of 1 15
The Second As A Universal Characteristic Time!
By!
Ian Beardsley!
Copyright © 2025"
of 2 15
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
⊙
: 2.8314E33J ⋅ s
M
e
: 5.972E 24kg
R
e
: 6.378E6m
M
m
: 7.34767309E 22kg
R
m
: 1.7374E6m
M
⊙
: 1.989E30kg
R
⊙
: 6.96E8m
r
e
: 1.496E11m = 1AU
r
m
: 3.844E8m
K E
m
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 3 15
Abstract: I present my findings that show a characteristic time of one second is in a wave
solution of the Solar System and the atom’s proton in common. As well I show it describes the
hydrocarbons, the skeletons of biological life chemistry. I show the unit of a second is not
arbitrary, as history would have it, but rather comes from the wise decision of the ancient
Sumerians to have developed an Earth day (rotation period) of 24 hours and a base 60
counting system. I also find the characteristic time of 1 second is in the protoplanetary disc
and in the formation of the proton in the Big Bang that gave birth to the Universe.!
In the expression on the left of the equation:!
!
it has units of mass times time divided by the mass of a proton, . This means we have
divided into a mass times a time giving one second. You want it to be because we have:!
!
Thus we can equate the these two equations to get the radius of a proton:!
!
Where is the golden ratio. The most recent value is 0.833E-15m. This is very close to the
radius of a proton, and may actually be the radius of a proton because is the golden ratio
which optimizes things in many systems, and the tendency over the years in measuring the protons size is
that it is getting smaller. The historic value was 0.877E-15m, 4% larger than the current value. That value
was measured twice by two different methods a long time ago. I find it can be approximated by
introducing the fibonacci ratio of 2/3 that approximates the golden ratio by replacing the golden ratio with
it, thus using:
The proton is thought not to have a precise radius but rather is changing by small amounts around a
central value due to it really being a fuzzy cloud of subatomic particles. Thus
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
m
p
6m
p
6m
p
ϕ ⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1secon d
r
p
= ϕ
h
cm
p
ϕ = 0.618
ϕ = 0.618
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
of 4 15
are both close to a second, just the first is a little over a second and the second is just under a second, But
let us ask what
is. It is action ( ) or energy over time as it occurs due to the nature of space, reduced by G, the pliability
of space, and the speed of light, c, the speed at which things interact, for the size of a proton ( and its
mass . is introduced for the surface area of a proton. We find then, this gives the characteristic
time for the proton, which happens to be nicely one second, But why the 6 in . Well this says six
protons give a characteristic time of one second. This means six protons which is carbon, the core element
of life, is described by one second. You want that because it has been found carbon is the most stable
element in terms of which to study other elements because a configuration of 6 protons, or six electric
fields, comes out mathematically stable. This can be thought of loosely as the six sided figure, a regular
hexagon, has its sides equal in length to its radii thus resulting in dynamic stability associated with six-
fold symmetry. It is why the bee’s honeycomb is tessellated regular hexagons. The same goes for the six
electrons orbiting, and attracted to, the six protons. Thus you want the basis unit of our physics, 1 second,
associated with carbon, 6 protons. But further, since you want your physics to be based on the six-fold
you want six seconds to give the proton, or hydrogen atom, because it is the basis unit of chemistry; 1
proton, 1 electron, which it does. We have
is carbon (C)
is hydrogen (H)
A very interesting thing here is, 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 life chemistry.
And indeed for our solar system the characteristic time is one second as well. The Planck type constant
for the solar system I found is is given by the Earth orbit meaning the Earth quantizes angular
momentum for the system
That is, it is 1 second times the kinetic energy of the Earth. But, the ground state is given by the Moon
orbiting the Earth
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
h
r
p
)
(m
p
)
4π
6m
p
1
6pr o t on s
⋅
1
α
2
⋅
r
p
m
p
4π h
G c
= 1sec on d
1
1pr o t on
⋅
1
α
2
⋅
r
p
m
p
4π h
G c
= 6sec on d s
ℏ
⊙
ℏ
⊙
= (1secon d )K E
e
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
of 5 15
Where is the mass of the Moon. That is,
is the analogy to the ground state of the hydrogen atom:
We know our Planck constant for the Solar System is correct, , because we have
Is 99.5% accurate, the radius of the Sun, the radius of the Moon:
=
=2.727E33J
The kinetic energy of the Earth is
Which is very good, about 100% accuracy for all practical purposes.
