Stability of light nuclei

by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
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Abstract
Dr. Wilfried Nörtershäuser of the Helmhotz Center for Heavy Ion Research at the University in Mainz says on the 2009 experiment which had detected a neutron halo in 4Be11 with distance 7fm from the cluster:
“By studing neutron halos, scientists hope to gain further understanding of the forces within the atomic nucleus that bind atoms together, taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics”[ 2 ]
In the case of 4Be11, the halo neutron and the nuclear core are separated by the distance of 7fm, and so such isotope represents the experimental proof that the cohesion of nucleons within the light isotopes cannot be promoted by the strong nuclear force.
Such experimental discovery published in 2009 had been predicted years ago, because according to the new nuclear model proposed in Quantum Ring Theory, published in 2006, the cohesion of the nucleons within the light nuclei is not caused by the strong nuclear force.
Here in this paper the new nuclear model is submitted to a scrutinity so that to verify whether from its structure it’s possible to explain the stability of the light nuclei and to reproduce the nuclear properties as nuclear spins, electric quadrupole moments, and magnetic moments. Nuclear magnetic moments are calculated from two different and independent methods.  In the second, named “method of equilibrium between nucleons”, it’s presented the Lagrangian for nuclei with Z < 8.  The results obtained from them agree each other, and are corroborated by nuclear spins and electric quadrupole moments suplied by nuclear tables.
In this Part One are presented calculations on magnetic moments for the isotopes of lithium, beryllium, and boron. In the next paper Part Two will be exhibited  calculations for carbon, nitrogen, and oxygen.  In the paper Part Three the author will exhibit calculations for electric quadrupole moments.
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485 comments to Stability of light nuclei

  • John

    Dear Andrea — Congratulations on your outstanging progress and breakthroughs!

    If I understand correctly the activator operates 35% of the cycle at slightly over 1 COP and this triggers the e-cat or Tiger to operate the remaining 65% of the cycle at 100 – 200 COP. Therefore it would appear the e-cat/Tiger average COP over the entire cycle is between 65 and 130 COP.

    Is the e-cat/ Tiger powered by any external energy source other than the activator?

  • Giovanni Guerrini

    Dear Dott Rossi,
    if the rat and the tiger are never active together and the first runs for the 35% of the time with a cop 1.1 and the second for the remaining 65% with a cop 150,is correct say that the total average cop is 97.88 ?

    Regards G G

  • Hank Mills

    Dear Andrea, 

    Thank you tremendously for sharing information with us about the new mouse/cat configuration before the report is issued. You have no obligation whatsoever to share any information with us, but the fact you do is very much appreciated. 

    1) Does the cat and mouse both utilize separate electrical resistance based heating elements, or is the resistance element utilized by the mouse/activator the only source of external heat in the entire ECAT?

    2) Does the mouse/activator provide the RFG (radio frequency generator) stimulation for both itself and the cat, or only itself?

    3) When the activator/mouse is being driven or turned on, is there any power being supplied directly to the cat?

    4) I can’t help but wonder if there might be something being transferred from the mouse to the cat — other than heat and radio frequencies — that helps in the activation process. Is only heat and radio frequencies used to activate the cat?

    Thank you.

    Hank

  • barty

    Dear Mr. Rossi,

    your recently said your US Partner is now producing the 1MW Plants.
    When do you expect they will finish the first plants, and when will they be delivered to their customers?
    I hope your “mass production” will start soon 😉

    Greetings from Germany!
    barty

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    All our products are going to be mass produced, but you must make a distinction between R&D and production. R&D does not leapfrog the mass production, it prepares the future mass productions.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Don Witcher:
    They are never active together.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Andrew:
    Yes, this is the path we are tracking. But for industrial applications, so far.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Dr Joseph Fine:
    1- yes
    2- yes
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Steven N. Karels:
    It is correct to make the average between the COP of the mouse and the E-Cat. In this sense, the answer is yes. We will be more precise, though, when we will sell the product.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Tom Conover:
    No, it is not right. The math implies only what I wrote, nothing more.
    Warm Regards,
    A.R.

  • Tom Conover

    10 amps. Oops.

    T.C.

  • Tom Conover

    Dear Andrea,

    The math seems to imply that the Tiger can be powered by a standard household outlet @ 120 volts with a power cord from a desktop computer, pulling about one ampere. Is that right?

    Is that what you use to power the Tiger?

    Warm regards,

    T.C.

