Anomalous mass of the neutron

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

2a = λ/2     (4)

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

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

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

T = π²2h²/4mπa² ≥ 180 GeV      (6)

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

Uπ = T =  180 GeV     (7)

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

UN = 40 MeV     (8)

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

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

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

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

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

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

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

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

759 comments to Anomalous mass of the neutron

  • Andrea Rossi

    Dear Martin:
    1- not yet
    2- still working
    3- what do you mean?
    Warm Regards,

  • Andrea Rossi

    Dear Andrzej Wasowski:
    Let me know when you have a proposal.
    Warm Regards,

  • This is an introductory statement to a complex subject. I believe a question has to asked, “does a product regardless of its ability, require a detailed understanding at its causal level?’. I believe the answer is yes because if something cannot be explained then its system of reality is technically flawed in the eyes of certain individuals. Therefore, although Andrea knows how his E-Cat works his inability by the active none disclosure/confidentiality laws currently in place prevent a necessary aspect in the belief system of the E-Cat. I shall attempt to explain: There is in every reality system i.e. that which exists, 3-Aspects regarding an outer shell activity. 1st Mathematician – plane of mind (cubic dimension/square). 2nd Architect – plane of mind (sphere/circle). 3rd Engineer – plane of mind (cone/pyramid/triangle). This 3rd plane is the base dimension being the most dense and is responsible for the final embodiment of an idea. Could be said to be three types of necessary hats, worn by either one triune individual or worn individually by a group of three people to form a team. These shapes provide for the geometry in the understanding of a neutral with regards its established electro magnetic field. They could be said to be written in stone because of their association to physical reality or the tangible world which is only a condensate from a fission dimension. Therefore, if one of these Aspects is missing with regards a good idea, even though the idea may be embodied and performing as intended, the idea can still lack credibility due to the missing information pertaining to its 1st Aspect. At the end of the day it could be said “does it matter whether so many people understand it or not, just so long as the good idea is being used in a productive way?”. This I believe is the crux of the matter, with regards its secrecy, nobody wants the E-Cat buried together with its inner working knowledge and therefore, like so many others, I look forward to its commercialization and its use in the ever day life of the general public. Regards Eric Ashworth

  • Luca M

    Dear Mr Rossi, just an information.
    It is known from literature that a particular nickel-copper alloy, a dimer, has a great catalytic power for hydrogen dissociation. More precisely it was found that, among several other alloys with catalytic properties, the highest catalytic power with respect to the hydrogen dissociation process is exhibited by NiCu alloys (see link below).
    So apperently it seem that the e-cat “fuel” could also be used as catalyst.
    Did you ever try this possibility ?
    Thank you for your attention, kind regards

  • Wladimir Guglinski

    Is it possible to detect the mass of photon by experiments?

    Some years ago I wrote two papers concerning the mass of the photon, and submitted them for publication in Infinite Energy.
    They declined to publish them, because the papers are not suitable for the readers of that magazine.

    Now I decided to published one of them in Peswiki:

  • @Icarus
    Yes, but Mr Rossi is looking for a prompt and viable solution for small ecat.
    They do not use steam and 230VAC comes from bigger converter, those produced in Europe. The small ones where in NZ.
    The main issue remains: 3bar seems to be the bottom limit to get electricity for home appliances through a Stirling and the small ecat generates 1,5bar only.
    That’s why things are getting toughter and side effects as thermoacustics et similia could play a role.
    Mr Rossi looks for a solution and we are all doing our best to try to give help, if we can.

