Anomalous mass of the neutron

by Wladimir Guglinski Mechanical Engineer graduated in the Escola de Engenharia da Universidade Federal de Minas Gerais- UFMG, (Brazil), 1973 author of the book Quantum Ring Theory-Foundations for Cold Fusion, published in 200

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A new model of the neutron n=p+s is proposed, where s is the selectron, a particle postulated by the Supersymmetry.  The model n=p+s belongs to the author’s “Quantum Ring Theory-Foundations for Cold Fusion”, which is composed by 26 papers  published in a book form in 2006 by the Bauu Institute Press.
The Nuclear Physics works with two models of the neutron.  The Yukawa’s model has several disadvantages (the most grave is the violation of the mass-energy conservation, although the theorists tried to justify it through the Heisenberg’s uncertainty principle), because his model cannot explain some phenomena.  The quark model (d,u,d) also cannot explain other sort of phenomena, and then the theorists use the two models, sometimes they use the Yukawa’s model, and sometimes they use the quark model.  However, they are two incompatible models, and it is difficult to believe that Nature works through the use of two incompatible models for the production of phenomena.
The old Rutherford’s model of neutron has been abandoned by the theorists because it seems that it cannot be reconciled with some principles of Quantum Mechanics.  Nevertheless, herein it is shown that Rutherford’s model can be reconciled with the principles of QM when we introduce the hypothesis of the helical trajectory.

Keywords:  new version n=p+s of Rutherford’s neutron, Borghi and Conte-Pieralice experiments, Natarajan’s helical trajectory incorporated to n=p+s, Borghi and Conte-Pieralice experiments suggesting a new Planck’s gravitational constant, deuteron’s quadrupole moment, neutron’s magnetic moment, deuteron’s magnetic moment.

This paper was submitted to several peer reviewed journals of Nuclear Physics.  All they rejected it.  In the last journal, the referee rejected it by claiming that a neutron cannot be formed by one proton and one selectron because the energy required to form a selectron is of about 20GeV.  However, 20GeV is the energy required from the current theories, which do not consider the helical trajectory of the electron.  So, a neutron formed by proton and selectron is impossible when it is considered by the current Nuclear Physics, but it is not impossible if we consider a model of electron with helical trajectory.

The model of neutron proposed in the Quantum Ring Theory does not violate the Fermi-Dirac statistics, as it is explained as follows:

  1. In the present theory it is proposed that the elementary particles move through a helical trajectory (HT).
  2. In the author’s paper [1], numbered No. 4 in his book,  it is shown that the HT has a property named Zoom-effect, according which the radius of the HT decreases with the growth of the velocity of the particle.  When the velocity is near to the velocity c of light, the radius of the HT tends to zero (which means that when an electron moves with relativistic speed, its motion approaches to a classical trajectory in the sense of Newton).
  3. In the author’s paper [2], numbered No. 5 in his book, it is proposed that the spin of the particles (in the sense of quantum theory) is a result of the intrinsic spin of the particle combined with the rotation of the particle about the line center of its HT.
  4. So, as due to the Zoom-effect an electron with relativistic speed does not move through the HT, then an electron with relativistic speed becomes a boson, because it loses its quantum spin (which is a property of the HT, which vanished with the relativistic motion).
  5. In the present paper it is calculated the velocity of the electron about a proton, within the structure of the neutron. Its velocity is 92% of the light speed, which means that within the neutron’s structure the electron becomes a boson.
  6. In the Supersymmetry it is postulated the existence of a particle with the same mass and charge of the electron, but with a null spin.  They call it selectron.
  7. So, we can consider that in the present theory the structure of the neutron actually is n=p+s, that is, the neutron is formed by one proton and one selectron.  Therefore the neutron actually is structured by one fermion (the proton) and one boson (the selectron).
  8. Then we realize that it is vanished the most grave restriction against the neutron formed by proton and electron, because now we can consider that the electron becomes a selectron within the neutron’s structure.  Thereby such new structure fits to Fermi-Dirac’s statistics, since in the new model n=p+s the neutron is formed by a fermion combined with a boson.

So, as from the model of neutron n=p+s there is no violation of Fermi-Dirac statistics, and since the other restrictions against n=p+s are eliminated in the present paper, then the theorists have no reason anymore for rejecting a model of neutron formed by one proton and one selectron.
The mechanism according which an electron becomes a selectron within the structure n=p+s has been named “spin-fusion” in the author’s theory.  Any lepton is subjected to be tied to a quark through the spin-fusion mechanism (within a structure with quark-lepton interaction we would rename the lepton by calling it “selepton”, which spin is zero).
A theoretical quark model of neutron n = (u,d,u-s) has been proposed by the author in a paper published by the Journal of New Energy [3], where it was shown that several paradoxes of Physics can be eliminated through the adoption of the new model.  As for example:

  1. From the proposal of the “spin-fusion” phenomenon the cause is found for the violation of the parity in beta-decay. NOTE: The spin-fusion mechanism is proposed in the author’s paper “Stern-Gerlach Experiment and the Helical Trajectory”[2], and it is based on the property of the helical trajectory of the elementary particles, as proposed in the author’s paper “Fundamental Requirements for the Proposal of a New Hydrogen Atom”[1].
  2. From the new comprehension of the cause of violation of the parity, it is possible to propose a new interpretation for the temporal reversion (an interpretation of Christenson’s discovery concerning the decay of some pions), in order that it is possible to eliminate the very strange hypothesis of temporal reversion in physics.

The new model of neutron (u,d,u-s) can also supply theoretical backgrounds for the explanation of several questions arisen from new experimental findings, as we may mention for instance:

  • a) Taleyarkhan[4] experiment cannot be explained from the old concepts of Quantum Mechanics, since the Suslick-Didenko[5] experiment has shown that the greatest portion of the energy of the sonoluminescence phenomenon is wasted in chemical reactions, and therefore the remaining energy is unable to yield hot nuclear reactions.
  • b) New astronomical observations [6], described in the journal Nature, are suggesting that Planck’s constant can have variation.  Such a hypothesis implies the breakdown of Quantum Mechanics, unless we show that for distances shorter than 2fm there are non-Coulombic interactions performed through a new sort of Planck’s constant, which nature is gravitational.

Before the acceptance of the model n=p+s by the scientists, there are several questions to be answered. Obviously the theoretical restrictions against the model n=p+e can also be applied to the model n=p+s (excluding the Fermi-Dirac statistics, as already explained before).  So, let us remember what are the restrictions against the model n=p+e.
One of the solutions proposed herein is concerning the anomalous mass of the neutron.
The repose mass of the proton and electron are:

Proton:  mP = 938.3 MeV/c²
Electron:  me = 0.511MeV/c²
Total mass: mT = 938.811MeV/c²

A structure of the neutron n = p+e would have to have a mass mN < 938.811 MeV/c², since there is a loss of mass.  However, it is known by experiments that neutron’s mass is mN = 939.6MeV/c².  This fact is one of the stronger reasons why the majority of the physicists do not accept the model n=p+e, although several experiments have shown that neutron structure is indeed n=p+e.  So, herein we will show why the neutron with structure n = p+e has such an anomalous mass mN>mp+me.
Another restriction against the model n = p+e comes from the Heisenberg’s uncertainty principle: such a model requires a force with magnitude 10³ stronger than the strong nuclear force, in order to keep the electron in the nuclei.  Herein we propose a solution able to eliminate such a restriction.
Considering the model n = p+e, the paper also exhibits the theoretical calculation for:

a)  the magnetic moment of the neutron
b)  the electric quadrupole moment of the deuteron
c)  the magnetic moment of the deuteron


