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by
Valeriy Y.Tarasov
E-mail: vytarasov@yahoo.com
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Abstract
The h-space theory is a variant of unified physical theory – a theory of everything.
This theory was built de novo, as the existing physical theories are incompatible and so unsuitable for unification.
A new approach is needed, and has been developed by re-evaluating the definitions of primary physical concepts.
The starting point for the re-evaluation was the following equation – Et = mvL, where energy – E, time – t, length – L, mass – m, velocity – v.
Analysis of these physical concepts resulted in the construction of a unique equation of the primary concepts such as space, length, energy and velocity.
From this, models could be developed that explain all well-known physical phenomena.
In addition, h-space theory predicts phenomena rejected by the current mainstream theories, such as limits to gravitational and electrostatic interactions, and the possibility of cold fusion (as a consequence of the electric charge definition, a modification of Coulomb’s law and the definitions of elementary particles in h-space theory).
The final section of this article describes a number of experimental tests that could be used to verify the h-space theory.
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Read the whole article
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Wladimir,
Do you think it is possible that gravitational fluxes n(o) also exist in the electrosphere, and not just in the nucleus, of an atom? They would be much weaker in strength since their radius would be much greater than the radius within the nucleus. This might explain similarities between the organizational structures of the electrosphere and those of the nucleus, such as the presence of shells.
All the best,
Joe
Herb Gillis:
Sincerely, I have not enough knowledge of Hot Fusion to answer you.
Warm Regards,
A.R.
Steven N. Karels wrote in December 25th, 2014 at 10:16 AM
Wlad,
Element % by Weight
Nickel 55.0
Iron 38.9
Aluminum 4.3
Lithium (total) 1.2
Hydrogen (no Deuterium) 0.6
Total 100.0
———————————————–
Dear Steven,
then you are right.
it seems 3Li7 generation is via capture the proton of the hydrogen.
So, it seems the hydrogen has 3 contributions:
1) it plays a role similar to a catalyst
2) The oscillation of the proton can also play a second role: to help the excitment of the nuclei 58Ni, 60Ni, 62Ni, 64Ni
3) 3Li7 is generated by the hydrogen
regards
wlad
Andrea Rossi:
Do you think it might be useful to study the reactions of LENR under “hot fusion” conditions- – by using a Farnsworth Fusor? It would be relatively easy to construct such a device with a mix of hydrogen and Ni + Li vapors, and then look for isotope changes. Do you think the reaction might follow a different course than under “cold fusion” conditions?
Merry Christmas; HRG.
On the missing of gamma emissions in Rossi-Effect and the Focardi Prophecy
Here is analysed the question concerning the nature of the reactions occurring in the Rossi-Effect , by considering the missing of gamma emissions by the eCat.
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1 ) How Quantum Mechanics, Standard Nuclear Physics, and Quantum Electrodynamics were developed
QM, Standard Nuclear Physics, and QED were developed after 1905, and therefore from Einstein’s concept of empty space (considered confirmed by the scientific community when an experiment of 1919 had measured the light deviation during an eclipse).
In 2011 the journal Nature published the paper A vacuum can yield flashes of light, proving that the space is no empty, but actually the space must be filled with a “substance” with structure, formed by particles and antiparticles capable to be transformed in photons.
http://www.nature.com/news/a-vacuum-can-yield-flashes-of-light-1.12430
So, it is reasonable to suppose that something is missing in the three theories mentioned above, and therefore they are incomplete.
Here we try to understand how this lack in the Standard Nuclear Physics reverberates in the field of cold fusion.
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2) Two sort of reactions for the energy production
There are two sort of known reactions able to produce energy :
– Chemical
– Nuclear
According to the Standard Nuclear Physics, nuclear reactions produce gamma emission via gamma decay.
As there is not gamma emission by the eCat , then there are only two ways so that to explain the Rossi-Effect:
a) FIRST WAY: it is by considering that Rossi-Effect is produced by nuclear reactions, however there is not yet a complete understanding of the Standard Nuclear Physics , and this is the reason why, in spite of in special conditions (as those occurring in cold fusion phenomena) nuclear reactions can occur without gamma emission, however the nuclear theorists consider impossible to have nuclear reactions without gamma emission, because they have not a complete understanding of the Standard Model.
