The proton, electron structure, its resonances and fusion products Dr. J. A. Wyttenbach Independent researcher juerg@datamart.ch; https://www.researchgate.net/project/Nuclear-and-particle-physics-20 Abstract: We here will show a fundamentally new method to model particles and dense matter. It's a departure from axiomatic point particle/masses physics. The new method will fnally enable us to express all properties of mass/particles as products of complex interacting EM fux/waves. At the current state of development we call the model SO(4) physics or SOP, as this, SO(4), is the symmetry group needed for attaining the frst deep understanding of EM mass interaction. SO(4) is used to the full extent of 5 rotations and thus there are no commonalities with standard model (3 rotations) physics mappings to SO(4)! The impact of this model will be without any historical precedent and will completely change some relevant parts of physics, that are based on dense matter interaction. We will show the exact relation(s) between all known forces and how the mass of standard particles is structured and forms simple nuclei like 4 He, 6Li. The model is exact and will challenge future measurements and methods to seek even more deep insight. The prerequisite for understanding SO(4) physics is basic knowledge of topology and EM-theory in higher dimensional space[31]. Further a deep understanding of rigid mass mechanics is needed for the correct perception of the evolving EM mass structures. Being able to think in at least 4(6) uniform dimensions will greatly ease your progress. Conversely trying to understand, compare the model from a QM/QED viewpoint will certainly hinder you. Obviously Nature can not be modeled by a simple force/feld approach. 1 Introduction Three year ago the author started a ground zero search for the basic laws of Nature, that are responsible for the formation of dense nuclear mass. The motivation behind this task was the fact that for 90 years physics has failed in explaining the fundamental relations between mass(es) and other frst order quantities like magnetic moment, charge and magnetic radius. The belief that mass and its structure can be understood at high energies only has led to a model [1] that looks like a stamp collection, that is of contemplative value only. The main obstacle to fnding a new solution was the common belief that general relativity is valid for all space and time and that the Maxwell equations cannot be used to model dense mass. As we will see, both assumptions are wrong. The universe is flled by EM mass – photons - and thus it is no surprise that dense mass is pure EM mass too. Obviously stable particles feel no time, as these exist eternally and if waves form the mass, then these “waves” must be strictly of rotation symmetric nature only, which is not given by the current standard Dirac (potential) like modeling. EM mass shows two forms of interaction, that are within some limits “symmetric”, as moving charge may generate a feld and a changing feld generates charge. The nature of this interdependence invalidates the use of a strict operator based solution for dense matter physics as its calculating power is restricted to closed system and we need a complete one (which includes all operation +,-,*,/). The frst/simplest suitable mathematical space for a successful description of dense mass is the symmetry group SO(4), with the center symmetry given by the full Cliford torus (all orthogonal construction, we will use the shortcut CT for Cliford torus) orbit structure. Energy (mass!) is stored in diferent numbers of rotations that are performed by coupling electromagnetic (EM)-fux. This EM mass strictly follows classic – higher dimensional - rigid rotator mechanics and couples Biot-Savart like with a topological and most of the time virtual (masslesstopological) charge. 1 Most solutions we show in this paper have a very high numerical precision, what might, by sceptic people, be accounted for as “numerology”. But at the end of the process the gravitation constant (chpt.11) will pop up by just using the basic electron/proton form factors of dense space. The gravitation mechanism thus has been found by using basic logic and experimental evidence for its derivation. We will show the internal structure of the most common particles (e, p, n) and their fusion products (n, 1H, 2H, 3He , 4He, 6Li, 7 Li, 12C in appendix A). Complementarily we also show the split of the proton into the so called resonant masses (Kaon, Pion, Muon). Of course we will devote a few lines to the accelerated proton resonances also known as Higgs masses as these resonances can be derived too. The beauty of the new model is fnally expressed in the naturally evolving wave factors e.g. of Hydrogen, that are 1, 1, 2, 3, 5 and show/explain the relationship with Fibonacci and the Golden ratio and fnally the Nature we all live in. ( - Real modeling starts in chpt. 3) 1.1 Historical models We here use SM as an acronym for Standard Model and SOP as a shortcut for SO(4) Physics. Over the last ninety years physicists did believe that we can explain matter by quasi “looking” into it, which is the standard method used in Biology. There was never given any reason why matter should behave like a living cell, that is composed of the very same matter. The end of this process is now given by a stamp collection of so-called particles, many of which are only resonances of interacting EM mass. The “concept” of the quark [2] as a building block of matter must today be called the most fringe idea of physics because an EM resonance, by defnition, cannot be a form factor of a particle. It's no surprise, that today it is still impossible to give the quarks a concrete mass, that deserves this name. The error bar is up to 30%. But interesting enough the sum of two down and one up quark - actual masses - is precisely equivalent to the SOP protons perturbative mass (owned by proton and neutron). Even more interesting the protons perturbative mass has a 3 rotation structure, what explains why SM calculations based on SO(3) logic give some useful results. So we can say that the current SM has been developed based on the interaction of electrons/protons with the proton perturbative mass only. The protons perturbative mass is about 1.2% of the total proton mass and seen from this perspective it is clear why SM failed. If you have no clue of the main proton mass structure, then you cannot draw any conclusion of the real physics of the proton! The main mass part of the proton is a 4 rotation structure, that is charge neutral and only interacts magnetically under a symmetric excitation, which sometimes randomly happens in collisions. This situation has never been modeled in modern accelerators as in fact, you don't need them to investigate the proton structure. This can be done with a very simple low cost experiment [3] [4] with an input energy in the single eV range... As you will learn, matter, under higher topological symmetry, can form out magnetic resonances, what allows the transfer of energy causing fusion or even fssion of a proton. Unluckily such topological symmetries are completely impossible at higher energies. This is one more reason why SM investigations e.g. in CERN failed to fnd any new physics as most data has been gathered from collision experiments. Nevertheless these SOP magnetic resonances do have a counterpart in SM, where these are called the virtual particle background, which usually is modeled by so called Feynman diagrams. The research in cold fusion or alternate [5] [6] fusion technologies will have a very deep impact on the future redirection of dense matter and particle physics. We will also see that magnetic resonances do explain all aspects of Rydberg matter and the new structure of dense hydrogen frst found – and “patented” .. - by Santilli [7]. Dense Hydrogen is “spin” based matter, that forms a condensate that people frst claimed to be a BEC (Bose Einstein condensate). But dense hydrogen is stable up to very high temperatures, what is not the case for a classic BEC. Even an optics efect - Goos–Hanchen [8] – the lateral displacement of an incoming ray (photon), under total refection, may be now explained by the SOP toroidal resonant electron orbit. 2 1.2 How can we find a new physics “standard” model? Initially only three questions needed an answer. 1) What is the exact dimensionality of the mathematical space we need for the new model? 2) What is the basic shape of particles? 3) How does mass compression work, when particles (proton, neutrons) join/fuse to form larger ones? The frst question very soon was reduced to the answer “at least 4 homogenous uniform ones”. The second question about the shape was easily answered by studying existing mass & radius data. The shape must be toroidal. The third question took much more time and brought the frst breakthrough, when the 3 mass compression factors (called 1FC,2FC,3FC) fnally could be identifed, where the strong force equivalent factor, called 3FC, did signal 6 a dimensional framework. (FC = fux compression) Today there is no longer any doubt that, SO(4) = SU(2) x SU(2) currently is the best suited mathematical space to describe dense condensed matter. This is based on experimental facts, that can be exactly reproduced by the model with highest possible precision. The three fux compression constants have their classical counterparts and refect the electro weak/electro strong and the strong force, legacy concepts of SM. The basic connection between mathematics and physics is given by the proton potential folding factor 2FC, that has an exact representation as potential mass disposed of at the proton De Broglie radius. The other two factors (1FC, 3FC) do depend on 2FC. Gravitation on the other side is a residual force of the electroweak force mediated by a potential that is driven by the electro strong force. To explain our calculations, relations and sums of contributions we, in this text, will use small spreadsheet like tables, that contain all steps in a condensed format. Real physics models, as explained, cannot be closed formulas – e.g. representing felds with a Hamiltonian density. Within this frst – non perturbative core model covering highly symmetric situations, we barely need any advanced mathematics. 1.2.1 What others did try Einstein himself was convinced that a future model must be based on a homogenous 4D surface with constant curvature. Unluckily he did stick to a single time dimension and ended up in 5D (4D,t) space. The model fnally was elaborated by Kaluza-Klein. Einstein[36] himself did also prove that general relativity GR is not a proper model for describing real world situations due to the fact that a feld only solution can only handle uniform point masses. Unluckily people never listened to him. More deep recent fndings based on classical reasoning were made by Klee Erwin [32] that found a modulo 6 structure in mass modeling and the 6th dimension being the golden ratio. This is exactly one result of our modeling. Of course there exist countless other models based on fantasies only, like string theory or ether like infnite small mass particles that all have the same common problem of not being self contained (internally consistent) and never showed any result. 1.3 What, who did contribute to the model The frst ideas of mass – space-time compression were developed by Randall Mills in his GUT-CP[13] book. The value for Mills so called “sec” factor is pretty close to 3FC but fnally Mills' “sec” is only measurement based – and did not reveal his origin. Mills also motivated the author to just use Maxwell- and Newton physics to fnd an advanced model for nuclear and particle physics as he, to some extent, could explain what exactly is neglected in QM like models. The input data for the SOP model is entirely based on public available data like NIST [9], IAEA [10] or the nucleon spreadsheet of [11]. All calculations where made on open source software and google did help to investigate more deep related papers. The LENR library [12] collection of papers was a constant source of inspiration. Online scripts like the one of Prof. Steven Errede (University of Illinois at Urbana-Champaign) about electro dynamics were helpful when doing calculations and modeling. Another important contribution were the LENR experiments in Essex with Russ George, I could be a part of, that fnally confrmed many fndings. Cold fusion is a magnetic resonance efect between magnetic masses, what explains again one reason why SM failed to predict it, as SM provides no magneto static solution. We should never underestimate the importance of experiments. A model without experimental backing has no value at all and gives the time you invest the same touch as gambling. So I'm grateful that I had the chance to combine modeling with “true research”. 3 2 What are the basic concepts of SO(4) physics ? All mass is equivalent to magnetic fux as in classical electrodynamics magnetic fux can own/express a moment, which defnes its energy/mass equivalence in interactions. Magnetic fux always expands into two dimensions, when it fows into a third dimension. Basically the concept of mass is assumed to be equivalent to the total rotating volume * density of the fux times the weight of its rigid mass form factor. The rotation frequency is given by the light-speed to radius relation and also involves topological factors like winding number and group measure. The core mass of particles, which builds nuclei, is formed by the relativistic fux of magnetic feld lines, where as the “non” relativistic so called excess mass is an efect of (dynamic/topological ) charge interaction with the self induced feld. We call magnetic fux relativistic, if the number of dimensions (rotations) covered is equal the number of energy Eigenvalues. If mass gets denser, fuses and forms nuclei, the non-relativistic fux will be converted into relativistic fux, or be disposed of. This allows a reduction of the space occupied by the total fux, which leads to the so called fux compression. Classically we can also say, that in fusion events a volume of space/time mass is released [13] 34.70 and is converted into photons. This efect has already been discovered about 30 years ago and did allow Randall Mills to predict that the universe must expand, a prediction confrmed one year later. A second means to increase the weight of dense mass is to acquire one more rotation = more rigid mass (e.g. the CERN measured 126 GeV proton resonance(s), the so-called Higgs particle(s)). A rigid mass doing one more rotation needs far less space to store the same amount of fux (=energy) as the original mass. Dense mass can do 2,3,4 or 5 rotations. 2.1 What is classical EM mass ? There are two possible picture for representing classic EM mass. For a “plane” EM wave in free space the energy density is given by the pointing power vector: The energy contained in an EM feld is given by the Volume/time integral of the pointing power vector. But inside dense matter a plane wave is fully contained and thus degenerates to a cylinder surface (-equivalent =torus) quasi stationary wave. The total momentum/mass stays in the CT surface. The other classic EM energy is given by the density of a magnet feld of an infnitely long coil is: w = ½ * H*B There is no contradiction between these two approaches as in the frst case we deal with a wave and in the second case with a “constant” feld inside a coil. Both samples represent an ideal situations as we e.g. assume the feld is inside the coil or wave energy can be harvested at any point in space. Further in both cases the EM mass has one rotation only and is not coupling with other EM mass. The crucial question is: Can a particle be modeled with such a simple ideal model? A super-conducting coil can maintain a current for an infnite time. But is the situation symmetric? Will the same feld produce the current? Of course not as we must be aware of the Meisner efect. Electrons migrating to super-conducting spin orbits release a certain amount of energy that lets the magnet rotate, what breaks the symmetry. Further, particles may or may not have a magnetic moment, what excludes simple waves and would need diferent coil like structures. We must also be able to defne a force between EM masses as atoms obviously are made out of individual EM mass packets. So the only thing we may use from classic modeling is: A coupling wave may transport energy. A constant feld produces a “constant” force. 2.1.1 How can EM mass interact ? Next we will look how we can construct a situation where a current/feld becomes self attractive like e.g. two parallel wires (currents). SO(4) has 6 dimensions and the space of minimal friction is the CT (Cliford 4 Torus) that covers 4 rotations. Further dense mass can only support a static E/B feld and hence most magnetic fux lines must be fully contained. Only the magnetic moment can extend in space. For the self attractive magneto static view we must assume that a bundle of magnetic feld lines is fully contained inside a closed n-dimensional cylinder like volume that is mathematically equivalent to a torus. This picture is similar to the infnite long coil as you can walk infnitely inside a torus. This picture of EM mass allows to explain why charge is generated and how charge and fux do interact. What happens when we place two current loops in 4 independent dimensions? We assume that the current fows along the circles and the feld stays inside the circles In Fig. 1 the two 2D orthogonal current loops mutually share – by producing a feld - one acting dimension. We moved the magnetic moment vector of the left loop into the (u,v) plane of the right loop just for explanation purpose. This means the generated felds mutually couple – over blue circle, a moving “charge” - in Biot-Savart like manner to the Lorenz force (by static feld,inside current loop) causing a rotation around the other non acting dimensions. This rotation is strictly orthogonal to the acting feld because only the charge vector (current) orthogonal to the B-feld is coupling Biot Savart like. B x,y plane B u,v plane Fig.1 shows just half of the 4 coupled rotations as in fact the magnetic fux runs on front and backside of the single sided Cliford Torus surface. You can also say that seen from the loop the situation is invariant. Now in your imagination you must add the (x,u) and the (y,v) plane to the picture what produces two additional rotation axes. Finally we end up with 4 magnetic fux directions that correspond to the 4 tangents of the CT. Fig.1 Coupled 4D rotation This picture explains why fully symmetric fux along the CT does 4 rotations. If you write out the Maxwell equations in 4 homogenous dimensions as done in [31] then you can easily see, that the fux of two 4D-coupled magnetic moments leads to a symmetric 4D rotation, because the feld must be transported by an orthogonal dimension that only can be reached after a rotation! For a possible 4D mathematical treatment of Biot Savart coupling look at [31] but the expansion to SO(4) – 6D -has not yet been done! What is so special if you leave classic 3D, t space and move to SO(4)? The Biot-Savart force is no longer symmetric the coupling is now of cyclic/circular nature. In 3D,t space such a coupling would end up in a “twisted” minimum potential position. Finally in SO(4) The coupling causes a ffth rotation that produces the coupling charge. In fact – for a deep understanding - you have to reverse classic logic. If we here talk of a current or a moving charge then in “reality” the EM fux is moving and the charge stays in place. But these two views are complementary and the physics (forces) remains the same. 2.2 Understanding the SO(4) situation The drawing in Fig. 1 (also Fig.4 below) refects only one side of the reality. This is a split view projection of the abstract “reality”. Magnetic feld lines are always closed and the induced current (charge!) must be of 1, 3 or 5 rotation symmetry to be able to couple with magnetic fux. We here frst focus on the 5 rotation situation as it is key for the structure of relativistic dense mass. The two (out of 4 on each CT side) projected current loops we see in Fig. 1 are in reality one connected flament that itself does one more rotation, that we can not show on 2D paper, as it is a virtual movement of the whole magnetic fux structure. 5 This postulates that in higher dimensional space (projective!) currents can produce a classic self-attractive situation. We later will assume that this “charge like” attraction is in fact the strong force that compensates the centrifugal force produced by rotating (EM) mass. As we will see, for the “simple” particles (e, p) that we look at, the situation can be treated as orthogonal which allows to simplify some well known formulas. In fact SM also did fnd 5 1/3 chunks of charge used for the quark charges (u: +2/3;d:-1/3) that logically match the SO(4) 5 rotation charge coupling situation. However the 3 SM quarks making up a proton are not particles, but they do own some abstract properties of real particles. 3 is also the number of the protons energy Eigenvalues (full rotations). 