  1. The helical trajectory of the elementary particles was proposed by Natarajan[7].  According to his proposal, “When we consider a particle at rest in the laboratory frame, it has no external motion (vCX = 0).  The internal velocity, however, is given by vIN= c (Postulate 4).  On the other hand, if the particle is observed to be moving with a uniform velocity v in the laboratory (vCX = v),  then vIN should be vIN = (c² –  v²)½  so that the result of these two velocities is still c (Postulate 3 and 4).”
  2. The helical trajectory appears in the Dirac’s theory of the electron.  In their book[8] Lindsay and Margenau say: “The only possible resolution of this apparent paradox is to assume that the electron performs, in a classical sense, a rapidly periodic movement with the speed of light, while it progresses uniformly along x in conformity with (12).  Schrödinger was the first to point out this peculiar trembling motion;  its actual significance is not clearly understood”.
  3. There is not any similar theory in the world.  The reason is obvious:  all the attempts of other theorists are made by considering the fundamental principles of quantum theory.  Nobody tries a model with a corpuscular electron, because all they consider that a corpuscular electron is incompatible with the Schrödinger’s Equation.

Unlike, within the neutron’s structure proposed here the electron is a corpuscular particle that moves through the helical trajectory, and so there is not any model of neutron similar to this model proposed herein.
OBS:  in the author’s paper [1] it is shown that a corpuscular electron that moves through the helical trajectory is compatible with the Schrödinger Equation.  This is the reason why the author can propose a model of neutron n=p+e where the electron is corpuscular, but other authors cannot do it.
Dr. Rugero Santilli and Dr. Elio Conte have proposed a model of neutron n=p+e, but in their theory the electron is not corpuscular.  Their models are unable to explain fundamental questions that arrive when we try to propose a model n=p+e, as for example the violation of Fermi-Dirac statistics, the anomalous mass of the neutron, the magnitude of the neutron’s magnetic moment (it would have to be in the same order of the electron’s magnetic moment).  These questions are explained from the model  n=p+s.

Anomalous uncertainly principle
According to current Particle Physics, the structure of the pion po is (d,d’), where d is a quark (d)–1/3 and d’ is its antiparticle (d’)+1/3. The pion po can have two sorts of decays:

χº → γ + γ
χº → e+ + e- + proton       (1)

The time decay has the order of 10ˆ-15s.
Let us calculate the binding energy necessary to pack together these two quarks d and d’, considering the following:

a) The quarks have a mass approximately 1/2000 of the proton’s mass
b) The Heisenberg’s uncertainty principle      Δx.Δp ~ h (2)

Consider the two quarks d and d’ into a rectangular well with a radius “a,” where “a” is the distance between the two quarks into the structure of the pion χº, in order that the uncertainty in the value of position is Δx ~ a.  From Eq. (2) the smallest possible value of Δp is given approximately by  Δp~h/a. So, the quarks placed in the potential well of radius a≤1fm would have kinetic energies, at least in the order of magnitude

T ~ Δp²/2µπ ~ h²/mπ.a² ~ 80GeV      (3)

where µπ = mπ/2  is the reduced mass of each quark.

Let us expound the matter in another more precise way, by considering the conditions necessary for the appearance of a standing wave. For the rectangular potential well of the radius a, this condition is:

2a = λ/2     (4)

where λ is the de Broglie wavelength. Substituting  λ = h/p ,  we have

2a = h/2p = h/2(2µπ T)½ = h/2(mπ T)½     (5)

where T is kinetic energy of the quark in the well.  From Eq. (5), with a ≤1fm, we have

T = π²2h²/4mπa² ≥ 180 GeV      (6)

Since the two quarks are into the potential well along a time with the order of 10ˆ–15s, it is necessary a depth of a well Uπ , as follows

Uπ = T =  180 GeV     (7)

Let us compare it with the depth of potential well UN of deuteron nuclei, since we know that into the deuteron the proton and neutron are tied by the strong force.  The depth of the well UN is:

UN = 40 MeV     (8)

Since Up /UN = 4×10³, this means that, for keeping the two quarks along the time 10ˆ–15s, it would be necessary to have a force thousands times stronger than the nuclear force.
Even if we consider the structure of the proton (u,d,u), two quarks ‘u’ cannot be packed by the strong force into the potential well with radius a = 1fm.  It is necessary a force thousands times stronger than the nuclear force.
Undoubtedly, this fact suggests that something is wrong with the uncertainty principle Δx.Δp ~ h into a potential well with radius a≤1fm .
Besides, the decay shown in Eq. (1) shows that the bound state to the two quarks cannot be 180 GeV, and this suggests that something is wrong with the relation  Δx.Δp ~ h when we apply it for a potential well with radius a£1fm.
We will see ahead other fact suggesting that we cannot apply  Δx.Δp ~ h into a potential well with a≤1fm .
Gravitational quantum of energy
There are two experiments where the model  n = p+e has been obtained.

