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

  • Jonatan

    You wrote about the ecat that -“before or later it will be copied. We are perfectly aware of this”. And you already sold some of this, so i assume that you close to getting patent on this?


  • Gillana Giancarlo

    Mi scusi la domanda ovvia, ma non esiste la possibilità che il cliente, qualunque esso sia, possa carpire il segreto del suo reattore, magari con uno spettrometro di massa o altre diavolerie, senza alcun brevetto che ne proteggano l’esclusiva?

  • insight

    Dear Silvano Mattioli,
    the single E-Cat device has to be warmed by electricity to start the reactions but of course with a series of E-Cats some of the output of one E-Cat can be used to warm the next. I hope that a multiple E-Cat system can be arranged to be fully self-sustaining too. Said that, I think there is no room for enthusiastic statements about “infinite” or “exponential” COP. There is the need of some input energy and the input energy has to be subtracted from the output, for each single device, that’s all. Still the E-Cat is an amazing device even with this necessity.

  • Wladimir Guglinski

    Hi, dear Mr. Insight

    Sorry the delay to replay. Yesterday I was busy all the day.
    Let me respond your questions now.

    1- I dont have all the answers.
    For instance, in my QRT there are magnetic massless particles m(+) and m(-), and they form magnetic fields [ induced by flux of electric massless particles e(+) and e(-) ], and into the magnetic fields the magnetic particles are glued by particles p(+) and p(-). Perhaps in the case of the gravitons they also interact in the particles p(+) and p(-) [or perhaps there are other particles that help the gravitons interactions).

    2- I suppose the gravitons g(+) into the flux n(o) move with the speed of light. Probably the gravitons G(+) are dragged in the surface of the flux n(o), but only in the surface. They do not penetrate within the flux.

    3- Let me tell you how I understand the meaning of the proposals of my Quantum Ring Theory.

    The New GravitoElectroMagnetic Theory – GEMT

    Suppose that Faraday should had not discovered the fundamental laws of the Electromagnetism. Well, in this case it would be impossible to develop a theory of electromagnetism by starting from the mathematical formalism.
    For the development of a theory, first of all there is need to discover the fundamental laws, and after that to apply the mathematics so that to get the equations.
    This is one of the problems of Quantum Mechanics. The quantum theorist have started its development from the mathematical treatment, and they inferred its fundamental laws from the interpretation of its mathematical formalism. That’s why some principles of QM are wrong, and must be replaced.

    A new GravitoElectroMagnetic Theory cannot be developed from the mathematics, as tried by Einstein and other theorists. In order to develop a new GEMT there is need to discover the structure of the aether, and the laws that rule the behavior of the particles that compose its structure.

    Faraday discovered the laws of Electromagnetism by making experiments. For the discovery of the laws of the Gravitoelectromagnetism, I followed the same procedure used by Faraday, with a difference: I did not make experiments, because it’s impossible to make experiments with the structure of the aether. Instead of experiments, I used the LOGIC, by trying to discover the structure of the aether taking in consideration my new nuclear model.

    Quantum Ring Theory actually is not a theory. QRT is actually a SYSTEM developed for a description of the microscopic world. It’s a system in the same sense as Newton developed a system for the description of the macroscopic world. Newton started from EXPERIMENTS and OBSERVATIONS. For instance, from the observation of the planets moving about the Sun, proposed by Galileo, Newton arrived to his theory of gravitation.
    I could not develop my system from the experiments, as Newton did. I had to use the LOGIC, together with the discoveries made in the field of Quantum Mechanics. So, in spite of I did not use direct experiments, however I had at my dispose several experiments made along the 500 years since from Galileo till Einstein’s Relativity and Quantum Mechanics. So, the discoveries made in the field of Quantum Mechanics and Relativity were fundamental for the development of my system.

    As I said, a new GravitoElectroMagnetic Theory must be developed by starting from the discovery of the structure of the aether and the laws that rule the particles that compose such structure.
    Obviously the theorists must have a point of departure, but it is also obvious that they cannot get such point of departure from experiments, and they cannot get it from the foundations of Quantum Mechanics.
    What Quatum Ring Theory proposes is just the point of departure.

