



J ⋅ d



S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y

of 24 39

We are now equip to do computations in proton-seconds. We use equation 7.6 from two to

three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal

proton that we found described its radius (My feeling is we introduce the factor of 2 because

carbon is 6 protons +6 neutrons and 2 times 6 is twelve):

Now we integrate from phosphorus is 15 protons=0.396 seconds to aluminum is 13 protons =

0.462 seconds which is to integrate across silicon (divide your answer 10 by 2 to get protons):

By what value would you like to increment?: 0.006

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

251.2490 protons 0.024000 seconds 0.248978 decpart

59.1174 protons 0.102000 seconds 0.117416 decpart

50.2498 protons 0.120000 seconds 0.249805 decpart

40.1998 protons 0.150000 seconds 0.199844 decpart

37.2221 protons 0.162000 seconds 0.222080 decpart

27.1621 protons 0.222000 seconds 0.162058 decpart

25.1249 protons 0.240000 seconds 0.124907 decpart

20.0999 protons 0.300000 seconds 0.099916 decpart

17.0338 protons 0.354000 seconds 0.033823 decpart

16.2096 protons 0.372000 seconds 0.209604 decpart

15.2272 protons 0.396000 seconds 0.227202 decpart

14.1549 protons 0.426000 seconds 0.154862 decpart

13.2236 protons 0.456000 seconds 0.223620 decpart

13.0519 protons 0.462000 seconds 0.051885 decpart

12.1084 protons 0.498000 seconds 0.108375 decpart

11.1666 protons 0.540000 seconds 0.166615 decpart

11.0439 protons 0.546000 seconds 0.043906 decpart

10.1515 protons 0.594000 seconds 0.151471 decpart

Equation 6.9.

Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus

enabling the construction of logic circuits that you can use to make computing machines. But

this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the

same group as aluminum, just above it. This results in a theory for AI elements as mathematical

constructs, that we will go into now.

h 4π r

∫

= proton s

h 4π r

∫

= − 6(6.376 − 4.685) = 5proton s /secon d = boron

of 25 39

7.0 Theory of Compounds

Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the

elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and

germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain

noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four

electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope

the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two

types of silicon we have a semiconductor for making diodes and transistors from which we can

make logic circuits for AI.

As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the

right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.

I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy

in the differential because boron B is actually used to make positive type silicon.

That the AI elements make a three by three matrix they can be organized with the letter E with

subscripts that tell what element it is and it properties, I have done this:

Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the

first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be

organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in

three dimensional space so the mechanics of the realm we experience are described by such a

matrix, for example the cross product. Hence this paper where I show AI and biological life are

mathematical constructs and described in terms of one another.

We see, if we include the two biological elements in the matrix (E14) and and (E15) which are

carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to

show not only are the AI elements and biological elements mathematical constructs, they are

described in terms of one another. We see this because the first row is ( B, C, N) and these

happen to be the only elements that are not core AI elements in the matrix, except boron (B)

which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a

mathematical construct, the harmonic mean. Which means we have proved our case because the

first row if we take the cross product between the second and third rows are, its respective unit

vectors for the components meaning they describe them.

of 26 39

The Computation

And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological

elements C and N being the unit vectors are multiplied by the AI elements, meaning they

describe them. But we have to ask; Why does the first row have boron in it which is not a core

biological element, but is a core AI element? The answer is that boron is the one AI element that

is out of place, that is, aluminum is in its place. But we see this has a dynamic function.

The Dynamic Function

The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon

(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic

(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)

and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they

need doping agents from group 13 (three valence electrons) like boron and gallium, and to have

negative type silicon and germanium they need doping agents from group 15 like phosphorus

and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a

different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)

is in boron’s place. This results in an interesting equation.

Equation 7.1.



A = (Al, Si, P)



B = (G a, Ge, As)



A ×



B =



Al Si P

Ga Ge As

= (Si ⋅ A s − P ⋅ Ge)



B + (P ⋅ Ga − Al ⋅ As)



C + (Al ⋅ Ge − Si ⋅ Ga)





A ×



B = − 145



B + 138



C + 1.3924



A = 26.98

+ 28.09

+ 30.97

= 50g/m ol

B = 69.72

+ 72.64

+ 74.92

= 126g/m ol



A ⋅



B = A Bcosθ

cosθ =

6241

6300

= 0.99

θ = 8

∘



A ×



B = A Bsinθ = (50)(126)sin8

∘

= 877.79

877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol

Si(A s − Ga) + Ge(P − Al )

SiGe

Ge + Si

of 27 39

The differential across germanium crossed with silicon plus the differential across silicon

crossed with germanium normalized by the product between silicon and germanium is equal to

the boron divided by the average between the germanium and the silicon. The equation has

nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the

equation have a higher accuracy):

Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic

mean between the semiconductor elements (by molar mass):

Equation 7.2

Now we move on to section 8.0 where I propose a theory for how the electric filed arises and

show that it is calibrated for life, for carbon, and its six-fold symmetry…

28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)

(28.09)(72.61)

2(10.81)