—————————————————
Note: it is of interest to call attention to the fact that from this FIRST WAY there is NO VIOLATION of any fundamental law of Physics, because the missing of gamma emission is consequence of some unknown mechanisms existing not considered up to now in the Standard Model, since the theory is incomplete.
b) SECOND WAY: it is by considering that Rossi-Effect is produced by chemical reactions. As chemical reactions do not emit gamma rays, a theory based on chemical reactions is acceptable for the explanation of the Rossi-Effect, from the viewpoint of the missing of gamma emission
Let us analyse those two WAYS from which the Rossi-Effect is possible to occur without the gamma emission.
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3) Rossi-Effect occurrence from nuclear reactions
As the space is no empty, there is need to consider that the space has a structure composed by elementary particles and antiparticles as electricitons, magnetons, gravitons, etc.
From a structure of the non-empty space formed by those particles and antiparticles there is no way to develop a concept of electric field with SPHERICALshape. In other words, by considering the non-empty space the shape of the electric field of elementary particles as the proton and the electron must be NON-SPHERICAL . The Figure 1 shows the electric and magnetic fields of the proton.
FIG. 1:
http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png
As the radius of the electric field has the magnitude of the Bohr’s radius 10^-11m, and the radius of the nucleus is 10^-15m, of course the Fig. 1 does not show the real proportion between the fields. The Fig. 2 show a better proportionality (but of course not real yet):
FIG. 2
http://peswiki.com/index.php/Image:FIGURE_2-_3_fields_in_real_proportionality.png
However, the proton is composed by quarks, and (due to their oscillation into the structure of the proton combined with the spin of the proton) the body of the proton acquires a chaotic rotation. As consequence, in average the electric field of the proton takes a SPHERICAL shape.
Note that the spherical shape of the proton is a statistical behavior of the actual non-spherical shape of its electric field.
The statistical spherical shape of the electric field of the proton (and also of the electron) is the shape considered as fundamental in the development of Quantum Mechanics, and also in the development of Quantum Electrodynamics. And as the Standard Nuclear Physics was developed from the principles of Quantum Mechanics and the fundamental concepts of considered in QED, we realize that the spherical shape of the electric field considered as fundamental by the nuclear theorists in the Standard Nuclear Theory actually is a statistical behavior, and so the spherical shape can change from some special conditions.
The structure of the electric field of the nuclei is similar to the structure for the proton shown in the Figure 1, and so in average the electric field of any nucleus has a sheperical shape as consequence of the statistical behavior, and this spherical shape is just considered in the Standard Nuclear Physics.
In another words:
– The electric field of the nuclei is non-spherical
– However, from the statistical viewpoint, the electric field of the nuclei is spherical, due to the chaotic rotation of the nuclei.
As we note from the non-spherical shape of the electric field of the nuclei shown in the Figure 1, there is a “hole” in the electric field, and in special conditions (as occurring in cold fusion) particles as protons and neutrons can penetrate into a nucleus via that “hole” with energy lower than that required from hot fusion process.
And, if a nucleus captures a low energy neutron via the “hole” in the electric field of the nucleus, and after the capture the nucleus decays , there is no emission of gamma rays during the decay, because the nucleus was not excited with high energy as happens in the cases when a neutron enters with high energy into a nucleus not via the “hole” in the electric field.
Therefore, from such new consideration of the non-empty space for the development of a new concept of electric field, the Rossi-Effect can occur without gamma emission, by considering some special conditions not considered in the Standard Nuclear Physics, because the Standard Model was developed from statistical considerations about the shape of the electric field.
From the explained above, we realize that the nuclear theorists have not a deep understanding on the Standard Model, because what they have is a statistical interpretation of the electric Coulomb barrier of the nuclei.
CONCLUSION 1:
The nuclear theorists have not an entire understanding of the Standard Model. And then we realize that Focardi Prophecy is correct: those ones who conclude that Rossi-Effect is impossible to occur via the Standard Model had inferred their wrong conclusion from their lack of comprehension of the Standard Model.
CONCLUSION 2:
It’s important to note that when we consider that Rossi-Effect occurs via nuclear reactions, there is NO VIOLATION of any fundamental law of Physics . Because the missing of gamma emission is not due to any phenomenon disagree to the fundamental laws of Physics, but actually the missing of gamma emission is due to a misunderstanding of the Standard Model, since the spherical shape of the Coulomb barrier considered in the Standard Model is due to a statistical behavior not conserved in the special conditions of the cold fusion phenomena, as Rossi-Effect.