2.3 How is charge generated ? The (magnetic-) fux – running inside a torus volume in a 3D projection - that fnally generates the internal binding charge must always have a diferent winding number (mass-energy Eigenvalue number of rotations relation) than the dimensions covered by the magnetic fux, it fnally (indirectly) binds by induced charge. From a mechanics point of view this happens quite naturally as e.g. in classic 3D space, our real space, you only can have 2 independent rotations to cover all possible points of a 2 sphere. But a 3D body can have 3 diferent energy Eigenvalues (rigid mass inertia-) factors if the shape is not symmetric. As we deal only with tori/spheres we can neglect this complication. So it is natural that we see a diference in independent rotation numbers and covered dimensions. Only under this condition we can have a situation with a constant average change in fux dΦ /dX (where X in SO(4) contains up to 6 dimensions) that produces a so called topological charge. Just for repetition: ∇×Eem = -∂Bem/∂t This is the general form of Faradays law that explains that charge is a consequence of a change in magnetic fux. If magnetic fux is symmetric and covers the full space(e.g. x,y,z,...) then always at least one dimension carries less momentum than the other ones and after a synchronous rotation this low momentum state (in 4D) fows to the next hyper quadrant. E.g. in 3D if the rotation axes stay in x and y then the fux through the “z” dimension is double the fux through x and y! We call this situation, where a (in overall) constant change in magnetic fux produces a locally varying but in total constant, self attractive charge, the magneto static solution of the Maxwell equations. Such a solution cannot be found in classic 3D,t space without using time as an acting dimensions, what leads to radiation. This inability to use classic space can be understood as we need at least 2 orthogonally rotating masses (needs 3D for holding the fux) for inducing magnetic coupling and one more dimension for the charge coupling fux tube that binds (adds one more rotation) the rotating masses thus in total need at least 4 homogenous dimensions. This – 4D - is the situation for the electron where we note the orbits as (1x1) x (0.5 x 0.5). The numbers in the bracket are not classic dimensions they denote rotating masses = number of rotations/dimension. So what we measure as electron charge is the virtual charge of a potential generating fux tube (right winded fux (3D)) that will bind to the closest left winded 3D -fux tube of a proton. We will show below how the protons external charge is generated.) We here do not postulate that more complex particles (nuclei) are free of time like acting internal behavior. We only will show that this “simple” electro static – time free setup for highly symmetric particles like e, p, n, 2H, 3He, 4He (alpha particle), 6Li, 7Li exactly (up to 10 digits) produces all known (4) forces and masses/energies we know from experiments. We also hope that more brilliant minds can fnd better pictures for the explanation of this complex situation. 2.3.1 Dense mass For understanding the following you have to be familiar with the Cliford torus topology and its embedding in SO(4). It is also obvious that there can be no current that produces the magnetic fux that is the base of any mass and that in fact the physics work the other way round: A changing fux produces charge that may look like a current. The (2 x 2) X (2 x 2) (or simplifed also 4 x 4) rotation structure is what we call the relativistic fux, as its energy Eigenvalues have the same dimensionality as the number of rotations seen on one projective half (4 6 “backside” 4 frontside) of the Cliford torus surface. This fux represent the core mass of 4-He the basic building block of structured dense matter. As the Cliford torus surface is a solution for a minimal Lagrangian[33] and in fact in the 6D space of SO(4) the relativistic fux stays infnitely close to the Cliford torus surface. Core dense mass ( so far we modeled it) does at least 3 full rotations (energy Eigenvalues e.g. proton) and at most 4. In both cases 1 “external rotation must be added for adding one the internal charge coupling that binds the magnetic fux. dimension Symmetric strong force coupling only occurs in the 4+1 rotation coupling situation. Unluckily there is no easy way to give you a 6D picture on 2D paper. The only thing we can do is showing projections of projections that always neglect/simplify one aspect of the problem. We think that the best way to understand everything in SO(4) is to relate/project the physical facts to the spheric plane Cliford torus “surface” or its 3D envelope (= 2 torii). Thus the 4+1 plane rotation coupling starts with a “rotating” Cliford torus. Fig. 2 Tubular fux segment The fux tube, that leads to a self-containment of the ( 4 rotations) magnetic fux lines (needed for the Maxwell induction law) is induced by a ffth rotation that involves the whole 4D Cliford torus structure. This ffth rotation usually contains no signifcant additional mass parts and only acts in a perturbative way. To understand why this 5th rotation leads to a closed tube you can make the following visualization experiment as a (simplifed for one of 5 coupled dimensions!) proof. If you rotate (at distance r from axis ) a tiny connected neighborhood of a higher dimensional single sided surface (manifold) – here of the CT - around a tangent axes, then seen perpendicular from the axes (along an intersecting radius) you always stay inside the tube that is formed by the trajectory of the connected small area. As the CT is a closed surface with no edges for all such small compartments the fux staying on the inner side of the torus is fully contained. This holds for all possible manifold sections (of the CT) that rotate around the “same” axes. The induction requirement of the Maxwell equation is thus upheld in all parts that contain magnetic fux. As there is just one side of the CT this holds for the whole torus. As you might know the torus is congruent to a cylinder. The cylinder (Fig. 2) rotation surface is congruent to a 2D rectangle (Cylinder surface!). Adding an orthogonal vector to the rectangle plane creates a new space with +1D. In the cylinder case its a tube where we are not interested into the inside of the tube rather than the tube segment that is formed by the two concentric circles (Fig.2 green). The same holds for the CT. If you add one more dimension to a manifold you generate a 3D volume equivalent structure. The general geometrical view of this picture is given by a manifold that runs in constant distance from the CT surface. Of course this (new) “torus” has a dimensionality +1 with respect to the classic CT! Summary: The CT is a minimal Lagrangian surface. The frst CT projection to 3D are 2 tori, that can be mapped to a cylinder. So 4x4 rotations go to 2x2 and with two Cylinders we get 1x1 along the diagonal of the surface area. This fact later will be used in the modeling as all these orbit are congruent – are able to share at least one rotation. 2.3.2 Problems with the understanding of the situation In Fig.5 below we show a projection to classic 3D space. In classic space e.g. gravitation is described by a point symmetric force. With each dimension you add the point expands by one order: 5D: point → line (circle) ; 6D point → manifold (CT). The center of mass in SO(4) is a manifold given by the CT surface! This also implies “that rotation” axes no longer can join in a single point and even the notion of an axes for more than 3D cannot be used. Best we can say is that each point of a fux line rotates around the tangent (at radius r distance) that is parallel to the momentum of the point. How is self attraction constructed: A current loop producing a self contained feld, that reproduces the current loop needs to do 2 x 2 rotations 7 (in a projection to 4D) two left/two right until the projected current loops are parallel and hence attractive! This is also the “classic” picture we have from a photon that is a self reproducing E/B feld entity! (Once more: In dense mass only the magnetic fux does move at light speed and generates a magneto static charge that externally looks like a current that produces e.g. a magnetic moment.) But efectively you have to add one more orthogonal dimension and only in this (Fig.1) simplifcation do the number of rotations match. 2.3.3 How to describe the rotations As we later will see the 5th (entire Cliford torus) rotation fux tube forms the charge for the strong force equivalent coupling of the 4 rotation relativistic mass that externally propagates as Gravitation. In the following we often use 3D/4D what means 3 rotations in 4 homogenous dimensions. This is always relative to the magnetic fux/mass only! e,g, the electron is 2D/3D particle. Dimension 1 x: 1 y: 1 u: 1 v: 1 1 0 0 0 0 1 0 1 0 0 1 0 5 1 0 0 1 1 1 1 0 7 0 0 0 0 0 1 1 1 1 1 0 0 1 0 1 1 11 0 1 1 0 0 0 0 1 1 0 1 0 1 1 0 1 0 0 1 1 16 0 1 0 0 The next mass-class is the 3D/4D rotating mass that fows along the Cliford torus surface. This mass is interacting classically and is the one e.g CERN measures in its experiments. Fig. 3 3D/4D fux rotation matrix Basically Fig. 3 shows all 16 diferent 180o 3D rotations for the 3D/4D fux, that walks through all 16 hyper quadrants! As you can see in Fig. 3, the opposite position is gained after 3 full rotations (column 7 : 6 * 180o) but the full covered path back needs 5 full rotations. This is why “intuition only” does not work here and the old SM logic fails as it misses the hidden 5 full 360 o rotations. From Fig. 3 you can also see that there is a possible short cut after 2 full rotations (4 * 180o - position 5) as you can write the starting column (1,1,1,1) instead of the given one. This ideally shows the connection with the Fibonacci sequence as the possible couplings/shortcuts evolve after 2,3,5,8 rotations. The coupling with 4 full rotations happens between the two symmetric shortcut points (position 5,13)! Thus the basic structure of dense matter (proton with in average 3 rotation energy Eigenvalues) is formed by a fux that in total does 8 full rotations in average in 3 of 4 rotating dimensions. In Fig. 3 where we did expand the rotations to all 16 possible (distinct!) Hyper-quadrants. 3D/4D fux does 3 rotations at the same time with one dimension being invariant. After 16 x 180 o the fux is at the same point again. The 3D/4D fux also has 4D diferent rotation axes but only 3 are active at the same time. This follows directly from Fig.1 where we can see that a current loop is transported orthogonally to the old loop and one dimension stays the same. Interesting is that (4x4) fully relativistic rotation structure does not express a magnetic moment where as the the part of a 3D/4D rotation structure that does not conform with relativistic mass structure defnes the magnetic moment. Modeling shows that we can calculate the magnetic moments of isotopes based on their 3D/4D fux. 2.4 Short summary Charge as we currently know it is a virtual efect of a constantly changing feld also in all 3 classic visible dimensions. Magnetic mass is bound by internal topological charge produced by the 3 or 4 rotation magnetic fux. The densest form of mass – doing 4 rotations - is given by the Cliford torus minimal Lagrangian surface. The symmetry concept in SO(4) is diferent from symmetry derivd from SM (SO(3)). SM uses very simple concepts for masses and rotations that only cover a tiny part of the full SO(4) reality. The Cliford torus (inside SO(4) ) is a single sided manifold and any fux closely following its “surface” will “circle” around to be once in the exact opposite dimension from where it starts. Thus Fig. 1 – two current loops - is a highly simplifed picture that corresponds to (1 x 1) x (1 x 1) situation of a double sided torus. The double torus rotation ((1x1)x(1x1)) X ((1 x 1) x (1 x 1)) or as given above (2 x 2) X (2 x 2) cannot be 8 simply projected to a 4D orthogonal rotation matrix. To do the same we would need 5 rows and 2 5 columns. But then we also would have to discuss what opposite dimensions means! Thus symmetry is a term you must forget as it does not simply ft the complex connectivity inside SO(4). Basically nature is asymmetric as E/B felds never show any strong symmetry. As you will see masses evolve according to the natural growth function (Fibonacci) what is given from the 3D/4D mass coupling that dominates all interaction of mass. 2.4.1 What happened to Time? (A change in ) Time can be classically viewed as the product of diference of energies and path it takes to measure it. In a rotating frame you can do an infnite number of measurements as the situation is constant or reoccurs with a known frequency. Based on this logic, on a closed circular orbit time is constant either 0 or 2π or an integer fraction/factor of 2π. This fact has been used in standard wave equations since a long time by reversing the logic – saying the delta in energy is 0 thus time is infnite, what means we would need an infnite amount of time/path to measure a change in energy of a wave like given in classic notation as e iθ. This allows us to do fancy (Fourier) transformation of waves from a frequency space to a time space or reverse. Physically, today, time is defned as wave number or frequency of an ultra stable oscillator. What directly shows that states classically have “no time” and the photon behaves time like. This explains why a static state can no be described time-like, - only by its energy, that fnally delivers a frequency of an emitted photon. Is not time a required part in general relativity? Of course if we believe in the proposition that gravitation is constant and independent of the mass structure then this is obviously the case. But neither propositions holds as already Einstein concluded that GR only is valid (exact!) in the close proximity of an (point-) object. Time is also a needed part in the electro dynamics equations, where we need to model the delayed path (tt') and fnally the same structure as in GR occurs. But already there we face the fundamental problem that a part of the equation is relativistic and the other half is time like. Now we will see that inside (stable) dense mass almost all parts are relativistic, what is immediately clear if dense mass is only based on rotating magnetic fux. The frst physicist that did explain the deep connection between free 3D,t mass and dense mass was R. Mills [13] when he found that the famous fne structure constant ( α) represents the length compression and 2π the relativistic mass increase (see fake Higgs particle chpt.11.1). This is also immanent in SOP's 2FC constant that classically represents nothing else than a change in reference frame from 3D,t to one SO(4) rotation. But the exact picture is as said: If a mass doing an SO(4) rotation stops one rotation, then it must unfold a potential or emit a photon. Unfolding a potential is done by adding a lower dimensional mass – typically two waves – and the corresponding energy. In the proton case the added mass is m p*2FC'. What is given by a mathematical identity from the Coulomb potential of the proton De Broglie radius. At the end when you see that most dense mass is formed by 4 rotation mass then you will also understand why GR – for mass - no longer can be valid even at short range. Conclusion: Dense mass must be treated “time-free” or fully co-linear with time in all dimensions! 2.4.2 The basic forces at work (1) B= Iµ0 /(2r) The feld of a ring current (magnetic moment) This is simplifed as we here have only orthogonal quantities. Biot Savart coupling force – 2 ring wires/currents integral! is : (2) FBI = 2I x B = I2 µ0/r Coulomb force torus norm: (3) Fc = e2/4π2ε0r2 = c2e2µ0 /4π2r2 = I2 µ0 Using relativistic ring current equivalence relation: (4) I = ec/2πr 9 This is just a starter to give you the forces at work under orthogonal conditions in a toroidal frame. In (3) the charge is given by the frequency that only depends on r and the speed of light. You see that the diference between the two forces is, as expected, exactly “r”. Thus fnally we have to integrate (2) over one radial dimension to get the structural equivalence. This also shows why the coupling rotation number is +1 relative to the energy rotating number of the magnetic mass. 2.4.3 The visible – interacting mass structure The mass structure that interacts with the environment e.g. - in the form a of tabulated gamma spectrum is formed by the so called 3D/4D mass that rotates in 3 out of 4 dimensions - with energy/mass Eigenvalues staying in 3 dimensions – along the Cliford torus. (But there are some other sources for gamma lines regarding neutron-proton interactions that are charge like.) External potentials usually are generated by a two wave structure, what is also the classical view how we model a surface charge. (Using two spherical harmonics as a cover of the sphere...) 2.5 The torus rigid mass (some mechanics basics) To be able to calculate “real” quantities, that can be validated in classic 3D,t space, we need to use projections. The natural projection of CT quantities is its 3D cover given by two tori. The radial measure of a projected dimension changes by the SO(4) group measure of 2 1/2 times the involved dimensions. The radius in CT is ½ of the classic radius (see also fg.5****) as we have two opposite rotating masses in one dimension. SO(4) calculations are easy to follow as long as the involved masses have some degree of symmetry e.g. a nucleus is formed by an equal number of protons/neutrons. But EM mass in SO(4) has no classic point symmetry as the coupling is both 1/r 3 (magnetic) and 1/r cyclotron like. Nevertheless we can show that the magnetic coupling under the condition of self contained fux flling a torus – using the torus surface metric – is equivalent to Coulomb coupling. The center of mass is the whole Cliford torus surface and hence there is also no classic central force as the force itself is always perpendicular to the CT surface. Nevertheless we can fnd a mapping of the acting forces, that we will have to integrate over the torus surface. Fig.4 Image from https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco To understand the fux torus rigid mass you must notice that the 3D torus has two diferent Eigenmodes based on the two possible rotation axes see Fig. 4. Basically the form factors given equal radii (a=b) are: 7/4 & 9/8. 7/4 is well known as magic mass factor for 28Si or 56Co(Fe) 84Kr. Here you can directly see the connection between nuclear physics and basic mechanical laws. We will later make use of the above. As we will see in the particle section the 7/4 rotation mode, where fux and “mechanical” moment cover the same dimensions expresses an excited form (e.g. Pion) of matter, whereas the “stable” proton follows the 9/8 form factor. What does 9/8 physically mean? Relative to a mass staying on the CT surface Mass staying outside needs mechanical extra energy. Fully relativistic magnetic fux rotates on circular orbits. SO(4) has 4 tangent dimensions and thus after 4 *2 (sides) circle like full turns on the CT surface magnetic fux is back at the 10 origin. So the rigid mass factor is 1. We will see that a proton has not enough mass to occupy all rotations and only the alpha particle does exactly fulfll the CT requirements. Further mechanically rotating mass Energy Eigenvalues are additive for orthogonal rotating dimensions. So the rigid mass form-factors usually are retained / are independent of the number of rotations that a mass does as long as their form factor stays the same. To compare the number of magnetic fux lines you can only take 8/9 of torus mass as 1/9 is mechanical! excess energy. This will be important, when we will look for a mass -equivalence formula e.g. between electron & proton. Because particles are typically formed by 2 or 3 connected rotating mass structures then during fusion/fssion the number of rotations changes but the fux line number usually stays the same. Then you have to correctly treat the 1/9 excess energy if the fux gets fully relativistic.. 2.6 fux compression factors Adding a rotation to an EM mass is logically the same as fux compression as more fux-lines/time do pass a location. The following 3 basic fux compression factors have been defned in SO(4) physics. Energy conversion constants: 3D/4D - 4D Flux compression 3FC 3D-3D/4D Flux compression 2FC = 1 - (α/2π) 2D-3D/4D Flux compression 1FC = 1 -16*(α/2π)2 For mass reduction = 0.99711307593398 = 0.99883859026758 = 0.99997841803894 3FC' = 2FC' = 1FC' = for fraction/amount 0.00288692406602 0.00116140973242 0.00002158196106 2FC is the Coulomb potential folding factor that defnes the mass loss if a proton binds over one dimension. The mathematical identity is given in section 2.7 – classic Coulomb potential. 1FC is the second torus radius “coulomb potential” folding factor. (structurally corresponds to electro weak force). You can also 2 understand it as the next order term of 2FC as 1FC' = 16*2FC' what is equivalent to taking its sum over all 16 4D hyper quadrants. 3FC is the metric factor that maps 5 rotations into 2x2 rotating magnetic fux. (structurally corresponds to strong force) Construction of 3FC: Eccentricity of 4D space (golden ratio excess normed to one rotation = 2π) Ex4D = (1.6180339887 – 1.6)/2π = 0.00287019845321 Z = 2FC5 = (1 - α/2π)5= 0.9942064244067 Z = Ex4D/(1 - 3FC) = 0.9942064244067= (1- α/2π)5 (1 – 3FC) = 3FC' = 0.0028869240660 Why do we see the golden ratio? Magnetic fux always expands in two dimensions. A synchronization between “magnetic fux masses” is only possible if these have an integer mass relation. The only possible quantization for a progressive split of an area that retains the length relations is the golden ratio! 