.
by
Stoyan Sarg Sargoytchev
York University, Toronto, Canada
.
Read the whole article
Download the ZIP file
.
Abstract
Advances in the field of cold fusion and the recent success of the nickel and hydrogen exothermal reaction, in which the energy release cannot be explained by a chemical process, need a deeper understanding of the nuclear reactions and, more particularly, the possibility for modification of the Coulomb barrier.
The current theoretical understanding does not offer an explanation for cold fusion or LENR. The treatise “Basic Structures of Matter – Supergravitation Unified Theory”, based on an alternative concept of the physical vacuum, provides an explanation from a new point of view by using derived three-dimensional structures of the atomic nuclei.
Dear Mr. Rossi you will be aware that Cold Fusion seems to be very close to some kind of acceptance and the possibility of the 25 years of neglect by main-line science to end.
Is there anything that yourself or your company can do or say at this time that would help to give it that final push into generally excepted Research.
For those in need of low cost power to help for instance in the supply of safe drinking water any more delay is life threatening to many people.
All best wishes.
orsobubu wrote in December 1st, 2014 at 7:18 PM
Dear Wladimir,
2 –—————————————–
Does the book include the arguments treated in your new paper waiting for publishing here on JONP?
———————————————
NO
3 –—————————————–
I understood the book will only be published as open access PDF file, by Dr. Prakash Somani publishing house. So there will not be a paper printed edition?
———————————————-
It seems Dr. Prakash wished to publish a small printed edition of about 100 volumes.
But I dont know his final decision.
Besides,
dear Orsobubu,
I dont understand what is going on.
Dr. Prakrash told the book would be published until the end of 2014.
The last exchange of email by us was the following:
====================================================================
On Saturday, September 20, 2014 8:34 AM, Wladimir wrote:
Hi, Dr. Prakash
did you give up to publish the books ?
regards
wlad
.
Date: Sat, 20 Sep 2014 02:53:14 -0700
From: psomani1@yahoo.com
Subject: Re: did you give up ?
To: wladimirguglinski@hotmail.com
Dear Author,
Thank you for reaching to me. I will live to my promise. Your books are under typesetting and I will get back to you soon.
Your books will be published before the end of this year (2014) SURELY. I will arrange to send you the proof – for corrections (if any) soon. We are still working with the Figures of your book.
With best regards
Dr. Prakash Somani
====================================================================
4 –——————————————-
You said that in the book will be something too on plagiarism of your discoveries in shape of nuclei and z-axis. But I remember that there was, here on JONP, an exchange of views with Stoyan Sarg in this regard, where he claimed to have made these discoveries prior to yours, and thus concluding:
“In 2001 I Submitted My BSM theory to the Canadian Office of Intellectual Property with a claim for discovery of new models of atoms. I received a document in 2001, but to be surer, in 2002 I archived the full electronic version of the theory and the Atlas of Nuclear Atomic Structures in the AMICUS database of the National Library of Canada, where it Obtained and international catalog reference number . Once done, the dated electronic version can not be changed, so it serves as a legal proof of the dates of my intellectual property. The deposited electronic versions are publicly accessible as I show below.All These archives have assign ISBN number like books. In May 2002, I published the first scientific article about my theory in the official physical archive operated by Cornell University http://lanl.arxiv.org/abs/physics/0205052”
But maybe I missed the follow. Can you summarize what is currently your position on this issue?
———————————————-
yes, I can summarize, dear Orsobubu.
If the nuclear model by Dr. Sarg is correct, he does not need to be afraid. Nothing can take off the merit of his theory. His prediction of the z-axis will be one among several correct predictions.
But if his nuclear model is wrong, his prediction on the z-axis loses any merit as prediction.
The same is applied to my nuclear model.
I did not discover an z-axis in the nuclei. Actually I have discovered a new nuclear model, where the distribution of protons and neutrons is along the z-axis.
5 –—————————————–
One last thing, what do you think of this article:
http://vixra.org/pdf/1408.0109v4.pdf
A post by Axil against Standard Model appeared on the blog of Mats Lewan says:
“From the summary: Mass generation via chiral symmetry breaking and the Higgs becomes irrelevant for two independent reasons. First, in the absence of the weak force there is no need for massive gauge bosons. And second, the chiral impedance is scale invariant, cannot communicate energy but rather only quantum phase, cannot deliver mass. Similarly, mass generation in QCD via dynamic chiral symmetry breaking is seen to be not possible in light of the scale invariance of chiral impedances. In the impedance approach the origin of mass is the energy in the fields of the coupled modes represented in the impedance network and connected by impedance mis-matches. The calculated mass of the electron is correct at the nine signi cant digit limit of experimental accuracy, the muon at a part in one thousand, the pion at two parts in ten thousand, and the nucleon at seven parts in one hundred thousand.
This says that there is no Higgs field. The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson. I have been postulating EMF condensation into particles as an important mechanism in LENR. This quantum impedance idea supports that belief. Energy gain in LENR is a energy balancing and transfer process in an EMF based energy transfer mechanism. The discovery of the “God particle” has been brought into question recently as a misidentification. I believe that when LENR is accepted, the standard model will need a rework to get rid of the Higgs field.”
———————————————————
Dear Orsobubu,
Higgs theory was debunked by the detection of the aether by experiments in 2011, published in the journal Nature.
Higgs developed his theory by believing that the space is empty, and therefore the space cannot have structure, so that to interact with the particles, in order to give them mass.
His theory is superfluous face to the existence of the aether.
But it’s good to see that Higgs theory is not satisfactory even from the theoretical viewpoint.
Even some academic physicists are already questioning the Higgs theory, saying that the boson found in the LHC is not a Higgs-Boson which gives mass to particles, but it is actually only one more particle as many other produced by the collision of protons.
regareds
wlad
Dear Dr Andrea Rossi:
Are the licenses that Leonardo Corporation bought back from the former licensees for sale?