In the 1980s, the physicist Don Borghi [2] et al. made an experiment where they obtained neutrons from protons and electrons at low energy.  At the end of the article they say, “Hence we may conclude that this experiment seems to confirm the possibility of observing directly the assumed non-Coulombic interaction between protons and electrons.”
In 1999 the physicist Elio Conte, together with Maria Pieralice [3], made an experiment where they obtained neutrons from the cold fusion between protons and electrons.
So, we have two different experiments where the researchers confirmed the structure n=p+e for the neutron.
The mass of the electron is approximately the same mass of a quark d, both having a mass approximately 1/2000 of the proton’s mass.  This means that, into the structure n=p+e, the electron would have to be confined into a potential well with depth Ue = 180 GeV, that is, if we consider that we must apply the Heisenberg’s relation (2).  And then it would require a kind of force thousands of times stronger than the nuclear force, in order to keep the electron in the structure n=p+e.
So, we have a dilemma:
  1. On one side, Heisenberg’s uncertainty principle  Δx.Δp ~ h imply that it is impossible a structure n=p+e.
  2. On the other side, two experiments are showing that n=p+e is the structure used by the Nature.
What have we to keep? We have two alternatives:
  1. We keep the relation Δx.Δp ~ h, and it means that we must reject the experiments. This is a betrayal to the scientific method.
  2. We keep the experiments, and this implies that we must analyze what happens with Heinsenberg’s uncertainty principle into potential wells with a≤1fm, because we must realize that something unknown by the physicists happens into regions with a≤1fm.
It is well to remember that in the beginning of the 20th Century several experiments suggested the structure n = p+e, as for example the neutron’s decay → p+e+ν’.  But Heisenberg rejected these experiments.  Since the Mathematics suggested that the structure n=p+e is impossible, Heisenberg decided to reject those old experiments.
But now new experiments are showing that n=p+e is indeed correct. We cannot neglect the experiments anymore, like Heisenberg did.  This indicates that we must propose a new interpretation for the Heinsenberg’s principle into a potential well with radius a≤1fm.
First of all, let us remember that Planck’s constant h =  6.6×10ˆ–34J-s  has electromagnetic origin, since he made his experiments with photons into a black body.  But into a potential well with radius a≤1fm, we have to consider the strong force. Then it is possible that Planck’s constant must be replaced by a new constant hG , by considering that hG is a smallest quantum of energy due to the interactions by the nuclear force.  In the last item we will show that electron’s bound energy into the neutron must have on the order of 0.1 MeV.  So, by considering that electron’s binding energy has the order of  0.1MeV, then, by introducing a correction, from Eq. (6) we get:
hG ~ [ h²/(180.000/0,1) ]½ = 1,3×10ˆ-37J-s     (9)
One argument against this proposal is to say that the electron has no interaction by the strong force. However, in past papers the author will show that there are evidences suggesting that the strong force has gravitational origin, when we consider a dynamic gravity (different from the static gravity of current Physics).
So, if we consider the quantum vacuum constituted by electromagnetic particles and by gravitons, through such a consideration it means that Planck’s constant h is due to interactions by electromagnetic particles of the quantum vacuum, while the constant hG is due to interactions by gravitons.
Pay attention that we are proposing here the constant hG through the same way as Planck proposed the constant h.  Indeed, Planck has been constrained to adopt the hypothesis of the constant h because that was the unique solution able to solve the paradox of the ultraviolet catastrophe into the black body.  By the same way, today we have two experiments, made by Borghi and by Conte, and these two experiments are showing that the neutron’s structure is n=p+e.  The unique way to explain this structure, obtained by the experiments, is through the adoption of the following hypothesis:
for a potential well with radius a1fm,  Heisenberg’s uncertainty principle is   Δx.Δp~h ,  where hG~1.3×10ˆ–37J-s  is the gravitational quantum of energy.
How to get the magnetic dipole moment of neutron
Magnetic moment of the electron is by three orders of magnitude larger than that of the neutron.  So, at first glance, it seems that the neutron could not be performed by the structure n= p+e.  However, as is shown in the author’s other paper [7] , the magnetic moment of the electron depends on its helical trajectory into the electrosphere of the atom.  In another paper [8] , the author shows that the radius of the helical trajectory has vanished when the electron’s speed approaches light speed c.  So, in the structure n=p+e the electron’s speed is 0.92c , as we will calculate herein, then into the neutron the electron loses its helical trajectory, and by consequence its magnetic moment into the neutron is very small, justifying the present theoretical calculation for the neutron’s magnetic moment.
Therefore the method of calculation is very simple:
a) The electron turning about the proton can be considered like a small spiral
b) The m of  neutron will be :  mNEUTRON =  mPROTON + mSPIRAL
Proton’s magnetic moment we get from experiments, µ = +2,7896µn
Spiral’s magnetic moment we have to derive from calculation. We need to know two data about the electron’s orbit:
  1. Spiral’s radius – we can get it from electron’s orbit about two protons , starting from the electric quadrupole moment Q(b) of deuteron. From experiments,  Q(b) = + 2.7×10ˆ–31m² , and from here we will get the radius R of the spiral.
  2. Electron’s speed – we can get it from Kurie’s graphic for beta-decay of neutron.
Proton’s radius
We will need proton’s radius with more accuracy than Nuclear Theory can give us. And we will get it from recent interpretations about recent experiments. From Nuclear Theory, we know two important facts about the nucleus:
  • 1st fact – protons and neutrons have the same distribution into the nuclei. This conclusion had been inferred from interpretation about the empirical equation shown in the Fig. 1.
  • 2nd fact – from the empirical equation, the physicists also concluded that all the nuclei have the same shell thickness  “2b” = 2 x 0.55F = 1.1F
From these two facts we can suppose that the protons and neutrons distribution into the nuclei is like shown in the Fig. 2, and thus we can get proton’s radius:
4 x Rp = 1.1F  →   Rp = 0.275F      (10)
The radius Rp = 0.275F is corroborated by the proton’s distribution of load, obtained from experiments, shown in Fig. 10.
We will verify that Rp = 0.275F can lead us to very good conclusions, according to the results of experiments.
Well-known calculation used by nuclear theory
Let us remember a theoretical calculation of electric quadrupole moment Q(b) used by Nuclear Theory.

Fig. 3 shows a nucleus composed by a  [ magic number  +  1 proton ].

For example, it can be the 51Sb123 = 50Sn122 + 1 proton. The magic number 50Sn122  has Q(b)= 0, because its distribution is spherically symmetrical.
The 51Sb123 will have
Q(b) =  ∫ρ [ – (r’ )² ].dτ =  -(r’ )². ∫ρ.dτ      (11)
∫ρ.dτ  =  + 1      (12)
because the ring (Fig. 3)  has 1 proton , and “ρ” is measured by proton’s units of load.
Q(b) =   -(r’)²     (13)
This is a well-known traditional calculation. The nuclear physicists know it very well.
Application to the calculation of Q8b)
Let’s apply this sort of considerations to the model of 1H2 shown in the Fig. 4, with one electron turning about two protons.
The two protons have Q(b) = 0 , because theirs distribution of load is spherically symmetrical. The electron can be considered like a proton with negative load, with punctual concentrated configuration, and therefore the electron produces a ring like shown in Fig. 5.
If a proton with positive load yields  ∫ρdτ  = +1 , the electron with negative load yields  ∫ρdτ  = -1. By consequence, the  electric quadrupole moment of  1H2 will be :
Q(b) = -(r’ )²∫ρdτ = -(r’ )².(-1) = +(r’ )²      (14)
But  r’= 2Rp (Fig. 4) , and Rp = 0.275F is the proton’s radius obtained in (10).
Q(b) =  +(r’ )² = +(0,55F)² = +3,0×10ˆ-31m²      (15)
But the radius Rp = 0.275F is not exact, because it is obtained by experiments ( b = 0.55F ).
If we consider  Rp = 0.26F, we will have  r’ = 0.52F, and then:
Q(b) = +(0.52F)Q(b)² = + 2.7 x 10ˆ-31m²      (16)
like inferred from experiments, and therefore we can take R = 0.26F (spiral’s radius).
NOTE:  Of course Yukawa’s model cannot explain Q(b) = +2.7 x 10ˆ-31m² of deuteron, because the two protons have Q(b) = 0, and the meson’s oscillation cannot be responsible by  Q(b) = +2.7 x 10ˆ-31m².  A deuteron performed by (u,d,u).(d,u,d) of current Nuclear Physics also cannot get the result Q(b)= +2.7×10ˆ-31m² of the experiments.

Electron’s speed
We will get electron’s speed from the neutron’s beta-decay (Fig. 9).