  • Steven N. Karels

    Dear Andrea Rossi,

    We understand the distintion between up-sizing the eCat modules and the independent action of changing the architecture of the eCat design.

    Independent of change in output power, you have implemented an Activator and Reactor combination. Obviously, you would not have done so except it improves the overall eCat performance and/or stability.

    What is of interest is how do we understand COP now that you have both the Activator and the Reactor? One way of understanding this is the average COP is a combination of the Reactor COP (100 – 200x) and the Activator COP (>1x). It is assumed the Activator only outputs a fraction of the output from the Reactor. If this is true, then the average COP must be the total thermal output of the reactor divided by a quantity bound between 1 and 200. Using the approximation that the Activator runs about 1/3 of the time (close to 35%), the apparent COP would be 3 if the Activator outputs nearly the same thermal power as the reactor. Thus my assumption that it must be much less. If the Activator output was 1/10 of the reactor, then the average COP would be around 30. Is this consistent with what you are seeing?

  • Joseph Fine

    Dear Andrea Rossi,

    Referring to the Tiger prototype, is its volume only about 0.2 m^3 (cu. meters)?

    Therefore, ten of these 100 kW modules assembled together (1 MW) would have a volume of only 2 m^3. Is that correct?

    Best regards,

    Joseph Fine

  • Andrew

    Dear Andrea,
    I am really excited by the prospects of a 100 kw cat/tiger
    with just 30% efficiency conversion to electricity it could provide enough heat and electricity to power a block of 20+ flats (or 12-15 houses) even in the UK I s this the sort of ‘distributed generation’ you have in mind? Even more exciting is the prospect of a (combined) COP exceeding 150, is this really true?

  • Don Witcher

    Dear Andrea Rossi

    Many congratulations on the very rapid progress you are now making toward the commercialization of the ecat family. The two stage activator-Ecat system has clearly been a major break through for you and we all appreciat the information you have provided. Could you, if possible, give us an idea of what the ratio of thermal output of the Ecat to thermal input to the activator is when both are active. Thank you for your consideration.

    Warm Regards DW

  • Bernie Koppenhofer

    Dr. Rossi: Sounds to me like you are leapfrogging yourself with your own research. When do you know when to stop and mass produce a product? Could you predict for us which E-Cat/Tiger will be the first mass produced product? Thanks again for sharing.

  • Andrea Rossi

    Dear Giovanni Guerrini:
    Actually, during these very days we are making exponential progress.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Steven N. Karels:
    First of all, we must make a distinction between the fact that we are working to make bigger modules ( “tigers”) from the fact that we have mofified the configuration of all our E-Cats, putting in a apparatus two components, one of which is the activator and the other the E-Cat. One thing is indipendent from the other. The power of the activator depends on the situations, but being its COP more than 1 the energy of the activator is paid by itself, indipendently from the E-Cat.
    The activator works for the 35% of the operational time of the system, while the E-Cat works for the 65% of the time.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Pekka Janhunen:
    1- No, it is not so. We will give a descriprtion as soon as the prototype will be a product.
    2- the Mouse runs 35 ( thirtyfive) % of the time, the E-Cat runs 65 % of the time.
    Warm Regards,
    A.R.

  • Robert Tanhaus

    Dear Dr. Rossi,

    I ment to say: Is this new version (the tiger) made because of a customer needed 100kW modules
    or was it made because of your teams considerations – to be more easy to handle for example?

    Best wishes,
    Robert Tanhaus

  • Dear Andrea Rossi,
    It is interesting to hear about your new 100 kW prototype development. Could you confirm the following:
    1. Is this correctly understood: a roughly 1 kW “Rat” (larger mouse) drives a 100 kW “Warm Tiger”, to make the overall average COP of the entire reactor 100-200?
    2. A detail: is the average duty cycle of the “Rat” 35% or 65%? I think that you answered 35% to “Andrea” and 65% to Steven Karels.
    regards, /pekka

  • Steven N. Karels

    Dear Andrea Rossi,

    Please, a little clarification on the Activator.

    a. If the Tiger eCat has a nominal output of 100kW, does the Activator have a lower output when it is running (e.g., 10kW)?
    b. After turn-on, is the Activator active 35% or 65% of the operating time?
    c. Would the average COP of the entire Tiger eCat be equal to the average total power output (eCat plus Activator) divided by the total input power going to the Tiger eCat plus the Activator?

