- Addendum: Nuclear Structure From Deuteron Pairing
- Addendum: Excited Nuclear Energy Levels
- Addendum: Particle and Dark Matter Bremsstrahlung
- Addendum: Neutron Half-life Discrepancy
- Addendum: Antigravity from Deuteron Oscillation
Addendum: Nuclear Structure From Deuteron Pairing
· Deuteron (H-2), alpha particle (He-4), Helion (He-3)
· Lithium-6 (Li-6), Triton (H-3), Lithium-7 (Li-7)
· Helium-5 (He-5), Beryllium-9 (Be-9), Beryllium-10 (Be-10)
· Boron-10 (B-10), Boron-11 (B-11)
· Carbon-12 (C-12), Carbon-13 (C-13), Carbon-14 (C-14)
· Nitrogen-14 (N-14)
· Oxygen-16 (O-16)
Quantum field theory models the strong force binding the deuteron as a sum of two main components: the S- and D- waves. The former is spherical, representing the deuteron in a ground state. It represents 96% of nuclear structure with the proton and neutron having no relative orbital angular momentum.
The D-wave component is a tensor arising from a state where the nucleons do have relative orbital angular momentum. It explains the remaining 4%.
Measurement reveals that the deuteron possesses a non-zero electric quadrupole moment (.286 e-fm2). Its charge distribution is ellipsoidal. This shape is attributed entirely to the small D-wave component.
With our view of deuteron structure, the charge distribution results from the alternating expression of each nucleon as a proton. We see the charge field as the average of these two alternating charge configurations. It does not arise from relative nucleon orbital angular momentum. The global clock rate implies the field we measure is an average, taken over a sampling frequency on the order of sextillion Hz.
From dynamical symmetry, the deuteron internucleon distance is fixed. In effect, the deuteron charge appears equally split between two foci. The result is an ellipsoidal charge distribution, matching measurements of the deuteron electric quadrupole moment. In our view, this stable ellipsoidal charge distribution arises from a symmetrical, time-averaged charge dipole.
Can our model of the deuteron, which we visualize as an oscillating proton-neutron pair, tell us anything about the structure of larger atomic nuclei?
Within a deuteron, nucleon pairing occurs as the alternating expression from one nucleon to the other of an electron and an antineutrino. These latter two components distinguish a neutron from a proton. We visualized how the equivalent quark structures of the proton and neutron provides the framework for this oscillation.
To apply this view to nucleons bound within larger nuclei, we must restrict our model to interactions like deuteron pairing. These are termed ‘low-level’ energy interactions.
For example, the experimental effects observed with high energy lepton probes are outside our model scope. Such probes don’t just alter nuclear structure, they alter the expression of mass for individual nucleons. Most of a nucleon’s static mass occurs in quarks, which we model as local entanglement energy.
When we considered quark formation, we saw how conservation of energy in the Vacuum Network (VN) requires particles of high mass to adopt a ‘lobed’ entanglement structure. In doing so, they express quarks.
The point-wise projection of quark geometry in the VN increases the effective surface area available to project local entanglement energy. Locally entangled QESTs express static mass. Without this geometric change, energy would be isolated in local spacetimes across ‘tick’s of the global entanglement clock. This would violate conservation of energy of particles localized to the VN.
High energy lepton probes targeting larger nuclei result in deep inelastic scattering. The probes momentarily alter the quark structure of target protons or neutrons, resulting in unstable hadrons (much as we visualized with neutral pion decay). Such high energy quark level interactions are outside the range of energies we want to consider here.
Lower energy scattering experiments with larger nuclei provide evidence of internal structure, binding energy levels, and momentum distributions. How much of this evidence can we explain with our model of the deuteron?
We visualized the ‘weak’ force binding a proton and neutron as the magnetic attraction produced by moving charge, mediated by quantum foam.
In our model of the quantum forest, quantum foam lacks a coherent metric, so magnetic fields propagate instantaneously. A momentary magnetic field is produced by the ‘tick’ to ‘tick’ projection of an electron, alternating between the two nucleons.
