of 1 14
A Theory For The Atom’s Subatomic Particles Utilizing A Characteristic Time Of 1-Second
By
Ian Beardsley
Copyright © 2025
of 2 14
Abstract Here the author develops a theory for the subatomic particles that make up atoms, the protons,
electrons, and neutrons. He finds they are described by a characteristic time of 1-second, which is
interesting because the second came from the ancient Sumerians dividing the rotation period of the Earth
(the day) into 24 hours from their base 12 counting (twelve hour day, twelve hour night) and from their
base 60 counting (each hour is divided into 60 minutes, and each minute is divided into 60 seconds). This
theory for subatomic particles was arrived at as a by-product of another of the author’s theories which
was a Schrödinger wave equation solution for the Solar System, which came out to have a characteristic
time of 1-second as well. The subatomic theory had to be developed to connect the microscales (proton’s
atom) to the macroscales (The Solar System). Now the theory includes the electron and neutron, which
both have characteristic times of 1-second as well. We begin with the proton, then move on the the
electron, then to the neutron.
of 3 14
Theory For Inertia I had two equations that gave the radius of a proton with characteristic times of one
second each. I had to break down the equations in terms of their operational parameters as described by a
geometric model. This is what I came up with, a proton is a 4d hypersphere who's cross-section is a
sphere. Of course occupying the dimension of time (4th dimension in drawing) is the vertical component
of the drawing. I have to draw this 3d cross-section as a circle (we cannot mentally visualize four
dimensions). The proton is moving through time at the speed of light (vertical component in the drawing)
it is a bubble in space. The normal force holding it in 3d space is proportional to the inertia
created by the pliability of space measured by G. So when we push on it (Force applied in drawing) there
is a counter force explaining Newton's action/reaction.
I think you could look at this another way: the cross sectional area of the proton moving against space is
in the opposite direction of the force applied and h is the granularity of space, G still its pliability. That is
to say, the flux of a normal force to a hemisphere is over the area of the cross-section of the sphere.
It is the goal of this opening section to provide a theory for inertia, that quality of a mass to resist change
in motion. We want the the theory to include not just the quantum mechanics constant for energy over
time Planck’s constant, but to include the universal constant of gravitation G, the constant the speed of
light from relativity, and the fine structure constant for theories of electric fields so as to bring together
the things that have been pitted against one another: quantum mechanics, relativity, classical physics,
electric fields, and gravitational fields. Towards these ends 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:
To get the ball rolling, I had found a wave solution to the Earth/Moon/Sun system where the Earth
orbiting the Sun is like an electron orbiting a proton with a quantum mechanical solution. I found this
solution had a characteristic time of one second. But, I found as well, I could describe the proton as
having a characteristic time of one second, and that this yielded the radius of a proton very close to that
obtained by modern experiments. So, it is now before me to come up with a theory for the proton in terms
of these characteristic times before I present my theory for a wave solution of the Solar System.
F
n
= h /(ct
2
1
)
h
c
α
of 4 14
The expressions for the characteristic times of 1-second for the proton that I found, were:
1.
2.
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, if you equate the left sides
of each, to one another:
3.
4.
To derive this equation for the radius of a proton from first principles I had set out to do it with the Planck
energy, , given by frequency of a particle, and from mass-energy equivalence, :
We take the rest energy of the mass of a proton :
The frequency of a proton is
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:
The radius of a proton is then
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 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
of 5 14
In order to prove our theory for the radius of a proton as incorporating , we will apply our model
outlined involving a normal force, to a 3d cross-section of a 4d hypersphere countering its direction
through time, t. We begin by writing equation 1 as:
5.
Where , the constant of gravitation measures the pliability of space, and the granularity of space, and
c the speed of propagation. measures the inertia endowed in a proton. We write equation 2 as:
6.
We now say that and that the normal force is
7.
This gives us:
8.
=
Since , we have
9.
