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A Theory Showing 1-second Time Invariance Across Vastly Diﬀerent

Scales, from the Microcosmos to the Macrocosmos.

October 11-16, 2025!

By Ian Beardsley!

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Compiler’s Notes

I wrote a paper titled: The Tel Aviv (Jaffa) Equation and a Theory for the Atom and Solar

System Linking Microscales to Macroscales; Where Cosmology and Archaeology Meet. I put it

up for discussion on academia.com. The paper provided theories for the Solar System and the

atom that bridged these scales from micro to macro through a characteristic time of 1 second,

showing the unit to be a structural part of Nature. It was here where I met Yochanan

Schimmelpfennig, and where he introduced new things into the theory. We decided to

collaborate on a theory and we started working on a paper. He started putting the theory into a

more sophisticated mathematical framework than my algebraic one. Here I have included some

of my more recent developments and and a couple of his of which you may have already seen in

his comments on the paper I put up for discussion. Most of his additions to the theory so far

were in putting my original theory into a topological mathematics that is new to me and to

suggest new insights from it. I won’t be able to put those developments of his in this paper, but

let me just say they were fantastic. I decided I wanted to restructure my paper, and make it more

concise and leave fewer questions. I would have rather wanted to wait for our paper to come out,

but I lost contact with Yochanan. At some point I stopped hearing from him. I hope he is okay,

but I have to continue with this project; I have put too much time and hard work into it to stop

now.

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Contents

List of Constants, Variables, And Data In This Paper………………………………4

Part A: The Solar System Theory…………………………………………………….5

We put forward a quantum wave solution for the Solar System with a basis unit of one second.

Part B: The Atomic Theory……………………………………………………………6

We put forward a theory for the proton, electron, and neutron with a basis unit of one second.

Part C: The Schimmelpfennig Formulation………………………………………….12

Here we talk about Schimmelpfennig’s alternate solution and its connection to mine.

Part D: The Delocalization Time………………………………………………………15

We see the delocalization time as given by the wave solution for a gaussian wave packet

distribution supports the determination of a solar system Planck-type constant.

Part E: The Equations Bridging Across Scales……………………………………….18

We show 1-second time invariance across enormously different scales, the scale of the atom to the scale

of the Solar System.

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List of Constants, Variables, And Data In This Paper

(Proton Mass)

(Proton Radius)

(Planck Constant)

(Light Speed)

(Gravitational Constant)

1/137 (Fine Structure Constant)

(Proton Charge)

(Electron Charge)

(Coulomb Constant)

(The Author’s Solar System Planck-Constant)

(Earth Mass)

(Earth Radius)

(Moon Mass)

(Moon Radius)

(Mass of Sun)

(Sun Radius)

(Earth Orbital Radius)

(Moon Orbital Radius)

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:

(Kinetic Energy Moon)

(Kinetic Energy Earth)

m

p

: 1.67262E − 27kg

r

p

: 0.833E − 15m

h : 6.62607E − 34J ⋅ s

c : 299,792,458m /s

G : 6.67408E − 11N

m

2

s

2

α :

q

p

: 1.6022E − 19C

q

e

: 1.6022E − 19C

k

e

: 8.988E 9

Nm

2

C

2

ℏ

⊙

: 2.8314E 33J ⋅ s

M

e

: 5.972E 24kg

R

e

: 6.378E6m

M

m

: 7.34767309E 22k g

R

m

: 1.7374E6m

M

⊙

: 1.989E 30kg

R

⊙

: 6.96E 8m

r

e

: 1.496E11m = 1AU

r

m

: 3.844E 8m

K E

m

=

1

2

(7.347673E 22k g)(966m /s)

2

= 3.428E 28J

K E

e

=

1

2

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

2

= 2.7396E 33J

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Part A: The Solar System Theory

The idea: We can model the Solar System and the atom such that across these scales, which are immense,

they can be shown as structured according the same physical concepts.