Electron’s repose energy ( E = m0.c² )  is  0.511 MeV.
From Kurie’s graphic interpretation, electron’s kinetic energy KeMAX when emitted in the beta-decay, corresponds to the binding energy 0.78 MeV , that is, electron’s kinetic energy turning about the proton.
0.78MeV > 0.511MeV,  by consequence  EKINETIC > m0.c², and therefore we need to apply Einstein’s Relativistic dynamics if we want to know electron’s “v” speed in the spiral.
The relativistic kinetic energy is  :
E = m0.c²[ 1/( 1 – v²/c² )½ -1 ]      (17)
Thus, we have:
0.78MeV = 0.511MeV[ 1/( 1- v²/c² )½ -1 ]      (18)
λ = 1/( 1- v²/c² )½ =  2.5264      (19)
1/( 1- v²/c² )   =  6.383      (20)
6.383 – 6.383.v²/c²  = 1       (21)
6.383 × v²/c²  =  5.383      (22)
v = c (5.383/6.383)½  =  2.746×10ˆ8 m/s   ~   91.83% c     (23)
A spiral with area “A” , a current “i” , and radius R , produces
µ = i.A = q.v.π.R²/ 2µR  =  q.v.R/2
and with relativistic speeds
µ = q.v.R      (24)
The magnetic dipole moment µSPIRAL of one relativistic spiral will suffer a correction proportional to:
λ = 1/( 1- v²/c² )½     (25)
because if  v→c  ,   then    µSPIRAL → ∞.
µSPIRAL = q.v.R/[ ( 1- v²/c² )½ ] ,   when   v → c     (26)
R = spiral’s radius  =  0.26F   (27)
q = -1.6×10ˆ-19C      (28)
v = 2.746×10ˆ8 m/s      (29)
µSPIRAL =  λ.[q.v.R]     ,    λ = 2.5264  in the present problem     (30)
µSPIRAL = 2.5264 x (-1.6 x 10ˆ-19C) x 2.746 x 10ˆ8m/s x 0.26 x 10ˆ-15m     (31)
µSPIRAL = 2.886 x 10ˆ–26 A-m² =  -5.715µn     (32)
Calculation of the magnetic dipole moment of neutron
The proton has µ = +2.7896mn , and then the magnetic dipole moment of neutron will be:
µNEUTRON = +2.7896 – 5.715 = -2.9254µn      (33)
and the experiments detected -1.9103mn.
This result is coherent, if we consider:
  1. The radius R= 0.26F has been obtained from the calculation of electric quadrupole moment, and therefore it is necessary to consider an external radius due to the electron’s orbit around the proton,
    Rext = 0.26F      (34)
    because the external radius is responsible by the measurement of  Q(b).
  2. In the spiral’s area responsible by the magnetic dipole moment, it is necessary to consider the internal spiral’s radius,
    Rint = Rext – Φe  (Φe = electron’s diameter)      (35)
    because the “internal area” of the spiral produces the flux of magnetic dipole moment.

The experiments already detected electron’s radius, which magnitude is smaller than 10ˆ-16m , and also proton’s radius, in order of 10ˆ-15m . Therefore, we can conclude that the density of their masses is approximately the same, because the relation between their masses is:

983.3MeV /c² / 0.511MeV /c²   =   1836     (36)
and the relation between theirs radii is:
Rp / Re = (1836 )ˆ1/3  =  12,25  ~ 10ˆ-15 /10ˆ-16m     (37)
Rp ~ 0.26F  →  Re ~  0.26 / 12.25  =  0.0212F     (38)
Thus, electron’s diameter is Φe = 2 x 0.0212F = 0.0424F  ,  and the internal radius of spiral will be:
Rint = 0.26F – 0.0424F  =   0.2176F     (39)
The correct magnetic dipole moment of electron’s spiral will be:
µSPIRAL = -5.715 x 0.2176 / 0.26 = -4.783µn     (40)
and we get
µNEUTRON = -4.783 + 2.7896 = -1.9934µn     (41)
which is a very good result.
Magnetic dipole moment of deuteron
The proton has µρ = +2.7896µn, and the neutron has µN = -1.9103mn.  Then let us see what magnetic moment for the deuteron we would have to expect from the current theories of Physics.
  1. From Yukawa’s model, as the meson has oscillatory motion between the proton and the neutron, it cannot produce any additional magnetic moment.  Therefore from Yukawa’s model the magnetic moment of deuteron would have to be mD = +2.7896µn – 1.9103µn = + 0.8793µn.
  2. From the model of Particle Physics (u,d,u)(d,u,d) there is no reason why an additional magnetic moment can be created.  Then we also would have to expect µD = +0.8793µn.
But the experiments show that the deuteron has magnetic moment µD =  +0.857µn.  So, from the models of neutron used in current Physics is impossible to explain the magnetic moment of deuteron.  Let us see if we can explain it from the present model of neutron n = p+e. In the formation of the deuteron, there are two protons with the same spin, so the spin due to the protons is i=1.  In the First Part of the paper New Model of Neutron [1] we already have seen that electron’s contribution is null for the total spin, as consequence of the spin-fusion phenomenon.  Therefore the deuteron has nuclear spin i=1.
Calculation of µ.
Fig. 6 illustrates the method:
  1. There are two protons each one with mp= +2.7896µn.
  2. We already obtained spiral’s  µS= -4.783µn.  But we will consider µS= -4.7mn , because 0.083 is due to error in the accuracy.
  3. When the electron of the structure n = p+e is situated between the two protons of the structure of the deuteron (see Fig. 6), it is submitted to three forces:
    a) The nuclear force of attraction with the proton into the neutron’s structure (proton at right side).
    b) The centrifugal force expelling the electron in the direction of the proton at the left side.
    c) The nuclear force of attraction with the proton at the right side.
Then there is an increase of area ΔA due to the electron’s deviation in the direction of the proton at the left side, which is responsible for an increase of Δμ .
We can approach the area ΔA of Fig. 6 from a rectangular area, as shown in Fig. 7, and the total magnetic moment will be performed as indicated in the Fig. 8.
We know that electron’s SPIRAL has a radius R = 0.26F.
Let us consider that ΔA is a rectangular area with dimensions 0.52F and 0.002F.  Then the area is:
ΔA = 0.52 x 0.002 = 0.001F²     (42)
The area of electron’s spiral is:
A =  p.0.26² = 0.212 F²     (43)
If the spiral with area A = 0.212 F²  produces m= -4.7µn , then an area  ΔA = 0.001F²  will produce:
Δµ = -4.7 x 0.001/0.212 = -0.022µn     (44)
and  the theoretical µ of  1H2, obtained from the model n = p+e, will be:
2.(+2.7896) – (4.7 + 0.022) = +0.857µn     (45)
Anomalous mass of the neutron
We will show that neutron’s anomalous mass is due to the growth of the electron’s mass, since the electron has a relativistic speed into the neutron, as we will calculate here. So, let us calculate the electron’s increase of mass.
The electron’s mass into the neutron n=p+e  is:
m = mo.γ      (46)
where γ we already obtained in (30):   γ = 2.5264
m = mo.γ = 0.511 x 2.5264 =  1.291 MeV/c²      (47)
Considering the electron’s increase of mass, the proton and the electron perform the total mass:
mp + me = 938.3 MeV/c² + 1.291 MeV/c² = 939.591 MeV/c² ~ 939.6 MeV/c²     (48)
Since mp + me ~ 939.6 MeV/c² , and the neutron’s mass is mN = 939.6 MeV/c², we realize that neutron’s binding energy is approximately zero, and this explains why it suffers decay.  However, with more accurate experiments, perhaps it is possible to discover the correct binding energy of the neutron.  So, by more accurate experiments, we can get the correct value of hG obtained in Eq. (9).
The first reaction of a physicist against the proposals of the present paper probably would be to claim the following: “It is hard for me to believe those difficulties raised in this manuscript will have escaped the scrutiny of all those prominent particle theorists. For instance, the author proposes a new Planck constant for the uncertainty principle in the femtometer scale.  Had this been true, the string theorists should have encountered the difficulty long time ago and even have proposed their own third different Planck constant.”
We must analyze such an argument from five viewpoints, as follows:
  1. First viewpoint: Up to know the theoretists have neglected the Borghi’s experiment, and this is just the reason why they never tried such a new theoretical alternative. Indeed, the proposal of a new Planck’s constant, proposed herein, is required by the results of two new experiments, made by Conte-Pieralice and Borghi. Even if the present new proposal is not a definitive solution, nevertheless any other different solution must be proposed by considering the results of Conte-Pieralice-Borghi experiments.  By neglecting their experiments is impossible to find a satisfactory solution.
    Moreover, it is well to note that the proposal of a new Planck’s constant is not able to solve the theoretical problems itself.  That’s why such an idea has never been proposed by the string theorists, since such new proposal actually must be proposed together with other new proposals, like the spin-fusion hypothesis, the helical trajectory, its zoom-property[8], etc.  The new Planck’s constant is not proposed here alone, actually it belongs to a collection of new proposals that performs new principles (which are missing in Quantum Mechanics).
  2. Second viewpoint: The recent new experiment made by Taleyarkhan, published by Science, has been explained by the scientific community as follows: “Theoretical explanations for the observation of neutrons in line with conventional theory do exist. Sonoluminescence is an observed and understood phenomenon. It is generally considered to be theoretically possible to generate fusion temperatures in imploding bubbles using sound. As for tunnelling through the Coulomb barrier at low temperatures, so as to achieve fusion at low temperatures, this could have been possible in principle, but experts who did the calculation say that, unfortunately, the rate will be far too slow to be observable, let alone be of any practical importance“. Nevertheless, Suslick and Didenko have repeated the Taleyarkhan experiment, and they have shown that the greatest portion of the sonoluminescence energy is wasted in chemical reactions. Therefore it is not possible to suppose that there are hot nuclear reactions in Taleyarkhan experiment. And since he obtained emission of neutrons (and therefore the existence of nuclear reactions is out of any doubt), we realize that these nuclear reactions cannot be explained by the old concepts of Quantum Mechanics. We must explain Taleyarkhan experiment from the hypothesis of non-Coulombic interactions, detected by Borghi’s experiment.
  3. Third viewpoint: In the present paper a new gravitational Planck’s constant has been proposed, taking in consideration the Borghi’s experiment.  A paper published in the journal Nature in August-2002, by Paul Davies corroborates such a hypothesis, in which he says that a new astronomical observation can lead to the conclusion that the Theory of Relativity may be wrong. The observation considered by Dr. Paul Davies is concerning the interaction between electrons and photons, and the results led him to consider two alternatives, as follows:
    a) FIRST HYPOTHESIS: The light velocity “c” is not constant
    b) SECOND HYPOTHESIS: The Planck’s constant can have some variation
    Well, it is possible that such a variation in the Planck’s constant, mentioned by Paul Davies, can be actually due to the interaction with the  new gravitational Planck’s constant proposed herein.
  4. Fourth viewpoint: It must be taken in consideration that the “spin-fusion” hypothesis is able to open new theoretical perspectives for the Particle Physics, through the establishment of a new Standard Model, as shown in the author’s paper “New Model of Neutron-First Part”,( 1 ) published by JNE, where it is shown that the lepton’s spin is not conserved in the beta-decay. Since the leptons are tied to the quarks through the spin-fusion, as proposed by the author, such a new proposal represents a new fundamental concept to be applied to Nuclear Theory and to Particle Physics.
  5. Fifth viewpoint: The theorists are trying since 1950 to find a satisfactory theory able to conciliate the several branches of Physics. Several genii as Einstein, Dirac, Heisenberg, and others, devoted their life to the attempt.  The problem has passed through the hand of several prominent physicists, among them several ones awarded the Nobel Prize and devoted their work to the question of the unification, as Salam, Gell-Mann, Weinberg , Glashow, t’Hooft, and others. All they have supposed that the rule of addition of spins, adopted in current Nuclear Physics, is the correct theoretical way. However, it is hard to believe that a satisfactory solution should have escaped the scrutiny of all those prominent theoretists, if such a solution should be possible by the way that they are trying (up to now there is not a satisfactory Standard Model in Particle Physics, which is incompatible with the Nuclear Physics, a theory itself not able to explain several questions). If a satisfactory solution via the Yukawa model should be possible, of course that it would have to be found several years ago.
A new model can replace an old one only if the new one brings advantages. The Yukawa’s model has several disadvantages, but the author considers that the most serious is the fact that in Modern Physics the description of the phenomena must be made through the consideration of two incompatible models: some phenomena must be described by the quark model of neutron, and others must be described by Yukawa’s model, but they are incompatible. It makes no sense to believe that in the Nature two incompatible models must describe the phenomena.  The author’s model (u,d,u-e) is able to describe all the phenomena and properties of the neutron, and perhaps this is the greatest advantage of the model.
Finally, we have to consider that, when a new experiment has a result that does not fit the current prevailing concepts of an old theory, the scientific criteria prescribes that the theoretists must try to find a new theoretical solution able to explain the result obtained by the new experiment, through the proposal of new concepts. This is just what the author of the model (u,d,u-e) is trying to do.  Nevertheless, nowadays the theoretists are trying to keep the old prevailing concepts of Quantum Mechanics by rejecting the Borghi’s experiment, and such a rejection does not fit the scientific criteria.
  1. W. Guglinski, “New Model of Neutron-First Part,”  J. New Energy, vol 4, no 4, 2000.
  2. C. Borghi, C. Giori, A.A. Dall’Ollio, “Experimental Evidence of Emission of Neutrons from Cold Hydrogen Plasma,” American Institute of Physics (Phys. At. Nucl.), vol 56, no 7, 1993.
  3. E. Conte, M. Pieralice, “An Experiment Indicates the Nuclear Fusion of the Proton and Electron into a Neutron,” Infinite Energy, vol 4, no 23-1999, p 67.
  4. R.P. Taleyarkhan, C.D. West, J.S. Cho, R.T. Lahey, Jr., R.I. Nigmatulin, and R.C. Block, “Evidence for Nuclear Emissions During Acoustic Cavitation,” Science, vol 295, pp 1868-1873 (March 8, 2002) (in Research Articles).
  5. Y.T. Didenko, K. S. Suslick, “The energy efficiency of formation of photons, radicals and ions during single-bubble cavitation,” Nature, vol 418, 394 – 397 (25 Jul 2002) Letters to Nature.
  6. P.C.W. Davies, Tamara M. Davis, Charles H. Lineweaver, “Cosmology: Black holes constrain varying constants,” Nature, vol 418, pp 602 – 603 (08 Aug 2002) Brief Communication.
  7. W. Guglinski, “Stern-Gerlach Experiment and the Helical Trajectory” J. New Energy, vol 7, no 2.
  8. W. Guglinski, “Fundamental Requirements for the Proposal of a New Hydrogen Atom,” J. New Energy, vol 7, no 2, 2004.

759 comments to Anomalous mass of the neutron

  • Andrea Rossi

    Dear Hank Mills:
    We are making preliminary tests, as you correctly write. The test of the Customer will be made on the 28th,
    Warm Regards,

  • Andrea Rossi

    Dear H. Visscher:
    The test will be on the 28th. But we are making our preliminar tests.
    Warm Regards,

  • Andrea Rossi

    Dear Mark Saker:
    Not yet.
    Warm Regards,

  • Mark Saker

    Dear Andrea Rossi,

    Has there been any interest from the United Kingdom, specifically from those in a position to make a difference, government perhaps?

    Many Thanks


  • H. Visscher

    Hello Andrea,

    Sorry to bug you again but something is not quite clear. You mentioned that you were testing the 1MW right now. Is this a different test than the one to preformed Friday or has the Friday test moved to today? Also you mentioned that you will transmit data of the test every hour. Is that from todays test or Friday’s test.


    H. Visscher

  • Hello Andrea,

    Could you please clarify if this is just preliminary testing, or if today’s testing is actually going to replace the test that was planned for the 28th of this month?


