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This paper completes the analysis of the 1/r (Coulomb) and a/r2 (nuclear) potential functions through a consequent analysis of the mass-energy equivalence using orbital energy formulas. The a/r2 function (modified Yukawa function with a as a nuclear scaling parameter) yields the desired crossing of potentials and the potential well one would expect to find. We also suggest a wave equation for the deuteron nucleus (p + e + p) based on a geometric interpretation of Schrödinger’s wave equation for the hydrogen atom (zero-spin model). Finally, we offer some thoughts on low-energy nuclear reactions (LENR or anomalous heat reactions) and the elasticity of space. The model has the advantage of (i) not introducing new fundamental constants or charges (the nuclear charge gN is identical with the electric charge), (ii) respecting relativity theory (no superluminal speeds), and (iii) confirming Planck’s quantum of action as the fundamental unit of physical action (h) and angular momentum (ħ) in particle-field exchanges of energy and momentum.
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The nuclear force and the neutron hypothesis
Jean Louis Van Belle, Drs, MAEc, BAEc, BPhil
1 February 2021
Abstract
This paper completes the analysis of the 1/r (Coulomb) and a/r2 (nuclear) potential functions through a
consequent analysis of the mass-energy equivalence using orbital energy formulas. The a/r2 function
(modified Yukawa function with a as a nuclear scaling parameter) yields the desired crossing of
potentials and the potential well one would expect to find.
We also suggest a wave equation for the deuteron nucleus (p + e + p) based on a geometric
interpretation of Schrödinger’s wave equation for the hydrogen atom (zero-spin model). Finally, we
offer some thoughts on low-energy nuclear reactions (LENR or anomalous heat reactions) and the
elasticity of space.
The model has the advantage of (i) not introducing new fundamental constants or charges (the nuclear
charge gN is identical with the electric charge), (ii) respecting relativity theory (no superluminal speeds),
and (iii) confirming Planck’s quantum of action as the fundamental unit of physical action (h) and angular
momentum (ħ) in particle-field exchanges of energy and momentum.
Table of Contents
Introduction .................................................................................................................................................. 1
The nuclear dipole and the Zitterbewegung (ring current) hypothesis ........................................................ 2
Binding energies, electromagnetic mass, and charge radii .......................................................................... 5
The neutron oscillation and elliptical orbitals .............................................................................................. 7
Analogy between the gravitational radial field and the Coulomb field ........................................................ 9
Orbital geometries ...................................................................................................................................... 10
The nuclear force and the range parameter a ............................................................................................ 12
Non-symmetric potentials and other considerations ................................................................................. 15
Force calculations and the concept of nuclear mass .................................................................................. 16
A wave equation for the deuteron nucleus? .............................................................................................. 18
Are low-energy nuclear reactions possible? ............................................................................................... 20
The E/m = c2 relation and the elasticity of space ........................................................................................ 21
1
The nuclear force and the neutron hypothesis
Introduction
In our previous paper
1
, we analyzed the nuclear force between the neutron and the proton inside of the
deuteron nucleus (D = n + p) as an electrostatic attractive force between an electric dipole field arising
from the (neutral) neutron dipole (n = p + e) and the (positive) electric charge of the proton. Let us think
of the dipole concept first, before we go on to think about the neutron. Figure 1 shows two opposite
charges and the equipotential and field lines of the dipole field they create.
Figure 1: Equipotential (V, solid lines) and electric field (E, dashed lines) in a dipole field
2
We have zero potential along the midperpendicular (line segment bisector) between the two charges,
which implies no work is done when bringing a charge from infinity to the midpoint along this line. Other
trajectories from infinity to the zero potential line will also involve no net work but consist of positive
work nullified by negative work. Such trajectories should be analyzed in the context of the minimum or
least action principle, which tells us a charge will follow a trajectory which lowers its total energy (kinetic
and potential) by moving along a path which minimizes (physical) action, which we may write as
3
:

The negative charge (electron) will, therefore, go and sit right on top of the positive charge (proton).
This problem (for the model, that is) may also be stated as the corollary: what keeps the positive and
negative charges separatenot approximately but exactly?
1
The electromagnetic deuteron model, December 2020.
2
The illustration was taken from a commercial sales site for fiber optic equipment.
3
For a full development of the least action principle both from a classical as well as a quantum-mechanical
perspective we refer to Feynman’s Lectures, Volume II, Chapter 19 (The Least Action Principle).
2
The nuclear dipole and the Zitterbewegung (ring current) hypothesis
The solution to the puzzle described above is the Planck-Einstein relation: the charges orbit around each
other (rather like a moon around a much larger planet in light of the mass of the electron and proton
respectively), and the (non-zero) energy and physical action in this oscillation explain why the charges
do not sit still.
4
Following additional remarks may be made:
The positive and negative charge cannot be a matter-antimatter pair because otherwise they
would annihilate each other.
5
The two charges are opposite and the current of their orbital motion may, therefore, be
compared to the like currents in two wires, which will repel each other.
6
Alternatively, we may
also think of two spinning charges with a magnetic moment that separates them magnetically.
7
Planck’s quantum of action (ħ) gives us the angular frequency and the radius of the oscillation
(we will calculate this in a minute), and we may assume the oscillation packs one unit (ħ) of
physical action.
8
We can now use a ‘mass without mass’ model or, to be more precise, an electromagnetic mass model
to calculate the radius (or radii) for these oscillations. The model is summarized below:





The mass factor in the a = ħ/mc equation is, obviously, the (equivalent) orbital energy. Let us first
calculate a radius for the n = p + e model. What is the orbital energy? It is the (positive
9
) energy
difference between the neutron (939.565 MeV) and the proton (938.272 MeV): about 1.3 MeV, which is
about 2.5 times the energy of a free electron. We, therefore, get the following charge radius
10
:



 
4
The reasoning here is pretty much the same as that in Feynman’s explanation of the size of an atom.
5
At first sight, matter and antimatter differ only by spin. See my paper on issues and gaps in the ring current model
of elementary particles.
6
A positive current in one direction is equivalent to a negative current in the other direction.
7
However, we will want to analyze things in terms of zero-spin particles. The comparison just shows we can swap
reference frames but physical reality remains the same.
8
The physical dimension of Planck’s quantum of action (Wirkung in German) is force times distance times time: [h]
= Nms.
9
This explains why a free neutron is not stable: the energy is lowered when its two constituents fall apart.
10
The ħc factor and its dimension can easily be verified from using the 6.5821016 eV·s value for ħ and the
299,792,458 m/s value for c and, of course, we should not forget to convert m into fm (1015 m):


3
This is an enormous value: almost 200 times the actually measured neutron radius (0.8 fm). Our
calculation can, therefore, not possibly be correctwhich is why we do not believe the electron and the
proton inside of a neutron are held together by the electromagnetic force: we must assume some
nuclear force.
We can repeat the calculation for the deuteron nucleus. The energy difference between the deuteron
nucleus (about 1875.613 MeV) and its two constituents (neutron and proton) in their unbound state
(939.565 MeV + 938.272 MeV = 1,877.837 MeV) is negative and equal to about 2.22 MeV. We can,
therefore, also think they must be in an orbital an energy-lowering orbital, this time
11
with an energy
of the order of 2.22 MeV per cycle. This value of 2.22 MeV is the sum of the potential and kinetic energy
in the oscillation, and the oscillation must also pack one unit of ħ. As shown in the equations for the
electromagnetic mass model, the assumption is that the two particles move around at the speed of
light, and that we may, therefore, write its equivalent mass as E/c2. The calculation yields this
12
:



 
This value is about 40 times the deuteron charge radius (2.128 fmwhich is the usual range parameter
a that is used in the Yukawa potential function
13
). This 88.7 fm value is, therefore, another rather
impossible value, which is why Yukawa and others before him
14
came to the conclusion that one
should invent some new force here: a nuclear force. At the same time, we do want to keep thinking of
the neutron-proton combination as an electric dipolea polar structure, or an electron blanket
oscillating back and forth between the two protons. We should, therefore, not necessarily invent a new
charge.
15
We must, however, quickly add another remark here:
The assumption of a nuclear force (separate from the electromagnetic force, that is) is quite convenient
because it offers an explanation for the neutrino. Indeed, if there is a separate nuclear force, then we
may think of the neutrino as its photon, so to speak: a photon carries electromagnetic energy and,
likewise, a neutrino would then carry nuclear energy.
If so, we should think of it as a zbw oscillation moving at lightspeed.
16
How does this work? We have the
mass factor in the denominator of the formula for the Compton or zbw (Zitterbewegung) radius, and
11
The deuteron nucleus is, therefore, stable (as opposed to the free neutron).
12
We equate this to the radius of oscillation but, of course, the usual caveats apply: this will be an average only
with elliptical orbitals, and the radius of interference (e.g. when illuminating the particle with photons) may be
larger because of the electromagnetic field surrounding the loop.
13
See: given in Ian J.R. Aitchison and Anthony J.G. Hey, Gauge Theories in Particle Physics (2013), section 1.3.2 (the
Yukawa theory of force as virtual quantum exchange).
14
Schrödinger suggested a Platzwechsel (change of place) model for the deuteron nucleus, according to which the
proton and the neutron continually swap positions (see the history section in the Wikipedia article on the Yukawa
potential). We would rather think of two positive charges moving around in an electron cloud.
15
It is instructive to do so, however. We suggested the dirac, and abbreviated it as Y so as to honor both Dirac as
well as Yukawa.
16
Interestingly, the latest hypothesis (according to Wikipedia) with regard to neutrinos is that they would have
some non-zero rest mass and, therefore, do never quite attain the speed of light. However, it mentions that
4
this mass must increase as the mass of our particle increases with speed. Conversely, the mass factor is
present in the numerator of the zbw frequency, and this frequency must, therefore, also increase with
velocity: we have a simple (inverse) proportionality relation here. The idea is visualized in the illustration
below
17
: the radius of the circulatory motion must effectively diminish as a linear component gets added
to the tangential component of the velocity of the pointlike zbw charge.
Figure 2: The Zitterbewegung radius must decrease with increasing velocity
The Zitterbewegung (zbw) model of an (elementary/stable) particle is that all of the energy is in the
oscillation of the charge (which itself has no rest mass whatsoever
18
). Needless to say, orbitals can be
circular or elliptical with v and c as tangential velocity vectors: E = mc2 and, unlike what the
illustration (Figure 2) suggests, the plane of oscillation need not be perpendicular to the direction of
propagation of the particle as a whole.
19
When the lateral component of the velocity goes to c, the
circumference of the oscillation must turn into a linear wavelength in the process, and we have a
existing measurements for MeV to GeV neutrinos provided upper limits for deviations of approximately 109, or a
few parts per billion and that, within the margin of error, this is consistent with no deviation at all. Hence, the
question may not have been solved yet.
17
We thank Prof. Dr. Giorgio Vassallo and his publisher to let us re-use this diagram. It originally appeared in an
article by Francesco Celani, Giorgio Vassallo and Antonino Di Tommaso (Maxwell’s equations and Occam’s Razor,
November 2017).
18
It acquires its effective (relativistic) mass from its velocity.
19
The reader may want to imagine that the plane of oscillation rotates or oscillates itself. He should not think of it
of being static unless we think of the charge moving in a magnetic field, in which case we should probably think
of the plane of oscillation as being parallel to the direction of propagation. Of course, the tangential velocity v will
be equal to c for circular orbits: when considering elliptical orbitals, lightspeed is reached only when the charge
passes the center (zero potential energy: all energy is kinetic). The reader should also note that, when treating
both rest as well as relativistic mass (m0 and m = m0) as electrostatic mass only, the Lorentz factor () can be
expanded to yield the following series expansion of the total energy:


Dimensions are easily checked when not confusing the mass symbol m (expressed in kg or N·m/s2 units) with the
distance unit (meter):