    Therefore Quantum Ring Theory is not ready. It is only the begginning of a New System. I have proposed its new foundations, new models, and some laws that rule the working of the models.
    So, QRT proposed the foundations of a New GravitoElectroMagnetic Theory, and it must be developed, like Maxwell and other theorists developed the foundations of Electromagnetism discovered by Faraday.

    Quantum Mechanics was not developed by one person. Along decades several theorists (among them some geniuses as Einstein and Planck (in the begginning), Bohr, Heisenberg, Dirac, Schrodinger, de Broglie, Pauli, etc.,) contributed for its development.
    The same must be made with Quantum Ring Theory. There are several experimental parameters to be obtained in laboratory, there are equations to be found, and new principles to be discovered beyond those ones already proposed by me.

    Till now the physicists fed a big hope to confirm the current theories in the Large Hadron Collider. But the LHC is showing that prevailing theories are wrong. There is not any chance for Supersymetry anymore. And the chance to find the boson of Higgs seems to be exhausted.
    Also, Rossi’s eCat is suggesting that there is need a New Physics, with new foundations.

    So, after the physicists arrive to the definitive conclusion that LHC proved to be wrong the current theories, they will be obliged to face the inevitable: to look for New Foudations for Physics, by starting from the ZERO.

    And the question is: where will they find the new foundations for a theory to be developed from the Zero ?
    I hope they will start to consider the proposals of Quantum Ring Theory, soon or later.

    I know that probably the most theorists will try to get the new foundations by keeping their old failed method of discovery. But as said Einstein, there is no sense to try to eliminate the crisis by keeping the same method which has created the crisis.
    Quantum Ring Theory proposes a new method. Perhaps it can eliminate the crisis of the Theoretical Physics.

  • Andrea Rossi

    Dear Alex:
    I willleave this to the sales department.
    Warm Regards,

  • Andrea Rossi

    Dear Alfonso Trois,
    We reached still stability up to series of three, so far, but we are resolving the problems toincrease the serialization.
    Warm Regards,

  • Alfonso Troisi

    Hello Mr. Rossi,
    How far are you on succesfully connect the e-cats in series?

    Thank you for what you’ve done.

    Best regards,

    Alfonso Troisi

  • Alex

    Dear Doctor Rossi,

    I was wondering if or when you have thought about creating a television commercial for the sale for your 1 MW ECat? I think that would open many customer opportunities for you, give you a working backlog to help if you were interested to apply for financing a larger start-up operation, and show the world a vision for hope and the future.

  • Andrea Rossi

    Dear Maarten:
    1- no
    2- 1 MW plants will be put in series and par.
    Warm Regards,

  • Andrea Rossi

    Dear FC:
    1- To make cheap energy
    2- Cheap and clean energy
    Warm Regards,

  • FC

    Dear Mr. Rossi,
    I think in the current era you may be able to obtain anything you want with enough time and the impressive invention you have created.
    So I’m just curious, would you like to tell Why did you want to build it at start ? and What would you like to get from the product of your efforts ?
    Don’t give up !!
    Best Regards

  • lenr4you

    Dear Mr. Rossi,

    how is the sound of the Nih reaction?
    Maybe you can, as the conductor of the Ni-H reaction, compose the Symphony of energy! 😉

    Thank you
    Warm Regards


  • Silvano Mattioli

    Regarding to my post, Mr Rossi can explain better then me i think 😀

    The idea that is behind my post is that if an E-CAT produce a COP of 6 (nominal) in term of thermal energy, this energy can be used by user or otherwise can be used to warm up others reaction E-CAT chambre (in series or parallel as writed in patent).
    For this reason y can warm up with electrical energy only one e-cat !
    The electric COP may be infinite…
    Is not exponential as writed before but grow up linearly: each e-cat reactors give energy without input, then the electic COP grow as y want.

    Is not an electric source but simple Ni-H source…

  • eernie1

    To all interested experimenters,
    So many atoms,so many isotopes,so many isomers,so many internal and external ways to create new isomers,so many ways to stimulate the isomeric decay mechanisms(external EM fields,internal particle interactions)like the laser stimulations,so many spontaneous possibilities after isomeric creation.I dont see how mainline physicists who are more aware of these possibilities than anyone else can declare that Rossi’s device cannot work in LENR form.I only wish that I was 22 years old again with the energy I had then for doing research.