Warm Regards,
DTravchenko
How are going the experiments with the gas fueled Ecat?
JCR
To the Readers:
is somebody able to explain what is in Physics the “Hierarchy problem” ?
Thanks,
W.G.
orsobubu,
Axil writes,
“The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson.”
I contend that there is no amount of “condensing” of a field that will form any property of a particle, including mass. Further, no type of manipulation of a field will ever create a particle since a field is of a wave nature. And wave and particle natures are mutually exclusive. I argued this point a little while ago here on the JoNP against Quantum Field Theory.
All the best,
Joe
Wladimir,
1. You write,
“The electron can change its intrinsic-spin simply changing its original position up to a new position upside-down.”
What you state here is essentially what I stated in my last post. The effect of being “upside-down” is that the movement of the orbit ring is now reversed.
2. I repeat my last question, what is the specific mechanism that flips the spin of the electron when it crosses the plane of the 1p1 rotation?
All the best,
Joe
<b<Efficiency of emitters and receptors in Rossi’s eCat
In the figures showing the structures of nuclei, the protons and neutrons are not at rest. They oscillate quickly, due to repulsions.
When the nuclei are aligned along an external magnetic field, and their z-axis is aligned with that field, the protons and neutrons start to oscillate toward the z-axis direction.
With the application of an oscillatory electromagnetic field, the oscillation of the neutron in the 3Li7 increases its amplitude, and so it helps the magnetic force (applied by the orbit of the electron p1 in the 3Li7) to extract that neutron to send it toward the 28Ni nucleus.
So, the efficiency of an emitter depends on some conditions of the cold fusion reactor, but the efficiency also depends on some properties, as follows:
– How the neutron is bound in the emitter
– The nuclear magnetic moment of the emitter
– The size of the emitter compared with the size of the of the receptor (because the radius orbit of the electron p1 increases with the size of the receptor)
The magnetic moment of the emitter helps to get quickly the alignment between the emitter and the receptor.
3Li7 has magnetic moment +3,26
5B11 has magnetic moment +2,68
7N15 has magnetic moment -0,28 , but excited it becomes -2,4
3Li7 is better than 5B11 from other viewpoint: after the 3Li7 decay, the 3Li6 has small magnetic moment, +0,822. Then suppose 3Li7 supplies a neutron to a nucleus 60Ni, and it transmutes to 61Ni. Having small magnetic moment, the newborn 3Li6 leaves away the 61Ni, and a new 3Li7 couples with the newborn 61Ni, and the 3Li7 supplies a neutron to it. A cascade reaction occurs:
60Ni + 3Li7 -> 61Ni + 3Li6
61Ni + 3Li7 -> 62Ni + 3Li6
62Ni + 3Li7 -> 63Ni + 3Li6
Unlike, 5B11 after decaying transmutes to 5B10 which has magnetic moment +1,8, and so it will not leave away the newborn 61Ni so easily as 3Li6 does.
7N15 after decaying transmutes to 7N14 which has magnetic moment +0,403 , weaker than that of 3Li6. Therefore from such viewpoint 7N15 is better than 3Li7.
3Li7 has also a big unbalance of masses, which can contribute for the extraction of the neutron.
Other factor wich can increase the efficiency is the quantity of stable isotopes.
However all the emitters have one unique stable isotope:
3Li7, 5B11, 7N15, 9 F19, 11Na23, 13Al27, etc.
But regarding the efficiency of the receptors, the number of stable isotopes can have a strong influence.
Let us analyse the elements suitable to be receptors, and their stable isotopes.
The elements marked with “(*)” are those ones which have more than 5 cascades.
For instance, 22Ti has an uninterrupted cascade from Ti46 until Ti50.
20Ca = Ca40, Ca42, Ca43, Ca44, Ca46
(*) 22Ti = Ti46, Ti47, Ti48, Ti49, Ti50
24Cr = Cr50, Cr52, Cr53, Cr54
26Fe = Fe54, Fe56, Fe57, Fe58
28Ni = Ni58, Ni60, Ni61, Ni62, Ni64
30Zn = Zn66, Zn67, Zn68, Zn70
32Ge = Ge70, Ge72, Ge73, Ge74
34Se = Se74, Se76, Se77, Se78, Se80
36Kr = Kr78, Kr80, Kr82, Kr83, Kr84
38Sr = Sr84, Sr86, Sr87, Sr88
40Zr = Zr90, Zr91, Zr92, Zr94
(*) 42Mo = Mo92, 94Mo, Mo95, Mo96, Mo97, Mo98
(*) 44Ru = Ru96, Ru98, Ru99, Ru100, Ru101, Ru102, Ru104
46Pd = Pd104, Pd105, Pd106, Pd108, Pd110
48Cd = Cd106, Cd108, Cd110, Cd111, Cd112, Cd114
(*) 50Sn = Sn112, Sn115, Sn116, Sn117, Sn118, Sn119, Sn120, Sn122, Sn124
52Te – Te120, Te123, Te124, Te125, Te126
As we see, 50Sn has a cascade uninterrupted from Sn115 until Sn120. Beyond the cascade, it has yet 3 isotopes able to have cold fusion: Sn112, Sn122, Sn124.
Therefore, it seems 50Sn would be the better element to supply the higher COP for the eCat.
Unfortunately 50Sn has a big quantity of protons, and perhaps it cannot get cold fusion by using 3Li7 as emitter (the orbit of the electron p1 of the 3Li7 will be very large, and the neutron will be thrown with too much energy, and the neutron will trespass the 50Sn nucleus).
But perhaps 50Sn can get cold fusion with other emitter, as for instance 11Na23.
The sodium 11Na23 has a structure similar to the 3Li7, as shown in the figure ahead:
http://peswiki.com/index.php/Image:3Li7-and-11Na23-SIMILARITIES.png
They both have a big unbalance of mass, which contributes for the expulsion of the neutron.