How: To do this we ﬁnd we can model the Solar System as having a quantum Shrödinger wave solution,

and the atom in terms of its proton, electron, and neutron components as having a solution as being

particles whose inertia is due to their rest motions through time.

The connection: The connection in scales is found to be through a basis unit of one second — making

this unit of time being not just a cultural development handed to us from the ancient Sumerians when they

divided the Earth day into 24 hours, with each hour being 60 minutes, each minute being 60 seconds —

but to be a structural one.

We start with the Schrödinger wave equation in spherical coordinates:

Whose solutions for the hydrogen atom are!

A1.

A2.

Instead of solving the wave equation we suggest since gravity from the Sun is an inverse square ﬁeld

governing the motion of planets around it and the proton is an inverse square ﬁeld as well with the

electron, the solutions would be of the same form except , the atomic number, would not be 1 proton but

the radius of the Sun. However, we ﬁnd for this to work we have to suggest its size is normalized by the

radius of the Moon, giving it a radius of 400. As such we see the Moon is the metric for measuring size,

and as well will see distance and mass as well. This comes to us from the condition for a perfect eclipse of

the Sun by the Moon, which is:

A3.

is the the orbital radius of the Earth, is the lunar orbital radius, is the solar

radius, and is the lunar radius. Thus the solutions are for energy and orbital radius of the

Earth (The energy can be taken as the orbital kinetic energy of the Earth, which matches better

than 99%) :!

A4. , A5.

−

ℏ

2

2m

[

1

r

2

∂

∂r

(

r

2

∂

∂r

)

+

1

r

2

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

+

1

r

2

sin

2

θ

∂

2

∂ϕ

2

]

ψ + V(r) ψ = E ψ

E

n

= −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

r

n

=

n

2

ℏ

2

Z k

e

2

m

e

Z

r

earth

r

moon

=

R

⊙

R

moon

r

earth

r

moon

R

⊙

R

moon

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

e

=

1

n

⋅

R

m

R

⊙

⋅

ℏ

2

⊙

G

⋅

M

⊙

M

e

M

m

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Where n=3 is the Earth orbital number. We notice it is in terms of the the masses of the Earth and the

Moon. Later, I will convert this to a solar solution in terms of the Sun and Earth. We get that the ground

state equation giving the unit of a second is (introducing to measure distance in time units):

A6.

(Follows directly from the Bohr atom ground state.) And, we ﬁnd the Earth/Moon/Sun system solution

above has a Solar System Planck-type constant given by:

A7.

Since the Earth day lengthens by small amounts as the lunar orbit grows because the Earth loses energy to

the Moon, we suggest the characteristic length of the Earth Day is of 24 hours because it is given by one

second in:

A8.

the kinetic energy of the Moon, the kinetic energy of the Earth, EarthDay is 24 hours (86,400

seconds, the rotation period of the Earth) and is the Earth’s tilt to its orbit around the Sun.

Part B: The Atomic Theory

Now we need to show that the electron, neutron, and proton have a characteristic time of one second that

we ﬁnd on the much larger scale of the Earth/Moon/Sun system.

We will suggest a proton is a 3D cross-section of a 4D hypersphere held in place countering its motion

through time by a normal force ( ) that produces its inertia (measured in mass in kilograms) much the

same way we model a block on an inclined plain countered by friction from the normal force to its

motion. The following is the illustration of such a proton as a cross-sectional bubble in space:

We ﬁnd the 1-second time invariant equations are:

c

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

ℏ

⊙

= (1secon d )K E

e

K E

m

K E

e

(Ear th Da y)cos(θ ) = 1.0secon d s

K E

m

K E

e

θ = 23.5

∘

F

n

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B1. ,

B2. , ,

, and so on…

Proton

B3. !

= !

=1.00500 seconds !

Neutron

B4. !

= !

=1.004779 seconds!

Electron

B5. !

= !

=0.99773 seconds!

B6. !

, , , !

