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

  • Neil Ferguson

    Dear Mr. Rossi,
    It has been a very exciting month. Since you expect the number of customers to increase rapidly (no surprise!), would you consider a announcing preference for those customers who agree to make their names public and to publicly describe the performance of their new power plants once their systems are set up and running? There’s the old saying: the best advertising is word-of-mouth.

  • John James

    Hi Andrea Rossi,

    Salute a cent’anni !!!

    Thanks, Yours,
    John James

  • John James

    Congratulations! Salute a cent’ nano!!! Questions, how many E-cats will it take to power a large city and all large cities?
    With design used in recent tests, seems about 50,000,000 to 250+million E-cats needed to supply
    electricity to planet?

    Thanks, Yours,
    John James

  • Paul Calvo

    Dear Dr Rossi,

    American homes need about 800 to 1000 KWH/month, what do expect your e-cat home units to support and what would the early price be for this unit?

    Kind regards,

    Paul Calvo

  • AussieGuy

    Dear James Bowery,

    Our calc was clearly based on the cost of the fuel to the cost of the delivered heat energy. What it shows is that the E-Cat fuel cost is so low as to be very insignificant. As to LCOE, which we know very well, we have done that study and post the following:

    Assuming a 1 MW E-Cat heat source, a 40% efficient conversion of heat into Ac kWhs, 25 years of life and a combined plant cost of USD2,000,000, which from Rossi’s posting will soon be deliverable or better.

    Simple LCOE (no interest)
    USD2,000,000 E-Cat capital cost including heat to Ac kWh conversion at 40% efficiency /
    25 years plant life /
    365 days per year /
    24 hours per day /
    0.95 PLF /
    400 Ac kWhs per hour = USD0.024 / Ac kWh.

    With carbon taxes in Oz at AUD0.023 / delivered Ac kWh (coal plant), RECS at ~AUD0.03 / AC kWh and the E-Cat system not generating any CO2, the effective cost of CO2 free E-Cat generated Ac kWhs are, in comparison to a new coal plant effectively – AUD0.030 / Ac kWh.

    This means the Aussie wholesale market will buy the E-Cat systems Ac kWhs at ~AUD0.10 / Ac kWh (carbon taxes applied), cost AUD0.024 / Ac kWh and get paid a REC value of AUD0.03 / Ac kWh.

    So a sale income of AUD0.10 (wholesale pool price) plus AUD0.03 (REC) = AUD0.13 / Ac kWh generated against a LCOE cost of USD0.024 per Ac kWh generated.

    Profit gross is then approx AUD0.10 or over the year AUD3,328,800. Payback should then occur in 7 months.

    Fuel costs over the 25 years of operation are then EUR1 / MW / 6 months = EUR50 / USD70 fuel costs over 25 years. Not even enough to be pocket change. The H can be generated on site using various E-Cat energy sources, with an also insignificant cost. With the E-Cats almost infinite COP, this is not a game changer but an entirely new game.

    As for the Heat to Ac kW conversion system, we have some interesting ideas to make that happen.

    Aussie Guy

  • Silvano Mattioli

    Small correnction: around 20nm… the other seems to be background… little mistake foar a beginner 😀

  • Robert Tanhaus

    Dear Dr. Rossi,

    thank you so much (so so much) for your efforts!
    We know the e-cats can have higher outputs.

    1. Is the customer able to drive 1MW plant to higher output than 1MW?
    2. If not 1. Is that barred by software and/or by contract?

    3. You could consider a multi module e-cat where there is one smal module that does not take too much energy to start up which then can give heat to the bigger module/s to start up. With that it could be possible to have the e-cat start up by a capacitor and load it again with the generated electricity (when the e-cat can do that in the future). The system would then be better for poor countries like Afrika where there is not everywhere electricity.

    4. You said (I think in your patent too) that e-cat is also possible with copper fuel but the output is not as good as with nickel. Are you able to enrich also copper isotopes as you do it with nickel?

    Feel free not to answer my questions!!
    Robert Tanhaus

  • propagare

    Dear Mr. Andrea Rossi,

    Congratulations! Another Huge Success! I am very happy for you and drink a glass of Champagne on you! 🙂

    But now, please let me suggest the following to you, because the time is maybe right for this:

    * Let us, the people of the world, be your only shareholder beside yourself or your company/partners.
    You and your company will get all the money you eventually need – from US. Here at the 8th continent – the Internet. You will get all the help if you let us allow to help.

