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

  • Dear Peter Heckert,

    I am not interested in advancing the experimental evidence. I have written this everywhere. Our results are entirely eclipsed by Rossi’s technology. If, on a purely theoretical level, you wish to explore this further – then by all means. There’s an email address given in that blog. There’s also an email address made available somewhere here. I’ll send you the papers in pdf format. I won’t subscribe to Scribd. They deleted a paper of mine – two thirds of which was written by me – on the spurious objections of a collaborator who is unable to string a sentence together let alone write a paper. He claimed that the paper was his work. I will not again go through that route.

    But this site is NOT about our thesis. I only published that explanation at Rossi’s specific request. It is a small contribution that proposes that magnetic fields may comprise particles. That’s all. No biggie. It is our academics who need to asses this. I only mention it as the debate related to LENR is likely to rage forever, when there’s a really simple explanation that conforms ENTIRELY to the standard model. But it’s an entirely ‘novel’ take on coalesced matter. But our findings conform to Rossi’s. We also get a strong heat signature at no measurable loss of energy from the supply. But we need batteries to show this. And they’re clumsy at best. Rossi has made his own battery at a cost of really little and really light materials. Our preferred route is to publish the thesis to allow our boffins to sort this out against the evidence. There are those collaborators who are qualified to comment. I certainly am not.

    Thank you for your interest, nonetheless.
    Kindest regards,

  • Peter Heckert

    Dear Rosemary Ainslie,

    I have read in your blog and found the theory interesting.
    I would like to have this in printable format and the images in better resolution.
    Possibly I try to replicate the experiment, I have the skills and experience needed.

    There are sites that host PDF documents and ebooks for free, for example
    If you upload it there, then it can be public downloaded without fees or registration procedures.
    Possibly this would be a way to bring your papers to a wider audience fast?
    You could publish the link here and in your blog and everywhere.

    best regards,

  • Peace be with the reader.

    Modern mathematics has skewed the equations:

    n = p + e

    should read:

    n/n = 1 = p/n + e/n

    = (can be written as)

    The faithful witness

  • Dear Andrea,

    This is – very broadly – the synopsis of the thesis that relates to our two papers.

    The thesis argues that magnetic fields are a primary force and that
    all the forces are the effects from varying dimensional structures of
    these magnetic fields. The fields themselves are structured from
    magnetic dipoles that naturally organise into ‘closed strings’. The
    fields are dynamic and they orbit at a velocity of 2C.

    These fields come in three dimensions. So, a 1 dimensional field (a
    binding field) is responsible for the weak nuclear interaction
    including the electromagnetic and the galvanic processes. 2
    dimensional fields, (having length and breadth) are responsible for
    the strong nuclear force. And finally, 3 dimensional fields (having
    length, breadth and depth) is the torus, which is associated with a
    complete magnetic field. This is responsible for gravity.

    Where it deviates from the standard model is only in this. It
    proposes that magnetic fields comprise this non-standard dipole that
    has a velocity that exceeds light speed. It also proposes that a
    magnetic field is a primary force underpinning all the known forces.
    In all other respects it conforms ENTIRELY to the standard model. It
    would explain the existence and operation of the forces – and it would
    account for the Casimir effect. and it would be precisely related to
    Plank’s constant. Because – in a field – they orbit at velocities that
    exceed light speed, then light cannot find these particles.
    Therefore they remain ‘dark’ – outside our abilities to detect it.
    Effectively they operate in a dimension of time that exceeds our own
    abilities to measure it. Therefore it relates precisely to the
    ‘dark’ energy that has been measured by our astrophysicists. I
    suspect that they’ll be very pleased with this evidence of yours.

    As this relates to your own system, the proposal is that these
    fields are responsible for binding coalesced matter. Read the
    appendix to the second part of that two-part paper. In a chaotic
    state which is when the particle is NOT in an orbital field formation
    – then the particles become as hot and as big and as slow as they were
    previously cold and small and fast. But in their ‘hot’ state, they
    are no longer ‘binding’ that coalesced material. Therefore the bound
    condition becomes compromised. This would enable the contamination of
    anything within range of those chaotic particles. For example,
    should copper be proximate – then the copper atoms would decouple
    from their coalesced condition and loosen from the structure.