  • Maarten

    Dear Andrea,
    First off, congrats! Thanks to you we might well be on the verge of the next industrial revolution. As a sociologist it’s a delight to be able to follow the process from this early stage.
    A few questions though:
    1. Are buyers allowed to make a setup in which the steam is being transformed into electricity? Or does the contract stipulate otherwise.
    2. We know there are plans for small household devices, but are there also plans for units bigger than 1MW, or can the 1MW units easily be put in a serial setup?
    I pose these questions to inquire if it is feasible for smaller, cooperative energy-providers (and big-scale organisations) to invest in the e-cat technology in this stage. Thanks a lot in advance for your answers, and good luck!

  • Andrea Rossi

    Dear Enzo De Angelis:
    1- sì
    2- no, completamente inutile
    3- bel problema: il brevetto ancora non mi è stato riconosciuto fuori dall’Italia e la teoria rivelerebbe molto. Vediamo cosa succederà, per ora non ho idee precise.
    Cordiali saluti,

  • Andrea Rossi

    Dear Fyodor:
    1- yes
    2- enough to guarantee at least COP=6
    3- yes
    4- depends
    Warm Regards,

  • Fyodor

    Dr. Rossi

    I hope that all is well with you and that your work continues to go well.

    When the reactor is in regular commercial use does it go into self-sustaining mode? How long does it stay in self-sustaining mode? Is there software in the device that periodically adds current to keep the reaction running? How frequently does that happen?

    Thank you again for taking the time to answer my questions.

    Best Regards

  • I’ve created an index page listing mainstream coverage of the Rossi E-Cat. News:Andrea Rossi’s LENR E-Cat

    So far we have:
    – Wired
    – Popular Science
    – Discovery
    – CBS News
    – Fox
    – Washington Times
    – PhysOrg
    – Slashdot
    – Yahoo News
    – Daily Mail
    – MSNBC
    – LiveScience
    – EE Times
    – Falls Church News Press
    – NyTeknik
    – Physics Procedia
    – Agoravox
    – Energy Digital
    – Gizmodo
    – Renewable Energy World
    – TechNewsWorld
    – Times of India
    – Motherboard
    – Corante
    – Tecca
    – Mother Nature Network
    – Zerohedge
    – TechEye
    – Repubblica
    – CNews R&D

    And appropriate to its name, the first was: Next Big Future

    I should also mention that the people at Wikipedia decided to go ahead and keep the Energy Catalyzer page, after putting it up for delection last week.

  • Enzo de Angelis

    Caro Ing. Rossi,
    Dopo la dimostrazione del 28 ottobre, che dai dati da lei riportati io ritengo molto positiva, ha ricevuto altri ordini di acquisto per la centrale da 1 MW?
    Non crede che ora una dimostrazione pubblica, controllata da ingegneri indipendenti, a beneficio dei media internazionali possa essere vantaggiosa per la divulgazione dell’E-Cat e per la sua conoscenza nel mondo?
    Lei aveva detto che dopo la dimostrazione della centrale da 1 MW avrebbe pubblicato la teoria che secondo lei ne è alla base in questo mese, è disposto a farlo o teme ancora che la cosa possa essere di pregiudizio al segreto industriale?
    Infine mi piacerebbe molto conoscerla personalmente, magari partecipando a un eventuale dimostrazione a Bologna organizzata a beneficio dei partecipanti al blog di Daniele Passerini.
    Cordiali saluti.

  • insight

    Dear Silvano Mattioli,
    The E-Cat is a source of energy, not an amplifier, I think: the COP means the ratio of output against the input necessary to sustain the reactions. If you pump more energy the temperature will raise too much so the nikel will melt down, stopping the reaction.
    Furthermore I do not think the ratios would be the same because the reactions themselves cause a raise in the temperature so the input is necessary just for the particular device the E-Cat is now like. I mean that it is because of the material contruction of the internal core and the spatial disposition and arrangement of components, and nikel and hydrogen, which cause a particular flow of energy from some part to another.
    Am I wrong?

  • Sorry, I meant to post this earlier.

    There has been an amazingly fair and even positive article posted at Wired magazine.

