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Analysis of the Fundamental Forces
Nonya1 ,
¤the internet :P
* E-mail: Corresponding ...

Abstract
The momentum of a bound electron gives rise to an emergent field of force that
restrains the position of the electron. A change in electron position is compensated for
by a loss in time for the relativistic component of the electrons standing wave. The
electron is pinned at a discontinuity, 2Rp that arises from the localized gravitational
flux of the electrons matter wave. These points are used to demonstrate a classical
origin for the quantization of momentum.

Author Summary
This paper explores the relation of gravitational and electromagnetic properties of an
electron to demonstrate how inertia compounds momentum into DeBorgie waves that
conserve gravitational energy within quantum systems. Configuration space has been
used for decades to factor forces out of the description of matter waves. This has led to
a general disconnect between the forces and the medium they act through. Particle
wave packets are treated as mathmatical functions which allow for computational
operations to describe observed effects in particles but does not address the means by
which these observed effects arise. Here I will describe the local pinning of fields of force
within the atomic structure of matter to explain the quantum phase transition that
occurs at impedance matched boundaries between emergent fields of gravitational force
and fundamental fields of electric force. Furthering the exploration of restraining forces,
I show that the thermal temperature of the quantum phase transition is proportional to
the inertial flux of the gravity in the local field of force through which the transition
occurs.

Emergent Fields of Force
The constant acceleration of an electron towards the nucleon of an atom produces a
frame dragging effect which is the induced inertial flux in the gravitational field of the
electron. The flux is expressed as the pinning of the electron to the nucleon. The
momentum of the collapsing electron gives rise to the displacement of the electron from
the nucleon. Special relativity provides the requisite understanding of inertial reference
frames to formulate the description of this phenomenon. To fully describe the
microscopic behavior of the pinned electron, the isotropic capacitance of the electron
and the capacitance of the visible universe must be used to calculate for the rest mass of
the electron. With this solution the inertial reference frame can be used to create a
relativistic description of the restraining forces that emerge to create the probabilistic
nature of the electron state. From the capacitance used to calculate the electron mass,
the harmonic motion of the electron bound to the nucleon can be formulated.

.

1/12

The relativistic energy of the electron is:
E = me c2

(1)

1
CV 2
2

(2)

The energy of a capacitor is:
E=
The charge of a capacitor is:
Q = CV

(3)

Substituting voltage for the quotient of charge and capacitance, the rest energy is:
Q2
2C

(4)

Q2
2me c2

(5)

me c2 =
Solving for capacitance yields:
C=

C = 1.5677e − 25F arads

(6)

The isotropic capacitance of a sphere with the radius of an electron bound to a
proton is:
C = 4π o re2

(7)

C = 1.5677e − 25F arads

(8)

The force between the charged surfaces of a capacitor is defined as:
Q2
4π o A

F =

(9)

An electromagnetic wave has two modes that are 90 degrees out of phase with one
another. The area of the wave is the square of the wavelength. Solving for force of an
electron:
F =

Q2
, λ = Re
4π o λ2

F = 29.05351N

(10)
(11)

Gravitational induction is defined by General Relativity as:
dp
g
×
2
c r
dt
The gravitational field of matter is defined as:
F ield =

(12)

GM
r2

(13)

dp
r
×
c2
dt

(14)

F ield =
Solving for mass:
M=

.

2/12

Solving for the mass of the electron using the force applied through an electron
bound to a proton at a displacement of twice the radius of the proton yields the
gravitational mass of the electron:
2rp
× 29.05351N = 9.1e − 31kg
c2
The electric force is 2.27e39 times stronger than the gravitational force.
Fg = G

M1 × M2
q1 × q2
, Fc = Ke
r2
r2
Fc
= 2.27e39
Fg

(15)

(16)
(17)

The isotropic capacitance of a sphere with the radius of the visible universe is:
Cu = 4π o rou , rou = 1.26e26M eters

(18)

From the capacitance of the observable universe the capacitance of a sphere with the
radius of an electron bound to a proton is calculated using the coupling constant
between the gravitational and electric forces. A coeficiant of 4π 2 arises as a result of the
doubly centric nature of the hypersphere:
Cu
= 1.5677e − 25F arads
(19)
4π 2 2.27e39
This equation is set equal to equation 5 where the same capacitance was arrived at
from the charge and mass of the electron.
4π o rou
Q2
=
2
4π 2.27e39
2me c2

(20)

Solving for mass:
me =

Q2 π2.27e39
2 o rou c2

(21)

Substitute the mass equation calculated using the inward force on the electron
bound to a nucleon yields:
2rp
dp
Q2 π2.27e39 dp
×
=
,
= 29.05351N ewton
2
c
dt
2 o rou c2
dt

(22)

Solving for force as the derivative of momentum with respect to time yields:
dp
Q2 π2.27e39 dp
=
,
= 29.05351N ewton
dt
4 o rou rp
dt

(23)

Substituting the force equation for the solution Coloumbs law using one unit
electron charge at a displacement equal to the radius of the proton:
e2
Q2 π2.27e39
=
2
4π o re
4 o rou rp

(24)

Solving for the quare of the radius:
rou rp
(25)
2.27e39π 2
The coupling coefficient between the gravitational and electric forces is described as
a ratio between the extent of the visible universe and the proton:
re2 =

.