    The thing is this. If there is an intrinsic molecular imbalance
    which occurs when more binding fields are available than required to
    ‘bind’ that material – then the condition of chaos can be perpetuated
    to become self sustaining. Our test 3 of the 1st part of that paper
    refers. As, indeed, do the results in your own experimental
    evidence. Then the requirement is to continually apply more material
    to reduce the rate of that interaction which, otherwise, will become
    catastrophically hot. We both applied water. You did this to much
    greater force and effect. LOL

    But here’s the thing Andrea. We have only defined a magnetic field
    as being ‘structured’ from a magnetic dipole. I’ve presumed to call
    this a ‘zipon’ because it sort of relates to a required function to
    ‘zip’ on and off atoms – or to ‘zip’ in and out of a field condition.
    And then it links to the thing that it is which is related to ‘zero
    point’ energy. But call it what you will. The minute you apply a
    particle to the magnetic field – then all those unanswered questions
    of our Greats – fall into place. It explains the existence of that
    ‘other’ dimension – which is required to explain many paradoxes
    including questions of ‘locality’. It also marries those diverse
    branches of physics including quantum and classical – dark forces and
    string theories. And it is ONLY based on an extension to Faraday’s
    Lines of Force. It is better explained in the appendix to our second

    Incidentally – we have been able to reconcile the mass/size ratio of
    the proton to the electron – using this field. And this was managed,
    through the simple means of analysing stable particles as composites
    of these fundamental magnetic dipoles. And, more to the point – it
    localises that ‘dark energy’ that our astrophysicists require. I
    suspect that they may be pleased to hear of this – as it explains so

    But most significantly – this explanation does NOT deviate – in any
    way – from the standard model. It’s only an extension. And the
    minute you apply this ‘extension’ then, as mentioned, everything sort
    of falls into place.

    Let me know if you want to discuss this further.

    Kindest regards,


    By the way. This has been extensively referenced in my blogspot and
    in the forum where I’m a member. The following links refer
    Dear Rosemary Ainslie:

  • Dear Dr Rossi,

    I am most anxious to reach you. We’ve written a 2-part paper – submitted to the IEEE in July this year. Still not reviewed, but this time, fortunately, nor were they rejected. 5 earlier papers submitted over an 11 year period – were rejected outright within 24 hours. The thing is this. We have similar heat signatures but at lower wattage levels (max 150 Watts) – from a simple switching circuit. Not anywhere near the elegant solutions that you have reached. But also with infinite COP.

    Where we hope you may be interested is that these experimental results were required and predicted by a non classical magnetic field model. I modestly propose that this may ‘fit’ your own results. The more so as they do not coflict with the with standard model. If so, then there would be absolutely nothing to prevent the patenting of your device – if and as required. We are not interested in going that route.

    I do not want to detract from your own technology. It is far, far too important. What you have done for us all is beyond words. We have been watching your progress with some considerable interest. But I would be very glad to forward you our papers that detail our own explanations – for your consideration. That may help as the solution has that dubious merit of not contradicting the standard model. And it would entirely answer the lack of emissions which are required for a nuclear process.

    Please do reach me. My email address is

    Kindest regards,
    Rosemary Ainslie

  • Andrea Rossi

    Dear Jon Sodeberg:
    Thank you, I agree with you,
    Warm Regards,

  • Jon Soderberg

    Dear Mr Rossi
    As a resident of Massachusetts and former renewable energy technician I was greatly excited to read of your visit with elected officials and Academia in Boston. I hope you look closely at our state and decide to locate your manufacturing and training facilities in the commonwealth. As you know we have the best Universities in Massachusetts including MIT. However we also have a very progressive renewable energy program, many technical high schools and colleges, and the world renowned Woods Hole Oceanographic Institute on Cape Cod. We have high tech companies outside Boston, ship building in Quincy, and a well trained workforce. Massachusetts would be the ideal location to introduce the Ecat into commercial maritime systems since we also host The Massachusetts Maritime Academy. Massachusetts also has several closed military bases that could be used as tech and manufacturing centers. We also has great seafood… and beautiful beaches…
    Good luck Mr Rossi

  • Andrea Rossi

    Dear Anthony Bolek,
    Thank you. Please contact our American Branch Ampenergo:

  • Mr. Rossi, as I have placed my name on the list of prospective buyers I am most interested in the timeline to production. Your timeless efforts in bringing this technology to fruition will be greatly rewarded. I read with interest about your visit to the Massachusetts State House. Your statement of concern about this being put to use for the benefit of mankind was music to my ears. However, as my website features in depth coverage of the present ills that afflict the United States and the greed of the people (known as the “the new world order”) that are responsible for the total collapse of the world banking system I simply point out that if you place this technology in the hands of the supper rich they will do what they do best, SCREW EVERYONE. I urge you to make this technology available to the common man. My good friends in Chicago would very much like to see a licensing of this technology to assure that it will be available to everyone at an affordable cost.The more people who can be exposed to this the better it will be for all, not just the super rich who will hold back this technology at all costs. Keep in mind these are the same people who own the oil stocks and the utility stocks that will take a beating because of this invention. My best wishes to you sir and continued success, Bravo!!!

  • Andrea Rossi

    Dear Daniele G.:
    Warm Regards,

  • Daniele G.

    Salve Rossi, Vorrei contattarla privatamente per proporle un idea per ridurre i spazi dei reattori e migliorare in controlle di essi. Sarei molto onorato avere una sua risposta.

  • Andrea Rossi

    Dear Eric Ashworth:
    Warm Reghards,

  • Dear Andrea, this is a quick question and maybe you have already anawered it. The used nickel that comes out of the core that has been transmuted does it revert back when under normal temperature, pressure and environment revert back to its original isotopic values by endothermic reaction?. Regards Eric Ashworth.

  • Andrea Rossi

    Dear Gherardo:
    Yes, the improvement you wrote about is very important.
    Warm Regards,

  • Gherardo

    I made some calculations using the stated costs of the 1 MW group on the website.
    My conclusion is that the next big production improvement step would be to avoid external electrical energy after the reactor startup since over 30 years of usage that component outcost everything else.
    The answer to this could obviously be related to the thermal-to-electricity conversion work you are already doing.
    Take care, Gherardo

    PS: how will you service the cores two times a year? swap? how long it takes?

  • Herb Gillis

    Andrea Rossi:
    You mentioned earlier today that you have intentionally produced controlled explosions without observing radiation. Why then would it not be possible to run a series of small explosions to generate heat in a closed container, or to generate pressure waves, and use the energy released to produce electricity?
    I wish you the best of luck with the ecat commericalization process.

  • Andrea Rossi

    Dear Jorge Jimenez:
    We are doing our best.
    Warm Regards,

  • Andrea Rossi

    Dear Jonatan:
    Same way you do it with a normal heater.
    Warm Regards,

  • Jonatan

    Dear Mr.Rossi

    If you have one source of thermal energy (heat) like the household ecat, then how you can transport the heat to all the rooms in the house, for example?

    Warm Regards

  • Buen día desde Costa Rica América Central!. Ciertamente que la especulación será la única respuesta de quienes persiguen de mala gana la tecnología de los Ecats. Señor Rossi, es muy importante que usted y los que le acompañan sepan mantener su altruismo para quienes en el futuro próximo deseamos poder aprovechar esta tecnología que se inicia con su invento. Es muy importante que quienes amamos la ecología veamos en su invento la primer posibilidad real para darle al mundo de hoy una cara rejuvenecida cuando empecemos a olvidar las formas retrógradas y todavía nocivas de hacer energía.
    Desde la Universidad y desde Nuestra Empresa generadora de Energía estamos siguiendo muy de cerca toda información que nos es viable, siempre a la espera de nuestro turno. La lucha apenas empieza, pero recuerda que, “nunca es más oscuro, que cuando va a amanecer”, saludos de Costa Rica, Dr. Rossi.

  • Italo A. Albanese

    Dear Insight,
    I was talking about instant temperature (and pressure) rise, caused by adiabatic compression of the hydrogen from the explosion shock wave. Explosion, by the way, non necessarily caused by the e-cat itself. If that temperature could make the e-cat release suddenly (milliseconds!) a very high energy, big troubles could happens. Andrea Rossi just said to have tested it, so very soon (I hope) we will safely install an e-cat in our basement.

    Best regards,
    Italo A.