With each clock ‘tick’, the electron establishes an orbital about one up quark in each nucleon. In the neutron, this alters the charge expression of an up quark to down. In the proton, it simultaneously restores the charge of the up quark. This effect produces an effective nuclear current flow with a corresponding momentary magnetic field. The neutron’s antineutrino is transparent to this magnetic flux.
Through ‘deuteron pairing’, a proton is altered to a neutron. Symmetrically, and simultaneously, the neutron ‘becomes’ a proton. This ‘tick’ to ‘tick’ interaction is our view of the ‘weak’ force that binds nucleons. It is the main nuclear binding force that we will be considering. Breaking this bond requires energy of about 2.23 MeV.
To enforce this energy constraint, we shall restrict our model to stable nuclei where all nucleons participate in deuteron pairing. We consider one notable exception: Helion, the nucleus of He-3. This a stable but rare isotope of helium that features prominently in fusion research. We shall begin with deuteron, the simplest example of nucleon pairing.
If we label the momentary expression of the neutron and proton forming deuteron as n1 and p2, we can trace the path of electron projection: (n1,p2)(p2,n1)(n1,p2)… This notation helps us track what is effectively a nuclear current flow, expressed on up quarks across alternate nucleons.
All nucleons paired in this manner oscillate their proton-neutron expression with perfect synchrony, driven by the global entanglement clock. Calculation of the deuteron’s rate of rotation from its small moment of inertia (obtained by viewing oscillation as rotation) suggests a clock rate on the order of sextillion Hz.
We will look for symmetries in nucleon configuration, and regard nuclear structure as the average over an immense sample. Scattering experiments reveal this as a coherent target structure.
For deuteron, the scattering radius (2.13 fm), which is more than twice that of a proton (.84 fm). This indicates the neutron is held (on average) well outside the charge radius of the proton.
We visualize this greater distance as a balance between a reaction force produced by momentum transfer, and local magnetic field attraction. With this view, we expect the inter-nucleon distance to not oscillate. As deuteron pair oscillation is perfectly symmetrical, in the absence of other forces there can be no asymmetry in nucleon separation.
How far can we go in visualizing larger nuclei? Can we infer anything about structure, such as charge radius, or binding energy, and momentum distribution?
As a start, we can see that there can be few stable nuclei hosting more protons than neutrons. With each clock ‘tick’, all protons are involved in stable deuteron pairing. Exposing an individual proton’s static charge across multiple clock ‘tick’s will make the host nucleus unstable from the force of Coulomb repulsion.
The locally attractive force binding protons to neutrons in pairs is not sufficient to bind a ‘free’ proton (that is, one not involved in pairing). Helion is the notable exception we will consider in more detail, later.
After deuteron, the next larger nucleus we consider is triton, with two neutrons and one proton. Unlike deuteron, triton is unstable. The source of this instability will help us in visualizing the instability of larger nuclei. Dynamic magnetic force interaction within neutrinos is the main contributing factor. While it stabilizes neutron half-life, it is not as stable an interaction as deuteron pairing.
We move on to the next stable nucleus, the alpha particle, comprised of 2 protons and 2 neutrons.
We will use the same notation as with deuteron to track electron flow. All nucleons participate in a cycle of pairing, producing a complete chain (denoted by {}), over 4 clock ‘tick’s ():
{ (n1,p2,n3,p4)(p2,n1,p4,n3)(n3,p4,n1,p2)(p4,n3,p2,n1) } { (n1,p4,n3,p2)… } …
The fifth configuration in the chain is the same as the first, from the perspective of the implied electron current. We visualize the chain as a repeating cycle of ‘deuteron pairing’.
All nucleons oscillate in complete synchrony, with a clock rate on the order of sextillion Hz. Distances between pairs may be flexible, but within pair distances are more rigid. We imagine the coherent charge and spatial structure of an alpha particle emerging as the average over the repeating chain.
Evidence of this symmetric oscillation is provided by calculation of the alpha particle’s rate of rotation from its small moment of inertia. It is extremely large, of the same order as the deuteron.
The charge radius of an alpha particle (1.68 fm) is very close to twice that of a proton, and much less than an individual deuteron pair (2.13 fm). This suggests that the two ‘momentary’, relatively rigid deuterons within the chain dynamically orient to maximize n-p interaction, over each cycle. The same momentum transfer that gives rise to large n-p separation distance contributes to closer p-p proximity.