2.6.1 Simple example for flux compression Deuterium is formed from p+n+e. The mass released in fusion is 2'224'572.8 eV (* 1995 neutron data!). A frst simple approximation for this can be attained by the multiplication of (p+n+e)*(2FC'+1FC') = 2'222'070.7eV. This removes mass from a certain volume of space as two symmetric waves do couple over both torus radii. But but the space left behind – we call it energy hole - is still connected to the mass what can be seen in the modeling of gamma spectra of higher Z nuclei, where we must fll the hole frst for the mass being able to interact with a gamma photon. Thus the next step is to compress the energy hole what releases additional space time energy. Mass of hole: (p+n+e)*(2FC'+1FC')*(2FC') = 2580.7 eV. Or, total 11 calculated mass after hole compression : 2'224'651.4 eV. (efective 2'224'572.8 eV) This method is simple and good enough for approximations and for explaining the basic wave structure of isotopes. In reality Deuterium is a bad example as it has a relatively complex symmetry that leads to orbitcoupling between diferent orbit types as we will see below. 2.6.2 Classic radius versus SO(4) or 4D radius Be aware that we always draw the torus with a tiny hole. In reality the tube touches in a 3D projection! Here we show the relation between the classic particle radius and the SO(4) Cliford torus based so called proton 4D radius ( rp4D). The red circle line Fig. 5 inside the torus section(s) indicates the Cliford torus surface where as the black circles indicate the rotating 3D/4D mass that fows along the surface. The classic radius corresponds to the fux tube diameter where as the 4D radius (rpm) is given by the rotating mass (black circle) diameter. Classic proton radius 4D proton radius This is a highly simplifed projection of a part of the proton structure just for the purpose of showing some link between existing reasoning and the new SO(4) physics one. To make the dimensions comparable we above defned the 4 rotation proton radius rp4D being 2 times the efective 4D radius of the proton which later in the proton magnetic mass equation leads to factor of “8” for the compensation of this change. Fig. 5 Connection between 3D and 4D radius Electron 1FC 1FC*2FC*3FC SO(4) Coulomb 2 Proton 3 Proton potential waves 3D/4D waves Current in u,v plane Current in x,y plane Fig.6 shows the proton electron bond – wave structure that explains the behavior of Hydrogen. The right side shows the classic view with a coulomb bound electron. The electron is defned by a two wave structures given by the (1x1) relativistic mass and 2D orthogonal bound perturbative mass (0.5 x0.5) – defned by the electron g-factor. The electron classically is attracted by the proton potential that is generated by a two wave structure – classically two spherical harmonics on a sphere. Fig.6 SO(4) structure of Hydrogen The left side of the picture shows the mirror part of the 2 x 2 ((1x1) X (1x1)) proton relativistic mass. Along the proton relativistic mass the 3D/4D mass – three waves - are running. Efectively the left and right torus are fully SO(4) connected and only in a projection they look as being separated. 3D/4D mass efectively can be drawn on both sides. The electron perturbative mass is interacting over the 1FC orbit of the electron with one free 3D-4D orbit of the proton. In SO(4) this interaction is perturbative only as the orbit contains no real mass. In classic 3D,t space this interaction looks like a symmetric potential that accepts a wave. What is not shown in the picture is the reduced mass (classic magnetic) interaction. Below in Section 2.8 we give some quantities for the waves/masses (energies!) we use. 12 2.7 Some measured & derived quantities we will use We discovered that particles can be split into relativistic mass and perturbative mass, where the later is defned by the classic anomalous magnetic factor. Charge e : 1.6021766208 e-19 C Speed of light c : 2.99792458 e8 m/s Fine structure constant α : 0.00729735256635 Gravitation constant G : 6.67408 e-11 m3/s2kg Electron g-factor eg : 1.00115965218091 Electron mass meeV : 510'998.9461 eV Perturbative electron mass mep : 1183.1037 eV 2D coupled Perturbative electron mass mep2 = ( me* (1-1/ eg2)) : 2366.2074 eV = 2* mep 1R relativistic electron mass mer1 : 509'815.8424 eV = ( me – mep) 2R relativistic bound charge mass mer2 : 508'632.7 eV = mer1- mep Bohr radius Rb : 52.9177210527 pm Hydrogen ionization energy mei : 13.59843449 eV Proton mass (eV) mpeV : 938272081.3 eV Proton mass mp : 1.67262189821 E-27 kg relativistic proton mass mpr : 926603083.13 eV Perturbative proton mass mpp : 11668998.16 eV Perturbative proton 3D/4D mass mpp34 : (1/(3FC*2FC*1FC)3 -1)*mpr = 11'396'588.4 eV or mpp - mppo Proton 4D/1D potential mass mppo : 272'409.8066 eV - (as factor: ((1-(α /(π*16)))2)* mp ) Proton de Broglie radius rpDBR : 1.3214098536 fm = h/( mp*c) Classic Proton de Broglie radius pot. EpDBR : 1'089'718.3271 eV = mpeV*(1-2FC) = mpeV*2FC' = e/4πε0 rpDBR Classic proton de Broglie radius potential 1D : 272'429.5818 eV = EpDBR/4 Proton magnetic radius rpm : 0.8408699161 fm (classic charge radius) Proton relativistic mass radius rp4D : 0.837653007404 fm → Proton 4D/3D de Broglie radius rpdbr : 0.841235640192 fm (4 times classic reduced de Broglie radius) Electron magnetic magnetic moment µe : 928.476462 10−26 J/T Proton magnetic magnetic moment µp : 1.410606787397 10−26 J/T rpr = 0.5* rp4D E-coulomb = e2/4πε0rB ; E (eV) coulomb = e/4πε0rB 2.8 NIST fudging constants Since 1999 NIST, to a certain extent, started to use QED-based fudging of mass values. The most severe cases are the neutron and the 4He masses. Especially the actual 4He mass value is based on fringe physics without any justifcation as QED is not a fundamental theory that can explain dense mass. In the neutron case the mass diference is given by missing knowledge of the coupling in measurement. The actual value seems to be OK if we deal with particles, but is of if we deal with nuclear masses. The delta in neutron mass is 6.4 eV what is quite large and for 4He its in the region of 3 eV depending on the year of the table … Thus, until this mess is sorted out, we must accept, that albeit we can give all calculations down to 1 eV, the uncertainty is given by the NIST produced mess... [11] [14]; The sources [15] [16] are even more messy. 13 3 The semi classic magnetic Bohr (Hydrogen) model As a frst SOP example we show a simple 4D Bohr model extensions frst using classic reasoning and secondly SO(4) based logic. If physics would work as expected conventionally then the total ionization energy of hydrogen should be the sum of formulas 1+2 – the potential and the stored magnetic energy. From Tab. 1 (below) it is easy to see that the calculated magnetic energy (base is classic Bohr radius!) is far too large. (1) E-coulomb(eV) = e/8πε0rB (2) Emagnetic(eV) = µB2*4π * µo/(rB3*e) The 4D model assumes that all energy is stored in magnetic fux. This implies that the electron orbiting a proton is not behaving just as a charge. The electron efectively behaves as magnetic fux. Thus, if we want to calculate magnetic coupling, we can use the orbiting weight(s). The electron mass in an SU(2) x SU(2) representation can be normalized into two parts. The magnetic core (relativistic) mass Fig.7 a left and the perturbative mass (mep) Fig.7 a right given by the magnetic moment perturbation, also known as electron g-factor. The coupling is given by a (1x1) x (1x1) mass/wave structure what implies that the perturbative mass is rotating Electron magnetic Perturbative (2D) orthogonal to the electron core mass. Mass 1'183 eV core mass But the perturbative mass is a split energy 509'815.8eV and only is equivalent to 0.5 orbit weights. It's more exact to note it as (1x1)x(0.5x0.5). Fig. 7a Electron 4D mass components Fig. 7b projected 4D orbits The proton potential is generated by the same wave topology (1x1) potential generating waves. Thus, frst order, this distant (1x1) – X – (1x1) p-e wave interaction has the classic Coulomb form and can be approximated by the potential. The frst 4 lines of Tab. 1 below show the classic electron ionization energy (light blue line a) - classic result) calculation based on the reduced mass. The assumption of a reduced mass is a core error of SM-classical physics as this concept only works in the “real” 3D,t world where the coupling is working in the center of mass. Already Mills [13] formula 1.253 showed that the reduced mass factor in reality is the radial magnetic coupling of the electron/proton system. Thus the classic assumption that the proton/electron pair (Hydrogen) is a two rigid mass coupled system is simply wrong. Rf: reduced electron factor Bohr radius (Rb) reduced mass Bohr radius (rRb)= Rb/Rf a) potential at reduced erRb classic error absolute (eV) classic error relative electron perturbation = eg factor m e = electron mass (eV) b) mep = me*(1-1/elg2) c) me bound meb = me - 0.5* mep d) mep /meb = (b/c) e) uncorrected magnetic energy at Rb (eV) f) (d*e) First adjusted ionization energy =(a + e) error absolute (eV) error relative g) magnetic correction ( (1 +1/9)* f) measured Ionization energy final corrected value = (a +g) 0.9994556794 52.9177210527 52.9465409443 13.5982871554 0.0001473346 0.0000108347 1.0011596522 510998.9461 1'183.1037038626 510407.394248069 0.0023179596 0.0572059311 0.0001326010 13.5984197565 0.0000147335 0.0000010835 0.0001473345 13.5984344900 13.5984344899 For the radial coupling we nevertheless can use the reduced mass instead of the stored magnetic energy as they are mathematically equal for - Hydrogen only! The classic reduced mass Bohr model ignores the second magnetic efect caused by the two additional 4D rotation axes of the electron perturbative mass. One axes resonances are well known from the Larmor precession of the electron, but they are dissipative. In SO(4) the axes is stable (2D orthogonal ) and thus in the ground state the wave energy (1/9) will be added orthogonal (in respect to the base magnetic energy produced by the electron orbit) This second momentum classically only couples with the non bound relativistic rest-mass of the electron. We show two results. Yellow adding the classic magnetic residual mass factor and then adding the additional wave with classic 1/9 weight. Tab. 1 4D magnetic Bohr model energies. 14 If you look at Fig. 6 then you see that the outside fux of the thin torus is bound to the outside fux of the larger (electron core mass) torus and thus is not free as they share a common orbit. (3) Mills [2] 2.243 For our classic calculations the magnetic energy correction is given by the standing wave (formula (3)) Larmor factor/angle = 60 degrees, which results in a factor 1/9 that gets added as a coupling weight to the source magnetic mass. A classic sample how to do this is given in Mills muonium fne structure formula Mills [13] 2.243. In a (1x1)x(1x1) rotation coupling system only the inner coupling of the outer mass does act. Below, section 3.1 Tab.3, we will show the analogue, much simpler model based on 4D orbits only. The reason for this inner coupling is that the Biot-Savart coupling works only from the rotation that is synchronous to the core mass. This small coupling force fnally gets added to the overall attractive force according to the relative masses at work. The coupling of the magnetic mass has already been included (via the reduced mass) thus the fraction of mass involved (see Tab.1) in coupling is 0.002317..times the weight of the magnetic energy Rf: reduced electron factor 0.9997276305 = 0.0572059..eV sum = 0.0001326..eV. The frst 4D radius (Rb) 52.9177210527 adjustment of the Hydrogen ionization energy is given by the Bohr reduced mass Bohr radius (rRb)= Rb/Rf 52.9321381531 potential at reduced e dark yellow feld fgure in Tab.1. It is about 5.6 digits exact 13.6019872382 error absolute (eV) 0.0001472818 which is quite good. Because classically this mass is added classic classic error relative 0.0000108278 electron perturbation = e factor according a wave we must do the additional correction by 1.0011596522 el. magn. excess mass = Me*(1-1/elg ) 1'183.1037038626 10/9 what results in 0.0000147335eV to be added. electron core mass+ ½ excess mass 510420.996382589 rRb g 2 The measured ionization energy (NIST) is given in the dark blue feld. The calculated Ionization energy is exact given current error bars. The same calculation also works for Deuterium with the same precision inside measurement error bar. Result (Tab. 2). Also the simple SO(4) orbit model Tab. 3 delivers the same exact result. Relativistic excess-mass/electron mass ratio uncorrected magnetic energy at Rb (eV) only coupling with rest-mass (eV) First adjusted ionization energy error absolute (eV) error relative magnetic correction (1 +1/9)*0.0001326010 measured Ionization energy final corrected value (spin/spin corrected) 0.0023178978 0.0572059311 0.0001325975 13.6021198357 0.0000146843 0.0000010796 0.0001473306 13.6021345200 13.6021345687 Tab. 2 deuterium ionization energy 3.1 The SO(4) Hydrogen coupling of the ground state Perturbations in SO(4) cyclic coupling rotating masses are proportional to neighbor coupling orbit forces. As a result the total force balance must be “1”. Lets assume that the basic e-p magnetic mass coupling M 1/M2 = me/mp = 0.999989165 = D is the deviation from experimental measurement. The deviation from 1 can be represented as D = f(v)/f(u) or for practical use as a fraction: D = 1- (1-f(v))/f(u). From the proton magnetic mass/moment formula (see below) we know that the radial force (e-p) factor for one dimension (f(u)) is given by (3FC*2FC*1FC), the spin pairing (1x1) electric force (electron-electron self interaction) is given by 1FC, thus the change in mass is given by 1FC' (f(v)) because we use it as additive factor. The 1FC coupling weight has already been discussed above and is ½. The resulting coefcient is 0.9999891650..(Tab. 3: row b) ). The matching is within the tiny measurement error of +-1 to the last digit. You can either interpret the 1 - (1- f(v))/f(u) factor as a change of the reduced mass factor or a change that afects the magnetic coupling mass. This “wave energy” gets added to the electron perturbative mass, what is equivalent to increasing the electron potential. Bohr radius Bohr potential classic reduced mass factor Bohr Hydrogen potential a) Corrected by reduced mass 1D attach to proton/1D electron b)1-0.5*1FC'/(3FC*2FC*1FC) corrected potential (a)/(b) measured potential error: none 52.9177210527 27.2113860282 0.9994556794 13.6056930141 13.5982871496 0.999989165 13.5984344892 13.5984344900 within measurement This simple example shows two facts: Classic modeling based on SO(4) assumptions leads to the same absolute exact result as the fundamental orbit based solution given by the SO(4) magnetic form factors. In the SO(4) orbit based sample we make no mention of the magnetic energy as per defnition in SO(4) all mass is magnetic mass and interacts by magneto-static orbit-orbit coupling. Tab. 3. Orbit based Hydrogen model 15 At the beginning it usually is not easy to keep track which factor leads to what reaction. In the Hydrogen case the combined force factor 3FC*2FC*1FC leads to an orbit/energy expansion because the strong force factor (3FC) only in the 4 rotation case is attractive. (see also below magnetic moment) 1FC on the other side is a binding orbit an releases energy. Using the same orbit correction for Deuterium gives a sightly larger value (13.60213461 instead measured 13.60213457) than the semi classic solution, because deuterium is not symmetric and we did not account for the internal neutron orbit binding. The use of structurally quasi symmetric perturbations as correction coefcients needs some experience and discipline. For convenience we used as a perturbation fraction x/y always a value < 1, what implies that the perfectly matching multiplicative correction must be y/x or > 1 which also can be realized by dividing by a standard value (<1). That way you can catalogue all standard perturbations, which sometimes helps in fnding the correct one. In the next chapters we will show that for symmetric particles the SO(4) model is much simpler than classic reasoning and gives a highly accurate picture of the internal structure of the particles of interest. The fusion of Hydrogen to deuterium and Helium can, in all details, be explained and exactly calculated by joining magnetic orbits. This also holds for the splitting of a proton into Kaon, Pion, Muons which will be explained using the Holmlid reaction.. 3.2 Base states of Hydrogen with n > 1 We here do not show all details for higher n-states. This will be done in a chapter we will add later. (A frst solution has been given in [35.] ) The only point of interest is the evolvement of e-p bond. Classic physics assumes that the higher excitation states of Hydrogen are defned by Coulomb interaction only, which is not the case. Both involved particles own a wave structure that obviously does multiple coupling. The proton owns 3 3D/4D mass waves also called the proton perturbative mass. The states n = 2, 3, 4 evolve with a reduced coupling of 3 → 2,1,0 waves attaching to the electron orbits. Thus starting from n=4, the level can possibly be modeled by a single formula that uses 1 common magnetic interaction term. There is still some detail work ahead and it looks like the frst excitation (n=2) is based on the 7/4 torus form factor that usually goes in line with an excitation. 4 Magnetic mass and moments Why is all mass electro magnetic mass? First, to calculate the “Bohr potential” of the e-p system you don't need the Bohr model and of course also no Bohr radius. Yo just need the electron mass and α! We can just use EBohr = me* α2! ( me in eV!) This is the (1 x 1) magnetic resonance energy of the electron! The second answer is even more simple and has been known for about 90 years. The Planck quantum “h” has been defned by the electron mass/ light speed relation that fnally has been used to defne the Bohr magneton & electron de Broglie radius together with charge (e) “α” has been defned: A simple change of the connected parameters (e, c, me, h, α) shows: Electron “magnetic mass” : (4) = 510'998.946eV - redbr electron reduced de Broglie radius = 0.3861592675486 pm ! 16 Thus our framework of dense particle physics is based on the electron magnetic mass formula. (See also magnetic mass formula from Mills [13] 32.32b. Mass equivalence.) This too explains why SM/QED fails to calculate anything relevant for dense (=nuclear/particle) mass as usually the Coulomb-gauge (=charge potential) is used. If we later use the identical formula for the proton, then we will only get the (1x1)x(1x1) orbiting mass – the so-called relativistic core mass. Due to the simple 2D orthogonal (two orthogonal tori) wave structure the electron has no classically measurable internal structure. 2D orthogonal waves (live in (x,y),(u,v)!) do act like one single wave packet = photon. But the classically calculated 3D,t electron g-factor contains one term ( the so called dissipative term of the electron-g factor = (1-α2/(3*π2) see also [13; 1.213]) with a quadratic factor for alpha. This term can defne a potential like fne structure bound to the perturbative mass only. Thus it is not a surprise that in solid state physics experiments some “strange” behavior of the electron can be measured that did lead to a plethora of speculative models [17] [18] [19]. An other efect we will see is that the weight of the perturbative mass of an electron (= charge mass!) can change by the number(0,1,2) of “nuclear” bindings it undergoes. The mass change always happens (is quantized) in number of perturbative mass units. 4.1 Magnetic moment of the Proton Here we do calculate the magnetic moment of the proton. For that purpose we will use a simple 3D physics formula for a magnetic moment. A magnetic moment is produced if a net ring current exists. The magnetic moment is the product of the Area with the intensity of the current that fows along its border. SO(4) is the product of 2 circles and if the mass distribution is 4D symmetric then no net current results. But (almost-) all mass in SO(4) consists of a small part called the 3D/4D mass. This 3 Waves are running along the 4 rotation core mass with a reduced number of energy Eigenvalues staying in 3 dimensions. This topological diference is the efective source for the charge excitation that generates the magnetic moment. (5) a) Magnetic moment of Proton Exper. chg. Radius 0.84087 fm. b) Measured µp 1.4106067873 c) 3D µp = (a*c*e/2) 2.0194353567 d) Metric change is 2(1/2) 1.4142135624 e) µp perturbed =(c/d) 1.4279564349 f) First error ratio (b/e) 0.9878500162 0.9959334913 g) Error(1/3) of 1D = (f1/3) 0.9959335244 h) 4D correction : 3FC*2FC*1FC j) calculated moment 1.4106069281 absolute error meas(j-b) 0.0000001408 relative error 0.0000000998 (6) (7) Tab. 4 proton magnetic moment and perturbation If we assume that the proton charge is equivalent to the electron charge, then the current can be calculated by the photon equation. This is adequate as the generating fux itself is fowing with light speed. Equation (5) gives the current – without the metric correction for the SO(4) radius. We frst use the measured proton radius of 0.84087 fm that also is used in (6) to generate the area. The combination of (5,6) leads to a simplifed formula (7) for the nuclear magnetic moment. Table 4 shows all steps in detail where the fat letters (c,e) are physical constants! The result of formula (7) is shown in the row labelled c). As explained above, the efective 4D radius is ½ of the 3D radius multiplied by the SO(4) group measure of 2½ thus row (e) shows the topologically corrected perturbed moment. The frst derived value is the light blue feld (e). The uncorrected result for the calculated proton magnetic moment is only 98,8% exact. Today everything is obvious as we know that the generating wave structure does 3 rotations what leads to perturbation in the from of µp*n3. The frst green (g) feld shows the deviation (perturbation) for one out of the three waves (rotations). This deviation is exactly caused by the 3 basic forces (1FC, 2FC, 3FC) that try to compress all magnetic fux further. 