  • Andrea Rossi

    Dear Insight:
    The safety issue is perfectly under control.
    Warm Regards,

  • insight

    Dear Italo A. Albanese ,
    I think that the system temperature could not suddenly rise during one explosion like you say. I mean, an explosion caused by the system itself. Indeed, it is the system itself that produces the bulk of the energy.

  • Andrea Rossi

    Dear Italo A. Albanese:
    1- during our safety tests we produced on purpose explosions to test and measure the consequenses: no relevant differences have been measured from the background.
    You are right: all the issues connected with safety have to be addressed with extreme attention.
    Warm Regards,

  • Andrea Rossi

    Dear Hampus:
    Soon, but remember that such R&D will be closed doors made and not public. I repeat: no more public tests will be made. We will make only closed doors R&D and tests for our Customers made along the test protocols agreed upon the purchasing contracts. No more information will be released until proper patent protection will be granted. Too many vultures fly around, ready to steal critic info. Look to what is going on around the Balcans: there are clowns saying they have a technology copied from us, actually they have just a moke up, waiting for the piece of info they need to make a real copy. They believed we would have been selling in October the small E-Cats, so announced they would have made a demo in october ( buying a model, disguising it as a copy made by them). But it was just a trap we made. Conclusion: from now on we will be more sealed than ever, and we will be open exclusively with our Customers.
    To put for sale the small unts we need:
    1- safety certification
    2- granted patents
    We are working on both the issues and I think they will be addressed within 1 to 2 years from now.
    Warm Regards,

  • Andrea Rossi

    Dear Gherardo:
    Core swap, takes 30 minutes
    Correct your analysis
    Warm Regards,

  • Gherardo

    a simple question: how is servicing performed every 6 months? Core swap?
    How long is it for each ecat (3 reactors)?
    Thanks, Gherardo

  • Gherardo

    I made some calculations using the stated costs of the 1 MW group on the website.
    My conclusion is that the next big production improvement step would be to avoid external electrical energy after the reactor startup since over 30 years of usage that component outcost everything else.
    The answer to this could obviously be related to the thermal-to-electricity conversion work you are already doing.
    Take care, Gherardo

  • Hampus

    Hi Rossi

    When will the experiment in Bologna and Uppsala university start?

  • Italo A. Albanese

    Dear Andrea Rossi:
    I’m sorry I have to insist on this point but safety is very important.
    You said that when nickel melts, reaction stops so the system is intrinsically safe. But in a explosion, the temperature can rise, for a short moment, much more than the melting point of nickel. What could happen in that moment, just before nickel starts melting (or vaporize, if the explosion is strong enough)?
    You said also you can control the reaction by varying the hydrogen pressure. Have you found an upper limit for it? Maybe there is a theoretical upper temperature/pressure limit for the reaction? Or an absolute energy/time maximum?
    You don’t have to answer to question that can discover industrial secrets, but please say you have considered these points.

    Best regards,
    Italo A.

  • Andrea Rossi

    Dear Italo A. Albanese:
    Should we have a hydrogen expolsion the reactions would stop immediately because the parameters of pressure and temperature would not be fit for the operation of the reactor. The system is intrinsecally safe.
    Warm Regards,

  • insight

    Dear Wladimir Guglinski,
    I read your paper about the photon.
    I would like to ask what do you think about the fact that, when e+ and e- annihilate, the resulting photon has hv frequency exactly the sum of the energy equivalent of the two masses plus their motion energy; and about the fact that, if a gamma photon yields a e+ e- couple, the mass plus the motion energy of the two particles are the same of the original hv?
    You say that there is a behaviour of photon as regards to momentum and another as regards to energy, that is, the two have to be described with a dual model of massless/not massless photon. The behaviour as regards to momentum is described with the mass.
    But, what remains of the momentum behaviour you wanted to account for, if you say the photon is made of e+ and e- with total rest mass 0, being that no experiment can detect it, as you say “it’s impossible to detect the mass of a photon with that sort of experiment, because the indidual masses of the particle and antiparticle are absorbed by the aether, and so no mass is deposited in the surface where the photon had fallen down”. I mean, from which phenomena comes the need to account for the momentum behaviour if you drop its causes at the moment of the interaction, that is when you need them most for your theory’s sake?
    In other words you end up demonstrating that photon has m=0 as to momentum and m=1 as regards to energy (see part 4 of your paper) , that is just the opposite of your original goal.

