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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Abstract
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.

Introduction
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

NOTES:

  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 .
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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.
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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.
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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)
But
∫ρ.dτ  =  + 1      (12)
because the ring (Fig. 3)  has 1 proton , and “ρ” is measured by proton’s units of load.
Consequently
Q(b) =   -(r’)²     (13)
This is a well-known traditional calculation. The nuclear physicists know it very well.
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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).
Thereby:
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)
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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
So
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).
.
Conclusions
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.
References
  1. W. Guglinski, “New Model of Neutron-First Part,”  J. New Energy, vol 4, no 4, 2000.
  2. C. Borghi, C. Giori, A.A. Dall’Ollio, “Experimental Evidence of Emission of Neutrons from Cold Hydrogen Plasma,” American Institute of Physics (Phys. At. Nucl.), vol 56, no 7, 1993.
  3. E. Conte, M. Pieralice, “An Experiment Indicates the Nuclear Fusion of the Proton and Electron into a Neutron,” Infinite Energy, vol 4, no 23-1999, p 67.
  4. R.P. Taleyarkhan, C.D. West, J.S. Cho, R.T. Lahey, Jr., R.I. Nigmatulin, and R.C. Block, “Evidence for Nuclear Emissions During Acoustic Cavitation,” Science, vol 295, pp 1868-1873 (March 8, 2002) (in Research Articles).
  5. Y.T. Didenko, K. S. Suslick, “The energy efficiency of formation of photons, radicals and ions during single-bubble cavitation,” Nature, vol 418, 394 – 397 (25 Jul 2002) Letters to Nature.
  6. P.C.W. Davies, Tamara M. Davis, Charles H. Lineweaver, “Cosmology: Black holes constrain varying constants,” Nature, vol 418, pp 602 – 603 (08 Aug 2002) Brief Communication.
  7. W. Guglinski, “Stern-Gerlach Experiment and the Helical Trajectory” J. New Energy, vol 7, no 2.
  8. W. Guglinski, “Fundamental Requirements for the Proposal of a New Hydrogen Atom,” J. New Energy, vol 7, no 2, 2004.

755 comments to Anomalous mass of the neutron

  • Andrea Rossi

    Dear Eric Ashworth:
    Thank you: these puppetteers and their puppett snakes (and the corollary of thieves moving around them like moskitos around a pork) have no idea of the battlefields I come from: they for me are just paper tigers, or plumbers trying to stop the Niagara Falls. The more they hit, the more we get hard.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Ermanno Peciarolo,
    Thank you for the suggestions we will think about.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Hank Mills:
    The persons like you confirm that my choice to prioritize USA in this business has been the right choice.
    Warmest Regards,
    Andrea Rossi

  • Andrea Rossi

    Dear Repi Exec:
    1- I prioritize nobody
    2- which dates? Reports have already been published with all the information we can give before a contract.
    Warm Regards,
    A.R.

  • REPI_EXEC

    Dr Rossi,

    Two questions:
    1) Are you planning to prioritize sovereign clients over industrials?
    2) What is your prediction for availability date(s) to the wider audience?

    Motive for asking is that many of us are in sourcing mode and our clients are contemplating putting their original plans (fuel cell based) on hold until this becomes available. Somewhat disruptive as it were.

    For those opposed to the eCat, the first stage is denial, second is anger, third is…(?)

    For those applauding the eCat and eagerly awaiting availability, the first stage is overwhelming joy, second is simmering enthusiasm, third is growing impatience…(!)

    Cheers, AB

  • Robert Mockan

    >fab70

    Correction: I meant to say 1/160 of total mass of 16 grams of nickel. That would be 1/10 gram active nickel, and thus would give the 500 watts per gram that I stated.
    When Rossi uses nano-particles to increase the surface area of the nickel, and does whatever else he does to stabilize it and enhance the output, that is good progress since Focardi and Piantelli were using solid rods 17 years ago.

  • Robert Mockan

    >fab70

    You said:”You are the only one who achieved a very elevated COP with a LENR reaction.”