  • Giovanni Guerrini

    Dear Dott Rossi,
    congratulation! As you said,you started with a Ford T,now you are getting a Ferrari.

    Regards G G

  • Andrea Rossi

    Dear Marco:
    Yes, also.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Giovanni Guerrini:
    Please see the answer to Steven N. Karels few minutes ago,
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Steven N. Karels:
    a- yes, but it drives the cycle of the E-Cat too
    b- the activator works for about the 65% of the operational time of the system
    c- yes, 100-200
    d- still a prototype, and yes
    e- the volume is about 0.2 m^3
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Andrea:
    The average COP of the activator is 1.02 – 1.1; the average COP of the E-Cat is from 100 to 200.
    Margin of error about 10%.
    The activator is turned on for about the 35% of the operation time.
    This is what we are getting from prototypes working in these days in the USA.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Robert Tanhaus:
    I do not understand exactly your question: can you kindly rephrase it, making it more specific?
    Warm Regards,
    A.R.

  • Robert Tanhaus

    Dr. Rossi,

    Thank you for the update of your and your teams work!
    Did you have specific requirements for the tiger or was it some proove of concept or test?

    Best wishes,
    Robert Tanhaus

  • Andrea

    Dear Andrea,
    may i ask you what is the approximate average COP of the low temperature tiger(activator+E-Cat)?

    Many thanks,
    Andrea

  • Steven N. Karels

    Dear Andrea Rossi,

    Interesting information on the “Tiger”. You discussed an Activator and an eCat, with COPs of greater than one and greater than 100, respectively. Can you please clarify the following:

    a. Does the Activator, as its name implies, primarily function during cold start-up?
    b. After stable operation is achieved, is the Activator primarily inactive?
    c. As suggested, it appears this eCat has its own control system, with an observed COP of 100 or greater. Please confirm.
    d. The 200 kilogram “Tiger” is a single reactor outputting thermal power around 100 kW?
    e. What are the rough dimensions of the “Tiger”?

  • Joe

    Wladimir,

    I do not know all the nuclear models that are in existence. Maybe there are some that could perhaps explain the excited 6C12 but have yet to be acknowledged or recognized by the general establishment. This would include QRT. Even if the most accepted current model is deficient in this respect, that does not mean that it needs to be COMPLETELY rethought, since it has already demonstrated its predictive potential. It would only need small, well-thought alterations in order to conform to new experimental realities. This does not mean that QRT should not be considered as a potential ALTERNATE model that can contribute in original ways to explaining phenomena that is perhaps being neglected by mainstream scientists.

    But, in order to know that QRT can explain the excited 6C12, I need to have more information. Hence, all the questions that I keep asking you. (And I am grateful to you for answering them.)

    As far as Nature is concerned, it does not follow a model. Models are abstractions of Nature and therefore LESS than Nature. A multitude of models will always exist for this very reason. There will always be new questions and therefore new explanations needed. And these new explanations might not all fit into one model.

    While it is true that QRT predicted a non-spherical shape for light nuclei with an even number of protons and neutrons, the total number of options for shape is really just two: spherical and non-spherical. Mainstream scientists expect more than just a game of chance. This does not mean that QRT is wrong, but that the bar of truth is set much higher. And the same logic applies to the agglutination of nucleons. They are either tightly set, or they are not. And it is not a stretch of the imagination to envisage nucleons at slightly greater distances from each other, since the concepts of clustering, shells, and mean free paths were already part of the vocabulary of nuclear scientists for quite some time. And again, this does not mean that it is just a mere coincidence that QRT called it right. But, scientists expect much more, especially in terms of quantitative predictions with great accuracy – and many of them.

    Scientists will always be wrong because a model is never Nature itself. That is why science is progress. Science only ends when we stop asking ‘why?’ – and not because we have reached the ultimate model.

    All the best,
    Joe

  • Giovanni Guerrini

    Dear Andrea,
    what do you mean exactly with “COP in the hundreds”?
    Plese explain,because I am getting an heart attack !

    Regards G G

  • Marco

    Very interesting both the E-Tiger, both the new COP of the E-Cat…
    Is the hundreds COP E-Cat the one delivered for testing to your USA partner/client some day ago?

  • Andrea Rossi

    Dear Tom Conover:
    We are testing low temperature tigers, for now, of 100 kW. All our reactors now have activator and E-Cat, allowing us an activator with a COP more than 1 and E-Cat with COP in the hundreds.
    Warm Regards,
    A.R.

