Does the quantization of the proton magnetic moment explain LENR? Dr. J. A. Wyttenbach Independent researcher
[email protected]; https://www.researchgate.net/project/Nuclear-and-particle-physics-20 About one and a half year after the start of Nuclear & Particle Physics version 2.0 there are now several experimental confrmations for the new model. After a careful extrapolation of the proton 4D radius (0.8376530063/2 fm), we were able to deduce the magnetic base mass of the proton. After the correction of the mass by the known magnetic moment perturbation a tiny amount (272'406 eV) of the proton mass energy was left unexplained. It turned out that this amount is exactly defned by a SO(4) = SU(2) x SU(2) conform alpha quantization. Table of content 1 Short overview of NPP2.0 (nuclear & particle physics 2.0)...........................................................................2 1.1 Flux compression/expansion constants............................................................................................... 2 1.2 Energy................................................................................................................................................... 2 2 Why/ how to fnd a new model..................................................................................................................... 3 2.1 What is stable data we can use to fnd a new model...........................................................................3 2.2 What need we to do frst ?................................................................................................................... 3 2.2.1 Intermediate step 2.1.1 LENR relation.......................................................................................... 3 3 SO(4) The true physical space...................................................................................................................... 4 3.1 Energy in SO(4)..................................................................................................................................... 5 3.2 Properties of 4D space......................................................................................................................... 6 3.2.1 Magnetic fux compression in 1;2;3 Dimensions..........................................................................6 3.2.2 Sample mass calculation.............................................................................................................. 7 4 The 4D Neutron and Proton.......................................................................................................................... 8 4.1 Magnetic moment of Proton................................................................................................................. 8 4.2 Proton mass calculations..................................................................................................................... 9 4.2.1 Radius discussion....................................................................................................................... 10 4.2.2 The potential free neutron radius ............................................................................................... 10 4.3 The Neutron mass.............................................................................................................................. 11 4.3.1 Conclusion.................................................................................................................................. 11 5 The magnetic Bohr (Hydrogen) model........................................................................................................ 12 5.1 Used formulas for stored magnetic energy of an electron orbiting a proton......................................13 5.2 Discussion.......................................................................................................................................... 13 5.2.1 The corrections for states n=2,3,4..............................................................................................14 5.2.2 Conclusion.................................................................................................................................. 15 6 Experimental fndings................................................................................................................................. 16 6.1 The frst LENR experiment with constant gamma ray production......................................................16 6.2 CERN prove of magnetic proton mass...............................................................................................17 7 Relevance for LENR.................................................................................................................................... 17 7.1 LENR heat transport........................................................................................................................... 17 7.2 What is needed for successful sustainable LENR ?...........................................................................18 7.2.1 Strong stable nuclear feld.......................................................................................................... 18 7.2.2 The LENR trigger........................................................................................................................ 18 7.2.3 What spectrums show................................................................................................................ 18 7.3 What is the halve live of deuterium fusion?........................................................................................18 8 Outlook....................................................................................................................................................... 19
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1 Short overview of NPP2.0 (nuclear & particle physics 2.0) The following base assumption are made: – – – – – –
Dense space is homogenous and has at least 4 dimensions Almost all energy is stored in rotations = magnetic fux Magnetic fux can be compressed/removed to release energy Magnetic fux can be expanded/added to increase the energy Stable particles have a base magnetic mass and carry (small part of) additional excess-energy The mathematical (base )space for the description of NPP2.0 is SO(4)
– –
In 4D space time is homogeneous and of periodic nature with a maximal duration of 2*π. To increase a relativistic magnetic mass = adding one more fux-rotation, you must multiply the base magnetic mass by 1/α To convert (v → c) a non relativistic mass to a relativistic one, you must multiply it by 2*π To fnd a non relativistic rest-mass you must divide a relativistic mass by 2*π 1/α corresponds to the classic length contraction, 2*π to the maximal relativistic mass increase.
– – –
These rules are not complete as e.g. a relativistic mass is only once afected by the time parameter (2*π) and further increases only involves length contraction by 1/α.
1.1 Flux compression/expansion constants Energy conversion constants: 3D/4D - 4D Flux capture 3D-3D/4D Flux capture 2D-3D/4D Flux capture
3FC 2FC = 1 - (α/2π) 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
Excess-energy is fowing(rotating) around the core mass with diferent number (2,3,4) of rotations. The numbers ( 1,2,3) prefxing FC denote the base number of rotation the “fux compression” works on. E.g. 1FC converts a one dimension fux/potential in a two dimensional rotation. 2FC converts fux from 2 → 3 rotations. 1D/2D-/3D relativistic photon fux capture (Mills [4]-) γ* = 1/(1+πα2) = 0.9998327339.. In NPP2.0 only the above constants are used to relate the Eigenvalues for fux-capture/expansion or to express the space like perturbations.
1.2 Energy Classically particle energy is modeled by waves and the associated spherical harmonics. Because nuclear fux is confned in a very narrow range, we can also use mechanical analogues of (force free) rigid rotating masses. In the symmetric case the mass is given by the sum of the eigenvalues of each rotating dimension. This can be irritating as the waves may cover e.g. 4 dimension but the independent energy Eigenvalues only cover 3.
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2 Why/ how to fnd a new model The mathematical analysis of the standard model for nuclear & particle physics shows, that it, to a great part, has nothing in common with the physical reality. The standard model is mostly potential based, albeit in reality 99% of the mass is magnetic fux equivalent, that does not couple potential like. The inner structure of any nucleus/particles has no conclusive standard model explanation beyond some basic structural properties like spin/charge. Exact energies (quarks etc.) are unknown – not given by the model!
2.1 What is stable data we can use to fnd a new model The best experimental base are: - nuclear/isotope masses, including proton/electron masses known with about 10+ digits. - Magnetic moments of proton/neutron 9+/8+ digits. - c,h,alpha - Charge radius are only 3-5 digits exact. - Gamma lines are known between 3-7 digits.
2.2 What need we to do frst ? We know that energy always is a scalar, that usually can be expressed as a sum or a product of Eigenvalues. In a highly symmetric system it is easy to “guess” the eigenstates. Thus step one was to fnd a relation that generates the nuclear masses from the particle masses. This step one has been completed one year ago with a model that delivers between 6-10 digits precision. The model does explain the periodic system and why only isotopes up to Z=81 (Pb) are stable. The model also consistently explains the gamma levels of isotopes. Step (2) was to fnd a relation with the above model and the well known magnetic moments. This was successful one halve year ago and led to the NPP2.1 version.
2.2.1 Intermediate step 2.1.1 LENR relation The NPP2.0 model [6] has been driven by LENR and the lack of it's physical explanation. In the intermediate step NPP2.1.1 the exact magnetic mass of the proton has been derived and the magnetic coupling of LENR active isotopes has been calculated. A more complete understanding of the role of the neutron has also been gained. The analysis of the frst spectra of a long running LENR reaction fully confrm the base assumptions of the model.