  • Enrico Billi

    Dear Eric Ashworth, about the Q-Drive i am not sure it works and i don’t know the final thrust you can obtain from this device, but it is interesting in any way because superconducting technologies have been applied on microelectronics, particle accelerators, but no one before thought to use an unbalanced magnetic field to create a thrust force.
    Luckly for the founder of Cannae, he was able to test a good solution of his problem developing a lattice simulation using the well known properties of EM fields and superconducting alloys. Unlucky for my friend Andrea Rossi, there is not such good knowledge of the nuclear physics, so after a long “lavolale lavolale” he found a good solution for his technology.
    Every time there is a new discovery, new opportunities become available, so because imagination is free, i will like to share such ideas with the readers of Journal of Nuclear Physics.
    Thanks to Andrea Rossi, who open this virtual place to let people discuss about physics, engineering and beyond.
    Best regards,
    Enrico Billi

  • insight

    Dear Wladimir Guglinski ,
    let’s stay focused for a moment on the neutrino question, I would like that you answer it because I want to understand.
    You say that “a neutrino needs special conditions to be formed by positron+electron” so, under those conditions, while they approach each other, 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. It does not matter what you think descends from your theory itself: it is not good to make a theory and demonstrate it by its own thesys (science is not so easy).

  • Dear Andrea!
    In this week – after “garage-test” – my prototype goes to Technical University on montage, verification and tests…

    I will inform You about all…

    Best and warm regards from Poland

  • RH

    to Neil Taylor:

    this article says that Løvvik’s device has an efficiency of 15%.
    This is too low to make the E-Cat practical as an electricity generator, you would need much higher COP that 6.

  • Wladimir Guglinski

    Neutrino experiment repeat at Cern finds same result

    BBC News – The team which found that neutrinos may travel faster than light has carried out an improved version of their experiment – and confirmed the result.

    Full article:

    Read also: CERN Experiment Excludes 1 Error In Faster-Than-Light Finding

    GENEVA — The chances have risen that Einstein was wrong about a fundamental law of the universe.

    Scientists at the world’s biggest physics lab said Friday they have ruled out one possible error that could have distorted their startling measurements that appeared to show particles traveling faster than light.

  • Wladimir Guglinski

    Gillana Giancarlo wrote in November 19th, 2011 at 1:00 PM

    L’esperimento di Michelson Morley avrebbe quindi portato ad una deduzione sbagliata, ma se le conclusioni di Einstein sono errate come fa a funzionare cosi bene tutta la relatività speciale! E’ una pura coincidenza?”

    Dear Gillana,
    I think it is not a pure coincidence. First of all, you have to take in consideration that the space is not Euclidian, becauses the aether has contraction near to the Sun. So, it does not occur as happens in Newton’s theory.
    Besides, when you apply the mathematics, you can get a correct result if you find the suitable postulates to be used as the starting point for your mathematical development.

  • Wladimir Guglinski

    Mr. insight wrote in November 19th, 2011 at 10:25 AM

    Dear Wladimir Guglinski ,
    How do you explain that, usually, when an electron meets a positron they annihilate and do not form a neutrino?”

    Dear Mr. Insight,
    Why the proton and the electron do not form a neutron when they meet together?
    Under special conditions, as occur in Don Borghi experiment (and many cold fusion experiments), the proton and electron fuse and form a neutron, at low energy.
    Why it does not allways occur ? Why they form the hydrogen atom, instead of the neutron?
    It’s because there are need special conditions so that to occur the fusion proton+ electron = neutron.
    So, a neutrino needs special conditions to be formed by positron+electron. It can be formed only into the nuclei.

    I would like to know, while they approach each other, 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 Insight,
    this question is applied also to the case of the photon, since it is formed by particle-antiparticle.
    Before to start to consider the hypothesis of repulsive gravity, I used to think about such problem along many years, because I had no answer for the question: why within the photon the particle and the antiparticle does not anihilate one each other?
    Only in the begginnig of 2011 I started to consider the hypothesis of repulsive gravity, and I finally discovered the answer for the stability of the photon.

    I analyse this problem by considering my model of photon and my new nuclear model, in the paper on the Rossi-Focardi experiment, submitted here for publication.

  • Wladimir Guglinski

    raul heining wrote in November 19th, 2011 at 10:26 AM

    As you know, light moves in null geodesics which are straight lines in special relativity.
    In general relativity, null geodesics are not straight lines because of a no null energy momentum tensor in the field equations. It means you have a source of curvature and this is one of the brilliant aspects of relativity theory and does not imply existence of mass for the foton which
    follows a curve close to massive objects where the space time has a no zero Riemann tensor.”

