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

  • Robert Mockan

    Dear Mr. Dingman,

    The short answer to your question about why those three elements, and inter-metallic compounds of them, should be investigated is derived from data in “Hydrogen-Metal Systems Databook”, by Galaktionowa, c 1980, and in “Smithells Metals Reference Book (8th Edition)”, c 2004, that seem to indicate their material properties and ability to protonate hydrogen interstitially while maintaining lattice integrity at elevated temperatures make them potential candidates as LENR activators for hydrogen, and hydrogen isotope, fusion. Possibly because of mention that there is transmutation in the E-Cat of nickel, the general public may be unaware that “cold fusion” research began with the hypothesis that the lattice structure of a host metal (palladium), loaded with interstitial heavy hydrogen, might result in the nucleons coming close enough together to fuse, under conditions where the released energy would be dissipated by phonon (lattice structure) oscillations. A variety of hypothesis were made to describe how this could happen, while maintaining quantum number equivalence between reactants and products. The transmutation phenomena was discovered later, and appears to be a possible, but not necessary, consequence of the nuclear active conditions that prevail in LENR reactors. That is the short reply to your comment. A more cognizant reply is that the Heisenberg indeterminacy converges to certainty of nuclear processes within nuclear action event horizons when conjugate variable values of QM observables are constrained by lattice rigidity and high electron density lattice bounded constants. Again, the properties of cobalt, vanadium, and manganese material properties enable such constraints. And the long reply?
    I’m going to be posting that eventually on my own web site (presently working on other projects at the moment). Meanwhile if you are interested there is also my blogspot where the results of experiments with the materials mentioned will be discussed. I do not know if there is enough interest to attract viewers to make comments, but it is just another place made available to people who wish to network ideas about experiments to make LENR nuclear active compositions. Not necessarily for the E-Cat, because the goal is to eventually have a host lattice ( or ideally a “fuel” composition using hydrogen, that does NOT transmute the activator nucleus), operating at a much higher temperature than the E-Cat.

  • Congratulations on the inspiring work!

    We’re the two people who created the “I Believe in the Ecat” song. Would you be interested in being interviewed on our radio show at some point?

  • Tony McConnell

    Dear Andrea,

    Thank you for your incredible contribution to the future of the human race!

    It will be wonderful news when the Uiniversity of Bologna announces the start of their research, and will be great news for Italy in these difficult times.

  • Bhagirath Joshi

    The cobalt, vanadium, and manganese can be by product in a LENR , but you have to start with Ca, Cr or Tin.

  • Bob Dingman

    Dear Robert Mockan:

    You mentioned in a previous post that cobalt, vanadium, and manganese are elements that could be used in a LENR similar to the reaction Senor Rossi has developed with Nickel. I have suspected other LEN reactions could be discovered based upon the work of Rossi and Focardi. Can you elaborate a bit on your theorey?


    Bob Dingman

  • Jon Soderberg

    RE: Bedrich Pola

    The design of a household ecat system has many design possibilities. Id like to build one that use reactors in both series and parallel. Id use 2 12 Volt batteries that would be able to start a single reactor which would then power the startup of other reactor cores. 12 volt because you could jump start it from a car and the battery is already in production. In fact Id try and build my ecat with as much off the shelf components as possible.

  • Roberto Mucciarini

    Una curiosità:

    Ha notizie sul motivo dell’assordante silenzio dell’agenzia AP presente il 28 Novembre a Bologna ma che non mi risulta abbia scritto una sola riga in merito?

  • Argon

    Hi Mr Andera Rossi,
    First of all congratulation for successful sale and acceptance test by reliable customer.

    I have been patiently silent to wait for October tests and exited about results.

    As I proposed earlier we should get organized helping you in Open Source way (as successfully demonstrated with HUGE Linux effort). Instead of repeating same questions again and again about valve positions and refuelling procedures and prices we (impatient readers) should browse through your old answers and concentrate more on helping You to succeed.