5
photon-like particle!
20
This rather remarkable geometric property relates our zbw electron model with
our photon model, which we will not talk about here, however.
21
All sounds nuts. Yesagreed, but is there an alternative way of thinking? We do not see one. Let us try
to organize our thoughts.
Binding energies, electromagnetic mass, and charge radii
Of course, we do not think of the neutron as a linear structure. In fact, the charge radius of a free
neutron (Rc) is assumed to be zero: it only has a magnetic radius (Rm), whose rms value is about 0.8 fm.
Measuring the magnetic radius of a free neutron is difficult because a neutron is stable only inside of a
nucleus. Indeed, we already mentioned how the instability of the neutron might be reflected in a
positive energy difference ( 1.3 MeV) between the neutron ( 939.565 MeV) and the proton ( 938.272
MeV). This energy range (1.3 MeV) is about 2.5 times the energy of a free electron but is only a tiny
fraction of the energy of a π meson (139.57 MeV) or the muon-electron (105.65 MeV), which is why we
find the reference to a π cloud in the illustration below (Figure 3) rather misplaced.
Figure 3: The neutron model and the concepts of charge and magnetic radius (Rc and Rm)
22
It is probably good to mention some more assumptions of our zbw model (of the neutron as well as of
the deuteron nucleus):
The model does not incorporate any spin angular momentum of the charge itself: all angular
momentum is orbital. The Planck-Einstein relation then expresses the quantization of the orbital
angular momentum and can, therefore, be written as a vector equation as well: E = ħ·ω. The
angular velocity for circular as well as elliptical orbitals is given by ω = dθ/dt = v/r or, in
vector notation, ω = rv/r2.
20
We may, therefore, think of the Compton wavelength as a circular wavelength: it is the length of a
circumference rather than a linear feature!
21
We may refer the reader to our paper on Relativity, Light and Photons.
22
Illustration from Christoph Schweiger’s presentation on the electron-scattering method, and its applications to
the structure of nuclei and nucleons (8 January 2016). Schweiger took these illustrations from Robert Hofstadters
1961 Nobel Prize Lecture, which has the same title. However, we find Schweiger’s added neutron model and the Rc
= 0 and Rm = 0.76 fm formulas very didactic.
6
The radial force is in an electromagnetic ‘mass without mass’ model, at least (we will soon
introduce a different force: a nuclear force) the electrostatic Coulomb force, of course, and the
energy in the orbital is, therefore, an energy per unit charge, and the energy equation for the
orbital must, therefore, also be written in terms of E/q.
Hence, for the (physical) dimensions to make sense, we must also write the (orbital) kinetic energy as
energy per unit charge. The orbital energy equation (and its physical dimensions) can then be written as:










 


We believe the energy per unit charge formula is relativistically correct because the kinetic energy uses
the velocity v along the orbital (which is denoted as escape velocity ve
23
). We request the reader to note
the m/q factor, which is the inverse of Bohr’s magneton q/m. Indeed, we think of elementary particles
as spin-1/2 particles. Hence, their gyromagnetic ratio
24
is ½ because of the following identity
25
:







The important thing here is the angular momentum L, which we can write as:


How do we reconcile this with the view that the total angular momentum of such orbitals packs one
ħnot a half-unit ħ/2!? We can only assume the angular momentum captures half of the energy only
the kinetic energy and that the other half must be in the electromagnetic field.
23
See the MIT OCW reference course on central force motion.
24
The gyromagnetic ratio is usually denoted as g. Symbols may be confusing: we use gN or eN for the presumed
nuclear charge, and the μ in the g = μ/L equation is the magnetic moment, but we will use the same symbol ()
also for the standard electromagnetic parameter, as well as for the other standard parameters. The reader should
be able to distinguish the meaning of these symbols from the context.
25
See our paper on a speculative but realist interpretation of quantum physics based on the ring current model.
7
Logical? You tell me. We may need the main angular momentum equation later
26
, so let us write it down
once more:


The neutron oscillation and elliptical orbitals
The (electric) dipole field of a single charge from a binomial expansion of the terms of the potential in d,
which is the distance between the two opposite charges +q and q, which we will want to equate with a
pointlike but massive proton and a Zitterbewegung (zbw) electron consisting of a pointlike negative
charge moving in and out.
The pointlike zbw charge reaches maximum velocity (c) when passing through the center of the radial
field and overshooting it. Its motion, therefore, combines a radial and a tangential component. The
image of a 3D polar (one may also think of the IAEA’s logo
27
) rose comes to mind here (Figure 4).
Figure 4: Polar rose : r() = a0·cos(k0 + 0)
The phase = ·t is given by the (angular) frequency = E/ħ, in which E must represent the total energy
of the oscillationkinetic + potential which, using the binomial theorem, can be written as
28
:


The relativistically correct formula for kinetic energy defines kinetic energy as the difference between
the total energy and the potential energy: KE = E PE. The potential energy must, therefore, be given by
26
When calculating eccentricities of elliptical orbitals, for example. However, we will probably do that in later
papers.
27
We find it interesting the IAEA did not change its logo at the occasion of the 60th anniversary of its existence as
an organization.
28
See footnote 19 also. The total energy is given by E = mc2 = m0c2 which can be expanded into a power series
using the binomial theorem (Feynman’s Lectures, I-15-8 and I-15-9 (relativistic dynamics). He does so by first
expanding m0:

This is multiplied with c2 again to obtain the series in the text.
8
the m0c2 term. This term is zero for r = 0 but non-zero because of the potential energy in the radial field
at distances r 0. The total energy of a charge in a (static) Coulomb field is given by
29
:

The potential itself is equal to V(r) = U(r)/qe:


We could define the kg (mass) in terms of newton (force) and acceleration (m/s2). Can we do the same
for the coulomb? Rewriting the energy equation as a function of the relative velocity and the radial
distance r does the trick:


We may say this defines the mass of the pointlike charge as electromagnetic mass only, which now
consists of a kinetic and potential piece. The energy in the oscillation, therefore, defines the total mass
m = E/c2 of the neutron electron (n = p + en). The kinetic energy is thus given by
30
:





The first term in the series gives us the non-relativistic kinetic energy

. In line with the usual
convention for measuring potential energy, we will now set the reference point for potential energy at
zero at infinity, and the potential energy will, therefore, be defined as negative, going from 0 for r
to − for r 0. This makes for a negative total energy which is in line with the concept of a negative
ionization energy for an electron in an atomic orbital which, for a one-proton atom (hydrogen), is given
by the Rydberg formula.
29
U(r) = V(r)·qe = V(r)·qe = (ke·qe/r)·qe = ke·qe2/r with ke 9109 N·m2/C2. Potential energy (U) is, therefore,
expressed in joule (1 J = 1 N·m), while potential (V) is expressed in joule/Coulomb (J/C). Since the 2019 revision of
the SI units, the electric, magnetic, and fine-structure constants have been co-defined as ε0 = 1/μ0c2 = qe2/2αhc.
The CODATA/NIST value for the standard error on the value ε0, μ0, and α is currently set at 1.51010 F/m, 1.51010
H/m, and 1.51010 (no physical dimension here), respectively.
30
In line with the usual convention for measuring potential energy, we will set the reference point for potential
energy at zero at infinity.
9
Analogy between the gravitational radial field and the Coulomb field
We established an analogy between gravitational masses and electromagnetic mass above. We will soon
introduce Kepler’s third law, which gives us the cycle time of a mass of 1 kg (the mass unit) in orbital
around a mass M
31
:



The formula uses the concept of the standard gravitational parameter μ = GM. The physical dimension
of G and M cancel out nicely when writing the gravitational constant G as (approximately)
6.67410−11 m3kg1s2, which is what most textbooks or general articles on the topic do. So if we are
going to describe a charge in orbit, we should use the electric constant k = 1/4πε0, whose physical
dimension is equal to [k] = [1/4πε0] = N·m2/C2: N·m2 divided by the square of the charge.
How comes m3/kgs2 does not look like that? The answer is: it amounts to the same. We can relate the
unit of mass (expressed in kg) to a force (expressed in newton) through Newton’s second law: a force of
1 N will give a mass of 1 kg an acceleration of 1 m/s2. Hence, 1 kg = 1 N·s2/m. Hence, a squared kg is
equivalent to [1 N·s2/m]2 Inserting this into the structure of the proportionality constants for all inverse-
square force laws, and then converting N back to kg so as to get that we get that m3kg1s2 expression,
we get a consistent dimensional equation:





The formula for the cycle time then works out nicelydimensionally speaking:



So how does that work out when we have electromagnetic mass? We just need to think differently
about the force: a force is that what changes the state of motion by changing the energy of the charge.
Hence, we just use Einstein’s mass-energy equivalence relation: m = E/c2. Hence, 1 kg is 1 J·s2/m2 = 1 N.
Hence, we think of the inertia of the energy rather than the inertia of the mass. Putting the same
squared mass or energy factor in the denominator of the denominator , and just mass/energy in the
numerator of the denominator of that square root function for the cycle time, we get:
31
We gratefully acknowledge the MIT OCW course on central force motion for the formulas. We consider orbital
angular momentum only and we are, therefore, essentially modelling spin-zero particles neglecting the spin
angular momentum (S) in the (vector) spin-coupling equation J = L + S. We assume orbital angular momentum
respects the Planck-Einstein relation: L will, therefore, be an integer multiple of Planck’s quantum of action ħ: L =
n·ħ = n·E/ω. We note the IAEA logo shows elliptical orbitals (closed trajectories) too !
10

That was not too difficult, was it?  The question now is: how are we going to write our standard
electromagnetic parameter μk? Look at our formulas with the series expansion of energies: the mc2
factor gets replaced by

and that works for both kinetic and potential energy (the radial force
changes both). Remarkably simple. Hence, we can just write μk as μk = keqe, just like we wrote μG = GM
and that is it.
32
So let us summarize what we have so far. We substituted an energy per unit mass for an energy per unit
charge concept. We need the total energy, which depends both on the velocity v as well as distance r.
We also noted that we believe this formula to be relativistically correct because the kinetic energy uses
the velocity v along the orbital (which is denoted as escape velocity ve
33
). The kinetic energy is,
therefore, equal to KE = mve2/2 = m(ωr)2/2 = mω2r2/2. So let us think about this. We use the mass
concept as a measure of the inertia to a change of motion as per the relativistically correct expression of
Newton’s second law F = dp/dt = d(m·v/dt for a centripetal force, whose magnitude can also be written
as the following function of the acceleration a:


 
 

But so we will think of a force as that what changes energy. We will show we need to choose our
reference point carefully so as to make sense of μk.
Orbital geometries
We must now get into the nitty-gritty of elliptical orbitals so as to get a better feel of what we are talking
about. The general formulas for orbital motion in a non-zero potential assuming closed and, therefore,
elliptical orbitals with negative energy (E < 0) and an eccentricity less than 1 (0 < e < 0) are the
following
34
:


32
The difference between the m for mass and the m for meter is obvious from the context.
33
See footnote 23.
34
We refer, once again, to the MIT OCW course on central force motion for the formulas. We consider orbital
angular momentum only and we are, therefore, essentially modelling spin-zero particles neglecting the spin
angular momentum (S) in the (vector) spin-coupling equation J = L + S. We assume orbital angular momentum
respects the Planck-Einstein relation: L will, therefore, be an integer multiple of Planck’s quantum of action ħ: L =
n·ħ = n·E/ω. We note the IAEA logo shows elliptical orbitals (closed trajectories) too !
11

We have a base state for n = 1 in the E = n·ħ equation (Planck-Einstein) and excited states for n > 1. As
for the 
law, this is just Kepler’s third law which calculates the frequency of any
orbit around a large mass as a function of (i) the so-called standard gravitational parameter
35
μ = G·M
and (ii) a, which is the length of the orbit's semi-major axis:



Hence, for an orbital assuming all mass is electromagnetic, we get


We already showed the dimensional analysis of this works out OK
36
, so let us now use the formula. We
should use a trick here. We have, in effect, a very useful point to evaluate potential and kinetic energy is
at the periapsis, where the distance between the charge and the center of the radial field is closed. The
θ angle is there set at 0, which allows us to define r = 0 and v = c as complementary limits:




However, to avoid the division by zero, non-limit values for rπ and rπ are used, which we can obtain from
the general orbital formulas:







Using the L = m/q formula, we can then do calculate the eccentricity e from the orbital energy formula
37
:


35
When googling this, you will find plenty of references, but the Wikipedia article on elliptical orbits is OK.
36



37
The eccentricity e will be smaller than 1 (0 < e < 1) for a closed trajectory and we, therefore, prefer to write 1
e2 rather than e2 1. The expression only makes sense if the total energy is negative, which is the assumption (or
convention, we should say) that we started out with.
12

This differs rather substantially from the formula for the orbital energy in a gravitational field
38
:




Why are these so formulas so different? We are not sure but the use of the E/q and E/m factors in these
orbital equations makes a difference. While it complicates the structure of the formulas, it is obvious
that this mass/charge ratio must, therefore, come into play somehow. Also, the structure of the two
radial forces (gravitational and electrostatic) are not same. The electromagnetic force is two-
dimensional, as the electrostatic (time-independent) field and magnetic (time-dependent) field always
go hand-in-hand.
We believe this formula establishes an equivalence between gravitational and electromagnetic mass
through the mass-energy equivalence relation, which we will write as c2 = E/m in the next section.
The nuclear force and the range parameter a
The mass as inertia and the energy per unit mass hold the key to writing the orbital energy per unit mass
as follows for the three forces and charges we have been consideringgravitational, electrostatic, and
nuclear:
1. Gravitational force, with the masses (M and m) as charge and G as physical proportionality constant:


One can easily the dimensions work out to the required m2/s2 dimension:







2. Coulomb force, with the electric charges (qe) as charge and ke as physical proportionality constant:

We can check the physical dimensions once more:




38
See the referenced MIT OCW reference course.
13
3. Nuclear force, with the nuclear charges (gN) as charge and kN as physical proportionality constant and
a as the range parameter to ensure dimensional consistency:


 
One can easily ascertain a nuclear range or distance scale parameter (aN) has to be introduced so as to
ensure the physical dimensions come out OK:





For n = 1, we get a nuclear force field based on a Yukawa-like nuclear potential, but one that follows an
inverse-cube rather than an inverse-square law
39
:
If we give the same numerical value to both qe and gN about 1.61019 C and Y respectively
40
we , we
might say the nuclear permittivity or standard nuclear parameter is determined by the scaling
parameter a.
Combining the 1/r and a/r2 potentials, we now get the desired crossing of potentials and the expected
combined potential well function (Figure 5)
41
:
We can now think of calculating the range parameter a. Indeed, we mentioned that we should not
invent new concepts. We made the case for a nuclear force hopefully convincingly so but there is no
such thing as a nuclear charge: the electromagnetic and nuclear forces act on the same electric
chargeswith opposite sign but we can use the same numerical values and the same physical
dimensions for the charge: about 1.61019 C. There is, therefore, no need to invent a dirac or some
other new unit for the nuclear charge (Y?). We can, therefore, write the (magnitude of the) nuclear force
per unit mass as:
Again, the idea is that the nuclear permittivity or the standard nuclear parameter if you want is
determined by the scaling parameter a.
Now, the (magnitude of the) Coulomb force is, obviously, equal to:
39
We will soon say something more about the implications of field energy conservation.
40
For a new force, one should propose a new charge and a new unit to measure it. We suggest the Dirac, which we
abbreviate as Y so as to honor Yukawa.
41
We leave it to the reader to play with the sign conventions. Needless to say, the C in UC stands for Coulomb and
the N in UN stands for nuclear.
14
We may say the range parameter a is unity for the Coulomb force. We can now calculate the range
parameter by setting r equal to r = a, and equating the forces. We get:
What does this mean? It just means we have a tautology here: we defined a as the distance r for which
the forces are equal, and we get that one equals one. No surprise here! Of course, one can continue to
play with the assumption that there might be a nuclear charge which is different from the electric
charge, and one can also alter the form factor for the nuclear potential by, for example, inserting an
additional e-a/r factor (as Yukawa did). However, these are rather random additional assumptions to get
a result no matter what. We, therefore, do not want to dwell on this.
42
Figure 5: The nuclear potential well arising from 1/r and a/r2 functions
42
The interested reader may want to look at our paper on the electromagnetic deuteron model, where we did
insert Yukawas e-a/r factor. Because we then have a non-trivial derivative of the nuclear potential function (which
we use to calculate the force), we do find a definite value for a. To be precise, we get a value that is equal to a =
2.885 fm. However, such calculations are of no value whatsoever. In yet another paper of ours, we were even
more flexible and did define a new nuclear charge and concomitantly a nuclear permittivity factor. It yielded an
impressive-looking formula which we jokingly referred to as Van Belles formula. However, closer observation
makes clear this formula is nothing but a definition of the electric constant.  As for our appreciation of Yukawas
contribution to physics, we think it is where physics went wrongnot because of trying to model the nuclear force
(a good thing) but because it led to quantum field theory and the associated multiplication of concepts (e.g. quark
and gluon theory), which we do not think of as being productive (of course, we respect the academics who think
otherwise).
15
Non-symmetric potentials and other considerations
We should point out something rather striking about the inverse-cube law for the nuclear force: it
implies the potential in 3D cannot be symmetric. That should be fine because we are modeling plane-
or disc-like orbitals/oscillations, but it does require further thinkingwhich we will not engage in here
(this paper is already lengthy enough!).
43
Such further analysis may also usefully focus on:
An analysis of the qe/m and gN/m charge/mass-energy ratios.
An analysis of the different nature of the singularities at the center: limits (r 0) for 1/r and
a/r2 functions yield infinity () but these two infinities are, clearly, not of the same nature.
Indeed, the different type of singularity one gets at the center from an a/r2 function (a more
‘massive’ infinite potential energy, so to speak) may explain the different (Coulomb versus
nuclear) charge/mass ratios and the rather enormous proton/neutron energies/masses.
An analysis of how our simplified model may or may not fit with the interesting work analyzing
non-integer or higher-powers of distance functions and effective nuclear field theory, which all
rely heavily on expansion into power series, taking into account the energy conservation law
must put geometric constraints on them.
44
Developing the associated vector potential functions (AC and AN) using the electromagnetic
Lorentz gauge:


For a time-independent scalar potential, which is what we have been modeling, the Lorentz gauge is
(·A = 0) because the time derivative is zero: φ/t = 0 ·A = 0.
45
The magnetic field, therefore,
vanishes. The time-dependent magnetic field should absorb half the energy in accordance with relativity
theory
46
and it should then be easy to develop the equivalent of Maxwell’s equations for the nuclear
force field using the theorems of Gauss and Stokes.
The model has the advantage of not introducing new fundamental constants (except for the nuclear
charge gN), respecting relativity theory (no superluminal speeds), and confirming Planck’s quantum of
43
One might think this may have something to do with the asymmetry of a dipole field (see our paper on the
deuteron as a dipole) but, no, we are modeling a nuclear force between two charges here. The asymmetry is,
therefore, puzzling.
44
We will leave it to the reader to google and pick some articles: the list is pretty much infinite, and we are only at
the beginning of this exploration ourselves!
45
The Lorentz gauge does not refer to the Dutch physicist H.A. Lorentz but to the Danish physicist Ludvig Valentin
Lorenz (we are grateful to a reader to have pointed this out to us!). We just want to note one thing here: it is often
suggested one can choose other gauges. We do not agree: in a relativistically correct analysis, it seems to use that
we do not have a choice: the Lorentz gauge is one and the same for time-dependent and time-independent fields,
but it vanishes with time-independent fields (electromagnetostatics). See our remarks on the vector potential and
the Lorentz gauge in our paper on the electromagnetic deuteron model.
46
When using natural units (c = 1), the relativity of electric and magnetic fields becomes more obvious.
16
action as the fundamental unit of physical action (h) and angular momentum (ħ) in particle-field
exchanges of energy and momentum.
To conclude this excursion in logic and dimensional analysis, we offer some final remarks on forces and
mass factors.
Force calculations and the concept of nuclear mass
If one thinks in terms of some force holding the pointlike charge in its orbit, then we calculate this force
for the electron as being equal to about 0.106 N. This is the formula
47
:

That is a huge force at the sub-atomic scale: it is equivalent to a force that gives a mass of about 106
gram (1 g = 103 kg) an acceleration of 1 m/s per second!
48
Puzzled by the ½ factor? Feynman was puzzled too: where is the other half of the energy (or the mass)
of the electron?
49
Our ring current model shows the ½ factor is quite logical: Feynman is assembling the
zbw charge herenot the electron as a whole. Hence, the missing mass is in the Zitterbewegung or
orbital/circular motion of the zbw charge. We can now derive the classical electron radius from the
formula above:




The electron also comes in a more massive but unstable version: the muon-electron.
50
The muon energy
is about 105.66 MeV, so that is about 207 times the electron energy. Its lifetime is much shorter than
that of a free neutron
51
but longer than that of other unstable particles: about 2.2 microseconds (106 s).
Now that is fairly long as compared to other non-stable particles all is relative! and that may explain
why we also get a sensible result when using the Planck-Einstein relation to calculate its frequency and
radius
52
:  



47
We have derived this formula elsewhere. The ½ factor is there because we think of the zbw charge as having an
effective (relativistic) mass that is 1/2 (half) of the total electron mass.
48
We could also calculate field strengths and other magnitudes here, see our paper(s) for more detail.
49
See Feynman’s Lectures, Vol. II, Chapter 28 (Electromagnetic Mass).
50
You may also have heard about the tau-electron but that is just a resonance with an extremely short lifetime, so
the Planck-Einstein relation does not apply: it is not an equilibrium state. To be precise, the energy of the tau
electron (or tau-particle as it is more commonly referred to
50
) is about 1776 MeV, so that is almost 3,500 times the
electron mass. Its lifetime, in contrast, is extremely short: 2.91013 s only. We think the conceptualization of both
the muon- as well as the tau-electron in terms of particle generations is unproductive: stable and unstable
particles are, generally speaking, very different animals!
51
The mean lifetime of a neutron in the open (outside of the nucleus) is almost 15 minutes!
52
This presumed longevity of the muon-electron should not be exaggerated, however: the mean lifetime of
charged pions, for example, is about 26 nanoseconds (109 s), so that is only 85 times less.
17
The CODATA value for the Compton wavelength of the muon is the following:
1.1734441101014 m 0.0000000261014 m
If you divide this by 2 - to get a radius instead of a wavelength, you get the same value: about 1.871015
m. Hence, our oscillator model seems to work for a muon as well! Why, then, is it not stable? We think it
is because the oscillation is almost on, but not quite.
53
In contrast, the exercise for the tau-electron does
not yield such sensible result: the theoretical a = ħ/mc radius does not match the CODATA value.
54
We
think this confirms our interpretation of the Planck-Einstein relation as modelling stable particles.
Why this digression? We can also use our model to calculate the centripetal force which must keep the
charge in its orbit for the muon-electron and the ratio of this force for the electron and the muon:



If a force of 0.106 N is pretty humongous, then a force that is 42,753 times as strong, may surely be called
a strong force !
What about the force inside of a proton? Let us first calculate its Zitterbewegung radius. When applying
the a = ħ/mc radius formula to a proton, we get a value which is 1/4 of the measured proton radius: about
0.21 fm, as opposed to the 0.83-0.84 fm charge radius which was established by Professors Pohl, Gasparan
and others over the past decade.
55
We only get the right radius from using a modified Planck-Einstein
relation (E = 4hf = 4ħω) for the orbital frequency of the charge:
 