They both have spin 3/2 (in Na23 the neutron is also weakly bound, because it is bound via spin-interaction with the deuterons D-2 and D-3, and they are far away.
11Na23 has also a good magnetic moment: +2,21
After emitting a neutron, 23Na transmutes to 22Na, which is no stable, and transmutes to 22Ne. 22Ne has magnetic moment zero, and therefore it leaves away the newborn receptor 50Sn(+n), which can start a new cascade with another 11Na23.
Joe wrote in December 1st, 2014 at 5:49 PM
what mechanism is responsible for reversing this movement of the electron along its own orbit ring?
—————————————–
Joe,
I think you did not understand how the electron changes its spin.
With the electron does not occur “reversion” of movement, like for instance occurs with the four wheel of a car, when it is moving ahead, and the driver stops the motion, and he puts the car moving in a reverse gear (in this case the four wheels of the car gyrate in reverse).
The electron can change its intrinsic-spin simply changing its original position up to a new position upside-down.
regards
wlad
Joe wrote in December 1st, 2014 at 5:49 PM
Wladimir,
You write,
“The two orbits continue being parallel.
The intrinsic-spin of the electron in the neutron changes from up to down.”
In QRT, the intrinsic spin of a particle is defined as its rotation about its central axis. Therefore, what mechanism is responsible for reversing this movement of the electron along its own orbit ring?
——————————————–
Joe,
the intrinsic-spin of the electron can be changed by several sort of interactions. Even the interaction with the flux n(o).
See Figure 9 in the page 12 of the paper Stability of the Light Nuclei:
http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf
regards
wlad
Dear Wladimir,
in March I asked you about the publication of the translation of your book “The Missed U-Turn”, treating the story of the evolution of physics, according to your vision of alternation of Newton and Descartes methods, up to include Rossi’s E-Cat and your Quantum Ring Theory models.
1 – You said more recent discoveries have been added, about aether, the mass of particles and neutron halo, together with references to plagiarism cases by Nature and European Physical Journal. You said that you do not intend to include proposals explaining LENR. It seems to me that in these last months you elaborated quite a significant amount of material on LENR, so I wonder now if you plan to include something about this.
2 – Does the book include the arguments treated in your new paper waiting for publishing here on JONP?
3 – I understood the book will only be published as open access PDF file, by Dr. Prakash Somani publishing house. So there will not be a paper printed edition?
4 – You said that in the book will be something too on plagiarism of your discoveries in shape of nuclei and z-axis. But I remember that there was, here on JONP, an exchange of views with Stoyan Sarg in this regard, where he claimed to have made these discoveries prior to yours, and thus concluding:
“In 2001 I Submitted My BSM theory to the Canadian Office of Intellectual Property with a claim for discovery of new models of atoms. I received a document in 2001, but to be surer, in 2002 I archived the full electronic version of the theory and the Atlas of Nuclear Atomic Structures in the AMICUS database of the National Library of Canada, where it Obtained and international catalog reference number . Once done, the dated electronic version can not be changed, so it serves as a legal proof of the dates of my intellectual property. The deposited electronic versions are publicly accessible as I show below.All These archives have assign ISBN number like books. In May 2002, I published the first scientific article about my theory in the official physical archive operated by Cornell University http://lanl.arxiv.org/abs/physics/0205052”
But maybe I missed the follow. Can you summarize what is currently your position on this issue?
5 – One last thing, what do you think of this article:
http://vixra.org/pdf/1408.0109v4.pdf
A post by Axil against Standard Model appeared on the blog of Mats Lewan says:
“From the summary: Mass generation via chiral symmetry breaking and the Higgs becomes irrelevant for two independent reasons. First, in the absence of the weak force there is no need for massive gauge bosons. And second, the chiral impedance is scale invariant, cannot communicate energy but rather only quantum phase, cannot deliver mass. Similarly, mass generation in QCD via dynamic chiral symmetry breaking is seen to be not possible in light of the scale invariance of chiral impedances. In the impedance approach the origin of mass is the energy in the fields of the coupled modes represented in the impedance network and connected by impedance mis-matches. The calculated mass of the electron is correct at the nine signi cant digit limit of experimental accuracy, the muon at a part in one thousand, the pion at two parts in ten thousand, and the nucleon at seven parts in one hundred thousand.
This says that there is no Higgs field. The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson. I have been postulating EMF condensation into particles as an important mechanism in LENR. This quantum impedance idea supports that belief. Energy gain in LENR is a energy balancing and transfer process in an EMF based energy transfer mechanism. The discovery of the “God particle” has been brought into question recently as a misidentification. I believe that when LENR is accepted, the standard model will need a rework to get rid of the Higgs field.”
Many thanks.
Wladimir,
You write,
“The two orbits continue being parallel.
The intrinsic-spin of the electron in the neutron changes from up to down.”
In QRT, the intrinsic spin of a particle is defined as its rotation about its central axis. Therefore, what mechanism is responsible for reversing this movement of the electron along its own orbit ring?
All the best,
Joe
Daniel De Caluwé wrote in December 1st, 2014 at 7:35 AM
Dear Wladimir,
Interesting hypothesis based on your theory, but, unfortunately, except A. Rossi, who signed a non disclosure agreement, there’s nobody else at the moment who can test it.
—————————————————–
Dear Daniel
along years Andrea Rossi had tested many elements, and he knows what emitters do not produce cold fusion.
So, Rossi already has intuition on the question whether my theory can be correct, or not.
Obviously he will say nothing, not only because he decided to disclosure nothing about his reactor , but also because if he says the theory can be correct, by this way he is giving a tool for his competitors so that to look for the best combination of emitters and receptors.
regards
wlad
Dear Wladimir,
Interesting hypothesis based on your theory, but, unfortunately, except A. Rossi, who signed a non disclosure agreement, there’s nobody else at the moment who can test it.