    Here is my (first) take on ‘eCat community effort’:

    1) How about taking next step by designing dCat (Digital eCat). Consider installing inside one bigger chamber into one common ‘rail’ 1, 2, 4, 8, 16 kW eCat reactor cores. Then depending on energy need, you fire up, with heating resistors, most suitable combination of reactors (I.e 16+4=20kW and so on). So you can achieve any power level from 0-31 kW with one kW steps! Embedding all reactors inside one chamber in my quess could keep ‘passive’ reactors in preheated standby state, being ready to fast kick up for full operation? That would help on sudden variable energy needs, like heating a house in winter is ‘base load’ and then when whole family going into Sauna causes additional sudden energy need.

    2) If you succeed miniaturizing eCat, consider dCat in auto-mobiles, where you can leave 0.1/0.5 kW module keeping chamber standby temp ‘on’ over the night and car would be ready to kick up in the morning with no preheating delays. (0.1kW enough in warm countries, and 0.5kW in cold climate to warm interior also)

    3) I believe you know photovoltaics better than me, but I think you too quickly overlooked proposal made by

    Harry Veeder
    November 3rd, 2011 at 10:55 PM

    I think to was specifically about advancements on material sciences that was the obstacle during times you last time researched the subject.
    Improvements in the materials used in these latest devices — possible in part because researchers can modify the material structure at the nanoscale — are now making much more efficient systems, Kassakian says.

    Last but not least. Please keep up the marvelous work and persistence, but prioritize on most relevant tasks. Remember also to take breaks to play tennis or take walk in nice surroundings every now and then. It took 20 years to invent steering wheel after steering stick on automobiles, so lower pace is good enough in such important invention like eCat.

  • XY

    I am not an expert. I wonder about the impact of this invention to social, economy and security issues. Experts in those areas might give a clue. And maybe should – something like eCat will happen in 10 years anyway. Do not take me seriously, though you never know. Imagine stuff like: Money itself might loose sense since they represent energy, in a way; Knowledge will become the only value; But private oil-industry army could bombard the whole Bologna, if they wish; Imagine the anger of the politicians (not the people) of Iran and Russia when they loose all oil income; Many industries will go bankrupt, even progessive ones (solar); Many people will loose work – but they will not have to “work” much anymore; Water desalination and indoor planting might turn many difficult regions nicely habitable and independent; Politicians might loose great part of reasons why they exist. This world is more intense from the one when sceptics refused the fact that light can be emmited from a bulb on copper wires and very different from when sceptics refused to believe you can start fire by slamming two stones. I pray that you have a bulletproof plan with some strong-but-fair men of the planet or then I pray that God is with you (It seems like he is pretty much). Sorry for this unscientific comment. Just had to share.

  • eernie1

    Dear readers,
    Science has always depended on the imagination of thinkers to lead them to discoveries.Many of the ideas proposed in science fiction stories in the past have become realities.Flying,space exploration,television,computers and many more technical realities were once only imagined thoughts.I am amazed by the present lack of ideas about the possible mechanisms for Rossi’s device that can be generated by scientists using available information about the structure and behavior of atoms and their nuclei.There are so many possible paths to excess energy that present discussions should have many more proposed solutions.Here is an example of a possibility that came to my mind that has some scientific viability.
    The hydrogen in Rossi’s device is broken down to a negatively charged ion using his secret catalyst.This ion then energized by his heater is introduced into the nickel lattice imperfections.The enhanced EM field created by the ion then appears to be an additional electron to a nickel nucleus.This stimulates a nuclear decay mechanism that produces an additional proton(changes a nickel atom to a copper atom)a small amount of gamma radiation,an electron type of particle(muon)and a positron.This is a kaon-pion branching phenomonon that is produced and seen in particle experiments.The muons exit the nucleus and interact with the lattice electrons of the nickel crystals creating the excess heat of Rossi’s device.This idea of a stimulation of decay processes can be compared to the process which makes a laser function.The long half life of the kaon decay is drastically shortened by the stimulation by the EM field of the hydrogen ion.I have simplified this description of a possible process,but the details can be argued from known experimental evidence from past efforts.
    I can imagine other processes that may have a possibility of occuring.The complexities of atomic structure can contain a large variety of interactions.We have uncovered a few such as fusion and fission and Rossi’s device may be another step towards converting the energy in the nucleus to a usable technology.Let your imagination expand!!!