Now, the proton mass is about 8.88 times that of the muon, and it is about 2.22 times smaller. So here
we get this strange 1/4 factor too. Any easy fix? No. The Planck-Einstein relation gives us a frequency, but
it also gives us the angular momentum of a (stable) particle. Hence, a proton is also definitely not some
more massive or stable antimatter version of a muon-electronif only the Planck-Einstein relation
suggests its momentum is four times that of an electron or a muon-electron!
To calculate those forces, we can use the same oscillator model
56
:
53
The reader can verify this by re-calculating the Compton wavelength from the radius we obtain and the exact
CODATA values (with or without the last digits for the uncertainties) for the constants and variables. The reader
will see that the value thus obtained falls within the uncertainty interval of the CODATA value for the Compton
wavelength. The reader think about this result and its meaning as part of the exercise.
54
CODATA/NIST values for the properties of the tau-electron can be found here: https://physics.nist.gov/cgi-
bin/cuu/Results?search_for=tau. To go from wavelength to radius and vice versa, one should divide or multiply by
2π.
55
For the exact references and contextual information on the (now solved) ‘proton radius puzzle’, see our paper
on it: https://vixra.org/abs/2002.0160.
56
See our paper on the oscillator model of an electron.
18



A force of 4,532 N inside of a muon and a force of 89,349 N inside of a proton? We get nonsensical results
here, don’t we?
Maybe. Maybe not. First, a few back-of-the-envelope calculations reveal we should not be too worried
we are modelling a black hole here.
57
Second, we assumed a pointlike charge with zero rest mass here.
Our remarks on the different qe/m and gN/m charge-mass ratios and the different nature of the
singularities one gets at the center (limits for r 0) for 1/r and a/r2 functions suggests this assumption
needs further examination.
58
However, we have not made much progress on this.
A wave equation for the deuteron nucleus?
Erwin Schrödinger’s thinking on the wave equation back in 1922-1926 started with an “analysis of
electron orbits in an atom from a geometric point of view.”
59
Can we interpret Schroedinger’s energy
operator in the context of the orbital energy equation, and can we apply to the oscillation of the two
protons around or within the electron cloud?
The following remarks may be offered here:
1. The potential function in Schrödinger’s equation is Coulomb’s 1/r potential: V(r) = e2/r.
60
We can
easily substitute this for the nuclear a/r2 potential. We should not worry about electric or magnetic
permittivity coefficients because these are taken into account by the scaling parameter a.
The problem is with the kinetic energy in the orbital energy equation. Indeed, to prove the second term
in the energy operators corresponds to the kinetic energy, we think one needs to prove the following
identity:





We have tried this before but did not meet with success.
61
2. The ½ factor in Schroedinger’s equation is not the relativistically incorrect ½ factor in the classical
kinetic energy equation (it also has nothing to do with the ill-understood concept of effective mass: he
basically models the orbitals for two (paired) electrons, assuming their magnetic moments are coupled.
57
We leave it to the reader to calculate the Schwarzschild radius of a proton. He or she may also calculate the
related energy densities and electromagnetic field strength. They are humongous (definitely!)but not
impossible.
58
We calculated several other ratios, such as the (0.106 N)/(2.821015 m) ratio for the electron and the (89,349
N)/(0.211015 m) ratio for the electron, which should, perhaps, help to explain the range parameter a in the force
formula.
59
See the Wikipedia article on Schrödinger.
60
See: Feynman’s Lectures, Vol. III, Chapter 19 (The hydrogen atom and the periodic table).
61
See our paper(s) on a possible geometric interpretation of Schrödinger’s equation.
19
In short, he models spin-zero electrons. He also omits the magnetic coupling between the proton at the
center and the electrons orbiting around (hyperfine structure of the hydrogen spectrum).
3. The mass in Schroedinger’s equation (2m, cf. the electron pair) is the electromagnetic mass/energy of
the free electron (two times 0.511 MeV/c2), which yields average classical (average) tangential velocities
vn = αc/n, with n the principal quantum number. As such, Schroedinger’s wave equation is just a
generalization of the Bohr-Rutherford model (circular orbitals, gross structure of the hydrogen
spectrum) to also include elliptical/asymmetric orbitals (the subshells of the hydrogen atom, consistent
with the fine structure of the hydrogen spectrum).
62
4. Modeling two paired protons orbiting around/within an electron cloud (the deep electron) assumes:
(1) the nature of the orbital energy is nuclear (cf. the fm scale of nuclear oscillations versus the
nanometer/angstrom scale for electromagnetic oscillations) and (2) the mass/energy of the two protons
is nuclear as well. Consistent thinking in terms of E/mC and E/mN and qC and gN charges (with mC, qC and
mN, gN representing electromagnetic and nuclear mass/charge respectively) should do the trick.
5. In a future extension of the model, one may try to combine the electrostatic and nuclear potential in
the energy operator, but that would get a set of two highly complicated wave equations, and I do not
immediately see what math would be required to solve such set of equations (finding solutions to
Schroedinger’s equation is already quite complicated, and the a/r2 potential in a ‘nuclear’ Schrödinger
equation will not make it easier).
The remarks above thus yield the following trial suggestion for a wave equation for the nuclear orbitals
of the proton pair in a deuteron nucleus
63
:



Needless to say, the eN2 factor is a shorthand for the gN2/4πε0 in the nuclear potential/force formula. The
reader may wonder why we prefer to write the scaling parameter a as a function of the distance but we
refer to our remarks on the nuclear permittivity factor, form factors or to put it more simply the
asymmetry in the non-spherical, planar or disk-like orbital. We may also add this wave equation
becomes easier to understand when substituting ψ for
64
:

Schrödinger’s equation for the nuclear oscillation(s) inside the neutron and/or the deuteron then
becomes:



Delightfully simple and beautiful, but we have made no start with solving this equation.
62
I may refer to a very early paper here (which may have some mistakes), which argues complex numbers
represent the two-dimensional orbital oscillation of the (paired) electron(s).
63
The gN2 here is
64
One can use complex numbers to represent the two-dimensional orbital oscillation of the (paired) protons.
20
Are low-energy nuclear reactions possible?
If the binding energies in the deuteron nucleus and in the neutron (n = p + e) are nuclear, one wonders
how low-energy electromagnetic oscillations can trigger transmutation.
65
Indeed, the reasoning
becomes quite complicated when trying to explain how low-energy (photon) stimulation
(electromagnetic irradiation) might break the nuclear bond between the two protons and/or the
electron. How could this happen? Integer fractions of the (high) nuclear frequencies leading up to a
build-up of the necessary electromagnetic energy so as to break the nuclear bond? Another possibility is
that such reactions basically revolve around the magnetic couplings and/or their nuclear equivalent.
Of course, one must hold a dynamic view of the fields surrounding charged particles. Potential barriers
or their corollary: potential wells should, therefore, not be thought of as static fields: they vary in time.
They result from two or more charges moving around and creating some joint or superposed field which
varies in time. Hence, a particle breaking through a ‘potential wall’ or coming out of a potential ‘wellis
likely to just use a temporary opening corresponding to a very classical trajectory in space and in time.
66
At the same time, it would seem that (too?) many hypotheses need to be fulfilled for such resonances to
occur.
We hold a simpler view. We have two bonds in the deuteron nucleus: the bond between the two
protons, and the bond between the protons and the nuclear electron. One may, therefore, assume that
the electromagnetic stimulation breaks the bond between the electron and the nearby proton, followed
by a neutron decay into a proton and an electron. Where does the electron go? And where does the
proton get the (new) electron to form a neutron once more? If the energy reaction is to occur
transmutation the decay must be permanent, and then we have to account for not one but two
electrons ! They must be pushing one or more electrons in an atomic orbital out.
What about net energies? The decay of the neutron into a proton and an electron yields a positive
energy of about 1.3 MeV. Hence, we may effectively assume this gives the electron a kinetic energy of
about 1.5 times the free electron energy (0.511 MeV): enough of push out one more electrons out of
the atomic orbitals (electromagnetic binding energies in the atomic orbitals is a few keV only). If the
remaining two protons stick together (their 2.2 MeV binding energy is negative and they will, therefore,
want to stick together), we would, therefore, get a net energy in the range of 0.7 to 0.8 MeV.
Not very explosive
67
but enough to work with and surely more energy than the energy one gets out of
chemical reactions: we are still talking MeV rather than keV, right? Right. However, such considerations
do not explain how one could possibly get net energy out of such reactionsif they occur. LENR
theorists need to provide better answers and/or more productive assumptions for physicists to work on.
65
We think of transmutation in, say, palladium, by laser stimulation, for example.
66
It is like shooting bullets through the blades of a fast-rotating airplane propeller: if the bullets are ‘fast enough’,
they will get through, and higher-speed bullets (speed relative to the angular velocity of the propeller) are more
likely to get through.
67
The non-runaway character of LENR or anomalous heat reactions has nothing to do with the low energy values
or with the targeting of the electromagnetic beam but with the fact that, apparently, no follow-on reactions
occurunlike high-energy nuclear reactions (hot fusion), where slow or fast neutrons produce more neutrons,
resulting in more (slow or fast) neutrons producing the same reaction which is, therefore, explosive rather than
just self-sustaining.
21
As there is a lot of interesting literature around
68
, we will focus on researching that rather than try to
come up with our explanations.
The E/m = c2 relation and the elasticity of space
One of the most remarkable things in the energy equations for the electromagnetic and nuclear (and
gravitational
69
) orbitals is that the E/m = c2 holds for all of them. In fact, it is this equation combined
with the Planck-Einstein relation (E = hf = h/T) which yields the orbitals in the first place!
Combined with the oscillator model (E = ma22 = mc2 c = a), this makes us think of c2 as modeling
an elasticity or plasticity of space
70
, with the two (or three) force oscillations representing different
oscillatory modes.
Of course, this triggers the question: why these modes only? We do not know. Perhaps the fine-
structure constant has something to do with it. To find out, one needs to examine the proportionality
coefficients in the potential and force equations much more closely, and see how they work together
with the scaling parameter a.
Brussels, 1 February 2021
68
See, for example, Péter Kálmán and Tamás Keszthelyi, On low-energy nuclear reactions, June 2019.
69
Thinking of the gravitational force as some kind of residual force is not easybut the reader should keep trying!
70
We consciously do not use the term ‘spacetime’ here: we are talking of an oscillation in space and in time. Of
course, this is a ‘spacetime oscillation’, but one cannot get rid of the concept of charge: something is oscillating,
and it is not the aether! 
ResearchGate has not been able to resolve any citations for this publication.
Article
Full-text available
Low energy nuclear processes that are strongly hindered by Coulomb repulsion between the reacting nuclei, are investigated in solid environment. It is shown that the hindering effect may be significantly weakened (practically it diappears) if one takes into account the Coulomb interaction of one of the reacting particles with the surroundings. It is obtained that if the modification of the wave function due to Coulomb interaction with charged constituents of the environment is taken into account applying standard perturbation calculation of quantum mechanics then wave components of high momentum with small amplitude are mixed to the initial wave of small momentum. To these partial waves of high momentum much higher Coulomb factor can be attached that can drastically increase the cross section. The mechanism (called recoil assistance) opens the door to a great variety of nuclear processes that so far have been thought to have negligible rate at low energies. The recoil assisted nuclear pd reaction is investigated like a sample reaction numerically. Low energy nuclear reactions allowed by recoil assistance and leading to nuclear transmutations are partly overviewed. Critical analysis of Fleischmann-Pons type low energy nuclear reaction experiments is presented too.