    Dear Raul,
    Einstein imposed several postulates in relativity.
    Postulates in Physics are only a sort of solution used when the author cannot find the true mechanisms.
    Why does the light has to moves in null geodesics?
    Probably because, as the light has mass, then an incomplete theory that considers the light without mass must impose a postulate to the light behavior: it must move in null geodesics.

    If Einstein’s theory should be correct, his attempt to get the unification would have been successfull.

    I don’t need to say that this is verified by experiment.”

    And I dont need to say that it does not prove that photon has no mass.

    Anyway, mixing quantum systems with relativity is not very good idea because as you know in quantum mechanics, time has a different treatment from space coordinates.”

    Of course, when God created the Universe, He took your wise words in consideration.
    Otherwise, He could not succeed to make the Universe to work.

  • Martin

    Dear mr Rossi,

    I have some questions, i hope you can answer them.
    1 is The first1 Mw. Plant already working again?
    2 are you still working with this plant, or advicing?
    3 during THE test of THE plant some gaskets were broken. THE plant was direct after THE test taken with THE costumer. If They fix this problem self they have to open THE ecat. Can you tel shag happende with this?

    Thanks for taling some time for this!

    Best wishes,


  • Mr. Rossi,

    This appears as a possibility for converting the thermal output of the e-cat into electrical relatively cheaply. If GM is in talks with them – they must have something close to ready to go…

    Good Thermoelectric Devices Coming Soon
    November 18, 2011

    The team is already in talks with a major automotive manufacturer (GM in the U.S.) with a view to placing the material in the exhausts of cars.

    See full article at:

  • icarus

    stefano wrote:

    “ uses diesel to produce heat to convert into hot water and 12-24V. They lost their factory by an earthquake on february 2011.”

    This is only your small selection of the truth.

    Whispergen burns other liquid fuels and gases also. The main product line has a grid-tie inverter integrated and produces 230 VAC. Their main factory for this product is in Europe (Spain) and was obviously not affected by an earthquake in New Zealand. The small operation producing DC product for remote applications is in Christchurch. All of this information is available on the website.

    Theoretically thermo-acoustic Stirling generators may work too, theoretically.

  • Dear Enrico Bill, Regards Cannae Q Drive technology. Thanks for the web connection. What an amazing site. Those people who doubt the E-Cat technology now have another one to go at, what a laugh. With regards the E-Cat and I hope Andrea does not mind me going on about how I believe his E-Cat works. I say this because together with this new Q Drive technology nuclear physics will be able to show itself as an understandable subject. I am going to put some material together that some people could find interesting regardless of an academic background. I am going to try and keep it short and to the point and section it up so as not to be too confusing. Regards Eric Ashworth.

  • Gillana Giancarlo Prof. Wladimir Guglinsky
    Grazie mille! Molto Interessante davvero, quindi secondo la sua QRT si dovrebbe concludere che i fotoni non sono deviati, in prossimità di grosse masse (come il sole), dalla deformazione spazio-tempo secondo la geometria einsteniana ma avendo una massa sono soggetti ad una attrazione gravitazionale newtoniana!
    L’esperimento di Michelson Morley avrebbe quindi portato ad una deduzione sbagliata, ma se le conclusioni di Einstein sono errate come fa a funzionare cosi bene tutta la relatività speciale! E’ una pura coincidenza?
    Vs. Gillana Giancarlo

  • raul heining

    Dear Guglinski, you said this:
    “The Sun deviates the light trajectory. If the photon should have no mass (as considered today in current Modern Physics), the trajectory of photon should not be deviated by the Sun.”

    As you know, light moves in null geodesics which are straight lines in special relativity.
    In general relativity, null geodesics are not straight lines because of a no null energy momentum tensor in the field equations. It means you have a source of curvature and this is one of the brilliant aspects of relativity theory and does not imply existence of mass for the foton which
    follows a curve close to massive objects where the space time has a no zero Riemann tensor.
    I don’t need to say that this is verified by experiment. Anyway, mixing quantum systems with relativity is not very good idea because as you know in quantum mechanics, time has a different treatment from space coordinates.