Kind Regards,
Daniel.
Joe wrote in November 30th, 2014 at 4:02 PM
Wladimir,
Are you saying that, before the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is parallel to the orbit of the 1p1 electron; and that after the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is anti-parallel to the orbit of the 1p1 electron?
—————————————–
of corse not.
The two orbits continue being parallel.
The intrinsic-spin of the electron in the neutron changes from up to down
regards
wlad
ERRATA:
In my last commment, where it written:
But with the aim of verifying the present theory, would be of interest to test 7N15 as receptor.
the correct is:
But with the aim of verifying the present theory, would be of interest to test 7N15 as emitter.
Substitute for 3Li7 and 28Ni in Rossi’s eCat
The fuel used in the eCat is composed by an emitter 7Li of neutrons together with Ni isotopes receptors of neutrons.
The emitter 7Li has three electron orbits:
a) two orbits of electrons s1 and s2, which cancel each other their magnetic fields
b) one orbit of electron p1, perpendicular to the z-axes of 7Li, whose magnetic field extracts the neutron of the nucleus 7Li
FIG. 6:
http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png
The radius of the orbit p1 depends on the element used as receptor.
a) By using a receptor with big quantity of protons, the orbit will be larger, and the force of extraction is stronger.
b) By using a receptor with low quantity of protons, the orbit will be smaller, and the force of extraction is weaker.
.
1- Beryllium 9Be isotope used as emitter
Beryllium has only one stable isotope, the 4Be9.
Figure 7 shows the orbits s and p.
FIG. 7:
http://peswiki.com/index.php/Image:FIGURE-7-substitute3Li7-28Ni.png
The orbits s1 and s2 cancell each other their magnetic moments.
The orbits p1 and p2 also cancell each other.
CONCLUSION: Cold fusion cannot be obtained by using Be as emitter
Any element having a pair number of protons cannot be used as an emitter of neutrons in cold fusion reactors.
.
2- Boron isotopes used as emitter
The stable boron isotopes are 5B10 and 5B11. They occur naturally as follows:
5B10 – 19,9%
5B11 – 80,1%
The isotope 5B10 is not of interest to be used as emitter, because it has not a neutron bound with a deuteron via spin-interaction.
The good candidate to replace 3Li7 is 5B11, because they have similar structures, since they both have a neutron bound with a deuteron via spin-interaction, as shown in the Fig. 8 ahead.
FIG. 8:
http://peswiki.com/index.php/Image:FIGURE-8-substitute3Li7-28Ni.png
The structure of the 5B11 is also shown in the page 60 Fig. 44 of the paper Stability of Light Nuclei
http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf
The magnetic force for the extraction of the neutron will be due to the orbit of the electron p3, shown in Figure 9.
FIG. 9:
http://peswiki.com/index.php/Image:FIGURE-9-substitute3Li7-28Ni.png
As the neutron in the 5B11 is strongly bound than the neutron in the 3Li7, a stronger magnetic force due to the electron p3 orbit will be required.
The 28Ni can be replaced by other receptors, in order to verify if the COP of the eCat can be increased.
The following receptors could be used if their partnership with 5B11 would be able to extract the neutron from the 5B11 emitter:
20Ca
22Ti
24Cr
26Fe
30Zn
32Ge
34Se
36Kr
38Sr
However, probably the radius orbit of the p3 electron will not be sufficiently large so that to induce a magnetic force capable to extract the neutron.
There is need to choose elements with higher number of protons.
We have to expect that a good receptor to work with 5B11 must be found between Z=40 and 50:
40Zr
42Mo
44Ru
46Pd
48Cd
50Sn
.
3- Nytrogen isotopes used as emitter
The two stable isotopes of nitrogen are 7N14 and 7N15.
The isotope 7N14 is not of interest to be used as emitter, because it has not a neutron bound with a deuteron via spin-interaction.
The structure of the stable 7N15 is shown in the Figure 10, and it could be used as emitter.
FIG. 10:
http://peswiki.com/index.php/Image:FIGURE-10-substitute3Li7-28Ni.png
Unfortunatelly, 7N15 naturally occurring is only 0,37% , while 7N14 in 99,63%, and so nytrogen is not commercially viable to be used as emitter.
But with the aim of verifying the present theory, would be of interest to test 7N15 as receptor.
Wladimir,
Are you saying that, before the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is parallel to the orbit of the 1p1 electron; and that after the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is anti-parallel to the orbit of the 1p1 electron?
All the best,
Joe
Andreas Moraitis wrote in November 30th, 2014 at 11:48 AM
Dear Wladimir,
In your reply to Joe you wrote: „[…] my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).”
Does the difference between the mass of the neutron and the sum of the masses of the proton and the electron (about 782 keV/c^2) result from the relativistic speed of the electron?
————————————————
Yes,
and it is calculated in my paper Anomalous Mass of the Neutron, published in JoNP:
http://www.journal-of-nuclear-physics.com/?p=516#more-516
regards
wlad
Ing. Michelangelo De Meo:
Thank you for the interesting information.
Warm Regards
A.R.
Dear Wladimir,
In your reply to Joe you wrote: „[…] my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).”
Does the difference between the mass of the neutron and the sum of the masses of the proton and the electron (about 782 keV/c^2) result from the relativistic speed of the electron?
Best regards,
Andreas Moraitis
Hello Dr. Rossi,
Italy is investing in the construction of expensive hydroelectric power stations in Montenegro (Albania ) for the production of electricity . Via a high-voltage submarine cable , connecting the two shores of the Adriatic Sea , will lead the electric current in Abruzzo and Puglia . Italy continues to squander money on ” clean energy ” . I hope that his plant Hot Cat is the solution to remedy this waste impactful . With the hope that it goes as soon as possible on the world market .
http://www.primadanoi.it/news/mondo/554816/Elettrodotto-in-Abruzzo–la-lunga.html
Joe wrote in November 29th, 2014 at 10:54 PM
Wladimir,
In your model, does the neutron ever change the direction of its spin? If so, at what point does it do so in its travel from source nucleus to target nucleus?