    That is not exactly correct. 20 years ago the research team of Liaw, Tao, and Liebert, operated a palladium and heavy hydrogen system at 460 C, with 1500% power gain. Their output was 52 watts per gram of palladium and heavy hydrogen fuel. That was published in the Journal of Electroanalytic Chemistry, “Elevated-temperature excess heat production in a Pd & D system”, JEC 319 (1991) 161-175. It was the first documented high power LENR system. Now it is true that Rossi is using nickel and hydrogen, not palladium and heavy hydrogen, but again, if we look at the original research of Focardi, Habel, and Piantelli published 1994 in IL NUOVO CIMENTO NOTE BREVI, Vol. 107, 163-167, we find they measured up to 50 watts of thermal power from a nickel rod weighing about 16 grams. Given that the reaction is very near the surface of the nickel, the mass of the active area of the nickel rod used would have been the surface and near surface, thus the actual power out would not have 50 watts divided by 16 grams, but 50 watts divided by a fraction of 16 grams. If we take the fraction as 1/10 of the total mass of 16 grams (not an unreasonable assumption), then the power output per unit mass of nickel in the original work published 17 years ago was already in the 500 watts per gram of nickel, and at greater than 400 C. So they were getting high power 17 years ago, also. A study of the literature reveals other systems operating at high power out per unit of fuel mass. This does not distract from what Rossi has accomplished. And if anything, he should be getting much more praise than he is receiving for fighting and winning against pathological skepticism from the scientific community and the news media at all levels about this technology that has kept it undeveloped for so many years. I will feel a lot more confident that he will succeed against the powers working against him after he sells a few more systems. Conceivably he might still be stopped. Unlikely as it seems, worse things have happened in the world.

  • Hello Andrea Rossi,

    I agree with you that the snakes (extreme cynics and your competitors) are getting desperate, due to the recent successful sale and test of the one megawatt plant. In my opinion, they are truly pathetic. The fact is they have not achieved a power output anywhere even remotely close to what your technology can produce. I cannot forget how Dr. Levi reported that during one of the E-Cat tests he conducted, he was producing 130 kilowatts of output — from a single reactor core with only one liter of internal volume! If I remember correctly, this was with less than a hundred watts of constant input (would have to look for his exact statement). Also, you have stated that the reactor cores with an internal volume of 50 centimeters have a maximum safe output of 10 kilowatts. Such a power density is incredible!

    What have the competitors produced? From what I hear their best result may have been fifty or a hundred watts of output power, maximum. In every way your technology is superior to theirs! They know this, and because they cannot control their emotions, are lashing out by making up lies and false statements.

    I think every test you perform and allow to be made public, is secretly like throwing hot coals on their heads. They may lie and say the tests prove nothing, but in reality the results make them furious.

    I think some of them also know they are in “too deep” with their bashing of your technology, and when the world does wake up to it’s reality, their careers will be over. Their words are plastered all over the internet, and no one can erase them!

  • Dear Eenie1, I do not need an example of mathematics in physics, there has been too many used throughout history. The E-Cat is not about maths, its about physical interactions. Andrea is tied up with a none disclosure so until he is able to disclose the knowledge he knows, we are all in a guessing game. The supporters using their postulates ( if they have any) prove nothing. it’s realy about whether you believe Andrae Rossi understands LENR and whether the E-Cat has proven a principle. As Andrae states, according to you, the proof lies in the taste of the pudding by the customer. However, if the customer determines the pudding to be sour although it may be excesively sweet. Maybe they will still pay Andrea a good price for producing a unique pudding and I hope they do. Although as you must be aware we the public will still not know how the E-Cat works but we have got a good idea. Regards E.A.

  • Gentile signor Rossi
    Mi scusi per l’arroganza, non essendo io titolato a darle dei consigli, ma personalmente sono dell’opinione che che la sua e-cat da 1 MWatt non dovrebbe essere ceduta solo ad aziende private ma anche al mondo intero.
    Purtroppo in assenza di contributi dell’Europa, dello Stato Italiano e del ministero di Mariastella Neutrini per reperire altri fondi con cui finanziare l’industrializzazione e le ulteriori ricerche, io penso che noi comuni cittadini potremmo sostenerla con il lancio di una sorta di “azionariato popolare” che abbia un obiettivo di sottoscrivere 10 milioni di euro coinvolgendo almeno 200.000 persone disposte ad investire 50 euro sullo sviluppo di questa nuova tecnologia; dovrebbe anche considerare che, in aggiunta o in alternativa a ciò e finalizzata alla prenotazione di una e-cat per la versione “residenziale” tipo Defkalion, potrebbero essere trovati altri fondi per il supporto al suo progetto, in quanto considero che se io personalmente sono disposto ad investire quale anticipo di fornitura 1000 euro, allo stesso tempo come me potrebbero esserlo tante altre persone.