The radius of the Sun in lunar radii, plays the role of the number of protons, Z, for the atom. That is for
the atom:
M
m
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22k g)
3
= 3.0281E8m
r
1
=
ℏ
2
k e
2
m
e
r
1
≈ 0.529E − 10m
ℏ
⊙
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
E
n
= n
R
⊙
R
m
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
6.96E 8m
1737400m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg )
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
K E
e
=
1
2
(5.972E 24kg )(30,290m /s)
2
= 2.7396E 33J
2.727E 33J
2.7396E 33J
100 = 99.5 %
R
⊙
R
m
→ Z
2
of 6 15
The characteristic time of about one second gives us, as well, the rotation period of the Earth, 24 hours, in
terms of the kinetic energies of the Moon and the Earth:
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:
Even though the second came to us historically because the ancient Sumerians divided the Earth day
(rotation period) into 24 hours, and because each hour and minute got further divisions by 60 because of
their base 60 counting system by the ancient Babylonians that inherited it from the ancient Sumerians, the
division is not arbitrary, and hence neither is the second, because I have found this system is given by the
rotational angular momentum of the Earth in terms the solar system Planck-type constant:
Where the angular momentum, , is given by the mass of the Earth, the size of the Earth, and its
rotation frequency.
The value is 2.5 which by modeling our Solar System is found to be the exponent in the pressure gradient
for the protoplanetary disc from which our Solar System formed. That is I found
the pressure of the disc as a function of radius. Which suggests that the structure of the protoplanetary
disc could be governed by the same fundamental time of one second in the Earth’s rotation and that the
Earth’s formation process may be encoded in the same number we developed since ancient times to
describe time (24, 60). This is the solution to:
E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
K E
m
K E
e
(Ear th Da y) = 1.1 − 1.3secon d s
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg )(30,290m /s)
2
= 2.7396E 33J
L
earth
ℏ
⊙
24 = 60
L
earth
=
4
5
π M
e
f
e
R
2
e
L
earth
P(R) = P
0
(
R
R
0
)
−
L
ear th
ℏ
⊙
d P
dr
= − ρ
(
GM
⋆
r
2
−
v
2
ϕ
r
)
of 7 15
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. It is the pressure gradient of the disc in radial equilibrium balancing the inward gravity and
outward centripetal force. 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:
The surprising result I found was, after applying it to the stars of many spectral types, 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 it is good for life chemistry. Here
are just a few examples using the data for several spectral types:
F8V Star
Mass: 1.18
Radius: 1.221
Luminosity: 1.95
F9V Star
Mass: 1.13
Radius: 1.167
Luminosity: 1.66
…
v
2
ϕ
/r
GM
⋆
/r
2
ρ
R
planet
= 2
R
2
⋆
r
planet
r
planet
M
⋆
= 1.18(1.9891E 30kg) = 2.347E 30kg
R
⋆
= 1.221(6.9634E8m) = 8.5023E8m
r
p
= 1.95L
⊙
AU = 1.3964AU(1.496E11m /AU ) = 2.08905E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.5023E 8m)
2
2.08905E11m
=
6.92076E6m
6.378E6m
= 1.0851Ear th Ra d ii
M
⋆
= 1.13(1.9891E 30kg) = 2.247683E 30kg
R
⋆
= 1.167(6.9634E8m) = 8.1262878E8m
r
p
= 1.66L
⊙
AU = 1.28841AU(1.496E11m /AU ) = 1.92746E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.1262878E 8m)
2
1.92746E11m
=
6.852184E6 m
6.378E6m
= 1.0743468Ear th Ra d ii
of 8 15
G0V Star
Mass: 1.06
Radius: 1.100
Luminosity: 1.35
G1V Star
Mass: 1.03
Radius: 1.060
Luminosity: 1.20
As you can see we consistently get about 1 Earth radius for the radius of every planet in the habitable
zone of each type of star. It might be that radius is right for life in terms of gravity and densities for the
elements. I got these results for the stars from spectral types F5V to K3V.
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:
Also, the theory utilizes the fact that the Moon as seen from the Earth perfectly eclipses the Sun as a
possible condition for optimal habitability of the planet, which is
Orbital radius of the planet to that of the moon is radius of the star to that of the moon. It is known that
the Moon has a lot to do with the conditions for life on Earth being good because its orbit holds the Earth
at its inclination to the Sun its orbit preventing temperature extremes and allowing for the seasons.