——————————–
Joe,
we are obliged to conclude that yes, the neutron needs to change its spin.
It changes its spin just after crossing the cross section of the p1 electron orbit.
I think the neutron do it as consequence of the least action principle.
As you know, my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).
The proton has magnetic moment +2,793
The neutron has magnetic moment -1,913
So, the orbit of the electron about the proton yields the magnetic moment -(1,913+2,793) = -4,706.
Before crossing the cross-section of the electron orbit p1, the neutron is moving having its electron moving with a parallel orbit with the orbit p1, and intrinsic spin up.
When the neutron crosses the cross-section, the electron of the neutron changes its intrinsic-spin to down. So, the proton of the neutron is constrained to also change its spin, and therefore the neutron starts to move with a contrary spin (contrary to the spin the neutron had before to cross the orbit p1).
Saying the contribution of the least action principle in other words: it is most confortable for the neutron to change its spin after crossing the cross-section of the orbit p1.
regards
wlad
Herb Gillis wrote in November 29th, 2014 at 11:09 AM
Wladimir Guglinsky:
Would you necessarily need to have an even number of protons if you could otherwise obtain an even number of electrons? For example; many elements form stable ions which have odd numbers of protons but even numbers of electrons (for example; halides like chlorine). Do you foresee the possibility of LENR in an ionic solid (or liquid) containing one of these ions?
——————————
Dear Herb,
positive ions with even quantity of electrons will probably capture one electron, and they will have odd quantity of electrons. So cold fusion will not occur.
In the case of negative ions with even quantity of electrons cold fusion would be possible if the excess electron is not expelled
However we have no guarantee that the excess electron of the ion will not be expelled due to repulsion with the electron p1 of the other nucleus aligned with the ion, since the excess electron is weakly bound to the ion.
regards
wlad
Wladimir,
In your model, does the neutron ever change the direction of its spin? If so, at what point does it do so in its travel from source nucleus to target nucleus?
All the best,
Joe
Joe wrote in November 29th, 2014 at 3:51 PM
And even if it would continue successfully past the plane, the neutron will experience the force of attraction when its South is pulled by the North of the electron’s orbit.
The situation is worse if the neutron’s vector is allowed to change its orientation at any time during its travel. The much stronger magnetic moment of the electron’s orbit will ALWAYS end up PULLING the neutron towards it, therefore effectively preventing any neutron transfer from occurring.
———————————–
Joe
what you say makes no sense
If there is force of attraction on the neutron when its spin is in clockwise direction, then if the neutron changes its spin in counter clockwise direction the force cannot continue to be of attraction
http://peswiki.com/index.php/Image:ATRA%C3%87AO_PROTON_NEUTRON.PNG
regards
wlad
Wladimir,
If both vectors, that of the neutron’s magnetic moment and that of the electron orbit’s magnetic moment, always stay in the same direction, there will be attraction at the beginning. But when the neutron is almost halfway across the plane of the electron’s orbit, repulsion will be experienced due to the North of the neutron’s vector approaching the North of the electron’s orbit vector. And even if it would continue successfully past the plane, the neutron will experience the force of attraction when its South is pulled by the North of the electron’s orbit. These two conditions should prevent a successful neutron transfer.
The situation is worse if the neutron’s vector is allowed to change its orientation at any time during its travel. The much stronger magnetic moment of the electron’s orbit will ALWAYS end up PULLING the neutron towards it, therefore effectively preventing any neutron transfer from occurring.
All the best,
Joe
Wladimir Guglinsky:
Would you necessarily need to have an even number of protons if you could otherwise obtain an even number of electrons? For example; many elements form stable ions which have odd numbers of protons but even numbers of electrons (for example; halides like chlorine). Do you foresee the possibility of LENR in an ionic solid (or liquid) containing one of these ions?
Regards; HRG.
Dear Dr. Rossi:
It is necessary to be fast with the delivery of your plants.
The Global Carbon Project of the Tyndall Centre for Climate Change Research published in the journal Nature the results of a study of 2012. This study shows that among the countries and regions that have emitted more carbon dioxide in 2011 are: China ( with the 28 percent of emissions ) , the US ( 16 percent ) , the European Union ( 11 percent ) and India ( 7 percent ) . It is estimated that , in 2012 , global emissions have increased by 2.6 percent , reaching a record high of 35.6 billion tons . China is the country with the largest amount of emissions , but on a per capita has a relatively low level , with only 6.6 tons per person , well below the 17.2 tons per capita issued by the United States . The European Union has issued 7.3 tons of CO2 per capita .
Italy wants to buy electricity from Montenegro and take it with a submarine cable .
Terna is developing plans for new submarine interconnections with the Balkans with the aim of contributing to the diversification of sources and areas of energy supply and the reduction of the price of electricity in Italy and to increase the safety levels of safety Italian electricity system .
http://www.terna.it/default/Home/AZIENDA/chi_siamo/terna_estero/Terna_nei_Balcani/interconnessioni_balcani.aspx
Joe,
I think only the nuclei with pair number of protons are suitable to transmute by cold fusion, because they also have pair number of electrons in the electrosphere.
Because of the pair number of electrons, each pair of electrons cancell each other their magnetic moment, and so they do not influence the penetration of the neutron.
While the nuclei with odd number of protons, have odd number of electrons, and therefore they always have one unpaired electron, and its magnetic moment do not allow the penetration of the neutron coming from other nucleus.
For instance, suppose Andrea Rossi had used a fuel composed by 29Cu and 3Li7 in his E-Cat. The neutron after exit the 7Li would not succeed to enter within the 29Cu nucleus, because the magnetic moment due to the orbit of the unpaired do not allow it.