The alpha particle has 0 net magnetic moment. This implies a geometric symmetry in the synchronous expression of its two component deuteron pairs. They must align so that their individual magnetic moments are oriented in precise opposition. We can visualize this as the pairs lying parallel in a plane, with a neutron from one pair always in opposition to a proton in the other. This results in a rotation of 180 degrees of one magnetic field relative to the other. With each 'tick' of the entanglement clock, the entire configuration rotates 180 degrees.
We can turn now to the helion, a nucleus that appears to be an exception to our model of ‘deuteron chaining’. It possesses two protons, but only one neutron. But our model constraint still applies: all nucleons participate in pairing.
As only one neutron is available, only one pair is expressed with each clock ‘tick’. We imagine a repeating 4-‘tick’ chain:
{(n1,p2,p3)(p2,n1,p3)(p2,p3,n1)(p2,n1,p3)} {(n1,p2,p3)…
In terms of inter-proton distances, the 4 repeating combinations may be written:
{(p-p)(p-dp)(p-p)(p-dp)}
where (p-p) is the distance between a ‘free’ proton and proximal paired proton, and (p-dp) is the distance between a ‘free’ and distal paired proton. We can estimate (p-p) as .84+.84=1.68 fm, and (p-dp) as .84 + 2.13 =2.97 fm.
Overall, we estimate the average proton separation is 2.33 fm. The measured scattering radius is 1.97 fm. Our estimate is significantly higher.
The (p-p) distance is not likely to be smaller than 1.68 fm. This suggests our estimate of the effective (p-dp) distance is too high. What might draw the ‘free’ proton closer to the proton bound by deuteron pairing?
The answer may be the effect of the neutron’s negative magnetic dipole, which is strongly opposite to that of the proton. For deuteron, the magnetic moment is approximately their vector sum. But for helion, the magnetic moment is negative, not more positive, as the additional proton might imply.
This is evidence that the ‘free’ proton is not bound by deuteron pairing. It is momentarily ‘flipped’ when not paired.
A strong magnetic flux must be produced from the dynamics of nucleons involved in deuteron pairing. The proton to neutron transition would produce a strongly negative momentary flux. This may be sufficient to flip the ‘free’ proton, resulting in the net negative flux of the helion. It may also draw the ‘free’ proton closer to the deuteron, reducing the effective (p-dp) distance.
The next stable nucleus to consider is Lithium-6, possessing a balance of 3 protons and neutrons. We again apply our model constraint: all nucleons are involved in deuteron pairing.
Li-6 has a scattering radius of 2.39 fm and a magnetic moment approximately equal to that of deuteron. We can visualize how this will result if all the protons from the three deuteron pairs simultaneously align in a plane. Their centers form an equilateral triangle in that plane. The neutron centers also form an equilateral triangle in the plane but rotated by 45 degrees.
The magnetic moments of the three deuteron pairs will assume a low energy configuration with two pointing out of the plan, and one into the plane.
With each ‘tick’ of the entanglement clock, the proton-neutron expression alternates. The net result is a balanced net magnetic moment equivalent to one deuteron, pointing out of the plane.
The scattering radius expressed by deuteron oscillation in the plane is almost perfectly circular. This would result from a circular charge with twice the radius of a single proton, producing four times the scattering radius.
Viewed ‘edge on’, the scattering radius is slightly less than that of two protons. On average, oscillation will produce a radius somewhat less than three times that of the proton (3x.84 = 2.52 fm). The observed value is 2.39 fm, in very good agreement.
Li-6 will show a very marked anisotropy in scattering radius (by a factor at least two), when oriented in a magnetic field. Anisotropy has been observed in experimental measurements.
If we return now to tritium, we can examine the simplest scenario for nuclear Beta decay. Triton, the nucleus of tritium, has one proton and two neutrons, and is radioactive with a half life of 12.3 years. Its Beta decay products are He-3, an electron and an antineutrino.
We recognize similar products from neutron Beta decay, with a single proton in place of the He-3 nucleus. In contrast to triton, the neutron half-life is on the order of 14 minutes. We will see that the dynamic structure of triton contributes to its relative stability.