  • Robert Mockan

    Dear Mr. Rossi,

    To sum up my previous post, I suspect your company is going to be the target of industrial espionage, industrial sabotage, and legal entanglements. There are TRILLIONS of dollars in potential business at stake, and your competition is going to fight dirty. They will probably do whatever it takes to win the war, to insure they make the money.

  • Robert Mockan

    Dear Mr. Rossi,

    You have commented many times you know the E-Cat design and fuel will be reversed engineered “sooner or later”. That may be very soon, as all that is needed for fuel analysis is a sample in an inductive coupled plasma (ICP) mass spectrometer to give atomic percent of composition, and a scanning electron microscope (SEM) for determining morphological configuration of the fuel. The major atomic percentage will be the metallic or inter-metallic active fuel component (when used with hydrogen) and the minor atomic percentage will be the reaction promoter(s). One then tests duplicated sample composition in the morphological configurations in a calorimeter under hydrogen gas pressure cycling at different temperatures to determine power output. There are literally thousands of laboratories in the country (USA) that do material analysis on a regular basis, and use the kind of equipment needed. The first E-Cat sold will most likely be taken apart, if it has not already been done, to reverse engineer all components. The chances the fuel “secret” is already known outside your company is probably very high. Your marketing plan, developing a product brand name, and sales to insure a revenue stream, might need to be implemented much faster than the time frame you have stated for introducing this technology, if you plan to stay in business. Have you considered merging with a (friendly) large corporation that has the deep pockets that would help insure this technology is developed as quickly as possible, and can provide the legal resources you will undoubtedly need to maintain intellectual property rights when your competition start their attack? Overcoming skepticism and getting people to accept the idea of the E-Cat is one thing, but when the big corporations go after the money, they spend millions of dollars to take out their competition. Their legal departments work all day and every day dedicated to maximize the revenue stream for the companies they work for. That is what they get paid to do. That is why corporations have legal departments. Your competition does not need to develop anything, they simply need to take it from you. And they are experts at doing business that way. Merging with a (friendly) corporation that would develop the technology, might be better than going up against many corporations as competition.

  • Dear Mr. Rossi,
    I congratulate you most sincerely on your marvelous invention, hoping it can soon spread over the whole planet.
    I was wondering if you have ever considered donating even a sub-sized version of your 1MW plant to any ONG for humanitarian applications, e.g. warming facilities for refugees camps or so on. It would be a huge publicity for your machine and it would draw attention on low costs and high benefits that e-cat would bring to humanity.
    Best regards,
    Cristiano S.

  • Bedrich Pola

    Dear Dr. Rossi,

    If the price of energy production (i.e. Ni + H) is negligible, it would make sense to buy a rather more powerful unit (e.g. something around 150 kW instead of 100 kW) and use this increase to generate the electricity for the E-Cat internal purpose (e.g. by Stirling machine + alternator). With this approach the power from the public net would be required only for E-Cat start. Could you please let me know how long it takes to start the reactor and how long then the reactor runs in power generation mode? If the start time is negligible in comparison to the power generation time, the above arrangement should help to reduce the total cost of the E-Cat generated energy.

    Thanks and kind regards.

  • Angelo B

    Egr. Ing. Rossi, io seguo ogni articolo riguardante questa invenzione sin dal primo esperimento del gennaio 2011. Ho notato spesso che gli articoli sui giornali sono fatti da persone incompetenti, sbagliano unità di misura, confondono potenza con energia… e così via, molto spesso sono gli stessi che sollevano dubbi, gli stessi che ipotizzano “accumulatori a ioni di litio” o qualche altra diavoleria con calcoli da fruttivendolo.
    Volevo chiedere come mai non avete, sin dall’inizio, utilizzato lo scambiatore e operare così con acqua e non vapore? Questo tipo di prova ha alimentato molti dubbi sulla valutazione dell’energia generata. Altra domanda: mi sembra aver letto che la reazione avviene ad una temperatura di circa 450°C o K? Se lavora a 450°C vuol dire che il sistema sarà in grado di generare vapore ad alta temperatura e questo permette una trasformazione di energia? Lei in alcuni punti dice che il sistema non è molto stabile, cioè è un sistema che può avere un aumento rapido di temperatura e un possibile scoppio?
    Aspettando il domani…
    Cordialmente e i migliori auguri