    We put this bullet in our news at

    What to make of Andrea Rossi’s apparent cold fusion success – This friendly article debunks the accusations of fraud by pointing out Rossi’s fraud-inconsistent requirement that the 1 MW plant customers be “qualified,” large companies (who have lots of attorneys and legal jargon in contracts), who first validate the technology for themselves before the purchase is complete. (; November 6, 2011)

    Note: quite a bit of recent mainstream coverage of the E-Cat.

  • Andrea Rossi

    Dear Bedrich Pola:
    1- From 1 to 2 hours
    2- the E-Cat can go on 6 months without stopping.
    Warm Regards,

  • Bedrich Pola

    Dear Dr. Rossi,

    Could you please let me know how long (typically) it takes to start the E-Cat reactor and how long then the reactor runs in power generation mode?

    Thank you in advance for your reply and kind regards.

  • Andrea Rossi

    Dear Silvano Mattioli,
    You are right. We have to resolve serialproblems, yet.
    Warm Regards,

  • insight

    Dear Wladimir Guglinski ,
    I would like to point out something:
    – If “The flux n(o) is formed by gravitons g(+)” and “the flux n(o) produces a field of particles G(+) around it” and ” There is attraction between g(+) and G(+)”, so why the G(+) do not collapse onto the n(0) flux?
    – In electromagnetism a charged particle produce a static field, but if another particle interacts with that field then energy is exchanged; in certain cases we can say that a photon is exchanged. But no photon is exchanged or sent before the interaction, so a charged particle does not produce a flux of quanta around itself. Variations of the field can produce e.m. waves or photons. In gravitation someone says that the same is with gravitational waves, but when the gravity source is not moving no gravitons are exchanged. You wrote about a sort of spontaneous flux, but this would imply energy flux too.
    – With “the flux n(o) produces a field of particles G(+) around it” you seem to make a similitude with the way a magnetic field is generated by a flux of moving charged particles. Can you provide some equations describing the relationship between the two fields (like Lorentz’s force formula or Maxwell’s equations)?

  • Silvano Mattioli

    Al solito complimenti.
    Ho letto delle caratteristiche della sua invenzione.

    Ha condiserato che il suo reattore è fondamentalmente un AMPLIFICATORE di energia di fattore circa pari al COP ?

    Se utilizzasse in serie più e-cat (come mi pare indicato nel brevetto) potrebbe moltiplicare tra loro i COP attivando “una fila” di ecat a partire dal primo che raggiunge le temperature di reazione in un circuito chiuso di riscaldamento.

    In linea di principio:
    RESISTENZA –Riscalda–> 1° E-CAT –>Acqua Riscalda–> 2° E-CAT –>…–> n E-cat COP*…*COP=COP^n

    Insomma se invece di tirare fuori il vapore dal reattore utilizzasse questo calore per scaldare una cella adiacente di reazione senza utilizzare energia elettrica (per conduzione alla fine le celle arriverebbero alla stessa temperatura) il guadagno crescerebbe in modo esponenziale.

    E’ un po’ come in elettronica mettere in serie gli amplificatori, il guadagno sarebbe il prodotto dei singoli guadagni, al netto delle perdite.

    Il COP non sarebbe quindi il nominale 6 ma 6*6*…*6 essendo i reattori stessi a generare il calore necessario per attivare la reazione.

  • Wladimir Guglinski

    Dear Mr. Insight,
    I have inferred the properties of the gravitons from the analysis of what sort of behavior they must have within my new model of nucleus.

    In my new nuclear model there is a flux of gravitons, which I named n(o). Such flux is produced by a central nucleon 2He4 existing in all nuclei. The flux n(o) is formed by gravitons g(+) in the proton and g(-) in the electron.

    The attractive gravitons are g(+) and g(-), and the repulsive are G(+) and G(-).

    Properties of particles g(+) and G(+):
    1- The fields of particles G(+) are induced by the motion of the particles g(+) into the flux n(o).
    CONCLUSION: the flux n(o) produces a field of particles G(+) around it.

    2- The fields of particles g(+) are not induced by the motion of the particles g(+) into the flux n(o).
    CONCLUSION: the flux n(o) does not produce a field of particles g(+) around it.