It may be the ancient Sumerians accidentally gave us the second, which turns out to be Natural, not
arbitrary, because they chose base 60 as their counting system because it was evenly divisible by the first
M
⋆
= 1.06(1.9891E 30kg) = 2.108446E 30kg
R
⋆
= 1.100(6.9634E8m) = 7.65974E 8m
r
p
= 1.35L
⊙
AU = 1.161895AU(1.496E11m /AU ) = 1.7382E11m
R
p
=
2R
2
⋆
r
p
= 2
7.65974E 8m)
2
1.7382E11m
=
6.751E6 m
6.378E6m
= 1.05848Ear th Ra d ii
M
⋆
= 1.03(1.9891E 30kg) = 2.11E 30kg
R
⋆
= 1.060(6.9634E8m) = 7.381E 8m
r
p
= 1.20L
⊙
AU = 1.0954AU(1.496E11m /AU ) = 1.63878589E11m
R
p
=
2R
2
⋆
r
p
= 2
7.3812E 8m)
2
1.63878589E11m
=
6.6491E6 m
6.378E6m
= 1.0425Ear th Ra d ii
r
planet
r
planet
=
L
⋆
L
⊙
AU
r
planet
r
moon
=
R
star
R
moon
of 9 15
six integers 1, 2, 3, 4, 5, 6, the smallest number that does this, and this is to introduce six-fold symmetry
that is at the heart of stable dynamics.
How we derive the solar system Planck-type constant:
Where
=
=
=
=
I find we can write a solution for the kinetic energy not in terms of the Moon and the Earth, but in terms
of the Sun and the Earth, which is
ℏ
⊙
= (hC )K E
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
C =
1
3
⋅
1
α
2
c
1
3
⋅
2 π r
p
G m
3
p
1
3
⋅
18769
299792458
1
3
⋅
2 π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.55976565E 33
s
m
m
kg
3
⋅
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
⋅
s
kg ⋅ m
=
1
kg
⋅
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= en erg y
hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s
hC =
(
kg
m
s
2
⋅ m ⋅ s
)
(
1
kg
⋅
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
⋅
s
2
m
2
)
= secon d s
K E
earth
=
1
2
(5.972E 24kg )(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
of 10 15
To do this we have to use the condition of a perfect eclipse:
And, use the orbital velocity of the Earth given by
And, redefining as
, which would come from
This gives the two different Planck-type constants for the solar system are related by
However, this again yields the characteristic time of one second by equating the solar and lunar solutions:
Where the square root of 3 is the square root of the Earth orbital number.
The Protoplanetary Disc If the characteristic time for both the solar system
and the proton
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
R
⊙
R
m
=
r
e
r
m
v
2
m
=
GM
e
r
m
ℏ
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
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
L
p
→ ℏ
⊙
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
1secon d = 2r
p
v
m
v
2
p
M
3
m
3
M
2
e
M
⊙
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
of 11 15
are 1 second, then the characteristic time of 1 second should be in the protoplanetary disc from which the
planets formed. I would guess it would be in the time between collisions of particles in the protoplanetary
disc. Ultimately, the planets form from these collisions. The time between collisions in the protoplanetary
disc are given by
1. Particle number density n ( the number of particles per unit volume).
2. Relative velocity between particles .
3. Particle cross-section (related to particle size).
For micron to millimeter sized grains in a dense inner region of the protoplanetary disc (like about 1 AU
from the star, which is the Earth orbit) the range of these values are:
1. particles per meter cubed (from disc models)
2. Particles sizes are meters.
3. Relative velocities of particles are as driven by Brownian motion, turbulence, and
gas drag.
We can imagine a scenario where this yields 1 second by using typical values:
Our equation
suggests that the proton’s fundamental structure encodes a natural unit of time, the presence of G, h, and c
may emerge from a balance between gravity, quantum mechanics, and relativistic effects. Since the
equation for the characteristic time of the solar system
indicates that the solar system is quantized by planetary formation processes in a way that maintains a
fundamental unit of periodicity of 1 second. So we are connecting the formation of entire planetary
systems to the fundamental structure of matter itself (the protons). A micron-sized dust grain at about
10E-12 grams, has about 10E11 protons, a millimeter sized dust grain, about 10E-6 grams, has about
10E17 protons. But where did these dust grains come from? They formed in the protoplanetary gas cloud
from elements, mostly hydrogen and helium, and elements like C, O, Si, and Fe, that formed from
hydrogen and helium in stars by nucleosynthesis. The heavier elements are made from hydrogen and
helium in stars then later expelled into space by supernovae. But where did the hydrogen (and some of the
helium) come from? They were made in the Big Bang, so we should be able to track the characteristic
time of the solar system, and proton back to the Big Bang and the formation of the Universe.