  • Tom Conover

    Dear Andrea,

    Could you power the larger Tiger with a 10Kw E-Cat activator, perhaps, or are you staying with resistors? Would the Tiger be 350 degrees too, for a larger turbine?

    Very exciting concepts! Thank you for sharing.

    Roaring Regards,

    T.C.

  • Andrea Rossi

    Dear Toussaint francois:
    Yes, we are working also on it, even if it is not easy: a “cat” that weights 200 kg changes name into “tiger”.
    Warm Regards,
    A.R.

  • Toussaint françois

    Dear mr Rossi

    I have heard your recent interview , and

    talking about the faboulus potentiel power

    with 1 gr of Nikel.

    Do you forsee in a near futur individual ECAT

    with more power 100 Kwh , 1000 Kwh ?

    It would mean less ECATS in a container.

    Thank you

  • Wladimir Guglinski

    From: wladimirguglinski@hotmail.com
    To: johna_6@yahoo.com
    CC: epja@itkp.uni-bonn.de; helayel@cbpf.br; jyeston@aaas.org; prc@aps.org; apr-edoffice@aip.org; nature@nature.com; cjp@fzu.cz; ver@cisp-publishing.com; pnj@bauuinstitute.com

    Subject: posted as comment in Andrea Rossi’s blog
    Date: Wed, 8 May 2013 06:21:08 -0300

    Dear Dr. John Arrington
    the comment of mine ahead had been posted in Andrea Rossi’s blog Journal of Nuclear Physics, in May 7th, 2013 at 6:58 AM

    http://www.journal-of-nuclear-physics.com/?p=802#comments

    Wladimir Guglinski
    May 7th, 2013 at 6:58 AM

    Dear Joe

    In 9th August 2012 John Arrington, from the Argonne National Laboratory, had sent an email to me, where he had written:

    “I have not looked at your theory in any detail, in part because detailed nuclear structure is not my
    main focus, in part because I see no deficiencies with the present models”

    In 3th October 2012 he had sent other email saying:

    “Dear Wladimir,
    What I said is correct.
    1) I am not aware of any deficiencies in the current models, and in particular, not in the context of our recent measurement.”

    In 16th April 2013 John had enphasized to me again that according to his opinion there is nothing wrong with the current nuclear models:

    “I said that I was not aware of any significant deficiencies in the current models describing the structure of light nuclei. […] So no, I have not changed my mind.”

    What do you think, Joe?

    For instance, consider that question in the Fig. 29-A, page 40, of the present paper.

    It’s very easy to understand that (by considering the classical current nuclear models) it’s IMPOSSIBLE to explain why the excited isotope 6C12 has spin i=2 and µ = 0.

    Please look at the Fig. 29-A again.

    Tell me:

    1- do you think is it possible to find an explanation for i=2 and µ = 0 of the excited 6C12, by considering the current nuclear models ?????

    2- do you think is it possible to explain i=2 and µ = 0 of the excited 6C12 without to consider the flux n(o), as shown in Fig. 29 (page 40) ????

    3- Any honest scientist has the obligation to respond:

    A) it’s IMPOSSIBLE to solve the paradox of 6C12 by considering the current nuclear models

    B) it’s IMPOSSIBLE to solve the paradox of 6C12 without considering the flux n(o), as proposed in my theory.

    If the nuclear theorists cannot understand such trivial question, then there is no honesty among the nuclear theorists.

    And then,as there is no honesty among the nuclear theorists, do you think is it possible to find a satisfactory solution so that to eliminate the inconsistencies of the current Classical Nuclear Physics ?????

    Regards
    WLAD

  • Wladimir Guglinski

    Joe wrote in May 8th, 2013 at 7:23 PM
    Wladimir,

    I did answer your question in the best way that I thought possible.

    You asked me to review Fig.29-A (page 40) as an example of the superiority of QRT to Classical Nuclear Physics, and that is exactly what I did. But since I do not know QRT as well as you do, I had to formulate questions in order to know more about it. And I need to know more about it, ensuring that there are no inconsistencies in it, before I promote it as the next advance in nuclear science.

    COMMENT 1:
    Joe, but you can at least answer to the following question:
    Please look at the Fig. 29-A again.

    Tell me:
    1- do you think is it possible to find an explanation for i=2 and µ = 0 of the excited 6C12, by considering the current nuclear models ?????