F

n

=

h

ct

2

1

t

1

= 1secon d

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

m

e

=

π r

2

eClassical

F

n

G

m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

π r

2

p

= Ar eaCr ossSect ion Pr oton

1secon d =

r

p

m

p

⋅

πh

Gc

1

3α

2

(0.833E − 15)

(1.67262E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1secon d =

r

n

m

n

⋅

πh

Gc

1

3α

2

(0.834E − 15)

(1.675E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1secon d =

r

eClassical

m

e

⋅

πh

Gc

(2.81794E − 15)

(9.11E − 31)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

1secon d =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

r

e

= r

eClassical

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These were derived from the formulations of the time invariant second that are the forms

B7.

B8.

Where is the golden ratio, is the radius of a proton, and is the mass of a proton. We ﬁnd

these produce close to the most recent measurements of the radius of a proton (0.831E-15m), if you

equate the left sides of each, to one another:

B9.

B10.

To derive this equation for the radius of a proton from ﬁrst principles (thought we have done it

geometrically, above) I had set out to do it with the Planck energy, , given by frequency of a

particle, and from mass-energy equivalence, :

B11.

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

B12.

The frequency of a proton is

B13.

We see at this point we have to set the expression equal to . So we need to come up with a theory for

inertia that explains it:

B14.

B15.

The radius of a proton is then

(

1

6α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

ϕ = 0.618

r

p

m

p

r

p

= ϕ

h

cm

p

r

p

= 0.816632E − 15m

E = h f

E = m c

2

E = h f

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

ϕ

m

p

c

2

h

⋅

r

p

c

= ϕ =

m

p

c

h

r

p

m

p

r

p

= ϕ ⋅

h

c

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B16.

Something incredible regarding the connection between microscales (the atom’s proton) and macro scales

(the solar system) if you want to get very close to modern measurements of the proton and to exactly a

characteristic time of one second. The radius of a proton is not constant, but depends of the nature of the

experiment, because protons are thought to be a fuzzy cloud of subatomic particles. Therefore, not using

in our equations for protons and the characteristic time of one second, but the right ratio of terms in the

ﬁbonacci sequence, we ﬁnd that ratio is 5/8:

=0.6303866

If

0, 1, 1, 2, 3, 5, 8, 13,…

is the ﬁbonacci sequence whose successive terms converge on , the golden ratio, then the two terms that

come closest to this are 5/8=0.625.

This is a characteristic time from

B17.

that has a value of

B18.

1.0007seconds

Combining

B19.

with

r

p

= ϕ ⋅

h

cm

p

ϕ

r

p

= ϕ

h

cm

p

ϕ =

r

p

m

p

c

h

=

(0.833E − 15)(1.67262E − 27)(299,792,458)

6.62607E − 34

ϕ

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1.0335secon d s

5

8

⋅

(352275361)π (0.833E − 15m)

(6.674E − 11)(1.67262E − 27)

3

1

3

⋅

(6.62607E − 34)

299,792,458

=

5

8

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1.0007secon d s

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B20.

Gives the radius of a proton to be

B21.

B22.

With this, while we get very close to one second (1.0007 seconds) with the ﬁbonacci ratio of 5/8 we also

get something very much in line with the most recent measurement for the radius of a proton

(0.831E-15m).

We ﬁnd that for our solutions of the Solar System, that is derived from the Fibonnaci approximation

ratio to that is 2/3, where as for the subatomic particles solution they take on the Fibonacci ratio 5/3

approximation to .

The electron has no size, its classical radius is determined by

B23.

This comes from the electron not having mass or size, from the idea that all of its mass is bound up in its

electrostatic energy. We see that from our 1-second time invariant equations that they are all one equation.

The equations for the proton and neutron have and for the electron is , because for the

electron there is no spacial impedance because it has no size or mass. Again, equations B3-B6:

Proton!

!

Neutron!

!

Electron!

!

Our equation for the subatomic particles is (we say )!