  • Italo A. Albanese

    Dear Andrea Rossi,
    I try to rewrite my former question. Can you exclude that at high temperature and pressure caused by an hydrogen explosion, nickel and hydrogen could react much faster, releasing suddenly a big energy? In a 10 kW module, assuming 6 months time, there is more energy than in 250 TNT tons!

    Best regards,
    Italo A.

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    Thank you really much. You have understood perfectly the situation.
    Warm Regards,

  • Andrea Rossi

    Dear Monti:
    Maybe you are right, but what I wrote is the truth, and I am tired to be absorbed in useless issues after a 16 hours work per day. Again: my Customers are testing our plants, and we do not need other “testers”.
    Thank you for your kindness.
    Warm Regards,

  • Gregory Goble


    I have been pondering sound, or sonic harmonic phenomenon (like cavitation), as being the catalyst in the Rossi – E Cat reaction chamber. Knowing your interest I am sending you this. You will find enlightening leads in this piece I have been doing flash crowd education with.

    A bit of research and I come across this item. Rossi is associated with these folks who are experts in the cavitation phenomenon.

    Different frequencies of sound may cause cavitation and focus heat within the nickel and hydrogen lattice, specific frequencies oscillating through a harmonic may create standing waves weakening the lines of resistance between subatomic particles. Is sound the catalyst in the Rossi E Cat reaction chamber? Related information see Sonofusion

    Are the radio frequencies creating a micro-cavitation within the nickel hydrogen lattice? This is the first post I found where Rossi mentions sonic frequency generation as part of the reaction process. “In the self sustained mode, a one megawatt plant can operate at full power, while consuming a miniscule amount of electricity to operate fans, pumps and radio frequency generators.”

    Further sonic speculation in a recent CBS News article “Cold Fusion Debate Heats Up After Latest Demo” ByNatalie Wolchover
    “Peter Hagelstein, an MIT professor of electrical engineering and computer science and one of the most mainstream proponents of cold fusion research, thinks the process may involve vibrational energy in the metal’s lattice driving nuclear transitions that lead to fusion.”

    “Not so far in the distant future, the next generation… will look back at our generation..
    and know that the term ‘energy shortage’ was a term for unenlightened minds” gbgoble-2008



  • Monti

    Caro Andrea,

    davvero, le vogliamo bene. Ma tenga duro anche nelle esternazioni, non le lasci scappare. Capiamo la pressione enorme, e non sa cosa daremmo per poterle dare un po’ di sollievo, ma non ha nessun vantaggio ad apparire stizzoso.

    Un grandissimo in bocca al lupo.

  • Bernie Koppenhofer

    Mr. Rossi: It is my opinion you are on the right course. These people who want more testing are in two groups: 1) Those who want to steal secrets 2) Those who do not understand entrepreneurship/economics and the current state of our patenting process. Stay the course, concentrate on producing more E-cats for as many different parties and uses as you can and as fast(safely) as you can. Let your friends on this site help you as much as possible.

  • Andrea Rossi

    Dear Ivan:
    We are no more in the mode of public tests, the times of public tests are over. We are manufacturing plants for our Customers, and our Customers will test the plants they have bought.
    The proposal of Celani is just a provocation, and an insult to all the people that already made tests.
    We are working for our Customers, not for the curiosity of our competitors.
    By the way: Celani is using since decades the money of the taxpayer to make his apparatuses: he should employ his time to make worth the public money he is using, instead of going to try to “test” the work of others.
    Warm Regards,

  • Ivan

    Dear ing. Rossi,

    I’ve been following your e-cat development from months, now the scientific community is asking for more proof, either from Italy and UK.
    Please reconsider the idea of not doing new tests in public, if you are in good faith.
    Especially the one of dott. Celani, with the industrial secret safe. He promise to publish results on scientific papers.
    Do this for the few people that believed in you and in your technology against all the others.