17 The values of rows (b, e) Tab. 4 could also be used (as in the electron case) to defned a proton charge gfactor. The combined force factor of 0.99593349 is within the error of the measured charge radius equivalent to calculated value 0.99593352 (g) After applying the correction the resulting proton magnetic moment Tab. 4 (j) has a higher precision than the radius measurement. If we do a reverse 3D radius calculation then we get 0.840869916fm instead of the experimental 0.84087fm. The combined force factor of 0.99593349 (1FC*2FC*3FC) will occur in many calculations and the reverse charge radius can be used too (in the excited particle mass calculations for Pion,Muon). For the deeper understanding: Here too the factor (1FC*2FC*3FC) is expanding the proton magnetic fux/mass. But the magnetic moment depends on the efective number of fux lines and not on the induced coupling mass. 4.2 Proton mass calculations The proton magnetic mass formula (10) below can be derived from the electron magnetic mass formula (4). In the electron formula the radius to use is r = “electron de Broglie radius”. In the proton case, for a frst derivation, we used the 3D equivalent 4D radius derived from the 4-He charge radius (1.6753fm). Keep in mind that the efective radius in 4D is ½ of the 3D equivalent. Thus, in formula (10), we must divide 1.6753fm by two → r=0.83765.. what gives the 3D equivalent (-4D) radius of the proton. For the 4D proton radius we must divide once more by 2 (because 4He has 4 times more fux than a proton) and fux is proportional to r2 or one can multiply the result of (8,9,10) by 8! (1.6753/2 fm. from Russian database with correct electron measurement) (8) Mpr(eV) = µp2*µo/(α * π* rpr3 *e) If you replace µo in formula (8) by 4* π * 100000 then the factor π can be crossed out. A better frst approximation can be made based on the knowledge gained from the magnetic moment. (9) Mpr3D4D (eV) = µp2*µ0/(α* π* rpr 3*e*(3FC*2FC*1FC)3 ) The factor 3FC*2FC*1FC is perturbative and always leads to an expansion of the more dense mass. In this step we did neglect the mass, that is responsible for the generation of the external potential. In section 2.6 we did tabulate the 3D potential energy EpDBR of the proton at the correct 4D De Broglie radius ( EpDBR = . The mass weight for one radial dimension is ¼ of EpDBR . With this approach, adding the potential generating mass, you get a very good estimate for the proton 3D relativistic radius = 0.8376530133fm. 1'089'718.3271 eV = mpeV*(1-2FC)) The only correction we must now add is an SO(4) conform potential generating perturbation. The best ftting function – instead of 2FC' (1-2FC) - so far is (1-(α/(π*16)))2that is a combined term of 1FC and 2FC that correctly reproduces the potential. (1-(α/(π*16)))2 is the relation of alpha to the whole 3D/4D surface (4 inner/outer spheres 8 rotations) of 4D space. The delta in proton mass by using the 1D - 3D potential (only based on 2FC) is about 20 eV. With the correct potential factor the proton mass formula looks as follows: (10) Mp (eV) = µp2*µ0/(α* π* rpr 3*e*(3FC*2FC*1FC)3 *(1-(α/(π*16)))2) (Always using ½ of the 3D radius!) If you do start with the 4He radius then the calculated proton mass (Tab. 5a.) is of by about 10 keV what is large compared to the usual SO(4) physics precision. 18 µproton 3D/4D radius from 4-He (fm) magnetic energy uncorrected 4D correction µproton correcting with µp perturbation Top down mass using 4D potential Error ratio proton mass Alpha quantization for 3D/4D (1-(alpha/(PI()*16)))^2 Mass corrected by above factor proton mass mass difference relative error 1.4106067873 0.837650000000 926'613'063.470 0.9878501147 938'009'774.596 937'999'671.493 0.9997204364 262306.703831434 0.9997096686 938'282'187.337 938'272'081.300 10'106.037 0.0000107708 µproton 3D/4D radius revers calculation magnetic energy uncorrected 4D correction µproton correcting with µp perturbation Top down mass using 4D potential Error ratio proton mass Alpha quantization for 3D/4D (1-(alpha/(PI()*16)))^2 Mass corrected by above factor proton mass mass difference relative error Tab. 5a Proton magnetic mass-calculation 1.4106067873 0.837653007404 926'603'083.121 0.9878501147 937'999'671.495 937'999'671.493 0.9997096686 272409.804500937 0.9997096686 938'272'081.302 938'272'081.300 0.002 0.0000000000 Tab. 5b Optimized radius The second perturbation or potential mass of the proton mass (1-(α/(π*16)))2 is exactly 272'409.8 eV if derived from the proton mass. The result shown in Tab. 5b is the reverse calculation starting with the proton mass and the known mass/wave structure. Formula (10) is the fnal proton magnetic mass formula that shows a possible α-quantization due to the α2 term that can be co-linear with a 3D,t excitation. This allows the prediction that the proton mass based on the magnetic moment can undergo a quantization. (r=0.837653007404/2 !) The frst fve (unperturbed) levels of the proton quantization are the following: (2'002.34, 4'034.33, 6'096.64, 8'189.95 ,10'314.96 eV using ((1/α) -n); n= 1,2,3,.. ). In [20] [21] [22] the experimenter(s) found that at 1 keV particle (proton) stimulation energy strange resonances do occur. 1 keV is half of the frst alpha quantization. This is the correct 3D,t resonance energy as in a SU(2) x SU(2) quotient only one half (outside running mass) can kinetically interact! The cut-of of the spectrum seems to ft the quantization. Similar results can also be measured for dense Hydrogen (see below chpt. 8.1). Another interesting aspect is that the proton quantization (1-(α/(π*16)))2) delivers very exactly ¼ of the de Broglie radial potential energy, that can be further refned by the known (missing) 1FC 3 radial perturbation. Seen from this perspective, we can say that the quantization energy (with high precision) is directly coupled with the classic potential energy as seen in experiments [20][21][22]. As you may see in tab 5a already a “small” deviation in the radius leads to a “large” error in the overall ft. Using the best experimental radius approximations (virtual deuterium model [23]) gives errors in the 200eV range but with a much larger error bar! This also explains why experimental radius data (< 5 digits known!) could never contribute to fnd a new physics model. 5 SO(4) orbit based modeling In the frst modeling phase we did use the simple averaging (of p,n) approach, that works fne for nuclei with N=Z. But latest after the discovery of the neutron 3:2 wave asymmetry it was clear, that for a deep understanding we need a more exact method. The orbit based modeling does only look at protons and electrons as basic building blocks. The orbit forces are proportional to the 3 force constants depending on the topology of the orbit. The basic building block of all higher nuclear mass is the p-n = p-e-p bond we see in Deuterium. The term bond is simplifying the fact that due to added rotations charge (the potential energy) is converted into magnetic fux and thus the potential gets folded what leads to an increase of the fux mass and hence an increase of the binding force. 5.1 The orbit structure of Neutron,Deuterium, 4-He To understand the neutron the best way is to look at Deuterium and the neutron in parallel see Tab.6 a/b below. The Deuterium orbit model (Tab.6b) is simple and the dark blue feld Tab.6b shows the formation energy released when deuterium is formed from e+n+p. The Deuterium internal charge from the neutron shows a 5 rotation charge based coupling. This can be seen from lines Tab. 6b(d) and Tab 6a(d). The neutron internal charge bond did free one electron perturbative mass whereas in Deuterium 4 more electron perturbative masses mep must be discounted, what gives a total of 5. This perfectly fts the three 3 rotation proton perturbative mass waves and the two electron mass waves that together form a 5 wave virtual 19 charge coupling. The charge coupling orbit is the proton 1FC orbit. Basically the two protons inside Deuterium share the 3 rotation orbits and form a new 2 rotation 1FC orbit that together, analogue to QM reasoning, assume a hybrid state in reference to the coupling manifold. a) neutron excess mass b) Neutron excess energy (a – me) c) Proton 4D potential 1D mppo d) electron perturbative mass mep electron mass (eV) e) electron relativistic mass mer remaining mass : b - c – e f) 2* 4D 1FC potential = (2*mp*1FC') g) 2*f*2FC' h) 2*f*1FC' i) g + h Rest excess (delta 3D/4D pot!) k) repulsive potential 2*mppo*1FC' eV 1'293'332.000 a) DD bond (n+p+e) delta E b) 2*proton 3D pot = (2*mp*2FC') c) 2* 4D potential (second radius) d) 4* electron perturbative mass e) Delta before repulsion (a-b-c-d) f) 2*3D pot of 4D pot mass g) 2*4D pot of 4D pot mass delta mass (e+f+g) 782'333.054 272'409.807 1'183.104 510'998.946 509'815.842 107.405 40'499.503 94.073 1.748 95.821 11.584 rest error absolute 2'224'572.773 2'179'436.654 40'499.503 4'732.415 -95.798 94.073 1.748 0.023 Tab. 6b Deuterium orbit mass a) 4-He from deuterium delta E b) 2* perturbative proton mass mpp c) dense charge mass merb e) sum a – b – c f) 2*1FC'*m p g) 2*f*2FC' h) 2*f*1FC' I) e + g – h 11.758 0.174 Tab. 6a Neutron orbit & coupling masses 23'846'533.869 23'337'993.471 508'632.739 -92.341 40'499.503 94.073 1.748 -0.016 Tab. 7 4He orbit masses Mass modeling in SO(4) is relatively simple as we only need to explain the diferences of masses, because these are linear dependent on the known masses and their coupling. If you only look at the neutron then in Tab. 6a) you see that in the formation process one mep is released line (d) and 2 mer do couple and share one mep. During the formation of Deuterium 4 more mep get released, which gives in total 5 mep, that are released, if we form Deuterium from proton an electron only. The coupling mass weight - (1x1) orbit/ mass - remains the same as inside the SO(4) frame the charge mass is mep for every involved 1FC coupling dimension. (This also holds for the electron alone!) Thus Deuterium primary is formed by one additional 1FC bond (1x1) between the two proton masses. Tab. 6b, line b). In the joined fux plane = 2 dimensions the de Broglie radius potential**** gets released and the orbits do join. The two symmetric (neutron-) charge masses change from a 1x1 orbit to 1x1x5 orbit. The two corrections Tab. 6b(f,g) are the “missing” coupling of the 1x1 orbit with the electric forces (1FC, 2FC) and the released potential Tab. 6b b). The neutron owns two proton potential masses whereas Deuterium has 3. Now it is easy to understand the free neutron mass Tab. 6a). The symmetric charge masses mer in Deuterium and the neutron are the same, the coupling mass/dimension is also mep . But the neutron must cancel the proton charge this is done by an “anti symmetric” proton 4D potential generating mass. The sum of these two masses mep + mppo (Tab.6a, rows b -c - e) is already 107eV close to the neutron excess mass. The small corrections are the same as for deuterium. The remaining 11.6 eV are given by the repulsion between the two proton 4D potential masses ( mppo). May be repulsion is not the appropriate term as in fact this is an excess-energy “bond” that only classically looks like a repulsive terms as energy is needed to form it. One solution to the question why the free neutron is unstable is: SO(4) orbits must be symmetric in relation to the “center” of mass. In the neutron p-e bond the 1FC orbit only covers 2 rotations what corresponds to the standard 3D,t space electron situation. The two proton potential wave structures contain excess energy that looks like a hidden potential. “Normal” nuclear mass has 3,4,5 rotation coupling that is not conform with radiative mass. The other point is that the internal binding charge should be virtual only. But in the neutron case it already contains a part of the free electron mass. ****The release of the two de Broglie radius potentials can also directly be seen in the frst 6Li gamma line (2186.2keV) that is given by the sum of the 2 de Broglie radius potentials and the 5 charge perturbative masses mep .(and tiny perturbations due to the stronger binding of the 4He core mass). 6Li is more or less a 4 He core with an orbiting Deuterium with two “magnetic bonds”. (see chapt. 6.3) 20 5.1.1 4-He Orbit structure The formation of 4He from Deuterium is straight forward to explain (see Tab. 9). The full 3D/4D fuxes of the 4 protons join their orbits and migrate to a radius with double the proton relativistic radius. As we will see later if the radius doubles, then the internal charge doubles too, thus half of the charge generating 3D/4D fux gets released and the same happens to the associated dense charge mass. The resulting error in mass is -92 eV. But this is only a summary view with no orbit details! From the modeling we know that the formation of 4He releases a 4D quantum of space-time/mass. We also cannot see what happens to the “internal” deuterium bonds. The initial model for calculating the 4He mass defect was using the averaging method of n-p bonds and the 2FC/3FC compression of fux. This did neglect that a small amount of compression in fact was mass neutral as the two neutrons internally do balance the hole and excess waves. In Tab. 8 we show a summary of this classic approach that initially did help to fnd 3FC – the strong force factor. Classically we know that proton and neutron inside 4He form out 4 (2x2) connections and one additional 3FC (strong force) quantum gets released. This a) 4-He mass 3'728'401'292.0028 b) proton mass 938'272'081.3000 slightly overestimates the compression you see c) neutron mass 939'565'413.3000 in Tab. 8 line (h) by about 2keV. This d) electron mass 510'998.9461 compression efectively happens without a mass e) 4-He particles mass 2*(a+b+c) 3'756'696'987.0922 f) compression =(e*(3FC'+4*2FC') 28'297'556.7110 defect because the neutron hole wave and the g) measured compression (e-a) 28'295'695.0894 excess wave exchange mass behind the scene. h) over estimated (f-g) 1'861.6216 j) Neutron excess (c-b-e) 782'333.0539 This hidden compression is calculated in Tab. 8 k) one wave = (j/5 ) 156'466.6108 lines (l,m). From the alpha particle mass we also l) 2 waves full compr. =(2*k*(3FC'+2*2FC') 1'628.5757 know that the potentials get added. m) 1 waves 2FC compr. =(k*2FC') 181.7218 n) 4-He second potential o) total hidden compression (l+m+o) Missing (o-j) 54.4153 1'864.7128 3.0912 This again is an approximation only that shows the now “classic” picture. Tab.8 classic 4He compression with averaging But what happens when we look a the orbits only? For this we have to start with Deuterium where we 4-He from deuterium 23'846'533.869 already have an exact orbit model. First thing we do in Tab.9 a) 4-He from deuterium 23'846'533.869 is adding the additional 6 n-p (p-e-p) bond orbits what leaves b) 2*mp*(2FC') 2'179'436.654 about one 4D 3FC quanta of excess energy Tab.9 (c) 6*(b) 13'076'619.923 c) (a-6*b) 10'769'913.947 unexplained. If we subtract the weight of one 4D 3FC quanta d) 4D quantum 10'845'298.896 from the excess Tab.8 (e) then we see that we overestimate e) (c-d) -75'384.949 the compression. But from the classic formula “we know that f) Deuterium 1FC bond 40'499.503 inside 4He there exists no 1FC compression that works inside 80'999.006 g) 2*mp*(1FC') freed bond h) free 2* 1FC bond of DD (g+e) 5'614.057 Deuterium. After subtracting the no longer existing j) me p 1'183.1037 compression we have a remaining excess energy of 5.6 keV k) 5 charge bonds free -(h – 5*j) -301.461 Tab. 9 (h) If we assume that the ffth rotation analogue to the (f)*(3FC') 116.919 th 4 causes one mep to be released/dimension (in total 5) then (f)*4*(1FC'+2FC') 184.650 Sum added mass from 301.569 we overestimate the compression by about 300 eV Tab.9 (k). unexplained 0.108 Tab.9 4He compression with orbits A possible explanation could be a coupling of the now virtual (carrying no mass/fux ) 1FC orbit with 1 dimension. This is reasonable as the energy of 4He stays in 4 rotation dimensions (5 real ones) and 1D is not occupied. But we still do not know - what is a 3FC quantum? How does the 3FC orbit look like? (Precision: If we don't use 1FC in Tab.10b(f) neutron has the same size. ) we get 3.6 eV unexplained more or less identical to Tab.9, But the NIST fudging for the 21 5.1.2 The true 4He story Here we start with the 4 protons only Tab. 10 (a). Compared to 4He they have a large excess mass. We already know that 4 of the 8 potential waves must vanish Tab.10 (g). Further we know that the 3D/4D fux must undergo a stronger compression that only works in 1 rotation dimension. The coupling of the internal masses changes from 3 to 4 dimensions and thus the force metric changes by 2 1/2. The proton force factor is given in Tab.10 (b) the 4He force factor is Tab.11(d). The absolute compression induced by the force factor is given in Tab.11 (e). If we subtract the compressed fux from the proton excess (h) and the 4 line (g) lost ( 2 from protons 2 from neutron excess mass) potential waves about 2MeV - line (j) - are left. We did not yet account for the 4 electrons that are originally contained in the neutrons. If we subtract the 4 electron masses the we get Tab. 10 (k). From the modeling above we know that that the particles (deuterium) internal binding charge mass in total is 5*mep2 . This gives the fnal result Tab.10 (l). As we know the alpha particle owns an excess mass of 83.9 eV! Thus this is within the current measurement error as the unmotivated NIST fudging already is about 3 eV... What is the interpretation of the above calculation? This looks like the four electrons 2 shell / 2 inside original neutrons do only contribute with their mep2 . All relativistic charge mass of the electrons is released a) 4 proton masses 3'753'088'325.2000 during the formation of 4He. As expected due to the b) force factor 1FC*2FC*3FC 0.9959335244 SO(4) 5 rotation (5 x 2) coupling we see exactly 5 m ep2 c) metric factor 21/2 1.41421356237310 as binding charge mass. Now the virtual orbit is fully d) folded force factor ( bc) 0.9942539823 occupied by (1x1) x (5) mep2. 4 e) 1 wave He pert. (a*d) f) 4He mass g) 4 proton potential waves h) proton mass excess (a-f) j) (h-e-g) k) (j - 4*me) 21'565'311.8980 3'728'401'292.0028 Thus the mass of 4He can be found from the proton 1'089'639.2263 mass and the correct compression factor and the 24'687'033.1972 release of halve of the potential mass (g) + 5 mep2 . The 2'032'082.0729 remaining excess compression (l) is exactly compensating the measured alpha particle excess -82.6745 mass! (That is slightly larger than the sum of the 83.8894 electron potentials 79 eV = 24.6 + 54.4 eV) -11'913.7115 l) (k+ 10 * mep) alpha masse excess measured Tab.10 4He compression with 4D orbits From the modeling of the Neutron and Deuterium we know that originally 4 mer + 4 mppo (the neutron excess mass waves) stay inside the particles if we start from Deuterium, that fuse to 4He. The modeling of Tab. 10 shows that efectively 4 additional charge masses mer2 + mppo get ejected (+ 2*mep). This indicates that indeed the forming of a strong force orbit is quite a complex process as all involved masses do undergo a signifcant transformation. From the electron, neutron etc. we know that a charge coupling bond in 1 rotations frees 1 mep . 4He has a fve rotation bound over two dimensions so efectively 10 mep must be released if we start from protons. In line k Tab. 10 we show the result of subtracting 4 full electron masses. Here again we started with 4 protons only and did not account for the 2 electrons in the shell. If we now add the two electrons the 2 out of the 4 ejected electron correspond to the shell electrons and two charge masses still stay inside 4-He. It is obvious & clear that 4He is not containing any internal neutron structure as 4 charge masses are freed that would stay in the neutron if we start from Deuterium. Also the 4 additional potential masses from the neutron get released. Inducing a neutron structure to 4He needs about 21'010 keV of added energy. This is about the energy of the perturbative mass wave Tab. 10(e) minus “an electron mass plus the induced 1FC bonds” for the internal Deuterium structure. During this process the internal electron mass of the neutron is regenerated from the perturbative mass wave. 5.1.3 Is this picture now complete ? One important thing we must learn: We are dealing with 6 dimensions of mass/feld interaction, thus all groups of projections to 3D,t we fnd, that are exactly conform with measured data, just show one specifc 22 aspect of mass (trans-) formation. If we see that the proton fux compression globally is exactly given by the value in Tab.10(d) then this still allows that this compressed mass part Tab. 10(e) has an inner structure! The only requirement is that the sum of the structure must be conform with the now known compression factor. We don't see any simple change that points to a change in the original internal Deuterium 1FC bond, that still could be hidden inside the structure of Tab. 10(e). If you do not trust the applied 2 1/2 factor, then you will see that this is exactly, what the strong force equation – given below - demands. Coupling charge Q 2 is proportional to the rotating mass and radius (see next chapter). Relative to the proton this factor increases by 8 this implies that the force of the charge does (must!) increase by (8/4) 1/2! where 4 is the efective internal! 