It also seems that the best nuclei to get cold fusion is those ones with biggest nuclear magnetic moment, because they can have their z-axis aligned more quickly, and so the reactions will occur faster, and by this way a higher COP will be obtained.
From this viewpoint, 24Cr would be best than 28Ni.
But 24Cr has 24 protons, and therefore its positive electric field is smaller than that of 28Ni. Therefore the orbit radius of the electron p1 of the 7Li will be shorter (since the orbit of the 7Li is influenced by the electric fields of 24Cr and 7Li working together), and by this reason the magnetic moment of the p1 orbit will not be so efficient to extract the neutron of the 7Li.
In the case of a fuel 46Pd-7Li be used in the E-Cat, the radius orbit of the p1 electron of the 7Li will be too much large, and the neutron will be captured from the 7Li with too much energy (high velocity). Then, instead of being captured by the 46Pd nucleus, the neutron actually crosses the 46Pd without to be captured, and the 46Pd will have not transmutation.
Probably this is the reason why, after trying many elements, the tests made by Rossi have pointed that 28Ni is the best fuel to react with 7Li and to get the higher COP
regards
wlad
Frank Acland:
Thank you for the information,
Warm Regards,
A.R.
Dear Andrea,
You may be interested to see your work discussed in this article on the Science Blog at the Huffington Post:
Low Energy Nuclear Reactions: Papers and Patents
http://www.huffingtonpost.com/david-h-bailey/low-energy-nuclear-reacti_b_6189772.html
Best wishes,
Frank Acland
Other reactions in Fleischmann-Pons and Mosier-Boss experiments
After the formation of tritium by the first sandwich Pd-1H2-1H2, other sandwiches can be formed, as follows:
1) Pd-1H2-1H3
2) Pd-1H3-1H2
3) Pd-1H3-1H3
In these sandwiches 2He4 is formed
As it is not used an oscillatory magnetic field so that to excite the nuclei, 2He4 is never captured so that to forma sandwich, because 2He4 has no magnetic moment
Pd is not transmutted, because as the proton has charge, it is deviated by the electrons of the Pd electrosphere, and so the proton never hits the Pd nucleus
In Fleischmman-Pons experiment they did not use an external magnetic field, and so the nuclei were aligned by the Earth’s magnetic field.
In the days of magnetic storms of the Sun they did not succeed to replicate the results.
In Mosier-Boss experiment she used an external magnetic field, and so the nuclei were aligned by that field.
,
Regarding Rossi’s E-Cat
In Rossi’s E-Cat, probably in the first experiments he applied an oscillatory magnetic field unable to excite the nuclei, and therefore only Ni61 had transmutation, because it has magnetic moment -0,75.
But 58Ni and 60Ni had no transmutation, because they have magnetic moment zero, and so they cannot align their z-axis, and therefore they did not capture 7Li nuclei.
By this way Rossi obtained a very low COP.
By improving the oscillatory magnetic field, Rossi did succeed to excite the nuclei, and by this way 58Ni and 60Ni started to have transmutation, and by this way he got to increase the COP of the E-Cat.
Andrea Rossi could confirm it to us, but as he never gives any information about the reactor, he will tell nothing.
Fleischmann-Pons and Pamela Mosier-Boss experiments finally deciphered
Fleischmann-Pons experiment and Mosier-Boss experiment are practically the same.
But Mosier-Boss experiment has a special interest, because in her experiment he measured the energy of the neutrons emitted, in order of 10MeV, and such result is paradoxical, since the binding energy of the deuterons is only 2,2MeV.
Let us see the mechanism of the cold fusion reactions in the Mosier-Boss experiment.
1) A Pd nucleus couples with a deuteron, and they align their z-axes.
Therefore both the Pd and 1H2 lose their spherical Coulomb barrier.
2) The electron s1 of the deuteron takes an orbit perpendicular to the z-axes of the two nuclei Pd and 1H2
3) The deuteron has only one proton, and so its positive electric field is small. Therefore the radius orbit of the electron s1 is short. By consequence, the magnetic force produced by the s1 orbit is not able to remove the neutron tied to the proton in the deuteron.
4) The couple Pd-1H2 is stable, and nothing happens.
5) Then the couple Pd-1H2 capture a second deuteron, and then a sandwish is made: Pd-1H2-1H2
6) The electron s1 of the second deuteron takes an orbit perpendicular to the z-axis, between the two deuterons.
7) Therefore the first proton-neutron of the first deuteron (sandwihed by Pd and the second deuteron) start to have a high oscilatory motion, because:
7.a) The first deuteron is between two parallel orbits, one orbit of the electron s1 of the first deuterion, and other orbit of the electron s1 of the second deuteron
7.b) the positive charge of the proton of the first deuteron is attracted by two contrary directions along the z-axis direction
8) The neutron is tied to the proton in the first deuteron by the spin-interaction, which binding energy is 2,2MeV. When, in the oscillatory motion, the proton stops and changes the direction of its motion in contrary direction, due to the inertia the neutron has the tendency to continue its motion, and so the 2,2MeV binding energy is broken, and the neutron exits the first deuteron, and is captured by the second deuteron, and they form a tritium.
9) The capture of the neutron causes an unbalance of masses in the newborn tritium. So, the z-axis of the tritium starts to gyrate chaotically, and it gets back its spherical Coulomb barrier, and it leaves away the couple Pd-1H2-1H3
10) But in the instant when the neutron left away the first deuteron, at that the same time the proton is accelerated toward the contrary directionBut in the instant when the neutron left out The proton gets a high velocity, because free of the link with the neutron the proton is pushed by the repulsion with the positive electric field of the tritium.
11) The proton enters within the Pd nucleus with 2,2MeV, helped by the orbit of the electron s1 pulling it against the Pd nucleus.