    La raccolta pubblica di fondi potrebbe essere utilizzata per far confluire nel suo “Team” le università di Bologna e Uppsala, più altri ingegneri e specialisti dei processi industriali e dei materiali; in tal modo lei potrebbe attingere know-how da professionisti di ciascun settore specifico, che, anche a partire dalle linee di produzione di attuali macchine termiche (per la parte elettronica di gestione non penso abbia problemi), selezioni e testi i singoli componenti adatti a sopportare pressione, temperatura e durata inalterata negli anni.
    Nel complimentarmi per la “grinta” e la determinazione che sta mettendo nel perseguire i suoi obiettivi, le porgo ancora le mie scuse e la saluto
    Ermanno Peciarolo

  • Dear Andrea, Thanks for the update with regards the imbecile. I agree with you he has to be a puppetteer. Although I have no knowledge of him, it comes as no surprise. What you have is intelligence and these people are frightend of it. Stay calm and keep everybody informed. I believe you are challenging the future and its not an easy road. I am amazed that you have got to where you are. You obviously have an amazing degree of determination both you and Focardi. Keep up the pressure. Best Regards E.A.

  • Iggy Dalrymple

    Dear Dr Rossi,
    I know you are currently occupied with manufacturing heat producing machines but when you get ready to mate your device to a steam turbine, you might consider the Cyclone Engine from Cyclone Power Technologies. Cyclone is located in Florida near you. http://www.cyclonepower.com/works.html

    Sincerely,
    Iggy Dalrymple

  • Andrea Rossi

    INFORMATION:
    An imbecile is going around sending “confidential” letters saying that our plant tested on the 28th did not have safety valves. Of course everybody with a minimum of knowledge of the matter knows that it is not possible not to put safety valves in a steam generator , but let me confirm the obvious: the plant has 104 safety valves, one per every vessel, regulated to open at 3 bars. The imbecile who is expanding this and other falsities is not a puppett, he is a puppetteer. Before or later I will publish the story of our relationship with this guy, as well as tapes in which he and his fellows have been videotaped while trying to steal samples of powder in my factory during a visit, as well as a draft of a contract which was a fraud. Desperate of the fact that we started the manufacturing of our 1 MW plants the puppetteers are scratching the bottom of their bullshit barrells, and teaming up with other gangs of thieves too.
    Andrea Rossi

  • Wladimir Guglinski

    Violation of thermodinamics law, Maxwell’s theory, and cold fusion

    The operation of the eCat in self sustained model during 5,5 hours has stimullated the reaction of the physicists, as we may see in some mainstream journals, as Discovery News:
    “Scientists say the method — cold fusion — is patently impossible. They say it defies the laws of physics.”

    Of course cold fusion does NOT violate the fundamental law of thermodinamics: the energy comes from some “place”. The problem with the current prevailling theories is because from their foundations there is no way to explain what sort of “place” it can be.

    However, there is another mystery that current theories cannot explain. Because beyond the quantum spin of elementary particles, they also have an intrinsic spin: they gyrate about their axis (as the Earth gyrates about its axis in 24 hours). Such rotation of the particles create their magnetic fields. For instance, within a nucleus the protons gyrate, and their rotation is responsible for the nuclear magnetic field of the nucleus (the nucleus also has a rotation, which increases the magnetic field created by the protons).

    But from Maxwell’s theory an electric charge with rotation must irradiate eleoctromagnetic energy (photons). And so, a proton gyrating would have to emit electromagnetic energy, and it would have to stop to gyrate after its kinetic energy of rotation is over.

    Therefore the current theories cannot explain why the elementary particles like the proton and electron violate the Maxwell’s law.

    In Quantum Ring Theory the proton has two fields: one inner principal field (composed by a flux of gravitons), and its rotation induces an external secondary electromagnetic field (composed by electric massless particles of the aether).
    The electromagnetic field does not gyrate (and therefore it does not violate the Maxwell’s law).

    But the inner gravitational field of the proton gyrates, and so it must emit gravitational energy.
    But the proton’s gravitational fields does not stop to gyrate.
    And therefore we have to conclude that, in spite of its field of gravitions emits gravitational energy, obviously it must receive gravitational energy from “somewhere”, in order to replace the energy emited.

