A Proposal For A Universal Particle Equation
Ian Beardsley
March 10 - March 30, 2026
ABSTRACT
A Normal force where , which we suggest is Lorentz invariant, is
introduced that determines a master equation for the proton, neutron, and electron, a kind of
Universal Particle Equation. It is suggested that when we push on a particle we rotate some of its
temporal velocity into spacial velocity and resistance to this rotation is experienced as the normal
force pushing back creating inertia that we experience as mass.
The Proposal
We can form two equations where the proton radius to its mass produces about 1-second [3, 4]:
1.
(Proton Mass)
(Proton Radius)
(Planck Constant)
(Light Speed)
(Gravitational Constant)
1/137 (Fine Structure Constant)
Using equation 6, equations 1 and 2 directly yield [3, 4]:
where . Here we see in equation 3, the cross-sectional area of the proton
is exposed to the normal force, mediated by the 'stiffness of space' as measured by ,
producing the proton mass, .
We suggest equations 1 and 2 are correct because they yield the proton radius closely. They give
it as (by equating them) [3, 4]:
F
n
= h /(ct
2
1
)
t
1
= 1 second
ϕ ⋅
π r
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1 second
2.
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1 second
m
p
: 1.67262E − 27kg
r
p
: 0.833E − 15m
h : 6.62607E − 34J ⋅ s
c : 299,792,458m /s
α :
3. m
p
= κ
p
⋅
π r
2
p
F
n
G
4. F
n
=
h
ct
2
1
5. t
1
= 1 second
κ
p
= 1/(3α
2
)
A
p
= π r
2
p
F
n
G
m
p
The CODATA value from the PRad experiment in 2019 gives
With lower bound , which is almost exactly what we got.
We can see equation 6 may be the case because we get it from Planck Energy ,
Einsteinian energy, , and the Compton wavelength when we
introduce the factor of , which is the golden ratio conjugate, where the golden ratio,
. [3, 4].
I explain this factor by invoking Kristin Tynski, her paper titled: One Equation, ~200 Mysteries:
A Structural Constraint That May Explain (Almost) Everything [2].
Tynski shows that for any system requiring consistency across multiple scales of observation has
the recurrence relation:
Which she says leads to
Whose solution is .
For the proton radius in our computations we will use
"A measurement of the atomic hydrogen Lamb shift and the proton charge radius"
by Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019).
It has a value of
The 1-second verification follows from the resulting Universal Particle Equation [3]:
Proton: , :
Neutron: :
6. r
p
= ϕ ⋅
h
cm
p
r
p
= (0.618) ⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.8166E − 15m
r
p
= 0.831f m
±
0.014f m
r
p
= 0.817E − 15m
E
p
= hν
p
E
p
= m
p
c
2
λ
p
= h /(m
p
c) = r
p
ϕ
Φ = 1/ϕ = ( 5 + 1)/2 ≈ 1.618
7. scale(n + 2) = scale(n + 1) + scale(n)
8. λ
2
= λ + 1
Φ
r
p
= 0.833f m
±
0.012f m
9. t
1
=
r
i
m
i
⋅
πh
Gc
⋅ κ
i
κ
p
=
1
3α
2
α = 1/137
t
1
=
0.833 × 10
−15
1.67262 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00500 seconds
κ
n
=
1
3α
2
t
1
=
0.834 × 10
−15
1.675 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00478 seconds
Electron: :
We suggest for the electron may be because it is the fundamental quanta (does not consist
of further more elementary particles).
. (Neutron radius)
. (Classical electron radius)
Equation 9 is a Natural Law. In general we can write the universal law as [4]:
Discussion
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 may be these different sizes are predicted by Fibonacci approximations to [4]. If
such an approximation is given by it could be that the proton radius is as large as
Which it was nearly measured to be before 2010 in two separate experiments. One using
hydrogen spectroscopy, the other electron scattering. In 2010 The recommended CODATA value
was . Then came the shocking 2010 measurement that was 4.2%
smaller using the new Muonic hydrogen result, which was . This resulted in the
famous “proton radius puzzle”.
