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.

755 comments to Anomalous mass of the neutron

  • Paolo

    Buongiorno, mi ha sempre risposto gentilmente a tutte le mie domande tecniche, ma ora le vorrei chiedere alcune cose che mi stanno a cuore.
    1) Non crede che ci sia il rischio concreto che l’acquirente prenda tempo e come spesso accade acquisti per poi celare tutto ?
    2) Non le sembra molto strano che il vostro acquirente non voglia divulgare l’evento con una prova aperta ai media e ai ricercatori , ciò non comprometterebbe nulla della segretezza del brevetto .
    3)Quando pensa che verrà superata questa riservatezza e entreremo finalmente in una nuova era ?
    Le persone come me che l’hanno sostenuta e tutti in generale, hanno bisogno di vedere gli sviluppi nella pratica quotidiana ; di rompere gli schemi e grazie a lei entrare e costruire un mondo nuovo.
    La prego sia prudente, ma non indugi.

    Un abbraccio e grazie

  • Andrea Rossi

    Dear Prof. Lino Daddi:
    Dr Bianchini, of the Bologna University, has only measured the radiations outside the reactors, for safety issues, and compared them to the background: there have not been significant differences between the background and the measured radiations outside the reactors during the operation.
    I do not give any information about the radiations inside the reactors, because such radiations are confidential.
    Warm Regards,

  • Andrea Rossi

    Dear James Bowery:
    The dissipator has been designed by me and the person who leaded the test, an engineer of NATO ( a Colonel) who has 30 years of experience in thermopower plants and thermodynamical systems.
    It has been designed to dissipate 4 times the necessary energy. All in the tested plant was redundant: we had 2 power generators, so that if one should fail the second would automatically had been put in operation, we had 4 pumps instead of 2, so that each pump would have a reserve, we had 2 control panels, so that ant electronic shortcoming would have been overcome by a reserve, etc: I simply could not fail, so we foresaw any possibility of failure and prepared a back up for any of it. The day before I wrote in this blog “we are ready”: when we say a thing you can bet on it.
    Warm Regards,

  • Andrea Rossi

    Dear Manik Sahai:
    We have started the manufacturing of 1 MW plants. Who wants to buy them whatever its Nation, can contact us at:
    Warm Regards,

  • Manik Sahai

    Dear Andrea Rossi,
    Many Congratulations on your Historic Breakthrough!
    How fast do you think will the world adopt this unprecedented energy technology? When can we know about the details of your commercial plans?
    Please keep India and other developing nations in your attention too – not only USA and Europe. The whole world needs to benefit from this source of energy.
    Manik Sahai.

  • Mr. Rossi, If you are liberty to disclose, what was the model of heat dissipator you used?

  • Lino Daddi

    Dear Andrea.
    Many congratulations to you and Sergio.
       In the report on the 28 October test there is scant information on the produced nuclear radiation; perhaps it is in the (not received) attachment 1 ?
       I want to know in what position the nuclear radiation was measured, and with what equipment. Should compare the background and the produced activity in specific energy bands; for example, background and activity in a gamma range around 511 keV. Activity may slow growth over time!
      Thank you.

  • Luca Salvarani

    Dear Andrea Rossi

    1- Can you confirm the second sale of the 1 mw plant to another customer?
    2- So far have you sold 2 plants, or even more?
    3- Could you reveal the identity of this second customer or it’s confidential (if so I can undestand)?

  • Wladimir Guglinski

    Dear Mr. Eernie 1,
    actually I did not note you wrote my name incorrectly.
    If you should had not told me, I would not have known it.

  • insight

    If not already known, yesterday 30/10/2011 italian 2nd public broadcaster RAI2 gave news about the 1MW E-Cat system demonstration inside TG2 and possible impact of the technology.

  • Dear Eenie1, Your reply to eng. Gulaginski, Quote “I would like to add my two cents worth of philosophy” and your reference to “you can prove anything with a few mathematical postulates”. I would like to say that you can prove nothing with a few mathematical postulates. I use Einstein as a prime example. Engineers are able to prove principles, mathematicians can only speculate. I agree with using the same logic regarding the rejecting of all existing gods being human because of the immediate limitations but you have to define god to put it into a context of understanding. On the ‘other side of the coin’, if there are humans who consider themselves ‘prime movers’ devoid of emotion, require no devotion or prayer, do not offer any reward or punishment and have no requirement for anyone to believe in them. What would their only reason for existence to be?. Could it be to control the mechanism that allows for the funtioning of the physical universe?. It’s not exactly what you call nature, It’s the control of and I am not saying it’s a bad thing. It’s more a necessary integral part of. Mathematics and probability I don’t think enter into it. It’s more to do with an aspect of intelligence that’s related to a time dimension/duration/distance travelled. With regards to the Hindu philosophy of reincarnation and the infinity of time, all events reoccur due to an individual learning process i.e. what big brother knows today due to his past mistakes, little brother will find out tomorrow by the same means i.e. a re-occurance of events due to a none understanding because of his affinity with his particular time and space. Fortunately, due to evolution humans move into the future where all is eventually revealed by an attribute of intelligence.
    Regards, E.A.