This gives
10.
is the cross-sectional area of the proton countering the normal force, , against its motion through
time, this is measured by the constant of gravitation. It is to say that
11.
r
p
= ϕ ⋅
h
cm
p
ϕ
F
n
m
p
=
1
6α
2
4πh
G c
⋅
r
p
1secon d
G
h
m
p
1 =
ϕ
9
⋅
π r
p
α
4
G m
3
p
⋅
h
c(1secon d )
2
⋅
h
c
t
1
= 1secon d
F
n
=
h
ct
2
1
1 =
ϕ
9
⋅
π r
p
α
4
G m
3
p
⋅
h
c
⋅ F
n
π
9α
4
⋅
F
n
G
⋅
r
p
m
2
p
(
ϕ
h
cm
p
)
r
p
= ϕ
h
cm
p
1 =
π
9α
2
⋅
F
n
G
⋅
r
2
p
m
2
p
m
p
=
1
3α
2
⋅
π r
2
p
F
n
G
π r
2
p
F
n
G
m
p
∝
AreaCrossSect ion Pr oton ⋅ F
n
G
of 6 14
And, the coupling constant is
12.
Let us see if this is accurate:
We used the experimental value of a proton . And we have demonstrated that our
model of a proton as a 3D cross-section of a 4D hypersphere countering the normal force against its
motion through time gives its inertia that can counter a force at right angles to its motion through time and
the normal force.
It is thought that the proton does not have an exact radius, but that it is a fuzzy cloud of subatomic
particles. As such depending on what is going on can determine its state, or effective radius. It could be
that the proton radius is as large as
Which it was nearly measured to be before 2010 in two separate experiments. Or as small as
Which is closer to current measurements, which have decreased by 4% since 2010, and could get smaller.
In which case the characteristic time, , could be as large as
Using 2/3 as a fibonacci approximation to . Or, it could be as small as
C =
1
3α
2
F
n
=
h
ct
2
1
=
6.62607E − 34J ⋅ s
(299,792,458m /s)(1s
2
)
= 2.21022E − 42N
m
p
=
18769
3
⋅
π (2.21022E − 42N )
6.674E − 11N
m
2
kg
2
(0.833E − 15m) = 1.68E − 27kg
r
p
= 0.833E − 15m
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
t
1
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1.03351secon d s
ϕ
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= (0.618)
(352275361)π (0.833E − 15m)
(6.674E − 11)(1.67262E − 27kg)
3
⋅
1
3
⋅
6.62607E − 34
299792458
of 7 14
=0.995 seconds
Or perhaps more often it is in the area of:
But, what this tells us is that the unit of a second might be a natural constant. And, since the second comes
from dividing the Earth rotation period into 24 hours, and each hour into 60 minutes, and each minute
into 60 seconds, which ultimately comes to us from the ancient Sumerians who first settled down from
hunting, wandering, and gathering and flaking stones into spearpoints to invent agriculture, writing, and
mathematics, that this might be related to the mechanics of our Solar System. We find if we take the
second as natural we have a wave mechanics solution to our Solar System with a characteristic time of
one second that is connected to the characteristic time of the proton, thus connecting macro scales (the
solar system) to micro scales (the atom).
Why Is Used In The Equation For The Radius Of A Proton
We ask why the golden ratio is used to derive the radius of a proton. We start with our equation
1:!
!
This can be written!
13. !
Where . We notice is the force between two protons separated by the
radius of a proton. Of course two such protons cannot overlap by current theories. But it would
seem this gives rise to the proton’s inertia. We will call it . We also notice is the
normal force that gives rise to the proton’s inertia, . We have!
14. !
Now we look at equation 2. It is!
1
6α
2
m
p
h 4π r
2
p
G c
= 1.004996352secon d s
ϕ
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1second
Gm
2
p
r
2
p
=
h
c
⋅
1
t
2
1
⋅
4π
36α
4
t
1
= 1second
Gm
2
p
r
2
p
F
pp
h
c
⋅
1
t
2
1
F
n
F
pp
= F
n
⋅
4π
36α
4
of 8 14
!
It can be written!
15. !
We see that is the inverse of the potential energy between the two protons
separated by the radius of a proton, we will call such a potential energy . We write 15 as!
16. !
Where !
!
Is the normal potential.!
17. !
