Copyright, Harold Aspden, 1974

This Essay was first published in 1974 by Sabberton Publications, P.O. Box 35, Southampton SO16 7RB, England.


The atomic nucleus is shown to have a form determined by the quantum structure of a Dirac-style vacuum. Nucleons occupy a series of holes in the structured vacuum forming a shell about a core region of unoccupied holes. These nucleons are linked by electron-positron chains. The lattice spacing can be related to the binding energy of the nucleus in precise quantitative terms. The special position of Fe 56 in the nuclear packing fraction curve is explained in terms of the cubic symmetry of the lattice system, the optimization of interaction energy with the core charge and the energy minimization of the chains.


This paper has been prompted by recent developments in elementary particle research having bearing upon a theory published in 1969. Chapter 7 of the author's work 'Physics without Einstein' incorporated some new ideas about nuclear structure. It was argued that nucleons are located at fixed lattice positions in a cubic structure and are linked by chains of electron-positron pairs. Each chain had association with what are now called partons. The mass deficit due to the negative interaction between a proton-sized parton and a pion-sized parton was deemed to balance the mass of the chain of electrons and positrons. Indeed, it was the energy of combination of these two heavy particles to form a nucleon at a nuclear lattice position which was the source of energy creating the electron-positron chain.

These ideas have progressed over the past five years and it is appropriate now to publish some of these developments. The author is indebted to Dr D M Eagles of the National Standards Laboratory, Sydney, Australia for helpful communications and encouragement. Dr Eagles recently drew to the author's attention a paper entitled 'Parton Chains in the Nucleus' by Wojciech Krolikowski, at p. 2922 of Physical Review D of 1 November 1973. It is this which has stimulated the publication here of some interesting advances of the chain nucleus theory at this stage of its development. The theory proposed offers scope for very detailed computational analysis of the structure of individual atomic nuclei.

A preliminary note about quark theory is appropriate before the structure of the atomic nucleus is analysed. This is important because it is the author's contention that the proton does, indeed, comprise three particles as demanded by quark theory. Such a structure of the proton was presented in 'Physics without Einstein' but in the form of a positive particle having the charge of the positron and associated with an electron-positron pair.


From a study of electron and neutrino scattering from protons Feynmann, writing in Science at p. 601 of the 15 February 1974 issue, has been able to show that protons have structure as if they comprise a plurality of particles of more fundamental nature, the so-called quarks. His paper entitled 'Structure of the Proton' has the introduction:
Protons are not fundamental particles but seem to be made of simpler elements called quarks. The evidence for this is given. But separated quarks have never been seen. A struggle to explain this seeming paradox may be leading us to a clearer view of the precise laws of the proton's structure and other phenomena of high energy physics.

Feynman explains how, on quark theory, there are three kinds of quark denoted u, d and s. The s and d quarks have charge -(1/3) and the u quark charge +(2/3) that of the positron. The s quark has higher mass than the d and u quarks which have the same mass. From this he presents a diagram showing how three quarks can combine to produce ten different particles:
Quark Theory
-3 sss
-2 ssd ssu
-1 sdd sdu suu
0 ddd ddu duu uuu
Charge -1 0 +1 +2

Now the unsatisfactory feature of quark theory is this concept that charge can be quantified in units which are one third or two thirds that of the electron or positron. It would be so much more satisfactory if Nature gave us a system of basic particles based exclusively upon charges which are measured in terms of the unit charge of the electron or positron. A little speculation shows how this is possible, provided we pay attention to some of the ideas presented to us by Dirac. It is well known that Dirac has proposed that the vacuum state is an aether permeated by quantum states filled by negative mass electrons. This implies that the vacuum has states with which particles can be associated and in which a negative charge of -1 electron units will pass undetected, being somehow neutralized by the vacuum medium. In these states the vacuum appears to add the charge +1. A particle can exist independently and not occupy such a state. Then we need add no charge to its own charge. On this basis, consider the following diagram:

Quark Theory
True Charge -1 -1 +1 +1
State 0 +1 0 +1
Effective Charge -1 0 +1 +2

Given a combination of three charges, each of which can be -1 or +1, and recognizing that stability criteria forbid three negative charges and three positive charges from combining, we must have a net charge of +1 or -1. Also, if we can have a free particle or one occupying a vacuum-polarized position, effectively adding +1, we see scope for four different charge entities. It follows that if the s, d and u quarks have charges +1 or -1, but masses as assumed on normal quark theory, we can have ten particles satisfying the observed charge system, but without recourse to the fractional charge features of quark theory.