    Hank Mills

  • Andrea Rossi

    Dear Franco Morici:
    Good Sunday also to you, while I write this comment we are testing the 1 MW plant, which is working well so far. I am very sorry of the very restricted attendance, due to the particular kind of our Customer, but during the test on this blog we will transmit the main data of the test every hour, while at midnight we will publish the full report and a video. The direct transmission is not possible because the officers of the Customer want not to be videotaped.
    Warm Regards,

  • Franco Morici

    Good day Ing. A. Rossi,

    it seem to me no logical that the experiment of 28/10 can not be followed by many, many people that in this period, with passion and interest, have encouraged and supported You in this effort and should only hope in the availability of a Report, written by who do not know, and perhaps disclosed in the days after the experiment.

    We are in front of an event that can change completely the future of humanity and I think that it is not only a private affair. Probably is the greater invention of this century!

    Please allows all us to follow the 28/10 demo by means of journalist present to the event at least by means of tweet, keeping save contract statements with your US customer.

    My best wishes for full success.

    Franco Morici

  • Andrea Rossi

    Dear Alfonso Troisi:
    With outsourching systems we can afford any kind of request, since the sole part that we have to produce mandatorily in our factory is the reactor core. We already have an ousourching network ready to work.
    Warm Regards,

  • Alfonso Troisi

    Caro Sig. Rossi,
    I think your work will help heal the world, so I want to thank you and hope that your work inspires us to be better persons

    can you say how many 1mw plants are you prepare to build in 1 year?

    Thank you and your crew. I’m sure they also work hard along with you.

  • eernie1

    Dear Mr Tengzelius,
    As I have stated before,it may not be necessary to penetrate the nucleus with a particle to achieve nuclear decay.We know that the EM fields of the atomic electrons and the nuclear protons influence each other and maintain a one to one ratio in the stable atom.By distorting these fields with additional EM fields provided by the hydrogen ions,any number of branching decays may be activated resulting in the emission of particles and transformation of inter nuclear configurations such as neutrons to protons.

  • Lou Tengzelius

    Dear Eric Ashworth,
    We accept the concept of Nickel (Finely Divided) as a substance known to facilitate separation of H molecules into the more reactive H atoms. We accept the formation of Hydride ions composed of a proton and two electrons under circumstances where electron donation by an element is well known. Take the case of Sodium metal and hydrogen gas reacting in the presence of Nickel powder. Now, instead of considering a Neutron n= p + e, consider a Hydride ion h*= e + p + e. One can consider this structure in terms of the classical central nucleus with two electrons within the spherical shell volume exhibiting confinement to classical energy wells depicted by the Bohr theories. In order to tidy up according to the New Nuclear Physics of Wladimir Guglinski consider each electron having a helical notion of opposite direction as they orbit in a circular path. Assume that within the Rossi reaction chamber we add ferromagnetic Iron powder and Magnesium powder to the mix. The introduction of heat and Hydrogen under pressure initiates the in situ formation of hydride ions. A suitable induction coil activated by an RFG creating magnetic B-fields which when properly placed and operating below the 100 kHz range will provide an alternating magnetic force within the chamber. Knowing the unstable nature of Hydride ions and Neutrons a mathematical and geometric analysis predicts tunable configurations to include electrons each confined to a donut shaped area as noted by R. M. Santilli on opposite sides of the proton, each having both a circular path around the center axis and a helical trace of the exterior surface of the donut. The alternating field changes the direction of the helical trace which can produce instability and result in what one may call a virtual neutron Nv. Nv is propelled into the Nickel nucleus as the extra electron migrates into an orbital mode within the Nickel electron shell. The exothermic nature of this fusion has been well described.
    Warm Regards,
    Lou Tengzelius

  • eernie1

    Dear Mr Mockan,
    If you have read my previous blogs,you know that I am a proponent of nuclear decay mechanisms as the provider of the excess energy in LENR(cold fusion)reactions.I base my theories on the fact that all radioactive isotopes involve a decay mechanism with a subsequent emission of some sort of energy or particle.The fission of Uranium and other atoms in its family also involve nuclear decay as we all know with huge amounts of particles and energy.In the case of Uranium,the decay is triggered by neutrons entering the nucleus.Fusion of Hydrogen isotopes also involve a decay of internal particles with the emission of energy.In all cases of fusion or fission the amount and control of the subsequent output is very difficult to achieve and produce many dangerous results.The decay mechanisms produced by injecting Hydrogen isotopes into the vicinity of lattice bound nickel atoms are dependent in my theories,on the EM fields created with enough magnitude to trigger decay in the nickel nuclei to produce emission of Beta like particles which then heat up the nickel core by interacting with the lattice electrons.Most radioactive decays produce Alpha particles which then result in isotopic changes of atomic numbers of 4(two protons and two neutrons). However Rossi’s device produces copper which is a change of one proton.This implies Beta like emission,which is the second most prevalent radioisotope emission with a change of nuclear neutrons to protons.Most of these changes also produce a small amount of gamma radiation which I do not consider important to the production of excess heat in the Ecat.The future in the development of nuclear mechanisms,I propose lies in the discovery of other methods of achieving these nuclear decay systems in other materials.What this achieves is a method of more easily controlling the output of the devices both in quantity and substance.Nature has shown us how to achieve this by producing natural fusion,fission and radioisotopes.The biggest problem is that there are so many ways that decays can occur and so many nuclear particles that can be involved in the atomic systems available to us.What should we look at.I think the past work in LENR and Rossi’s work are a good starting point.

  • Bhagirath Joshi

    Dear Robert Mockan
    The paper brings to light possible low energy interactions. Mr. Rossi is planning to publish that paper in December.

  • Robert Mockan

    Dear Bhagirath Joshi,

    The earlier you can publish the better. The only hypothesis I’ve come across that might explain whole nucleon fusion (for example sodium plus oxygen to give potassium) reported in mid 20th century experiments is wave mechanical tunneling into nucleus. As a fundamental approach to also solving LENR reactions it would be topology of the “catalyst” that would enhance the tunneling by changing the nuclear field in space inside the atom. Just going by the description of your paper “excess neutron shell model of Nuclei” it would satisfy the criterion that space in the near nuclear field might be altered to effect the probability of nuclear processes. I’m way over my head with the mathematics to describe this, but if topological structure can account for “nuclear active catalysis” that would be consistent with catalysis theory in general, especially to also explain Kervan research on transmutation in living organisms. A “shotgun” approach to developing enhanced catalytic materials that function like what ever Rossi has discovered might be a feasible approach given the lack of mathematical rigor to model exactly what is happening.

  • Andrea Rossi

    Dear Casey:
    The familiarity with materials did help, even if it’s a totally different issue.
    Warm Regards,

  • Casey

    Dear Mr. Rossi,

    I was wondering how your earlier work such as Petroldragon influenced your current work with the e-cat and cold fusion. I am not familiar with the techniques that you used during your Petroldragon days, but some hydrocarbon recycling processes involve heat, pressure, hydrogen and nickel. Did you notice anomalous amounts of heat from that which prompted you to explore it further?



  • Bhagirath Joshi

    Dear Eric Ashworth,

    In december My Paper “excess neutron shell model of Nuclei ” will be published here. Everyone than realize that the nuclei and hence elements behave differently in solids. As there was a need for the branch of physics “Solid state physics” to explain electron behavior, in the solids and liquids. THere is a need for “Solid state nuclear Physics” to explain the behavior of free protons and neutrons of nuclei in the solids. As we already know, event hough there is a meantime for radio active disintegration, all nuclei do not go through radio active decay at the same instant in time. There are empirical formulas to obtain half life of a radio active elements, but there is no theoretical base to calculate that. Also there is anomaly that even though neutron is neutral, we do not have isotope of Hydrogen with 1000 neutrons, not even with 100 neutrons. This shows that the present theories of Nuclei fails to explain that.

