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This paper shows how one can use potentials to build up a spin-zero model of the deuteron. The spin-zero model consists of a proton, and another proton plus an electron which combine in an electrically neutral particle which we refer to as the neutron. We treat all particles as spin-zero particles because we assume their magnetic moment is zero. As such, it may complement Paolo Di Sia's model of the nucleus (2018), which we give due attention. In contrast to Di Sia, we think of neutrons-or the electron cloud that surrounds the proton inside-as electric dipoles. The model does so by interpretating Yukawa's potential function as a dipole potential. Instead of predefining the range parameter a, we calculate it from the equilibrium condition (equal but opposite magnitudes of the Coulomb and nuclear forces). We find a very acceptable value of about 2.88 fm for a, and find an equally acceptable value for the distance between the positively charged center of the neutron and the center of the electron cloud which, in a deuteron nucleus, must shift it center of charge towards the proton so as to ensure stability-not unlike the sharing of valence electrons in chemical bonds.
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The electromagnetic deuteron model
Jean Louis Van Belle, Drs, MAEc, BAEc, BPhil
31 December 2020
Abstract
This paper shows how one can use potentials to build up a spin-zero model of the deuteron nucleus. The
spin-zero model consists of a proton that is attracted by a neutron, which we think of as a proton-
electron combination. We treat all particles as spin-zero particles because we assume their magnetic
moment is zero. As such, it may complement Paolo Di Sia’s model of the nucleus (2018), which we give
due attention. In contrast to Di Sia, we model the nucleus using the electric dipole potential.
The model does so by re-interpretating Yukawa’s potential function as a dipole potential. Instead of
predefining the range parameter a, we calculate it from the equilibrium condition (equal but opposite
magnitudes of the Coulomb and nuclear forces). We find a very acceptable value of about 2.88 fm for a,
and find an equally acceptable value for the distance between the positively charged center of the
neutron and the center of the electron cloud which, in a deuteron nucleus, must shift it center of charge
towards the proton so as to ensure stability not unlike the sharing of valence electrons in chemical
bonds.
Table of Contents
Introduction .................................................................................................................................................. 1
Di Sia’s model: nuclear force as magnetic force ....................................................................................... 1
Intermezzo: physical dimensions .............................................................................................................. 1
Di Sia’s model: zero-charge nucleons ....................................................................................................... 3
Evaluation of Di Sia’s model ...................................................................................................................... 3
Another electromagnetic explanation for interproton attraction ............................................................ 3
The potential of an electric dipole and Yukawa’s nuclear potential function .......................................... 5
Energies and radii ...................................................................................................................................... 6
Potential, potential energy, fields, and inverse square/cube force laws ................................................. 8
Yukawa’s nuclear potential and force ........................................................................................................... 9
Yukawa’s potential .................................................................................................................................. 10
The nuclear permittivity factor υ0 ........................................................................................................... 12
Force and force range calculations ......................................................................................................... 13
Numerical example ..................................................................................................................................... 15
Conclusion ................................................................................................................................................... 18
Annex: Force fields and higher-order potentials ........................................................................................ 19
Introduction ............................................................................................................................................ 19
The asymmetric force field of an electric dipole .................................................................................... 22
Feynman’s field equations ...................................................................................................................... 24
Another interpretation of the 1/r and 1/r2 dependence on distance? .................................................. 25
The magnetic field................................................................................................................................... 26
Feynman’s derivation of Feynman’s field equations .............................................................................. 27
1
Introduction
Di Sia’s model: nuclear force as magnetic force
Paolo Di Sia (2018
1
) showed the presumed nuclear attraction between nucleons might be explained in
terms of the interaction between the magnetic moments of nucleons. This magnetic moment arises
from a presumed internal current in one nucleon, which will generate a magnetic field whose strength is
given by the Biot-Savart law, which is illustrated below (Figure 1).


Figure 1: The magnetic field from a current is obtained from an integral around the circuit
2
Di Sia assumes the current itself is electrically neutral (his nucleons have spin but zero charge
3
) and
there is, therefore, no electrostatic field. Hence, if we place a charge q at point 1, the qE term of the
Lorentz force F = qE + q(vB) charge is zero and the charge will experience a magnetic force q(vB) only.
Intermezzo: physical dimensions
An intermezzo? Already? Yes. Because we are talking real things here by describing them using a mix of
physical and mathematical vectors and because we attach great importance to the physical
interpretation of the equations as part of our realist interpretation of quantum physics
4
we request
the reader to carefully check the physical dimensions and the nature of the magnetic force field:
1. Current is measured in A = C/s (ampere). A current is usually neutral: electrons move but the
wire as a whole is usually thought of as being neutral.
5
2. The integral itself does not change the physical dimension of the integrand
6
but the vector
cross-product
7
involves a spatial rotation by 90 degrees or a rotation by i and, therefore,
1
Paolo Di Sia, A solution to the 80-year-old problem of nuclear attraction, October 2018.
2
We borrow the illustration (and formula) from Feynman’s Lectures (II-14-6).
3
This is not stated explicitly in the paper. However, we must assume this because Di Sia does not mention the
electrostatic potential.
4
All mysteries can be explained: see https://vixra.org/author/jean_louis_van_belle.
5
For a good analysis of the magnetic field vector as a relativistic effect of the motion of charges in a neutral wire,
see Feynman’s Lectures (II-13-6).
6
The energy integral F·ds, for example, yields an energy expressed in the appropriate N·m = J dimension, which is
just the dimension of the physical vectors F (N) and ds (m).
7
Understanding the physical dimension of a vector cross-product e12ds2 = n·e12·ds2·sinθ is not too difficult but
one tends to forget what is what. Sine and cosine functions are absolute numbers in the [1, +1] in the interval.
Likewise, the length of unit vectors (e12 and n) is an absolute number between 0 and 1 but they have very different
2
involves a multiplication of the dimension of the physical vector and the s/m dimension. Think
of the F = qvE equation here, which must yield a physical force vector with the force
dimension (N), which works out alright because the m/s dimension of v and the added s/m
dimension because of the cross-product cancel each other out: C·(m/s)·(N/C)·(s/m) = N.
3. The physical proportionality constant is usually just k = 1/40, but we have an added c2 in the
Biot-Savart law which gives us the magnetic permittivity μ0 = 1/0c2, so its dimension is
[1/40c2] = [μ0/4] = [1/40]/[m2/s2] = (N·m2/C2)·(s2/m2) = N·s2/C2.
8
We are finally ready to multiply all physical dimensions here:



We get an interesting s2/m2 dimension here: it is, of course, the dimension of 1/c2 once more, and it is
consistent with the other way of writing the Biot-Savart law:






S is the cross-section area of the current (wire) and dV2 is, therefore, an infinitesimal volume element.
The reader can check the physical dimensions once again: he should find the s2/m2 dimension once
more.
Why are we attaching so much importance to dimensional analysis? Because Hideki Yukawa did not
when he wrote down his (in)famous equation for the nuclear potential back in 1935, that was and
we find it, therefore, totally incongruentand we will explain why, of course. As far as we are
directions: e12·n = e12·n·cosθ. Depending on your convention, you may allow for negative lengths but this will
screw up the right-hand rule you need to apply very scrupulously apply when dealing with vector cross products
(the vector cross product is not-commutative (the vector dot product is, however) and, hence, keeping tracks of
signs is often quite a headache). We will, therefore, not associate the unit vectors themselves with a physical
dimension: they are mathematical vectors only, as opposed to physical vectors, such as force (expressed in N) or
velocity (expressed in m/s). The added s/m dimension when vector cross products are involved is quite mysterious:
it suggests the argument of the sine and cosine function (the angle θ) relates distance and time: we may effectively
express both distance as well as time units in radians when using equivalent units (c = 1). Perhaps a connoisseur of
relativity can tell us how all of this rhymes with relativity theory.
8
The electric and magnetic constants are related through the c2 = 1/ε0μ0 relation. The dimensions here work out
alright too: [c2] = [1/ε0μ0] = (N·m2/C2)·(C2/N·s2) = m2/ s2. Note that the 2019 revision of SI units resulted in a defined
value for h, c and qe2 (and, therefore, for the N, m, s, C units as well as all units that depend on them), but that the
vacuum (electric) permittivity (ε0), (magnetic) permeability (μ0) and fine-structure constant are defined as a
function of each other:

The estimated standard deviation (SED) of is, therefore, the same: 1.51010 (expressed in F/m or N/A2 for ε0 and
μ0, respectively, but in absolute units for ). This gives the fine-structure constant = kqe2/ħc consistent physical
interpretation. We think the 2019 revision of SI units put physics firmly back on the road to reality. We think it was
about time.
3
concerned, quantum physics left the road to reality there and then. We will try to make sense of his
equation again by interpreting it as a potential for a dipole field.
Di Sia’s model: zero-charge nucleons
Back to Di Sia’s model. Apart from modelling neutral currents, we also note Di Sia does not put a charge
in the field: he puts another zero-charge nucleon with spin (magnetic moment) and, therefore, he
models the magnetic force between two zero-charge nucleons only. The numerical example which he
provides is for nucleons with an approximate size of 0.5 fm (we take this to be the radius of the current
loop) which are separated by the typical interproton distance (about 2 fm), which corresponds to the
usual range parameter in Yukawa’s formula for the nuclear potential.
Interestingly, Di Sia also considers the phase of the currents, which may effectively be in or out of phase
and conveniently calculates energy levels for the magnetic binding. We can, therefore, immediately
compare with relevant values for nuclear binding energies. For the mentioned values (0.5 and 2 fm) he
gets an energy range between 3.97 KeV and a more respectable 0.127 MeV (the latter value assumes
in-phase currents). While this is, without any doubt, significant, it is only 5% of the 2.2 MeV 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).
The values get (much) better when changing the parameters (nucleon size and internucleon distance)
significantly (2-3 MeV) and, better still, considering paired nucleons creating dipoles acting on other
paired nucleons (values up to 5 MeV), but we would wish the author would offer a model of a nuclear
lattice showing how currents and nucleon pairing actually works in 3D space.
Evaluation of Di Sia’s model
We like the modelif only because it provides a consistent interpretation of the internucleon force as
an electromagnetic force. However, we are hungry for more detail. Di Sia models full-spin nucleons
without a charge, and thinks of a nuclear lattice as a lattice of magnetic dipoles.
In this paper, we want to do the opposite: we want to develop a model for charged nucleons without
spin, and a small nuclear lattice based on the concept of an electric dipole. The basic idea is this: we
think of deuteron as consisting of two protons and a deep electron
9
and we, therefore, have a classical
three-body problem, for which no general closed-form solution exists.
10
Another electromagnetic explanation for interproton attraction
Special-case solutions for three-body problems exist, and one of these special cases is the case where
two of the three bodies are very massive relative to the third one: think of the EarthMoonSun system,
for example. Our deuteron model will basically be the same: we think of one of the two protons being
closely bound with the electron. However, we will not think of its magnetic moment but of its electric
9
We are grateful to Andrew Meulenberg (Meulenberg and Paillet, Highly relativistic electrons and the Dirac
equation, 2019-2020) for having drawn our attention to the work on deep electron orbitals. In fact, we write this
paper as input for his forthcoming paper, which should generalize the concept and show full consistency with all of
classical electromagnetic theory.
10
We may refer the reader to the Wikipedia article on the three-body problem, which we find to be intuitive and
well written (not all of them are like that, but we find it has gotten a lot better).
4
dipole moment resulting from two opposite charges being very close and creating an (electrostatic)
dipole field.
Of course, the informed reader will immediately object that a neutron has a significant magnetic
moment about 9.661027 A·m2, to be precise
11
but, in contrast, it has no noticeable electric dipole
moment. In fact, now that we are here, we should quickly give you the (approximate) value of the
magnetic moments of all these particles so you can check what adds up and what does not:
Electron (e)
9.281024 A·m2
muon-electron μ)
0.451027 A·m2
proton (p+)
+ 14.111027 A·m2
neutron (n0)
9.661027 A·m2
deuteron (D+)
+ 4.331027 A·m2
As you can see, the magnetic moments of the proton and neutron add up but very approximately only
to the magnetic moment of the deuteron, but the magnetic moments of the electron and the proton
do not add up to that of the neutron: we have a very different order of magnitude for the electron mm
(961 times the proton mm, and 658 times the neutron mm), and thinking of it the neutron-electron as a
muon-electron does not help either. Hence, the neutron electron must be a very different beast than
the electron we are used to. We might come back to that later.
12
What about the electron electric dipole moment (oft abbreviated as EDM)? Is our model dead even
before we developed it? Maybe. Maybe not. Experiments do measure a EDM, but it is very weak, and
successive experiments have pushed the upper limit down to an order of magnitude of 1026 qe·m. The
discussion is bound up with discussions on symmetry-breaking in Nature
13
and other complicated
matters which are of no interest to us here because they would only confuse us even more. The point is:
if the electron cloud around the proton in the neutron is spherically symmetric, then the neutron EDM
should, effectively, be zero. So why would we even pursue the hypothesis?
Because we are thinking of a deep electron orbital that is part of a deuteron nucleus and we, therefore,
think the presence of the (other) proton will result in a shift in the electron blanketjust like in a water
molecule, and causing the same kind of polarity.
14
11
We prefer the A·m2 unit to the equivalent J/T tesla because it reminds us of the μ = I·A formula for the magnetic
moment of a current loop, with I the current of the loop and A the surface of the loop. The minus sign gives us the
direction of the magnetic moment when applying the usual right-hand rule. We should, of course, relate this to the
angular momentum (spin) of the particle but we will not do this here. For a short overview of how this works, see
our paper on the radius and magnetic moment of electrons and protons. As for the n = p + e model, see our paper
on proton and neutron reactions.
12
We offer a comprehensive analysis of particle spin and magnetic moment based on the ring current model
which, in essence, is an application of Wheeler’s mass without mass model in our paper on classical quantum
physics.
13
You can google various popular science articles on this, but the Wikipedia article on the nEDM is a good starting
point.
14
In order to ensure our analysis is dimensionally correct, we will actually first assume genuine nuclear charges
may exist (besides or combined with electric charges), but we will soon substitute this hypothesis for a simpler
one: we think the presence of the proton results in a shift of the neutron electron cloud, thereby creating an
5
Indeed, two protons do normally not bind in any stable way. A diproton the nucleus of 2He isotope is
extremely unstable: a neutron is needed to glue them together in a more stable configuration: 3He. So,
yes, we do think of a neutron as consisting of a proton and a deep (nuclear) electron which binds both. It
is an idea which Rutherford thought of when he first hypothesized the existence of neutrons and which
would explain proton-neutron reactions as well as the instability of the neutron outside of the nucleus.
Does that make sense? Maybe not. We will soon see when we try to put all of this into equations, which
we will do soon enough. However, let us first present the basic formulas for an electric dipole field.
The potential of an electric dipole and Yukawa’s nuclear potential function
An electric dipole moment is measured by multiplying the magnitude of the two opposite charges ( q)
by the distance that separates them (d) and the electrostatic potential it generates is approximated by
the following formula
15
:


This formula tells us the magnitude goes down with the square of the distance (as opposed to the 1/r
function for electrostatic potential
16
) and also depends on the angle between the axis of the dipole (the
line of charges) and the radius vector.
17
Also note the dipole moment p·cosθ = q·d·cosθ factor versus the
charge q in the numerator of the potential function for a single charge which, for the convenience of the
reader, we reproduce below:

We can now compare both functions to the Yukawa potential
18
:

electric dipole moment with a potential which diminishes with the (radial) distance as 1/r2, as opposed to the
electrostatic 1/r (Coulomb) potential.
15
See: Feynman’s Lectures, II-6-2 (the electric dipole). The formula is derived from (1) a calculation of the two
(opposite) potentials (this yields the 1/r function), (2) expanding the (d/2)2 term in the distance calculation using
the binomial expansion, (3) neglecting all terms with higher power than d2 (because d is supposed to be very
small), and then (4) adding the two simplified potential functions (or, because of the opposite sign, subtracting
them). Assuming the distance d is measured along the z-axis, this yields the formula below, which can easily be
rewritten as the formula we use above:


16
Note that we will write the potential as V(r) = kqe2/r2 rather than as V(r) = kqe/r2. We mentioned the subtle
difference between potential and potential energy already, and that we would not always respect these subtleties
ourselves! We are in good company here, however, because Aitchison and Hey do the same!
17
For a succinct analysis, see: Feynman’s Lectures, Vol. II, Chapter 6 (The Electric Field in Various Circumstances).
18
The Wikipedia article uses a mass factor but we prefer the original formula given in Aitchison and Hey’s Gauge
Theories in Particle Physics (2013). It is a widely used textbook in advanced courses and, hence, we will use it as a
reference point. We may also refer the reader to Feynman’s remarks on it (II-28-6, the nuclear force field) because
these are online and free.
6
Check the dimensions: there is no physical proportionality factor (only a mathematical 1/4π factor
19
),
and what is the physical dimension of the nuclear charge gN? It is not qe because otherwise Yukawa
would have used that, right? He also introduces a scaling constant a, but only in the exponential decay
factor e r/a: if the idea is to provide a scaling parameter, then one should also use it to rescale distances
in the 1/r or 1/r2 factor for a potential or a field.
In short, a lot of fixing is needed to give Yukawa’s formula some respectability as a proper formula for a
nuclear potentialand we will do that soon enough!
OK. We are almost done with our introduction to the real thing. Before we dive into it, let us quickly give
you the basic data on energy, radii, and other experimental data that will need explaining in a full-blown
model for the deuteron nucleus, towards which we want to make a first step only.
Energies and radii
The electron-proton scattering experiment by the PRad team at Jefferson Lab measured the root mean
square (rms) charge radius of the proton as rp = 0.831 ± 0.007stat ± 0.012syst fm.
20
The root mean square
radius of a neutron, as calculated from electron scattering experiments, was also measured to be
around 0.8 fm.
The results from electron scattering experiments are precise enough but a straightforward
interpretation is not easy because the electron-proton interaction is an interaction between charged
particles and will therefore involve the Coulomb force mainly. In contrast, electron-neutron interactions
are (mainly) magnetic. Hence, such experiments use form factors which are then used to interpret the
scattering data and calculate what we would rather refer to as the radius of effective interference. We
copy a slide from a presentation on form factors which we rather liked
21
because it shows Rutherford’s
model a neutron combines a proton and a nuclear electron is actually being used to calculate the
neutron form factor. It should be noted that the 1961 measurements of Hofstadter and Rudolf
Mössbauer
22
then yielded a neutron charge radius Rc = 0 and a magnetic radius equal to Rm = 0.76 fm.
19
Gauss’ Law can be expressed in integral or differential form and these spherical surface area and volume
formulas pop up when you go from one to the other. Hence, you should not think of this 4π factor as something
weird: it is typical of spherically symmetric fields.
20
See: https://www.jlab.org/news/releases/new-measurement-yields-smaller-proton-radius. A root mean square
statistic involves (1) the squaring of measurements before adding and dividing by the number of observations
before (2) taking the square root again to calculate the mean value. The approach ensures positive and negative
deviations from a zero point do not cancel out while summing, but gives large values more weight. The mean of 1,
2 and 3 is 
, but the rms calculation yields 
. The arithmetic mean is, therefore, referred
to as an unbiased estimator. Experimenters obviously take care of these things and we should, therefore, not
worry about it.
21
Christoph Schweiger, 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. We find Schweiger’s added neutron model and the Rc = 0 and Rm = 0.76 fm formulas very didactic,
however.
22
Schweiger took the illustrations from Robert Hofstadters 1961 Nobel Prize Lecture, which has the same title. We
laud Schweiger for adding the neutron model and the Rc = 0 and Rm = 0.76 fm formulas, which we find very
didactic.
7
Figure 2: The neutron model and the concepts of charge and magnetic radius (Rc and Rm)
The neutron model shows a proton surrounded by a meson cloud: n = p + π. The negative meson (π) is
unstable and disintegrates into a muon-electron while emitting a neutrino: π μ + νμ. The energy of a
meson is 139.570 MeV, which far exceeds the energy difference between a neutron and a proton
(939.565 MeV 938.272 MeV 1.3 MeV). The (rest) energy of the muon-electron is about 105.66 MeV,
which is about 207 times the electron (rest) energy (0.511 MeV). We, therefore, think the neutron
consists of a proton and an electron rather than a proton and a π or μ. Why? Because one can better
account for the 1.3 MeV energy difference between a free neutron and a free proton by a particle
whose energy is 0.511 MeV only.
Of course, the electron accounts for about 40% of the energy difference only but a (free) neutron is
unstable. A free neutron effectively disintegrates into a proton and an electron
23
: n0 p+ + e + νe. The
presence of a neutrino suggests the binding energy is nuclear rather than electromagnetic.
24
As we are
going to model the nuclear force as a force of attraction between like charges (as opposed to the
electrostatic Coulomb force, which causes like charge to repel each other), we are fine explaining the
remaining 60% as positive nuclear binding energy between the proton and the electron. Negative
nuclear binding energy is illustrated by the 2.2 MeV energy difference between the deuteron nucleus
(about 1875.613 MeV) and its two constituents (939.565 MeV + 938.272 MeV = 1,877.837 MeV).
23
It takes an extraordinarily long time to do so, however: its mean lifetime is = 879.6 ± 0.8 s, so that is a rather
amazing 14.96 minutes! Assuming exponential decay, this means we are still left with a fraction of 1/e 0.37 out
of the initial number of neutrons after 15 minutes!
24
All these reactions respect conservation of charge, energy as well as linear and angular momentum (spin).
Adding the overbar or not for the neutrino is a matter of convention: we believe neutrinos and antineutrinos differ
in their angular momentum (spin) only because they do not carry charge. To be precise, we think neutrinos are the
photons of the nuclear force: just like photons in electromagnetic interactions, they account for the nickel-and-
dime in the energy, momentum, and spin conservation equations. As lightlike particles carrying no charge, they
must travel at the speed of light, exactly. We, therefore, any experiments which supposedly measure neutrino rest
mass are erroneous.
8
Potential, potential energy, fields, and inverse square/cube force laws
To get an intuitive grasp of the nature of the potential and force functions, it is probably good to review
the basic conventions and definitions:
1. The scalar potential is the potential energy of a charge, say Q, per unit charge: V(r) = U(r)/Q. This
explains the rather subtle difference between the [kqe2/r] and [kqe/r] dimensions: N·m versus N·m/C.
however, because we will be talking unit charges only, we will not distinguish between potential and
potential energy functions.
2. The field is, likewise, nothing but the force per unit charge and, hence, it is expressed in N/C. For the
magnetic field (B or F = qvB), we have an added s/m factor because of the geometry of the magnetic
field and force.
25
3. The (electrostatic) force (or, when thinking in terms of the unit charge (see remark 1), the force field)
is the (negative of the) gradient of the (scalar) potential U or φ:




This is quite wonderful: we have a scalar function U(x) or U(r) from which we can derive the electric field
and, therefore, the electrostatic force F = q·E on a charge. We have three components here, which we
may write as three equations: Ex = −U/x, Ex = −U/y, Ex = −U/z. The minus sign depends on the
nature of the force: the Coulomb force causes like charges to repel each other. We will assume a nuclear
force which attracts like charges in order to explain the mutual (static) attraction of the two protons in
the deuteron nucleus.
4. The magnetic force (or, when thinking in terms of the unit charge once more (see remark 1), the force
field) is the curl of the (vector) potential A:




We again have three components and, therefore, three equations hereone for each of the
components (Bx, By, and Bz) of B:

 


 


 

One can see that an analysis with both E and B vectors is somewhat more difficult because of the
simultaneous presence of two vector fields.
5. The Lorenz gauge connects both the scalar and vector potentials:
25
See the remarks on physical dimensions in footnote 2.
9


For a time-independent scalar potential, which is what we are going to model, the Lorentz gauge will be
zero (·A = 0) because the time derivative is zero: φ/t = 0 ·A = 0.
26
The B field, therefore,
vanishes.
If anything, this shows that modeling electric and magnetic fields simultaneously might, after all, not be
so difficult: all we need to do, is to model the time dependence of the scalar potential (φ/t) and
through the Lorenz gauge we then get the time-dependent B field for free, so to speak! What is the
connection here? Maxwell’s equations, of course! Why? Because Maxwell’s equations connect the E and
B fields through (relatively) simple vector algebrathe theorems of Gauss and Stokes, basically!
27
6. The electrostatic potential decreases with 1/r. A dipole field, however, follows an inverse square law
and, therefore, decreases with 1/r2. The respective force or force field will, therefore, follow an inverse
square and inverse cube law, respectively. We need to watch the signs here: the Coulomb force should
repel the two protons, while the nuclear force should attract them. The forces will, therefore, have
opposite signs.


 
 







 
 





OK. Let us get into the meat of the matter nowliterally.  What is the charge radius of deuteron
again? Right. About 2.1 fm.
28

Yukawa’s nuclear potential and force
We gave you the potentials and forces we need to ensure a neatly separated near and far electronuclear
field (Figure 3). The nuclear force will be larger than the Coulomb for r 2 and, vice versa, smaller for r
2. Note, however, that the potentials equal each other not at r = 2 but at r = 1.
Zones
Forces
Potentials
Near field: FC FN
Far field: FC FN
26
The Lorentz gauge does not refer to the Dutch physicist H.A. Lorentz but to the Danish physicist Ludvig Valentin
Lorenz. The reader should not think we have a choice here: the Lorentz gauge is one and the same for time-
dependent and time-independent fields, but it vanishes with time-independent fields (electromagnetostatics).
27
This should probably be the next development of our deuteron model, then!
28
For a discussion, see: O.J. Hernandez et. al., The deuteron-radius puzzle is alive: A new analysis of nuclear
structure uncertainties, . Note this is a radius but it is, of course, also a measure of the internucleon distance.
10
Figure 3: Opposing inverse square and inverse cube force laws
So what is the assumption here, really? The assumption is that the neutron electron cloud does not
shield the proton from the electrostatic force between the two protons in the deuteron. However,
because the proton also pulls the neutron electron blanket towards itself, so to speak, an electric dipole
is created, whose potential follows an inverse square law. The proton, therefore, experiences an inverse
cube force field which, within the region r 2, is sufficiently strong to counter the electrostatic Coulomb
force.
Let us see if we can capture that with Yukawa’s equation.
Yukawa’s potential
The Yukawa potential is usually written with a 1/r dependence on the radial distance. Transforming the
mathematical proportionality factor into a proper physical proportionality factor by adding a nuclear
permittivity factor (υ0)
29
:

To make sure you understand what Yukawa tried to model, we remind you of the formula for the
electrostatic (Coulomb) potential:

However, Yukawa’s potential function cannot do the trick: one never gets a nuclear force field that is
stronger than the Coulomb force: the 1/r and er/a/r functions never cross, regardless what value we use
29
The Wikipedia article uses a mass factor but we prefer the original formula given in Aitchison and Hey’s Gauge
Theories in Particle Physics (2013). It is a widely used textbook in advanced courses and, hence, we will use it as a
reference point. We may also refer the reader to Feynman’s remarks on it (II-28-6, the nuclear force field) because
these are online and free. As for the symbol, υ0 is an upsilon, so that is the lowercase form of the capital letter Y,
which we want to reserve for a possible physical dimension of the nucleon charge: if it is not Coulomb, it has to be
something else, isn’t it? 
Note that we write the potential as V(r) = kqe2/r2 rather than as V(r) = kqe/r2. We mentioned the subtle
difference between potential and potential energy already, and that we would not always respect these subtleties
ourselves! We are in good company here, however, because Aitchison and Hey do the same!
11
for the range parameter a. Hence, the associated force functions do never cross either!
30
Hence, we
must boldly decide to re-write the Yukawa potential as:

We mentioned a already: it is a range parameter. Aitchison and Hey present it as some kind of natural
distance or scale unit, which implies we would measure all distances in units of a, so we should write the
1/r2 as 1/(r/a)2 and also check the implications for the proportionality factor accordingly. Not doing this,
is pretty mortal and shows no or little understanding of the physicality of the situation. However, it is
what it is, and we just have to soldier on. At the very least, we should agree on a mathematical scale,
and so that is what we shall do: we shall assume all distances are measured in fm (1015 m). So what
value could this range parameter possibly be?
According to Aitchison and Hey, we should use a value around 2.1 fm, which is about the size of
deuteron, i.e. the nucleus of deuterium, which consists of a proton and a neutron bound together.
However, that does not seem to make much sense if we think of er/a as the distance d in the dipole
moment p = qd: the order of magnitude is then more like 0.1 or 0.2 fmor, say, 0.5 fm at most! Why?
Because these are the distances that come out of actual experiments, and they are also the values which
Di Sia used to get relevant nuclear binding energiesi.e. energy values in the MeV range. Have a look at
the dipole potential formula once more:

So what can we say about a right now? The e r/a in Yukawa’s formula must, somehow, correspond to the
d in the dipole formula, but we do not know how, yet.
[…]
So. Right. Nothing much. Just note a cannot be zero (we should avoid divisions by 0) and that the factor
becomes the e r/a = e r/a = 1/e ≈ 0.37 factor we encountered when discussing the exponential decay of
neutrons
31
for a = e 2.71which, from a mathematical viewpoint, would seem to a suitable fit for
aif we have to define one, that is, which we do not have to do right now! Indeed, unlike Aitchison and
Hey and others writing on the topic, we will not want to predefine a. We repeat:
The goal is to only find r but also a, and we will measure both in fm (1015 m).
Let us look at the structure of these two formulas once more. They are exactly the same, except for (1)
the e r/a function (which substitutes the distance between the two charges in the dipole potential
formula by a variable) and (2) the υ0 factor, which is usually forgotten. We explained the former (e r/a as
d), so let us now look at the latter (the proportionality constant with the nuclear permittivity υ0).
30
See the presentation and development of the Yukawa function in our previous paper: Moving charges,
electromagnetic waves, radiation, and near and far fields, December 2020. For a more authoritative graph of the
problem, see Fig. 28-6 in Feynman’s Lectures (II-28-6).
31
See footnote 23.
12
The nuclear permittivity factor υ0
We think we need it for the time being, at least to ensure the physical dimension of both sides of the
equation is the same. It is, therefore, similar to the physical dimension of the electric constant ε0
32
:
instead of C2/N·m2, we write: [υ0] = Y2/N·m2. It does the same trick as ε0 for the electrostatic Coulomb or
dipole potential it gives us a U(r) expressed in joule or N·m and it would, therefore, probably be a
mistake to leave it out.
Why? Well… We started off by saying that the idea of a nucleon charge is something new: we associate
some potential with it. However, we should not think of it as electrostatic charge. We have no positive
or negative charge, for example: all nucleons positive, negative, or neutral
33
share the same charge
and should attract each other by the same (strong) force. We, therefore, think we should define some
new unit for it.
Now, we thought of the Einstein, but that name is used for some other unit already.
34
In my previous
papers on the topic of the Yukawa potential
35
, we suggested the Yukawa but I now think there is too
much association between that name and the presumed unit of the Yukawa potential.
36
We, therefore,
propose the dirac.
37
However, for reasons of consistency we will continue to use the charge symbol we
used in previous papers: Y.
38
[…]
OK. Now that we have fixed the dimensions, what numerical value should we take for υ0? We have no
idea, but now that we are discussing these things in very much detail, should we wonder about the 4π
factor? Do we need it? It is common to both potentials (and to the forces, which we calculate in a
minute) because it is the 4π factor in the formulas for the surface area (4πr2) and the volume (4πr3) of a
sphere. Feynman often substitutes qe2/4πε0 by e2, which is a unit with a strange but exceedingly simple
physical dimension: the C2 in the numerator and denominator cancel out and we are left with N·m2 only,
which is great because we need the force to be expressed in newton, of course!
39
So we will do the same
32
Note that ε0 for ε0 = qe2/2hc since the revision to the 2019 revision of SI units here, which we think of as being
very significantmore significant than CERN’s experiments on testing the quark hypothesis or the Higgs field.
33
Negative? We only have neutrons and protons, don’t we? Yes, but we can imagine anti-atoms and, hence, anti-
protons. Protons and anti-protons will annihilate each other, but two anti-protons should stick together by the
same nuclear force.
34
Believe it or not, but the Einstein is defined as a one mole (6.022×1023) of photons. It is used, for example, when
discussing photosynthesis: we can then define the flux of light or the flux of photons, to be precise in terms of x
micro-einsteins per second per square meter. For more information, see the Wikipedia article on the Einstein as a
unit: https://en.wikipedia.org/wiki/Einstein_(unit). If we would truly want to honor Einstein, I would suggest we re-
define the Einstein as the unit of charge of the nucleon.
35
See: The nature of Yukawa’s force and charge and Who needs Yukawa’s wave equation? (June 2019). Our
treatment here is a shortened and revised version of Neutrinos as the photons of the strong force (October 2019).
The main revision consists of the use of gN and qe instead of gN2 and qe2 in the potential and force formulas.
36
The Wikipedia article on the Yukawa potential associates the 1/m unit with it, but that makes no sense
whatsoever to us.
37
We note that Dirac’s colleagues at Cambridge seem to have defined the dirac as ‘one word per hour’ but we
think there is no scope for confusion here.
38
This matches the upsilon (υ) it is not a μ (mu)! we use for the proportionality factor.
39
See, for example, Feynman’s calculation of the Bohr radius (a) using the p·a = h relationa rather precise
expression of the Uncertainty Principle, that is! Note that we will effectively get force formulas both for the
13
here and we hope the reader will be able to distinguish e2 and e (Euler’s number). Writing gN2/4πυ0 as N2
(again, we hope this causes not too much confusion in the mind of the reader!), we get the following
formulas now:

OK. Let us move on.
Force and force range calculations
If we have a potential, we can calculate the force. In fact, we should calculate the force, because we
should not be thinking in terms of terms of equating potentials here but in terms of equating forces.
40
To
do this, we should use this force formula:

 
 
 

Let us think about the minus signs here. The forces should be opposite, right? Right, but the magnitudes
should be the same and the formula takes care of that. Because we should really keep our wits with us
here, let us remind ourselves once again of what we are trying to do here:
We are thinking of two protons here, and these two protons carry an electric charge (qe) as well
as what we vaguely referred to as a nuclear charge (gN).
The electric charge pushes them away from each other, but the nucleon charge pulls them
together.
At some in-between point, the two forces should be equal but opposite. So we should find
some value for a force expressed in newton.
Yes. That should work. A force is a force, even if we know it acts on two different unit charges: qe versus
gN. We express one in Coulomb units, and the other in this new unit: the dirac. Sounds good?
Let us go through the calculations, then. The Coulomb force is easy to calculate:


 

This is just Coulomb’s Law, of course!
Coulomb as well as for the nuclear force with 1/r2 in the denominator, so we get something expressed in newton
alright!
40
The 1/r and er/a/r functions do not cross anyway, so we should not try to equate them. In fact, the 1/r2 and (r/a
+1)·er/a ]/r2 do not cross either! Note that we can compare forces only because the nucleon carries both electric
charge as well as nuclear charge. The associated fields are, therefore, different: newton/coulomb versus
newton/dirac, to be precise.
14
The calculation of the nuclear force is somewhat more complicated because of the er/a factor
41
:


 
 
 








This gives us the condition for the nuclear and electrostatic forces to be equal but opposite:



What can we do with that? We know what e2 is (we know what the electron charge is and we can,
therefore, calculate it), but what about N2? We have one equation and two unknowns here, so we
cannot calculate anything, right? Should we convert back to qe and gN? Not sure, but let us see if we get
something more meaningful by rewriting the condition above like this:



Eureka!
42
We know r must be equal to a if the two forces are equal, right? Right. And we know that is
the case when r = 2. So let us just forget about the dirac and the nuclear permittivity factor (that was
just to make you think) because we have the freedom to choose their units. We can then equate
gN2/4πυ0 and qe2/4πε0, and the condition above can then trivially be solved for a. Solve for a?
Yes. That is actually want we should do: remember we did not want to predefine a? But we should get
some value for it, right? Right:






That is very close to the mathematical value which we thought could make sense of Yukawa’s function:
Euler’s number e. The question is: does it make sense as an interproton distance?
We think it does! Indeed, the model offers a pretty plausible explanation of why typical nuclei ranging
from the deuteron nucleus which we studied here to more massive elements should be stable,
41
We need to take the derivative of a quotient of two functions here. Needless to say, we invite the reader to
carefully check all logic and double-check the calculations and if needed to email us their remarks and/or
corrections.
42
Archimedes is said to have exclaimed this in the bathtub when he found a way to distinguish fake from real gold
for the tyrant who paid him, but Scientific American thinks the story is fake news. We think Archimedes must have
had several aha moments. We do too.
15
although the graph below shows the formula for nuclear binding energies is not as straightforward as
you might think!
43
So, yes, OK. That is good enough for our purpose here, and that was to give you an explanation of the
nuclear force in terms of an electromagnetic rather than a nuclear force!
What more can we do? From the formula that we have found, we can also calculate the distance d
between the neutron proton-electron charges. It must be equal to:



Does this make sense as the distance between the electron and the proton inside of the neutron?
We think it does! In fact, we get very similar values to the internucleon and current radius values which
work in Di Sia’s analysis. This looks nice !
Unfortunately, when using a more complete equation for the dipole potential (see the Annex for the
derivation of Feynmans formula for it), we get reasonable separation distances again even better
values, actually but the associated energy values are ridiculously low. We must have done something
wrong !
Numerical example
Neutron and deep electron nearly cancel each other’s electrostatic potential by forming an electric
dipole. The reference frame is the equilibrium condition with a joint time-independent electrostatic and
dipole potential. Particles must have spin to explain dynamic equilibrium (oscillatory motion of the
proton and the dipole), which can probably be modeled by modeling the time rate of change of the
potential through the Lorenz gauge:


43
We borrow the graph from the Wikipedia article on Fe-56, which is probably the most stable element in the
periodic table. The same article tells us the size of the iron nucleus is about 4 proton diameters wideso that is a
lot more than our calculated 2.88 fm! Of course, there are more practical explanations for this than our theoretical
deuteron model. We will not go into the various formulas for calculating nuclear binding energy but LibreText
offers a nice explanation of nuclear (in)stability using magic numbers: we warmly recommend the read!
16
Static equilibrium assumes the sum of kinetic and potential energy is minimized and constant. More
than two potentials are probably present but the simplified model assumes all fields cancel each other.
Hence, if we have two force fields only, both forces must be equal and opposite. The electric dipole field,
which gives us the force on the unit charge (proton), is equal to:





We must first determine θ, because this will give us the axis along which we will want to calculate the
forces. The behavior of the  function is graphed below. The function
has only stable local minimum at θ = 1.081 rad π/3 = 30°. We should, therefore, only consider this line
and measure all distances r along this axis.
The p = qd and  factors are parameters. However, we may consider d to be a variable which is
determined by the system as well. We, therefore, write the dipole force function as:
 


From the two functions for Edipole, we gather that the distance d must be equal to:

The Coulomb force is given by:


17
Both forces will be opposite and equal if:






Is it possible that the separation distance between the neutron and electron would be equal to d
1.165, while the separation distance between the neutron dipole and the proton would be only 0.737?
The graph shows this as a solution and, hence, we suggest the reader reviews our calculations and try
other approaches.
The work that needs to be done to bring the proton from infinity to r = 0.737 in the dipole field can be
calculated as:
 

 




Likewise, the potential energy of the proton at a distance of 0.737 is equal to the work that must be
done to bring the proton from infinity to r = 0.737 in the (radial) Coulomb field. We calculate this
quantity as:
 

 



The difference yields:


 



  
This is a rather a nonsensical value. We must have done something wrong ! :-/
18
Conclusion
We showed how one can use potentials to build up a spin-zero model of the deuteron. As such, it may
complement Paolo Di Sia’s model of the nucleus (2018), which we gave due attention. In contrast to Di
Sia, we think of neutrons or the electron cloud that surrounds the proton inside as electric dipoles.
We did so by interpreting Yukawa’s potential function as a dipole potential. Instead of predefining the
range parameter a, we calculate it from the equilibrium condition (equal but opposite magnitudes of the
Coulomb and nuclear forces): we found a very acceptable value of about 2.88 fm for a, and an equally
acceptable value for the distance between the positively charged center of the neutron and the center
of the electron cloud which, in a deuteron nucleus, must shift it center of charge towards the proton so
as to ensure stability not unlike the sharing of valence electrons in chemical bonds.
Brussels, 31 December 2020
19
Annex: Force fields and higher-order potentials
Introduction
In one of our recent papers
44
, we introduced Feynman’s rather particular formulas for the electric and
magnetic fields of a point charge in motion
45
:








From these formulas, it is clear the charge goes from point (2’) to point (2) and its potential is measured
at point (1). Because no effect can travel faster than lightspeed, Feynman uses the concept of retarded
time and distance. Point (2’) is, therefore, given by a position vector x = x(t’) = x(t r’/c). The position at
point (1) itself is written as r = r’·er’ = r’·e2’1.
46
The retarded time t’ is, therefore, given by the t’ = t r’/c
equation. Also note that c·t’ = c·t r’ and r’ = c(t t’).
Figure 4: The concepts of retarded time, position, and distance (Feynman, II-21, Fig. 21-1)
Feynman likes his new expression: he writes that he “has not found anywhere in the published
literature, except in his own lectures” and that he worked it out around 1950 when thinking about how
to explain synchrotron radiation. The beauty of the formula is the very neat separation of the potential
in terms that vary with 1/r2, 1/r and 1/r0 = 1 respectively:
1. The first term (the retarded Coulomb field) falls off as 1/r2 and, therefore, follows the usual
inverse square law.
2. The second term, which Feynman thinks of as Nature compensating for the velocity v of the
charge), diminishes with distance following a simple 1/r’ potential
47
:
44
Moving charges, electromagnetic waves, radiation, and near and far fields, December 2020.
45
Illustration and formulas (including derivation) as per Feynman’s Lecture (II-21-6).
46
Conversely, we can write the retarded position x at point (2’) as the r’ = r’·er’ = r’·e2’1 = r’·e12 vector. We will
need to do so soon enough.
47
The expression assumes two like charges (both positive, or both negative) which, therefore, will repel each
other: the force has, therefore, a minus sign. The reference point for the potential of the second charge is zero at r
20
3. The third term does not diminish (we mean not at all here) with increasing distance r’ and is,
therefore, associated with radiation which, in turn, we associate with photons carrying an
energy E = h·f.
48
One should wonder here if higher-order derivatives of the velocity vector should be considered.
49
Let us
first think the other way around: can we imagine a force field following an inverse cube law? The answer
is: yes and no. Electric dipoles, for example, have at first sight, at least a potential that goes down
with 1/r’2. The relevant potential formula is
50
:
= (infinite distance between the two charges) and the potential at a finite distance r’ must, therefore, be
negative. It is always a bit of a headache to keep track of signs but if convention is followed, all comes out alright.
48
We write Planck’s quantum of (physical) action (Wirkung) as a vector quantity (h) in both expressions of the
Planck-Einstein relation: E = hf and h = pλ. The following remarks should be made here:
1. We assume an energy and/or momentum exchange between the field and the charge must also respect
the Planck-Einstein relation. Hence, besides energy, (linear and angular) momentum, we think physical
action must be conserved as well. The photon should, therefore, pack both the energy E = hf as well as
one unit of h = pλ = Eλ/c. We talk about the photon model in the introduction to our paper on moving
charges, waves, and radiation so we will limit ourselves here to some basic remarks.
2. The speed of light c gives us the direction of propagation of the photon. Writing the frequency as f should
not be too confusing if we think of angular frequency ω = 2πf as a vector too. Of course, we then also
need to write the phase θ as a vector too: θ = ω·t = (E/ħt, but this merely follows the convention to write
the unit vector as a vector too and thinking of the motion of the unit vector as having both a radial as well
as a transverse component as well.
3. Consistency in our concepts, then tells us we should either write the cycle time T or, else, the (linear)
wavelength of the photon as a vector too: c = λT = λ/f or c = λT = λ/f. This gives the h or (in reduced
form) ħ spin of photons (photons have no zero-spin state) a physical dimension. With c as the
propagation speed of the pointlike photon and, therefore, ω and ħ pointing in the same or the opposite
direction as c (depending on the plus or minus sign of the photon spin), we get: ω/c = (E/ħt/c =
(E/ħt/c = ω/c = (1/c)·(dθ/dt) ω/c = d(θ/c)/d(t).
4. The above shows that the phase θ and its time derivative dθ/dt are expressed in radians and radians per
second, respectively. Now, while we usually treat the radian as a distance unit, we may also express it in
equivalent time units by dividing both the angular frequency as well as the phase by c: m/(m/s). Note that
distance and time still have very different physical meanings. When using natural units (c = 1), distance
and time will effectively be measured in equivalent units, but a distance is and will always remain a
distance in 3D space: we can just measure it in equivalent time units because of the absolute velocity of
light c: T = 1/f = c/λ.
5. To sum it all up, we think of the photon as a pointlike electromagnetic oscillation traveling linearly
through space with a clock speed that is equal to ω = (E/ħt, a linear momentum that is equal to pλ =
Eλ/c, and an angular momentum that is equal to ħ = E/ω.
49
The first-order derivative of the velocity vector v is the acceleration vector a = dv/dt. The second-order
derivative is usually referred to as jerk (we do not like that word) or (our preference) jolt (j = da/dt = d2v/dt2).
Third- and higher-order derivatives carry names like snap, crackle, and pot (which are the cartoon mascots of
Kellogg's cereal Rice Krispies). Peter Thompson whose design of devices to control the moves of tools, printers,
amusement park rides, aircraft autopilots and telescope mounts thinks engineers are entitled easily remembered
mnemonics so we do not want to challenge the seriousness of these names.
50
See: Feynman’s Lectures, II-6-2 (the electric dipole). The formula is derived from (1) a calculation of the two
(opposite) potentials (this yields the 1/r function), (2) expanding the (d/2)2 term in the distance calculation using
the binomial expansion, (3) neglecting all terms with higher power than d2 (because d is supposed to be very
21





