Anomalous mass of the neutron

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

2a = λ/2     (4)

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

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

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

T = π²2h²/4mπa² ≥ 180 GeV      (6)

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

Uπ = T =  180 GeV     (7)

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

UN = 40 MeV     (8)

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

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

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

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

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

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

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

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

759 comments to Anomalous mass of the neutron

  • Andrea Rossi

    Dear Alex 2E:
    In this moment your question is very important, even if I have already answered to it many times in the past, but it is opportune to remember that:
    1- We made thousands of hours of measurements with specialists from the University of Bologna (Prof. Sergio Focardi, Dr David Bianchini, Prof. Pierluigi Rossi et al.) to detect the radiations out of the E-Cats and around: never have been found values significantly above the background
    2- microwaves ovens, cell phones, television sets have values not inferior than the E-Cats, and the laws regarding such appliances are the same valid for the E-Cats
    3- I spend not less than an average of 10 hours per day in proximity of operating E-Cats from the year 2008.
    Thank you very much for your important question.
    Warm Regards,

  • Alex 2E

    Dear Dr.Rossi,
    I am extremely interested in the use of the Home E-cat. Could you compare the radiation danger of his e-cat to a few commonly used home devices, such as Microwave Ovens, Cell Phones, TV’s, etc… It could put us potential customers at ease, and make safety concerns seem trivial. Yes we won’t be making calls from the e-cat, or watching it for 8 hours a day, but it almost seems like it would be safer than many things in our daily lives. I build, potentially dangerous linear accelerators, and I know that I am safe doing so. Please, if your team can help to compare these, in the form of milliroentgen per hour (mR/hr), and yearly, it may seal the deal on more views than my own.

    Alexander Tuohy
    Amherst, MA.

  • Andrea Rossi

    Dear Oliver,
    Just contact
    for commercial issues.
    Warm Regards

  • Oliver

    Hello Mr.Rossi,

    Fantastic! Please send me more Information later. I want to order some?

    Best regards from Germany,


  • Andrea Rossi

    Dear Pietro B.:
    The consume canm be half than that if the E-Cat has not to be used in continuous. At the end: you consume 1/6th of the energy you use. For example: if you use 6 thermal kWh, you consume 1 kWh of electric power.
    Anyway: we are preparing a plot where the ratios between energy produced and energy consumed are detailed, considering multiple cases of use, to avoid confusion.
    Warm Regards,

  • Pietro B.

    Egregio Ing. Rossi,
    ho letto di un presunto consumo elettrico medio dell’E-Cat stimato intorno agli 1.2Kwh, necessari a stabilizzarne/controllarne il funzionamento…
    Conferma o smentisce questa notizia ?

    Dear Mr. Rossi,
    I’ve read about a presumed average electric consumption of the E-Cat estimated in about 1.2Kwh, necessary to stabilize / control its operation …
    Confirm or deny this news?

  • Tim

    The website mentions that you are using a material to release and absorb hydrogen.

    I had been looking at materials that do this, and came across the website. They have information on a material that absorbs and releases hydrogen at pressures from 0.5 bar to 50 bar when heated from 25C to 175C. It is described at

    From your patent applications this appears to be the pressure range that the e-cat requires.

  • Andrea Rossi

    Dear Paolo:
    You can buy the E-Cat when you want. The price is not gonna change in real value. The 1 MW Customer is not yet working with the 1 MW plant, because we are still completing the control systems with National Instruments.
    Warm Regards,

  • Paolo

    Salve caro Rossi, sono uno dei primi che ha prenotato il tuo E-Cat e che crede nel successo della tua idea, anche se in realtà non so se in quel momento potrò realmente acquistarlo..Dipende dal prezzo. Potrò acquistarlo in seguito allo stesso prezzo ? Fornirete e pubblicherete via via tutte le informazioni per il collegamento domestico con tutti i dati e parametri necessari e la rete di assistenza ? Inoltre volevo sapere se i clienti dell’impianto da 1 Mw sono soddisfatti della resa e del conseguente risparmio che stanno ottenendo, perchè questo punto è di fondamentale importanza… Grazie