It is therefore submitted that, since no experimental evidence exists supporting the fractionally-charged quarks but since experimental evidence does support other features of quark theory, then the alternative is to accept that some features of Dirac's aether theory need scrutiny.


Bernstein writing in Annals of Physics, 69, 1972, p. 19 has recently pointed out the need to incorporate 'holes' as constituents of an atomic nucleus. His reason is coupled with the explanation of energy levels and the inadequacies of the existing shell models. The approach we will take here is to examine the possibility of substituting nucleons for electrons in the Dirac continuum. We will presume a hole structure forms around the charge core of the nucleus and that the holes are occupied by negatively charged nucleons. This imparts mass to the nucleus but the charge of these nucleons is merged into the continuum. Interesting quantitative verification of this principle is available.

It is generally believed that an isolated electric charge will attract an equal charge of opposite polarity and so one imagines that two equal and opposite charges will pair together and form a neutral aggregation. Yet Earnshaw's theorem denies that two equal and opposite electric charges can rest adjacent one another in stable equilibrium unless they are immersed in an enveloping electrical medium. Dirac's continuum would, in effect, be such a medium. The observed vacuum polarization adjacent an atomic nucleus supports the exception also. Therefore charge neutralization should occur. Why then is the atomic nucleus itself not a neutral entity?

The answer is found from classical electrostatic theory. Laplace proved that the outward forces due to mutual interaction of a surface charge on a conductor are only half the forces exerted by the field on similar free charge just outside the surface. Thus, when an electron is added to the surface of a conductor to charge it, a free electron migrates from the atomic lattice system of the conductor and joins the added electron. Together the electrons form a surface charge just outside an inner charge of opposite polarity and half the magnitude. This latter is the residual charge left by the ionized lattice. This is a displacement phenomenon. The field on each electron is zero because the displaced electron has created positive and negative influences which cancel. The field away from the conductor is that due to the single added electron. In our atomic case, however, we have no displacement. Instead, a spherical shell of charge can centre upon a core of opposite polarity of half its strength and be held stable. A core of Ze charge can and will form a stable aggregation with a surrounding shell of -2Ze charge. If these added charges are not electrons but are negative nucleons then the atomic mass number A should be 2Z. If the nucleons are uniformly distributed over the volume of a sphere because they form in a structure of some kind then the same principles of Laplace apply except that a charge of -2.5Ze can be aggregated and held stable. This tells us to expect the ratio A/Z to increase from 2.0 to 2.5 as an atomic nucleus formed in shells increases in size.

In line with Bernstein's ideas we need to recognize that 'holes' are part of the nucleus. These cancel the effects of the nucleon charge. From another viewpoint we might say that space is pervaded by an electrically-neutral continuum which nevertheless contains discrete negative charges (electrons or the like) in a positively charged background continuum. Heavy negatively charged nucleons can occupy holes from which the negative charges are displaced. However, these nucleons tend to nucleate, if only by stronger gravitational effects, in regions immediately surrounding the atomic core charges Ze. Thus the atomic nucleus is formed, and it may have structural form characteristic of the properties of this pervading medium.

The analysis relating A and Z just presented has bearing upon nuclear stability. Z sets a limit upon the value of A, but one may expect the exact relationships to depend upon the structural links between the nucleons.

This concept has already been presented in the author's 1972 book 'Modern Aether Science'. The relevant part of chapter 14 of this work is reproduced below.


The physics of the aether is to many minds the physics of the nineteenth century. The twentieth century has so far been concerned with the physics of the atom and its quantum behaviour. Physics has assumed importance in industry primarily because electrical technology in the semiconductor field has become the province of the physicist rather than the electrical engineer. Also, physics has now an undeniable place of importance because everyone is all too aware of the energy hidden inside the atomic nucleus. For this reason the minds of many research physicists are technology-orientated. Theoretical physics is complicated, the aether is dead and who has the time anyway to be concerned with such an antiquated topic! The more open minded may say that if the aether has a place it is in cosmology; it is certainly not in the field of the nucleus. But let us see if we can dispel this belief.