    3- Two particles g(+) have no attraction

    4- There is attraction between g(+) and G(+)
    The same is aplied to the particles g(-) and G(-).

    Laws of interactions:
    a) There is not any interaction between two particles g(+)
    b) There is attraction between g(+) and G(+)
    c) There is repulsion between G(+) and G(+)

    Concerning particles g(+) and g(-):
    1- The particles g(+) and g(-) are neutral
    2- They work only through a partnership with the particles G(+) and G(-)
    3- They don’t have attraction or repulsion with the particles e(+) and e(-)
    4- They are not attracted by the flux n(o)
    5- However the relative motion between g(+) and e(+) & e(-) induces the flux n(o)

  • Andrea Rossi

    Dear Nick Pourmi:
    Thank you, is interesting, but too prototypical. We are working on the application of the classic Carnot cycle. By the way: the 70% of the energy consumed in the world is thermal.
    Warm Regards,

  • Nick Pourmi

    Dear Andrea Rossi,

    You are a brilliant engineer. You realize that your Energy Catalyzer, as a source of thermal power, although with a tremendous perspective, is bound to have a sluggish growth economically. Making it a decisive efficient electricity generator would be a very different story.

    May I bring into your attention the work of Romanian physicists Liviu Popa-Simil, former Los Alamos National Laboratory nuclear engineer and founder of private Research and Development Company LAVM and Claudiu Muntele, of Alabama A&M University, US. There is an article in New Scientist relevant for their work: ”Nanomaterial turns radiation directly into electricity”. Contact Information E-mail:
    Maybe you could integrate the nanomaterial into the shielding of your E-Cat and produce electricity.

    I am not a specialist, only a fervent apostle in awe of your ingenuity and dedication.

    From Romania,
    Nick Pourmi

  • Andrea Rossi

    Dear Giuseppe Cola:
    We will contact you when we will be ready for electric power production.
    Warm Regards,

  • Andrea Rossi

    Dear Klaatu:
    Thank you for your imortant question.
    As I said, we will not give money to organizations, because most of them spend big part of the money in issues different from the core destination. I already had a bad experience on this in past.
    Our system will be totally different: we will give the money directly to the families to pay the healthcare, or, better, we will pay directly the structures which will make the healthcare. We are selecting the subjects. As soon as we will have the proceeds available, we will start.
    Warm Regards,

  • Andrea Rossi

    Dear Jouni Toumela:
    Interesting , is worth to study it.
    Thank you and
    Warm Regards,

  • Jouni Tuomela

    Dear Mr. Andrea Rossi.
    The linked article I found interesting, hope You could also read it. Perhaps not applicable in your system?

    Br Jouni

  • klaatu

    Dear Dr. Rossi

    Thank you again for the considerate answer to the previous question. Why the second one did not survive moderation is unclear; if it was in any way inappropriate then deepest apologies are extended, it certainly wasn’t meant to be. Perhaps you could wish to explain why by email, then further questions will clearly be tempered by your advice.

    You have said that 50% of the proceeds from the sale of E-Cats will be donated in support of children’s otherwise unaffordable medical problems. The merit of such a noble thing will be passed along to each purchaser, and so greatly increases the value of bringing the E-Cat to the world.

    Have you already made this donation from the funds received by the sale of the 1MW plant, and will you consider making public the organization(s) that have been able to receive this charitable contribution in support of their efforts to help families in such need?

    Kind Regards

  • RH

    to Oscar Galli:

    let us refine your numbers.

    I think your capex figure for other infrastructure much too high. What are you having in mind here?

    regarding opex figures (assuming 1MW plant full time operation):
    € 1000 per year for servicing.
    fuel cost €1/Mwh , or €8760 / year
    assuming COP=6, regulating input power needed is 1460 MWh, the price of which depends on your location.
    Although the E-Cat can last 30 years, I think it is reasonable to expect the capital investment to be paid back faster then that. Maybe you should assume 10 years?

    If the E-Cat is to be used for electricity production, rather than heat production, then of course you need to add the capital cost of generator but the levelised energy cost will most of all be affected by the generator’s efficiency.