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4) Rossi-Effect produced from chemical reactions
The energy produced from chemical reactions is calculated by taking in consideration the attraction between the positive charge of the nuclei and the negative charge of the electrons in the electrosphere of the atoms. Therefore the energy of chemical reactions is based a fundamental law of Physics: the Coulomb attraction between electric charges.
In the Lugano Report the experimetalists had concluded that the energy produced by the eCat cannot be produced by chemical reactions, a conclusion inferred from the Ragone plot of energy storage ( , page 26, Figure 13, where it is written: “The E-Cat, which would be far off the scale here, lies outside the region occupied by conventional sources“). In the Summary and concluding remarks at the page 30 it is written: ”We have a device giving heat energy compatible with nuclear transformations” .
So, from the Ragone plot there is not any chemical reaction able to produce the amount of energy produced by the Rossi-Effect in the eCat. Therefore, if an author of a theory on the Rossi-Effect discards the nuclear reactions as cause of the heat produced in the eCat, we have the following situation:
a) As nuclear reactions are discarded, then only chemical reactions can be the cause of the heat produced by the eCat.
b) But as no one among the all known chemical reactions is able to produce the quantity of heat produced in the eCat, then the author of the theory must propose the existence of a new sort of unknown chemical reaction, never seen before.
c) The existence of this new unknown chemical reaction proposed by the author of the theory violates a fundamental law of Physics: the energy produced by this new chemical reaction is beyond the calculation obtained via Coulomb law.
CONCLUSION 3:
Any theory which with the aim of explaining the Rossi-Effect by discarding nuclear reactions violates a fundamental law of Physics.
5) FINAL CONCLUSION :
5.1) A theory with the aim of explaining the Rossi-Effect by considering nuclear reactions does not violate any fundamental law of Physics, and such theory is according to Focardi Prophecy: those ones who claim that it is impossible to explain Rossi-Effect from the Standard Model actually did not understand in deep the Standard Nuclear Theory.
5.2) A theory which aim is of explaining the Rossi-Effect by discarding nuclear reactions violates a fundamental law of Physics: the Coulomb law.
Wlad,
You asked about the estimated fuel content. This is what I previously posted:
4. Page 28: “From all combined analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and H in addition to the Ni.”
5. Page 28: “… from the ICP-AES analysis which shows the mass ratio between Li and Al is compatible with a LiAlH4 molecule.”
Element % by Weight
Nickel 55.0
Iron 38.9
Aluminum 4.3
Lithium (total) 1.2
Hydrogen (no Deuterium) 0.6
Total 100.0
Bob:
All the data regarding the operation of the 1 MW plant will be published after the end of the test. As I said many times now, I cannot give any data before that.
Best wishes for the New Year to you and all our Readers,
A.R.
Frank Acland:
Also tomorrow I will spend some time with the 1 MW plant.
Merry Christmas to you and your team and, as usual, to our Readers,
A.R.
Wladimir,
I am sorry about the loss of your brother. My condolences to you and his family.
Joe
Dear Andrea,
Will you have to spend Christmas with your 1MW plant? Or can you leave it alone some days?
Merry Christmas to you and your team!
Frank Acland
Dear Andrea Rossi
Is some or all of the water heated by the 1MW plant returned to the plant for reheating?
If it is, can you tell us the temperature of the water before it returns to the 1MW plant?
Best wishes for the New Year.
Bob
Ing. Michelangelo De Meo:
Thank you for the information. Yes, our work has really moved the giants.
Warm Regards,
A.R.
Bill Gates Sponsoring Palladium-Based LENR Technology
Rossi you were right your work is moving the giants!!
http://www.kitco.com/ind/Albrecht/2014-12-23-Bill-Gates-Sponsoring-Palladium-Based-LENR-Technology.html
Italo R.:
8 400 straight hours of stable work with long ssm periods with the 1 MW E-Cat of our Customer.
Warm Regards,
A.R.
Dear Dr. Rossi,
have you asked Santa to give you some gift?
What do you hope for you for next year?
Kind Regards,
Italo R.
Wladimir Guglinski:
Please accept my condolences for your Brother Sacha.
Warm Regards,
Andrea Rossi
Dears Joe and Steven,
sorry the delay in answering your questions.
I was out of the internet along 4 days, because my brother Aleksander (Sacha) was very bad in the hospítal, victim of cancer. He died yesterday.