4He (Q2 ) charge. The coupling factor( as already shown in the neutron radius calculation Appendix A or 4He compression) goes into the exponent as in “reality” this is a coupling of harmonic waves, what leads to a logarithmic increase/decrease of the radial force. Finally after fnding the basic structure of 4He we must confess that we “luckily” - by intuition - always did make the right decisions, when we assumed that 4He is a derived structure from the proton and the two relativistic radii must be related by a factor 2, etc.. One more thing we can learn: In the proton case that only covers 3 rotations, the 3D/4D mass is expanded. In the 4He case 4(5) rotations we see a compression. Thus only waves that fully cover the Cliford torus surface doing 4 rotations (on one side) undergo compression. This also happens with the fake Higgs particle masses proton resonances. From a mathematical point of view all higher dimensional spaces imply a more complex internal structure. A 4D space has a 4D volume/surface and a 3D volume/surface. A 5D space has a 5D volume/surface, a 4D volume/surface and a 3D volume/surface. Further if e.g. the internal 3D space is not connected then we have diferent internal volume groups like 5D volume/surface, a 4D volume/surface and (a 3D volume1/surface1, 3D volume2/surface2). So spaces can decompose recursively. This is already the case for the Cliford torus 3D volume that is formed by two disjoint tori! 5.1.4 The 3He structure a) 3He mass b) 2H mass c) proton mass d) mass defect from 2H (b+c-a) e) single e-p 2FC bond f) single e-p 1FC bond g) 2 more deuterium 2FC e-p bonds (2*e) h) 2 more deuterium 1FC e-p bonds (2*f) j) discounting two 2H waves (d-g-h) k) proton potential wave l) 2 proton pot. waves (2*k) m) discounting 2x potential (j-l) n) electron perturbative mass o) refilling 2 mep due to 5-->3 rotations (2*n+m) p) 1FC-2FC coupling see ( 2H) =( 2*2*(1FC'+2FC')*f) q) subtract coupling r) neutron internal potential repulsion (2*k*1FC) rest error after freeing neutron repulsion 2'809'413'492.6119 3 1'876'123'913.2785 He is an exceptional nucleus as we claim that's 938'272'081.3000 the only stable one with with Z>N. If we add a 4'982'501.9665 2'179'436.6538 proton to Deuterium then 2 new rotations form 40'499.5031 out that are synchronous to the Deuterium 4'358'873.3075 “bond”. Tab.11 lines (g,h). The neutron structure 80'999.0061 542'629.6529 dissolves as the two internal proton potential 272'409.8066 waves (j) get released and also the repulsive 544'819.6131 -2'189.9602 energy Tab.11(r). Deuterium has a 5x1 or( 5x 1'183.1037 0.5x0.5)rotation charge coupled orbit and in 3He 176.2472 188.1461 -11.8989 11.7583 -0.1406 its obviously a three rotation coupled charge as 2 mep out of the 5 Deuterium orbits must be reflled. Tab. 11 3He orbits Line (p) Tab. 11 is just the usual Deuterium perturbation/p-p bond you see in Tab. 6(b) lines (f,g). It is quite astonishing that we only need to know the Deuterium orbit model and be aware of the fact that neutrons in strong bonds get partially – 3He - or fully – 4He – dissolved! 3He still contains the two neutron charge masses! 3He is superfuid. I will be interesting to see whether this is due to the fact that both, the mass and internal binding charge wave, do have a three rotation symmetry, whereas in 4He its a 5:4 relation. Summary: 3He is formed from 2H + p by forming out 2 more Deuterium p-p bonds this includes the perturbations (Line (p) Tab. 11) too. The internal charge mass coupling changes from a one dimensional symmetry to a full 3 dimensional (rotation) symmetry (of course in 6D!). The two 4D potential masses from the internal neutron get “dissolved” (released), together with the repulsive energy. This does still not tell how 23 the real structure (charge radius?) of 3He looks like and what happens with the two former internal charge masses ? One electron seem to replace the missing shell electron (we added only one p no electron!) . If one electron – from from the neutron - just moves to the shell then also 3He contains more or less no real internal neutron structure! 6 The proton and 4He inner force equation(s) In our model we assume that the magnetic fux in SO(4) is bound to the surface of the projected 3D torus and the “virtual charge” stays on the 5D torus center line. (Thus in classic 3D the magnetic fux is homogeneous inside the projected 3D torus.) This 4D model refects the diference in dimensionality of charge/magnetic fux. Normally magnetic fux occupies one more space dimension than charge. A classical pictures we can use: The torus surface that encloses the magnetic fux is the time horizon of the EM-fux/mass from which it cannot escape. Thus the frequency (in radians) that defnes the amount of current or fnally the mechanical centrifugal force on the mass is given by the radius rpr and the speed of light and the number of windings the magnetic mass takes. One problem we face is the proton's wave structure that is given by three entities doing 4,3,2 rotations each. As we already know, most of the mass is concentrated in the 4 rotation structure and if we know start we this mass part only, then we will get an equation that is slightly of balance. We also know that the charge orbit in SO(4) is doing 5 rotations and the two remaining masses in total do 5 rotations. Thus after fnding the slightly perturbed mass-force equation we will look for a symmetric 1x5 perturbation that is able to correct the miss balance. (11) mpr = µp2*4* π * 100000/(α * π* rpr3*e) = 926'603'083.294 eV Remember that rpr always is just rp4D/2 as it is the real 4D torus radius. Because in dense space all magnetic feld lines are contained inside the current loop (due to the complex 4D rotation) the Biot Savart force (integrated over the torus cross section) and the coulomb force are interchangeable. (Only under full orthogonal torus symmetry!). Thus we will look for a coulomb-coupling that compensates the mechanical force. For this purpose we will focus all mass in one point using the proper metrics The base frequency of the charge that fnally defnes the coupling current (charge) is given in (12). (12) ω = c/(4*21/2*π*rpr) = 0.2013871189 E23 On a torus the combined trajectory that covers both circles and radii has the following length that defnes ω = c/(4*21/2*π*rpr). 21/2 simply is the group measure of SO(4) for one radius. This is the true frequency and not a projected one. From this we can derive the projected mechanical (centrifugal “cf” (13) ) force on the EM point mass that in SO(4) has a constant distance rpr to the “center” of rotation. (To remind you once more: In SO(4) the efective center of rotation/mass is the total surface on the single sided Cliford torus boundary!) . But EM mass is “mechanically connected” by the induced charge that in this frst approximative approach stays at a constant distance of rpr from the Cliford Torus surface. (13) Fcf= mpr*mpr4D*ω2 rpr = 280.6647723036N Eq. (14) below is the metric/fraction (mpr4D)of the proton mass, where as mpr is the fraction of mass that is rotating in 4D. (eq(11)). We do sum here all orbits that are conform with the (1x1)X(1x1) symmetry also written simplifed as (2x2). The potential wave mppo is (1x1) and also the Bohr potential mei is (1x1) – its a negative value. We know from the alpha particle mass that the electron orbits must be added to the net mass. (14) mpr4D := (mpr + mppo + mei)/mp The following equations treats the charge as a point charge. Energy → the integral over the total torus surface. The electric force Fef4D (15) the 4D coulomb force over the same distance -using the torus norm - is given as 24 following: (15) Fef4D = e2/(ε04 π2 rpr 2) = 418.6431608349N If we make the simple quotient then we get: (16) (13)/(15) = Fcf / Fef4D = 0.67041528098495 This (16) is roughly 2/3. Why? The distance between 2 current circles (virtual ring currents is not rpr! Its (3/2)1/2 as the true distance is given by 3 components (vectors), because we added on more radius to model the fux tube. The center of the circles has a distance of r in the projection only but not the average path of attraction in SO(4). This again is somewhat simplifed as the true relativistic 4D radius is rpr / 21/2 given by the metric and all points stay on the Cliford Torus surface. And thus the distance between to “parallel” circles ( we need to use unit lengths for ri! to get the norm) is e.g. (rz2+ry2+ru2)1/2/ 21/2 =(3/2)1/2 rpr.- all having the same length. (The charge does not stay on the Cliford torus surface where we can only map 4 rotations without adding one more dimension!) The fnal derivation Including Eq.(12) ω of the simplifed force model is. (Due to r2 in eq (15) ). Also ε0 and c2 get replaced by 4 π*10E-5 (µo). m'pr is written for mpr*mpr4D. (17) Fcf / Fef4D =(3/2)*(rpr*m'pr/(4*π*64*e2)) = 1.0056229215 (18) Fec4D / Fcf = 0.99440851898129 Equation (17) is the reduced quotient that we now call the strong force equation and shows that “the internal electric” binding force indirectly increases with the radius – more charge is generated - , which is in agreement with the strong force behavior seen in experiments. This counter intuitive efect is due to the way the frequency is defned. The frequency factor ω2 decreases with r2, which implies that the centrifugal force decrease with increasing r (stronger dilution of magnetic density = coupling mass!). From the 4D physical point of view the explanation is that charge Q 2 (e2) produced is proportional to r and also to the product of radius*mass, thus in reality the central electrical charge force increases if we try to split the relativistic mass. We just want to remind the reader that we usually treat coefcients as being <1. This, (18) 0.9944.., is a very good match for this simplifed model that only respects the proton relativistic mass and the directly attached symmetric potential orbit. We also expected this value from the compression factor we did see in last 4He model. But in reality the center of mass coupling is the SO(4) charge radius what we already corrected. What is very difcult to model is the connection of the 5,4,3,2 rotation masses. The only (currently) feasible approach, that does not need a lot of modeling is looking at the orbit relations. 6.1 The perturbation of the strong force orbit The diference in rotations between the proton relativistic mass and perturbed mass is 1 rotation (1x1). This is also responsible for the unfolding of the proton potential (also 1x1). Thus the expected perturbation must be proportional to 2FC the potential folding factor for one dimension. Further we see two coupled torus (SO(4) inside outside) which will lead to a product of 2FC with 1FC (the torus charge coupling force). The coupling 3D/4D torus runs see Fig. 4 (left side) over both dimensions of the 4D torus, thus the coupling involves 2 times 1FC – the torus second radius force (derivation of 2FC) In fact the expected perturbation of 2FC*(1+2*1FC') does give the exact deviation for one dimension. The energy of an unperturbed orthogonal 5 rotation system is given by the trace of the corresponding tensor. Because only a linear (1 rotation) amount must be added per dimensions that is equal for all dimensions we can also add the summary value to the sum of the trace (5* perturbation for 1 rotation). The line (l) of Tab. 12 below gives the value for all 5 dimension. It is just the sum as only one radial dimension is involved (no r2,r3 coupling). Once more. 1FC' is the correction for the second radius torus force. Usually if we fnd a general solution that is conforms with the SO(4) modeling the chosen approach is 25 Rrp relativistic proton radius a) relativistic proton mass at Rrp b) 4D potential (1D) at Rrp c) coulomb potential d) mf=orbiting mass factor (a+b+c)/mp e) orbiting frequency c/(4*21/2*pi*Rrp) =w f) mechanical force mpw2Rrpmf g) toroidal Coulomb force e2/(ε0pi 2Rrp2) h) Ratio (f)/(g) j) correcting charge radius factor (h*3/2) k) Factor (1/j) l) 1-5(1-2FC(1+2*1FC')) matching (l/k) 0.837653007404 926'603'083.294 272409.80657182 27.2113860282 0.98785361342663 0.20138711898941 280.6647722861 418.6431607828 0.67041528102673 1.00562292154009 0.99440851891932 0.99440852029356 1.00000000138196 safe. In [24] we did frst show the all digits exact mass equivalence formula for the proton-electron particles. This formula was already based on the assumption that charge is running on 5 rotations. Also the Deuterium model shows a 5 rotation coupling and of course 4He. Thus this was the fnal confrmation of the 5 rotations coupling structure. Tab 12. Proton inner force summary The reduced force equivalence formula (17) above shows that the force quotient is proportional to mass. Mass is always the sum of all rotating masses = Eigenvalues of all 5 dimensions. Thus the found perturbation of the proton inner force equation works symmetrically over the full SO(4) space. Because SO(4) orbits work/couple circularly we could also argument from both sides. It is just a convention to model perturbations with values <1. E.g. the strong force equation shows that the compression of the 4x4 fux is only 1.0056229215. So the rest must be delivered from the 5 rotation coupling. In that case you have to multiply 1.0056229215 by the 5 rotation perturbation being 1-5(1-2FC(1+2*1FC')) = 0.994408520293555, what gives you a match of 1.00000000138.. (rounded). This is extraordinary as we just used a “crude” approximation. What is not used here is the small perturbation between the potential wave and the 3 rotation fux. 6.2 The 4He coupling force. The new model for 4He did show that basically 4He is made up of 4 protons. Here, Tab.14, we do look for all orbits that are conform with (1x1) or (1x1) X (1x1) or (((1x1) X (1x1)) X ((1x1) X (1x1))). Line (e) Tab. 14 is the sum of 4 proton relativistic masses, line (g) has the weight of 4 potential masses. efectively – if you look at the building of 4He only 2mppo do survive but the radius the waves runs on doubles and thus the weight increases by a factor of 2. The 10 perturbative charge masses (m ep) we use are still conform with 5 charge x 2 rotations. 4-He mass a) charge mass of deuterium b) 4D potential (1D) at R rp c) coulomb potential d) relativistic proton mass at R rp e) core 4D mass (4*d) f) mep g) 4* (b) h) 4-He measured mass j) ((a) +(e) + 10*(f))) k) (j)/(g) l) proton radial perturbation n) deviation (l/k) m) ((a)+(e) + 10*(f)-10*c)) o) deviation using (m) instead of (j) 3'728'401'292.003 40'499.503 272'409.807 27.211 926'603'083.121 3'706'412'332.484 1'183.104 1'089'639.226 3'728'401'292.003 3'707'554'302.251 0.9944085982 0.9944085203 0.0000000783 3'707'554'030.137 0.0000000049 The only exception is line (a) the occurrence of one 1FC bond from an internal Deuterium structure. This is what we hinted at above when we said that the whole 4He perturbative mass can be given as one “wave structure”. So far we did not yet look at it and this is as frst indication for further details in the 4He 3D/4D wave structure. From experiments its obvious that there must be a small asymmetric internal charge mass as 4He has a quadrupole moment! Tab.13 4He 4D coupling mass 6.2.1 Conclusion of 4He coupling force In line (l) Tab. 13 we gave the proton 4D mass perturbation factor we did correct with a 5 rotation SO(4) 26 physics conform coupling. Now we see the surprising outcome of the 4D 4He coupling mass factor. It's an exact mirror of the proton perturbation, what allows the conclusion that the 4D mass of 4He is an exact complement of the proton missing mass coupling. This is the second deep insight into dense matter structure what will allow to get an even better understanding of 4He. If we treat the 4 building protons as individually rotating systems and then couple this system with the 4He cluster then, at the end, the compound looks like a big fat proton with the standard proton 3D/4D mass perturbation given in Tab.13 (d). This also explains why the calculation in Tab. 10 works extremely well. An other note: The proton mass factor (EQ(14) = 0.98785361...) is 5 digits conform with the proton perturbation = (3FC*2FC*1FC)3. The remaining quotient of 0.9999964583 is close to the known quotient for the electron/charge perturbation (0.9999964461...) using the same logic. This is open work on the 8 th digit of the perturbation... Disclaimer..: We here just the frst time show some interesting relations inside a dense mass structure. 6 dimensional – 5 rotation reasoning is far from easy understanding. Further we work with projections that also would allow alternate paths to be taken. Even the 4D relativistic radius of the proton could have a very tiny correction : 8376530073518 instead of 837653007404. So everything beyond 8 digits precision looks nice but could change with a more deep understanding. 6.3 Modeling Lithium-6 a) mep b) mep rot1/9 c) electron photonic mass rel. ((me-b)/me) d) 6Li mass e) 2H mass f) 4He mass g) delta mass = (e+f-d) h) 1D pert j) 2/3 2FC'+1FC' proton wave k) 1D compression/charge 1/9 =(j*h/c) Rest error absolute (g-k) 1'183.104 131.4559670958 0.9997427471 5'603'050'886.4772 1'876'123'913.2785 3'728'401'292.0028 1'474'318.8041 0.9959335244 1'479'957.4379 1'474'318.4999 0.3043 Tab 14.a) 6Li mass formation from 4He and 2H 6-Li first line a) 2*proton 3D pot = 2*(mp*2FC') b) mep c) virtual mass = b) *1/9 d) charge rel. compression. ((me-c)/me) e) Flux compression 6Li in eV f*(1-d) f) Flux expansion a/d2 Total line energy e+f 2'186'200.0000 2'179'436.6538 1'183.1037 131.4560 0.9997427471 5'638.9380 2'180'558.4195 2'186'197.3575 Tab. 14b) 6Li frst gamma line Lithium-6 is straight forward to explain as we basically can say that one deuterium nucleus is orbiting a 4He core. What we must take into account is the diferent rotations number (in brackets) of the 4He core (4) and 2 H (3). As you can see from the energy of the frst gamma line of 6Li, is that its energy almost corresponds to the known p-p “bond” we above calculated for Deuterium, see Tab. 14b line(a). Each p-p bond splits into 3 waves = three rotations, what we already do know from the proton. The deuterium internal binding orbit does 5 rotations and the p-p bond only uses 3 of them, thus two more are “free” to join the 4He wave structure. Tab. 14a line(j) shows that 2/3 of a p-e-p (2FC'+1FC' 2H bond) bond would release about 5.639keV (diference of lines (j)-(k)) more than we see from the real fux compression line(g). So the the added 2/3 of p-e-p fux does acquire an excess mass. The excess mass perturbation for one dimension is known and = 3FC*2FC*1FC Tab. 14a line(h). On the other side the binding charge (m ep) has to migrate from a 3 → 4 rotation orbit what implies that the 9/8 torus excess energy rotation becomes and 8/8 Cliford torus surface bound rotation. Line c Tab14.a gives the related force factor for 9/8 – 8/8 of the binding potential. If we apply the two perturbations then the calculated mass is exact. It is nice to see that the frst gamma line confrms this fnding as the same amount is added to the p-p bond energy that gets emitted by the gamma ray. It also confrms that in fact only the p-p bond splits an the charge stays bound. One very interesting fact can be seen that we in Tab 14b(f) write the energy as proportional to the p-p bond and in Tab. 14 a(k) as compression. So what happens if the p-p bond breaks the 2/3 wave bond orbits suddenly become attractive and the compression reverses which frees extra energy. Even more impressive 27 is the symmetry in the two energies which is obvious if you understand how magnetic energies can go into resonance! 