    COMMENT 2:
    You dont need to say that QRT is correct or wrong.
    Tell us only if you think it is possible:

    2- do you think is it possible to explain i=2 and µ = 0 of the excited 6C12 without to consider the flux n(o), as shown in Fig. 29 (page 40) ????

    .

    The way that you organized your argument forces a person to choose between QRT and Classical Nuclear Physics. Although CNP has its deficiencies, QRT will have them as well, since science is never a finished book

    COMMENT:
    You are wrong.
    In the case of nuclear physics, it is not a finished book because the nuclear physicists did not discover the true model existing in the nature (and the principles by which the model works).

    Joe
    The nature work by physical models. When the nuclear model existing in the nature will be discovered, the Nuclear Physics will become a finished book.

    And if the nuclear model proposed in QRT is indeed the model existing in the nature, it will be able to explain all the nuclear phenomena.

    .

    Therefore, there might legitimately exist other models that solve the shortcomings of both CNP and QRT.

    COMMENT:
    If the model proposed in QRT is represents the nuclear model existing in the nature, there will no need other models.

    .

    You claim that QRT has made successful predictions

    COOMENT:
    I dont claim.
    The prediction that light nuclei with Z=N=pair have non spherical shape is in my book.
    In the page 137 is published the argument which explain why they have zero quadrupole moment.
    The argument was plagiarized in the article How atomic nuclei cluster, published by Nature in 2012
    This is a fact. I am not claiming

    According to my nuclear model, the aggregation of nuclei is not caused by the strong force.
    The experiment of 2009, which detected the halo neutron in a distance of 7fm corroborates the argument that nuclei aggregation does not occur due to strong force interactions.
    This is a fact. I am not claiming.

    .

    but CNP has, as well, in other areas of nuclear science (the masses of the W and Z bosons).

    COMMENT:
    Yukawa also predicted that the mass of meson, and proposed that it belongs to the structure of the neutron.
    Today we know that he was wrong.
    The prediction of the boson mass W is also a coincidence. The boson W is emitted when the neutron has decay (that’s why its mass is correctly predicted). But the phenomenon does not occur as the nuclear theorists believe.
    The boson W is created only in the instant when the neutron decays. It does not exist as cause of interactions within the neutron, as the nuclear theorists believe.

    .

    Many models can find solutions to different problems. We should not limit ourselves to any one model. And that includes QRT.

    COMMENT:
    You are wrong.
    What you say is what the scientists believe, because they did not discover the true structure existing in the nature.
    The nature yield the phenomena by one unique model. She does not use several models.
    So, one unique model produce all the nuclear phenomena.
    If such a model existing in the nature will be discovered, it will be able to explain all the nuclear phenomena.
    If the model proposed in QRT is the same model existing in the nature, it will be able to explain all the nuclear phenomena

    .

    As far as the lack of interest shown by some scientists with regard to QRT, there are any number of reasons for why this is so. This is an existential reality. It has always existed. You will keep promoting what you think is right, and the establishment will continue to be very slow in responding. But do not worry, science always progresses. And if QRT has any validity to it, it will find its rightful place in the big book of science.

    COMMENT:
    Yes, I know.
    Planck told us that:
    A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it.

    Regards
    wlad

  • Joe

    Wladimir,

    I did answer your question in the best way that I thought possible.

    You asked me to review Fig.29-A (page 40) as an example of the superiority of QRT to Classical Nuclear Physics, and that is exactly what I did. But since I do not know QRT as well as you do, I had to formulate questions in order to know more about it. And I need to know more about it, ensuring that there are no inconsistencies in it, before I promote it as the next advance in nuclear science.

    The way that you organized your argument forces a person to choose between QRT and Classical Nuclear Physics. Although CNP has its deficiencies, QRT will have them as well, since science is never a finished book. Therefore, there might legitimately exist other models that solve the shortcomings of both CNP and QRT.

    You claim that QRT has made successful predictions, but CNP has, as well, in other areas of nuclear science (the masses of the W and Z bosons). Many models can find solutions to different problems. We should not limit ourselves to any one model. And that includes QRT.

    As far as the lack of interest shown by some scientists with regard to QRT, there are any number of reasons for why this is so. This is an existential reality. It has always existed. You will keep promoting what you think is right, and the establishment will continue to be very slow in responding. But do not worry, science always progresses. And if QRT has any validity to it, it will find its rightful place in the big book of science.