    The question is: “where” the energy which replaces the energy emitted by the proton’s gravitational field comes from ?

    It is known that Tesla supposed that the energy of the radiactive nuclei comes from the Sun. And he has reason to be intrigued with the phenomenon. After all, the radiactive nuclei emit energy along billion years. It’s hard to believe that such energy was stored into a nucleus, which continues emitting it along billions years.

    I think Tesla was right. I suspect that the gravitational energy which replaces the energey emitted by the proton’s field comes from the Sun.

    I also think such gravitational energy coming from the Sun is responsible for the excess energy in cold fusion occurrence.

    As it’s impossible to have violation of the fundamental law of thermodinamics, then (when Rossi’s eCat will be definitivelly accepted by the scientific oomunnity) the physicists will be obliged to realize that the energy supplied by the eCat working in the self sustained mode comes from somewhere.

  • Andrea Rossi

    Dear Robert Tanhaus:
    I am not expert of hydrogen production, but maybe a good idea.
    Warm Regards,
    A.R.

  • Robert Tanhaus

    Dear Mr. Rossi,

    You said:
    Dear Rita Ekholm:
    up to 2 years, mainly for authorizations for heating systems; 20 years for cars.

    But: What do you think of producing hydrogen with a MW e-cat and use it as fuel for hydrogen powered cars? That should be possible right now! What do you think?

    Best regards,
    Robert Tanhaus

  • Andrea Rossi

    Dear Fab70:
    Thank you for your suggestions, of course we made deep meditations on this issue. We prefer to remain the manufacturers.
    Warm Regards,
    A.R.

  • fab70

    Dear Mr. Rossi

    An important element in evaluating an investment is the consistency of the different elements that are part of the big picture. In this case data are extremely promising but the operating mode is, at least, not the most effective. Few thoughts:

    – You are the only one who achieved a very elevated COP with a LENR reaction

    – The patent application at the international level is, if I understand correctly, blocked because of incomplete description. Resubmit the application making public the exact functioning of the E-Cat would most likely secure you the patent and also the economic benefits of this discovery.

    – The disclosure of how the E-Cat works would allow a quick confirmation by scientists around the world and then an extremely rapid adoption of this technology at a worldwide level

    – The rapid spread of E-Cat would mean for you a very rapid increase of patent-related rights, far superior to the gains that you could realize with the direct production strategy that you have chosen (we are talking about millions and not thousands of units).

    – In any case, even if patent process would not end positively (unlikely given its importance), fame and glory would make you both economically and morally one of the richest person in the world.

    So way following this difficult road (direct production) instead of following the easiest, more remunerative and quickest way (patent + licensing)?

    Grazie e in bocca al lupo!

  • Andrea Rossi

    Dear Al Potenza:
    The answer is yes.
    Warm Regards,
    A.R.

  • Mr. Rossi,

    I have no doubt your reactors will make it someday! Since when is the U.S. a land of conservatism and underachievers and not Godly visionaries full of hail mary’s (american football term, not the catholic term)? Oh, that’s right, the space shuttle was our last hail mary! As an American, I’m disappointed.

    I would enjoy meeting you face to face someday. I work as an engineer for a major US based airline, so if you have time to give some wisdom to a young engineer over some lunch, let me know where you are in the world and I’ll see if I can make a flight. I also worked at a PWR nuclear plant and wouldn’t mind some advice about switching back to a nuclear career.

    Also, I saw from a previous post that you are a Godly man. If that’s the case, everything will work out in your life, independent of your reactors. As a young man, I’ve at least discovered that much!

    Gods blessings,
    Seth Witte

  • Al Potenza

    Dear Mr. Rossi,

    Thank you for your response. I understand that you can not reveal R&D results that would compromise your trade secrets. But that’s not what I was requesting.

    When you deliver an ecat to a university, could you simply authorize the university to say publicly that they have received it and have started their work — as in a brief press release? That alone would help those of us who believe in you when we have to deal with skeptics. And it would not jeopardize any secrets.

    Best wishes,

    Al

  • Aussie Guy, thanks for the detailed calculation.

    So if you’re getting 40% efficiency, and since you’re limited to 110C (383.15 kelvin) by E-Cat safe operation range, you must be dumping into a cold medium which is at (1-0.40)*383.15 kelvin = 230 kelvin = -43.15 degrees Celsius. This heat sink must be your secret since that would require dumping into the antarctic in winter. Summer temperatures aren’t that low at the south pole.