Where is the golden ratio. Now we notice from equations
14 and 16 that!
18. !
Or!
19. !
And this should explain it. The gravitational force and its potential is in the normal force and
normal potential in time . The golden ratio is to divide a line such that the whole is to the
greater part as the greater part is to the lesser. What that means is the normal and the action
ϕ ⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1second
(
1
9
⋅
ϕπ
α
4
)
(
r
p
Gm
2
p
)(
h
2
c
2
⋅
1
m
p
⋅
1
t
2
1
)
= 1
(
r
p
Gm
2
p
)
U
pp
(
1
U
pp
)
(
U
n
)
(
1
9
⋅
ϕπ
α
4
)
= 1
U
n
=
(
h
2
c
2
⋅
1
m
p
⋅
1
t
2
1
)
4π
36α
4
1
9
ϕπ
α
4
= Φ
Φ = 1/ϕ = ( 5 + 1)/2 = 1.618...
F
pp
F
n
= Φ
U
n
U
pp
(
F
pp
)(
U
pp
)
=
(
F
n
) (
U
n
)
Φ
ct
of 9 14
are in the ratio of which being the most irrational number, there is no repetition over cycles,
there is minimal interference between the two for equation 19. Verifying using :!
!
The Electron We had equation 10 for the mass of a proton:
10.
Where is the cross-section of a proton and . We have
The electron has no radius, it is a one-dimensional point, all of its rest mass is in its electrostatic energy.
But we can speak of a characteristic size of the electron, not a size it has, but that it acts like its has. We
will call it . We will distinguish this from a characteristic size it already has, the classical electron
radius, which we will call . It is given by
20.
Where is the charge of an electron, is the permittivity of space, is the mass of an electron, a well
established natural constant, and c is the speed of light. Equation 10 for the electron becomes
21.
Putting in the mass of an electron we have the characteristic radius of an electron:
22.
=
We notice
23.
We consider equation 1:
Φ
2.7E − 34N
2.21E − 42N
⋅
2.92E − 57J
2.24E − 49J
= 1.6 = Φ
m
p
=
1
3α
2
⋅
π r
2
p
F
n
G
π r
2
p
F
n
= h /ct
2
1
F
n
=
h
ct
2
1
=
6.62607E − 34J ⋅ s
(299,792,458m /s)(1s
2
)
= 2.21E − 42N
r
e
r
eClass ical
r
eClass ical
=
e
2
4π ϵ
0
⋅
1
m
e
c
2
e
ϵ
0
m
e
r
e
= m
e
G
πF
n
⋅ 3α
2
(9.11E − 31kg)
(6.674E − 11)
π (2.21E − 42)
⋅
3
18769
4.5E − 19m
(4.5E − 19m)
1
3α
2
= (4.5E − 19m)
18769
3
= 2.8E − 15m = r
eClass ical
of 10 14
1.
If
24.
We have
25.
This should equal one second for the electron:
26.
Since we have from equation 21;
21.
We have
27.
=
Which is correct. Where we have lost the factors of in equation 21 when converting from a
characteristic electron radius to the classical electron radius, which is a characteristic length as well
because the electron has no size, all of its mass-energy is bound up in electrostatic energy. We have the
characteristic times for the proton and electron are each 1-second:
1.
28.
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
r
p
m
p
=
r
e
m
e
(
1
6α
2
4πh
G c
)
⋅
r
e
m
e
= 1secon d
1
6
18769
4π (6.62607E − 34)
(6.674E − 11)(299,792,458)
⋅
45.E − 19m
9.11E − 31kg
= 0.9968secon d s ≈ 1secon d
r
e
= m
e
G
πF
n
⋅ 3α
2
r
eClass ical
= m
e
G
πF
n
(9.11E − 31kg)
(6.674E − 11)
π (2.21E − 42)
= 2.82E − 15m
α
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
1
2
4πh
G c
⋅
r
eClass ical
m
e
= 1secon d
of 11 14
=
Finally,…
29.
% accuracy
This recurrent is the scaling factor in equation 10
10.
Where is the cross-second of a proton and is the normal force that gives rise to inertia.