    Don’t let “Hedge Funds” or big investors, which surely have a right to exist, take over in ways you will never all foresee.

    They can surely offer you a lot of things more BUT NOT this possibility:

    * To may teach people of different nations worldwide in practice to take individual responsibility for their gloabl energy source _without_ the need of a global government.

    You keep the majority of your stocks to decide whatsoever.

    If I got you right this shouldn’t be in the way of your already expressed (imo correct) thoughts about the “free market”, earn honest money, own a proprietary technology which is hopefully fastest patented worldwide to be disclosed.

    (The founders of the USA would surely be proud too. I can’t imagine a better way to recommend “by the way” all the values and the “Constitution of Liberty – from the people for the people”, which is at stake in my humble opinion because of the upcoming financial crisis if the Dollar crashes.
    We couldn’t afford to lose this oldest constitution of the world with the most freedoms & prosperity for their citizen.
    As an entrepreneur you know how important these rights are.
    Thomas Hobbes Leviathan as well neosocialism grows imho to fast these days and John Locke is unfortunately half forgotten.)

    I am truly afraid about possibilities like companies which e.g. filter gold from the seas, cause the energy is now available.

    If the shareholders let them, they will bear the consequences with you.

    What do you think about shares for prosperity and individual responsibility in the hand of many (nations)?

    All the best from Germany & warm regards from a true fellow!


  • Andrea Rossi

    Dear Rick Gresham:
    We soon will reach 400-500 Celsius using diathermic oil as a primary fluid. We are studying this throughly with our Customer for his next needs.
    Warm Regards,

  • Andrea Rossi

    Dear Paolo:
    To be ready to sell household E-Cats we need from 1 to 2 years.
    Warm Regards,

  • Sebastian

    Dear Mr. Rossi,

    could you please explain why on Oct 28th there were additional E-Cats installed on the roof of the container?

    Thank you very much.

  • Paolo

    Salve, vorrei sapere se possibile la sua strategia rispetto all’arrivo del sistema nelle abitazioni, nel senso dei tempi che prevede. Mirando per il momento alla produzione a fini industriali, pensa di preparare il terreno in quel senso ? Ha dei vincoli; ma la divulgazione vera e propria quando avverrà ? La prego , mi rassicuri e con me tutti coloro che la stimano tanto…Noi ci prenderemo tutto il tempo necessario, ma dobbiamo sapere da che parte guardare.
    Questo paese potrebbe non andare più a rotoli anche grazie a lei..

    Saluti Paolo

  • Paolo

    Dear ing. Rossi,

    we’re based in Italy, and we’re very interesting to buy single modules (for a total power of 20 – 50 KW) rather than 1MW plant.
    Is it already possible?

    Best ragards

  • Robert Mockan

    >Silvano Mattioli,

    The H(1/p) resonance at 13.6 ev is common to a number of other hypothesis, and thus might be applicable. If we were to assume the resonance per BLP theory, then some nickel transmutation may result from CAF (Coulomb Annihilation Fusion), also hypothesized by Mills of BLP. Yet wrong models can sometimes explain a whole series of correct predictions. The models may work, but may not apply.
    For example, the Widom-Larsen hypothesis is based on a physical model that describes the synthesis of ultra low momentum neutrons, that can initiate nuclear reactions and can explain energy and transmutation data from many LENR and CMNR experiments. Although the model may actually be valid, there is no theoretical need for it to explain the energy output of the nickel hydrogen energy process.
    What happened in this field the past few years is that the relevant factors explaining nuclear reactions in open space were found to be in error when applied to the condensed matter environment. When the Gamow factor, as one example, was re-calculated taking into account the metallic bonding environment and periodicity of lattice structure in solid matter, the probability of the phenomena of “cold fusion”
    happening was found to be, not just possible, but inevitable. The time frame of nuclear events happening dropped from a probability of a few in billions of years, to a certainty in picoseconds. Thus no new physics needed to be invoked, no “miracles” need occur. Skeptics of the LENR and CMNR phenomena are so far removed from the existing knowledge base in the field of LENR and CMNR their opinions should have been ignored years ago. The existing knowledge base will be expanded upon, but even presently there are ample reasons to expect other materials to operate at higher temperatures than nickel. For example nano-particle cobalt, vanadium, and manganese should be investigated, given that existing hypothesis have predictive value in describing substitutes for nickel.