Our model is again limited to deuteron chaining between the three nucleons. With only one proton, only a single deuteron pair is expressed with each ‘tick’ of the entanglement clock. A single deuteron pairing circulates in a cycle over the three nucleons. It forms a repeating chain from three consecutive pairings.
Each nucleon is expressed as a ’free’ neutron in one out of the three configurations. We visualize each pairing resulting in an identical configuration: roughly the shape of an isosceles triangle with the n-p pair as the base. All configurations lie in the same plane. Through oscillation, they express an apparent rotation. We would predict a very high rate for this apparent rotation, in view of the low moment of inertia.
Triton has a smaller scattering radius (1.60 fm) than deuteron (2.13 fm). This is very close to twice the radius of a single proton. The three nucleons must be held more closely than in deuteron, suggesting the pairing length is shortened.
Pairing length arises from a balance of energy transfer. The balance is between momentum transferred to the proton and an opposing attractive force from strong magnetic flux. The shorter length implies that less energy is momentarily transferred. The difference must be dynamically ‘stored’ with the ‘free’ neutron.
A revealing property is triton’s large positive magnetic moment (2.98 nM), which is more than three times that of deuteron (.86 nM). Clearly the magnetic moment of triton is not a vector sum of its components (2.8 nM – 2x1.9 nM).
As we saw with helion, a strong momentary magnetic flux from deuteron pairing must be ‘flipping’ the magnetic moment of a ‘free’ nucleon. For helion, the ‘free’ nucleon was a proton. For triton, it is a neutron.
The magnetic moment of neutron is only -1.9 nM. To produce the large positive moment of triton, 2.98 nM, the neutron must not only be ‘flipped’, but its magnetic field must be enhanced.
To transform the magnetic field of a proton into that of a neutron, the electron in orbit about a neutron’s up quark expresses a very strong flux. We can visualize how the extra energy from triton’s shortened deuteron pairing can enhance this orbital flux. In doing so, it stabilizes the ‘free’ neutron.
The net result leaves triton with a stronger positive flux than a proton (2.98 nM vs 2.8 nM). It is the vector sum of deuteron (.86 nM) and the momentarily ‘flipped’ neutron with its enhanced orbital electron flux (> 1.9 nM).
The “stabilized” free neutron contributes to a longer neutron half-life, when bound within triton. This is offset somewhat by triton’s lower dynamical symmetry, arising from the asymmetry in its deuteron pairing cycle. Together these effects contribute to triton’s relatively short half-life of approx. 12.3 years.
We can combine our models of triton and Li-6, and try to visualize the structure of Li-7.
The latter nucleus has 3 protons and 4 neutrons. It differs little from Li-6 in scattering radius, but markedly in magnetic moment: 3.26 nM compared to .85 nM for Li-6.
Considering the effect of the 'free' neutron in triton, we can picture Li-7 with a similar triple deuteron pairing ring structure as Li-6, but hosting a central neutron.
With each cycle of deuteron pairing, the central neutron engages in the same magnetic flux interaction that we saw with triton. This 'flips' the neutron's magnetic moment, and shortens the deuteron pairing bond. As all pairing occurs synchronously, their is always one shortened 'triton bond', making room for the central nucleon.
We predict a scattering radius very similar to Li-6. For each iteration of deuteron cycling, there is one bond expression similar to 'triton bonding' (2.98 nM) and two deuteron pairs (2x .86nM). This yields a net moment of 4.7 nM.
The observed moment is only 3.26 nM. This suggests one of the deuteron moments is rotated 180 degrees, leaving only the 'triton bond'. This must be a magnetically enhanced bond, which would also explain why Li-7 is stable, unlike triton.
Before considering our next nucleus, Be-9, which is stable, we must look first at He-5, which is extremely unstable.
He-5 is estimated to have a half-life on the order of 10 ** -22 sec. Its magnetic moment is not measurable, but we would predict that it may be negative, but smaller in magnitude than a neutron.
In its brief existence, He-5 has 2 protons and 3 neutrons. We saw from our model of the alpha particle (2p, 2n) that its two deuteron pairs form an extremely stable nucleus with 0 magnetic moment.