  • Andrea Rossi

    Dear Gabriele B.:
    We have protected very well the reactor’s core. Anyway, before or later it will be copied. We are perfectly aware of this.
    Warm Regards,

  • Andrea Rossi

    Dear Klaatu:
    Yes, we did, but it doesn’t work.
    Thank you for your very kind attention,
    Warm Regards,

  • A Casali

    Dear Dr. Rossi,

    let me congratulate you for the succesfull sale of the first 1MW plant.

    I have now a few questions and hope you can reply:

    1)did you already deliver the 1MW plant to the customer?
    2)if not, when do you think you will deliver it?
    3)did the customer perform other private tests on the plant (before or after the 28th) or did they buy the pland only based on the 28th test?
    4)when will the R&D with university begin? is it a matter of weeks or months?
    5)when do you think you will be allowed to release the name of at least one of your customers?
    6)are the e-cats mounted on the second plant going to be in the final industrial form or are they going do be still hand-crafted?

    Thanks a lot for your kindiness,

    Warm regards,


  • klaatu

    Dear Dr. Rossi

    Your courage, vision, and methodical persistence have brought our world & our children a gift beyond compare. You have done the right thing, to bring the E-Cat to market as you have. Thank you, and may the enormous gratitude we all feel protect you & bless you.

    If only it would be possible to work for you to help in bringing this new future about, but alas your facilities on the East coast will be a great distance away.

    In considering some of what has been said of your catalytic energy reactor, perhaps a question might be addressed, should you have the time & wish:

    The intermetallic alloy Ni3Fe, when freshly annealed or ordered, has been shown to dissociate molecular hydrogen to atomic. This mechanism may provide atomic hydrogen directly at the nickel surface for the LENR. Particle sizes for planetary ball-mill produced Ni3Fe can be a few nanometers, and the product is relatively inexpensive. Did you investigate this material in your fuel trials?

    Kind Regards

  • Gabriele B.

    Dear Ing. Rossi,
    how do you think you can protect your industrial secret when one or more plants are in possession of your customers so they can examine them and even try and make reverse engineering? How do you prevent them to discover the secret catalyzer?

    Best regards.
    Gabriele B.

  • Andrea Rossi

    Dear Stefano:
    I hope to be ready as soon as possible with a household.
    Warm Regards,

  • Italo

    Dear Ing. Rossi, reading your last messages it seems that you are more relaxed, and having less stress, and I really hope it is so.
    Surely, having started the selling of your plants is a fundamental step for your life, in all senses.
    Anyway, I hope that your Customers will inform the world as soon as possible about their new plants. A lot of people needs yet to be positively assured.
    And this positive informations will help you in all senses.
    Kind regards, Italo

  • Stefano

    Signor Rossi complimenti per la sua invenzione,se ieri ero dubbioso adesso grazie a lei posso dire di sentirmi orgoglioso di essere italiano,perchè sono le persone come lei che migliorano il paese.
    Aspetto trepidante l’uscita dell’E-cat in versione domestica,nel frattempo sono felice di sapere che questa scoperta aiuterà l’umanità intera.
    Ancora complimenti,e le ricordo che tutti i G.A.S. d’Italia(gruppi di acquisto solidale)fanno il tifo per lei da sempre!

  • Observer

    Hi Anrdea Rossi!

    Just my 2 cents; when earning money with a product like yours, the bottleneck is not the amount of money you get from one unit; the bottleneck is how fast and how many you can sell. Since the markets are virtually endless, you would get a lot faster (and most likely bigger during your lifetime) income if you licence your technology to as many other parts as possible.

    Of course, this requires that you can protect yout IPR for example wath a patent.

    Makeing an E-Cat car ; how about hybrid, ie E-CAT generating electricity installed to an Opel Ampera or Chevrolet Volt? Would not need 20 years..