  • Wladimir Guglinski

    Bhagirath Joshi wrote in October 30th, 2011 at 9:14 AM:

    “Let me tell you, if Einstein had proposed the “Theory of Relativity and gravitational waves” today, He would have been called lunatic.”

    Dear Mr. Bhagirath,
    any man with revolutionary ideas is considered crackpot by the contemporary academicians, when his new ideas destroy what they believe.

    Galileo was a crackpot because he had revolutionary ideas. He showed be wrong the Aristotlean millennial ideas considered untouchable by the academicians at that age.

    Newton was not a crackpot because he did not propose a revolutionary theory capable to destroy the academic dogmas. He proposed a new theory in an age when several scientists were looking for a new theory, so that to continue the revolution started by Galileo.

    Einstein was a crackpot because he had revolutionary ideas. He showed be wrong the Newton’s theory.

    All the scientists who developed Quantum Mechanics in the begginning of the 20th Century, as Bohr, Dirac, Schrodinger, Heisenberg, etc., all they where crackpots, because they were developing a revolutionary theory. They distroyed the foundations of the Classical Physics. They were never accepted by their old contemporary academicians, like Lord Kelvin.

    I am a crackpot because I have revolutionary ideas. I show that some fundamental principles of Quantum Mechanics are wrong, and must be replaced by new ones.

    Andrea Rossi is a crackpot because he discovered and developed a revolutionary technology. This is a mortal sin against what the academicians believe, and they cannot forgive him for that (perhaps mainly because he is not a scientist).

    Science needs academicians.
    When the science is in crisis (because new experiences comes tearing down the existing theories), the academicians use to protect it from a lot of speculators.

    There are the good speculations, and the bad ones.
    To separate the chaff from the wheat (to separate the good theories from the bad ones) experiences are made.
    Soon or later the best theory (the best speculation) survives.

    This is the way followed by the advancement of science.
    The academicians do their work (sometimes they exaggerate, and betray the scientific method, by rejecting some experiments, like they are trying to do with Rossi’s eCat).

    And we, the crackpots, we do our work too: we make the science to advance… with the help of experiments (like the cold fusion made by Andrea Rossi) which corroborate our theories…

  • Pietro Maiello

    Thanks Rossi I got the answer.

    Best Regards


  • Andrea Rossi

    Dear Peter Roe:
    You are right, we are close to use other fluids for the production of electric power. This is a work in progress.
    Warm Regards,

  • Andrea Rossi

    Dear T.R.:
    I publish all questions I receive. Many comments are automatically dropped in spam by means of our robot, if there are viruses, links to advertising or other exclusion of addresses, names or links we have put in the black list.
    Warm Regards,

  • eernie1

    Dear Eng.Guglinski,
    I apologize for spelling your name incorrectly.Please feel free to verbally castigate me.

  • Andrea Rossi

    Dear Hampus:
    1- From 30 to 100 for the first year is what we are already ready to make, but we will get exponentially faster in time
    2- done
    3- this is a more complicated issue, because we need complex certifications
    Warm Regards,

  • Andrea Rossi

    Dear Pietro Maiello:
    The test was not public, but still important, we tried to make a balance.
    Warm regards,

  • Dear Andrea Rossi, congratulation for your sale

    A few days ago I was dreaming of a nuclear reactor like the pebbles bed, with spheres with an e-cat inside. Activated by a gas burner and dropped in a tank of water. If the geometry is done right, the e-cat inside would produce a pre-determined quantity of energy and power, enough to keep the temperature inside high enough to self-sustaining and would dump in the water the excess. The water tank would be heated initially to say 50°C and then the spheres would be activated and dropped inside. As more spheres would be added to the pool, the (gas?) heater heating the pool would be dimmed down. To cool down the pool, fish the single spheres one by one and throw them in a pool of cold water that will steal more heat than the e-cat can produce stopping the reaction.