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3 SO(4) The true physical space SO(4) is far more complex and thus difcult to imagine than its related sub-spaces SO(3), SU(2) and all their derivations. This is because you cannot separate the time dimension meaning you must be able to think in (at least) 4 real, uniform space dimensions. Things become even more complicated as the common center of mass is a 4D surface known as Cliford torus, that is single sided. Thus, in any projection to a 3-D space, you must be aware of the front/back side nature of (EM-) mass-fux. (1) SO(4) = SU(2) X SU(2) This topological equation shows one connection to existing physics and already explains why the existing models for dense space fail. The cross product is not commutative, respectively at least the sign changes. The only exception are scalars like energies that are square sums. (2) SO(4) =
The Cliford torus is the “topological equivalent” of SO(4) namely the connection of two circles (staying in independent dimensions) with a 4 dimensional bundle of tangents. (3) (wiki) This is one possible representation of SU(2) as a 2x2 conjugate complex matrix. (4) curvature of Cliford Torus: F''(X1,X2,X3,X4) = const (5) Radial norm: x21+ x22+ x23+ x24 = 1
Graphical representations of Cliford Torus from wikipedia:
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Or topologically: SO(4) = SU(2) x SU(2)
Any projection of SU(2)xSU(2) to SO(3), SO(3)R, S(3) etc. leads to a radial change of measure: (6) R4 → R3 : r3 = r4 * (1/2)1/2. If two radii are involved, the the factor becomes ½ (or 2 in the other direction).
3.1 Energy in SO(4)
Fig 1: 4He nucleus as torus projection In dense space most matter/energy is represented by rotations. In SO(4) we have 4 independent rotations, that – for simplifcation - can be mapped to two disjoint 3D tori, where each rotation is represented by the individual base radii of the two tori. If we map everything to one single 3D torus then two rotations are given by the surface fux and the other two by the whole body rotation (green, black axis). This kind of simplifcation is only appropriate for highly symmetric nuclei like 4 He. Further this picture can only be used for scalar quantities like mass/energy of the nucleus. The 4 rotations energy structure of dense space is new, albeit it forms the core of any nucleus. Even more complex to understand is the 3D/4D fux of mass. In any 4D space you have a 3D subspace. This subspace contains the well known mass we know from a proton, but it performs one more independent rotation. This form of mass (the 3D/4D fux of mass) is new and is now spotlighted because time is becoming a uniform space-dimension. To imagine this movement just draw a proton (mass-fux represented by two spherical rotations) and add one more rotation given by the 4 th dimension. This three times rotating proton (in fact the three mass/charge waves) is now fowing along the Cliford torus (touching red line Fig.1 ) surface of SO(4). In the following the 3D/4D fux is always counted in 1/3 units what is the weight of one wave.
Fig. 2a. 3D/4D fux of mass
Fig. 2b. Field
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Classic proton radius 4D proton radius
In Fig. 2a the red line indicates the Cliford torus surface. The surface has two sides in a 3D projection thus the orthogonal (to Cliford torus surface) wave drawn as black circle is counter rotating on the front/back side. In addition we indicate the other two rotations as full body rotations. If we associate charge with one radius, Fig. 2b, which is logical given the magnetic moments, then we notice that in a perfect symmetric confguration (as in 4He) the magnetic felds vanish (at least macroscopically!) - green front/back arrow. We also can conclude, if charge(-density) has the same property as in 3D space, that the two front/back-fux charge waves must be attractive if they run in the same plane. A slight change in the angle between front/back-fux could be the origin for deviations of the third fux compression constant we found.
Fig.2c 3D/4D radius
3.2 Properties of 4D space In a rotating (3D,t) system, the base line is the equivalence “point” of forces/masses. In SO(4) this point is not the common center of mass it is the entire surface of the Cliford torus. In a perfectly balanced system the sum of back/front side mass/rotations must be equal. Expressed in mathematics: For a perfectly balanced system the quotient of front/back fux must be equal (=1) at any point of the surface. For the 4D rotations this implies that all radii must be equal. For the 3D/4D rotation fux a system is “balanced” if the resulting SU(2)XSU(2) quotient = 1 All perturbation is measured as deviation (factor!) from 1! Closed 4D space has the following metrics: (normed for r=1) 4D hyper volume = ½ π2. 4D hyper surface = 2 π2. Internal 3D volume = 16/3 π. Internal 3D surface = 16 π.
3.2.1
Magnetic fux compression in 1;2;3 Dimensions.
Basically long time stable fux reduction(compression) is only possible between proton and neutrons. Key for the n-p binding is then split nature of the neutron that can give or accept fux.
a)
b)
c)
Fig. 3 N-P fux reduction “bonds” between protons and neutrons The term bond is wrong as in reality the magnetic fux is unidirectional. Thus here double arrows are only illustrative. If we, in the following text, talk of a 3D/4D wave, then we mean a wave, that is “equivalent” to a 3D mass, but traveling/rotating along a 4D surface in 4D space! In 4D space most energy is stored in fux, what is a synonym for compressed magnetic feld lines. 3D/4D fux is a bundle of magnetic feld lines with 3 rotations that stays in 4D space. 4D fux makes 4 rotations and 2D/4D ( +,-) fux-potential seems to (radially?) balance the frst two. The base particle electron makes only two full rotations, because a large part of the disposable energy is
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stored in the radial feld. The proton mass has a “large” (compared to electron) excess mass that needs a third dimension to fow in. in the 4D world, radial energy is converted into rotational mass/energy or will be disposed.