Is there anything about the atomic nucleus we cannot explain? The atomic mass does not increment in proportion to the atomic charge. It seems that over a range of atoms of low atomic mass the number of nucleons is approximately twice that of the number of proton charge units in the nucleus. The nucleons comprise the protons and neutrons believed to form the nucleus. At high mass numbers the ratio of two increases roughly to about two and a half. An explanation of this would help our understanding of nuclear physics. Does the reader already have such an explanation? If not, perhaps the following analysis will have some appeal.

Consider an electric charge surrounded by a concentric uniform spherical distribution of discrete charges of opposite polarity. Now calculate the electrostatic interaction energy of such a system. This quantity will be found to be negative until the spherical charge distribution has a charge exactly double the magnitude of the central charge. Thereafter we would have positive interaction energy signifying instability, because the 'binding' energy associated with the negative polarity has ceased to 'bind'. We may expect, therefore, an entity to form as a stable aggregation in which the central charge acquires an enveloping double charge of opposite polarity, assuming the spherical distribution. If we consider instead a central charge with a uniform spatial charge distribution surrounding it, bounded by a sphere, then instability sets in when the surrounding charge is two and a half times that of the core. Between these two limiting examples, we could have, say, charge distributed in two concentric shells of unit and double unit radius, the charge content being proportional to the area of the spherical shell form. This gives a ratio of 2.166 for stability.

It needs little imagination to recognize the relevance of this to our nuclear problem. The atomic mass number is a measure of the number of negative nucleons clustered around a central core of charge. This charge has negligible mass compared with the nucleon mass contribution but the charge is the positive charge we regularly associate with the atomic nucleus. We need not speak of a combination of neutrons and protons to explain qualitatively the numerical difference between atomic number and atomic mass number. Somehow the charges of the nucleons are not detected, because we well know that the atomic electrons only react to the central charge. They ignore the nucleon charges just as they ignore charges in the aether medium. Indeed, the electrons may see these nucleon charges as they see the aether. In fact, the nucleons may be deemed to be arrayed in a structure and to have displaced negative aether charge so as to substitute themselves in the structured form of the aether itself. Their charge is undetected just as the mass of a buoyant body goes undetected in a fluid of equal mass density.

Hence, we need to invoke our aether. Also, we see support for the cubic lattice distribution of aether charge. An oxygen nucleus can be adequately populated by a single shell of discrete charges. There are 26 charges disposed in a regular cubic system about a central charge and 16 of these are presumably replaced by negative nucleons. The two to one ratio applies, because the oxygen atom has a atomic number of 8. Now take chromium, for example, which has an atomic number of 24. Here, we might expect charge to be distributed over another shell as well. The stability condition, calculated for idealized spherical distributions, requires 2.166 times as many nucleons as units of central charge. Hence an atomic mass number of 52, as is found. Similarly, for heavier atoms we find an appropriate relation between the two quantities conforming with this theory.

It has to be accepted from this that the nucleus consists of a central charge surrounded by a cluster of regularly spaced nucleons of negative charge. As the author has explained in his book Physics without Einstein, the nucleons form into a lattice structure with bonds joining the nucleons and, additionally, pions contributing to the energy of the bonds also derive their energy from an interaction with the nucleons. These features of the nucleus modify the mass and add some complication. Different isotopic forms may depend upon alternative structure configurations rendered possible by the different bond positions available. This is a matter for further analysis. When the above-mentioned book was published the author supposed the nucleons to be formed as a system of neutrons and protons, as is conventional. The later realization of the stable charge system introduced in this chapter, however, has led to a revision of the model. All the nucleons are the same. They are negative particles of mass approximating that of the proton.
The above text appears at pp. 139-141 of the book.

Contrary to established theory, the author's proposal is that the enveloping nucleons are neutralized by the occupancy of vacuum states. The mass of the atomic nucleus is essentially that of these neutralized nucleons and their related electron-positron chains.