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
v
rel
σ = π r
2
n ≈ 10
10
− 10
15
r ≈ 10
−6
− 10
−3
v
rel
≈ 1 − 10m /s
t
c
≈
1
(0.32 × 10
12
m
−3
)(π (10
−6
m)
2
(1m /s)
= 1secon d
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
of 12 15
The Big Bang In other words, if the planetary system inherits a characteristic time of 1 second from the
dust grains in the protoplanetary disc, and the dust grains inherit a characteristic time of 1 second from the
elements, and elements inherit the characteristic time of 1 second from the proton, then we would guess
that the proton inherits the characteristic time of 1 second from its origins in the Big Bang that gave birth
to the Universe. And, indeed it does:
Big Bang
(time=1second)
Around one second after the Big Bang, the universe had cooled enough that neutrinos decoupled, and
protons and neutrons were forming in equilibrium, this is the moment when baryons (protons/neutrons)
become stable, linking the 1-second time unit to matter itself.
(time=1-3 minutes)
At this time the first atomic nuclei (H, He, Li) form.
Thus the 1-second time unit marks when the proton’s number became fixed in the Universe, given by our
equations
which give the radius of a proton when set equal to one another
Or,
Where we say
Where
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
r
p
=
2
3
⋅
h
cm
p
r
p
= ϕ
h
cm
p
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
of 13 15
is a little over a second, and
is a little under a second. These equations contain the fundamental constants related to gravity (G),
quantum mechanics (h), relativity (c), and electromagnetism ( ), constants that cover the fundamental
forces that shaped the early Universe. G, h, c, and control the rate of interaction in the early universe,
including weak interactions, and gravity, which govern proton stability and neutrino decoupling. Thus,
our equations describe the fundamental physics at t=1 second, confirming that this time is deeply
embedded in the structure of the Universe. This means the 1-second characteristic time was imprinted at
the birth of the Universe then inherited by the proton through fundamental constants, planetary systems
(through particles interactions in the protoplanetary discs), galaxies and cosmic evolution because protons
make up most of the universe’s baryonic matter. My equations link proton properties to 1-second, and
protons were fixed in the Universe at 1 second, meaning we could be seeing a universal clock that has
influenced everything since the Big Bang.
How is it figured that the protons and neutrons stopped converting into one another and their numbers in
the Universe became set? The idea is that neutrino decoupling (neutrinos stop interacting with one
another) happens when the reaction rate of weak interactions falls below the Hubble parameter the
expansion rate of the Universe . The reaction rate per particles is given by
is the Fermi constant is about , and is the temperature of the Universe. The
expansion rate of the Universe is given by
Where is the Plank mass is about 1.22E19GeV. and have units of inverse time ( ). Neutrino
decoupling happens when
This happens when the temperature of the universe is which occurs at 1 second after the Big
Bang. We know this because temperature evolves with time as
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
α
α
Γ
H
Γ ≈ G
2
F
T
5
G
F
1.166E − 5G eV
−2
T
H ≈
T
2
M
Pl
M
Pl
Γ
H
s
−1
G
2
F
T
5
=
T
2
M
Pl
T
decoupling
= (G
2
F
M
Pl
)
−1/3
T = 1MeV
T ∝ t
−1/2
of 14 15
Meaning that the universe had cooled to 1MeV after 1 second. Before decoupling, the universe was so
dense that protons and neutrons were constantly interconverting but because weak interaction stops
maintaining neutron-proton equilibrium at 1 second, the proton to neutron ratio freezes in. Neutrons are
unstable and decay with a half life of about 10 minutes, however when a few minutes after the bang they
start forming, with protons, helium-4 nuclei, they become stable. The protons don’t decay rapidly so you
end up with, after freezing, 6 times more protons than neutrons in the Universe, this explains why 25% of
mass of the universe is helium. It is about 75% hydrogen.
The expansion rate of the Universe is governed by the Friedmann equation
Where is the energy density of the Universe. It is
The Hubble expansion rate is
Since
we have
We said protons and neutrons are set in the universe when it has cooled in its expansion to 1MeV. We
have
This was done in Planck units where time can be expressed in inverse energy. Since in Planck units
we have
H
2
=
8π G
3
ρ
ρ
ρ ∝ T
4
H ∝
T
2
M
Pl
M
Pl
≈ 2.4E18G eV
t ∝
1
H
t ∝
M
Pl
T
2
t ∝
2.4E18G eV
(1E − 3G eV )
2
= 2.4E 24 G eV
−1
1G eV
−1
= 5.39E − 25s
t ≈ (2.4E 24)(5.39E − 25)
t ≈ 1.3secon d s
of 15 15
The Author
