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.

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

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:

Feyman's quarks Strangeness -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:

Aether but no Quarks charge -1 -1 +1 +1 state 0 +1 0 +1 net 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.

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.

The Nuclear Aether

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 bookPhysics 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.

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.

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 must have contemplated this possibility. He derives the quantitative result that m

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 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 11 Nuovo Cimento, v. LXV. Referring to the problem of Schott energy (discussed by the author at p. 97 of his book

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=r

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.

A | Z | A/Z | r_{o} | s_{o} |

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 (eg. Larmor, Phil. Mag., xliv (1897) p. 503 is but a suimple one) have formulated the energy of the electron of charge e and radius b as 2e

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 1.32Z

One is led to suspect that the hydrogen nucleus will be the same size as a positron, which makes 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.

Nuclear Bonds

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 positroins 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. FirstIy, 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 (below) 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.

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 e^{2}/3a. This is the conventional electrostatic energy of interaction between two electric charges e of radius a in contact. Since 2e^{2}/3a is mc^{2}, as applied to the electron, this energy unit is 0.75mc^{2}. 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.

For N particles, with N even, the total interaction energy is:

(N-1) + (N-2)/2 - (N-3) + .... 2/(N-2) - 1/(N-1)

which is -Nlog2, if N is large. If N is odd, the last term in the above series is positive and the summation, for N large, is 1-Nlog2. To find N we need to know how many particles are needed for the chain to span a distance d. d can be related to m by eliminating r from equations (4.1) and (6.60).

These are equations in the author's bookThen d/a is found using 2ePhysics without Einsteinpublished in 1969:

r = h/4πmc .... (4.1) hc/2πe ^{2}= 144π(r/d) ... (6.60)^{2}/3a=mc^{2}. 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 Energy Calculation N 168 169 170 -Nlog2 -116.45 -117.45 -117.83 Binding Energy (units) -116.45 -116.14 -117.83 Binding Energy (mc ^{2})-87.34 -87.11 -88.38 Add Self Energy (mc ^{2})168 169 170 Total Chain Energy 80.66 81.89 81.62 Ground State Correction 0.61 0.62 0.62 Corrected Energy (mc ^{2})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 amc^{2}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 -e^{2}/a or 1.5 times the energy unit mc^{2}. 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:

2e ^{2}/3a = mc^{2}... (7.8)

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, e^{2}divided by the sum of their radii, is 3c^{2}/2 times the product of their masses divided by the sum of their masses. Let M_{o}be the mass of the meson involved and M be the mass of the H particle. The surplus energy is then:

3M _{o}Mc^{2}/2(M_{o}+M) - M_{o}c^{2}... (7.9)

Starting from this basis, we will now seek to improve this 1969 account. Firstly, a very important advance emerges if we take the equation (7.9) from the text and find the solution which gives maximum surplus energy. Thus we put the expression at a minimum with M set at 1836m and M

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.

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.

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.

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 extremp- 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 hoiv 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 config- uration. 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.

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

June 30, 1974

H. Aspden

Acres High,

Hadrian Way,

Chilworth,

Southampton,

SO16 7HZ,

England.

Readers interested in this subject should take note that the above was published in 1974, one year before the author collaborated with Dr. D. M. Eagles as co-authors of a paper giving a definitive account of the theoretical evaluation of the proton-electron mass ratio. It appeared in