  • eernie1

    Dear Eric AShworth,
    If you want a good example of the misuse of mathematics one does not have to go further than the discussion of the Ecat by supporters and critics.The supporters using their postulates prove to themselves that the device works. The critics using their postulates prove to themselves that the device does not work.As Rossi states the proof lies in the taste of the pudding by the customer.

  • Rick Gresham


    Congratulations on your successes.

    I have two questions, one a clarification; You have stated the 1 MW plants will be priced at 2,000 euro/kw. But it isn’t obvious how that 1 MW is to be calculated. The test of the first 1 MW plant produced only 479 KW. So would it be priced at 2,000 x 1 MW or 2,000 x .479 MW? In other words, is the priced based on predicted or measured output?

    The second question relates to the temperature of the coolant in the primary cooling circuit. To be useful in many industrial processes, the temperature will need to increase beyond ~100 C. Can provide any information on what higher temperatures will be available and when?

    Thank you.

  • Aussie Guy, you have one big error and one big omission in your calculation:

    Error: Sterling efficiency is limited by the highest temperature input provided by the 1MW plant. Do the Carnot efficiency calculation. As I’ve already pointed out to Mr. Rossi, the AVE (atmospheric vortex engine) can replace high temperature input with low temperature output and achieve the kind of efficiencies you are talking about with the 1MW thermal plant.

    Omission: You have not included the amortization cost of $2/W installed in your delivered energy cost. This is called “levelizing” in the energy industry.

  • Silvano Mattioli

    Ultima INTERESSANTISSIMA coincidenza è che le dimensioni delle nano-particelle di Nickel (come dichiarato nel brevetto) e la frequenza fondamentale di una di queste reazioni è simile alla lunghezza d’onda di emissione delle nuove linee di spettro dell’idrogeno (orbitali supposti 1/p) che sono dell’ordine delle decine di nanometi (22 nm se non erro la nuova linea di spettro dell’idrogeno).

    Insomma nano-polveri potrebbero entrare in risonanza e quindi dare l’energia mancante al sistema.

    A questo punto mi fermo; buon lavoro e speriamo che presto una università possa certificare il lavoro da Lei Svolto con tanta passione.

    Non so se è reale la fusione fredda o altro, (il tempo è il miglior giudice in queste materie) vorrei comunque ringraziarla per avermi fatto appassionare e solo per questo le dico GRAZIE.

    Per il resto se i risconti saranno positivi le assicuro che il coro sarà molto ma molto più forte !

  • Ulderico

    Congratulations, Ing. Rossi.
    Lavolale…lavolale..è la strada giusta!

    Kind regards


  • Bhagirath Joshi

    Dear Wladimir Guglinski
    I like your thoughts very much. You will enjoy reading my paper “Excess Neutron Shell model of Nuclei” to be published here in December. You may find it very interesting and useful. The model correctly predicts upper limit to isotope building, and shows why there is no possibility of stable nuclei after Bismuth. It also shows that , “so far neglected force of nature in considering the binding energy of nuclei may be playing the major role “

  • Andrea Rossi

    Dear Aussie Guy:
    You can contact our commercial channel here:
    We started the commercial strategy limited, for now, to the 1 MW plants.
    Warm Regards,

  • Andrea Rossi

    Dear Iggy Dalrymple:
    Already did. The charges are already made in the USA. Of course the work is in progress.
    Warm Regards,

  • Andrea Rossi

    Dear Alfonso troisi,
    Same answer of lenr4you.
    Warm Regards,

  • Andrea Rossi

    Dear lenr4you:
    Nobody would invest anything in a non proprietary technology.
    Warm Regards,

  • Andrea Rossi

    Dear Greg:
    1- The 350 power generator (not 500) has continued to work to give power to the heat dissipators and to the water pumps, which, of course, had to work also during the self-sustained mode
    2- In order to perpare the 1 MW test we charged nothing before the start up, but the hydrogen, of which the consultant of the Customer has taken due record, weighting the hydrogen tank before and after the test.
    Warm Regards,

  • Greg

    Dear Dr. Rossi,

    congratulations to the successful test and sale of the 1MW plant! Many thousands of us has been waiting for this moment for almost a year now. I admire your persistence and courage with which you are steadily going forward on the path of commercialization.