Some recent experimental evidence from research at the Brookhaven National Laboratory was reported by Bugg et al in Physical Review Letters, 31, 1973 at p. 475. This research indicates an abnormally-high probability that a tenuous halo of neutrons may surround the central charge of the atomic nucleus. This seems to add support to the role of the vacuum state in compensating charge effects due to nucleons and gives strength to the author's ideas concerning a Dirac-style aether. Also encouraging is the reported activity of Lee and Wick of Columbia University in studying the effects of the properties of the vacuum upon the atomic nucleus. This is mentioned in Science at p. 51 of the 5 April 1974 issue.


It is interesting to digress to examine a recent proposal by Ross writing in Il Nuovo Cimento, 9A, May 1972 at p. 254. Ross interprets the muon as an electron orbited by a massless spin-1 wave and we will contrast this with a classical electron concept.

Ross has suggested that a particle might orbit the electron at its classical radius. By regarding the particle as having zero mass and applying the principles of General Relativity, Ross then shows that this orbit would be a null geodesic and is able to calculate the energy involved. Though at pains to show that the massless particle is not a normal photon, Ross rnust have contemplated this possibility. He derives the quantitative result that:
Mμ = Me[1+ 3/2α],
where alpha is the fine-structure constant. This gives the muon mass mμ as 206.554 times the electron mass me, in comparison with the observed ratio of 206.767. It is interesting then to note that had we regarded the electron as a mere sphere of electric charge of radius b and presumed a disturbance of some kind to ripple around it at this radius but at velocity c, we would have reason to derive a disturbance frequency of c/2πb. Multiplied by h this could represent energy, particularly if we are alive to the possibility that the mechanism of the photon may be involved in this model. Such energy, in mass terms, when added to the mass of the electron, gives a total mass of:
me[l + e2/αbmec2], since alpha is 2πe2/hc.
Then one can see by analogy with the Ross result that the muon mass could be derived with the same quantitative success if the rest mass energy of the electron were 2e2/3b. It is interesting then to note that this is exactly the rest mass energy found in classical works from the study of the electromagnetic properties of the electron.

The purpose of this is to show that we need not appeal to General Relativity to derive quantitative results in accord with Ross discovery. On the other hand Ross has come to his result by careful qualitative analysis and has argued that his muon should not affect the applicability of quantum electrodynamic theory. Our object in this paper is not to treat the problem of the muon, but rather to take the classical model of the electron and, guided by the quantitative result emerging from this analogy with the Ross speculations, examine how the classical model can be tailored to suit larger particle structures, particularly the atomic nucleus. We can be encouraged also by a statement made by Dirac in Scientific American in May 1963. He wrote:
I might mention a third picture with which I have been dealing lately. It involves departing from the picture of the electron as a point and thinking of it as a kind of sphere with a finite size.... the muon should be looked on as an excited electron. If the electron is a point, picturing how it can be excited becomes quite awkward.

The method of reverting to a physical model of the electron also takes strength from observations made by Grandy on the classical Lorentz-Dirac theory of electrodynamics. Grandy was writing at p. 738 of the February 1970 issue of Il Nuovo Cimento, v. LXV. Referring to the problem of Schott energy*, he said that an insight into its nature was outside the scope of classical electrodynamics and also that 'no relief is to be found in quantum electrodynamics, either, which is totally unable to account for the structure of the electron'. However, Grandy's comments about the impossibility of quantum electrodynamics helping an understanding of electron structure prevail, though this does not preclude the photon-electron interaction or combination to account for elementary particles or atomic nuclei.

* The problem of Schott energy has been discussed by the author at p. 97 of his book Modern Aether Science.

The muon can behave as an atomic nucleus. In muonium a positive muon replaces the proton in an ordinary hydrogen atom. Also, the muon can replace the electron in normal atoms. A study of such so-called exotic atoms is reported at p. 148 in the March 1972 issue of Physics Bulletin by Kim who refers to evidence of vacuum polarization effects and data showing that the charge radii of nuclei are given by R = roA1/3, where ro is approximately 1.2x10-13cm and A is mass number. It is standard to relate the radius with the mass number, but since we are referring to charge radii it is very interesting to examine more detailed data and perform a conversion putting R proportional Z1/3, where Z is the charge number. Such data is available from Condon and Odishaw's Handbook of Physics, 2nd Ed. at pp. 9-13. According to these data, the core appears spherical and the charge has a root mean square radius R given by the formula in ro, where ro ranges between 0.91 and 1.05 in units of 10-13 cm as A varies between 12 and 209. We may instead express R as soZ-13 cm to find that so would vary between 1.22 and 1.32x10-13 cm, a variation of less than 4% about the mean, in contrast with ro, which varies more than 7% about the mean.