We can visualize that for the duration of a single cycle of deuteron pairing, the momentary magnetic field of one deuteron pair in an alpha particle can interact with a neutron. This interaction is in the manner of the free neutron interaction of triton. But the stability of the alpha particle precludes forming a stable nucleus. We would predict the half-life of He-5 to be on the order of 10 ** -21 sec, the duration of the deuteron pairing cycle.
How can we visualize the Be-9 nucleus, with 4 proton and 5 neutrons?
We note that the nucleus Be-8, with 4 protons and 4 neutrons, does not naturally occur. The stability of the alpha particle precludes its formation. To form Be-9, which is stable, the extra neutron must serve to effectively bond two alpha particles.
We can visualize from the momentary magnetic interaction in HE-5 how this Be isotope can form. We picture a ‘free’ neutron interacting with two alpha particles. With a single ‘tick’ of the entanglement clock, the ‘free’ neutron interacts with one deuteron pair in one alpha particle. On the next ‘tick’, it interacts with one pair in the other alpha particle. It forms a dynamically symmetric composite nucleus, bound by alternating, and aligned, momentary magnetic fields.
The stability of this nucleus arises from the constant interaction of the neutron with a deuteron pair. With each ‘tick’, it is stabilized by momentary magnetic fields arising from alternate alpha particles. This is not an example of deuteron pairing, nor of neutron ‘flipping’ in triton pairing.
The magnetic moment of Be-9, -1.17 nM, reflects this hybrid type of interaction. The magnetic field of the neutron is not ‘flipped’ as with triton. But it is not as strong as an isolated neutron, -1.9 nM. This reflects the contribution from the positive moments of the alpha particles.
The dynamical symmetry of this nucleus can be maintained if another neutron is added and participates in bonding of the two alpha particles. This nucleus is Beryllium-10, with a half-life of over 1.3 million years.
With each ‘tick’ of the entanglement clock, the two ‘free’ neutrons can interact with deuteron pairs in opposing alpha particles. With the next ‘tick’, they alternate their interaction with the opposing alpha particle, on separate deuteron pairs. The result is a nucleus expressing dynamical symmetry about the neutron bonding ‘axis’.
We would predict Be-10 to have a net magnetic moment of 0 nM, and a scattering radius similar to Be-9.The observed radius is smaller (2.36 fm versus 2.54 fm).
This provides evidence of the nature of the ‘hybrid’ bond formed by the neutron bridge between the two alpha clusters. The negative moment of Be-9 shows there is no ‘flipping’ of the 'free' neutron, as with triton. The shorter bridge distance of Be-10 suggests there is no momentum transfer, as in the deuteron pairing bond. The hybrid bond must be formed from the neutron's magnetic field, and magnetic flux from deuteron pairing.
Both Be-9 and Be-10 have a ‘barbell’ geometry, with dynamical symmetry of the internal, momentary, and opposing magnetic fields. This provide an environment for photon node formation. We explore this aspect of the quantum forest when we consider “spin-flipping” in Nuclear Magnetic Resonance.
We consider next Boron 10 and Boron 11, which are both stable isotopes having atomic number 5.
We first note that the isotope Boron 9 exists, but only briefly. By emitting a proton, it decays to Be-8. The latter in turn decays immediately into two alpha particles. This suggests that the ‘barbell’ geometry of Be-9 and 10 also applies to Boron. We visualize the fifth proton contributing to an axial bridge bonding two alpha clusters, in both Boron 10 and 11.
What type of bridging bond can this proton provide? The alpha clusters have 0 net magnetic moment, while a free proton has a large positive moment (2.8 nM).
In its brief existence, we would predict a large positive moment for Boron 9. With Be-9, the flux from the ‘free’ neutron can form an attractive bond with the momentary positive flux from deuteron pairing. With Boron 9, the positive flux from the free proton can not form a stable bond.
This situation is altered in Boron 10, which is stable. With an additional proton and neutron, it can form a separate deuteron pair, besides two alpha clusters. This pair provides the bridge between the clusters
An attractive flux arises between the bridge neutron and a proton from a deuteron pair in each alpha cluster. By symmetry, an attractive flux can also arise between the bridge proton and a neutron in the same alpha deuteron pairs.