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

r

p

=

5

8

h

cm

p

r

p

=

5

8

(6.62607E − 34)

(299,792,458)(1.67262E − 27)

= 0.8258821E − 15m

ℏ

⊙

ϕ

r

eClassical

=

e

2

4π ϵ

0

⋅

1

m

e

c

2

κ

i

= 1/3α

2

κ

i

= 1

1secon d =

r

p

m

p

⋅

πh

Gc

1

3α

2

1secon d =

r

n

m

n

⋅

πh

Gc

1

3α

2

1secon d =

r

eClassical

m

e

⋅

πh

Gc

r

eClassical

= r

e

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; , , !

We see, as we said, that the Solar System Planck-Type Constant can nicely be taken as;!

B24.

B25. where,

B26.

In order to apply this to other star systems, we have to be able to predict the radius of the habitable planet,

presumably in the n=3 orbit. I found the answer to be in the Vedic literature of India. They noticed that the

diameter of the Sun is about 108 times the diameter of the Earth and that the average distance from the

Sun to the Earth is about 108 solar diameters, with 108 being a signiﬁcant number in Yoga. So I wrote the

equivalent:

B27.

radius of the star. The surprising result I found was, after applying it to the stars of all spectral types

from F through K, with their different radii and luminosities (the luminosities determine , the

distances to the habitable zones), that the radius of the planet always came out about the same, about the

radius of the Earth. This may suggest optimally habitable planets are not just a function of the distance

from the star, which determines their temperature, but are functions of their size and mass probably

because they are good for life chemistry, atmospheric composition, and gravity when they are the size and

mass of the Earth.

In order to get , the distance of the habitable planet from the star, we use the inverse square law for

luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter

than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times

that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is

given by:

B28.

the luminosity of the star, and the luminosity of the Sun. We compute the orbital radius of the

Moon…

B29.

Which works for our Solar System, Ag and Au the relative masses of silver and gold atoms.

1secon d =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

ℏ

⊙

= (hC )K E

e

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

G m

3

p

ℏ

⊙

= (hC )K E

earth

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

R

planet

= 2

R

2

⋆

r

planet

R

⋆

r

planet

r

planet

r

planet

=

L

⋆

L

⊙

AU

L

⋆

L

⊙

r

m

= R

⋆

Ag

Au

= R

⋆

/(1.8) =

6.957E 8m

1.8

= 3.865EE 8m

of 12 20

Part C: The Schimmelpfennig Formulation

Schimmelpfennig made a big development that should really open up the theory. He started out by writing

the equation of the potential for an orbit:

C1.

He then deﬁned the effective action constant as the orbital angular momentum:

C2.

Then said the Bohr-like orbital radius is

C3.

He then observed by substituting into this last equation (equation 3) we have

C4.

He then observes that is the exact orbital equation, the identity. That, in other word his solution collapses

to the classical solution. I then show close to a second appears in his equation for the Moon:

C5.

That would be the distance to the Moon. I then suggested we divide that by the speed of light to measure

distance in time:

C6.

It is close to one second. We will see my ground state equations are:

C7.

Which is a little less than the Moon’s orbital distance (3.844E8m). We then divide this by the speed of

light and we get exactly 1-second for all practical purposes.

C8.

V(r) = −

GMm

r

h * := L = m vr

r =

h *

2

GMm

2

h * = m vr

v

2

=

GM

r

h *

2

GM

e

M

2

m

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

= 3.8179E 8m

h *

2

GM

e

M

2

m

⋅

1

c

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

1

299,792,458m /s

= 1.29secon d s

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E 8m

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

of 13 20

We saw my solution was the equation:

C9.

This would mean my equation for the orbit is, using and :

C10.

Where n=3 is the Earth orbital number. Thus Schimmelpfennig has

C12.

Where for my equation the Planck-type constant is:

C13.

And his is

C14.