  • insight

    Dear Wladimir Guglinski ,
    How do you explain that, usually, when an electron meets a positron they annihilate and do not form a neutrino? You say that neutrinos have very little mass because of “almost” cancellation of positive against repulsive gravity, so the neat effect of gravitation is very small. I would like to know, while they approach each other, 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.

  • Andrea Rossi

    Dear Enrico Billi:
    In this case: “la-volale”
    Warm Regards,

  • Wladimir Guglinski

    Dear Mr. Insight,
    I was thinking about the question on the mass of the neutrinos, by supposing they really have a structure formed by positron-electron.

    Probably the mass of particles is due to the interaction of the flux of gravitons (within the particles) with the gravitons of the aether.

    Suppose that there is indeed a repulsive gravity. In the case of a corpuscle formed by particle-antiparticle (as the photon and the neutrino), as their structure is symmetric, then there is a tendency the gravity of the particle to be cancelled by the antigravity of the antiparticle.

    In this way the total mass of the corpuscule will be “almost” cancelled.

    Consider for instance the neutrino formed by positron-electron.
    The mass of the neutrino is not null (because its mass is formed by the mass of the electron together with the mass of the positron). However the resultant of their masses will be near to zero.

    So, in spite of the electron has a mass “m”, and the positron has a mass “m”, the mass of the neutrino is not 2m, because the gravity and antigravity cancels one each other, and so the mass of the neutrino is near to zero (it is not zero because there is an assymetry in the intensity of gravity and antigravity).

  • Wladimir Guglinski

    Gillana Giancarlo ,
    we use to think about mass in the sense as Newton proposed it.
    Newton proposed the concept of mass for the matter.
    He did not know the existence of antimatter.

    A particle composed by matter-antimatter does not follow all the rules aplied to matter.

    Actually the massless particles also have mass.

    The photon has mass, but not in the sense which we know mass, as Newton proposed.

    The photon ia a combination of particle and antiparticle. So, it hehaves as it should have no mass (in the sense of Newton).

    The Sun deviates the light trajectory. If the photon should have no mass (as considered today in current Modern Physics), the trajectory of photon should not be deviated by the Sun.

    The prevailing theories consider the photon as massless because it is required by Compton equation and also by Einstein’s relativity.
    However those two equations need to require the photon as massless because when Einstein developed the relativity he did not know about the existence of antimatter, and so he could not think about a photon composed by particle-antiparticle.

    Because the photon is composed by particle-antiparticle, that’s why we need to consider the photon as massless in Compton equation and in Einstein’s relativity.

  • Enrico Billi

    Dear Andrea Rossi, surely i am flying very high with my fantasy, but thinking of Steve Jobs who told us with his motto “Stay hungry, stay foolish”, i was curious if in the future (possibly not too far) we could see the e-cat applied for these astronautics projects:

    1) Nautilus-X Interplanetary Starship has a plasma engine wich needs 200KWe for its superconducting acceleration chamber. NASA want to use solar pannels, but may be e-cat could be a cheaper choice

    2) Cannae Q-Drive: i found this website of an USA company who says they develop a new kind of engine for starships with no propellant using imbalanced superconducting cavity, i don’t know wich is the power needed but may be in future e-cat could supply the KWe necessary saving space, weight and money for future starships.

    Obviously in the while… “lavolale lavolale”
    Best regards,
    Enrico Billi

  • Wladimir Guglinski

    Mr. insight wrote in November 19th, 2011 at 5:36 AM
    Dear Wladimir Guglinski ,
    “You are wrong: the effect of mass, that is, the behaviour of the particle, does not depend just on the electrical charge and electromagnetic interactions. Other experiments could be lead to determine the mass of such a particle you described, for example sending a beam and measuring the parabolic motion due to the earth attraction.”

    OK. Did they do it with neutrinos ?

  • Andrea Rossi

    Dear H.Hansson:
    Thank you for your suggestion, but I think we are going through the right strategy. If wrong, we will see in the battlefield.
    Warm Regards,

  • icarus

    Dear Ing. Rossi,

    I’ve sent you a follow-up email to as it was of a more confidential nature.