7 Proton – Electron mass relation top down 4D proton radius a) (8/9) magnetic mass of proton b) reduced charge mass e/4π c) metric change 1D d) 4D charge mass to subtract =(b*d) e) weight of Mpr – charge = (a-d) 0.837653007404 eV 823'647'184.997 40'664.004 1.4142135624 57'507.586 823'589'677.410 f) electron mass 510'998.946 g) electron perturbative mass 1'183.104 h) charge expansion 2-3D =(f+2*g) 513'365.154 j) going from 2--> 3 rotations = a-3/2 1'604.176 k) Relativistic Mass electron 3D =(H*j) 823'528'042.048 l) metric factor for 2-->3(5) rotations ( 23/5) 1.5157165665 m) electron charge added 2-->3 rot =(b*l) 61'635.105 n) Rel. Mass electron 3D + charge (a-m) 823'589'677.152 delta projected mass 0.258 top down proton radius 0.8376530074046 (8/9) magnetic mass of proton 823'647'184.99473 charge flux expansion (23/5 +21/2)*me/4π 119'142.69075 a) 8/9 proton mg. mass + reduced charge 823'528'042.30398 513'365.15367 b) reducing : a) * alpha 3/2 c) electron pert. Mass 1'183.10370 calculated electron mass = b - 2*c 510'998.94626 electron measured 510'998.94610 Tab. 15a p/e Torus mass projection Tab. 15b electron mass from proton mass For the proton electron mass relation (Tab 15a) we use the proton relativistic mass that can be exactly calculated from the proton mass. The proton has the rigid mass form factor of 9/8. Thus to get the mass equivalence we choose 8/9 of the proton relativistic mass. This is equivalent to stopping one rotation that produces the 9/8 of mass. The charge-mass associated with this operation is stopping one rotation * group measure. From the electron side we must use the excess mass formulas. First we will extend (fll) the orbit with a second mep and then we have to add one more for the added 3thd excess rotation. Thus we must start with the opposite as the dense electron mass. Then we must speed-up the electron from 1 x 1 x (0.5x0.5)) rotation to (1 x 1) x (1x1) rotation (adds 0.5 rotations to both sides) and then from 2 → 3 rotations this factor of total 1.5 can be seen in the α exponent being -3/2. The associated charge mass for going from 2 → 3 (double sided) rotations (out of 5) is given by the factor 2 3/5. To repeat it once more: Already Mills found that α is the length contraction for a relativistic rotating frame and e/4π is the mass reduction if you stop a rotating charge-mass. (Mills missed the torus norm!) Relativistic mass increase/decrease in rotating frames is fnite! Mass - increase is given by adding more rotations as you also will see in the short section about the “fake” Higgs particle masses. The change in charge mass for 1--> 2 rotations did only afect the perturbative mass which is refected with starting at me+ 2* mep . The factor 23/5 = 1.5157.. has already been used to derive the proton relativistic radius from the neutron interaction radius (see Appendix A). It is the weighted sum of 3 rotations - out of 5 (waves) running along a single sided SO(4) manifold. In table 15b) the electron mass derived from the proton mass is shown. The only simplifcation we used is the pre-calculated electron perturbative mass (c), that depends on the highly precise electron g-factor. The top down calculation is of course much more precise because the proton mass is known with higher precision than the electron mass. 28 8 Spin or 1FC orbit pairing If two electrons on the same orbit couple then they join one of their (0.5x0.5) → (1x1) perturbative waves that forms the (3D,t) relativistic mass mer. The product of 1FC'*mer exactly delivers 11 eV. (Seen e.g. in 4He are 10.99). From a 3D,t perspective spins must be parallel to be attractive. Thus one key functionality of a so called dense Hydrogen catalyst must be supporting singlet Hydrogen with parallel spins. It looks like on catalyst surfaces (e.g. styrene catalyst) such small ensembles of H/D can form out large regions of spin paired orbits. Fig. 8a Electron fux torus Fig. 8b Joined fux torus. In SO(4) physics all waves have at least a (1x1) coupled orbit structure – also photons. Thus if (magnetic) fux joins, then always two rotation dimension are involved and the classic fux node (Fig. 8a green dot) in reality is a torus intersection circle like area see Fig 8b. At the node surface (circle frame) fux changes from outside to inside due to momentum conservation. In a circular confguration also called H7, H19, H37 these 1FC orbits can span the whole circumference. This Fig. 9 red orbit is a super conduction orbit and it is assumed that such a H7, H19 forms a stable molecular rotator. The fux released is proportional to the number of electrons and also defnes the frequency of the rotating mass. Fig. 9 Larger 1FC SC spin orbits There is good reason to believe that such long range coupled 1FC orbits in general are able to explain super conduction. The current research of Rydberg matter – dense hydrogen still is only based on a good heuristic approximation for orbitals and what people believe to be spin. In fact the standard model (SM) treats spin as a binary (natural number based) quantity, which is one reason it completely fails to understand complex spin-spin coupled masses. The other reason is the assumption that an electron is orbiting a nucleus and owns all classic mechanical properties like spin & orbital momentum. This view is a good approximation only in the cases where there are no longer magnetic resonances that perturb the “potential”. What future experiments must show is whether the 1FC orbits also do allow a quantization and whether it is possible, when more fux is running in parallel, e.g. in the from of two stacked (e.g. H7) rotators, that the free electrons (Rydberg electrons) do form two spin connections too. Because 1FC spin pairing delivers 11 eV for two electrons, this efect allows (in catalysts) to split H 2 bonds and also allows other electrons to migrate to higher (Rydberg) orbits. Now we do switch to existing Hydrogen spin matter. 8.1 Dense Hydrogen For more than 10 years Leif Holmlid has performed successful experiments [25][3][4] with laser ignition of so called dense or ultra dense hydrogen (UDH). Dense Hydrogen is the ultimate 1FC spin paired proton mass that is based on a pairing of 2 mpp. If an external resonance mass exists that is able to accept about 500 eV then two protons can join their 3D/4D excess fux on a nuclear 1FC orbit. In our terminology we call/note 29 the spin-spin coupled Hydrogen/Deuterium as H*/D*. As a physicist Holmlid grew up and still works on an SM basis, albeit his experiments completely invalidate most rules of SM. Initial explanations of UDH by Winterberg [26] are quite creative but are based on too many implausible stretches of classic physics. Protons orbiting electrons is from a symmetry point of view correct but the proton orbit radius must remain the classic tiny one. Also an electron at rest is a contradiction with classical assumptions and integrating a current inside dense matter using an unperturbed free space current is just fantasy. To make it short. No valid model for Rydberg matter and of course UDH exists so far as in fact SM is an over simplistic model of the physical reality of dense matter. a) mp p b) 1FC' spin pairing of 2*mp p = (2*a*1FC') 2D coupling with neighbor waves c) = (b)*(1FC+(2FC'*1FC'*2)) d) Bohr potential e) 2/3 wave- 1/3 freed potential = (1/3)*(d) Net energy gain H*-H* = (c-e) 11'668'998.0057 503.6797 504.8497 27.2114 9.0705 495.7792 The spin pairing of 2 mpp delivers Tab16 line (b) uncorrected 503.7 eV. This must be slightly corrected by the neighbor orbit forces. See Tab16(c). 2*2FC'*1FC' is the same perturbation we did see in the proton inner force equation. Because the potential bound 3D/4D fux couples now only with 2/3 the external visible potential is reduced by 1/3. Tab. 16 H*-H* dense hydrogen The value of 495.8 eV matches very closely with the value Mills measured for his pure “Hydrino” (H*-H*) condensate [27] Fig. 10 and slightly shifted due to Mo binding in Fig. 10 peek c. Also the peek halve width is in good agreement with the basic perturbation of Tab. 16 lines (c-b) with about (+-)1.2 eV. The orbit resonance values that are posted in a b c countless Holmlid papers are less reliable because such spin-paired H*-H* on surfaces do couple and form clusters of 3,4 and more atoms where some also can be in a normal Rydberg state and obviously do interact with the carrier catalyst nuclear mass too. The change in potential of the Bohr orbit – given by Tab16(e) (4.5 = 13.6-9.07 eV) Fig. 11 has been measured in a paper [28] about high temperature Rydberg matter for a wide range of temperatures! What we show here is just the underlaying physics of an overall efect that needs much more detailed modeling as in a catalyst many diferent interaction/resonances do work in parallel. Fig. 10 Mills XPS orbit resonance from [27] MO embedded H*-H* Efectively Mills [27] Fig. 10 peeks a,b, also measured the 503..505 eV peeks of Tab.15 lines (b,c). But these “peeks” were very small compared to the main orbit resonance. An even better match is seen by EDX Fig. 10. [29] below. If we would assume that only the proton 3D/4D fux joins – Tab. 16b, then we get a similar value but no explanation what happens with the H* Bohr potential. But again this helps to fnd the beginning of the peak in Fig. 8 around 492 eV a) mpp34 11396588.1990844 +- 1.2 eV b) only mpp34 coupling 491.9214455718 perturbation. c) proton one wave perturbation 3FC*2FC*1FC d) 2D coupling with potential = (b)/(c)2 0.9959335244 495.9467552028 Tab. 16b H*-H* dense hydrogen This is just the tip of the iceberg of a new domain in physics that just waits for more conclusive experiments & measurements. Changes of the desorption signal of the excited potassium species from cryptomelane at various temperatures. 30 Fig.11 from [28] Fig.12 [29] Shows the EDX version of H*-H* where the peak starts at 495 eV and tops exactly where we expect it. Fig.12 H*-H* EDX [29] spectrum detail around 495 eV. From the Deuterium orbit model it is clear that D*-D* can do up to 4 1FC orbit connections. In Tab.17 we use the same values as in Tab 16. If we look at the fnal energy balance then we may see the connection to the frst proton magnetic moment relaxation energy. This is no coincidence as the potential factor is similar to 1FC. Because the coupling weight of the potential doubles the overall potential of D*-D* is reduced by 18.1 eV. Unluckily we know no paper with Deuterium Rydberg matter so far that measures the desorption potential. 4*s-s energy 2* potential Sub 2* potential lost (9eV) 2'019.3987 If Deuterium does only two connections we get again the 18.1409 famous 1 keV proton resonance fgure. 2'001.2578 Tab. 17 D*-D* dense deuterium 9 The Holmlid Hydrogen to Kaon reaction(s) m4He 3'728'401'292.0 mp 938'272'081.3 a) 4 m p- m4He b) Kaon 0 mass c) Kaon+ d) sum (a+b) K+ + K0 e) mass of 2 K+ +ef) Mass excess to proton – (m p- d) 24'687'033.2 497'611'000.0 493'677'000.0 991'288'000.0 987'864'998.9 53'015'918.7 g) mass excess 2 K+ +e- (e -mp) 49'592'917.6 h) 8 m p- 2*m4He =( 2*a) j) Missing for 2 K+ split (g-h) k) Missing for K+, K0 split (f-h) l) mass of 8 electrons m) j using the electrons (l-j) n) k using the electrons (l-k) 49'374'066.4 218'851.3 3'641'852.3 4'087'991.6 3'869'140.3 446'139.3 The physics establishment is currently faced with many challenges. According to its conventional wisdom it should be impossible to produce Kaons, Pions from Hydrogen by just switching on the light in the lab room as it happened in Sweden. Of course normally a LASER is used with 532nm wave length. But why can an input of e.g. 1 Watt produce the same as the CERN accelerator with 100MW? Did not the most recent experiment (super Kamikade Japan [30]) show that the proton is absolutely stable and does not decay ? Tab.18 The proton pathway to Kaons,Pions,Muons We here did already reveal a large part of the “real” particle structure and if you look at Tab.18 (f) then you will note that a normal split of a proton into a K +, K0 pair needs about 53 MeV (f) of added energy. The production of two K+ + e- would need even less energy (g), but this is just fantasy as we do not yet know how charge evolves after the split. If 8 protons do fuse to 2 4He then Tab18 (h) 49.37 MeV are freed. In Holmlids experiments Hydrogen is used and thus the mass of 8 electrons can be used too. (See formation of 4He from protons only!) All together we have an excess of 0.446 MeV Tab.18 (n) above the threshold needed to form the K+,K0 pair. We here did not yet discount the formation energy of H*-H* that will slightly reduce the available energy. The overall reaction is: 9p + 8e+ → 8Be + p → 2 4He + K+,K0 + 466keV 31 Current in x,y plane Current in u,v plane Current in u,v plane Current in x,y plane Fig.13a H*-H* wave structure Current in x,y plane Current in u,v plane Current in x,y plane Current in u,v plane Current in u,v plane Current in x,y plane Current in u,v plane Current in x,y plane Fig.13b H*-H* - H*-H* wave structure From the modeling of Deuterium and H*-H* above we know that the frst orbit that forms out between two protons is a 1FC orbit. This orbits occupies the only free plane left by the 3D/4D fux and in Fig. 13 we did choose the (u,v) plane. Just as a reminder each rotation covers a plane that holds the momentum. Each rotation added to magnetic fux allows to store more energy in less space. The joining fux – inside dense matter - must be symmetric in respect to the other orbits. Only the very special condition of an ultra thin atmosphere 0.05 tor and a catalyst surface that induces an order does allow the condensation of Hydrogen into spin-paired clusters. Classically people did believe this can only happen in a BEC (Bose Einstein condensate). But this is old reasoning, based on a highly simplifed structure model for matter. What we do not yet know is: • How long must two Hydrogen atoms (protons) be synchronous to form out the two possible spin states? • Is Hydrogen electron-electron spin matter a precursor for H*-H* (dense Hydrogen)? • What happens to the electron(s) after the formation of H*-H*? • Or is it the other way round, H*-H* induces Rydberg matter? ( Indeed chemical reaction catalysts do typically produce up to around 1000 target molecules what is in line with H*-H* formation energy! ) LENR (cold fusion) experiments do show that H*-H* can be formed at any temperature (Fig. 11) and even at high pressure if locations (cavities) of reduced dimensionality are available. If we now go one step further and look at Fig. 13b then we can see that between multiple H*-H* pair fux can be joined what potentially will align a large set of H*. But this only works if the coupling is ”circular”. Such a macroscopic expansion of the 3D/4D fux will lead to a strong increase of the combined magnetic moment. But the resulting fux structure still is 3x(1x1) and only weekly synchronous with the center mass (2x2)- 4 rotation structure. Only if we manage to strongly synchronize with the two protons (2x2) fux then we can expect fusion to happen. So far we did neglect – not separately treat - the 2 potential waves that always are synchronous with the center 4 rotation fux. This potential wave fux is only loosely coupled with the 3D/4D waves and we can treat it as an (1x1) structure that is synchronous with the shell electrons. From optics experiment we know that polarized light ideally couples with both waves of an (1x1) orbit, what is measured as Goos - Hanchen [8] efect. If now, as in the experiment of Holmlid, (1x1) orbits get loaded with magnetic energy (LASER photons) that arrives in a shorter time than it can be refected/released, then the (1x1) orbit force (due to added EM mass) strongly increases, what will align 2 of the 4 central rotations of all protons in a cluster. 32 Modeling the exact proton split and energy transfer from 2 4He fusion events needs better experiments. But the origin of Kaons, Pions is now understood. 10 The particles Kaon, Pion, Muon For successfully modeling the nature of particle we must make some basic assumptions we can deduce from the above path given by the formation 4He from protons. All models we made so far did show that the Proton relativistic mass 9 or 7 waves = m 926'603'083.121 charge mass always splits the same way just a) pion 0 mass 134'976'600.000 depending on the number of couplings (1,2). The b) One proton planar wave 1/7 energy of m 132'371'869.017 internal wave number inside the proton are 2,3,4 c) 4/7 real mass part of wave 4/7*b 75'641'068.010 d) Pion virtual mass wave a-c 59'335'531.990 (2x2),5 where 5 is the strong force coupling. As e) going back to 9/8 real mass = 8/9 of d) 52'742'695.102 we can only use torus projections to simplify the (Is equivalent to proton real mass) SO(4) situation, we have to stick to the given f) Kaon 0 mass 497'611'000.000 mechanics base properties of rotating torus g) Kaon+ 493'677'000.000 991'288'000.000 mass. The allowed split of the torus is 9 or 7 error bar of Kaon sum +/30'000.000 given from the rigid mass momentum factors proton mass 938'272'081.300 (9/8 or 7/4). Either we fold 9 waves rotating h) measured Kaon +/o excess 53'015'918.700 j) proton potential wave 272'409.807 around a diameter or 7 (14) waves for a radial Kaon excess from pion/proton added mass excitation. pr pr Kaon mass difference from Pion (j+e) 53'015'104.909 Tab.19a. Proton-Pion Kaon relation As we did see above, the mass ( mpr 926'603'083.1 eV) of the core 4D waves seems to be invariant at least from a todays perspective. But we know that the Pion mass is close to 1/7 and the muon mass close to 1/9 of the proton 4D core mass. Thus we here do assume that all three particles originate from 1/7, 1/9 split of the proton relativistic mass. Lets look at Tab. 19a where we show an interesting relation about the structure of the Kaon excess mass. Tab.19a(b) shows a 1/7 base mass chunk of a proton. According to mechanic laws 4/7 of this chunk is “real” mass (c) and the rest inertial mass. If we subtract this real mass chunk from the neutral pion the we get the virtual mass of the Pion Tab. 19a(d). The Pion is a follow-up particle of proton destruction and thus we relate this virtual mass again with the proton regular mass partition being 9/8. The result is given in Tab.19a(e). This gives the excess mass in proton like units. As we wanted to know why the proton splits, the answer is given by the sum of (e) and one proton potential mass (j) that is exactly the energy needed to split a proton into a K +,K0 pair. This indicates that indeed the potential wave plays a crucial role in the Holmlid reaction. How does the charge coupling evolve between Kaon/Pion? Rows (k,l) of Tab. 19b give the Kaon/Pion 3'934'000.000 relative charge mass that can be derived as 4'593'580.000 1.5555555556 diferential mass from charged and neutral particle. 2'953'015.714 0.75063948 The frst thing we do is we reduce the pion charge mass to the proton frame as it is based on a 4'589'666.667 3'937'354.286 diferent rotation structure. The combined torus 3'354.286 factor is 1.555 (o) Then we form the quotient of the 139'570'180.000 3'913.333 two charge masses Tab. 19b (o). k) kaon charged kaon excess = (f-g) l) Pion so called charge mass = (s-a) m) conversion of waves 7/4 * 8/9 = 1.55555.. n) rotation reduced (l)/1.555.. o) pion/kaon charge mass ratio = (n/k) calculated Pion charge from Kaon p) = (3/4)*k*1.5555 q) calculated Kaon charge from pion charge r) Kaon charge error absolute s) "+-" charged Pion Pion charge error (l-p) Tab.19b. Pion Kaon charge mass relation Now we can draw the following conclusions. Kaons follow a 4 rotation charge coupling where as the pion only owns a 3 rotation charge coupling. Thus the 2 Kaons must be a direct follow-up of a core split of the proton relativistic mass as these own the same 4 rotation coupling but in total one “photonic” excess mass of 7/4 wave is now shared between the two particles. As most decays of Kaons go to Pions we must assume that the Kaon undergoes a complex reorganization that is driven by the 7/4 (14/8) wave structure. The 7/4 torus rotation is the only possible degree of freedom for the proton as the 9/8 rotations “waves” are already occupied by the coupling with the 3D/4D fux and also do represent the 3 out 5 rotating external structure of the proton. 33 10.1 Discussion about particle structure. a) Proton relativistic radius b) Proton charge radius c) Proton relativistic mass d) one wave = (c)/9 e) electron perturbative mass f) electron mass g) 4D bound charge mass = (f)-2*(e) h) Mills gamma (1/(1+π*α2)) j) (g) with photon correction (2*g*h) k) radius expansion =(b)/(a) l) Bohr potential correction (g)*α2 m) proton perturbation muon = k*(d + j – l)/(m3) measured delta Tab. 20a Muon structure 0.8376530074 0.8408699161 926'603'083.1211 102'955'898.1246 1'183.1037 510'998.9461 509'815.8424 0.9998327339 1'019'461.1350 1.0038403834 27.1484 0.9959335244 105'658'374.3734 105'658'374.5000 -0.1266 a) muon mass b) magnetic radius Rm c) relativistic radius Rr d) One proton wave 1/9 energy at 2) e) electron perturb mass f) relativistic bound charge= (f-2*e) g) subtracting relativistic charge (d-f) r 105'658'374.500 0.8408699161 0.8376530074 102'955'898.125 1'183.104 508'632.738 h) relativistic charge m e*2*pi j) Adding charge and base wave (g+h) k) muonmass error (b/a) l) proton electron dim ratio (c/b)3/2 m) 2FC5/1FC3 n) (l/m) o) matching (k/n) corrected muon mass error in muon mass 3'210'701.070 105'657'966.457 0.9999961381 0.994266955 0.994270798 0.9999961349 0.9999999968 105'658'374.839 0.339 102'447'265.386 Tab. 20b Electron like structure Detailed modeling of unstable particles is far beyond the current knowledge we currently have. (Thus here we discuss just two possible structures.) But we can use logic and basic knowledge we gained so far during our modeling. The muon has a very high afnity with the proton and we may treat it as a 1/9 mass particle. We will in both cases expand the mass from the 4D radius to the 3D proton charge radius. The frst solution Tab 20a looks more like a neutron structure with two (1x1) coupling charge masses added to the wave where the second proposal is based on the proton/electron mass relation and the experimental knowledge that the muon shows a 5 rotation charge coupling, what explains its high penetration power. The Mills gamma factor also occurs below in the neutron magnetic moment derivation. In Tab 20b we use a relativistic electron mass that gives back its weight bound at the charge radius and is added as fully relativistic (excited) mass . Further we do the logarithmic electron/ proton radius expansion with a 3/2 rotations relation we also did see in the proton-electron mass equivalence formula. Then we assume the charge is following a standard 5 rotation binding (2FC) with 3 rotation 1FC orbit coupling to the mass wave. To see the value of such an approach we have to look more deeply at experimental data, what has not yet been done in detail (but the 5 rotations structure can be found in some experimental muon mass descriptions too). There are other tries to relate e.g. the electron mass and the muon mass [13] by R. Mills based on semi classical 3D,t reasoning. But here too. There must exist many diferent internally consistent mappings depending on which resonances we do group. 10.2 Neutron magnetic moment The basic question is, why does the neutron magnetic moment deviate from the proton magnetic moment? The impact of the electron magnetic moment must be low, as we know from the neutron model where internally we see two charge masses and hence these two moments must more or less cancel. Because the neutron is a 4D ( 4 rotation) particle and a mirror mass of the 4He bond we must expand it by the radial group measure. The free neutron is an unstable particle and thus what we treat here is still a speculative derivation. The structure of the neutron has been given in Tab.6a, so we here use the known waves to reconstruct the neutron magnetic moment. The neutron owns 4 proton potential waves that occupy 4 out of the fve rotation charge coupling. The ffth charge coupling rotation is given by the e-p coupling that share a common 1FC orbit. Like in the Hydrogen case the weight of this wave is 0.5 what can also be seen in the mass model where 1 mep is shared by the two charge masses. 