The flux interaction is out of the parallel planes formed by all deuteron pairs involved in bridging. We would expect a reduced ‘in-plane’ positive flux from each of the three deuteron pairs (two from the alpha clusters and one from the bridge). These will combine to form a net positive magnetic moment for Boron 10.
An isolated deuteron pair has a moment of .86 nM. The measured moment of Boron 10 is 1.8 nM, suggesting the altered flux from the three bridging deuteron pairs is reducing the moments to about .6 nM per pair. The difference in flux is the basis of the bridging bond energy.
Boron 11 has an additional neutron beyond Boron 10. It also has a much larger magnetic moment and smaller scattering radius. We can visualize how the formation of a ‘triton bridge’ will produce these differences.
Triton has a large positive magnetic moment (2.98 nM) and a smaller scattering radius than a deuteron pair. This is a result of ‘flipping’ a free neutron expressed by each nucleon over 1/3 of the deuteron pairing cycle.
We can visualize how the ‘flipped’ neutron can interact with protons in the alpha clusters. This will form a stronger bridge bond than Boron 10, while stabilizing the free neutron. Because of this stabilization, Boron 11 does not decay, unlike triton.
The net magnetic moment of Boron 11 is large (2.69 nM), but less than triton (2.98 nM). This suggests less flux is contributed by deuteron pairs than occurs with ‘deuteron bridging’ in Boron 10. The free neutron flux must provide most of the net moment of Boron 11.
We consider next carbon 12, which is a stable isotope with 6 protons and 6 neutrons. It has 0 magnetic moment, 0 quadrupole moment, and a relatively small scattering radius of 2.47 fm. This suggests a symmetrical, compact nuclear geometry.
Some nuclear models propose a ring structure for carbon 12. Our model of deuteron pairing does not support this. Arranging 6 pairs in a ring structure, such as an equilateral triangle, results in a scattering radius larger than the measured value.
Each side of such a triangle would have 5 nucleons (counting the corners once), alternating as a pattern of {3 protons, 2 neutrons} and {2 protons, 3 neutrons}. The length of each side is sum of the lengths of 2 deuteron pairs and one proton, on average. This is over 5 fm (2 x 2.13 fm + .84 fm), more than twice the observed radius.
For our model, we extrapolate the nucleon arrangements from Beryllium and Boron, using Li-6 as our cluster component. It has a magnetic moment of .85 nM and a scattering radius of 2.39 fm.
The two clusters, when closely opposed, can express zero net magnetic moment. This occurs when the pattern of deuteron pairing for each cluster is in the opposite cyclical direction.
A very compact and symmetrical geometry results from the opposition of the two clusters. The between-cluster bond forms from ‘off-plane’ proton and neutron magnetic flux interactions in the opposing clusters.
As with Beryllium, isotopes of carbon should form by adding neutrons to form a bridge between clusters. This is observed in Carbon 13 and Carbon 14, the latter being radioactive, with a half-life over 5,700 years. Both have scattering radius similar to Carbon 12, suggesting the neutrons forming the bridge are tightly bound within the equilateral triangle geometry of Li-6.
Carbon 13 has a magnetic moment of .7 nM. Our cluster model estimates the total moment (from Li-7 at 3.26 nM, and Li-6 at -2.39 nM) as .87 nM, a reasonable first approximation.
Carbon 14 has zero magnetic moment, implying the moments from its two Li-7 clusters are in precise opposition.
Carbon 14 undergoes beta decay, sourced from the neutrons forming the bridge. Carbon 13 is stable, suggesting the magnetic flux supporting its ‘flipped’ bridge neutron is strong enough to stabilize it.
The beta decay products of Carbon 14 are Nitrogen 14, an electron and an antineutrino. This is consistent with one of the bridge neutrons undergoing beta decay to yield a proton. The proton can form a deuteron pair with the remaining bridge neutron. We can visualize Nitrogen 14 as two Li-6 clusters with a deuteron bridge.