  • Gillana Giancarlo

    Egr. Prof. Wladimir Guglinsky:forse la domanda che Le faccio Le è già stata rivolta e per Lei sarà certamente banale.
    Lei considera l'”etere” come la quintessenza del mondo fisico, che permea ogni cosa e da cui tutto deriva.
    Lei ipotizza 5 particelle prive di massa, 10 se consideriamo le cariche posive e negative (antiparticelle?), che darebbero origine ai quarks ai gluoni (considerati il substrato primordiale della materia), e su scala sempre più grande a tutte la particelle conosciute (fermioni, bosoni (force carriers) etc.).
    Partendo da entità prive di massa come si può originare la stessa e giustificarne le proprietà ed interazioni?
    Vs. Gillana Giancarlo

  • insight

    Dear Wladimir Guglinski ,
    you wrote
    “But the neutrino does not interact with the matter, because the neutrino’s total electromagnetic field is null, since the positive electromagnetic charge of the positron is cancelled by the negative electromagnetic charge of the electron.
    So, the neutrino behaves as it should have no mass.
    Of course the neutrino has mass, and its mass is the sum of the electron and postitron masses (discounting the packing loss). But as the neutrino does not interact with matter, there is no way to detect its mass by experiments, and so it looks like the neutrino had no mass.”
    You are wrong: the effect of mass, that is, the behaviour of the particle, does not depend just on the electrical charge and electromagnetic interactions. Other experiments could be lead to determine the mass of such a particle you described, for example sending a beam and measuring the parabolic motion due to the earth attraction.

  • H. Hansson

    Dear Mr. Rossi,

    Reading available observations of your operation,.. it seems that your company is very much still organized as a small start up company with you personally engaged in every small decision and issue, no matter how insignificant it may appear.. Of course with the expected exponential growth the current Management is not sustainable.

    When will you get your company organized to meet the challenge??.. You will of cause loose control over the small and medium size issues.

  • By converting e-cat steam into electricity through a Stirling engine, main problem is steam pressure: at least 3bar is required for 12-24V production.
    Probably, setting up ecats in parallel may help getting higher pressures?
    Theorically, even exist thermoacustic stirling engines, using standing waves and no displacer piston. Would than work just inside the ecat rector… uses diesel to produce heat to convert into hot water and 12-24V. They lost their factory by an earthquake on february 2011.

  • Andrea Rossi

    Dear Iggy Dalrymple:
    In the tank we utilize there is biatomic H. What happens in the reactor is confidential.
    Warm Regards,

  • Andrea Rossi

    Dear Icarus:
    OK, send a precise proposal, for 10 kW of power, specifying the integral of efficiency in function of pressure of steam.
    Warm Regards,

  • Wladimir Guglinski

    LHC experiments may lead to New Physics

    Researchers from Large Hadron Collider (LHC) have shown such results, which are out of our current understanding of the Physics. These conclusions are leading to the “new physics”.

    Matthew Charles, LHC-beauty (LHCb) told HCP 2011 meeting on Monday that D-mesons are found to decay in a slightly different manner than the antiparticles of mesons. The researchers found a difference of about 0.8% that is a significant one. This conclusion may help to explain why antimatter is present in lesser amount than the amount of matter in the universe. However, according to experts, further research is required in this respect.

    It’s interesting to note that physicists need to make experiments in the LHC, so that to arrive to the conclusion that there is need a New Physics, only because now the LHC experiments are showing that D-mesons do not follow what we expect from the Standard Model.

    Several particles do not follow what the Standard Model predicts, and the neutron is among them. For instance, look at this page in the Wikipedia:
    The Standard Model of particle physics predicts a tiny separation of positive and negative charge within the neutron leading to a permanent electric dipole moment.[9] The predicted value is, however, well below the current sensitivity of experiments. From several unsolved puzzles in particle physics, it is clear that the Standard Model is not the final and full description of all particles and their interactions

    It is obvious that the quark structure [d,u,d] of Particle Physics is wrong. From such quark structure, the neutron’s decay would have to have a time decay in the order of 10^-23 seconds, which is the order of the decay occurred by the strong force (because it is formed by quarks, according to Particle Physics).
    However the neutron’s time decay is 15 minutes. I said 15… minutes !!!!