3/12

rou
= 2.27e39
4π 2 rp

(26)

Substituting into the square electron radius equation:
re = 2rp

(27)

Thus the radius of the electron is defined by the impedance matched boundary
between the isotropic capacitance of the edge of the visible universe and the isotropic
capacitance of the edge of a proton. The electron is pinned in place by the emergent
restraining forces of the gravitational field. This is caused by the inertial reference
frame of the collapsing electron bound to a proton.
The potential and kinetic components of the restrained wave are displaced by 90
degrees. The wave is localized by the restraining forces of the capacitive boundary. The
energy of a wave is described as the product of momentum and it’s constant phase
velocity. In the case of the electron the phase velocity is that of light.
E = pc

(28)

The relativistic momentum of a traveling electric wave is:
p=

Mv
(1 − v 2 /c2 )1/2

(29)

Soliving for energy:
Ex =

M vc

(30)

2
1/2
( 1−v
c2 )

Within this model the traveling energy is the displacement along the complex plane.
The energy of the standing wave is:
Ey = M c2

(31)

The rest energy is the displacement along the real plane. The vector sum of the
traveling wave (Ex) and the standing wave (Ey) is the relativistic energy (Er) of moving
matter.
Er2 = Ex2 + Ey2
Er2 = (M c2 )2 + (
Er =

M vc
2
1/2
( 1−v
c2 )

M c2
2
1/2
( 1−v
c2 )

(32)
)2

(33)

(34)

The relativistic energy is the hypotenuse between the rest and traveling energies in
their respective planes.
The ratio of the standing to wave energy to the relativistic energy of the wave is:
Ey
1 − v2
= ( 2 )1/2
Er
c

(35)

The arc sin of the ratio is the phase of the wave:
φ = arcsin(

.

1 − v 2 1/2
)
c2

(36)

4/12

The phase angle represents the separation of the kinetic and potential energy of
matter. A loss in time by the relativistic component of the wave is compensated for by
a gain in position.
The ratio of the traveling wave to the relativistic wave reduces to v/c this is the
ratio of flowing energy to the total energy of the wave.
i

M c2
1

2

]2
[ 1−v
c2

φ

M c2

M vc
2

1

]2
[ 1−v
c2

Flux in electromagnetism is described by the permeability and permittivity of a
closed surface using the energy integral with respect to the derivative of time. Gravity
is presumed to possess the same properties of flux as electromagnetism. Within
electromagnetism permittivity and permeability have a c12 relationship:
1
(37)
c2
Gravity is taken to possess a similar relationship of two undefined constants.
However, as technology does not yet provide a means of directly measuring gravitational
permittivity or permeability of free space the constants will remain undefined for the
purpose of this supposition, the relationship is assumed and calculations will be made
with the assumption since the values of the undefined constants are reduced to their
ratio and factored out of the calculation before a measured quantity would impede the
utility of the equations being defined.
µ =

µg g =

1
c2

Flux in electromagnetism takes the form:
I
d
F = µ
E · ds
dt A

(38)

(39)

Restated in terms of gravity by substituting mass for the potential energy:
d(Gm)
dt
Substituting the assumed relation of g and µg c12 for µg g :
Fg = µg g

Fg =

G dm
c2 dt

(40)

(41)

Current is defined as:
I = qV /l

(42)

Solving for gravitational current:

.

5/12

dm mv
·
dt
l
Momentum is defined as the product of mass and velocity:
Ig =

(43)

p = mv

(44)

Substituting momentum into current:
dm p
·
dt l
Solving for flux in terms of momentum:
Ig =

(45)

G p
·
(46)
c2 l
Taking the derivative of momentum with respect to time yields the gravitational
acceleration where l is the length of the moving mass:
Fg =

G dp 1
·
·
(47)
c2 dt l
The solution arrived at is the acceleration of the induced inertial gravitational field
that opposes the acceleration of the mass.
Let us now consider the case of a box containing a particle, presume that each
impact of the particle on the inner walls of the box produces a gravitational field as
described in equation 47. The gravitational field induced varies as 1/r. We will
calculate the gravitational field for a point x at some displacement r from the box
containing the particle. The particle impacts one wall, B and approaches another wall
A, the walls are displaced by some distance L.
ag =

A

B

x
r

L
The gravitational field at point x is the vector sum of the fields produced by the
impact of the particle on walls A and B.
ag at x = A f ield − B f ield
ag at x =

G
dp
G dp
− 2
c2 (r + L) dt
c r dt

(48)
(49)

Simplifiy
ag at x = −

G dp
L
[ 2
]
2
c dt r + rL

(50)

Take the limit as r approaches L

.