This has me thinking if the moon is close to a light-second away from the Earth several things: ﬁrst, if we

are to measure distance with time, that is why we introduce the speed of light c into the equation for the

ground state. Second, in my solution the energy equation uses the radius of the Sun in place of the atomic

number, number of protons, but its size is normalized by the size of the moon giving it a radius of 400. In

the same way the Moon being close to a light-second away from the Earth, means its orbital radius

normalizes distance. This works great because the moon has a nearly perfect circular orbit, it is a near

perfect sphere, and its mass is very accurately known. Thirdly, the moon as seen from the Earth near

perfectly eclipses the Sun, this gives .This allowed me to make a Earth/Sun

solution as opposed to the Moon/Earth solution (In this paper) with a characteristic time of 1 second as

well. Fourthly, It might be good to ﬁnd how far the moon was from the Earth and when, when its distance

was exactly a light second from the Earth as opposed to the 1.3 light-seconds of today.

Equating my with his yields the relationship between our solar system Planck-type constants;

C15.

This can be written

C16.

We then consider my solar solution for energy from using the eclipse condition:

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

e

K E = (1/2)m v

2

v

2

= GM

⊙

/r

e

r

e

=

1

n

⋅

R

m

R

⊙

⋅

ℏ

2

⊙

G

⋅

M

⊙

M

e

M

m

r

e

=

h *

2

GM

⊙

M

2

e

ℏ

⊙

= (1secon d )(K E

e

)

h* = M

e

v

e

r

e

R

⊙

/R

moon

= r

earth

/r

moon

r

e

h*

2

ℏ

⊙

=

1

n

⋅

R

m

R

⊙

⋅

M

2

⊙

M

e

M

3

m

M

3

m

M

2

⊙

M

e

=

ℏ

2

⊙

h*

2

⋅

1

n

⋅

R

⊙

R

m

of 14 20

C17. , and ,

And using:

C18.

Which is:

C19.

And consider the equation for a second that followed from equating this solar formulation with the

original lunar formulation :

C20.

Putting the terms from equation C16

C21.

We see our theory has applications to archaeology because the second came to us historically from the

ancient Sumerians because they divided the Earth day (rotation period) into 24 hours, and, because each

hour and minute got further divisions by 60 because their base 60 counting system was inherited by the

ancient Babylonians who were the ultimate source of dividing the hour into minutes and the minutes into

seconds. I have found this system is given by the rotational angular momentum of the Earth, , in

terms of the solar system Planck-type constant, because, as I already pointed out:

C22.

C23.

This base 60 counting combined with dividing the day into 24 units is mathematically optimal because

the rotational angular momentum incorporates not just the day (rotation period of the Earth) but the mass

and size of the Earth. We can say we are touching on archaeoastronomy. This is because 60/24=2.5 and

the Scottish engineer, Alexander Thom, found ancient megalithic (stone) observatories throughout Europe

may have been based on a unit of length he called the megalithic yard and that the separations between

R

⊙

R

m

=

r

e

r

m

K E

e

=

1

2

⋅

GM

⊙

M

e

r

e

v

2

m

=

GM

e

r

m

ℏ

⊙

: L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

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

m

2

s

E

3

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

1secon d = 2r

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

M

3

m

M

2

⊙

M

e

M

3

m

M

2

⊙

M

e

=

ℏ

2

⊙

h*

2

⋅

1

n

⋅

R

⊙

R

m

1secon d = 2r

e

⋅

v

m

v

2

e

⋅

ℏ

⊙

h*

R

m

R

⊙

⋅

M

⊙

M

e

L

earth

L

earth

ℏ

⊙

24 = 60

L

earth

=

4

5

π M

e

f

e

R

2

e

of 15 20

stones, that align with celestial positions and cycles, are recurrently separated by 2.5 megalithic yards.

Like in Stonehenge.

You will notice the 4/5 in the above equation can be written as 2/5, meaning the equation becomes:

C24.

the rotational frequency of the Earth, Since , where is the angular velocity of the Earth, we

have:

C25.

We notice , the ancient Sumerian factors deﬁning the second from the rotation period of the

Earth, We see the rotation, mass and radius (size) of the Earth are optimally formulated with ancient

Sumerian base 12 and base 60 mathematics (they divided the day into 12 hours, and the night into 12

hours).