    Such time decay of the neutron is compatible with a neutron model formed by proton+electron (when the electron is captured by the proton and they form a neutron, the electron loses its spin, a phenomenon named spin-fusion in Quantum Ring Theory).

    Other particles also have a fermion in their structure, according to QRT.
    For instance:
    – the neutral meson pi has a quark structure [d,d’] .
    – the positive meson pi+ has a structure [d,e’,d’] , where e’ is the positron
    – the negative meson pi- has a structure [d,e,d’], where e is the electron

    Because of the spin-fusion, the mesons pi+ and pi- have spin zero, because the positron and the electron lose their spin within the structure of mesons.

    Several properties of the mesons (and other particles) do not fit to what we expect from the predictions of the Standard Model.
    And the reason is obvious: it’s because the spin-fusion is not considered in the Standard Model.

    It’s is also easy to understand that the existence of leptons within the structure of several particles is responsible for some differences in the behavior of the particle and the antiparticle, as the LHC is showing.

    But in spite of so many fails, incompressibly , the physicists now hope that LHC will give them evidences on why there is no antimatter in the Universe.
    Well, it seems the reason why there is no antimatter in the Universe is obvious: there must be some assymetry in the fundamental structure of the Universe.
    But it’s impossible to find such assymetry from the fundamental structure of the Universe proposed in Particle Physics.

    The physicits suppose that several particles participate in the formation of the fundamental matter. For instance, they believe that the boson W participates in the weak interaction, and they explain the neutron’s decay by considering the participation of the boson W. And they believe it just because the boson W was predicted, and later detected by experiments.

    But what is the true meaning of the detection of particles predicted in the Standard Model?
    Let’s see it.

    Suppose that you have a brick in your hand, and you want to throw it against the floor, so that to crash it in several pieces.
    Can you predict the size of the several pieces ?

    Of course you cannot.

    And now consider the collision of two protons, so that they crash in several smaller particles.
    Can you predict the size and the characteristics of the particles ?

    Yes, you can.
    Because the particles have several properties as mass, charge, spin, baryon number, parity, etc. And there are some rules for the formation of the particles. Such rules were discovered by the particle theorists.
    So, it’s possible to predict successfully the particles resulted from the collision of two protons.
    But such successful prediction does not mean that those particles have participation in the formation of the structure of matter in the Universe.

    The fundamental structure of the universe is formed by proton and electron, as follows:
    – The aether is formed by massless particles: electric particles (+) and (-), magnetic particles (+) and (-), permeability particles p(+) and p(-), attractive gravitons g(+) and g(-), repulsive gravitons G(+) and G(-).
    – The massles particles form quarks
    – And the quarks form the electron and the proton.

    All the other particles are only breaking up of collisions. They do not contribute for the structure of the Universe.

    As said, the explanation on the reason why antimatter does not exist in the Universe requires an antissymetry in its structure.

    It’s possible the antisymmetry existiing in the Universe’s structure is due to the repulsive gravity.
    In Quantum Ring Theory is proposed that there is a small difference between the force of interactions due to the gravitons g and the gravitons G.

    Also, in QRT is considered that gravitons g and G have interaction with the same magnitude of the electromagnetism. Besides, in QRT it’s proposed that strong force is not a fundamental force of nature. In QRT it is proposed that strong force is actually only a special kind of gravity (dynamic gravity).

    And so an assymetry in the structure of the gravity can have influence in the agglutination of the quarks, according to QRT.
    But such hypothesis is IMPOSSIBLE in Particle Physics, because according to current theories the force of gravity interaction is 10^38 times weaker than that of the electromagnetism, and 10^40 times weaker than that of the strong force.

    Therefore, according to the Standard Model, the gravity cannot have any influence in the agglutination of quarks, and so gravity cannot be the answer why the antimatter does not exist in the Universe.