The Neutron The neutron mass is approximately 1.675E-27 kg close to the proton’s of 1.67262E-27 kg.
And, its radius is anywhere from 0.8E-15m to 1.2 E-15m depending on the interpretation and method of
measuring it. Both size and mass are close to that of a proton. Let us suggest a radius for the neutron
based on our theory for inertia. Let the mass of a neutron be , and its radius be . First, we have
equation 1:
1.
We write, then, for the characteristic time of a neutron:
30.
=
The characteristic time of a neutron is thus a 1-second. The mass is given by
31.
1
2
4π (6.62607E − 34)
6.674E − 11)(299,792,458]
⋅
2.82E − 15m
9.11E − 31kg
= 0.998secon d s ≈ 1secon d
1
3α
2
r
p
m
p
=
r
eClass ical
m
e
18769
3
⋅
0.833E − 15m
1.67262E − 27kg
=
2.82E − 15m
9.11E − 31kg
3.0955E15
3.1157858E15
100 = 99.35
1/3α
2
m
p
=
1
3α
2
⋅
π r
2
p
F
n
G
π r
2
p
F
n
= h /ct
2
1
m
n
r
n
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
(
1
6 α
2
4πh
G c
)
⋅
r
n
m
n
= 1secon d
18769
6
4π (6.62607E − 34)
(6.674E − 11)(299,792,458)
⋅
0.834E − 15
1.675E − 27
= 1.004779secon d s
m
n
=
1
3α
2
⋅
π r
2
n
F
n
G
of 12 14
=
32.
=
Not much to do here where the neutron is concerned because it all follows from everything we did with
the proton.
Conclusion We have the characteristic times of the proton, the electron, and the neutron are all one
second:
And, that their masses and radii can be predicted by a theory for inertia where mass in space is due to
motion through time, a normal force ;
,
,
,
, ,
18769
3
⋅
π (2.21022E − 42N )
6.674E − 11N
m
2
kg
2
(0.834E − 15m) = 1.68E − 27kg
r
n
= m
n
G
πF
n
⋅ 3α
2
(1.675E − 27kg)
(6.674E − 11)
π (2.21E − 42)
⋅
3
18769
= 0.830E − 15m
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
1
2
4πh
G c
⋅
r
eClass ical
m
e
= 1secon d
(
1
6α
2
4πh
G c
)
⋅
r
n
m
n
= 1secon d
F
n
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
eClass ical
F
n
G
m
n
=
1
3α
2
⋅
π r
2
n
F
n
G
r
p
= ϕ ⋅
h
cm
p
r
n
= ϕ ⋅
h
cm
n
r
n
= m
n
G
πF
n
⋅ 3α
2
r
p
= m
p
G
πF
n
⋅ 3α
2
r
eClass ical
= m
e
G
πF
n
of 13 14
Where
And,
In the case of an electron, it has no size and all of its rest mass is bound up in its electrostatic energy and
is the fine structure constant, which measures the pliability or impedance of space. It was our goal in
this paper to explain the proton, electron, and neutron in terms of a theory based on a characteristic time
of one second, so we could not only have the basic components of atoms described in terms of a second,
but a theory for the atom that shows the larger scale of the Universe, the Solar System, is described in
terms of a characteristic time of a second and is a related idea on a larger scale. We want a theory for
atoms, H, He, C, N, O,.. also, because the planets formed from the collapse of a cloud made of these
elements. For my theory about the Solar System and its connection to the atom’s proton, which can now
include the electron and neutron, and thus atoms, see:
Theory For The Solar System And The Atom’s Proton; Linking Microscales
To Macroscales (Beardsley, 2025).
It is at:
https://www.academia.edu/143829793/
Theory_For_The_Solar_System_And_The_Atom_s_Proton_Linking_Microscales_To_Macroscales
In the end we will want extend this theory to the rest of the particles of the standard model of particle
physics.
r
eClass ical
=
e
2
4π ϵ
0
⋅
1
m
e
c
2
α =
e
2
4π ϵ
0
⋅
1
ℏc
α
of 14 14
The Author