  • Dear Mr Rossi, Your reply to Robert Moockan is, as you are probably aware, a part of the reason why you are not receiving the just accolades you deserve in the press. The E-Cat is not only able to make oil/gas, nuclear energy etc. obsolete but able to expose a principle in atomic physics that certain people consider must be kept secret. This I beleive could be a part of the problem with regards your E-Cat and its technology. If you can prove a theory in physics that helps explain a problem then many other problems will be able to be explained, its like a pack of cards falling down. Maybe, but I think the E-Cat does and thereby it could be termed a ‘technology before its time’. I eagerly awaite the production of your E-Cat and the explanation of its technology. Regards E.A. P.S. I am not academic but I do have contacts with regards timing.

  • Kevin

    Dear Andrea Rossi,

    Will your October 28th demonstration be the test to finally generate headline news of your amazing work?


  • Andrea Rossi

    Dear Robert Mockan:
    When we will disclose the theory and the operation of our reactors you will see that no one of the existing systems is close.
    Warm Regards,

  • Robert Mockan

    Dear Andrea Rossi,

    Experimental evidence for nucleon fusions (for example sodium with oxygen to give potassium), multiple particle fissions (for example to give titanium and copper from thorium), and fusion resulting in multiple positive charge increase ( for example many elements made from tungsten electrodes), continues to raise the question is the nuclear active catalyst (promoter, enhancer, accelerator?) enabling these alternate nuclear reactions fundamental to all the examples? You have stated you have a hypothesis about how the fuel in your reactor design works. I’m looking forward to your disclosure. Meanwhile, can you say if your hypothesis can be applied to explain the wide variety of nuclear processes that have been observed in research on these alternate nuclear reactions?

    Robert Mockan

  • Andrea Rossi

    Dear Lutoit:
    When I work I do not listen music. When I listen music I prefer Beethoven.
    Warm Regards,

  • Lutoit

    Scusa Rossi l’ OT, ma che musica ascolti mentre lavori?


  • Andrea Rossi

    Dear Davide C.
    Il sistema è disegnato per questo.
    Cordiali saluti,

  • Davide C.

    Salve Dott. Rossi,
    il 28 ottobre nell’impianto da 1 MW scorrerà un fluido salino termovettore oppure acqua demineralizzata? Nel secondo caso pensa che la pressione risultante metterà a dura prova il sistema?
    Grazie e buon lavoro

  • Robert Mockan

    The engineering of heat exchangers for reactors using this alternate nuclear process have not been emphasized, but probably should be. Other research into gas loaded fuels indicates the lattice structure needs to be above the curie temperature, thus is the limiting cold sink temperature for a Carnot cycle. The source temperature should be as high as possible for overall Carnot efficiency, but is limited by the lattice structure of the fuel composition deteriorating. The fuel composition needs to be distributed to avoid hot spots (and “cold” spots limiting the reaction kinetics) while passing thermal power to the heat exchanger(s). This is not a simple problem. Efficient promoters (or activators) in the fuel composition to increase reaction kinetics actually aggravates the heat exchange problem. A more stable reactor design might limit thermal power production to perhaps 1 watt per gram of fuel (or even less) in self sustaining mode. Overall heat to electricity conversion is only going to be a few percent maximum in any case.

  • Andrea Rossi

    Dear H. Visscher:
    I have not decided, yet.
    Warm Regards,

  • Andrea Rossi

    Dear John L Miller-George:
    The pressure in the primary was around 1 bar.
    Warm Regards,

  • Watching the High Def Production of Oct. 6th. This is the 11th and this thing flows like a movie made by Lucas or Spielberg.

    BRAVISSIMO A. Rossi. You are a “ROCK-A-THE-STAR”.

    Mark H.

  • John L Miller-George

    Dear Andrea Rossi,

    I’ve been reading all the accounts of the Oct 6th test in Bologna.

    I noticed several accounts of hearing boiling sounds in the primary loop, and that the E-Cat became too hot to touch, etc.

    I’ve read that the primary loop reached 110C, or 230F, definitely above the boiling point of water.

    Were any pressure measurements made of the primary loop during the Oct 6th test? If not, have you made any pressure measurements at another time or during another test that you could share with us?

    Thanks for any reply, sir.


  • H. Visscher

    Hello Andrea,

    Just a quick question; a while back you said that you will reveal your theory of what was happening inside the reactor with the demonstration of the E-Cat. Are you going to disclose your theory on October 28 too?


    H. Visscher

  • Andrea Rossi

    Dear Jouni Toumela:
    130 kW
    Warm Regards,

  • Jouni Tuomela

    Dear Andrea Rossi,
    just wondering, how much electricity will you be needing to get the 52 e-cats to the starting point of the reaction? Will they be started with the help of other e-cats?
    Or just by putting 130kW of electrical power in use?

  • Andrea Rossi

    Dear Renzo Scopelli:
    Since we guarantee 6:1 ratio, 1/6 of the time is made with the drive, as an average. Theoretically can be mere, during the test we made more self sustaining mode. We arrived up to 6 hours in our experiments, but the longer, the more dangerous. With 6:1 it’s totally safe.
    Warm Regards,

  • Renzo Scopelli

    Caro Ing.Rossi,

    le faccio innanzitutto i miei complimenti per i risultati fin qui faticosamente raggiunti.

    Mi preme farle una domanda, che credo possa essere LA DOMANDA che si pongono tutti:

    Quante ore lei stima sarebbe durata complessivamente a fase di autosostentamento se non fosse stata deliberatamente interrotta?

    Grazie per risposta.
    Renzo Scopelli

  • Dear wladimir, Again thanks for the info and web site address. Hope this is not too much information but I have tried to keep it brief. The neutron I think could appear in five different states. Two states when in an arrested state at the extremeties of a structure, proton/electron position with regards helical trajectory. One state when in the neutral position of a structure being midway within a helical trajectory and two states when commuting between structures. So what is a neutron?. It is, I beleive, the single potential required to form a structure and it is the single potential able to traverse between structures. Potential: It has the potential to be a positive a negative or a neutral. Allow me to explain.

    The neutron as a stand alone unit of identity is representative of a flat plate displaying a neutral or none existent potential. However, because there is no such thing as empty space the neutron is either within structure or commuting between structures and thereby is able to display a potential.

    When the neutron is within structure it forms the helical trajectories providing a proton position and an electron position. To connect inner to outer the neutron changes shape from that of a plate to that of a cone with its point innermost and its circumference outermost. This configuration is to conform to the disparity in area, outer surface area of a structure to inner area. The point of the cone when looked into represents a positive position with regards compression and its called proton. The outer circumference of the cone represents the outer circumference of the structure and its called electron. The helical trajectory, comprising of many cones, has a middle principle whereby the neutron becomes undecided and thereby takes on its true identity by vibrating/trembling as a plate to satisfy both dimensions. This position could be referred to as ‘the neutrons vanishing position’ and being responsible for photon emissions. Explanation.

    only a structure can create a structure and destroy a structure. The atom contains no structure. So in this instance, the helical trajectory is an integral part of a mass only. Thereby, the atom is not a creator force and neither is a planet. Both represent an internal mass being a product of an exterior structure comprised of four helical trajectories. A planet is formed from the outer edges of a helical projection from photons that have formed atoms that were drawn in on the helical projectory. As a planet is drawn within by the positive gravity value of the structure it begins to fuse down and fizz out. When it enters the vanishing point within the helical trajectory, the point where the neutrons display neutrality the planet being composed of atoms experiences a sudden expansion on its negative zones and a compression on its inner positive zones. This is a planet with mass. An atom with mass will likewise fuse down and expand out propelling a neutral from the vanishing point within the helical trajectory by a combined force of a vibration, centrifugal force and plus the ‘trigger mechansim’ this being the exterior fission dimension provided by the structure within which the atom resides, evolves and eventually decays by entering successive zones of increased potential. All planets eventually fuse down and fizz out.