    May I have two questions? Would you please put these two discussion points to rest which I have seen on several forums:

    1. The plant produced 479 kWh output per hour but (they say) a 500kW power generator has been working continuously during the self-sustained mode.
    2. In order to prepare the 1MW plant for the test and it had to be “charged” in advance. The energy input during this preparation period hasn’t been considered in the calculations.

    Thank you in advance.


  • lenr4you

    Hello Mr. Rossi
    thank you that you have brought the LENR technology so far.

    By spreading your invention over the internet, many people think about it and invent more new things.
    Can you imagine your LENR technology for licensing?
    I don’t think you can deliver millions of customers with hundrets of different LENR products.

    Sincerely your


  • Alfonso Troisi

    Sig Rossi e tutti,

    I follow the e-cat situation with passionate heart and mind. I want to thank Mr. Rossi and all who are involved.
    I would like to give my personal opinion on a matter: This invention does not save humanity. It will help with giving hope, but we and only we can save ourselves, changing our way of thinking and doing. If we are dirty inside ourselves, clean energy wont help us on that. We will find another way of making dirt outside.

    Thank you

  • Dear Eenie1, No I have not heard the phrase “liars figure and figures lie” but I have no doubt to its authenticity. Mathematics has never proved anything as far as I am concerned. Maybe it did try in the middle ages using religion as a support system. With regards lasers not being able to work I have no knowledge. Statistics are used everyday in an attempt to prove a point by politicians and salesmen. Whether they fail or succeed is dependent upon the skill of the statistician and his chosen line of enquiry. With regards a top electronics engineers opinion (1965) with regards costs and practicality of a high def television and computer what more could be impliede other than what was said by those Greek philosophers, according to yourself, ‘nothing human like could control the universe’. The point is taken but maybe they can be proven wrong.
    Regards E.A.

  • Iggy Dalrymple

    Dear Andrea Rossi,

    When do you intend to start production in the USA?

    Iggy Dalrymple

  • AussieGuy

    Dear Andrea Rossi,

    I do appreciate your quick reply.

    We may be able to help to make the heat to Ac kWh conversion work. Should we contact you via your sales channel? If not just email me via the email address I supplied.

    Your commercialization pathway here is 100% correct. We would have followed the same pathway. BTW we too are used to fighting game changing / disruptive technology wars. In the end it is satisfied customers that win wars. I suggest your war against those who will defame you, your E-Cat and just about everything else they can use to try to get stuck into you and stop / slow down the world wide roll out of your E-Cat technology, has just kicked into high gear. It is good to see you have grown a thick skin. Comes with the territory. Been there, done that. 😉

    Aussie Guy

  • Andrea Rossi

    Dear Fyodor:
    1- yes
    2- industrially
    3- yes, it is in our standard contract
    Warm Regards,

  • Clinton F

    Mr Rossi,

    Everyone wants to believe that this works. I hope you’re right in that time will tell.

    Good luck.

  • Andrea Rossi

    Dear Aussie Guy:
    Your numbers would be true should we be able to use efficiently a Sterling engine, but we still are not, yet. We are producing only heat production plants, so far.
    Warm Regards,

  • Andrea Rossi

    Dear Robert Mockan:
    Warm Regards,

  • Robert Mockan

    Dear Mr. Rossi,

    You have commented there “is a war” against you (your organization? your efforts to market the E-Cat?). Do you expect the successful sale of the megawatt demo module to
    stop the war, or is it going to get bigger?

  • AussieGuy

    Dear Andrea Rossi,

    Congratulations on the E-Cat. A total game changer. As an engineer and inventor I do know your road has not been an easy one. I admire that you never gave up.

    Based on your earlier supplied data, I have calculated the existing E-Cat, coupled to a 50% efficient Sterling Engine could produce Ac kWhs for around USD0.00065 / Ac kWh.

    USD1.40 – Fuel cost per 1 MW E-Cat. (1EU / MW as quoted earlier)
    4,320 – Hours of operational fuel life. (180 days at 24 hours as quoted earlier)
    50% – Heat to Ac kWh conversion efficiency. (Sterling Engine)
    2,160 Ac kWh generated – Per fuel operational life time per 1 MW E-Cat.
    USD0.00065 / Ac kWh – Fuel cost per delivered Ac kWh.

    Can you please confirm the numbers I have used are correct as if so, your E-Cat is more than a game changer in the world of electricity generation. It is a entirely new game and my children can now look forward to living on a beautiful planet with clean and abundant non polluting energy.