3.2.2
Sample mass calculation
Deuterium mass Summary: (2FC) 2D/3D-3D/4D Flux captured between neutron & proton = 2341.971 mamu (micro atomic mass units). (1FC) 2D/3D--> 2D/4D Flux captured = 43.520 mamu – gets added to 3D/4D. Fine-tuning correction: New Sum fux released = 2385.491 * 2FC = 2.771 second order correction due to fux hole compression. The diference between the measured mamu value and the calculated one is now 0.084mamu! Or the precision is 8 digits. Later we will show the hidden (asymmetric) compression of the neutron. In the deuterium case this given 0.0995 mamu of additional compression and the loss due to 13.6 eV of the electric potential gives -0.0176 mamu's. The remaining sum (delta = 0.002mamu) gives the exact mass of the deuterium. Deuterium calculation Neutron Proton + electron Sum particles Sum first order adjustments Correction by 2FC on the Induced flux calculated difference measured difference Calculated Deuterium mass Deuterium mass measured
mamu 2FC 1FC 1'008'664.923 0.0011614097 0.0000215820 1'007'825.032 Reduction amount 2FC Reduction amount 1FC 2'341.971 2'016'489.955 43.520
2'385.491 2.771 2'388.261 2'388.177 2'014'101.694 2'014'101.778
2.771
0.0505443282
relative error absolute error relative error
0.0000353391 0.084 mamu 0.0000000419
Table 1 Deuterium mass-calculation Deuterium (n-p) Fig.3 a) is only able to exchange 3D-3D/4D fux in one dimension (through one plane!). Two deuterium Fig.3 b) that stay in the same plane in 3D(3, 1) space, can only build up 4 nodes of fux exchange. (2 x 2D wave =4 nodes , 2 planes). To further double the number of “connections”, as needed to model 4He, at least 4 uniform space dimensions, where we can get up to 6 disjoint planes (4 disjoint planes are needed), that can be used for 3D-3D/4D fux exchange are needed. (If right/left associative math is used, then the number of hyper-planes (halve-planes) - potentially can double.) In 4D space Helium-4 builds out 4 more connections, than possible in 3D space, with magnetic fux going through in total 4 disjoint! planes. The resulting 3D/4D wave on a 4D curved surface releases 8 3D/4D (2FC) fux exchange quanta, (in total 18735.768 mamu), because it acquires one more degree of rotational freedom. (The surface charge motion of a standard 3D particle -proton/electron - can be modeled by two spherical harmonic waves, what corresponds to 2 rotations. In 4D space you need at least 2 hyperspherical harmonics for the 3D/4D mass.) Additionally the fux of 4-He starts a 4D rotation and releases the so called 4D quantum (3FC) of energy. 4D rotations can be modeled by mechanical analogues. A simple sample: The 6Li mass Helium base calc tot. mamu Neutron mamu Proton + electron mamu sum(particles)/flux reduction 2FC
2 2 4
1'008'664.923 1'007'825.032 4'032'979.910
3FC (use 1 -3FC) Used 4D He4 quanta
1
0.9971130759 3D/4D flow c. 11'642.907 0.0028869241
1 1
1'008'664.923 1'007'825.032
newly added particles mamu Neutron mamu Proton + electron Total Li6 particles sum Additional flux quanta released (5/3) mamu Li6 measured Charge correction by 1FC Delta mamu measured Delta mamu calculated calculated mass Absolute error Relative error total mass
micro amus
1.667 1 3
6'049'469.865 2'341.971 6'015'122.281 1'007'825.032 34'347.584 34'347.213 6'015'122.652 0.371
α/2π * 8 2D/4D flow c. 9'371.786 9'363.982 18'735.768
3'903.285 65.253 34'347.213
0.0000000617
Tab. 2 Li mass-calculation 6
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Li can be understood as a Helium (alpha) core that is orbited by a deuterium nucleus. The “deuterium” is bound by two fux reduction waves which fnally gives a total of 5 fux reduction waves. 6
The frst 5 lines of table 2 show the mass calculation for 4He core. Below is the added “deuterium”. The ft with known Eigenvalues for simple low Z nuclei usually is excellent.
All symmetric nuclei show a very high agreement with the above so called “fux compression constants”. The gamma wave structure can be explained by the free 3D/4D fux, what has been verifed by analyzing the gamma levels!
4 The 4D Neutron and Proton Almost all nuclear mass is built of protons and neutrons. But classic theory tells us nothing about the internal structure of neutrons/protons. The postulation of quarks is an oversimplifcation of the reality and has no predictive power for quantitive variables. 4D physics reveals some internal structure of the particles and allows to quantitatively calculate some properties.
4.1 Magnetic moment of Proton As a frst illustrative sample we will calculate the magnetic moment of the proton. For that purpose we will use a simple 3D physics formula for a magnetic moment. (7) (8) (9) The only parameter of interest in this example is the radius (a o) which is given by the latest measurement. Because the proton has a magnetic moment, in average charge must fow on one radius, that is the 3D Magnetic moment of Proton Exper. chg. Radius 0.84087 fm. projection of the measured 4D radius. If we use Measured µp 1.4106067873 the measured radius the moment will be to 3D µp direct calc from exp. Charge radius 2.0194353567 large because in the 4D torus (see Fig.2) the µp= rp*c*e/2 corrected 4D-->3D 1.4279564349 efective radius is ½ of the classic radius. If we Metric change is 2 (1/2) 1.4142135624 stick to the 3D model, then we have to divide First error ratio µp mes . /µp calc 0.9878500162 the result of formula (9) by 21/2. The other way calculated moment 1.4106069281 round is a bit more complex to understand. If Error(1/3) of one dimension 0.9959334913 0.9959335244 we use ao/2 as the input the we must multiply 4D correction : 3FC*2FC*(1FC) absolute error meas -clc 0.0000001408 the result by 21/2. relative error
0.0000000998
Table 3. Proton magnetic moment and perturbation The uncorrected result for the calculated proton magnetic moment is only 98,8% exact (see tab.3 light blue) because the proton mass is highly perturbed by its own magnetic feld that is fully expanded to 4 dimensions (that can be normalized to 3). Because, the proton can only acquire 3D/4D fux energy the number of involved radii is 3. According to our method we calculate the perturbation for one radius that is 0.99593349. The big surprise is that the perturbation is the exact product (0.99593352) of the well known 3D/4D fux compression constants. After applying the correction the result is far below the precision of the radius measurement. If we do a reverse 3D radius calculation the we get 0.840869916 instead of the experimental 0.84087.
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4.2 Proton mass calculations The proton magnetic mass formula (10) below can be derived for the exact electron mass, by change of constants by starting from E=mec2. In the electron formula the radius to use is r = “electron de Broglie radius”. In the proton case we use 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), you must divide 1.6753fm by two → r=0.83765.. what gives the 3D equivalent (-4D) radius of the proton. For the 4D proton radius you must divide once more by 2 or multiply the result of (10) by 8! (10) Mproton(eV) = µp2*4* π * 100000/(α * π* r3*e) (See also magnetic mass formula from Mills [4]32.32b. Mass equivalence.) In formula (10) 4* π * 100000 stays for µoand the adjustment of the dimensions to get electron volts but the factor π can be crossed out. According to Mills relativistic treatment, we know, if a mass is accelerated in two more dimensions, then you have to increase the energy by the factor 2*π/ α . Because magnetic fux already is at light speed, we only have length contraction by alpha. The formula 2*π/ α , for mass increasing has recently also be refound by N.Chiatti [3] using QM-related reasoning but assuming a “complex” time. Because the magnetic mass stays in 4 dimensions we use the extrapolated (from 4He fux 1.6753/2 fm. Radius from Russian database) 4D radius for the proton as a base to calculate the magnetic energy. To the intermediate result we apply the same perturbation correction we found for the exact 3D calculation of the proton magnetic moment, namely: (3FC*2FC*1FC)3 . (See chpt. 4.1 above) It is obvious that a formula that follows the proton magnetic moment has its perturbation. The frst correction is already close to the known proton mass because we added 0.000003 to the proton radius. µproton 3D/4D radius from 4-He (fm) magnetic energy uncorrected 4D correction µproton correcting with µp perturbation 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
µproton 3D/4D radius from 4-He (fm) magnetic energy uncorrected 4D correction µproton correcting with µp perturbation 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 0.837653000000 926'603'107.69 0.9878501147 937'999'696.37 0.999709695 272385.109185457 0.9997096686 938'272'106.18 938'272'081.48 24.70 0.0000000263
Tab. 4a proton magnetic mass-calculation
1.4106067873 0.837653006969 926'603'084.57 0.9878501147 937'999'672.96 0.9997096702 272408.34105432 0.9997096686 938'272'082.77 938'272'081.30 1.47 0.0000000016
Tab. 4b optimized radius
The second perturbation of the proton mass is (1-(α /(π*16)))2 (corresponds to exactly 272'409.8 eV if derived from the proton mass). (1-(α /(π*16)))2 is the relation of alpha to the whole 3D/4D surface (4 inner/outer spheres) of 4D space. The same perturbation can be calculated from the de Broglie radius potential of the proton. The result shown in Tab. 4b is very good because we already used the neutron 4D interaction radius and the efective 4He compression that are mathematical quantities known with 10 digits precision! Thus we have to discuss (4.2.1 below) the physical relation of the 4D radius of the proton to other particles. Formula (11) is the fnal proton magnetic mass formula that shows an α-quantization. (11) (Mass proton in eV) = µp2*4* π * 100000/(α * π* r3*e*(3FC*2FC*1FC)3 *(1-(α /(π*16)))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 [1][2] the experimenter(s) found that at 1keV particle (proton) stimulation energy strange resonances do occur. 1keV is halve of the frst alpha quantization. This is the correct 3D,t resonance energy as in a su(2) x su(2) quotient only one halve (outside running mass) can kinetically interact! The cut-of of the spectrum seems to ft the quantization. An other interesting aspect is that the proton quantization (1-(α /(π*16)))2) delivers very exactly ¼ of the de
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Broglie radial potential energy, that can be further refned by the known 1FC 3 radial perturbation. Seen from this perspective, we can say that the quantization energy (with high precision) is directly coupled with the potential energy as seen in experiments [1][2]. As you may see in tab 4a 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[6]) gives errors in the 200eV range but with a much larger error bar!