These data show that it is better empirically to look for dependence upon Z rather than A. This may well be the outcome as better measurement data are forthcoming.

Dimensions of Atomic Nucleus
A Z A/Z ro so
12 6 2.00 1.05 1.32
40 20 2.00 1.02 1.28
51 23 2.22 0.97 1.26
115 49 2.35 0.92 1.22
122 51 2.39 0.93 1.24
197 79 2.49 0.91 1.27
209 83 2.51 0.93 1.27

Numerous writers* have formulated the energy of the electron of charge e and radius b as 2e2/3b. In the author's book 'Physics without Einstein' it is shown at p. 209 that this indicates a uniform field within the radius b and corresponds to a charge density in cgs. units of e/2(pi)xb2 at radius x. The root mean square radius of such a charge distribution is b/(2)1/2. The value of b calculable from the rest mass energy of the electron 8.2x10-7 ergs, and the value of e of 4.8x10-10 esu, is 1.87x10-13 cm and its root mean square is 1.32x10-13 cm. There is a remarkable comparability between this electron radius and so particularly for smaller Z values.

* Larmor, Phil. Mag. , xliv (1897) p. 503 is but one example.

It seems obvious from this that if we take the classical formula given above for the size of the electron and then apply this also to the positron we have only to conceive the charged core of an atomic nucleus as an aggregation of Z positrons occupying the same volume as Z separate positrons and the root-mean-square radius of the resulting core is 1.32xZ1/310-13 cm. This fits the experimental data quite well.

One is led to suspect that the hydrogen nucleus will be the same size as a positron, which makes the Ross observations about the nature of the muon all the more intriguing. However, accepting the empirical implications just presented, there is need for caution in interpretation. One may wonder how the inner electrons screening the atomic nucleus really escape involvement with the measurement of the core radius.

Collectively the majority of the electrons associated with the atomic nucleus happen to exhibit an aggregate volume of just the right order to conform with the measurements of core size.

The interesting feature of the analysis is the applicability of the classical formula for the size of an electric charge. Also, the table above indicates a relationship between A and Z such that as Z increases A/Z varies from 2 to a value close to 2.5. This satisfies the theoretical proposal already made.


It is appropriate to reproduce next an extract from the author's 'Physics without Einstein', noting that some of the views expressed are subject to modification below. The text preceding this material involved a rigorous analysis of the structure of the vacuum and the computation of a lattice dimension d, which was found to be 6.37x10-10 cm. It is also noted that since that work was published, Dr D M Eagles and Dr C H Burton have made careful calculations using the computation facilities of CSIRO in Australia and the results reported in Physics Letters at p. 423 of the 23 October 1972 issue support the value just given for the lattice dimension d of the likely aether structure.

Nuclear Bonds

Fig. 7.8
What is the form of the nuclear bonds? Each of the six nucleons in Fig. 7.8, three protons, say, and three neutrons, identified by the full bodied circles, has a bond of its own providing one of the links. These bonds are the real mystery of the atomic nucleus. We will assume that their most logical form is merely a chain of electrons and positrons arranged alternately in a straight line. The reason for the assumption is that electron-positron pairs are readily formed in conjunction with matter, and we have seen how an in-line configuration of alternate positive and negative particles has proved so helpful in understanding the deuteron. Stability has to be explained. Firstly, the chain is held together by the mutually attractive forces between touching electrons and positrons. Secondly, it will be stable if the ends of the chain are held in fixed relationship. This is assured by the location of the nucleons which these bonds interconnect. In Fig. 7.9 it is shown how the bonds connect with the basic particles. In the examples shown, the nucleons are positioned with a chain on either side and are deemed to be spinning about the axis of the chain. Intrinsic spin of the chain elements will not be considered. It cancels as far as observation is concerned because each electron in the chain is balanced by a positron. In Fig. 7.10 it is shown how, for the neutron, for example, the spin can be in a direction different from that of the chain. Also, it is shown how another chain may couple at right angles with this one including the neutron. Note, that the end electron or positron of the chain does not need to link exactly with the nucleon. Therefore, it need not interfere with the spin.