  • Andrea Rossi

    Dear Nicolas Chauvin:
    1- the COP we guarantee is 6
    2- yes
    3- we need an efficiency of 35%
    Warm Regards,

  • Nicolas Chauvin

    Dear Dr Rossi,

    First of all, congratulations for all your work.
    Recently you have mentioned that the technology used in the domestic e-cat will differ from the one used in the 1MW plant.
    Some articles mentioned that you achieved a COP around 40 during 5h30 during the 1MW test in self-sustained mode.
    However, I assume that this COP would drop when the plant is working in continuous mode.
    Other articles are mentioning you talking of a minimum COP of 6.

    If we assume a COP of 10 for a domestic e-cat in continuous mode, a 20kW e-cat would consume 2kW of electricity 24h/day.
    You also mentioned that an electric generator will come about 1 year after the e-cat launch.
    Then if we want to have the e-cat producing electricity, it would need to produce more electricity than it consumes.
    At least a 15% efficiency in the thermoelectric conversion would be needed for a 20kW e-cat to produce 1kW of electricity continuously (2kW for the e-cat + 1kW electrical output).

    This raises the following questions:
    1) Is the assumption of COP of 10 correct for domestic e-cat, or is it expected to be higher?
    2) Will the domestic e-cat be able to shutdown itself to reduce the electrical consumption when no thermal energy is required?
    3) Do you expect a thermoelectric conversion efficiency of 15% or higher for the electrical generator?

    Thanks a lot, I wish you all the success you deserve.


  • Andrea Rossi

    Dear Keith W:
    We are making the certification .
    Attention: I did not say that we will begin to deliver in August, I said we can be ready to make order confirmations in Autumn; we are right now making the engineerization of the factory ( the production will be completely robotized to squeeze down the price as much as possible) and, even if we will be very good, the first deliveries will start perhaps within the year, but I cannot promise it; I can say that the price will be lower than expected and that it will be a revolution. We are making a tremendous work with top level Partners.
    Warm Regards,

  • Keith W

    Dear Andrea Rossi

    On 5th Jan you mentioned that the E-Cat home will be available in August 2012 to everyone in the world through a network of licensees.
    Does this mean you are already in the process of getting EU certification for the e-cat home? I ask because if I receive the ecats home which I pre-ordered will I have problems with using it in the UK and possibly my summer home in France.

  • Andrea Rossi

    Dear P.G. Sharrow:
    Your pre-order has been accepted,
    Warm Regards,

  • @ Wladimir Guglinski; I examined your paper a second time as well as all your comments on this thread and am impressed by your insight on the subject of Neutron creation and behavior. It shows a great deal of thought over sometime. I know as I have done the same thing over the last 50 years.
    On your views about gravity and Aether I feel there is need for more thought. Perhaps there is need to simplify. My view is you have too many moving parts. Einstein got lost into a study of light and relativity as he tried to understand gravity and mass/inertia. Your point on the creation of a proton, a real singularity, from aether or dark matter/energy is much as my own. but your description of aether is too complex, Consider all that is known about the behavior of Electro-Motive-Force in an aether universe and you will understand aether and the force we call gravity. Good luck in your investigations. pg

  • P.G.Sharrow

    @Joseph Fine
    January 7th, 2012 at 10:50 PM
    I certainly hope to install at least 1 E-Cat on my farm before the start of the next era. 😎 pg

  • Joseph Fine

    The Mayan calendar says on Dec 21, 2012, the Earth enters a new 5,125 year cycle.

    The previous era was the Pre-ECat Age. The next era may be called the E-CAT Age.

    Some of us will get their home E-Cats on Dec 22nd or 23rd.

    And that’s just the start.