(72.61 + 28.09)

0.213658912 = 0.21469712

0.213658912

0.21469712

= 0.995

(A s − Ga) +

(P − Al ) =

2SiGe

Si + Ge

of 28 39

8.0 The Cause of the Electric Fields

I construct the electric field as such: is one component of . Events in are

through time with components and where c is the speed of light. The electrons and the

protons and in the presence of one another cross into giving themselves acceleration

in the and directions. We have

We suggest

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

and that is . Thus g is frequency squared ( ) and we suggest it is

ℝ

(x, y, z)

ℝ

(x, y, z)

−x





0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

g =

⋅

⋅ s =

of 29 39

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from equation 3.16

And we have

We get

Is carbon. We see the six-fold symmetry that is the carbon is due to the cube having 6

faces. I think I can be more accurate using space expanding in a sphere as opposed to a cube.

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4π r

= q

h 4π r

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792,459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6pr oton s

1/6α

of 30 39

9.0 Natural Units

In total, the list of our constants are:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

One of the interesting things that happens is if we take our equation

And set q equal to the charge of a proton or electron and change to and 6 to :

And solve for k, we get:

Which is:

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G : 6.67408 × 10

−11

c : 299,792,459m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

(

⋅

)

h 4π r

= q

6α

(

3α

⋅

)

h 4π r

= (1.6022C )

(

6α

⋅ 1.6022E − 19C

)

h 4π r

= k

(

0.236

6(0.000053)

⋅ 1.6022E − 19C

)

h 4π r

= k

8.988E 9

299,792,459

(

0.236

6(0.000053)

⋅ 6.022E − 19

)

(6.67408E − 11)(299,792,459)

(6.62607E − 34) 4π (0.833E − 15)

of 31 39

We said the unit of a second was

Where is six proton masses is carbon the core element of life chemistry. Upon seeing this

we suggested the second was a natural unit, and that, if it was, should be in the orbital

mechanics of the earth because the second comes from the calendar, which is based on the

orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the

rotation of the Earth. We found that it was as the following:

Let’s show that…

To find the translational kinetic energy of the moon:

Distance from earth: 3.85E8m

Orbital period:

T=27.32 days=2.36E6seconds

v=1.025E3m/s~1000m/s

Mass: 7.34767E22kg

Use

E=3.67E28 Joules

To find the translational kinetic energy of the earth:

Distance from Sun: 1AU=1.496E11m

Orbital period: 1 year=3.1558E7 seconds

v=2.9785E4m/s, Earth mass: 5.9722E24kg

E=2.649E33 Joules

Earth day=(24)(60)(60)=86,400 seconds.

We have exactly

(4.2388E − 31)(1.861E 30) = k

k = 0.888m /s ≈ 0.9 ≈ 1

⋅

h 4π r

= 1.004996352secon d s

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

K . E . Moon = 3.67E 28J

K . E . Ear th = 2.649E 33j

2π (3.85E 8m) = 2.419E 9m

E =

2π (1.496E11m) = 9.399E11m

of 32 39

Since we have the second as a natural unit, and meters per second as natural, then we have

meters as natural. We have

That is

One of the other things I found in my work Logos, was

Where 2AU is the diameter of the Earth orbit and is the golden ratio conjugate 0.618. I found

this by arriving at the following

Consider minute times day:

Thus,..

The circle is divided into 360 units, each unit (each degree) is the distance the earth moves

around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are

360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).

We have:

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof Ear th

⋅ Ear th Da y = (6)1.2secon d s

⋅

h 4π r

= 1.004996352secon d s

(

6α

⋅ q

)

h 4π r

= 0.888m /s

⋅

h 4π r

(

6α

⋅ q

)

h 4π r

= 0.89m

100

360

∘

= 2AU

orbit rotat ion orbit m oon

29.786k m mi nute 1da y k ilom eter

secon d 27.83k m degree secon d

min ⋅ d a y

deg

⋅

k m

(min)(d ay) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec

min ⋅ d a y

deg

⋅

k m

86,400s

deg

⋅

k m

= 86,400 ⋅

k m

deg

of 33 39

1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always

close to that because its orbit is approximately circular. We have

This is approximately the diameter of the Earth orbit. We define our variables:

Earth orbits:

Earth rotates:

Earth orbits:

Moon orbits:

Earth completes a 360 degree orbit yields:

Where on the right it is in radians and is the radius of the Earth’s orbit. We have

This is (0.00618)360=2.225

0.00618 is

Where is the inverse of the golden ratio. .