    Want to go to higher temperature? Use glycole or molten salts and modify the geometry of the “pebble” accordingly.

    Am I dreaming or it could be a “reasonable” idea?

  • Andrea Rossi

    Dear H. Hansson:
    Probably we will have to deal with this problem, but now it is not a priority.
    Warm Regards,

  • eernie1

    Dear Eng.Gulaginski,
    As long as you are entering the field of metaphysics I would like to add my two cents worth of philosophy.I enjoyed your reference to Occam’s razor and agree that too many scientists make their theorys too complicated.I suppose they think people will dismiss their thoughts if they are too simple.Your remarks about the existence of a power that controls the universe were first formulated by the Greek philosophers Plato and Aristotle.They were unhappy about the fact that all the existing gods had human attributes.Because of this their logic told them the gods were not perfect since no human could be perfect and so they rejected them.We can I think reject any of the present religeous god using the same logic since they all have some type of human characteristic ranging from selfishness,wrath,vengence and rigidivity of thought. In their place the entity they proposed was called the prime mover that had the following characteristics.It was were devoid of any emotions,required no devotion or prayer,did not offer any rewards or punishment and had no requirement for anyone to believe in its existence.What was its only reason for existence was to control the mechanisms that allowed the functioning of the physical universe.This is what I think you call nature.If you are looking for something to describe its primary characteristic and you think it is a mathematical form,I would suggest the mathematics of chance and probility.QM theory is best described by the wave function which is just an application of probability considerations.An intriguing application of probabiility was used by Hindu philosophers to formulate their theory of reincarnation.their logic was as follows.In an infinity of time,any event that has occured has a probability of reoccuring equal to one which indicates that it will occur not only once but a multiplicity of times.Therefor you will be reincarnated many times until a higher force tekes you to a better place when you perform adequately on earth.If you apply the same logic to our existence we have to endure an infinity of life experiences which include any situations we can imagine. Isnt it fun to speculate on the basis of mathematical formulations?I have entertained myself many times with manipulating possible meanings of functional relationships.You can prove anything with a few mathematical postulates as has been done in the past.

  • H. Hansson

    Dear Mr. Rossi,

    In 1911 John D. Rockefeller lost his “Standard Oil” because it was the only supplier of oil (91% of all oil in the USA). He was forced to sell his company and it was breaked up into 34 independent companies.

    You will control 100%

    How do you plan to handle the antitrust laws in USA and EU??

  • Dear Wladimir Guglinski, ‘Is God Mathamematician or Engineer’?. Bravo! you have ‘hit the nail on the head’. In the sixteenth century, as you are no doubt aware, physicists believed that the cosmos followed mechanical law, which of course using common sense it does, and because it does both math and geometry exist and are able to be used as useful tools in the field of atomic physics. Einstein was a mathematician not an engineer and consequently, atomic physics today is treated as an abstract subject whereas the embodied material world is everywhere to be seen. EMFs are abstract yet detectable and behave as a solid with regards possessing an identity i.e. specific distribution and own specific location but besides this they are an integral part of all embodiments. It is quite literally the apparent magic of the subject that makes physics so fascinating a subject for those people with an enquiring open mind. It is the enquiring self reasoning ability that is a necessary key to the unraveling of any mystery. You Wladimir obviously not only do a lot of reading but possess an open enquiring mind. This, of course, can sometimes put you at odds with certain people when you deviate from what’s considered the norm in modern day physics. As you are probably aware by now, I know nothing about what is considered the norm (the norm being accepted theories) of which according to you, helical trajectories are not recognized. This theory of helical trajectories has actually been embodied into a mechanical design and although able to prove a principle in physics with regards the unified field of atomic physics was refused demonstration. Briefly, this complex subject involves understanding interactive curvature flow forces that form laminar flows or you could say the geometry required with regards the internal dynamics of the atom to produce a laser beam (crystals are nothing other than open atoms i.e. same structure less density). This specific mechanical design is also able to prove that it is the internal oscillations of the neutrals that form a static barrier layer/valency or maybe this is what you refer to as the coulomb barrier. I believe that there is potential energy and power energy. Potential energy possesses the potential to alternate polarity between power energy and constitutes the power of the mass with regards the make-up of the fission fusion dimensions of the mass. Einstein’s famous equation relates to power energy not potential energy unless you lump all the potentials together. I believe math and geometry is a good idea born of a mind that understands energy. You probably think what I am stating is too simple but energy follows a basic geometric pattern, nature works in a complex way because of the number of permutations involved in the make-up of the four dimensions of physical reality of both the seen and the unseen with regards helical trajectories. Regards E.A.