  • Andrea Rossi

    Dear Italo:
    I don’t even know what is the Maya prophecy, but I really hope to be able to start the sales of the domestic E-Cats before the end of the year. About the end of the world, I think is right the “prophecy” of the great astrophysic Margherita Hack: it will take 2 billions of years, the time enough to make the Earth enough close to the Sun to vaporize.
    Warm Regards,

  • Italo

    Gentile Ing. Rossi, io conosco esattamente la data che lei sceglierà per la commercializzazione dei suoi E-Cat!!
    Sarà il 21 Dicembre 2012, come da profezia Maya!!

    Sarà la fine del mondo come lo conosciamo.
    Ma sarà l’inizio di una nuova Era per l’Umanità.

    Mi scusi, è solo un gioco.
    Io non credo affatto alle profezie Maya.
    Ma credo assolutamente all’E-Cat!!
    Cordiali saluti,
    Italo R.

  • Andrea Rossi

    Dear Alex:
    Warm Regards,

  • Alex

    Dear Mr. Rossi,

    I understand you are in discussion with Home Depot for distribution of the Home E-Cat. I was wondering:

    1) Will Home Depot be requiring you to submit safety, and efficiency testing through the Underwriters Laboratory (UL)?
    2) Are you also In discussion with other companies for distribution such as Sears, or Loews, etc.. or will Home Depot have and exclusive deal with you for a given period of time?
    3) Will the purchaser be allowed to take a certification course through Home Depot to service their own Home E-Cat?

    Thank you,


    Alex Tuohy

  • Joseph Fine

    BJSchaeffer, Andrea,

    There are many issues about the neutron that have not yet been resolved. One is why the neutron has 0.782 Mev more “mass” than a combination of an electron and a proton.

    If you believe in quarks, that may help explain it. If you don’t believe in quarks, you may be in the Hadronic Mechanics or the Santilli school.

    Still, yours is an excellent question or observation on the excess “mass” (of the neutron). This surplus might somehow be partially available as energy when neutrons decay into electrons and protons. Fortunately, neutrons tend to stay together in (most) nuclei for a very long time.

    Some radioactive isotopes decay by electron capture which doesn’t make sense to me. When a proton in a nucleus becomes a neutron (???) by capturing an electron, this decay seems to absorb energy rather than release energy. This probably goes on at the same time with other processes which release more energy.

    I’m always fascinated when things do not yet have a complete explanation. And, if you believe in dark matter and dark energy, then 96% of the universe is beyond our current dogma. (“My karma ran over my dogma.” – Unknown source.)

    Joseph Fine

  • Andrea Rossi

    Dear bjshaeffer:
    I do not give any information about our theory; when a comment is published, regarding hypothesys of theories, the publication does not imply that we agree.
    Warm Regards,

  • There is a simple explanation explaining why the neutron is heavier than the proton is that the proton contains one positive charge +e and the neutron one +e and one -e. The mass of the electron being 0.5 MeV, the mass of the neutron should be 0.5 MeV larger than the proton. In fact it is 1.3 MeV heavier, apparently too much but this is not a precise calculation and other effects could be added. How do you explain the difference?

  • Dear Andrea,

    Thanks. I thought your own early view of the paper was all that was needed.

    I’m holding out great hopes for 2012 – on just so many levels. My personal mission is to get the E-cat to our own SA media. It’s been uphill but I’ve found a sympathetic editor. Who knows? We may yet get there.

    Kindest regards,

  • Andrea Rossi

    Dear Rosemary Ainslie:
    Your paper is under peer reviewing, as all the papers received from the Journal Of Nuclear Physics. As soon as the reviewer will pass it the Journal will publish it. There are many papers under peer reviewing, so maybe it will take time.
    Warm Regards,

  • Dear Andrea

    Wishing you all the very best of the season. I’ve been away. We’re still awaiting publication of our paper. Please advise us soonest.