We have:

The important thing about this is just as we had the second in the earth-moon orbital

mechanics:

86,400 ⋅

k m

deg

⋅ 360

∘

= 311,040,000k m

311,040,000k m

149,598,000k m /AU

= 2.079AU

= 29.786

k m

27.83k m

min

27.83k m

min

⋅

min

60sec

= 0.4638

k m

sec

1d a y

deg

(24)(60)(60)

deg

= 14,400

sec

deg

1k m

sec

⋅ θ

⋅ v

⋅

360

∘

1AU

= 2π

⋅ θ

⋅ v

⋅ r

= 1AU

(29.786)(14,400)(1k m /s)

(0.4638)(149,598,000)

360 = 2.225AU

100

ϕ = 0.618 =

100

360

∘

= 2AU

of 34 39

We have our meters per second multiplied by 1000 is the lunar orbital velocity:

Or the velocity of the moon is

Thus in total we have…

In the 17th century Christiaan Huygens suggested a definition for the meter noticing pi squared

is approximately the gravitational constant for earth, g. That is:

And, if we were to redefine g as 9.87m/s2 instead of having it 9.81m/s2 then from the equation

for the period of a pendulum

Then if a the length of the arm of the pendulum is 1 meter we have

That is the period of the pendulum would be 2 seconds a swing left and back right, or vice versa.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

= 1k m /s

= 1000m /s

⋅

h 4π r

= 1s

K E

moon

K E

earth

(Ear th Da y) = 1s

(

6α

⋅ q

)

h 4π r

= 0.888m /s

= 1000m /s

100

360

∘

= 2AU

⋅

h 4π r

(

6α

⋅ q

)

h 4π r

= 0.89m

= 9.87

T = 2π

T = 2

(1m)

m /s

= 2secon d s

of 35 39

Just how close is this suggested definition of a meter to the meter we have today? It depends on

the value you use for earth gravity, g. The nominal average is , and at the

equation it is . I find the latter more natural because it is pointed to by the

rotation of the Earth. We have

So we see the meter as we have it today is very near something we could call a natural meter. We

want the kilogram as natural. To see this we write out in terms of the mass of the Earth and its

radius.

Since

We have

Since

We have

We have that seconds and meters are natural units, meters in terms of the constants other than

this above in terms of Lunar and Earth kinetic energies, We have

= 9.80665m /s

= 9.78033m /s

= 0.990955054m

1secon d

= 1secon d

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

(

moon

ear th

(Ear th Da y)

)

= 1m eter

of 36 39

In order to define the SI system of units as natural, we only need an expression for kilograms. To

do this we take our equation 3.15 on page 16, of section 3.0 for intermediate mass:

We change to and change to , and have

In total we have a natural system of units:

Natural System of Units

And we might include in terms of the Earth and Moon:

⋅

h 4π r

= 1s

⋅

h 4π r

(

6α

⋅

)

h 4π r

= 1m

ℏ

1/2

3/2

α /2

ℏ

1/2

= 1.137kg

⋅

h 4π r

= 1s

⋅

h 4π r

(

6α

⋅ q

)

h 4π r

= 1m

1/2

= 1kg

(

moon

ear th

(Ear th Da y)

)

= 1m eter

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10.0 Abstract Cosmology We have our natural system of units that falls conveniently out of

our theories here:

While the first and the third are close to 1s and 1kg respectively, the second is actually 0.8924

meters. If we consider the the fourth

We see since where is given by the Earth rotation period is

(24hrs)(60 min)(60 sec)=86,400 seconds, that the period is

We have redefined the period as one swing left or one swing right of a pendulum as opposed to

the convention of a period as a swing left and then right. So instead of

We have Or

Where g is earth gravity actually given by

Thus we have:

⋅

h 4π r

= 1s

⋅

h 4π r

(

6α

⋅ q

)

h 4π r

= 1m

1/2

= 1kg

(

moon

ear th

(Ear th Da y)

)

= 1m eter

(

K E

)

= 1m eter

(

K E

)

3.67E 28J

2.649E 33

⋅ 86,400sec = 1.197secon d s

T = 2π

2T = 2π

T = π

g = G

(6.674E − 11)(5.972E 24kg)

(6,371,000m)

= 9.82m /s

a = T

= 1.4256m eters

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We want 1.00 meters…but . But we have said our

equation for the meter as a natural is 0.8924m, so when we equate the fourth and second

equations, we introduce the factor of .

We find if we add the factor of in our equation for meters number two and equate it with our

equation for earth and moon, we have the relationship between meters of protons and their

charges and how they are affected by gravity (G) and the speed of light (c). This is the approach

of Abstract Algebra more a subject studied by mathematicians than by physicists. We can call it

Abstract Cosmology, The idea is we can connect the atomic world with the world of the solar

system through the physics of the pendulum as an abstraction because it has to abide by it

because the units work by dimensional analysis. We are then left with an equation we can

analyze to figure out what it represents. Our hypothesis here will be that it is in the formation of

the moon around the Earth because there are periodic forces as found in the pendulum, like the

number of rotations of the Earth per unit time, and atomic forces involving the charge of

electrons and protons and their is gravity acting on them in terms of their masses and radiation

pressures from the Sun, when the accretion discs accrete into the moon orbiting the Earth and

the Earth orbiting the Sun. Our equation is:

Where we have introduced a concept that is a variation on momentum

Is a length.

ϕ(1.4256m) = 0.618(1.4256m) = 0.88m

(

πϕ

6α

)

⋅

⋅ 𝕄𝕃 =

(

K E

)

𝕄 =

= kg

𝕃 = 4πr

= m

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The Author!