6/12

G dp
L
[ 2
]
2
r→L
c dt r + rL
The result is the gravitational far field of the particle:
ag at x = lim −

(51)

G dp L
(52)
[ ]
c2 dt r2
The field varies inversely with with the square of displacement to the point of
measure, x.
The momentum of a field that propogates with a phase velocity equal to that of light
is given as:
ag at x = −

E
c
where E is energy and c is the phase velocity of light, thake the derivative of
momentum with respect to time.
p=

dp
=
dt

2E
c
2L
c

=

E
L

(53)

(54)

Substituting into equation 52:
GE
c2 r2
Substituting in Einstein’s relationship between matter and energy:
ag at x = −

(55)

GM
(56)
r2
Equation 56 is Newtons gravitational equation. Forces arise from the restraining of
energy, these forces induce the gravitational field of matter.
ag at x = −

Energy Quantization
Now that we have established the gravitational field of matter with the forces that arise
from restraining energy we can begin to analyze the structure of the atom to
understand the origin of energy quantization from the perspective of momentum pinned
at impedance matched boundaries within the nuclear structure.
Hooke’s Law defines force as the product of a constant and a displacement:
f = kx

(57)

Using the inward force, calculated in equation 11, through the surface of a nucleon
the spring constant is:
29.05351N
(58)
x
The spring constant for an electron bound to a proton is calculated with using a
displacement of twice the radius of the proton:
k=

k=

29.05351N
2Rp

(59)

The frequency of a harmonic oscillator is:

.

7/12

r
1
k
f=
(60)
2π m
Substituting the sprint constant of a bound electron and the mass of a bound
electron the frequency results in:
s
29.05351N
1
(61)
f=

2rp me
This frequency converted to a wavelength is the Compton wavelength of the electron
λ=

c
1


q

29.05351N
2rp me

(62)

The velocity of the harmonic oscillator is calculated as the product of the frequency
and the displacement of the oscillator.
r
k
(63)
v=x
m
Substituting from equation 61 into 63 with a displacement equal to the median
displacement of the nucleon we arive at:
s
29.05351N
v = Rn
(64)
2rp me
The photo electric effect states that energy is the product of Plancks constant and a
frequency:
E = hf

(65)

Capacitance is defined as the product of the permittivity of free space and the
quotient of area to distance:
o A
(66)
D
In the case of the electron or a photon the area is the square of the wavelength. As
the electromagnetic field has two modes that are 90 degrees out of phase with one
another the area is twice the square of the wavelength. The displacement is the
wavelength. Thus the energy of an electromagnetic wave is:
C=

C = 2 o λ

(67)

The energy of a capacitor is:
1
1
× Q2 ×
(68)
2
C
Substituting the capacitance of an electromagnetic wave into the energy equation:
E=

E=

1 Q2
2 2 o λ

(69)

The wavelength is proportional to the frequency of the wave:
λ=

.

v
f

(70)

8/12

Substituting for wavelength in the energy equation yields:
E=[

1 Q2
]f
2 2 o v

(71)

The photo electric effect is the product of Plancks constant and a frequency. The
terms in the brackets above are equal to Plancks constant.
h=

Q2
4 o v

(72)

The velocity of the harmonic motion solved for earlier is now substituted for the
velocity in Plancks constant to produce a classical description of the constant:
h=
4 o Rn

Q2
q

29.05351N
2rp me

(73)

The velocity is equal to to the half the product of the phase velocity of light and the
fine structure constant:
v=

ca
2

(74)

Analysis
Constant Acceleration of Electrons
Hawkings and Unruh introduced thermal equations in the 70’s that took the same form
as one another [1], [2]. Hawkings was describing the radiation of thermal energy at the
boundary of a black hole[3]. Unruh was describing the thermal energy observed by an
accelerating reference frame[4]. According to the Unruh effect an accelerating observer
will experience space as having a temperature proportional to its acceleration.
T =

~a
~c3
,T =
4πckb
8πkb GM

(75)

Modeling the Unruh equation as a standing wave with inward force providing
constant acceleration that is diametrically opposed by the gravitational force of the
inertial frame the equation can be directly related to a black hole with the mass of an
electron. The equation is solved using a classical acceleration calculated from the force
of an elementary charge displaced by the classical radius of an electron, a temperature is
R
produced that is 2Rpu times the Hawking temperature for a black hole with the mass of
an electron.
The inertial frame is derived from the Lorentz forces of a standing wave collapsing
on its self at a constant acceleration. The inertial mass of an electron is considered
when performing the calculation. A constant acceleration is found by considering the
electrical properties of the electrons standing wave when bound to a nucleon. The
electron has a inward force that can be applied through it’s surface before it is displaced
to the boundary of the nucleon and a phase transition arises. This force is calculated
from the constant electric charge of one unit electron displaced by the radius of the
electron. Using Coulumbs law the force is calculated as:
F =

.

e2
4π o re2

(76)

9/12


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