    Kindest regards,

  • Andrea Rossi

    Dear Neil Ferguson:
    1- We registered” E-Cat, Energy Catalyzer, The New Fire”.
    2- yes: when it will be re-engineered by the Customwer and NI it will be unrecognizable inside. Outside the same, just a container, to make it easy to be transported.
    3- You are right: many particulars of the version tested on October 28th have been a makeshift.
    Warm Regards, and thank you for your attention: and Merry Christmas!

  • Neil Ferguson

    Dear Mr. Rossi,

    Merry Christmas. Perhaps for Christmas you could give us stocking-stuffers by way of a few tidbits of information. For example, your preferred nomenclature: you have mentioned it previously no doubt, but it wouldn’t hurt to repeat. Is “Energy Catalyzer” proper? E-Catalyser”? Is “ECAT” okay? Or “E-Cat”? “E-cat”? “Ecat”? Help us get the formal name and brand right.

    Next, is your first 1MW plant going to be rehoused (given NI’s advise, perhaps)? If you will pardon me for mentioning it, the arrangement of numerous extra E-Cat’s on top of the container housing on October 28 (as reported) seemed makeshift. Was it because the plant operation was intentionally suboptimal for that particular test? Will the production 1MW plants eventually fit into the 52 unit container housings already publicly revealed? Or will they each consist of two housings? Or perhaps you are planning a completely different housing arrangement?

    As always, any information you find suitable to share is eagerly awaited.
    Neil Ferguson

  • Andrea Rossi

    Dear Jan Srajer:
    National Instruments is the best Partner a manufacturer of plants of high technology can have. We are learning from them, and thanks to them the 1 MW plant we are finishing for our Customer is growing up a magnificence. Useless to say, all our plants will be supplied by the control systems of N.I. We are honoured to be supplied by them. The most important thing of their working way is that they do not just supply their technology, but they teach to you to grow up in its knowledge to make you able to peer-interfacing with them, like they do not just sell you good fish, but also teach you how to fish, so that fishing is empowered.
    Warm Regards,

  • jan srajer

    Mr. Rossi
    You are new Edison.Your success will be depended on the technological impovements of the E-cat claim.
    Question: Is the National Instruments good enterprise for you and how many jobs make for you.
    Merry Christmas

  • Andrea Rossi

    Dear Maxmanetta:
    Thank you, but to make this technology able to really give a substantial contribution to the global energy issue we have to work, work, work…
    Warm Regards and Best Wishes to you,

  • maxmanetta

    Gentilissimo Ing Rossi volevo augurarle i nostri migliori auguri di Buone Feste e che l’anno che verrà sia ricordato come l’inizio di una nuova era, dove i problemi energetici siano ormai un brutto ricordo grazie all’intuito di un una persona eccezionale (uno a caso: Lei) 🙂
    Spero di aver un giorno l’occasione di incontarla di persona per farle i miei personali complimenti per tutto quello che sta facendo.
    Ancora tantissimi auguri.
    Ventura Massimiliano

  • Andrea Rossi

    Dear Giorgio Adorni Francia:
    Best Wishes to you !
    Andrea Rossi

  • Giorgio Adorni Francia

    Carissimo Ing. Rossi,vorrei fare gli auguri a Lei ed ai suoi Collaboratori come ai nostri amici più cari …
    G. Adorni

  • Andrea Rossi

    Dear Antonello Lai:
    Merry Christmas!

  • Andrea Rossi

    Dear Giuseppe Cola,
    Thank you,I will remember,
    Warm Regards,

  • Dear Mr. Andrea ROSSI,

    as I wrote you at 6th November, we are a little factory in Veneto region ,
    we are always at your disposal for any collaboration.
    We trust in you and in your precious discovered, all our resources are
    at your disposal for any help.
    Don’t hesitate to contact us.
    Merry Xmas and happy new year

    with deference,
    Giuseppe COLA

  • Antonello LAI

    Caro Babbo Natale quest’anno ti chiedo 5 piccole cose:

    1. Che chi di dovere si svegli e supporti come in un mondo serio dovrebbe essere il Dott. Rossi e la SUA e sottolineo SUA e non di altri INVENZIONE
    2. Che nel 2012 gli sia dato il premio Nobel
    3. Che le t.d.c che stanno cercando di mettergli i bastono tra le ruote con pseudo cloni si vaporizzino
    4. Che io mi possa comprare un bel e-cat presto
    5. Porta i miei auguri al Dott. Rossi e digli che il popolino è con lui anche quando non si vede.