    Either that or you’ve figured out a way around Carnot efficiency.

  • Andrea Rossi

    Dear lenr4you:
    No, not possible,
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Al Potenza:
    As I said already many times, puppett snakes are not an issue for us. Our Customers are an issue for us. R&D results are confidential, and will remain confidential.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Brad:
    Not so far.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Rita Ekholm:
    up to 2 years, mainly for authorizations for heating systems; 20 years for cars.
    Warm Regards,
    A.R.

  • Rita Ekholm

    Best Dr. Rossi
    First, I congratulate you for your wonderful work
    Have followed your work with great interest and hope that E-cat may soon be in sales.
    How long do you think it will take before we here in the cold north to warm our houses with E-cat
    What if E-cat could also be used in cars

  • Brad

    Dear Mr. Rossi,
    Do you currently have relationships with any US universities?

    Will you be looking for physicists in the Manchester, NH location?

    Best regards,
    Brad

  • Al Potenza

    Dear Mr. Rossi,

    I understand why your work with the universities needs to be secret to protect your invention. But could you authorize the universities to tell the public when they have received an ecat? And can they also announce when they have completed preliminary tests of the ecat and confirmed that it works well? That would not compromise any secrets but it would immediately demolish the pathological skeptics, including the snakes and clowns! Could you do that? And if so, how soon?

    Best wishes for your success,

    Al

  • lenr4you

    Dear Mr. Rossi,

    I have one simple questions:

    Is it possible to cool the NiH Reactor with his own hydrogen gas inside?
    You could split the Reactor core from the cooling system.
    Temperatures can be higher and the cooling power and density will be much greater.

    Thank a lot to you and your team

    Warm Regards
    LENR4you

  • Charlie Zimmerman

    Mr. Rossi,

    I understand completely. You can’t fault a guy for trying:) Again, congratulations on your work and best of luck in the future.

    Best Regards,
    Charlie Zimmerman

  • Andrea Rossi

    Dear Charlie Zimmerman:
    You are entering in the mined field of industrial secrets.
    Warm Regards,
    A.R.

  • Charlie Zimmerman

    Dear Mr. Rossi,

    These are very exciting times indeed! Thank you for your attention to my previous questions.

    Regarding the remaining NI58, am I correct in assuming that NI58 does not react at all? I ask because if it reacts similarly to the NI62 and NI64 I see no way to avoid the radioactive byproduct of NI59 decayed from CU59.

    Is the reason for the reduction in NI58 simply to have a higher ratio of the reactive isotopes NI62 and NI64 or does reducing it somehow reduce it’s reactivity?

    Warmest Regards,
    Charlie Zimmerman

  • Andrea Rossi

    Dear Charlie Zimmerman:
    1- just reduced
    2- not eliminated
    3- no
    4- no
    5- yes
    6- yes
    7- I cannot answer to this question, until I will disclose the theory of the effect we get.
    Warm Regards,
    A.R.

  • Charlie Zimmerman

    Dear Mr. Rossi,

    Congratulations on the demonstration and sale of the 1MW plant. I am sure many great things are to come for you and the world. I am also super excited to hear more about the theory that you have developed regarding this process. I think you mentioned that you would be revealing this after the 1MW demonstration.

    I has a few isotopic questions.
    1) You said that NI58 is depleted. Does this mean that it is eliminated or just that the ratio is reduced?
    2) If NI58 is eliminated, why is it eliminated? Does it react and you are eliminating it to avoid long half life byproducts (NI59 decayed from CU59)?
    3) Is (2) inconsistent with your statements that only NI62 and NI64 react?
    4) Significant enrichment of the Nickel for NI62 and NI64 is necessary to produce 30% transmuted copper. Do you agree?
    5) I have argued that you are not claiming cheap isotopic enrichment but rather that you are saying that the isotopic enrichment is not expensive relative to the overall costs of the production of the powder. Is this correct?
    6) Is Leonardo Corp doing the enrichment?
    7) Finally, Prof. Focardi in a recent interview talked about all nickel reacting and a series of decays which seems inconsistent with your statements of only NI62 and NI64 reacting to produce stable copper. Are you guys in agreement about the process?
    Thanks,
    Charlie Zimmerman

  • Andrea Rossi

    Dear Gabriele B.:
    USA and Europe.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Marco:
    Not so far. We need a couple of years to complete the authorizations.
    Warm Regards,
    A.R.