Joe,
when a particle enters within a nucleus, there is need two sort of energy for its penetration into the nucleus:
1) The energy required to win the Coulomb barrier
2) The energy necessary to perforate the flux n(o) of the nuclei ( this is the reason why also a neutron requires an energy so that to enter within a nucleus).
Energetically low neutrons cannot enter within a nucleus. It this would be possible, there would not exist stable nuclei with small quantity of neutrons, as for instance 58Ni, 60Ni, 62Ni.All they would be converted to the stable 64Ni, by capture of low energy neutrons, because low energy neutrons exist in the enviroment all the time (background of neutrons).
This is a fundamental questions not responded by the Standard Model
From the principles of the Standard Model, 58Ni, 60Ni, 62Ni could not exist in the nature
In the case of the hydrogen and helium isotopes, there is need 3 sort of energy:
1) The energy required to win the Coulomb barrier
2) The energy necessary to perforate the flux n(o) of the nuclei
3) The matching energy. This energy promotes the overlap between the two principal fields Sp(p) of two protons.
These 3 sort of energy are shown in the TABLE 6.1 at the page 121 of my book Quantum Ring Theory.
The calculation of the binding energy of the isotopes 1H2, 1H3, 2He3, 2He4 is made in the page 114 of the book, by taking in consideration the 3 sort of energy mentioned above.
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So, dear Joe
obviously the puzzle: why gamma rays are not emitted in cold fusion ?
cannot be responded by the principles of the Standard Model.
The answer for the puzzle requires my new nuclear model.
That’s why the subtitle of my book is FOUNDATIONS FOR COLD FUSION
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Joe,
I would like to hear from the other authors of cold fusion theories the answer for your question.
For instance, from Stoyan Sarg, Hagelstein, Widon-Larsen, Robert Godes, etc…
Rober Godes said in a thread:
“I am glad to see that Cu63 does not appear to have changed much between start and finish seen on page 52. As I had predicted, as noted above, Cu65 is actually higher than Cu63 in the ash after the experiment”
http://www.e-catworld.com/2014/12/21/robert-godes-of-brillouin-on-beta-decay-of-nickel-in-the-lugano-test/
But I cannot see how Rober Godes can be glad if somebody asks him why there is not emission of gamma rays in the nuclear reactions proposed by him
regards
wlad
Joe wrote in December 22nd, 2014 at 2:45 PM
Wladimir,
How would your model account for the unexpectedly small amount of gamma rays present in the Rossi Effect?
————————————————————–
Joe,
let us consider the decay:
60Co -> 60Ni + gamma-rays
http://en.wikipedia.org/wiki/Cobalt-60
First of all, we need to note that 60Co has a half-life of 5,2 years.
Then the first question is:
why 60Co decays so fast to 60Ni in the Rossi-Effect?
Let us try to find answers for those questions.
The 60Co is produced artificially by neutron activation of the monoisotopic and mononuclidic cobalt isotope 59Co:
59Co + n → 60Co
However,
such artificial activation by 59Co + n → 60Co is made via neutrons with high energy (because the neutron do not enter within the 59Co via the hole in the Coulomb barrier of the 59Co.
So, the newborn 60Co is highly energetically excited, and this is the reason why its half-life is 5,2years.
But in the Rossi-Effect, when occurs the neutron capture in the reaction 59Co + n -> 60Co , the neutron enters within the 59Co nucleus via the hole in the Coulomb barrier of the 59Co.
The neutron captured by 59Co in Rossi-Effect has lower energy than that required when the neutron is captured by 59Co in the artificially by neutron activation of the 59Co made in laboratories.
Therefore the 60Co obtained in the Rossi-Effect is not so excited as in the case of the 60Co obtained in the laboratories.
This is the reason why in the Rossi-Effect we have:
1) in the decay 60Co -> 60Ni there is no emission of gamma-rays.
2) when the 60Co is formed by 59Co + n , the half-life of the 60Co is very short, because the newborn 60Co is not highly excited as occurs in the reaction 59Co + n occurring in the laboratories.
regards
wlad
regards
wlad
Stefano Landi:
1- Sometimes understanding is a subjective process
2- I already answered to this question
3- Merry Christmas to you and to all our Readers, as always
Warm Regards,
A.R.
Dear Andrea.
1) so according, to your answer, I understand that there are not specific improvements of the e-mouse/e-cat system in these last months?
2) Please would you mind to tell the official start date of the 1 MW plant?