Nitrogen 14 is also a compact nucleus, with a scattering radius of 2.64 fm, suggesting the two Li- 6 clusters interact strongly with the deuteron pair bridge. The net magnetic moment is only .40 nM, implying a strong ‘off-plane’ interaction that reduces the deuteron moment of .85 nM.
If we ask Google Gemini to build a nuclear model for Oxygen-16 based on deuteron pairing and the preceding models as reference, it does quite a nice job:
The nuclear model for Oxygen-16 based on the principle of Deuteron Pairing is constructed using the Alpha Cluster Model, which is the logical extension of deuteron pairing for N=Z nuclei beyond Helium.
This approach views the O-16 nucleus not as an inert spherical shell (as in the Shell Model), but as a structure composed of tightly bound sub-units of alpha particles in a specific geometric arrangement.
Composition via Deuteron Pairing
The Oxygen-16 nucleus contains 8 protons and 8 neutrons, making it a doubly-magic nucleus.
1. Deuteron Pairs: The fundamental unit in the referenced model is the deuteron pair (one proton and one neutron, p-n). The O-16 nucleus is composed of 8 deuteron pairs.
2. Alpha Clusters: These deuteron pairs are assumed to combine tightly to form alpha clusters (He-4), which consist of two protons and two neutrons (2 p-n pairs).
Oxygen-16 (O-16)=4×Alpha Particles (He-4)
Total Nucleons=4×(2p+2n)=8p+8n
Geometric Model (Tetrahedral Cluster)
The model predicts that the four alpha clusters arrange themselves to achieve the greatest possible symmetry and stability, maximizing the binding forces between the clusters.
Ground State (JP=0+)
• Structure: The four alpha particles occupy the vertices of a regular tetrahedron.
• Symmetry: This configuration, with Td symmetry, naturally explains the observed properties of the ground state of O-16, particularly its zero spin and positive parity (JP=0+).
• Binding: The strong binding energy comes from the six "bonds" between the four alpha clusters (each cluster is connected to three others, forming the six edges of the tetrahedron).
First Excited State (JP=02+)
For comparison, this model also offers a structural explanation for the first excited 0+ state (the 02+ state at 6.05 MeV), which is known to be structurally different from the ground state:
• Structure: The alpha clusters are theorized to be arranged in a less symmetric, planar square configuration.
• Binding: This configuration has only four alpha-alpha bonds, making it less stable than the tetrahedral ground state and accounting for its higher energy level.
Addendum: Excited Nuclear Energy Levels
The relative motion of nucleons determines the internal energy of a nucleus. The nuclear shell model treats this as independent motion producing a mean energy field. Within this field, discrete energy levels are occupied by nucleons with unique quantum numbers, much as with the electron shell model.
The shell model describes the nucleus in its ground energy state. It predicts the inner nuclear structure, arranged in stable shells, and outer “valence” nucleons. In contrast, we focus here on the nucleus when it is in an excited energy state. We must revisit our model of electron orbitals, aspects of which apply directly to the excited nucleus.
We visualized electron orbitals as the momentary projection of electrons about a central charge. In an atom above its ground energy state, electron energy levels are determined by absorbed photons. These establish “nodes” through the interaction of photon magnetic fields with two other sources of magnetic fields present in the atom.
The first source is the momentary field that propagates the central protons’ charge across QESTs in the Vacuum Network (VN), at light speed. The second source of momentary magnetic field is that of projected electrons, expressing their charge in a similar fashion. By random projection, electrons form an orbital about the central charge. For s-orbitals these projections are spherical, with no net current flow.
Electrons that are excluded from such photon nodes establish orbitals further out from the central charge. The ‘excited’ energy state of the atom reflects the energy of the absorbed photons. The photoelectric effect confirms the role of the photon’s magnetic field in establishing nodes. In addition to wavelength, photon absorption is dependent on polarization. The magnetic field orientation of the photon restricts the formation of orbital nodes.
By forming nodes, electron orbitals isolate photon energy within atomic structure. How can we adapt this model to the scale and energy levels determining excited nuclear energy states?
We view the binding energy from deuteron pairing as the main source of internal energy held in the nucleus. How can the pairing of nucleons isolate higher energy?