    Best regards

  • Andrea Rossi

    Dear Magnus Holm:
    Thank you for the precise comment. I want to add that this policy has been agreed upon between Hydrofusion, our North Europe Commercial Branch, and me. I totally agree with the policy of Hydrofusion. The waiting list will allow us to have an economy scale good from the beginning of the distribution.
    Warm Regards,
    Andrea Rossi

  • Statement:

    The pre-order list for ECAT Home units on is clearly, as is stated on the product page (ECAT Home Units), a non-binding waiting list only. Once the product is ready for the market, or new information on the product is available, this information will be sent out to all people on the waiting list and then real orders will be processed by a first-come, first-serve policy from the waiting list. hopes the interest list will help Andrea Rossi lowering the end product price for the ECAT Home units.

    Magnus Holm, (operated by Hydro Fusion)

  • Andrea Rossi

    Dear Christor Stremmenos,
    Your comment should merit an English tramslation. I have not the time, in this period I am working, as usual, 16 hours per day.
    Thank you anyway for your usualintellectual honesty.
    Warm Regards,

  • Andrea Rossi

    Dear Dr. Paolo Bruschelli:
    Thank you!

  • BRAVI !!!

  • Ch. Stremmenos

    Caro Andrea
    Riporto c.c. del mio intervento relativo ad un articolo “CACIATORI DI FUSIONE FREDDA” pubblicato ieri al giornale greco Tovima. Il commento ha il titolo “più serietà e più onestà”e non si rivolge solo alla redazione del Giornale:
    “Mi sento personalmente responsabile per il trasferimento e il destino dell’invenzione (Cold Fusion) di Rossi in Grecia, quindi custode dei
    valori in essa contenuti:Con motivazioni scientifiche e idealistiche,coinvolgendo i miei amici, colleghi e partner a lungo termine: ing. A. Rossi e Prof.S. Focardi protagonisti di questa innovazione epocale, abbiamo tentato insieme a trasmettere in Grecia, la scienza e le prospettive tecnologiche per un futuro promettente in questo .Paese.. .. Luogo di nascita di
    Democrito e di Leucippo e tramite questo simbolismo, promuovere la nuova era energetica a beneficio di tutta l’umanità.
    Per quanto riguarda la società greca “Defkalion Green Technologies SA”, purtroppo e con grande rammarico, mi trovo in coscienza contrastante per il frivolo e incoerente a dir poco comportamento
    generale e per la documentata violazione degli obblighi contrattuali finanziari con A. Rossi che dura circa quasi da un anno.
    Concludo con un saggio detto popolare della mia terra natale l’Evrytania: “l’agnello del povero, non diventa mai montone …!.”
    Prof.Ch. Stremmenos “

  • Donald Chandler

    I’m wondering if these ultralow density (<1mg/cc) nickel microlattices (described in the latest issue of Science, Nov 18 2011) might have some use to your e-cat? Huge nickel surface area, light weight…

  • Wladimir Guglinski

    Mr. insight wrote in November 20th, 2011 at 6:27 AM :

    “Dear Wladimir Guglinski ,
    … when is it the moment when the two particles lose their mass as a whole? I ask this because the two particles both have attractive gravity before.”

    Dear Mr. Insight,
    I think that I finally understood your doubt.

    Consider the emission of a photon by an atom. First is created the particle. The creation of the particle produces a “hole” in the aether, and so at once the antiparticle is created, and they form the photon. Due to such delay, there is a distance between the particle and the antiparticle. And thanks to such distance of them, it is possible to have polirization of the light.

    In the case of the neutrino it happens the same. The positron and the electron does not approach one each other, so that to create the neutrino, as you supposed. The electron and the positron do not exist “before” the nucleus decay as free particles within the nucleus, as you supposed. They are created together within the neutrino’s structure, in the instant of the decay of the nucleus.

    For instance, when the neutron has a decay, the pair positron-electron is created in the form of an antineutrino. The antineutrino is born without mass, because in its creation the positron and the electron are created together, performing the structure of the antineutrino.

  • Wladimir Guglinski

    Mr. insight wrote in November 20th, 2011 at 6:27 AM :
    Dear Wladimir Guglinski ,
    … when is it the moment when the two particles lose their mass as a whole? I ask this because the two particles both have attractive gravity before.

    They have repulsive gravity in their fields too.

    You have to realize, dear Mr. Insight, that it’s hard to explain it here. The best would be to discuss after the publication of the paper of mine.
    Besides, as I already said before, I have not all the answers.

  • Andrea Rossi

    answer: no.
    Warm Regards,

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