  • Dear ing. Rossi,

    in addition to 1MW plant, are you selling single modules (e.g. with total power from 5 – 20 KW)?

    Best ragards
    Marco

  • Andrea Rossi

    Dear John James:
    As I already said , we so far are ready only to sell 1 MW plants to make heat (steam, hot water, hot air, cool air): for electric power production and for household applications we are not ready yet, respectively for technological and for certification problems. We need from 1 to 2 years to resolve these problems.
    Warm Regards,
    A.R.

  • Gabriele B.

    Dear Ing. Rossi,
    You told further customers are buying your 1MW E-cat. Can You tell us what nations they are from? Are they Americans too or what else?

    Thank you.

  • Andrea Rossi

    Dear Propagare:
    The scientist who made the radiations control in our plants is Dr David Bianchini, from the University of Bologna.
    Warm Regards,
    A.R.

  • propagare

    Dear Mr Rossi,

    Please tell us/me :

    Who is the radiation scientist from Univ. Bologna. He has glasses, grey hair?

    Here he is in an important video about the E-Cat: http://www.youtube.com/watch?feature=player_detailpage&v=Kl6FOVnnIro#t=376s

    Thank you in advance

    propagare

  • John James

    Andrea Rossi,

    If about a billion E-cats, at x billion dollars, are needed to supply electricity to planet,
    how soon can orders be delivered along with retrofits to convert current electricity generation systems over to E-cats?
    Also, are there any plans for updates to designs of E-cats, like perhaps larger scale E-cats??

    Thanks, Yours,
    John James

  • Lou Tengzelius

    Dear Eng. Rossi,
    Any experienced individual familiar with heated hydrogen pressurized reaction chambers uses the finest powdered reactants in most pure form available with carbon powder as an oxygen scavenger to clean up and sensitize surfaces of contents and powdered iron to promote reactive species of hydrogen. Work is always done within dry boxes with inert atmospheres. Ball milling is also accomplished with special care. Your success builds upon previous state of the art. Congradulations Sir. Lou Tengzelius

  • The following article has been posted to PESN.

    Rossi’s E-Cat Victory on Cold Fusion’s Emergence Day — E-Day

    The date October 28, 2011 will be recorded in history as the day when Andrea Rossi’s cold fusion E-Cat technology emerged victoriously into the commercial marketplace, after an important test by a yet-undisclosed customer. Test parameters discussed. Move over oil, coal, and uranium… Ni-H cold fusion technology is coming to town!

    Click the following link to read the article.

    http://pesn.com/2011/11/02/9501943_Rossis_E-Cat_Victory_on_Cold_Fusion_Emergence_Day–E-Day/

  • Andrea Rossi

    Dear Paul Calvo:
    The price can drop to 500 $/kW if we will have to produce millions of pieces.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear John James:
    Thank you, same to you!
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Neil Ferguson:
    The names of Customers unavoidably will pop up, before or later.
    I agree with you.
    Warm Regards,
    A.R.

  • tj

    To James Bowery —

    The first hand built Mega-cats are priced at 2,400 Euro per KW.

    By comparison, new coal power plants cost about $3,500 per kW including the generator.

    If Rossi is able to reduce the cost of his E-Cat plants by an order of magnitude (and this is very likely given the simplicity of the E-Cat cores) then the capital cost of mass produced E-Cat power plants, even with your generators, will likely be just a fraction the cost of coal power plants.

    But its the E-Cat’s very low fuel costs and absence of externalized costs that makes it shine.

    Coal to generate 1 megawatt of power for a year currently costs $145,858.90 (3.41200 Btu/W x $2.44 per MMBtu of coal x 2x for best case 50% efficiency x 24 x 365 = $145,858.90)

    By comparison a year’s supply of nickle and hydrogen fuel for a 2 megawatt heat E-Cat would cost a tiny fraction of that $145,858.9o price for coal.

    Coal also inflicts externalized costs of mining, air pollution, climate disruption, and toxic ash disposal. All of these externalized costs severely damage human health and planetary sustainability.

    According to a recent Harvard study the real cost of using coal is several time higher than market prices would indicate.

    Stable climate, clean air, and abundant energy really have value beyond calculation.

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