34 Neutron µn proton µp a) - µn*21/2/µp b) Proton 4D potential 1D c) Proton 4D 1FC potential (2*1FC'*m p) d) (c) for one dimension(1FC'*m p) e) coupling of (b),(c) = (2*b)/(2*b +d) f) ratio (e)/(a) g) 4D force factor = 1FC*2FC*3FC h) Mills gamma (1/(1+π*α2)) j) 4D photon expansion =(f)/(h)4 k) 1D proton perturb (g)/(j) n) potential wave repulsion n) = 1/(1+2* 1FC') precision = (k/n) m) using 1/1FC2 instead of (h) o) reverse neutron µn -0.9662365000 1.4106067873 0.9687070664 272'409.807 40'499.503 20249.7515269443 0.9641641314 0.9953103108 0.9959335244 0.9998327339 0.995976516 0.9999568348 0.9999568379 0.9999999968 0.9999999986 -0.9662365031 These 5 waves do couple over a (1x1)x(5) charge coupling orbit. In table 21(e) we calculate the average weight of the 5:4 Coupling. In the neutron model we did already see that the 1FC orbit only works perturbative. The way to get a magnetic moment of a nucleus is usually to fnd frst the total weight of the coupling mass = 2*mppo + 1FC'*mp and second to relate it with the acting= momentum generating mass ( 2*mppo). Please also remind that 1FC'*mp is a virtual mass = resonant potential of the binding. Tab 21. Neutron magnetic moment We know that the proton magnetic moment has a 3D perturbation and we also know that the neutron is a 4D particle with a 4D perturbation. This you can also see in step Tab.21(a) where we do expand the neutron magnetic moment µn by the SO(4) group measure before we relate it with the proton magnetic moment. (g) is the standard proton perturbation for one rotation dimension (the 5 th of the neutron). This is the universal force factor that in the 4 rotation case is attractive. The only “new” thing is the usage of the Mills gamma (1/(1+π*α 2)) factor for the photon like excitation state of a neutron. The exact same factor has been used in the first neutron model to correct the mass gain by the 4D expansion (excess energy mode) of the neutron. Line (n) is obvious as one rotation perturbation (1x1 coupling) is given by the 1FC orbit - the proton 1FC mass resonance. Both values in lines(n,k) – applied to- mppo gives obviously the same result (-11.758.. eV) for the repulsion term we found in Tab.6a. This is obvious as one rotation charge coupling is given by the 1FC orbit - the proton 1FC mass junk. Line Tab.21(o) gives the reverse neutron magnetic moment. Mills gamma (1/(1+π*α2)) is also used in the gamma ray energy calculation [23] chpt.9. This factor defines the conversion of photon energy to 3 rotation energy without compressing the energy hole. This only indicates that the neutron is already in a state where it can directly send out a photon or release its energy. 11 The SO(4) based deduction of the gravitation constant SO(4) has 6 dimensions that allow 5 independent rotations. If we apply a fux compression constants to a SO(4) mass, the result is mass -gain/loss. Thus we have to fnd a relation between known SO(4) form factors and a potential (caused by a known rotating mass) that is the same as the one induced by gravitation. We know that the central coupling is given by a 5 rotation charge coupling. This coupling is coulomb like. The base topology following from the torus second radius force is 1FC, thats why we also can note the charge coupling as proton mass : Mp 1.6726218982 e-27 (1x1)x(5) relative to the basic 1FC spinelectron g-factor eg: 1.00115965218091 spin coupling. So what we assume is Bohr radius (R b) 52.917721052700 e-12 that gravitation is the residual force of 4D relativistic radius of proton (R p) 0.837653007340 e-15 gravity G 6.674080000000 e-11 the strong force as the combined 4.682249193831 e-24 a) 1FC'5 strong force orbit is a 1FC orbit doing 5 Rp4D/Rb 1.582934772466 e-5 rotations. b) (Rp4D/Rb)2 c) scale factor of force = (a)*(b) d) 1/2FC2 e) 2D potential correction for (Rp4D/Rb)2 : (c*d) 2.505682493883 e-10 1.173222983698 e-33 1.002326872358 1.175952923829 e-33 f) Gravitation potential Mp2*G/(Rp*e) f) Rest potential of gravity at Rp in eV coupling mass (f)/(e) electron perturb mass : me*(1-1/eg2) ratio reverse gravity relative error 1'391.273210044450 e-33 eV 1'183.102811220150 eV 1'183.103703862580 eV 0.999999245508 6.674085035733 0.000000754520 Tab 22. Gravitation potential – mass equivalence 35 The best radius to start with we have is the proton relativistic mass radius that is known with 10+ digits precision. Thus we defne the base line of gravity as the mutual attraction between two proton masses at the center of mass distance of the above mentioned radius. The value is given Tab.22(f). The second step of the search is to fnd the mass in the proton/electron system that generates the same potential of 1391.27 e-33 eV. Because gravity is mediated over the whole free space, what is outside the relativistic frame of our base (proton) masses, we have frst to do a proper scaling from the relativistic proton frame to the non relativistic electron shell. Because in SO(4) “potentials” and masses are equivalent, we can do the scaling with the known form factors, what allows to calculate a mass that is equivalent to the calculated gravitation potential. Because in SO(4) all masses are generated by rotation (are of magnetic origin) the transportation (scaling) of magnetic fux involves two radial dimensions – fux diameter. The radial scaling factor is given by Tab.22 b). It's the quotient2 of the proton relativistic mass radius and the electron Bohr radius. The change of reference frame from the relativistic proton mass frame to the non relativistic Bohr radius electron frame involves the 2FC factor for each of the two involved dimension (Tab.22 d). The weak force factor is 1FC'und must be used for 5 coupled dimensions (Tab.22 a). The multiplication of all scale factors is given in dark yellow (e). To fnd the acting mass, we have to divide the calculated p-p gravitation potential by the scale factor. As expected, the coupling mass is the electron perturbative mass, a value known with 9+ digits. The electron perturbative mass is defned by the fux (potential mass) induced by the electron g-factor (=4D perturbation) part of the magnetic moment. This mass stays exactly at the Bohr radius. The match for reverse gravity is 6.7 digits or inside the known precision for the gravitation constant. Reverse gravity is given in light blue and formula (19) below. Reverse gravitation from basic constants: (19) G = mec2*(1-1/eg2)(rp4D3/a02 )*1FC'5/(2FC*mp)2 = 6.6740850357 e-11 - m3/kgs2 From 4He we know that the center of mass charge orbit does 5 rotations. We also do know that the charge binding factor is 1FC. 1FC,2FC,3FC deal with mass conversion in rotations where as 1FC',2FC',3FC' are the absolute force constants that defne the potential= mass at the boundary of the reference frame. So we can say that gravitation is the mirror of the strong force! 11.1 Repeated: The relativistic proton radius and gravitation/Higgs mass (8) Mpr(eV) = µp2*µo/(α * π* rpr3*e) This formula has been introduced above. The radius rpr that has been use to explain all particles and is highly predictive and also allows to calculate the CERN found proton resonance (frst announced mass claimed as Higgs!) mass as 125.95 GeV. (calculated as Mpr*(3FC*2FC*1FC)2/α. ) The second so called Higgs mass involves one potential mass added with compression. 1FC, 2FC, 3FC can also be seen/understood as integrated scaling constants of the basic forces of nature. The forces behind 1FC, 2FC, 3FC are always present and work on magnetic fux that would like to collapse in 4D rotations. This e.g. cannot happen in the CERN fake Higgs situation because in a crash symmetry never will increase and the formation of 4He needs a 4 fold symmetry. As we, in a collision, can only stimulate the 3D,t visible site of a particle we also do only see a 2D compression. But as soon as CERN was aware that this simple derivation would end their happy live they started to fudge the measurements and to suppress the second (higher) proton resonance mass. In the most recent papers they no longer talk of the Higgs doublet – albeit expected... If we do reverse the logic and apply it to gravity, then we can say, that the electron perturbative mass generates a potential of the same magnitude as gravity that is felt by all relativistic rotating masses that contains protons. In our case down to the proton radius (rpr) = border of dense mass. If this mass stays in the middle of two protons, then the potential from the two proton up to the electron perturbative mass is in total equivalent to the p-p gravitation potential between two protons. Because the universal charge number relation between protons/electrons always is very close to 1:1 It is obvious that the electron must be the intermediary of gravitation. The beauty of this solution is it's foundation on the most deep structures, that we know with very high precision. But this does not automatically mean that the universal gravitation is based on a true constant, because we do not have the absolute proof that the relativistic proton radius is the front of all gravitating dense matter. But there is good reason to believe so as for e.g. 4He the radius is the same as it is formed by 4 protons only! A second reason for a possible variation of the gravity force could be the change of the 36 electron perturbative mass in atomic orbits. (See 6Li model!) But these changes always cause a change of the radius scale too so that there is good reason to believe that the force works on the radius to preserve the original ratio. The only serious challenge will be electrons inside neutrons that themselves stay inside nuclei. The modeling of e.g neutron, deuterium, 4-He shows that the electron structurally does not change but in larger Z nuclei topological efects could possibly change the weight of the electron perturbative fux. (see also 6Li!) A good idea would be to compare the gravitation of solar objects (made mostly of H, He)with objects that are built of stone and thus larger Z nuclei. 11.2 The 4-He hint Question: What indications did we have to fnd the electron as an intermediary for gravity? Answer 4.884eV! For decades people have accepted and were taught that electrons do orbit nuclei and may acquire relativistic speed, what is a contradiction with the fact that most electron mass already is at light speed. This old reasoning was based on the undisputed fact, that in a conservative feld the potential energy and the kinetic energy must match. For example an electron joining 4-He would then be heavier – acquiring 4He gravitational mass 3'728'401'292.003 more kinetic mass - after falling to its alpha mass 3'727'379'378.000 potential being about 54.4 eV. Unluckily difference 1'021'914.003 nobody tried to understand experimental Subtract 2 electrons -83.889 data, which shows a net excess mass of orbital reduction +54.4 +24.587 79.005 error (missing) -4.884 4.884 eV in the 4-He case. Tab. 23 Helium mass and orbits Tab. 23 shows the mass of the alpha particle compared to the 4-He mass, minus 2 times the electron mass. Then, if the potential energy of the two electrons is subtracted, this leaves -4.884 eV (missing mass in gravitational 4-He) , that cannot be explained by relativistic mass gain and is not measured or seen to be dissipated! This extra mass can only be explained by the 4D spin pairing, that defnes the elevated frst ionization energy, something that does not follow classic (SM) potential rules. But again how is this mass dissipated in the bound case? The answer it is not! All SOP reasoning, so far, is based on the fact, that only the feld generated by the electron/proton charge pair contains gravitational mass. To get the last digits you must always subtract the lost potentials. The disappearing of 4.884 eV can only be explained by the fact that a part of the electron does not gravitate. As you may know the two electrons of the neutral 4-He undergo spin-pairing. This mass of the spin-paring (measured: 24.597 – 13.598.. = 10.989eV) is not given by the classic potential, but it is refected in the measured Helion (alpha particle) mass too! If we calculate the second radius potential-dependent mass (given by 1FC') of the spin-pairing feld mass, we get the same amount for the spin-pairing energy as calculated from the measured ionization energies. ( m e-2*mep)*1FC' = 10.99) This is given with no fne tuning. The rigid mass of a torus is defned as (r 2m/8)*(4+5) assuming equal radii (Fig. 2). From this it is easy to see that 4/9 (4.884eV *9/4 = 10.989) of the total spin-pairing energy gets attached to the electron (follow electron 4D torus projection) and in fact vanish because the electron perturbative mass does not gravitate! Of course our picture of a torus is just a projection from 4D to 3D but this is not important for the rigid mass ratio. What we need to know is the fact that the 3D proton potential couples with 2 rotations and the electron also with two. Thus the 4D-fux generated by the two torus radii does split into two: One radius fux gets attached to the relativistic proton fux, the other radius fux generated mass gets attached to the electron fux. 11.2.1 Discussion of hint But... Today you will no longer notice this 4.884 eV discrepancy because NIST [14] defnitely, after 2010, 37 started to use theory (QED 36Ar) based fudge factors to correct the “wrong” 4-He mass. The old (1995), very precise measurement with mass 4He being 4002603.2497 + -0.001 mamu's suddenly changed to 4002603.25413. This 4.2 fold above the confdence interval change to a new fudged value has therefor no real direct measurement (experimental!!) background. The very same thing happened with the 4He charge radius. NIST claimed that muon measurements, as done in the proton case, are much better than the old classic ones. In the proton case it is obvious that the muon can bind much better because of its proton conform 3D/4D wave structure. But 4He has a 4D wave structure – defnitely an odd ft with a 3D wave given by the muon. Thus in the 4He case the muon should not be used as an ersatz electron! So we must note that todays physics has reached a state where wrong models start to destroy that base fro future progress. 12 The electron g-factor equation For a long time the author tried to fnd a relation between the proton strong force equation and the electron masses. In the last try we assumed that mep the electron perturbative mass can be treated as a potential equivalent mass at the electron De Broglie radius. We started with right side of equation (15) chpt. 4.2 (3/2)*(rpr*m'pr/(4*π*64*e2)) and replaced m'pr with me and did no (3/2) orbit correction as the electron fux mass is 1x1 = 2 rotation orbit where as the proton is a 3 rotation particle. Thus the start was : 1= rpr*m'pr/ (4*π*64*e2)). This is for a fully CT symmetric fux. Tab.24 line (a) is the value you get when you divide both sides by the radius we search line (b) is the radius. Now we expand and convert the 4 rotation CT (circle orbit!) inertial mass value to an 9/8 torus diagonal a) strong force factor for electron 0.0004412436 rotation in 3 fo 5 dimensions. Basically b) 1/a) gives radius (in pm) 2266.3220771537 the same we did in the electron proton c) torus form factor expansion (8/9) (3/5) 0.9317694917 mass relation formula. So line (c) is the d) expanded radius Rfp =(b)/(c) 2.4322776152 e) de Broglie radius electron 2.4263102361 expansion metric. If we apply this to (b) 592.0231 we get (d), a radius value that is similar to f) Reverse Bohr potential E=e/(4 π*ε0*Rfp) g) for two rotations 2*f 1184.0462 the electron de Broglie radius. Now we 1.00115965218091 h) electron g-factor elg = j) removing 0.5 of e.g = g/h(1/2) 1183.3603 assume that mep is a potential, that k) electron perturbative mass mep = 1183.1037 attaches in 2D (g). We remove the l) deviation mep /(j) 0.9997831996 electron g-factor for 0.5 rotations (j). Full m) Mills factor 1-π*α2 0.9998327059 electron orbits are (1x1)X (0.5x0.5) as we n) excitation expansion by m =l/m 0.9999504853 only can compare real fux not charge o) 2D 1FC orbit compression = n/1FC2 0.99999364852163 perturbation added mass. Then we add p) electron g-factor strong force coupling p) = 1/(elg (5/2)*3FC) 0.99999364852982 the photon excitation (n) and couple it for q) matching of factors o/p 0.99999999999181 two radius with 1FC (o). This is based on the fact that the electron is an excitation of the proton. The Mills factor Tab. 24(m) compensates the change in mass when we separate the electron from the proton. The same factor is used in the calculation of gamma energies to make the photon energy comparable with the proton mass. The astonishing result was a known factor for the mep coupling with the strong force (p). (p) is the weighted force for 2 out of 5 rotations of a charge coupling. Tab.24 strong force equation and electron mep is obviously an excitation what explains the photon like behavior of the electron. This experimental – speculative derivation was only based on known rules and orbit relations. It looks like fantasy but.. if you consolidate all formulas into a single one then all non metric factors disappear! What remains is a cubic equation for the electron g-factor (given as “x”). 0 = (x3-x)*430.414778081467-1 430.414778081467 = ((3FC*2*π*64)/ ((8/9)(3/5)*((1-π*α2)*(1FC)2))) 38 The numeric result for the electron g-factor is – given by Maple – is : 1.0011596521804564116. The deviation is small and the error can be tracked down to an additional, known correction in one factor. But this makes no sense as the precision of α is to low. This would indicate that the electron g-factor is just given by basic metric factors. Whether this makes sense remains to be seen. The more serious result would be: The electron is just an excitation of the proton as all factors ( me ,electron g-factor) can be derived from the proton structure and the known metrics. 13 Conclusion We did derive and fnally show that a consistent set of basic form factors with an exact physical meaning are enough to explain the formation and structure of mass. The assumption that dense mass is EM-mass has been strongly confrmed. All masses & particles we know can be derived using a small set of basic form factors that can be described by the metric and freedom of rotations within the symmetry group SO(4). The SO(4) magneto-static solution of the Maxwell equation is the origin of the strong force where as the other forces can be derived from the coulomb potential folding into a rotation with a coupled second torus radius rotation. Gravitation is a residual force of the strong force coupling. All these derivations are consistent within the current measurement precision. Further we could prove that the Holmlid neutron free 4 H → 4He fusion is the natural outcome of the SO(4) modeling of 4He. The fnal densifcation step of mass, associated with the strong force factor 3FC, that locks 4 protons into an alpha particle needs to expel the exact equal amount of charge bound mass we see in 2 neutrons! 14 Outlook Currently the new model SOP or NPP2.0 looks much simpler than SM. But this might be the wrong perception. As soon as we will start to model larger nuclei with N > 84, the search for valid projections will become very demanding. There are still open questions like: Why exactly do electron/proton have the actual form factors? How shall we model unstable particles that need? a perturbative model for the internal coupling? As we now know the detailed structure of the basic building blocks of dense matter and the principles of matter-matter interaction, we are able to start developing more deep math – mainly developing a geometrical orbit model for the action of magnetic fux. One possible future path is to look at octonions as there the Golden ratio (base of 3FC) occurs naturally [32]. It is also possible that in a higher dimensional space we will fnd even better partitions and equivalence relations for the particles and also a more natural – algebraic relation (right/left ring like structure) to explain the action of the EM coupling. What we did not show here is 3H, 7Li and 28Si where we also have exact models, that also have allowed us to calculate exact gamma energies like in 6Li! A sample how to do magnetic moment calculations (Deuterium) has been given in [35]. But measured charge radii are of low quality and we have to work on a method to derive the working radius of the 3/5 rotation coupling mass radius. We also did not discuss in detail how we can model a gamma spectrum or other type of gamma radiation. To make it clear: 1 man – handicapped - can not redevelop the whole dense matter physics.. there is a labor shortage. 39 15 Appendix A - more about mass modeling Mass modeling of higher Z/N nuclei is a very challenging task. Already from the frst version of SOP (originally called NPP 2.0) we do know, that the so called 4D quantization of the period table reveals, that all stable nuclei up to lead have a 6x2x7 4D orbit structure that is defned by the strong force quanta released. 