  • Pietro Maiello

    Dear Mr Rossi,

    1- Could I ask you if there is a precise intention from you, behind the decision of making your tests public, especially the last one for the 1MW plant (even though in the 1MW one people couldn’t actually see the test running)?

    2- Don’t you think that could have been far better dealing with just scientists and private investors since the very beginning and once demonstrated and validated thoroughly your invention making a big announcement to the mankind?

    I don’t understand why causing all this certainly unavoidable noise(snakes, possibilists, deniers, beleviers, non beliviers and so on).

    I hope you get my questions as genuine questions.

    Very Kind Regards


  • Hampus


    Great work.

    1. How many 1 megawatt plants can you sell in one year do you think?

    2. When is the next sell?

    3. When will the home plants be available for pursued?

    I know time is of the essence, that’s why my questions are so rash. Thanks for you answers the mean alot for me.

    Hampus Ericsson

  • T.R.

    Dear mr Rossi
    I’ve noticed that you don’t publish uncomfortable questiones. Why?

    warm regards and best wishes for you your technology

  • Oscar Galli

    Numerosi sono i commentatori che criticano l’invenzione di Andrea Rossi adducendo diverse motivazioni, spesso basate sul sospetto che alla base di tutto ci sarebbe un colossale inganno.

    Ma per quale motivo l’invenzione di Andrea Rossi non dovrebbe essere vera?

    Non è forse vero che nel 1847, Ascanio Sobrero, a seguito di un primo insuccesso ad opera di un chimico tedesco Christian Friedrich Schönbein, ripeté in Italia l’esperimento della sintesi di nitrocellulosa, con la variante dell’aggiunta della glicerina? Sobrero mise due gocce in una provetta e la riscaldò, ma la piccola esplosione che ne scaturì durante l’esperimento danneggiò il laboratorio, così decise di interrompere gli esperimenti e non volle più saperne; in seguito questa mistura prese il nome di nitroglicerina, altrimenti detta trinitrina. Sobrero, per esibire agli altri scienziati la consistenza della nitroglicerina, ne poneva una goccia su di un’incudine e la batteva con un martello, mostrando che questo, per lo scoppio, veniva lanciato via.

    Non è forse vero che nel 1867, Alfred Nobel scoprì che la nitroglicerina, miscelata con la farina fossile, avrebbe trasformato il liquido in una pasta che poteva essere plasmata in canne di dimensioni e forma idonea per l’inserimento nei fori di perforazione, brevettando poco dopo questo materiale sotto il nome di dinamite?

    Per quale motivo il dispositivo inventato da Andrea Rossi non potrebbe rappresentare una nuovo modo di produrre energia pulita, sconosciuto fino ad ora?

    Forse ci vorrà ancora un po’ di tempo (siamo solo agli inizi), ma credo che con l’invenzione di Andrea Rossi sia iniziata una nuova era, nella quale l’energia non verrà più prodotta inquinando l’ambiente.

    L’unica preoccupazione è che ora questa nuova invenzione possa restare appannaggio di pochi e che non venga resa disponibile a tutti; su questo punto le persone che hanno seguito questa vicenda sperano che ciò non accadrà, speranza che fino ad ora Andrea Rossi si è sempre impegnato ad onorare.

    Oscar Galli

  • Dear Andrea Rossi

    Firstly, I would like to thank you for continuing to respond to those who follow your progress, but who have no part to play in your success. Many people hope that the Associated Press reporter who attended the 1MW test will bring the news to a wider audience.

    Before the test, I speculated on a blog that the 1 MW unit would use oil or glycol cooling to feed an oil boiler so that high pressure steam could be generated for turbines, etc. I was obviously wrong on this occasion, but do you have any plans to use such a system?

    With best regards,

    Peter Roe

  • Andrea Rossi

    Dear Prof. Serdar S. Celebi:
    1- About 200 Wh/MW
    2- 1 Euro/MW
    Warm Regards,

  • Andrea Rossi

    Dear Christopher:
    Thank you!
    Warm regards,

  • Andrea Rossi

    Dear Bhagirath Joshi:
    Thank you!
    Warm Regards,

  • Kim Patterson

    Like the cavemen
    at the First Fire with flint and kindling.
    Every one has run to the corner of the cave
    scared, confused, and looking like a deer
    with the light in the eyes.
    Let Humanity know that a new fire is among us.
    Lets not be Children any more.
    But for the Children


  • Andrea Rossi

    Dear Franco Morici:
    for safety issues.
    Warm regards,

  • Andrea Rossi

    Dear LBG:
    Thank you!
    Warm Regards,

  • Andrea Rossi

    Dear Andy Wu:
    1- 50 cc is the volume per 4 kW
    2- maybe
    Warm Regards,

  • Andrea Rossi

    Dear Max:
    I did already.
    Warm Regards,

  • Andrea Rossi

    Dear Eric Ashworth:
    Thank you!