We might suggest that the proton radius might get still smaller, closer to something using the
Fibonacci approximation of . In which case we would have:
κ
e
= 1
t
1
=
2.81794 × 10
−15
9.10938 × 10
−31
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 1 = 0.99773 seconds
κ
e
= 1
r
n
= 0.84E − 15m
r
e
= 2.81794E − 15m
10. m
i
= κ
i
⋅
π r
2
i
F
n
G
F
n
=
h
ct
2
1
F
n
=
6.62607015 × 10
−34
J·s
(299,792,458 m/s)(1 s)
2
= 2.21022 × 10
−42
N
t
1
= 1 second
ϕ
ϕ ≈ 2/3
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
= 0.8775f m
±
0.0051f m
r
p
= 0.84184f m
ϕ ≈ 5/8
In such cases, in equation 1 takes on different ratios between successive Fibonacci ratios.
The normal force has a relationship to the Planck force, the maximum gravity for the minimum
mass. It links the normal force to a full rotation ( ). We have the normal force
We have the Planck force for gravity
Where, is the Planck mass, and is the Planck length. They are given by:
And, Planck time is:
We form the ratios between the normal force and Planck force:
Divide by Planck time squared and we have:
r
p
=
5
8
h
cm
p
r
p
=
5
8
(6.62607 × 10
−34
)
(299,792,458)(1.67262 × 10
−27
)
= 0.8258821 × 10
−15
m
ϕ
2π
F
n
=
h
ct
2
1
= 2.21022E − 42N
F
Planck
= G
m
2
P
l
2
P
= (6.674E − 11)
(2.176434E − 8kg)
2
(1.616255E − 35m)
2
= 1.21020E44N
m
P
l
P
m
Planck
=
ℏc
G
= 2.176434E − 8kg
l
Planck
=
ℏG
c
3
= 1.616255E − 35m
t
Planck
=
ℏG
c
5
= 5.391247E − 44s
F
n
F
Planck
= 1.826326E − 86
That number is . We have the final equation:
From the Planck units we have:
So, it can be written:
We can write
is a full rotation, so we can define an angular frequency, :
Integrating one more time gives the angle over 1-second:
F
n
F
Planck
1
t
2
P
= 6.2834743s
−2
2π
t
1
= 2π
F
Planck
F
n
⋅ t
P
= 1.00seconds
F
Planck
= G
m
2
P
l
2
P
=
c
4
G
t
1
= 2π
c
4
GF
n
⋅ t
P
F
n
= 2πF
Planck
⋅
t
2
P
t
2
1
2π
ω
F
n
= F
Planck
⋅ t
2
P
⋅
dω
dt
F
n
F
Planck
⋅
1
t
2
P
∫
1second
0
dt = ω
1
ω
1
=
2π
secon d
F
n
F
Planck
⋅
t
1
t
2
p
∫
1second
0
dt = θ
1
Thus, the normal force is the force that, when scaled by the Planck force and the Planck time,
gives a full angular displacement in one second. This geometric origin explains why
appears as a natural invariant. We see the second arises naturally from Planck-
scale physics through a factor of .
It might make sense to say: One second is the time it takes for the ratio to accumulate a
full of angular phase, closing a loop in the temporal dimension – out of the temporal and
back in again.
This is reminiscent of the idea in some quantum gravity or pre-geometric models that time
emerges from a cyclic variable. The equation may be hinting at exactly that: the normal force
(which was previously linked to inertia and mass) is the “restoring force” that makes the cycle
close after exactly one second.
References
[1]Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019) "A measurement of the atomic
hydrogen Lamb shift and the proton charge radius"
[2]Tynski, K. (2024). One Equation, ~200 Mysteries: A Structural Constraint That May Explain
(Almost) Everything.
[3]Beardsley, I. (2026) How Physics and Archaeology Point to a Natural Constant of 1-Second,
https://doi.org/10.5281/zenodo.18829259
[4]Beardsley, I. (2026) The Sublime and Mysterious Place of Humans in the Cosmos; A Work in
Exoarchaeolog, https://doi.org/10.5281/zenodo.18715148
F
n
F
Planck
⋅
t
2
1
t
2
P
= θ
1
θ
1
= 2π
F
n
2π
t
1
= 1 second
θ
1
= 2π
F
n
F
Planck
2π