  • Andrea Rossi

    Dear John L Miller-George:
    Very interesting insight, thank you! We will calculate this.
    Warm Regards,

  • Andrea Rossi

    Dear RH:
    1- 1000 Euros/year for a 1 MW plant
    2- I expect 30 years, because foundamentally is a static apparatus. Just will need subsitution of some electric motor, some spare part.
    Warm Regards,

  • Andrea Rossi

    Dear Harry Veeder,
    I know very well the Seebeck Effect, also did a patent in the nineties. The issue is that the efficiency drops very low when you industrialize. I got 20% efficiency when I made myself the directional fusion of the semiconductors, but when I needed to produce industrial quantities the efficiency dropped below 5%.
    Of course if someone industrializes a thermocouple able to reach 20% of efficiency, that would be great. I worked for three years on that, but unsuccessfully at last.
    Warm Regards,

  • Andrea Rossi

    Dear James Arness:
    The water pumps push the consensed water through the circuit .
    Warm Regards,

  • James Arness

    Dear Andrea Rossi,

    In your answer to James Bowery you say that the feed water pumps bring the water from the dissipator. Since the water tanks in between are open, how is this possible?

    Kind regards, James Arness

  • Harry Veeder

    Dear Rossi, have you considered applying thermophotovoltaic systems to your heat producing eCats. These systems convert heat into light and then into electricty.

    Imagine using heat from an eCat to make light directly without the need to generate electricity! You could design special eCats for the specific purpose of illumination.

  • Paul

    MSNBC now has an article on the E-Cat and it’s pretty upbeat, looks like main stream media is picking this up:

  • RH

    Mr. Rossi

    Forgive for wasting your time with my idle questions, specially if they have already been answered.

    1) I understand the E-Cat has a charge that can run about 4000 hours, and at that time it will need servicing by your representative. Do you have an estimate for the cost (to the customer) of this servicing?

    2) Do you have any estimates on how long the 1Mw unit can operate (how many times it can be “recharged”)? will it last 10 years at fulltime operation? 20 years?

    I ask these question to try to make clear what the levelized energy cost is likely to be.

  • Paul

    MSNBC now has an article on the E-Cat, and it’s pretty upbeat and it looks like the e-cat is starting to catch on to the main stream media:

  • John L Miller-George

    Dear Andrea Rossi,

    Please take some time (or have a trusted subordinate do it) to compare any radiation ‘flux’ during the operation of an E-Cat to a microwave oven. There are hundreds of millions of microwave ovens out in ‘consumer land’.

    If the E-Cat generates a similar amount of temporary radiation then it should have similar consumer safety requirements.

    This argument should help with the idea being peddled that somehow E-Cats are dangerous in the home. Very few people with an IQ higher than their shoe size are terrified of microwave ovens.

    I don’t want to wait years for government certificates for E-Cats that are routinely handed out for microwave ovens.

    Warm Regards,


  • Andrea Rossi

    Dear Roberto:
    Thank you very much,
    Warm Regards,

  • Andrea Rossi

    Dear Tony Naebauer:
    1- yes, just see all the rests reports about this issue. We run thousands of hours controlling radiations. We have always a radiation control instrumente applied to every E-Cat under testing.
    2- We did not publish the analysis of spent fuel because they contain confidential data.
    3- It is proven that we do not use radioactive materials, we do not produce radioactive wastes, we do not have radioactive emissions outside the reactors during the operation.
    Warm Regards,

  • Toni Naebauer

    Dear Mr. Rossi,
    congratulation and appreciation for your extraordinary work and your sucessful demonstration of the ECAT. Many people are waiting eagerly for a clean and cheap energy source. But for many people everything assosiated with nuclear reactions is a missunderstood and dangerous thing.
    Any who wants to promote yout invention wants be sure, that he promotes a superior and clean energy generation system. And here I ask:
    1. Is it proven, that in your process there are no other radio active elements generated which need to have a costly disposal procedure afterwards.
    2. Is anywere published a analysis of the nuclides of the analysis of the consumed fuel.
    If not Do you plan to publish these data in the near future?
    3. Is it proven that there are no fission products in the end product after the reaction?