4.2.1 Radius discussion The ideal 4He compression is (1-5*2FC)=0.9941929513381.51.. = 0.9961649819 ( 1.51... = 23/5 ). Derivation mamu He4 He4 from particles compression ratio (CR) torus area r*r → r * CR1/2 Ideal compression
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
see table 6 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.
Table 5. Possible approximation of ideal 4He fux compression. The base assumption is that in 4He there is hidden fux compression happening between the two neutrons. Basically one neutron (see 4.3 below) can release three hole wave equivalents of fux and accept two more waves. The frst column of table 5 shows the unchanged 4He compression (0.99622 for torus fux area) based on measured data. The next two columns 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 4He compression. In the last column we did add the 3 waves with the corresponding***** weights multiplied by the expected 2FC/3FC compression, what gives 8 digits (green feld). This is just to show that there is a physical explanation for the factors we fnally used. *****Used weights: 500.929=336.541*3FC2*2FC4 + 168.271*2FC2 . 4D excess mass must frst be once compressed by 3FC*2FC2 to be again plain mass and the once more compressed by 3FC and 2FC 2 to get the 4He mass density equivalent mass. The hole needs only a 2FC 2 compression! 4D potential free radius 3D radius Quotient 1-5*2FC 23/5 (Quotient)1.5157165665
0.840877788500 0.837653006969 0.996164981909 0.994192951338 1.515716566510 0.994192951338
The 4D potential free radius (0.840877885fm) is 10 digits exact because it can be exactly derived from the neutron mass. Thus the 3/4D radius is 10 digits exact too, because it is found by a mathematical relation. The quotient of R 4D/R3D is the ideal (real) 4He compression of the involved particles..
Table 6. Logarithmic radius/compression relation The factor (1-5*2FC) can also be found in the 4D mass build up because ½*(1-5*2FC) gives the exact amount of 3D/4D energy that is converted into additional 4D energy in nuclei starting with 11B.
4.2.2 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. Details for Neutron see chapter below. The 4D excess-energy of the neutron is “neutron mass” * 3FC'=2'712'454eV. 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 above radius of 0.8408777885fm is just the coulomb radius where the diference of 4D-3D pot is the quotient base. In [6], one year ago, we already used this radius for the virtual deuterium model and found a 7 digits agreement between the magnetic moments of proton/neutron deuterium radius and the above radius. The problem with experimental data is the low quality/precision of any radius measurement.
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4.3 The Neutron mass The neutron is a 4D excess Energy particle. This is obvious as the source of all neutrons is the nucleus where they “live” in a 4D environment. A combination of an electron and a proton is only possible with maximally two coupling rotations. The so called 4D-excess-energy part of the neutron is calculated by 3FC' applied to the neutron mass. This gives a value for 4 rotations. According to Mills we can reduce one dimension by applying the above (see 1.1) mentioned γ* factor. The result (tab.7) , by applying the factor Tabulated neutron excess mass 782'332.96 twice is the so called “electron sec factor” that Neutron mass in eV 939'565'413.21 can also be calculated by using the Mills formula * Neutron mass * 3FC = excess 4D flux 2'712'454.00 for mass equivalence. The efective neutron 4D compensation for electron excess 1.0003346161 excess mass has the same mass density as the ** (1+PI()*D2^2)^2 : Mills 36.15 3D → 2D 3FC** charge normalized by* 4D--> 2D 0.9974467260 4D excess flux gained by neutron-mass * 3FC** 2'398'967.91 electron and is 2'398'967... eV (yellow feld). Corresponds to electron 4D->2D/3D flux loss (M:36.4)
313'486.10
Table 7. The neutron 4D energy The values of table 7, above and table 8 below, are calculated as 3D externalized energy amounts of the neutron decay. If a neutron decays the potential energy 1.7MeV (tab 8) must be built up again. The potential is calculated at the calculated magnetic radius (0.840869916fm). After this step about 686505 eV of 4D energy remain. The fux captured by the electron can be calculated according to Mills formula as rest-mass of electron divided by “2π”. During the decay one spin of a down quark is fipped. This energy has been calculated by Mills. Further the de Broglie energy of the neutron changes to the proton's. This are the terms marked green. They together constitute the so called kinetic energy terms and are, in their sum, exactly the ones measured in the aSPECT experiments[10]. The green marked energies are directly coupled with the ep bond. The blue marked energies are the 3D/4D and 4D excess energy parts of the electron rest mass. These magnetic proton charge radius in fm. 0.8408699160 energies stay in 4 dimensions. Summed up, electron potential at exp p-radius 1'712'470.04 the neutron excess energy is (5 + digits) (4D flux gain) – (potential to overcome) 686'497.87 correctly calculated. Some small parts are still Freed excess electron flux of neutron e0/2pi() 81'328.01 unknown like the role of the 13.6eV at the rebuild 1FC potential -11.03 Freed excess electron flux 3D/4D (2FC) induced 94.46 Bohr radius. Do we have to account them in Freed excess electron flux 4D (3FC) induced flux 234.79 the electron rest mass? The 1FC correction Spin flip energy N-->P (Mills: 39.7) 15'691.94 (second torus radius potential) might then also (less gain) – de Broglie wave correction Rn--> Rp -1'502.11 Neutron kinetic excess mass 4D--> 2D/3D 782'015.71 be slightly diferent Missing to measured Neutron excess Adding 3D & 4D flux to 81238.01 eV e excess
317.25 318.21
Table 8 Neutron 3D/4D “excess-energy” parts mamu Neutron excess in mamu Neutron 4D hole Freed energy neutron->4D
eV 839.891 336.541 503.350
782'353.507 313'486.098 468'867.409
Table 9 Relevant amounts of neutron energies. Table 9 shows the two junks of energy we must take into account if a neutron stays inside a nucleus. The neutron 4D hole (313'486 eV = 2 waves) consists of two missing, uncompressed waves that initially contain no mass. The neutron excess energy has the weight of 3 waves and consist of matter with a reverse 4He compression. These amounts of energy can directly be seen in mass diferences of isotopes or in the 4D internal wave structure. In measured isotope data we may also see cases (14-C, 10-Be), where halve of the hole – one wave - gets flled. In total the neutron can make a 5 wave connection, with the above shown small diferences.