3) Marry Christmas
Stefano
Steven N. Karels wrote in December 22nd, 2014 at 5:26 AM
Wlad,
If hydrogen is not involved in the fusion reactions — then there is insufficient 7Li to account for the heat generation. A means of regenerating 7Li is needed. It seems to me the only way to resolve this need is to propose the fusion of the hydrogen with 6Li. Comments?
————————————————-
Dear Steven,
I did not understand the table in the end of the last page of the Lugano Report:
Results as weight present of the samples
2 fuel 2.13 mg 50ml
Li 670nm % = 1.17
Did you?
what is the total percentage of littium in weight in the fuel?
regards
wlad
Alexis:
To give data regarding the 1 MW before their consolidation would expose us to dangerous situations should the data change or worse. For this reason it would be not appropriate, under a professional point of view, to give data when the result could be transitory or substantially wrong.
Warm Regards,
A.R.
Pietro F.:
The data of the performance of the 1 MW plat will be published at the end of the tests, when they will be consolidated.
Warm Regards,
A.R.
Pietro F.:
Please see my answer at 9.24 p.m. of Yesterday,
Warm Regards,
A.R.
I’m sorry I did not see the answer to KeithT.
What is the the average COP of the 1 MW plant delivered to the Customer of IH?
Regards
Pietro F.
JC Renoir:
I am very sorry, but it is not possible, and I am sad for this, because it is a magnificence. It is necessary we end the testing period before data and images of it will be communicated.
Warm Regards,
A.R.
Curiosone:
Apparently you should be right, but it is not so. As a matter of fact, concatenating Feynman diagrams, is possible to theorize virtual massive elementary particles build up in the interacting fields: such virtual massive particles ( nothing other than resonances of waves, like waves of the ocean that make up tsunamies piling up one upon the other) can indirectly connect Higgs bosons with photons; virtual particles are bookkeeping devices that indicate how quantum fields are vibrating on course of stable elementary particles interactions, but the resonances make massive virtualities for about 10^-23 s that are enough for the Higgs field to indirectly resonate with the electromagnetic field: imagine a piano in a room that can resonate only with another specific piano and not with an arp, but this specific piano is able to resonate with an arp, so that the first piano indirectly resonates with the arp and from the sound of the arp you can understand that in another room there is the first piano, due to the fact that the specific piano can be put in resonance only by the first piano before resonate with the arp.
Warm Regards,
A.R.
Frank Acland:
Thanks to our great Team, I can work on both the fronts with my Teammates!
Warm Regards,
A.R.
Dear Andrea,
Are you able to personally devote much time to the Hot Cat R&D? Or are you still tied up with the 1MW plant at the customer site?
Many thanks,
Frank Acland
Keith T.:
Yes, of course, it is in the thread of the Hot Cat R&D: gas fueling is an evolution of the Hot Cat.
Warm Regards,
A.R.
Dear Andrea,
Both the electrically driven low temperature Ecat and electrically driven high temperature hot cat have been seen in tests, you have stated that you are developing a gas powered high temperature hot cat; are you also developing a gas powered low temperature Ecat ?
Kind regards,
Keith Thomson.
Dear Dr Rossi:
I read that the Higgs boson has been discovered from photons that decayed from it: but in other sites I read that the Higgs interacts only with massive particles, so how can degenerate into photons, that are massless?
Thank you for your generous patience.
W.G.
Dear Andrea Rossi:
Can you give us as a Christmas gift a photo of the 1 MW plant?
JPR
KeithT:
I will answer to this question after the R&D we are making on both of them will have been completed: for the low temperature the R&D is on course by means of the 1 MW plant delivered to the Customer of IH, for the Hot Cat intense R&D is on course by means of new prototypes upon which we are making exponential progress by the day. Within about 1 year we will have consolidated data. I deem meaningless to give transitory data.
Warm Regards,
A.R.
Wladimir,
How would your model account for the unexpectedly small amount of gamma rays present in the Rossi Effect?
All the best,
Joe
Dear Andrea,
Between the electrically driven low temperature Ecat and electrically driven high temperature hot cat, which has the higher COP?
Kind regards,
Keith Thomson.
Frank Acland:
Thank you for the information: no, I am not familiar with it, even if I contacted them a year ago or so and they were not ready with a product for our needs.
Warm Regards,
A.R.