  • Alex or Bernie

    The petition is a great idea but will have the most effect if “We the People…” start it.

    Start the petition and let us know about it please.


  • Alex

    @Bernie Koppenhofer
    December 16th, 2011 at 12:07 AM

    “I think it is important that we all write our government representatives about the outrageous patent situation that is confronting Andrea Rossi.”

    Bernie, I suggest you open a petition in accordance to the FOIA at:!/petitions

    In addition to asking the state reps, we can ask Obama to repeal the unpatenable state cold fusion is in, through a petition of 30,000 votes. Tell news outlets once you have started your petition.

  • Andrea Rossi

    Dear Dr Joseph Fine:
    Thank you, and Merry Xmas and Great 2012 to you!

  • Joseph Fine

    Wladimir, Andrea,

    The following paper is available in ArXiv (September 2011).

    Electromagnetic Composites at the Compton Scale
    Authors: Frederick J. Mayer, John R. Reitz

    The paper was brought to my attention via Giuliano Bettini. It is about composites of charged particles called Tresinos or Quatrinos at the Compton size scale of 10^-12 m. On page 2 of the paper and subsequently, it discussed zitterbewegung, so this might be of interest here.

    Do you think such composites could exist? If so, how long could they remain stable and how could they be detected. With so much attention and expense directed to the search for the Higgs Boson, some effort should be made to detect the Tresino/Quatrino or similar particles.

    I wouldn’t call Tresinos (or Tres-Ions??) a form of dark matter, but if they interacted readily with normal matter – they should have been detected by now.

    Merry Christmas and a Happy and Healthy New Year to all.

    Joseph Fine

  • Bernie Koppenhofer

    I think it is important that we all write our government representatives about the outrageous patent situation that is confronting Andrea Rossi.

  • MissGeschick

    Dear Mr. Rossi,

    i am hopefully and all my trust is on your E-Cat!

    how could we know if the project is really unhealthily that it is? will you tell it befor you see the money?

    excuse the question, but to see all the gentlemen lying who are ordered to to rule us to get into better lifes, driving me so hopeless

    best wishes for you, your family and everything you love 🙂


  • Andrea Rossi

    Dear Aussie Guy:
    1- Not yet
    2- I will give this info as soon as possible.
    3- I will,
    Warmest Regards,

  • AussieGuy

    Dear Andrea Rossi,

    Can you please share in which US state the E-Cat modules and the 1 MW E-Cat plants are being built and tested?

    When do you expect potential clients can view a working 1 MW plant?

    Be sure to take a day off now and then to play tennis, play your drums, listen to Jazz and to enjoy your family and friends. In the end, family and friends are all we really have.

    With all my hopes for your success,
    Aussie Guy

  • Andrea Rossi

    Dear Fyodor:
    Warm Regards,

  • Fyodor

    Mr. Rossi

    Has production begun yet in your US facilities? Or are they still starting up?

    Thank you for taking time to answer my question

  • Peter Heckert

    Dear Rosemary Ainslie,

    if the battery cannot been replaced by a big electrolytic cap and a high efficiency switching supply, then I do indeed not see too much chances for me to make it reasonably work.

    I read through the theory but Im far from understanding it, but liked the ideas that where shining through.
    I myself are more interested in working hardware and its application and evaluation than in theories.

    The e-cat has indeed raised the bar a lot for all new energy devices. This is good, it separates the wheat from the chaff.

    Wish you a lot of success with this!


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