4.3.1 Conclusion The neutron hole and excess waves are a fact seen already in the mass structure. The kinetic & 4D excess energies are used in [6] to calculate the exact neutron halve live. Thus we know the neutron structure with about 5 digits precision.
11
5 The magnetic Bohr (Hydrogen) model (11) E-coulomb = e2/8πε0rBh (12) Emagnetic(eV) = µB2*4π * µo/(rB3*e) If physics would work as expected then the total ionization energy of hydrogen should be the sum of formulas 11+12. From table 10 (below) it is easy to see that the calculated magnetic energy (base is classic Bohr radius!) is far to large. The 4D model assumes that all energy is stored in magnetic fux. This implies that the electron orbiting a proton is not only behaving as a charge. The electron factually behaves as magnetic fux. Thus if we have 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 mass and the perturbative mass given by the magnetic moment perturbation, also known as electron g-factor. It looks like the core mass is not tightly bound to the charge and thus is not directly interacting with the Electron magnetic Perturbative magnetic feld. The correct magnetic energy is given by the ratio of Mass 1'183 eV the perturbative mass/total-mass-perturbative mass. Just remember core mass 509'815.8eV that the magnetic mass (= energy) must be equivalent to the efectively rotating mass that is orthogonal to momentum axis. Fig.4 Electron 4D mass components Following Mills [4] formula 1.253 given below as(16) we can – for radial coupling - always use the reduced mass instead of the stored magnetic energy. Thus in the frst step we just use the known Bohr model with the reduced mass. This optimized Bohr model ignores the second magnetic efect caused by the “second 4D rotation axes” well known from the Larmor precision of the electron. In the ground state there will also be energy added to the orthogonal (in respect to the base magnetic energy produce by the electron orbit) Larmor momentum of the electron. This second momentum seems only to couple with the non relativistic rest-mass of the electron. The measured ionization energy (NIST) is given in the dark blue feld. The reduced mass only ionization energy is given in the light blue feld. The frst 4D adjustment of the Hydrogen ionization energy is given by the dark yellow feld fgure in tab.10. It is a bout 5.6 digits exact what is quite good. Rf: reduced electron factor Bohr radius (Rb) reduced mass Bohr radius (rRb)= Rb/Rf potential at reduced erRb classic error absolute (eV) classic error relative electron magnetic excess mass electron core mass+ ½ excess mass 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)
0.9994556794 One thing has been neglected. The electron 52.9177210527 core mass and the perturbative mass are spin 52.9465409443 coupled and do interact. Mills has calculated 13.5982871554 0.0001473346 0.0000108347 1'183.1037038626 510407.394248069 0.0023179596 0.0572059311 0.0001326010 13.5984197565 0.0000147335 0.0000010835 0.0001473345 13.5984344900 13.5984344899
Table 10 4D magnetic Bohr model energies.
12
the muonium fne structure energy, that he also used to calculated the 4He ionization energy. The formulas (13,14) to derive the coupling energy are given below. They are valid for the muonium. Because we already did the relativistic correction, when using the correct reduced (splited) electron mass, we only do compensate for the “Larmor energy” given by the cosine term of (14). The fnal calculated value is given in the orange feld. The correction factor is given proportional to the Bohr magneton ( µB2) used in the magnetic energy.
For our calculations of the magnetic energy corrections, we used formula (13) below, that after integration leads to the cosine factor in Eq. (14). The coupling of the cosine factor with the base magnetic mass (always given at the Bohr radius) is 1 + 1/9. To mention it once more: The rules for 4D physics are not based on a wide agreement and are are given based on logical derivations. Thus it looks like the center of mass for the formula is the center of the perturbative electron excess mass. But it is very unlikely that this derivation is wrong. Good luck by using a correction (with a reasonable derivation ) at the 6 th digit that exactly delivers 5 more digits can be excluded!
5.1 Used formulas for stored magnetic energy of an electron orbiting a proton. The following equations of R.Mills are given for the muonium. (13) Mills 2.243 This is the energy of a spherical harmonic dipole of the magneto static “Larmor” feld caused by the spin/spin interaction. The spherical harmonics is represented by SIN( θ). We use this formula to fnd the coupling if the SU(2) x SU(2) interaction of the core magnetic X perturbative electron mass with the proton magnetic moment. The solution of the integral delivers the cosine term (14) of the correction. (14)
Mills 2.244 Delta Emag muon:
(15)
Mills 1.162
(16)
Mills 1.253
Force-fux equation for the p-e system including the magnetic energy.
5.2 Discussion To fnd out how this formula develops for higher (exited) states n>1 we made some additional calculations. 1'183.1037038626 1172.9048978026 1171.0162090026 1170.3551709726 1170.0492057926 The precision is always 510407.394248069 510412.493651099 510413.437995499 510413.768514514 510413.921497104 0.0023179596 0.0022979549 0.0022942504 0.0022929538 0.0022923536 better than 6 digits higher 2s 3s 4s 5s states 7 digits. But this is 1/(1+(e-10.1988eV)/mp) 0.9994556903 0.9994556923 0.9994556930 0.9994556933 also true for the Rydberg reduced radius 52.9465403691 52.9465402626 52.9465402253 52.9465402080 radius at n*r 105.8930807382 158.8396207877 211.7861609011 264.7327010401 formula with states n>2 E2r. potential at n*r 6.7991436516 4.5327624435 3.3995718350 2.7196574689 that is given by E(n)= E10.1986786807 12.0873501650 12.7483851845 13.0543499649 n*Er - 1/n E base 2 2 E mag base 0.0001473345 0.0001473345 0.0001473345 0.0001473345 Rydberg *(n -1)/n . The Adding Emag base 10.1988260152 12.0874974995 12.7485325190 13.0544972994 “problematic” states are E mag at r=n 0.0071390711 0.0021152803 0.0008923839 0.0004569006 coupling with rest-mass (eV) 0.0000164053 0.0000048530 0.0000020462 0.0000010474 n=2,3 where the new magnetic correction at n*r 0.0000199347 0.0000020898 -0.0000001991 -0.0000007484 formula above delivers -0.9325911781 delta pot/rest pot. 0.3827740304 -0.0972931197 -0.7145224033 electron first Q state Rest potential 1.5109396300 0.8499016000 0.5439364200 the largest relative error 10.1988060805 12.0874954097 12.7485327180 13.0544980478 but for n=2 is still much measured 10.1988060600 12.0874948600 12.7485328900 13.0544980700 better than the Rydberg relative error -0.0000000020 -0.0000000455 0.0000000135 0.0000000017 formula. 2
Table 11. Turning point of e-p 4D mass coupling & some test corrections What we noted is the following: Between the states n=3,4 the 4D magnetic correction just reverses. If we, for the electron, make the same assumption as for the proton, that one part of its energy (perturbative
13
mass) undergoes a quantization, then we can calculate the frst electron quantization step being -0.93eV. What we must notice is that the frst electron rest-mass quantum is slightly larger than the rest-potential of the electron at state n=4. This also implies that the sign of the force between electron core mass and restmass in relation to electron 4D rest-mass to magnetic + potential energy changes after n =4 is reached, what certainly inverts the correction. It looks like the correct formula can be derived by including the additional quantization of the electron rest-mass.