    All the best,
    Joe

  • Andrea Rossi

    Dear Brian:
    The plant

      you saw in the photo has been made in the factory of Bologna, but recently it has been transferred to our factory of Ferrara, bigger and well supplied, to be modified with the application of new technologies, to be delivered to our USA Partner, who tested it in a 24 straight hours test on April 30 and May 1.
      At the same time in the USA has been started the manufacturing process. All the next plants will be delivered from the USA factory, while in Ferrara will remain the production of experimental prototypes and relative tests.
      Warm Regards,
      A.R.
  • Wladimir Guglinski

    Dear Joe,
    you did not respond my questions posted in May 7th, 2013 at 6:58 AM:

    Please look at the Fig. 29-A again.

    Tell me:
    1- do you think is it possible to find an explanation for i=2 and µ = 0 of the excited 6C12, by considering the current nuclear models ?????

    2- do you think is it possible to explain i=2 and µ = 0 of the excited 6C12 without to consider the flux n(o), as shown in Fig. 29 (page 40) ????

    3- Any honest scientist has the obligation to respond:

    A) it’s IMPOSSIBLE to solve the paradox of 6C12 by considering the current nuclear models

    B) it’s IMPOSSIBLE to solve the paradox of 6C12 without considering the flux n(o), as proposed in my theory.

    .

    If the nuclear theorists cannot understand such trivial question, then there is no honesty among the nuclear theorists.

    And then,as there is no honesty among the nuclear theorists, do you think is it possible to find a satisfactory solution so that to eliminate the inconsistencies of the current Classical Nuclear Physics ?????

    Regards
    WLAD

  • Wladimir Guglinski

    Joe wrote in May 8th, 2013 at 5:16 AM
    Wladimir,

    1. In QRT, what causes the sign of the spin of a nucleon to change abruptly and work against the Least Action Principle?

    RESPONSE:
    Where did it occur ?

    2. What exactly is the effect of the gravitational flux n(o) on the sign of the spin of a nucleon, that an abrupt (working against the Least Action Principle) change in that sign should be IGNORED and treated as if there were NO change at all that had occurred, thereby affecting NEITHER the signs of the magnetic moments (intrinsic and induced) of that nucleon, NOR the F(SI) between that nucleon and any other nucleon?

    RESPONSE:
    The interaction responsible for the changing of the spin of the electron is not its interaction with the flux n(o).

    Consider the two protons right (R) and left (L) in the deuteron, and consider that they both gyrate in the counter clockwise direction (because the deuteron has spin i=1, and so the two protons gyrate in the same direction).
    So, they both have spin-up, and move in the counter clockwise direction.

    Consider the electron orbiting with spin-up about the proton R, moving in the counter clockwise direction (therefore the same direction of the rotation of the proton R).

    When the electron leaves out the proton R (in order to continue in its orbit with the shape of the infinite symbol), the electron changes the direction of its motion: it will orbit the proton L in clockwise direction.

    So, when the electron starts to move about the proton L, the electron starts to move in the contrary direction of the proton L).

    Therefore the electron needs to change its spin from spin-up to spin-down, because it had changed the direction of its motion (from counter clockwise direction to clockwise).

    Regards
    wlad

  • Brian

    Mr. Rossi

    Can you clarify where the 1MW plants are being made and tested? I had understood you to have said earlier that the 1MW plants were being made in the USA. The caption of your picture, however, said that the 1MW plant was being shipped from Bologna to Ferrara. Can you clarify this for your readers?

    Thank you for taking the time to answer my question

    Brian

  • Steven N. Karels

    Koen,

    My sense is that the LENR development would have continued without Andrea Rossi but at a much slower pace. Eventually, patents would be applied for and obtained. I see that Andrea Rossi brings the passion, the energy and the financial execution to make this happen, bypassing the normal scientific drama and going straight to industry for validation.

    None of us are irreplaceable, but we each add our uniqueness. If eCat is accepted by industry and society, Andrea Rossi will be honored. “The things we do in life, echo throught eternity”

  • Andrea Rossi

    Dear Giuliano Bettini:
    In our new factory of Ferrara we will maintain prototypes and capacity to produce them, besides all the scientific instrumentation to make the tests. A huge work is going to be made there.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Koen Vandewalle:
    I have the sensation you are right.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Ing. Michelangelo De Meo:
    Thank you for the infrmation,
    Warm Regards,
    A.R.

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