Because of the 1/r2 potential, we have a force field which will follow an inverse cube law.
However, this field will not be radial: it will be asymmetric.
An electric dipole field assumes a neutral charge ensemble (two charges +q and q separated by the
distance d), with θ measuring the angle between the axis of the dipole and the radius vector r’. Let us
analyze this geometrically. The magnitude of the force field E will then decrease as 1/r’3:








What about energy conservation here? The field will be non-radial and, therefore, asymmetric. For θ = 0
and θ = π/2 respectively, we have magnitudes equal to respectively:








This is consistent with writing the dipole potential in terms of the z-distance, which is zero for θ = π/2
and, therefore, for z = 0:



What are we doing wrong here? We only measured the z-component of the field Ez and forgot about the
transverse component E.
Let us work it all using Feynman’s illustration of the asymmetry of the field (Figure 5) and already give
you the relevant formula for the magnitude of the total field, which is given by the Pythagorean
Theorem:
small), and then (4) adding the two simplified potential functions (or, because of the opposite sign, subtracting
them). Assuming the distance d (i.e. the separation between the +q and q charge) is measured along the z-axis,
this yields the formula below, which can easily be rewritten as the formula we use above:



22
Figure 5: The electric field of a dipole (Feynman, II-6, Fig. 6-4)
Note that we have an (electro)static field here (the +q and +q charges do not move), there is actually no
need to use retarded distance and time: r = r, and so we will simplify notation for the time being by
dropping the prime (’) everywhere. We will use r’ and other retarded concepts soon enough again.
The asymmetric force field of an electric dipole
We can write p·cosθ as the vector dot product p·r = p·r·er = r·e21 = r·e12 when defining the dipole vector
p as a vector pointing from +q to q with magnitude p. The p·cosθ factor in the numerator can then be
written as p·cosθ = p·er, with r = r·er er = r/r. Hence, the electrostatic potential can also be written as:




This potential formula is valid for a dipole with any orientation and position r = (x, y, z). We can now
calculate the gradients in the x-, y- and z-directions by using our 


 formula.
The force field in the z-direction is directly given by the 

 formula:
















Now, when generalizing again from θ = 0 to any angle θ (so that is for any r with non-zero x- and y-
components, in other words), we can write z/r = cosθ z = r·cosθ. Bringing the minus sign into the
brackets, we get the following formula for the Ez component of the potential:



Feynman then calculates the x- and y-components of the force field as
51
:
51
We are intrigued by the detail, which we could not reproduce. Feynman must be using the Pythagorean
Theorem once more (r2 = x2 + y2 + z2) once more, but we are interested in the results here only and we, therefore,
assume Feynman made no mistake.
23


This, then, finally allows us to write the transverse component


Remembering the
formula and the

, we finally have the 1/r3 field
we thought we would get
52
:




We should get back to Feynman’s field equations and properly derive them from Maxwell’s equations.
Before we do so, we sum up our results so far:
1. The force field of a single charge has two terms which vary inversely with the distance according to
the 1/r and 1/r2, respectively. The energy conservation law tells us these fields must propagate planarly
and spherically, respectively.
2. The force field of a single charge also has a radiation term which does not depend on distance.
Energy here is conserved by the photon which carries the pointlike electromagnetic oscillation traveling
along a straight line with the speed of light (c)
3. The force field of a dipole follows an inverse cube law (E 1/r3) but is non-radial (asymmetric) and
the distribution of its intensity should, therefore, also respect the energy conservation law.
4. We do not immediately see any obvious 1/r4 or higher-order fields and/or potentials. However, the
formulas for the dipole field were based on a power expansion using the binomial theorem but an
approximation under which powers higher than d2 are not being considered. Hence, higher-order fields
and potentials are there but have not been modelled.
5. It is rather obvious that the superposition principle allows for any waveform and, therefore, any
dependence of fields on distance. Such fields will likely be asymmetrical so as to respect the energy
conservation principle.
6. In short, the energy conservation principle is the only real constraints we can put on the coefficients
when expanding a field (and the corresponding potential) into a power series.
52
The formula may, perhaps, be simplified further, but in this form it shows E(r) is not zero for θ = 0!
24
The graphs below show the various fields will define various zones where one or more force field will
dominate the other providing the signs of these force fields are opposite: if not, they just add all up to
create one joint force on any charge we would bring into the field. We showed how that works in our
rather primitive electromagnetic model of the deuteron nucleus.
53
Let us now study Feynman’s particular field equations more in detail.
Feynman’s field equations
Let us jot them down once more:








Let us now briefly re-discuss the dependence of these terms of the distance from the moving charge so
as to point out the rather remarkable properties of all these terms.
1. The first term (
) is the (retarded) Coulomb force field and follows the inverse square law. We
request the reader to also think about waveshapes by slowing moving point (1) around a sphere r = r’·er’
= r’·e2’1 centered around x(t r’/c).
54
We think of a spherical wave here. A spherical wave radiates
energy radiates energy in all directions (producing the spherical wave) and its intensity therefore
decreases in proportion to the squared distance from the object.
2. Feynman writes the second term (


as being (also) inversely proportional to the
squared distance r’2 but the r’ and r’2 in the numerator and denominator make us write it with a 1/c
coefficient and an 1/r’ factor only. However, Feynman prefers the

form because we can then
easily think of it compensation for the retardation delay, as it is the product of:
(i) the time rate of change of the retarded Coulomb field

multiplied by
(ii) the retardation delay: the time needed to travel the distance r’ at the speed of light c is
effectively equal to t t = r/c.
In other words, the first two terms correspond to computing the retarded Coulomb field and then
extrapolating it (linearly) toward the future by the amount r’/c which is right up to time t.
55
3. Finally, we have the third term. Feynman effectively writes the change in the retarded velocity vector
v = dx/dt, which is nothing but the retarded acceleration vector a = dv/dt = d2x/dt2 and, therefore, the
second-order derivative of x also in terms of the unit vector the unit vector er’ = e21 = e12’. Indeed,
53
An electromagnetic deuteron model, December 2020.
54
As mentioned in footnote 46, we are thinking here of the motion of x as the motion of r = r’·er’ = r’·e2’1 =
r’·e12’ centered around point (2’) = x(t r’/c).
55
We apologize for quoting quite literally from Feynman’s exposé here, but we could not find better language.
25
the third term in Feynman’s expansion of the retarded E vector the second-order derivative d2(er’)/dt2
is an acceleration vector which because of the origin of the unit vector er’ is fixed at point (2’) – can
and should also be analyzed as the sum of a transverse component and a radial component too.
56
Needless to say, this second-order derivative of d2(er’)/dt2 will be zero if the charge moves in a straight
line with constant velocity v. In other words, the third term will vanish (be zero) if there is no
acceleration. As mentioned above, this is why the third term is associated with radiation, which does not
vary with distance.
Another interpretation of the 1/r and 1/r2 dependence on distance?
We think there is another interpretation possible: Newton’s force law equates a force on a charge to the
time rate of change of the state of this charge. This state of motion is modeled by its momentum p = m·v
and we, therefore, write Newton’s force law as F = dp/dt. We know the relativistic mass concept can be
expanded into a power series using the binomial theorem
57
and we will probably want to compare those
1/2, 3/8,… coefficients in it with the coefficients of any second-, third- etc. order potentials ( we may
want to distinguish:

Multiplying with c2 and defining the kinetic energy as the difference between the total energy of the
charge and its rest mass (which should include potential energy), we get the relativistically correct
equation for the kinetic energy:

We can see it will be more convenient to write velocity v as a relative velocity β = v/c [0, 1]
58
and,
therefore, rewrite the mass and kinetic energy in equivalent units:






56
In the chapters where Feynman uses these equations (Vol. I, Chapters 28 and 29 as well as Vol. II, Chapter 21),
he assumes the transverse piece is far more important than the radial piece, but such statement crucially depends
on the assumption that the charge is moving at a more or less right angle to the line of sight, which is not
necessarily the case. Feynman corrects for this assumption in Chapter 34 of Vol. I, in which he gives the reader a
fuller treatment of the ‘relativistic effects in radiation’.
57
See: Wikipedia, Binomial theorem.
58
The β vector is a mathematical vector only (it has no physical dimension) because the (m/s)/(m/s) dimensions
cancel out.
26
If we then also measure time in equivalent distance units c·t = 299,792,458 meter (light-seconds)
59
, we
will write first-, second-, third- etc. order derivatives of the retarded position x = x(t’) = x(t r’/c) as
follows:









No surprises here: v is and remains a retarded velocity vector. However, we can now see we should
distinguish between the transverse and radial components of the velocity vector: v = v + v. We are
using polar coordinates here (the velocity vector and the line of sight form a plane) and we should,
therefore, distinguish between the radial and transverse components of de12’/dt
60
:


  
 
 
 
 

In the chapters where Feynman uses these equations (Vol. I, Chapters 28 and 29 as well as Vol. II,
Chapter 21), he assumes the transverse piece is far more important than the radial piece, but such
statement crucially depends on the assumption that the charge is moving at a more or less right angle to
the line of sight, which is not necessarily the case. Feynman corrects for this assumption in Chapter 34 of
Vol. I, in which he gives the reader a fuller treatment of the ‘relativistic effects in radiation’.
The magnetic field
In fact, both the second and third term in the expression for E(1, t) will, obviously, equal to zero if the
charge is not moving, in which case the charge comes with a static (i.e. non-varying in time) Coulomb
field only: in this case, the retarded field is then just the Coulomb field tout court.
However, when assuming a field that varies in time, we will have a magnetic force field, which Feynman
writes as: 
The vector cross product which gives us the magnetic field whose magnitude will be equal to the
product of:
(1) the magnitude of the unit vector er’, whose origin is (2’) and which points to (1) and whose
magnitude is equal to 1;
(2) the magnitude of the electric field vector at point (1) at time t;
(3) the sine of the angle between er’ and E(1, t).
Of course, a vector cross product will also involve the normal vector n, which will be orthogonal to the
plane formed by the vectors er’ and E(1, t). In addition, the Lorentz force is given by F = qE + qvB and
59
Since 1983, both distance and time units have been defined on the basis of the speed of light (c). The speed of
light is an absolute physical constant, just like the elementary charge ( qe) and Planck’s quantum of action (h),
which were defined as absolute physical constants in the 2019 revision of SI units only. This fixing of physical
dimensions should bring physics back to the road to reality.
60
We already made some remarks on Feynman’s assumptions in regard to the motion of the charge vis-à-vis the
line of sight. If x = r’·er’ = r’·e12’, then v = dx/dt = d(r’·er’)/dt = d(r’·e2’1)/dt = d(r’·e12’)/dt and the radial and
transverse piece of the velocity can effectively be written like we write it above.
27
the final calculation will, therefore, also take into account the velocity of the charge on which the
magnetic potential is acting and should, therefore, involve yet another normal vector and the right-hand
rule.
Needless to say, this business of Nature trying to anticipate and/or compensate for a retarded effect is
rather weird and we must, therefore, check Feynman’s derivation of his rather particular rendering of
the E and B field equations.
We will do so hereunder. In fact, our initial enthusiasm gave way to skepticism: we find Feynman’s
assumptions very specific and the general applicability of the field equations we started off with is,
therefore, rather questionable. At best, they are useful to effectively make sense of the concepts of the
near and far field, which is what we did in our paper.
Feynman’s derivation of Feynman’s field equations
Feynman’s derivation starts of the simplifying assumptions
61
:
1. The moving charge is a tiny blob of charge whose motion (v) is non-relativistic, and whose charge
density is equal to (x, y, z). The motion of the charge blob results in a current j(x, y, z) = v·(x, y, z).
62
This is depicted below (Figure 6).
Figure 6: The potential at (1) is given by integrals over the charge density (Feynman, II-21-4, Fig. 21-2)
2. While moving, the point charge is shaking up and down (wiggling) in very small motion which we
might hope might lead to a consistent interpretation of a particle as a ring current, but Feynman does
not introduce Planck’s quantum of action and, hence, one should not think of it as a Zitterbewegung
interpretation of charge.
63
Indeed, soon enough Feynman introduces the usual transient sinusoidal
61
We will refer the reader for most of the derivation to Feynman (II-21-4), but focus on comments that may or
may not be useful in light of the idea of modeling nuclear fields as combining electrostatic repulsive forces as well
as magnetic or electric dipole fields. As mentioned, our initial enthusiasm gave way to skepticism here.
62
The current definition here is very classical and is defined by taking the zero limit for a surface area: 

.
The physical dimension of IA is the ampere (C/s), while a surface area is expressed in m2. Hence, the j(x, y, z) =
v·(x, y, z) only yields the expected C/s dimension if and when the m2 dimension would shrinks to m. See the
Wikipedia article on current densities. In light of the need to keep track of energies, we would suggest a more
suitable definition of current in terms of energy per second. This yields a J/s = N·m/s dimension which would solve
this apparent dimensional problem.
63
See, for example: The meaning of uncertainty and the geometry of the wavefunction, October 2020.
28
motion, which should explain a drop-off of the radiation term and interprets atoms as transient
oscillators ().
Figure 7: The z-component of the vector potential (Az) from an oscillating dipole (Feynman, II-21-4, Fig. 21-3)
3. Finally, the assumption is that the charge is actually part of a neutral dipole. The wiggling charge q
must, therefore, be located near an equal and opposite charge at rest. One can think of the electron
cloud centered around a positively charged nucleus here or, what we tried to do, a neutron consisting of
a proton and a deep orbital electron. This gives rise to a derivation in which the first-order time
derivative of the dipole moment p can be written as p/t = qv.
Then follows a rather maddening series of calculations and derivations which we will not repeat here.
The electric field of an oscillating dipole should be time-dependent and, hence, the time rate of change
of the scalar potential in the Lorenz gauge cannot be zero:


However, Feynman derives the electric field from the magnetic field by using a gauge that may or may
not be equivalent to the Lorenz gauge:



More calculations then give the following result, which is supposed to be equivalent to the field
equations we started off with:






Monstrous? Yes. I now understand why Feynman’s representation of the field equations did not find
huge following.
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