Fig. 7.9
We will now calculate the energy of a chain of electrons and positrons. For the purpose of the analysis we will define a standard energy unit as e2/2a. This is the conventional electrostatic energy of interaction between two electric charges e of radius a and in contact. Since 2e2/3a is mc2, as applied to the electron, this energy unit is 0.75 mc2. On this basis a chain of two particles has a binding energy of -1 unit. If there are three particles the binding energy is the sum of -1, 1/2 and -1, since the two outermost particles are of opposite polarity and their centres are at a spacing of 4a and not 2a.

Fig. 7.10
For N particles, with N even, the total interaction energy is:
-(N-1) + (N-2)/2 - (N-3)/3 + .... 2/(N-2) - 1/(N-1)
which is -Nlog 2, if N is large. If N is odd, the last term in the above series is positive and the summation, for N large, is I - Nlog 2. To find N we need to know how many particles are needed for the chain to span a distance d. This distance d can be related to m by eliminating r from equation (4.1) (in 'Physics without Einstein'), namely:
r = h/4πmc
combined with equation (6.60), namely:
hc/2πe2 = 144π(r/d)
Then d/2a is found using 2e2/3a = mc2. It is 54π, so N may be, say, 169, 170 or possibly 168, particularly if N has to be even and there has to be space for any nucleons. For our analysis we will calculate the binding energy of the chain and the actual total energy of the chain for all three of these values of N. The data are summarized in the following table.

Chain Binding Energy
N 168 169 170
-N log2 -116.45 -117.14 -117.83
Binding Energy (units) -116.45 -116.14 -117.83
Binding Energy (mc2) -87.34 -87.11 -88.38
Add Self Energy (mc2 168 169 170
Total Chain Energy 80.66 81.89 81.62
Ground State Correction 0.61 0.62 0.62
Corrected Energy (mc2) 81.27 82.51 82.24

In the above table the binding energy has been set against the self energy of the basic particles and a correction has been applied of αmc2 per pair of particles to adjust for the fact that mass is not referenced on separation to infinity, as was discussed earlier in this chapter. The total mass energy of the chain is seen to be about 81 or 82 electron mass energy units, depending upon its exact length.

This shows that while the electron-positron chain proposed will provide a real bond between nucleons linked together to form an atomic nucleus, it will nevertheless add a mass of some 81m per nucleon. This seems far too high to apply to the measured binding energies. Furthermore, it is positive and the nature of binding energy is that it must be negative. This can be explained by introducing the π-meson or pion, as it is otherwise termed.

The Pion

When an electron becomes attached to a small but heavy particle of charge e, the interaction energy is very nearly -e2/a or 1.5 times the energy unit mc2. This means that the mass of the heavy particle is effectively reduced when an electron attaches itself to it and becomes integral with it. If we go further and seek to find the smallest particle which can attach itself to a heavy nucleon to provide enough surplus energy to form one of the above-mentioned electron-positron chains, we can see how this nucleon plus this particle plus this chain can have an aggregate mass little different from that of the initial nucleon. This can reconcile our difficulties. The fact that an electron can release the equivalent of about half its own mass indicates that to form the chain of mass 81m we will need a meson-sized particle of the order of mass of the muon or pion. To calculate the exact requirement we restate the inverse relationship between the mass m of a particle of charge e and its radius a:
2e2/3a = mc2
This applies to the electron, but it can also be used for other particles such as the meson and the H particle. It may then be shown that if two particles of opposite polarity charge e are in contact, their binding energy, e2 divided by the sum of their radii, is 3c2/2 times the product of their masses divided by the sum of their masses. Let Mo be the mass of the meson involved and M be the mass of the H particle. The following table then shows the value of the surplus energy Es:
3MoMc2/2(Mo+M) - Moc2
in terms of units of mc2, for different values of Mo/M and a value of M of 1836m.