5.2.1 The corrections for states n=2,3,4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
1'172.905 510'412.494 0.0022979549
1'171.016 510'413.438 0.0022942504
1'170.355 510'413.769 0.0022929538
2s 3s 4s 1/(1+(e-Epot lost)/mp) 1/(1+(e-10.1988..)/e) 0.9994556903 0.9994556903 0.9994556903 reduced radius 52.9465403691 52.9465403691 52.9465403691 radius at n*r 105.8930807382 158.8396211073 211.7861614763 E2r. potential at n*r 6.7991436516 4.5327624344 3.3995718258 n*Er - 1/n2E base 10.1986786807 12.0873501376 12.7483851475 Emag base 0.0001473345 0.0001473345 0.0001473345 Adding Emag base+ e mag state 10.1988260152 12.0874974721 12.7485324820 static 4D correction 10.1988096099 12.0874926191 12.7485304358 measured 10.1988060600 12.0874948600 12.7485328900 error 0.0000035499 -0.0000022409 -0.0000024542 E mag at r=n 0.0071390711 0.0021152803 0.0008923839 coupling with rest-mass (eV) 0.0000164053 0.0000048530 0.0000020462 dynamic 4D correction 10.1988060867 12.0874948745 12.7485328887 Two waves at n*r 0.0000199284 0.0000025977 -0.0000004067 Abs. Error dynamic correction 0.0000000267 0.0000000145 -0.0000000013 delta pot/rest pot calc 0.7256782113 0.3827749458 -0.0972899512 delta pot/rest pot effective 0.7256784977 0.3827740304 -0.0972931197 Rest potential 3.3996284300 1.5109396300 0.8499016000
In the table (12) left, the reduced radius (5) frst was always slightly corrected by the lost energy. But the results are best if just the frst ionization energy (n=2) is used. (Why?? New base at frst excitation state?) The same is done for the electron rest-mass & relativistic core mass. But all these small correction vanish/ have virtually no efect after n=3. The frst calculated energy (9) is the standard reduced mass Bohr level energy with a slightly adapted radius. Then (10) the magnetic base mass is added and the new static – coupling magnetic mass (16) of the state is subtracted. This gives the static 4D magnetic mass (12). The error is in the last 3 digits where as the frst 7 do agree.
Tab.12 state 2,3,4 corrections
5.2.2 Using the delta of the potentials First quantization step of electron eq1:-0.93259eV. Ionization energy: 13.59843449eV The corrections of line (16) are of because we use the same coupling as in the base state. In the base state only the non relativistic mass of the electron couples (in 2D) with the orbital stored magnetic mass. In the excited states (2,3,4) also the spin spin coupling ads (releases energy). In the base state the added mass corresponds to the hyperfne/spin pairing energy (factor 1 + 1/9). In the excited states (2,3,4) also the spin/spin energy gets added (factor 1 + 4/9 + α - Mills [4] 2.224 angle “0”). In the state n=2 it looks like the fne-split factor is also added to the new added spin/spin mass and thus is coupling twice this gives a total weight of 15/9. Further as the two potentials – electron frst quantization and electron/proton rest-potential are close the coupling is proportional to the ratio (20) of the potentials (= 1 + e q1/rest-potential). (20) gives the correct sign for the correction and also the correct weight for all states n >= 2. The coupling factor (20) is used for all higher states. coupling factors fro n=2: (1-0.93259/(13.59843449-10.1988096099))*(1 + 1/9 + 1/9 + 4/9+ α) For the states N=2,3 the spin/spin coupling decreases with the radius. But from state n=2 to n=3 the change is not the usual 1/22 because the coupling waves stay in 4 dimensions. The ratio is 1/(n-1) 2/3 = 1/22/3 or (n=4) 1/32/3. Thus for the state n=3 the coupling factor (18) is : (1-0.93259/(13.59843449-12.0874926191))*(1 + 1/9 + 4/(9*2 2/3) + α) n=4:
14
(1-0.93259/(13.59843449-12.748530435891))*(1 + 1/9 + 4/(9*3 2/3) + α) For all other states the magnetic correction is the same as in the ground state. Only the coupling with the potential must be added. n>4 (1-0.93259/(13.59843449-rest-potential))*(1 + 1/9). Further improvement can be made for the last two digits by refning the magnetic coupling constant (1 + 1/9) that fts better with a decreasing factor depending on n. But frst experimentalist should understand why so called p orbits should have less energy than S orbits... n= 15 is currently the last measured state: The error with the standard magnetic correction in average is 1.2 digits. 1 1'170.049 1'169.883 1'169.783 1'169.718 1'169.673 1'169.641 1'169.600 1'169.566 2 510'413.921 510'414.005 510'414.055 510'414.087 510'414.110 510'414.125 510'414.146 510'414.163 3 0.0022923536 0.0022920276 0.0022918311 0.0022917035 0.002291616 0.0022915535 0.002291472 0.0022914053 4 5s 6s 7s 8s 9s 10s 12s 15s 5 0.9994556903 0.9994556903 0.9994556903 0.9994556903 0.9994556903 0.9994556903 0.9994556903 0.9994556938 6 52.9465403691 52.9465403691 52.9465403691 52.9465403691 52.9465403691 52.9465403691 52.9465403691 52.9465401808 7 264.7327018454 317.6792422145 370.6257825836 423.5723229527 476.5188633218 529.4654036909 635.3584844290 794.1981027114 8 2.7196574606 2.2663812172 1.9426124719 1.6997859129 1.5109208115 1.3598287303 1.1331906086 0.9065524901 9 13.0543499236 13.2205530118 13.3207682320 13.3858117643 13.4304053959 13.4623029583 13.5038537303 13.5378498649 10 0.0001473345 0.0001473345 0.0001473345 0.0001473345 0.0001473345 0.0001473345 0.0001473345 0.0001473345 11 13.0544972581 13.2207003463 13.3209155664 13.3859590988 13.4305527304 13.4624502928 13.5040010648 13.5379971994 12 13.0544962107 13.2206997402 13.3209151848 13.3859588431 13.4305525508 13.4624501619 13.5040009891 13.5379971606 13 13.0544980700 13.2207013500 13.3209166400 13.3859599700 13.4305534900 13.4624509400 13.5040015400 13.5379976000 14 -0.0000018593 -0.0000016098 -0.0000014552 -0.0000011269 -0.0000009392 -0.0000007781 -0.0000005509 -0.0000004394 15 0.0004569005 0.0002644100 0.0001665089 0.0001115480 0.0000783437 0.0000571126 0.0000330513 0.0000169222 16 0.0000010474 0.0000006060 0.0000003816 0.0000002556 0.0000001795 0.0000001309 0.0000000757 0.0000000388 17 13.0544980896 13.2207013354 13.3209165674 13.3859600614 13.4305536390 13.4624511446 13.5040018117 13.5379978211 18 -0.0000008315 -0.0000009891 -0.0000010009 -0.0000009627 -0.0000009086 -0.0000008519 -0.0000007469 -0.0000006217 19 0.0000000196 -0.0000000146 -0.0000000726 0.0000000914 0.0000001490 0.0000002046 0.0000002717 0.0000002211 20 -0.7145165427 -1.4689049092 -2.3604551488 -3.3891673788 -4.5550417320 -5.8580783609 -8.8756391424 -14.4307145565 21 -0.7145224033 -1.4689154308 -2.3604727699 -3.3891906572 -4.5550728080 -5.8581176038 -8.8756967572 -14.4308267365 22 0.5439364200 0.3777331400 0.2775178500 0.2124745200 0.1678810000 0.1359835500 0.0944329500 0.0604368900
Table 13. Higher states n> 4
5.2.3 Conclusion The basic 4D magnetic Bohr model is logically well understood. For the ground state the delivered absolute precision for all known measured digits is breath taking! Whether it is possible to fnd a highly exact closed form description for all higher n states is a theme for future work. The new 4D magnetic model explains, why the potential has a small turning point (between n=3,4), what is an other contribution to our basic physics knowledge. The error usually stays at the last two measured digits and can be improved by detailed modeling (e.g. n=3,4). The state n=2 is not yet fully explained, but it looks reasonable. May be it will be possible to fnd a common magnetic coupling factor (function) for all states. But it looks like the coupling after the turning point fundamentally changes as there is no longer a spin/spin term. 4D physics just starts. We are sure that the understanding will arrive when more people can accept that the old model is invalid.