I have found that the pressure gradient of the protopanetary disc, as a function of radius, that gave birth to

our solar system, is given by:

C26.

Because we can know the exponent in C26 for the protocloud from which a star system forms, given by

where the radius of the disc where pressure is , we can determine for the star

system. This is the solution to:

C27.

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. This becomes useful for modeling other star systems because is the exponent in the

equation for the pressure gradient of the protoplanetary disc.

Part D: The Delocalization Time

In order to show that our hypothesis is right, we solve the wave equation for a Gaussian wave packet and

determine the delocalization time, . If it is about six months, the time it takes the Earth to delocalize

(travel its orbital diameter), using the Moon playing the role of the mass of an electron and our as

above to describe the Earth, then the hypothesis can be taken as correct, and we can solve the whole

system for the Earth/Moon/Sun system from the rest of the equations in the hydrogen atom solution to the

Schrodinger wave equation.

L

earth

=

2

5

M

e

2 π f

e

R

2

e

f

e

2 π f

e

= ω

e

ω

e

L

earth

=

2

5

M

e

ω

e

R

2

e

2/5 = 24/60

P(R) = P

0

(

R

0

)

−

L

ear th

ℏ

⊙

p

p = L

earth

/ℏ

⊙

R

0

P

0

ℏ

⊙

d P

dr

= − ρ

(

GM

⋆

r

2

−

v

2

ϕ

r

)

v

2

ϕ

/r

GM

⋆

/r

2

ρ

L

earth

ℏ

⊙

τ

ℏ

⊙

of 16 20

The delocalization time of a particle, molecule, or mass in general, , is the time it takes a particle to

delocalize. If we want to apply our wave equation theory of the Solar System to this concept, then the

delocalization time should be the time for the Earth to travel the diameter of it’s orbit, which would be

half a year (about six months). In order to derive the delocalization time we must consider a Gaussian

wave packet…

The Gaussian wavefunction in position space is

D1.

It’s Fourier wave decomposition is

D2.

We use the Gaussian integral identity (integral of quadratic)

D3.

τ

ψ (x,0) = Ae

−

x

2

2d

2

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2 π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

− a x

2

+bx

d x =

π

a

e

b

2

4a

of 17 20

We ﬁnd via the inverse Fourier transform. It is

D4.

Substitue :

D5.

The solution is standard and is:

D6.

D7.

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 uses 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

D8.

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

D9.

Now let’s compute a half a year…

D10. (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

D11.

Conclusion We see we the quantization of the Earth/Moon/Sun system is with a characteristic time of one

second and the mass of the Moon.

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

d

τ

ℏ

⊙

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22k g)(3.844E 8m)

2

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

of 18 20

Part E: The Equations Bridging Across Scales

In Schimmelpfennigs Formulation

C5.

That would be the distance to the Moon. In units of time, this is close to a second:

C6.

Beardsley’s ground state equations are:

C7.

C8.

Which is a little less than the Moon’s orbital distance (3.844E8m), but gives the light-second accurately.

The 1-second time invariant equations are

Proton

B3. !

= !

=1.00500 seconds !

Neutron

B4. !

= !

=1.004779 seconds!

h *

2

GM

e

M

2

m

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

= 3.8179E 8m

h *

2

GM

e

M

2

m

⋅

1

c

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

1

299,792,458m /s

= 1.29secon d s

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E 8m

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

1secon d =

r

p

m

p

⋅

πh

Gc

1

3α

2

(0.833E − 15)

(1.67262E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1secon d =

r

n

m

n

⋅

πh

Gc

1

3α

2

(0.834E − 15)

(1.675E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

of 19 20

Electron

B5. !

= !

=0.99773 seconds!

B6. !

, , "

1secon d =

r

eClassical

m

e

⋅

πh

Gc

(2.81794E − 15)

(9.11E − 31)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

1secon d =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

of 20 20

The Author