    Do you have any Australian customers? If not, I would be interested in becoming your first Aussie install site.

  • Eyal

    Dear Rossi.

    Thank you for your great work and persistence!

    I have a simple question.. Why don’t you just release to the world your entire research? As this information starts changing the world rapidly – everyone will know that you are the one behind it, and in the new and better world that we will transform into – you will be an unimaginable hero – responsible for probably the most practically important invention of our age. The one who saved humanity from itself! The entire control grid that currently hunts you will collapse. Now is the time. As of now, enough people know about e-cat and Rossi. Your gift to humanity will never be forgotten.

    I can’t explain to you how excited I am.

    Thank you.


  • Fyodor

    Mr. Rossi

    I hope that all is well with you. I had a few questions I was hoping you might answer if you have the time.

    1. From my understanding, the testing of the 1 megawatt plant involved reasonably active supervision from you during the test. Can the commercial products run successfully without supervision?

    2. Can you tell us more about how the new products are being assembled? I understand that you made the first unit by hand. How are the newer units being assembled?

    3. Have your new customers also requested a similar verification process for their 1MW reactors (i.e. all day tests)?

    Thank you for taking the time to answer

    Warmest Regards

  • Dear Mr Rossi

    The Associated Press science writer Peter Svensson and his photographer were present at the 1 MW test, but the story (if he submitted one) seems to have failed to appear on the ‘news wires’.

    Do you have any comment on this apparent suppression of the news of the success of the 1 MW test? Would you prefer it if the story did not ‘break’ worldwide for a while?

    With regards,

    Peter Roe

  • Andrea Rossi

    Dear Strat: we sell the 1 MW plants at 2000Euro/kW.
    Warm Regards,

  • Andrea Rossi

    Dear Camillo:
    We didn’t yet consider coupling with compressed air engines.
    Warm Regards,

  • Vorrei chiedere se è stato considerato il collegamento con il cogeneratore ad aria compressa di Guy Nègre e con la tecnologia Eolo (auto ad aria compressa).
    In particolare il motore a cogenerazione dovrebbe trarre grandi vantaggi da un riscaldamento intorno ai 100° A mio parere dato che per il momento non site pronti per la produzione di energia elettrica da turbina a vapore, quella tecnolocia consistente nel riscaldare l’ultima decompressione dell’aria compressa, dovrebbe sposarsi come il cacio sui maccheroni.

  • strat

    Dear mr. Rossi,
    your job is incredible!
    Can you tell us the price of the 1MW module.
    We are interested!

  • Andrea Rossi

    Dear GP:
    1- we are ready for 30-100 units per year
    2- no
    Warm regards,

  • G P

    Dear Mr. Rossi,

    I have followed for some time with passion and hope your endeavours, and hope you will be able to change things for the better.
    If you have time could you let me know:

    1- How will you be able to build between 30 to 100 plants in 2012 if it is going to take 3 months to deliver the second one.

    2- Is the second customer requiring the same level of secrecy as the first one? I am asking you this because all the skeptics/snakes will argue that you are just buying more time here (without any real buyer in sight), so if you don’t have a nondisclosure agreement as strict as the one with the first customer, you would be doing yourself and all people that are supporting you a great favour giving a little more detail.

    Thank you in advance for the consideration.

    G P

  • Andrea Rossi

    Dear T.R.
    3 months,
    Warm regards,

  • eernie1

    Dear Eric Ashworth,
    You have heard the phrase”liars figure and figures lie”.Here are some uses of mathematics to prove someones point.In the middle ages mathematicians used formulas to prove that the sun revolved around the earth to preserve religious beliefs.Lasers could not work because of thermodynamic noise considerations found in the laws.Statistics are used every day to prove a point by politicians and salesmen.A top electronic engineer told me that analogue to digital conversion(used today in high def television and computers)would be too costly to be practical(1965).I could go on but I think you get the point.The Greek philosophers implied that nothing human like could control the universe.

  • T.R.

    Dear mr. Rossi
    when the second customer will recive his plant?

  • Andrea Rossi

    Dear Luca Salvarani:
    1- yes
    2- more
    3- I want not our Customers assailed by the puppetts moved by their puppetteers. Let the plant go to work, then they will reveal themselves automatically. Probably you did not understand that there is a war against us.
    Warm Regards,

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