12) Within the electrosphere of the Pd nucleus the proton captures an electron, and they form a neutron. That electron in the Pd electrosphere was moving with helical trajectory. As the electron loses its helical trajectory when it is captured by a proton and they form a neutron, the energy of the helical trajectory is transferred to the neutron. So, the neutron gets a portion of the energy 10MeV from the helical trajectory of the electron captured by the proton.
13) When the proton entered within the Pd electrosphere, it was accelerated due to its attraction with the electrons of the Pd nucleus. If the proton was to leave away the Pd electrosphere, it would leave it with the same 2,2MeV, because leaving it its velocity would be decreasing by the attraction with the electrons. However, when the proton becomes a neutron, it loses its charge, and when the neutron starts to leave away the Pd nucleus it is not decelerated by attractions with electrons of the Pd nucleus. Therefore the neutron receives the other portion of the energy 10MeV from the acceleration of the proton before to become a neutron.
Joe wrote in November 28th, 2014 at 7:30 PM
Wladimir,
Problem 1.
If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue.
===========================================
Dear Joe,
I already explained it to Eric:
————————————
However, in the case of the neutron, when the neutrons is moving toward the Ni nucleus, it can happen the following:
1) Before to arrive to the cross-section of the electron’s orbit, the magnetic field induced by the electron’s motion applies a force of ATTRACTION on the neutron (the two magnetic vectors point to the the same direction).
The neutron is PULLED by the electron’s orbit toward the Ni nucleus.
2) After crossing the cross-section, the magnetic field of the electron starts to apply a force of REPULSION on the neutron (because the two magnetic vectors continue with the same direction).
And so the neutron is PUSHED toward the Ni nucleus.
However, I did not mention it because I am not sure if the neutron changes its magnetic field regarding the electron’s orbit after crossing the cross-section of the electron’s orbit (in this case, if the neutron changes the magnetic field vector in the contrary direction, then the electron begins to apply a force of ATTRACTION on the neutron, after it crosses the cross-section, and therefore decreasing the speed of the neutron going by inertia toward the Ni nucleus).
———————————-
,
Problem 2.
Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)
=====================================
Joe,
due to Pauli’s Principle, only one electron can take the orbit taken by the electro p1 of the 7Li in the Figure 6 (orbit perpendicular to the z-axis)
FIG 6:
http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png
As the inner electron p1 of 7Li is more energetic than the outer electrons of the 58Ni, that orbit in the FIG 6 is taken by the electron of 7Li
So, the electrons of the 58Ni do NOT take an orbit PERPENDICULAR to the z-axis, and they actually cancell each other their magnetic moments (the vector magnetic moment for the electron p1 has contrary direction of the vector for the electron p2, and the vector for the electron d1 has contrary direction of vector for the electron d2, etc)
regards
wlad
regards
wlad
Dears Joe and Steven Karels
58Ni and 60Ni have magnetic moment zero
http://www.webelements.com/nickel/isotopes.html
Therefore, the oscillatory magnetic field applied within the reactor is used not only for shaking the nuclei, but also for exciting them, because a nucleus with magnetic moment zero cannot be aligned by a magnetic field
https://www.google.com.br/search?q=nucleus+excitation&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:pt-BR:official&client=firefox-a&channel=sb&gfe_rd=cr&ei=WzF5VIi6E43dwAT_rIKQDw
Excited 58Ni has magnetic moment -0,1, while excited 60Ni has magnetic moment +0,2
regards
wlad
Franco Sarbia:
Thank you for your suggestion. The hydrogen produced that way is too expensive due to the very low efficiency of the electrolysis. I have made production of hydrogen with water electrolysis, therefore know well the issue.
Warm Regards,
A.R.
Dear Dr. Andrea Rossi,
still about the Hydrogen fueled Hot Cat on which returned Dr Joseph Fine. There are on the market electrolysers with size of a computer case, secure and low price, which store hydrogen in metal hydride cylinders. In case of using batteries of Hot Cat for cogeneration,it would be enough a hydrogen reserve to start the system.Once in operation the electrolyzer would provide the gas needs to drive it. The whole system would therefore be independent of any supply network of cables or pipes. The strategic interest of this hypothesis is too important because you will not have even thought.
Warm Regards
Franco Sarbia.
Wladimir,
Problem 1.
If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue. The same force that pulled the nucleon out of the source nucleus will have a much easier time to stop the nucleon’s progress toward the target nucleus when the nucleon crosses the plane of rotation of the 1p1 electron of 3Li7. And when we add to this the matter of distance, like you mentioned, which makes the attraction to the target nucleus even weaker for the nucleon, then I do not see much success for this model.
Problem 2.
Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)
All the best,
Joe
I am trying to determine an approximate specific thrust using a thermally (1500 C ) pressurized gas.
I am trying to ballpark a given mass for the Rossi Ecat and electrical power source, say about 1000 Kg
and a liquid gas propellant that has a good expansion Coeffiecent with a contained mass also of 1000 Kg.
Once in space and turned on, what would be the approximate velocity at the depletion of the heated expanding gas?
And what is the resulting Specific Impulse value compared to say an ION engine….
Thanks to those who can calculate on this end of the spectrum.
Jeff Smathers
Tom Conover:
Very interesting, but not commercial yet.
It is R&D of the DOE.
Warm Regards
A.R.
Christopher Calder:
I cannot give this kind of information.
Warm Regards
A.R.
Dear Mr. Rossi,
I believe you stated years ago that you tried different metals to create the LENR effect, and only nickel really worked for you. I believe you also tried titanium powder, which has an interestingly high melting point of 1,668°C. That was with your old fuel formula used in the Warm-Cat (joke). Have you retried titanium powder with your new fuel formula in the Hot-Cat? Maybe you could play around with ways to create more nano sized cracks in the titanium lattice? Some scientists claim luck with titanium, and it would be fun to see if it reacted differently with the other elements of the new fuel mixture. Would it work at all? If it did work, would it be more stable?