  • Andrea Rossi

    Dear Andy Wu:
    Thank you!

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    You mean the Puppett Snake? You bet, he will never stop shooting at me, he has been paid to do it from the well known Puppetteers. Until they use a puppett snake they are not dangerous. Soon they will use more efficient shootings. Nevertheless we will continue to work.
    Warm Regards,

  • Kim Patterson

    I think what people are needing
    is support from you that the
    new fire will not be lost, or will
    not go out.

    For you the noble peace price.

    Strike while the iron is hot!

    Humanity is waiting.


  • Bernie Koppenhofer

    The snake will not let us email him, I think he has crawled back into his hole. Congratulations Mr. Rossi!!

  • Andy Wu

    Dear Mr. Andrea Rossi,

    Congratulations to you on your Oct 28 test. It was a great success. You are making a history!

    You did it! Thank you!

    Andy Wu
    from Canada

  • Dear Andrea Rossi, Great to hear the test was a complete success on the 28th. I never had any doubts and now look forward to future developments in the power industry and a possible Nobel prize for you.
    Regards E.A.

  • Max

    Dear Dr. Rossi,

    Can you tell us anything about your future business plans, perhaps sketch them in a brief exposé?

    Kindest regards,


  • Andy Wu

    Dear Mr. Andrea Rossi,

    Congratulations to you on your Oct 28 test. It was a great success. You are making a history!

    I have a couple of questions:

    1. does the container have to be 50 Cubic centimetre or it can be bigger. It seems too small if you make 100MW or bigger power plant, is it?

    2. Can it be used in Fuel Cell? It would be another big area for application if it can.

    This would be a huge benefit to Oil Sand industry in Canada on environment protection.

    You did it! Thank you!

    Andy Wu
    from Canada

  • LBG

    Dear Dr. Rossi,

    Congratulations on your epoch achievement for humankind and mother Earth. You have made the future safe for my children.

    Yours sincerely,


  • Bhagirath Joshi

    Dear Wladimir Guglinski
    very good comment. What nuclear science suffers from is the power of incremental scientists. It is hard to find willing scientists to even debate a new idea constructively without a bias. Let me tell you, if Einstein had proposed the “Theory of Relativity and gravitational waves” today, He would have been called lunatic.

  • Franco Morici

    My congratulation for the success of testing of Your 1MW plant.

    I would like to ask if the reasons You preferred to built a 1 MW plant using an array of a lot of 10 KW E-Cat modules instead of to experiment bigger E-Cat reactors are due to be, in a short development time, ready for commercialization of the 1MW plant or You think that many small reactors are better then few large.
    Kind Regards


  • Bhagirath Joshi

    Congratulations, Dr. Rossi.

  • Dear Dr. Rossi,
    We, Turkish cold fusion team – in 1989, are interested in your E-Cat system and specially in the test results of 1 MW E-Cat unit. According to the first evaluation of your results on 28th October 2011, it seems that you are successful even if you get approximately the half of the suggested amount- 1MW- . But at this point, I have some questions to make the matter clear.

    From the viewpoints of engineering and economics, as you know the evaluation of the efficiency of the system should be made overall. I mean,

    1.What is the magnitude of the energy that is consumed for providing Ni/Catalyst system (probably including the supplement of electromagnetic wave) which is used in your process? This energy should be considered in the calculation of overall energy efficiency of the system if the amount of that energy is not negligible compared to energy input -as heat- to the unit at initial. This point can be very important if a special synthetic isotope of Nickel is used or formed in your process at initial and also if the activity life time /stability of Ni- Catalyst is low.

    2.What is production/operation cost of Ni/Catalyst system and also what is it’s activity life time? And specially the cost of this catalyst on the base of it’s active life time (hydrogen cost can be neglected) should be taken into account for unit cost of energy output in your unit.

    Thank you very much for your response in advance.
    Kind regards,
    Prof. Serdar S. Celebi (Ph.D. in Chemical Eng.)

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