    When a neutron is ejected from a mass or a structure because of either the immediate position of the mass within its structure of existence or because of gravitation attraction by an external mass/ structure with a more positive gravity the ejected neutron will vanish at a neutral position as it commutes to its new destination as a photon created by the reversal of the neutron cone to accomodate the structure into which the neutron will assimilate. Photons can be considered protons when viewing into the cone of a mobile neutron. Four helical trajectories comprised of photons make an atom. Electrons are neutron cones when viewed from the opposite direction.

    In conclusion: The neutron is a potential that is only able to be recognised because of the existence of structure. Without there being positions such as an inner and an outer with regards structure the neutron would not be able to display any potential and thereby would be unable to express itself. However, because it is able to express itself both math and geometry are able to serve a useful purpose by being able to establish the anomalous mass of the neutron. Again this is my simple understanding of a complex subject. Regards E.A.

  • Enrico Billi

    During the test Andrea Rossi and scientists weight everything, heat up, operate, cool down, re-weight, disassemble, analyze.
    Aluminum, silicon, or anthracite??? Why couldn’t be kriptonite?
    I attended the test, i saw the device open, nothing strange chemical elements have been found anywhere, just only warm water vapor, and i didn’t smell any kind of chemical smelling.
    Enrico Billi

  • Andrea Rossi

    Dear Stephen T.:
    Maybe that’s a possibility, in future. Now the baby-Cat still needs the father.
    Warm Regards,

  • Andrea Rossi

    Dear H. Hansson:
    Thank you for your info: I didn’t even know what the hell is thermite/Al, all I know is that my E-Cat weighted more after the test then it weighted before the test. You want not to know where the snake can put, along my suggestion, the thermite/aluminum.
    Warm Regards,

  • H. Hansson

    Dear Rossi,
    Good answer to Jan, may I add that using a fuel such as thermite/aluminum would probably burn a hole in your fat-cat and would be very hard to control, a engineering challenge.

    30g of thermite/aluminum are enough to create a hole in 2 cm of steel plate. Mixing water with thermite/aluminum or pouring water onto burning is very dangerous because it can cause a phreatomagmatic explosion, spraying flaks in all directions. IN SHORT: a fat-cat with that kind of fuel would be IED (improvised explosive device).

    (above info snapped from diffident sources)

  • Stephen T.

    Dear Andrea Rossi,
    Truly, you are self sustaining. You are wonderful! All else is jabberwacky. Pricing and economy of scale: Can you make a percentage and turn the work over to the big players? Can you parcel out the contracts so they compete against each other and you take your percentage? You have done enough. Can you find a way to engage the enormous abilities of the global engineering and production? We need them, they need your secrets. Make them compete! Humbly and uncertainly presented for your thoughts.
    Stephen Taylor

  • Andrea Rossi

    Dear Jan:
    Not considering all the other considerations that make ridiculous the supposition of the snake, just, please consider this:
    Warm Regards,

  • Jan

    Dear Andrea Rossi,

    I consider myself to be an unbiased person, meaning I won’t rule out any possibilities which can be reasonably argued for. I found an analysis, written by someone you might refer to as a ‘Snake’, claiming that:

    “NyTeknik indicated that the inner box that was never opened was at least 9.9 liters, possibly more. The detailed report says: “Inside the heat exchanger there supposedly was a layer of about 5 centimeters of shielding, and inside the shielding the reactor body, supposedly measuring 20x20x1 centimeters and containing three reactor chambers.” If there was no gap between the reactor and shielding, then the unopened box would be just the reactor plus shielding on all sides, which is 30x30x11 cm, which is 9.9 liters. It’s possible that there was a gap, and so the box was bigger. One poster on this blog estimated 18 liters from the photos. But let’s be conservative and assume the lower limit of only 9.9 liters.

    The estimated heat production here was 101.3 MJ – 33.6 MJ = 67.7 MJ. Energy density in MJ per Liter can be found on the vertical axis of this chart from Wikipedia:

    According to that chart, 67.7 MJ could be released by less than a liter of aluminum, silicon, or anthracite. Some chemicals, such as thermite (which is based on aluminum), consume and generate no gasses, and simply stay solid. They would be cheap and easy to put inside the black box.”

    I do not wish to doubt your honesty, I am simply pointing out that even though the 3.5h run in self-sustained mode was impressive as it is, an even longer run could have ruled out all chemical sources of energy.

    I hope I did not offend and wish you all the luck in the world!


  • Andrea Rossi

    Dear Simon Knight:
    The test started at 9,30 and finished at 24.
    Inside this timeframe we had to: weight everything, heat up, operate, cool down, re-weight, disassemble, analyze.
    The 3 anf half hours of self sustaining mode have been more then enough for unbiased People.
    Warm Regards,

  • Simon Knight

    Dear Andrea Rossi,

    According to Mats Lewan’s report from the October 6 demo he considered the test proper to start when the self-sustained regime was entered. This means that the effective testing time was limited to 3:15 hours whereas prior to the test you announced that 12 hours would be the minimum testing time.

    Because of this premature shut down the Fat-Cat was only allowed to cross the border to the improbable. It would have been so much more impressive if you had allowed the Fat-Cat to also cross the border of what the nasty skeptics consider to be the impossible.

    Please forgive me for asking, but why did you miss out on such an opportunity?

    Kind regards, Simon

  • Andrea Rossi

    Dear Vinnie Jones,
    Here is the list of the Scientists who attended the test of October 6th:
    Prof. Petterson Roland – Uppsala University
    Prof. Campari Enrico (Univ. Bologna)
    Prof. Bonetti Ennio (Univ. Bologna)
    Prof. Levi Giuseppe (Univ. Bologna)
    Prof. Clauzon Pierre (CNAM-CEA Paris)
    Dott. Bianchini David (Univ. Bologna)
    Ing. Swanson Paul D. (Space and Naval Warfare Systems- US Navy)
    Prof. Focardi Sergio (Univ. Bologna)
    Prof. Stremmenos Christos (Univ. Atene)
    Prof. Jobson Edward (Univ. Goteborg)
    Ing. Vandevalle Koen (Belgio)
    Dr Enrico Billi (Fisico, Ricercatore, CINA)
    This list does not include many other techicians who attended.
    Warm Regards,

  • Joseph Fine


    I apologize in case I asked my question about Lead shielding before. I recently read about the material ‘DEMRON’ (TM) and wondered if this material can be used instead of, or combined with, a smaller amount of Lead to conform with licensing requirements that prohibit or minimize the use of Lead. Of course, the question (preventing the melting of Lead) still remains. Tungsten would solve the problem, as you won’t melt Tungsten easily.

  • Andrea Rossi

    Dear Vinnie Jones,
    You are right, I will prepare it tomorrow (now is 1 a.m. in Italy and I am tired, working from 6 this morning-and is Sunday!).
    Warm Regards,

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