Best Regards, Christopher Calder
Dear Andrea Rossi
Have you heard of this “Low-Cost Microchannel Heat Exchanger”?
Fabricated electronic cooling and high pressure MCHEX units to prove manufacturing approach:
Tests showed 400% higher heat transfer rates
80% reduction in volume
5,000 psi pressure capability
High effectiveness > 90%
80% lower estimated external heat loss
60% estimated lower cost
http://energy.gov/sites/prod/files/2014/06/f16/A2%20Poster-Altex%20AMO%20RD%20Project%20Peer%20Review%202014.pdf
Low-Cost Microchannel Heat Exchanger
DOE Grant DE-EE0004541
2013-2014
Hope this helps,
Tom
Dr Joseph Fine:
Prices of fuels are very unstable. So far gas is cheaper than any other fuel, included hydrogen. Besides, the use of hydrogen as a fuel makes more complicated the certification issue. Neverheless, your information is interesting.
Warm Regards,
A.R.
Andrea Rossi and Readers:
Ross Koningstein and David Fork, engineers at Google, described their work on the RE<C project in the Nov 2014 edition of "IEEE Spectrum".
http://spectrum.ieee.org/energy/renewables/what-it-would-really-take-to-reverse-climate-change
The RE<C project goal was to find methods to improve usage of Renewable Energy by reducing the cost of Electricity while reducing the amounts of CO2 gas and other pollutants going into the atmosphere by the burning of Coal to produce that Electricity.
Google shelved the RE<C project in 2011 when they realized that their goals were unrealistic. While renewable energy costs had been reduced, they were not reduced below the costs of burning Coal. Nor would it have had significant effect on Global production of CO2.
( Note: Instead, Energy Policy has raised the costs of burning coal rather than reducing costs of other sources. The result has been to increase the cost of Electricity. )
Google would probably be interested in helping develop applications that use "New FIre". (If they haven't contacted IH or you already.)
1) Can you run/drive an E-Cat with Hydrogen rather than Natural Gas? If so, 2) Is it cheaper to produce natural gas or Hydrogen?
Renewed regards,
Joseph FIne
Dear Andrea Calaon,
You responded:
“I see that silicon is present in the ash. Could the following reaction be possible?
27Al + 7Li + e -> 28Si + 6Li
…
Let me say. Hehehe. You noticed those lines at 28 on Figure 9 (lone) and 11 (with Al as well) of the ITPR.
Let me first say that the reaction you wrote is impossible…”
The reaction I was considering was in two steps:
27Al + 7Li + e -> 28Al + 6Li
while 27Al is stable, 28Li has a half-life of 2.24 minutes and decays to 28Si.
So, I think, this might explain the removal of the 27Al from the fuel to the ash.
Orsobubu, Koen Vandewalle:
Without that ” damned spam robot” this blog could not work. We receive about one thousand of spam-messages per day: can you imagine how could we check a thousand messages per day to select the spam?
Warm Regards,
A.R.
Koen Vandewalle, you said:
>PS: Andrea, soon or later, that damned spam-robot wille erase the entire J.O.N.P. and all knowledge about E-cat. I have no idea why you keep that monster.
hehe because the artificial intelligence of the little monster is under the strict surveillance of the IT guy inside Rossi’s Team, and the IT guy is firmly in charge and he rules he commands he knows everything and he makes tons of backups… I’m sure that, even in case of an ultimate catastrophical event, like the one depicted at the end of the film “Planet of the Apes”, the knowledge of the E-cat could not be erased forever, since the IT guy has organized everything to bring it up held tight in the hand of the Liberty Statue, so that a new Charlton Heston could discover it in this way hehe:
http://4.bp.blogspot.com/-OIF9h3JRpTM/Ti1kM3kFqQI/AAAAAAAAE8M/GIkgM4zivAY/s1600/planet-of-the-apes-statue-of-liberty-blu-ray-disc-screencap-hd-1080p-05.jpeg
Anyway, from a human science point of view, to me Artificial Intelligence, robotics, Quantum Computing, Singularity etc are even more fascinating than new energy technologies and also physics in general, but I don’t care what skeptics say, Rossi’s personal saga is unmatched!
Based on the Lugano Report the fuel consists of three components.
This analysis speculates on those components.
Known facts:
1. Page 43: “Figure 1. Three different types of particles from the fuel material”
2. Page 43: “Figure 2. SEI of two different types of particles from the ash material”
3. Page 41: “Note that Li cannot be detected using EDS”
Analysis
On page 44 of the report, the EDS analysis of the different fuel particles is shown.
Particle 1: Mostly nickel with some carbon and oxygen and a small amount of aluminum are seen.
Particle 2: This particle consists of mostly aluminum with some carbon and oxygen and small amounts of nickel and chlorine.
Particle 3: This particle contains mostly iron (Fe) with some oxygen and smaller amounts of carbon, silicon, chromium and manganese.
Particle Sizes:
From Figure 3, the particles sizes may be estimated. Particle 1 is around 50 to 100 microns; particle 2 is around 75 to 150 microns and particle 3 is 100 to 200 microns.
Hypothesis
Particle 1 is the nickel components with some cross contamination from the aluminum from particle 2. The oxygen is assumed to be from the air (oxides). Perhaps a trace of carbon either from the air (natural carbon dioxide) or as a small additive to remove the oxides at higher temperatures.
Particle 2 is assumed to be LiAlH6 (recall the lithium and hydrogen are not detectable using EDS). The nickel is assumed to be cross contamination from particle 1. Unknown where the chlorine came from but it is small relative to the aluminum – it could be an impurity from the LiAlH4 source.
Particle 3 is mostly iron with the same observation on the oxygen and the carbon. The silicon could be from the adhesive using to hold the sample while the chromium and manganese might be part of the iron sample (an alloy?).