15
6 Experimental fndings After fnding the SO(4) conform proton quantization, we tried to fnd and fnally support experiments that could delivers an answer how magnetism afects LENR. In the following table 11 we give the frst 32 -unperturbed - quantization energy steps of the proton magnetic moment based mass. 1 2 3 4 5 6 7 8
2'002.337 4'034.328 6'096.637 8'189.947 10'314.963 12'472.410 14'663.038 16'887.616
9 10 11 12 13 14 15 16
19'146.941 21'441.834 23'773.139 26'141.732 28'548.514 30'994.416 33'480.400 36'007.458
17 18 19 20 21 22 23 24
38'576.618 41'188.941 43'845.523 46'547.499 49'296.042 52'092.367 54'937.731 57'833.434
25 26 27 28 29 30 31 32
60'780.825 63'781.300 66'836.307 69'947.345 73'115.971 76'343.798 79'632.500 82'983.818
Table 14 Proton 4D α - quantization: (1-(α /(π*16)))2 Proton magnetic base mass: Mproton(eV) = µp2*4*100000/(α * rp4D3 * e) = 926'603'086.8eV Proton magnetic perturbation “p-1Dimension” = 0.9959335244; For full moment : (p-1D ) 3 = 0.9878501147 Rest-mass of perturbative proton potential: 272409.8eV. Quantization with ((1/α) – n) : n = 1,2,3,4,.... A frst review of old experiments gave two hits: Both Iglev[1] & Lipinski [2] reported the highest proton resonance at 1000eV. Lipinski(s) did fnd/confrm this in several experiments. The above tabulated values are 4D equivalent energies that are valid for emitted radiation energy. In a kinetic experiment with non relativistic protons only halve of the SO(4) = SU(2) x SU(2) responds to a proton event, because in 4D physics the “kinetic mass” is fowing inside and outside of the center of mass surface. Thus the 1000eV perfectly matches the frst proton quantization step.
6.1 The first LENR experiment with constant gamma ray production Two months ago there was a big surprise ! We, the frst time, had access to a along running LENR experiment that allowed to measure gamma radiation over weeks. At frst sight we had no explanation for the seen lines, that were nowhere conform with known lines. Then I detected that the Neutron central peak exactly waves corresponded to the predicted neutron 4D energy hole wave resonance. After that we started to count the peaks between 20 & 80 keV and found that they exactly correspond to the expected number of lines due to proton quantization. We thus see a modulation of a Neutron wave by the proton momentum quantization. Fig.5 Spectrum from a running LENR experiment measured by Russ George (Atom-Ecology) There are other modulations visible with magnetic gamma states too, that will be (possibly) explained by Russ George. The above spectrum has been collected by a very long run. The total energy of the measured gamma-lines
16
is far less than 10-6 if the total energy produced. Thus gamma radiation only delivers a signature of the involved magnetic moments.
6.2 CERN prove of magnetic proton mass The last prove for the proton magnetic mass has recently been given by CERN's LHC. CERN is doing collision between highly accelerated protons. Thus we expect to see resonances of proton base waves (Muon, Kaon, Pion) and the proton itself. The primary masses seen (measured) in collisions are not excess energy particles, they obey the compression rules. The CERN claimed Higgs event at 126GeV is an exact resonance of the proton magnetic mass. If you multiply the above magnetic-mass Mproton with 1/α and the (1D moment-perturbation)2 then you get (125.95GeV). CERN measures exactly what is expected from 4D physics rules, that tell how magnetic mass is converted (See also [3] for an alternative derivation of the rule to get the next deeper magnetic mass. 1/α is the 4D relativistic length contraction if a magnetic mass acquires one more rotation. ). The perturbation is only active in two dimensions because the two magnetic base fux masses couple over the other two dimensions.
7 Relevance for LENR In [5] we already did show that LENR only can be explained, if the magnetic moment is polarizing the nuclear charge & coulomb-cloud and the “incoming” fusing particle is “hitting” the magnetic axes. Today we are 99.999% convinced that the basic force behind LENR is of magnetic nature. Calculations show that the decay of LENR energy can be explained by the momentum quantization and magnetic transport of transient excess feld energy. The later has been confrmed by secondary LENR reactions, we found in a deeper investigation of several spectra.
7.1 LENR heat transport The experiment of the spectrum above did show a large amount (25 watts ) of excess energy from a tiny mass at temperatures between 200 and 500 C. The Boltzmann constant is defned (Wiki) as following:
With a surface of about 1-2 cm2 we can roughly estimate the energy that can be emitted by the fuel as: Lower bar 200C : E = 5.67*10-8*5004*0.0001 = 0.354watts. It can be excluded that the rest of the energy has been transported by normal heat conduction because only a tiny part of the fuel is in direct contact with the quartz wall, the rest stays in air or deuterium vapor. The nuclear felds that occur in LENR reactions are greater 10 10 T. These felds are fast oscillating and are able to interact with all orbital electrons of distant nuclei. An external feld will superimpose with the radial coulomb force of the nucleus what leads to a smaller binding radius for the electrons. Further the orbits will assume a toroidal shape, what induces a secondary feld, that is aligned with the “LENR” produced feld. This process, for its nuclear neighborhood, fnally has the same characteristic (shaking electron cloud) as a fast thermal movement of a nucleus. Thus, after a critical distance, we see induction heating by strong nuclear felds. In adjacent nuclei the amplitude of the induced thermal movement is tiny because of the high
17
frequency. The turning point is the distance where the magnitude of the magnetic force and the coulomb force are close. K-electron will be the frst that try to fallback into classic orbits what leads to turbulent movents and fnally thermal oscillation of the coulomb-cloud. Without any additional perturbations the overlying felds would be non radiating/dissipative. But fnally the expansion/reduction of the nuclear radius is doing work in the grid. One future task is to fnd the optimal coupling material for the most efcient – distant - heat transport.