  • Giuseppe COLA

    Dear Mr.Andrea ROSSI,
    I’m a holder of a small factory in Veneto region.
    We have over ten years experience in energy production by gensets.
    I’m very impressioned of your amazing innovation .
    I’m embarassed to write you this message, but a deep sensation tells me that
    I must do it
    I’m at your disposal for any work outlook .
    Don’t esitate to contact me for any help or collaboration.

    with deference
    Giuseppe COLA

  • insight

    Dear Wladimir Guglinski,
    I wonder, and I would like to ask you, whether attractive and repulsive gravitons, responsible of a force of the magnitude of electromagnetism, attract each other like positive and negative charges and like magnets, or they are neutrals like photons, quanta of electromagnetic interaction. Thanks

  • Andrea Rossi

    Dear Oscar Galli:
    It’s far less than that.
    Warm Regards,

  • Andrea Rossi

    Dear Marvin:
    Warm Regards,

  • Wladimir Guglinski

    Dear Alan DeAngelis

    but the skeptics will change the name of the eCat.

    They will call it eMouse

    ha ha ha

  • Marvin

    Dear Mr. Rossi,

    Did you take appropriate precautions so that the knowledge of this technology is not lost to mankind, should some catastrophic event happen?

  • Humanity&Friends

    Thank You!!

  • Oscar Galli

    Dear Mr. Andrea Rossi
    I made some estimates of the final cost of energy based on the information provided in this blog.

    I have considered the following data:
    – CAPEX = € 3,000,000 (€ 2,000,000 for the purchase of E-cat + € 1,000,000 for other infrastructure)
    – OPEX per year = € 100,000 (€ 1.000 for maintenance and € 100.000-Cat + for other operating costs and maintenance)
    – Duration = 30 years plant
    – Interest rate 5%

    I would be finished at a cost of € 0.0310 per kWh produced, I ask politely if you could give me confirmation of the order of magnitude of this value.

    Many thanks,
    Ing. Oscar Galli

  • Wladimir Guglinski

    Oh my God !!!

    Dear Alan DeAngelis, I predicted what is the Rossi’s eCat secret in my paper ”How repulsive gravity contributes for cold fusion in Rossi-Focardi experiment”, submitted to his journal.

    I’m afraid some skeptics hackers have stolen the secret from my paper !!!!
    Now the skeptics will get the patent, and they will make money selling the eCat.

    Well, but at least the world will have cheap energy anyway…

  • Andrea Rossi

    Dear Alan De Angelis:
    I am devastated: really the Skeptics have discovered my secret of what really happens inside the E-Cat !!!
    I hope nobody will see the youtube video you gave the link of to our Readers.
    Warm Regards,

  • Alan DeAngelis

    Dear Ing. Rossi:

    Many of us have been hoping for a LENR breakthrough since 1989 and the megawatt reactor is more fantastic than anything I could have imagined. I’m glad I lived to see it.

    Nothing can stop you now.

    Alan DeAngelis

    P.S. BREAKING NEWS: The skeptics have discovered the hidden secret to the E-Cat.

  • Sa

    Ing. Rossi lei è un genio. Il catalizzatore potrebbe essere idrossido di nichel.
    L’ idrogeno immesso inizialmente serve solo per innescare la reazione : idrogeno ed idrossido di nichel producono rame e idrogeno, l’idrogeno si lega con il nichel e produce idrossido di nichel e così a catena….segue produzione di calore.

  • Wladimir Guglinski

    New astronomical observation suggests that laws of Physics are not the same in the whole Universe

    The article:
    Indications of a spatial variation of the fine structure constant

    Such conclusion corroborates the hypothesis that macrocospic gravity is resultant of the mixture of repulsive and attractive gravitons (they both interacting with forces of the magnitude of the electromagnetism).
    Such hypothesis is proposed in Quantum Ring Theory.

    Indeed, by supposing that the space (aether) is filled by a “soup” of repulsive and attractive gravitons, we may expect that such “soup” may have different concentrations (of repulsive and attractive gravitons) in some places of the Universe.

    So, the density of the aether can be different in many places of the Universe, so that some constants of Physics can change.

  • Andrea Rossi

    Dear Thomas Florek:
    Your song is delighting!
    Of course I would like to talk with you.
    Just send your phone number to
    Warm Regards,

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