6x2 is easy to explain as this is the number of orthogonal plans in 4D/front/back. But initially the question why does “7”, as an orbit grouping number, occur could not be answered. After many models e.g. Pion etc. it is now clear that magnetic fux can transform from a 7/4 rotation structure to an 9/8 rotation structure, what will release a signifcant amount of excess energy. Also do most models show that the 7/4 excited form of EM fux, where fux and rotation are co-linear, is running on the proton charge radius and not on the so called 4D radius. This gives a small extra contribution to the released fux. The frst question to answer was when (which nucleus) does the 7/4 rotation to 9/8 rotation structure start? It looks like 14N and 16O could be the frst candidates, whereas 14C shows an other interesting efect where and added charge leads to 3x5 (3x4 strong coupling weight) particle structure. But frst we look how the core nuclear mass formed by alpha particle evolves. To show this we will use Carbon (12C). Modeling 12C C classically is formed by 6 neutrons 6 protons and already the simple averaging method to calculate the mass shows that in fact 12C forms out 10 1/3 proton waves. 12 a) 12C mass b) 3 4He masses c) force factor 1FC*2FC*3FC d) metric factor 21/2 e) folded force factor ( cd) f) Excess to 3 * 4He g) deuterium bond 2mp*(2FC'+1FC') h) 10/3 waves j) 10/3 waves expansion h*e 2 k) excess after compression l) one 1FC bond 2*mp*1FC' m) missing rest excess n) first gamma 12C o) Mills gamma p) Corrected 2*g*o q) error in gamma line s) second gamma energy t) 3 p-p bonds =3*mp*2FC' – l u) t*c*o v) delta second gamma line 11'177'929'142.0403 11'185'203'876.0084 0.9959335244 1.4142135624 0.9942539823 7'274'733.9681 2'219'936.1568 7'399'787.1894 7'314'992.8899 40'258.9218 40'499.5031 240.5812 4'438'910 0.9998327339 4'439'130 220 3'214'830 3'228'655 3'214'988 158 But how does the fux connect? Do we see 6x2 waves or 3x4 waves? Fig a) C core orbits 12 The left side of Fig.a) shows the 3x4 – 4He core structure and if we we cut it half way then we get the 2x6 structure. This is indicated by the light-blue dotted circle. If we would have 3x4 then the metric compression factor should be 3 1/2. The result will be shown below in table b) Table a) 12C from 3 4He cores With the 2x6 structure it is easy to explain the excess waves. We in Fig b) gelow just show 5 out of ten connections and you can see that always two protons (arrows are left out. So in total 4 are once left out and thus only a) 12C mass b) 12 protons c) force factor 1FC*2FC*3FC d) metric factor 31/2 e) folded force factor ( cd) f) delta mass from 12p =b-a g) wave from 12p =b*(1-e) h) Rest excess from 12p j) proton proton bond 2*mp*2FC' k) 2D compression = j*e2 l) missing excess Table b) 12 11'177'929'142.0403 11'259'264'975.6000 0.9959335244 1.7320508076 0.9929671451 81'335'833.5597 79'184'777.0573 2'151'056.5024 2'179'436.6538 2'148'889.1274 2'167.3750 C from 12 protons with 3x4 orbit Fig. b) 12C 10 excess waves 3 full proton (p-p) waves (10-4)/3 can form out a deuterium bond 2m p*(1FC'+2FC') structure and as only 6 40 protons are fully connected we, in total, do see only two Deuterium bonds. This is also given by the frst gamma energy of 12C, that is pretty close to 2 Deuterium bonds! Thus you see in Table a), to get a good approach for the 12C mass, we must – as expected - subtract a p-p 1FC' bond as we calculated (lines h,j) with 3 Deuterium bonds. The same happens with the second gamma energy. 12C is instructive as you here too can see the astonishing simple relation between gamma energy and basic bond energies! But also the 3x4 wave approach now based on 12 protons where we have, as expected, to change the overall compression delivers a good approximation for the 12C mass. But here only one full p-p bond would be needed to complete the mass formula what could – best case – deliver a crude approximation for the frst gamma and none for the second. But for higher z nuclei the mass structure will strongly increase in complexity as we will encounter many diferent interactions between all possible internal wave forms. Modeling 14C a) 14 proton masses b) force factor 1FC*2FC*3FC c) metric factor 31/2 d) folded force factor ( bc) e) 3 wave 4He pert. (a*(1-d)) f) 14C mass g) proton mass excess (a-f) h) delta waves base mass (g-e) j) mer = me - mep k) + relativistic electron mass m er l) 6th 14C shell electron m) delta in mass 13'135'809'138.2000 0.9959335244 1.73205080756888 0.9929671451 92'382'239.9002 13'043'937'225.0590 91'871'913.1410 -510'326.7592 509'815.8424 -510.9168 -489.9932 -20.9236 C is a linear factor of 28Si what we already did show in the frst release of NPP2.01. So we expect that it can be modeled straight forward. In fact if we start to model 14C from protons – it already contain 3 4He cores, that are formed from protons only – then we see in the table left line h that only a neutron charge (line j) is missing, what is exactly what it should be. After subtracting, as usual, the released shell electron potential the error is tiny. The most interesting fact is that following the triple 4He fux the binding force of (line d) now is following 31/2, what indicates that the core has 3x3 symmetry instead of 2x2. 14 Of course we do not know the exact mass defects of higher Z/N nuclei. For the 4He core it was a bit larger than the sum of the potentials. But all nuclei with more than 3 electrons show strong paramagnetic/ diamagnetic efects and diferent forms of spin pairing. It's completely unclear how this will afect the core nucleus mass. So currently all error bars are at least as high the sum of all shell electron potentials. Modeling 14N a) 12 proton masses b) 14 proton masses c) force factor 1FC*2FC*3FC d) metric factor 21/2 e) folded force factor ( cd) f) 12 proton 3 wave 4He pert. (a*(1-e)) g) 14 N mass h) proton mass excess (b-f) j) delta waves base mass (g-e) k) 7/4 wave factor = ( 1 -9/14) l) charge radius expansion = l/m m) Rpr 11'259'264'975.6000 13'135'809'138.2000 The frst nucleus that could be able to show 7/4 wave 0.9959335244 compression is 14N as it owns 2x7 protons. It is formed by 1.41421356237310 4 He cores and two more protons. But 14N owns an 0.9942539823 4 64'695'935.6941 incomplete (3.5) number of He cores and thus we should 13'043'780'749.8864 not expect a perfect ft. Line (f) gives the 12 proton 92'028'388.3136 compression and line q the 7/4 wave compression if we 27'332'452.6195 4 0.3571428571 assume a virtual (3.5) number of He cores mass. This of 0.9961743087 course must lead to a small asymmetry as one p-p must n) Rpc o) 4He mass p) 3.5* 4He masses q) 7/4 compression k*p*(1-e)/l1/2 r) Delta mass after 7/4 wave s) deuterium bond t) repulsive potential (2/3)*s*(1-e) u) delta after 7/4 wave = j-q v) mer = me - mep w) delta in mass u-v 0.8408699161 3'728'401'292.0028 13'049'404'522.0098 26'830'696.8727 501'755.7468 2'219'936.1568 8'503.8616 510'259.6084 0.8376530074 509'815.8424 443.7660 3 acquire some excess energy see line (t). Also one more neutron bound charge mass must be released. The only negative point is that the 7/4 excess wave does not start at the charge radius as we must use square root of the expansion factor. Also the Deuterium, bond mass only shows up as a perturbation and thus this wave split cannot explain the gamma spectrum of 14N. 41 Modeling 16O from protons With 16O we the frst time encounter the 7/4 torus wave compression. We thus here frst model 16O as 2x7 strongly coupled protons plus an added deuterium. (lines b,c, left) If you line up the fux of 2x7 protons in a row then this corresponds to a 7/4 torus excitation, that after changing to the 9/8 wave structure is able tor release a large chunk of energy. The factor 1 -9/14 is delta of fux that gets released if you change from 7/4 torus mode to 9/8. He mass proton bond a) 16 O mass b) 14 proton masses c) 2H mass d) 16 O mass excess e) force factor 1FC*2FC*3FC f) 21/2 g) folded force factor ( e f) h) 7/4 wave factor = ( 1 -9/14) j) 7/4 compression b*h*(1-g)/k k) charge radius expansion = l/m l) Rpr 4 m) Rpc n) alpha excess =4*4He-mass*(1-g) o) after basic waves =d-j-n p) 2/3 proton bond compression if 2H q) 1 perturbative mass r) 8th electron potential s) rest difference = o -p -q -r 3'728'401'292.0028 2219936.15681298 14'899'168'523.1417 13'135'809'138.200 1'876'123'913.278 112'764'528.3368 0.9959335244 1.4142135624 0.9942539823 0.3571428571 27'060'163.7044 0.9961743087 0.8376530074 0.8408699161 85'693'839.0746 10'525.5578 8'503.8616 1'183.104 871.41010000 -32.8176383316 We thus will see the classical 4He compression for the “2x7” part of the coupled proton (line n) and the new 7/4 compression (line j) for the “2x7” part of the coupled protons. The other new thing we see here is the change of the radius from the so called 4D radius to the 3D charge radius! - Given as factor by line k. Line p shows the additional compression of the 2/3 Deuterium bond with the strongly coupled nuclear mass we already did see in 6Li. The only diference is that here the bonds are co-linear with the strongly coupled fux and thus the compression changes from line(e) → line(g). As the 7/4 compressed fux runs on the 3D radius doing 3 rotations it is obvious that the Deuterium part that contains 5 m ep has to release one more mep to get 3 mer2. So the 7(14) wave fux at the end looks like 2x2x3. Whether we have to write this as 2x(2 + 3) will be seen. O has two base gamma lines (6049.4, 6129.899 keV) that slightly difer. It looks like the second line is a resonance of a Deuterium bond with the 7/4 wave. If you expand the Deuterium bond energy by the factor of line(h) the result is pretty close to the second line. But it is obvious that adding 6129.899 keV will signifcantly change the internal structure. This is also confrmed by the relative long live time of the two states what usually indicates a signifcant fux reorganization. This will certainly also afect the Deuterium bond 1FC' mass. 16 Modeling 16O from 4He He mass proton bond a) 16O mass b) 3.5* 4He masses c) 2H mass d) 16O mass excess b+c'-a e) force factor 1FC*2FC*3FC f) 21/2 g) folded force factor ( ef) h) 7/4 wave factor = ( 1 -9/14) j) 7/4 compression b*h*(1-g)/k k) charge radius expansion = l/m l) Rpr 4 m) Rpc n) after 7/4 wave = d-j o) adding one charge = n +me p) Excess 2/3 proton bond compression of q) rest 3'728'401'292.0028 2'219'936.1568 14'899'168'523.1417 13'049'404'522.0098 1'876'123'913.278 26'359'912.1466 0.9959335244 1.4142135624 0.9942539823 0.3571428571 26'882'167.5845 0.9961743087 0.8376530074 0.8408699161 -522'255.4379 -11'256.4918 8'503.8616 -2'752.630 An other try to model 16O is starting from 4He cores. Here instead of 14 protons we use 3.5 4He core masses. How to model the split of 4 4He cores to a 3.5 4He, 2H structure we currently can only guess. Of course here we only will see the 7/4 rotation to 9/8 rotation fux transformation. The mass excess line(d) and the transformed wave ft very well as we only need to add one more charge, what makes senses as 2H contains 2 charge masses and the excess wave couples with 3 rotations. So this charge is produced by the above mentioned split. Also the addition (line p) of the 2/3 2H bond wave mass expansion factor (see also 6Li) makes sense. (Here we did not add the additional charge as it makes no sense to speculate further). 42 16 Appendix B Here we show a historical path about how we could derive the proton 4D radius. The knowledge is very useful if you do cold fusion as the structures we identifed are active resonances that can be used to plan reactions. 16.1.1 Radius discussion Our initial research started from the base assumption that the 4He radius is the same as the proton relativistic radius (4x fux → 2x radius!). We also did fnd a very good neutron radius that delivered a very precise virtual deuterium model [34]. See table 27 below for the radius relations. Based on the neutron radius we got a tiny diference for the current proton relativistic radius that is equivalent to about a 1.47eV bottom up delta in mass. Compared to standard model precision SOP is miles better but we should never believe, if we see an “exact” result, that it is the fnal explanation of the physical reality. These 1.47 eV are a reasonable error for all what we do, even 10 x more would still be miles better than what “we” (with SM) currently have. The following should deepen your understand of how fux compression works and it should also show how close the relation between the proton and 4He in reality is. If you understand the above shown real 4He structure you will understand that all we do here is a simplifcation that only works for one possible abstraction of 4He. The following discussion is based on the classic assumption that 4He is formed/contains 2 protons and 2 neutrons. But “nature=experiments” shows us that we can form 4He by fusing 2 Deuterium or 4 protons (in fact 9 protons are needed for two 4He - see chpt. 9). Both processes do run at rest and are so called low energy nuclear reactions. But in nature the deuterium reaction is much more likely. Under space conditions the Hydrogen reaction will dominate. The ideal SO(4) 4He compression – all 5 potentials folded - would be (1-5*2FC) “=” 0.994192951338; This compression is relative to the original n+p+e masses that form 4He. Because this factor is based on 5 rotations we have to convert it into the base particle mass 3 rotation space. This can be done by quaternion like logarithm that maps harmonic functions. 0.994192951338 1.51.. = 0.9961649819 ( 1.51... = 23/5 ). This corresponds to the folding of 5 dimensions of potential energy, 5 1x1 rotations, into 3 coupled rotations. This would be the ideal mathematical compression if nature is totally SO(4) symmetric. The “natural” 4He fux compression can be derived from the 4He mass/”base particles” ratio as it is the sole nucleus that has no free 3D/4D fux (no gamma spectrum!). As fux can be modeled/measured as units of energy passing through a boundary in a plane (manifold) the square root of the natural compression of 4He gives a frst approximation (0.99622.. column 1 in Tab.25 – versus ideally 0.996164..) of the efective (magnetic density) fux compression. As 4He has an internal structure the overall value is not an average. In fact 4He owns the same type of waves as the proton but all three wave parts (with diferent rotation number) are subject do a slightly diferent compression. mamu He4 He4 from particles compression ratio (CR) torus area r*r → r * CR1/2 Ideal compression column 1 4'002'603.25 4'032'979.91 0.9924679367 0.9962268500 503.35 4'002'351.58 4'033'231.59 0.9923436058 0.9961644472 weighted waves of 503.35 336.541, 168.271 4'002'603.25 4'002'603.25 4'033'483.26 4'033'480.84 0.9923440836 0.9923446793 0.9961646870 0.9961649860 0.9961649819 Detailed derivation see table 25 below. From the 4D model of the neutron we know that the neutron has a 4D fux hole and also the ability to release 4D excess fux. Tab 25. Possible approximation of ideal 4He fux compression masses in mamu units. The base assumption is that inside 4He there is hidden internal fux compression – what leads to a reduction in volume - happening between the two neutrons that explains the calculated mass compression diference given in Tab 26. Basically the neutron (see 8.4 ) is a 5 rotation dense mass particle. If we explain mass structure as n-p bonds then a neutron can do up to 5 bonds and the proton 3 bonds. We also call the two additional, neutron bonds energy hole bonds as these stay in 4(5) rotation bound dense mass space. The sum of the fve bonds is very close to the neutron excess energy of 782'333 eV. 43 Table 26 shows the two junks of energy we must take into account if a neutron stays inside a nucleus. The neutron 4D hole (168.271mamu = 313'486 eV => 2 waves) consists of two missing, uncompressed waves mamu eV that initially contain no mass (=fux). The neutron excess 839.869 782'332.965 energy has the weight of 3 waves and consists of matter Neutron excess in mamu Neutron 4D hole 336.541 313'486.098 4 Freed energy neutron->4D 503.328 468'846.867 with a reverse He compression. Table 26 Relevant amounts of neutron energies. This neutron wave structure is immanent in the periodic system of elements and can directly be seen in the mass build up of e.g. 9-Be, 10-B, 14-C, 15N,.. with one hole wave or: 10-Be, 15-O, 56-Co,57-Fe,.. with two hole waves. Or: 3-H,3-He,13C,17-O, 21Ne etc. with 3 excess waves. The next two columns of Tab.25 show adding hidden mass (three waves) compression symmetrically (column 2) and on top (column 3). With this (blue feld) we already see 6 digits agreement with the optimal 4 He compression. Adding on top is obviously the right thing to do as the real mass of 4He is given and adding on top just reproduces the amount of mass that would ft a homogenous compression. In the last column we did add the 3 waves with the correct weights (500.929=336.541*3FC 2*2FC4 + 168.271*2FC2 ). This is given just to show that there is a physical explanation for the factors we fnally used. As you could see in the 4He model, the neutron is a mirror excess energy particle of the 4He structure. Thus 4(5) rotation (neutron) excess mass must frst be once compressed by 3FC*2FC 2 to be again plain mass and the once more compressed by 3FC and 2FC2 to get the 4He mass density equivalent mass. The 4D hole needs only a 2FC2 compression as it is already inside the 4D mass but not yet bound. The neutron 4D potential free radius (0.840877885fm see [35] or below) is about 10 digits exact because it a) 4D potential free neutron radius 0.840877788500 can be exactly derived from the neutron mass. Thus the b) 3D radius 0.837653006969 3/4D radius is 10 digits exact too, because it is found by a c) Quotient (b)/(a) 0.996164981909 mathematical relation. The quotient of R4D/R3D is the ideal 1-5*2FC 0.994192951338 (real) 4He compression of the involved particles.. With this 1.515716566510 23/5 '= (c)1.5157165665 0.994192951338 radius (Tab. 7 line b) the proton mass is of by 1.47eV Table 27. Logarithmic radius/compression relation The factor (1-5*2FC') can also be found in the 4D mass build up because (1-2.5*2FC') gives the exact amount of 3D/4D energy that is converted into additional 4D energy in nuclei starting with 11B. For people interested in basics math: Relations of frequencies are relations of energy. They (e.g. sin((3/5) *x) can be mutually expressed in quaternion math by exponents and logarithms. That's exactly what is used above. 16.1.2 The potential free neutron radius Because the neutron is a proton with excess mass, we did look for a consistent interaction radius for the neutron, that is slightly larger than the proton radius. The 4D excess-energy of the neutron is “neutron mass” * 3FC'=2'712'454 eV. The coulomb-potential for e.g. the largest possible proton 3D radius (0.8408739) is 1'712'462 eV. The diference of the two potentials is 999'992 eV. In 4D physics we usually build quotients to compare quantities. The quotient base of the 4D/3D potential is 1'000'000. This is coincidentally the same base we also use for µ0.(radius in denominator!) The 3FC' added mass is magnetic fux mass/energy whereas the potential mass is electric energy/mass. The above radius of 0.8408777885 fm is just the coulomb radius, where the diference of 4D3D pot is the quotient base. In [23], we already used this radius (0.8408777885) for the virtual deuterium model and found a 7 digits agreement between the magnetic moments of proton/neutron and deuterium radius with the above radius. The problem with such experimental data is the low quality/precision of any charge radius measurement. Why exactly this works and reproduces the correct experimental data is still not fully understood. That the neutron is potential free is obvious and the occurrence of the factor µ0 – without surface norm , is no surprise, but why does it ft the 3FC excess mass? 44 This was the historical path to fnd the proton structure. Only the extreme high coincidence - 8 digits precision - of the diferent models did convince us that the chosen approach is correct. [1] Mary K. Gaillard,1, Paul D. Grannis,2, and Frank J. Sciulli,3 1,University of California, Berkeley, 2,State University of New York, Stony Brook, 3,Columbia University The Standard Model of Particle Physics,1998, https://arxiv.org/pdf/hep-ph/9812285.pdf [2] http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html [3] Leif Holmlid, Mesons from Laser-Induced Processes in Ultra-Dense Hydrogen H(0), PLOS one doi:10.1371/ [4] Leif Holmlid, Emission spectroscopy of IR laser-induced processes in ultra-dense deuterium D(0): Rotational transitions in D(0) with spin values s 1⁄4 2, 3 and 4,http://dx.doi.org/10.1016/j.molstruc.2016.10.091 [5] Fleischmann, M., S. Pons, and M. Hawkins, Electrochemically induced nuclear fusion of deuterium. J. Electroanal. Chem., 1989. 261: p. 301-308 and errata in Vol. 263, 187-188. [6] Safre project, https://aureon.ca/science-of-safre [7] Aringazin; Toroidal confguration of the orbit of the electron of the hydrogen atom under strong external magnetic felds, http://www.arxiv.org/abs/physics/0202049v1 [8] F. 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Newell c) National Institute of Standards and Technology, Gaithersburg, Maryland 20899-8420, USA Recommended Values of the Fundamental Physical Constants:, 2010https://physics.nist.gov/cuu/pdf/JPCRD2010CODATA.pdf [15] CODATA “fudging” 2010 [http://dx.doi.org/10.1063/1.4724320] [16] fle : mass_rmd.mas95 * ATOMIC MASS ADJUSTMENT DATE 29oct95 TIME 14: 9 A= 0TO273 Which is one fle out of a series of 6 corresponding to : "The 1995 update to the atomic mass evaluation" by G.Audi and A.H.Wapstra Nuclear Physics A595 vol. 4 p.409-480, December 25, 1995. [17] Wonjae Lee,1* Andrei H. Gheorghe,1† Konstantin Tiurev,2 Tuomas Ollikainen,2 Mikko Mottonen,2,3 David S. Hall1‡ Magnetization oscillations and waves driven by pure spin currents V.E. Demidov1, S. Urazhdin 2, G. de Loubens 3, O. Klein 4, V. Cros 5, A. Anane5, and S.O. 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