Dear Andrea,
Are you familiar with a product being developed by the US Utility NRG called the ‘Beacon 10’? It is the invention of Dean Kamen (of Segway fame) who is working in partnership with NRG on the product.
Basically it is a Stirling engine-based generator that is designed to produce electricity from a home’s natural gas source.
The first model will apparently be a 10kw model which will be suitable for commercial buildings. A smaller 2.5 kw for home use is in the plans.
I see no reason why, technically, this could not work with E-Cat heat.
You can read more about it here: http://www.forbes.com/sites/christopherhelman/2014/07/02/dean-kamen-thinks-his-new-stirling-engine-could-power-the-world/
I thought you might find this interesting, if you were not already aware of it.
Kind regards,
Frank Acland
Dear Steven N. Karels,
you have to consider that 1 gr of Lithium is what has been input but you know nothing of what was inside reactor before charging it with 1 gr powder.
I mean there could be a bit Lithium deposited on the internal surface of the most internal cylinder of the reactor and this could have taken part in the reaction.
Ing. Michelangelo De Meo:
Thank you for the interesting link.
Warm Regards,
A.R.
Dear Dr. Rossi, I send the link to an interesting interview engineer Mats Lewan that followed the ‘ E cat.
I would like to do to her readers and best wishes for a Merry Christmas .
http://www.ecat-thenewfire.com/blog/interview-mats-lewan-why-ecat-works/
Pietro F.:
Thank you, and as always the Wishes are also to you and all our Readers,
Andrea Rossi
Dear Steven N Karels,
In the Don Borghi experiment, where the hydrogen into the reactor is submitted to similar conditions of those occurring in the Rossi’s eCat, the hydrogen is ionized, and a plasma of free protons and electrons is formed.
Along the time there is fusion p+e -> n
In order to discover if the hydrogen in the eCat is converted to neutrons as occurs in the Don Borghi experiment, there is need to perform an experiment, by putting hydrogen gas into the empty Rossi’s reactor.
This experiment would be a replication of the Don Borghi experiment, in the case the hydrogen be converted to neutrons.
In the case the hydrogen be converted to neutrons in the Rossi’s eCat, the hydrogen cannot play any role in the cold fusion reactions… I guess…
regards
wlad
Wlad,
If hydrogen is not involved in the fusion reactions — then there is insufficient 7Li to account for the heat generation. A means of regenerating 7Li is needed. It seems to me the only way to resolve this need is to propose the fusion of the hydrogen with 6Li. Comments?
ERRATA:
In my last post, where it is written
b) Due to the increase of the temperature within the reactor, the hydrogen molecule tends to move very fast and chaotically along all directions, having collision with the walls of the reactor and also with the elements of the fuel.
the best is:
b) Due to the increase of the temperature within the reactor, each of the hydrogen atoms tends to move very fast and chaotically along all directions, having collision with the walls of the reactor and also with the elements of the fuel.
Auguri e buon lavoro.
Pietro F.
Keith T.:
No.
Warm regards,
A.R.
Orsobubu:
he,he,he..
Warm regards,
A.R.
Dear Steven N Karels,
I was thinking about the role played by the hydrogen in the eCat, and perhaps finally I understood it.
My conclusion is:
1- The hydrogen is not involved in the fusion reactions
2- The hydrogen does not play a role as a catalyst
3- However it is possible the hydrogen plays a role similar to a catalyst, as follows:
a) The proton has magnetic moment +2,793
b) Due to the increase of the temperature within the reactor, the hydrogen molecule tends to move very fast and chaotically along all directions, having collision with the walls of the reactor and also with the elements of the fuel.
c) However, due to the magnetic pulse produced by the coils, the proton also gets another sort of motion: a fast oscillation toward the axis of the reactor (as consequence of the interaction between the magnetic moment of the proton and the magnetic pulse created by the 3 coils).
d) Such component of the proton’s motion, oscillating along the axis of the reactor, causes the alignment of the z-axes of the nuclei Ni and 3Li7 toward the axis of the reactor. Perhaps this is the reason why the eCat does not need to have a source so that to create a constant vector magnetic field along the axis of the reactor, so that to align the z-axes of the nuclei Ni and 3Li7 along the axis of the reactor (and it can be the response for the question: why any DC current was not found, according to the Lugano Report).
e) The oscillation of the proton can also play a second role: to help the excitment of the nuclei 58Ni, 60Ni, 62Ni, 64Ni, helping the magnetic pulse induced by the 3 coils.
regards
wlad