    Actually it seems to be impossible to explain why antimatter does not exist in the Universe from the foundations of the current theories. And Matthew Charles, mentioned in the beggining of this article, is right: there is need a New Physics.

    But of course such New Physics must consider an assymetry in the structure of the Universe. If this assymetry will not be discovered, even the New Physics will not be able to explain why the antimatter does not exist in the Universe we live.

  • Wladimir Guglinski

    Mr. insight wrote in November 18th, 2011 at 5:32 AM:

    Dear Wladimir Guglinski ,
    what about the gravitational interaction? It exists even if the two charges make an electromagnetically neutral entity.

    Dear Mr. Insight,
    I did not understand what do you mean

  • Wladimir Guglinski

    Dear Vinicius M Reinaldi

    But QRT shows that some phenomena interpreted to have statistical cause (without a physical cause) in Quantum Mechanics can be explained by having a physical cause. For instance, the emission of alpha particles according to current Nuclear Physics occurs without a physical cause.
    In QRT the emission of alpha particles are emitted statistically by having a physical cause. This is shown in my paper “Contribution of gamma-rays as the cause of alpha-decay”.

    From the nuclear model of Nuclear Physics it’s impossible to consider that gamma-rays cause the emission of alpha particles.

  • icarus

    Below post should have included link to technology development company contact for Stirling engine (Whisper Tech Limited) who are separate from production and marketing of WhisperGen product:



  • icarus

    Dear Ing Rossi,

    for converting heat to electricity in a compact domestic unit there is already a commercial technology in production. The Stirling engine of reputable New Zealand company Whispergen

    Currently it uses gas, diesel, kerosene fuels for a combined heat/electricity (1 kW) output. I see no reason why the core wobble-yoke technology (for external heat source) could not be integrated with an E-cat heat source for a similar combined heat and electricity domestic unit.

    Hope this helps with your priority 1 research efforts.


  • Iggy Dalrymple

    Dear Dr Rossi,

    Same question as Zimmerson:

    Does the E-Cat utilize atomic hydrogen, as opposed to di-atomic hydrogen?


    Iggy Dalrymple

  • Andrea Rossi

    Dear Hendrik: is the website of our North Europe commercial branch. I like the website. If you have suggestions to make it better, or corrections, please be specific. Undefined criticism is totally useless.

  • Andrea Rossi

    Dear Andrzej Wasowski:
    Interesting. When you will have an operating prototype, let me know.
    Warm Regards,

  • Andrea Rossi

    Dear Frank Acland:
    We have organized a good outsourching system.
    Warm Regards,

  • Andrea Rossi

    Dear Charlie Zimmerman:
    1- no
    2- What do you mean exactly?
    Warm Regards,

  • Charlie Zimmerman

    Dear Mr. Rossi,

    Dr. Focardi mentioned in a recent interview that there is gamma radiation while the device is running but that it is shielded (something I recall you mentioning as well). I hadn’t heard you comment on this before but…

    1) Is the water in the gamma ray flux?
    2) Do you have any issues with dissociated hydrogen? Have you checked?

    Best Regards

  • Frank Acland

    Dear Andrea Rossi,

    From all accounts you have already received many orders for you 1 MW plants — this must be very gratifying for you. How quickly can you expand production facilities to meet this high demand, especially if demand grows exponentially?

    Best wishes,

    Frank Acland

  • Dear Sir!
    It will be very nice to get Your oppinion about my invent: rotary Stirling engine.
    Short movie about it:

    Best and warm regards
    Andrew Wasowski

  • Hendrik

    Dear Mr. Rossi – thank you for making us hope 🙂

    Please note that claims to present ECAT plants and home units IN ASSOCIATION with you and Leonardo Corporation.

    For your own, your business and your scientific reputation I very much urge you to review that site (or have somebody you really trust to do so) and and make a statement if necessary.

    Best regards

  • Vinicius M Reinaldi

    Caro Wladimir Guglinski

    Is this related to the QRT?

    The quantum state cannot be interpreted statistically

    Grande Abraço

  • insight

    Dear Wladimir Guglinski ,
    what about the gravitational interaction? It exists even if the two charges make an electromagnetically neutral entity.

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