  • Robert Mockan

    There have been a few web sites started up the last couple months that appear to be legitimate, but I’ve noticed any body posting specific details that could result in successfully determining how the E-Cat is built, or how to make the E-Cat fuel, never pass the moderator. They get deleted.

    The web sites are:

    The first has an administrator with a book for sale.
    The second has an administrator who is unknown.

    It is possible that competitors of Rossi are attempting to control speculation about the fuel because they still intend to either profit from it after taking him down, or they are planning to just bury the whole technology again. They could do this by forming web sites that interested people would converge on, then try to limit their interaction.

    As I’ve said before if any attempt to stop Rossi is successful, then me and thousands of engineers like me, will not only publicize how to make the E-Cat, but how to make the fuel, to everybody. Call it the nuclear option. This time the technology will not be delayed, again.

    The ONLY web site about the E-Cat that should be trusted is this one.

  • Roberto

    Stimato signor Rossi
    Volevo solo inviare una email con i miei complimenti e parole di Apollo per il duro lavoro
    tanti saluti

  • Roberto

    estimado señor Rossi
    solo le queria mandar un email con mis saludos y palabras de apollo por su gran trabajo
    muchos saludos

  • Andrea Rossi

    Dear Roberto, Dear Jorge Jimenez:
    Thank you for your attention: for commercial issues please contact
    Warm Regards,

  • Roberto

    Dear Andrea Rossi
    I want send you a email
    A questo indirizzo email posso inviare ??

  • Andrea Rossi

    Dear James Bowery:
    The pump which pumped the water to the reactor.
    Warm Regasrds,

  • Mr. Rossi, thank you for providing the figure of 20mm water column as the steam pressure at the output thermocouple. I presume this to be the pressure above atmospheric.

    With such a low steam pressure, what pumped the condensed water through the return pipe to the tank?

  • Bob Dingman

    Dear Signor Rossi:

    Although I am sure you have seen or done a cost analysis on a home E-cat, I thought you might be interested to see an independent estimate on the subject.

    As someone who has used Electric power as the sole source for heat in the winter for the last 20 years, I would like to share the following data. Two story house, built in 1945, average insulation, approximate 2200 square foot volume, Southwest Michigan location, and electric baseboard heat. Target temperature of 68-74 deg Fahrenheit for roughly 1200 square foot of the house.

    From October 1 2008 to May 1 2009 the house consumed roughly 30,000 kiloWatt hours of electricity. The highest months are Dec, Jan, and Feb when 6000 kW/hour/month is used. This equates to 8 kW per hour. Since I have time of day metering, my rate is heavily discounted to an average rate of $ .055 per kW hour. With an E-cat that delivers 9 kWatts of heat, only 1500 Watts of electricity is required. With a 9 kW E-cat my typical $1600 winter heat bill would then be one sixth of $1600 or $267. A modest savings of $1333 per year.

    Most people pay $0.12 per kW hour of electricity or more. The heat bill is then $3600 for seven months. A 9 kWatt E-cat would reduce the electric bill by 5/6 to $600 for seven months. This scenario predicts a savings of $3000 per year.

    People in the Midwestern United States use propane, natural gas, oil, or wood as fuel for heat in the winter. Propane can easily range from $4000 to $6000 per year. Cord wood is among the least expensive yet highly time consuming, cumbersome, and fouls the air we breathe. A 9 kWatt E-cat would require 1500 Watts, or 12.5 Amps at 120 Volts. Basically the thermal output of a pellet or wood stove for the electrical cost of a space heater. Who wants one?

    Warm Regards,

    Bob Dingman

  • Andrea Rossi

    Dear Andrea Ran.
    Too soon for household applications.
    Warm Regards,

  • Andrea Ran.

    Carissimo ing. Rossi,
    crede sarebbe una bella idea proporre ad un sindaco di istallare un apparecchio su di un immobile comunale.
    Io credo sarebbe un’ottima testa d’ariete per rompere diffidenze ed insabbiamenti.

  • Andrea Rossi

    Dear Iggy Dalrymple:
    Interesting. I gotta study this too.
    Warm Regards,

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