Meson Energy of H particle
Mo/m *** Es
230 76.4
240 78.3
250 80.0
260 81.5
270 83.0
280 84.5
The above-quoted text appears at pp. 147-150 of 'Physics without Einstein'.
Starting from this basis, we will now seek to improve this 1969 account. Firstly, a very important advance emerges if we take the latter equation and find the solution which gives maximum surplus energy. Thus we put the expression at a minimum with M set at 1836m and Mo variable. Simple analysis then shows that for this condition Mo is M(3/2)1/2 - M or 0.225 M or 413m. This is higher than the pion mass contemplated above. The energy released is found to be (0.225)2M or 93m. Thus subtracting the chain energy of about 81m we find that each chain together with the parton pair represented by that equation will contribute mass some 12 electron units m less than that of the proton.

If our atomic nucleus comprised simple chain bonds and had one per nucleon we should find that the mass of a nucleus would be 1824 times the number of nucleons when measured in terms of electron mass units. In fact this mass varies. As the number of nucleons increases the unit mass rapidly decreases through a minimum of about 1820 for iron and then rises gradually until it is 1823 for the largest nucleus.

There is a very interesting explanation for this effect. Note that the energy of a chain is proportional to its length. Then ask how three nucleons arranged as below can be linked by chains. Three configurations are shown in Fig. 1.
Fig. 1(a)
Fig. 1(b)
Fig. 1(c)

We now assume that the configuration adopted will be that of minimum energy, that is minimum total chain length. Simple analysis shows that 2x+y can be less than 2d. The minimum value is 1.933d when z is approximately 0.2d. This means that at the corner of the nuclear lattice the energy of a normal chain of length d is effectively reduced to 0.967 of its normal value, that is, from 81m to 78m. There is a decrease of three electron mass units whenever a chain is able to cut a corner so to speak as in Fig. 1(b).

Now consider a nucleus of iron and let us suppose that the charge of the nucleus is due to 26 vacancies in the vacuum structure, an absence of 26 electron-sized charges which normally neutralize the vacuum state. This core will be surrounded by nucleons occupying other lattice sites, 56 in number. Now note that a 3 by 3 by 3 array of a cubic lattice system comprises 27 sites and that there are 6 faces to this cubic array each having a 3 by 3 array in adjacent lattice planes. This is 54 sites. We thus see how iron can be close to an optimum state of symmetry. Also note how most of these 54 sites are associated with a chain of minimum energy. This is evident from Fig. 2.

Fig. 2

It seems likely that in the iron nucleus of atomic mass number 56 there are 6 arrays of 8 nucleons as depicted in Fig. 2 and that four of these arrays have, as illustrated in Fig. 3., central nucleons linked both to a nucleon in an outer lattice position and to one of the nucleons at P in Fig. 2.

Fig. 3

In every respect, therefore, iron with an atomic mass number of 56 is the nucleus for which every chain is at the low energy. Hence it is not surprising that it appears to be a most stable nucleus. Also, our theory has shown the unit mass to be three electron rrass units below the extreme of having all chains lie on the lattice lines. Such an arrangement can be expected to be more nearly applicable in very large nuclei where multiple shells of nucleons exist and we have seen how such large nuclei have a unit mass higher by three electron masses.

But it is of interest to ask about the Helium 4 nucleus. This appears to have four normal chains in its most natural configuration. The unit mass of the Helium 4 nucleus is about 12. 5 electron mass units below that of the proton. This compares with the figure of 12m deduced on the basis of the chain energy of 81m.


From such analysis it is concluded that we are arriving at results which encourage rigorous calculation of detailed structure. The fact that the value of 80.5m is indicated from the Helium nucleus as the mass contribution of a chain of standard lattice length checks very well indeed with the data given in the reproduced tabulation from the author's 'Physics without Einstein'. By analysing the atomic nucleus and the dependence of its mass upon its size we can deduce the lattice dimensions of the structured vacuum state and check a theory which has independently afforded an exact evaluation of Planck's constant, as reference to the above-mentioned paper by Eagles will show, and an exact evaluation of the Constant of Gravitation. For the latter refer to the full text of 'Physics without Einstein' or a new work 'Gravitation' due to be published by the same author early in 1975.

The author is, of course, interested in any work which may advance the ideas presented above and invites correspondence.

June 30, 1974 H. Aspden
This concludes the text of that 1974 paper. It will be of interest to some readers to examine how that theory has evolved from that point, especially in respect of the development of the theory of the proton from 1975 onwards.
Harold Aspden
27 May, 1998