We can visualize this process by combining our model of deuteron pairing with our model of photon energy isolation within electron orbitals. To do so, we will need two key properties of the local spacetimes that are formed by neutrinos.
The first is that local spacetimes can isolate charge from the VN. The second is that they do so while remaining transparent to magnetic flux. Quantum Foam enables both by expressing momentary magnetic fields between ‘tick’s of the global entanglement clock. Conservation of energy will again be a key constraint. No additional model components or assumptions are required.
Experimental evidence, such as charge conservation in excited nuclear states and nuclear absorption spectra as revealed by the Mossbauer effect, will provide the necessary clues. We again assume all nucleons participate in deuteron pairing. We add one additional constraint: we consider only changes in nuclear energy levels that do not alter the type of nucleus. We will see that this limits our model scope to ‘gamma transitions’, either absorption or emission.
Our model rests on the dynamic interplay of magnetic fields in two separate spacetimes: the Vacuum Network (VN), and the local spacetime of the antineutron (AN) hosting the neutron. While energy is exchanged between these spacetimes, the exchange does not persist beyond a single ‘tick’ of the entanglement clock. There is complete dynamical symmetry, which preserves conservation of energy. Dynamical symmetry will help us visualize how photon energy is isolated in the nucleus.
We need to review the properties of particle and photon projection in these two spacetimes. All projections are momentary and are indexed by the same global entanglement clock. All forms of energy hosted in local spacetimes must be projected into the VN, to maintain conservation of energy in the VN.
Within a local spacetime such as the AN, particles can be expressed as point projections with the static properties of mass, charge, spin and magnetic moment. By establishing momentary magnetic fields, a particle’s charge is propagated across locally entangled QESTs. This occurs at the speed of light relative to the local spacetime.
If these charge fields are localized, and do not interact with QESTs in the VN, charge can be ‘masked’ from the VN. Energy conservation dictates that the energy from this localized charge must project into the VN. This is accomplished by a momentary magnetic field expressed through Quantum Foam. In this manner, spacetimes like the AN hosting the neutron shield particle charge from the VN. For the neutron, this applies to its two component particles: a proton and an electron.
In the case of photon projection, charge from photon propagation can manifest in two ways: charge propagating on boundary QESTs, or charge propagating on QESTs interior to the AN.
How do we visualize these projection properties with deuteron pairing? The oscillation of proton-neutron expression requires ‘masking’ of charge with each ‘tick’ of the entanglement clock. To provide this ‘masking’, entanglement energy expressing the AN spacetime must alternate in projection about each participating nucleon (both are protons).
We can visualize how the AN hosts the alternate projection of the two protons. Simultaneously, it hosts the projection of the electron, also a point particle, which establishes an orbital about an up quark, ‘flipping’ its charge to down. The entire process is dynamically symmetrical. All masked charge energy is expressed back to the VN as momentary magnetic flux.
There is a precise balance of opposing forces: (1) an attractive force from the ‘unmasked’ proton charge interacting with momentary magnetic flux; and, (2) momentum exchanged between nucleons, sourced from the low mass but extremely rapid displacement of an electron and the AN itself.
By contributing to this balance of forces, but at higher energy, a photon can be isolated by absorption in the deuteron pair. To see this, we apply our model of photon node formation in electron orbitals. In this case, the orbital is extremely small and localized to an orbital about an up quark in the AN spacetime.
To participate in this node formation, the photon must have a very specific and short wavelength, such as a high energy gamma photon. We would predict it must also have specific polarization.
The strong magnetic flux expressed by the photon, as it participates in alternating projection, increases the attractive force produced through momentary magnetic flux. This is balanced by the momentum transferred to the ‘unmasked’ nucleon.
The higher energy interaction increases the distance between the paired nucleons. This distance is constant, a result of the dynamical symmetry of the process. As this greater pairing distance is expressed by deuteron chaining throughout the nucleus, the nucleus structure will be altered. It will enter an “excited” state.
Evidence for the process we are visualizing is provided by the Mossbauer effect. Gamma photons of very specific frequency are absorbed by nuclei, without altering the nuclear species, except for its internal energy. The opposite process also occurs, where a gamma photon is emitted, returning the nucleus to its previous configuration.
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