7.2 What is needed for successful sustainable LENR ? There are two prerequisites:
7.2.1 Strong stable nuclear feld A long time stable strong nuclear feld that is able to polarize the coulomb cloud of reacting atoms can only be generated by nuclear magnetons. This can only be guaranteed by so called long living gamma states we see in (109Pd, 61Ni, 107/109Ag, 111Cd , ..), the so called most active LENR nuclei. The 117Sn magnetic gamma state, living for 14 days!, is a double neutron hole wave equivalent state! Fusing nuclei generate an EM-pulse (adding radial magnetic fux of up to 16..24 MeV) that aligns neighbor spin axes, what leads to potential attraction of new atomic pairs. This efect is increased by the resonance of magnetic gamma states. These moments (long living magnetic gamma states) may also serve as a temporary storage of magnetic excess-energy. The lower their level, the easier you can induced the needed feld to get a resonance. This also can explain sudden cool down efects seen in recent experiments! Because long living magnetic gamma states obey an exact probabilistic behavior, it should be possible to model the reaction kinetics of a given isotopes mixture by a coupled Markov (chain-) process.
7.2.2 The LENR trigger In [5] we did explain that the short time change of magnetic fux is responsible for triggering a LENR reaction. The strongest feld change you get, when the e.g 107Ag long living state decays to the 107Ag base state. In this case also the polarity of the magnet feld changes what induces an additional rotation! Even more efcient is the pulse that a higher laying magnetic state is able to deliver, because of its much shorter live time! E.g.: Based on the 107Ag 324.81 & 423.15 gamma states.
7.2.3 What spectrums show In the collected spectra of LENR reactions we mostly see lines of magnetic gamma states. This proves that LENR excess energy temporarily is stored in magnetic gamma states. Higher magnetic states usually decay to lower states with the same magnetic polarity, if the path is available. We never saw a non magnetic decay from a magnetic state!, if it was available too. We also often see magnetic state gammas, that are modulated by the deuterium quanta energies and thus can be found with a specifc +- delta!
7.3 What is the halve live of deuterium fusion? In [6] we did show that the well known radiation formula for particle state “live time = energy/power” can be successfully transformed (= energy/power 1/sqrt(2)) to 4D physics space. Because classically time occurs in the exponent and space & time are indistinguishable on the nuclear level, the radial metric change occurs
18
H fusion energy 23'846'533.88 Input energy(power) Live time = energy/power1/sqrt(2) 2 x 2000 67'668.14 2 x 8000 25'390.05 2 x 78000 5'073.78 2
in the exponent. The formula is a good approximation for other 4D particles (e.g. muon) too.
Table 15 sample live time(s) for DxD energy released If we assume that the triggering power is the D x D momentum energy released, then we see that in systems with low input energy (2k x 2k) the reaction is running very slowly. In fact you will have to wait about 19 hours to see the energy coming out of halve of the reactions! This very well correlates with the measured ramp-up time in the the above experiment. The other astonishing fact of this formula is that it can explain why long heat after death phases can be seen – in case the input energy is low. It also explains that increasing the input energy is not coupling linearly to the output energy. What is not yet completely understood is the upscaling of phonon/input energy to a frst (2keV) proton momentum relaxation. We know, since a long time, that it happens spontaneously in the Holmlid [7] dense Hydrogen experiments, even just triggered by the laboratory light!! The other open question is whether the above formula always works in both directions - what is the case in normal spectroscopy. But here, on the nuclear level we have no experience. The weight of one neutron bond is about 156keV. Could it be that states of magnetically connected D-D with 78 x 78 keV deep moment mark the point of no return? If you look at the above spectrum it looks like this is happening. But there could be a second, even more important efect, we can deduce from the above spectrum! The neutron 4D hole wave is an efcient storage/emitter of nuclear excess energy, what fnally gives the neutron it's center place in LENR. ,
8 Outlook To get a successful LENR reaction you need the right mixture of magnetic isotopes. The typical released fusion energy of an A-D fusion is about 16MeV's and for D-D direct fusion 23.8MeV. This reaction energy can be used to induce stable nuclear magnetons, that are able to polarize the fuel. A strong external feld is also able to squeeze/stabilize/align fusing proton/deuterium nuclear charge orbits, what potentially leads to new fuel. Further you must guarantee that the activation energy of the basic D-D fusion is high enough to get a useful amount of energy. The frst step of “massifcation” is the transformation of potential energy into rotating energy. The externally visible condensed matter – EM-felds – make three rotations. To convert a potential into a three times rotating fux you need either an Alvén condition[2, explained in 5] with a sheer wave or a 2D stimulation with quadratic EM wave forms and a polarity change to induce the third rotation. Most critical is the timing of the overall reaction chain, that must be in high agreement with the used magnetic isotopes. We also see the need for isotopes that can mediate a neutron hole wave! It looks like the frst nuclear bond made is equivalent to one neutron wave. Because the NAE should remain stable, we strongly urge you to use only D-D fusion, that decays magnetically and you should also avoid secondary D-A and A1-A2 fusion, because of kinetic/ destructive outputs! A direct corollary of this is: Try to stay in medium temperature ranges. We will also have a deeper look at deep magnetic orbits and how to induce them. Table 16 shows a 4'702.141 comparison of electron fux using the magnetic mass -667.577 4'034.563 formula of the proton at the Holmlids deuterium radius -4'034.320 (2.15..pm) with lost magnetic momentum, potential energy. 0.243 It looks like the energy balance is very close.
nuclear magnetic energy at 2.157pm lost potential Energy in orbit energy of two coupling protons Summ
Table 16 deep magnetic orbits.
[1] B.I. Ivlev Conversion of zero point energy into high-energy photons [2] : Lipinski WO2014189799 united gravity about LiP (H*) fusion.
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[3] Leonardo Chiatti, Quantum Jumps and Electrodynamical Description [4] 2016: Mills, Randell L., The GRAND UNIFIED THEORY of CLASSICAL QUANTUM MECHANICS;ISBN 978-0-9635171-5-9 (2016) online. [5] J.A.Wyttenbach About LENR, researchgate, (2016, online),https://www.researchgate.net/project/Low-energy-nuclear-reactions-2 [6] J.A.Wyttenbach NPP 2.1, researchgate, (2018, online), https://www.researchgate.net/project/Nuclear-and-particle-physics-20 [7] 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
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