by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
.
.
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.
.
.
Dear Mr. Rossi
Could you possibly give us an update on how your efforts to get the hot-cat certified for operation is going?
Many thanks.
Victor
Koen,
I understand you are talking about the environmental and social impact of refilling eCat modules. I make the following assumptions:
a. eCat modules require replacement every 180 days
b. A large (~1GW output) electricity generation plant has 180,000 10kW eCat modules.
Therefore, in one typical day, 1,000 eCat modules are replaced. Or about 40 per hour or about 1 per minute. This is a manageable rate.
The eCat modules are returned, cleaned, re-fueled and shipped back to this theoretical plant.
Assuming a Million eCats can be produced in a single factory per year, that factory could easily support two such electricity production plants.
Compared to a current coal-fired plant that consumes 100 tonnes of coal per day, the environmental impact of the 300,000 to 500,000 eCat modules per year is insignificant.
Joe wrote in April 26th, 2013 at 4:42 AM
Wladimir,
1. Deuterons have spin and can interact with each other, creating 2He4. And 2He4 is extremely stable. So how is it possible that the two deuterons in 4Be8 can break the very stable 2He4? (2He4 has no net spin, so it would seem more reasonable for the two deuterons to interact with each other, form 2He4, and exit.)
RESPONSE:
I dont know what really happens. Perhaps the two deuterons interact each other, and form 2He4.
However, it’s possible each deuteron interact with the central 2He4.
According to the nuclear model proposed in QRT, the nuclei have a “hole” in their secondary field. See Fig. 4 in the link:
http://www.journal-of-nuclear-physics.com/files/How%20repulsive%20gravity%20contributes%20for%20cold%20fusion%20occurrence.pdf
Thanks to such “hole” the central 2He4 can exit, for example, the nucleus U238 with energy lower than the energy required for exiting the nucleus (Gamow’s paradox).
http://peswiki.com/index.php/Article:Cold_Fusion_and_Gamow's_Paradox
So, nucleons also can enter easily within a nucleus thanks to that “hole”.
They enter through moving along the z-axis.
Joe, you are trying to understand the behavior of the new nuclear model proposed in Quantum Ring Theory by considering the old nuclear models of classical nuclear physics.
2. In 6C8, why is it impossible for the two diprotons to exit at the same time? (The central 2He4 has no net spin, so it should not interact with the 4 protons in any way. And since they are already coupled, they should leave as pairs rather than singles. There is no force that is breaking them into their two constituent parts.)
RESPONSE:
A) They do not exit at the same time:
6C8 -> 2p + 4Be6
4Be6 -> 2p + 2He4
B) the central 2He4 has TOTAL null spin. However each deuteron of the 2He4 has spin i=1.
Besides, in order to enter within a nucleus, the proton needs a kinetic energy, which helps it to interact with the central 2He4.
3. How can QRT explain nuclei with null magnetic moment, since the (charged) nucleons are always in the fluxes n(o) which pass through the rings of 2He4 which are always DISTANT from the axis (z) of rotation? (A magnetic field will always be generated.)
RESPONSE:
In the nuclei with Z=N=pair there is always two symmetric deuterons regarding the z-axis.
Look for instance the structure of oxygen 8O16 in the Fig. 2 (page 5 of the present paper):
D-2 and D-4 cancel each other
D-1 and D-5 cancel each other
D-3 and D-6 cancel each other
Each other cancel the spin and magnetic moments
Regards
wlad
Dear Yona:
No, I cannot.
Warm Regards,
A.R.
Dear Rèmi Andrè:
I cannot give this information.
Warm Regards,
A.R.
Dear Mr Rossi,
I don’t know if this question has been already asked before so excuse me if it is the case. Do you think that the shape of grains used in your powder is important ( I don’t mean size but shape)
Thank you
Rémi
Dear Andrea Rossi
Can you reveal the names of the Professors of the Indipendent Party that made the report, and from which Universities they are?
Warm Regards
Dear Koen Vandewalle:
1- all the plant has to me reviewed every 6 months
2- we have informatics working with us
Warm Regards,
A.R.
Dear Franco:
We cannot give information about this topic.
Warm Regards,
A.R.
Wladimir,
1. Deuterons have spin and can interact with each other, creating 2He4. And 2He4 is extremely stable. So how is it possible that the two deuterons in 4Be8 can break the very stable 2He4? (2He4 has no net spin, so it would seem more reasonable for the two deuterons to interact with each other, form 2He4, and exit.)
2. In 6C8, why is it impossible for the two diprotons to exit at the same time? (The central 2He4 has no net spin, so it should not interact with the 4 protons in any way. And since they are already coupled, they should leave as pairs rather than singles. There is no force that is breaking them into their two constituent parts.)
3. How can QRT explain nuclei with null magnetic moment, since the (charged) nucleons are always in the fluxes n(o) which pass through the rings of 2He4 which are always DISTANT from the axis (z) of rotation? (A magnetic field will always be generated.)
All the best,
Joe
Dear Andrea,
Dear Steven N Karels,
For everyone that follows Mr. Rossi and the E-cat, it is common knowledge that an E-cat needs maintenance every six months. So it is not only the production capacity of new modules that is important, but also the time, transportation and labour needed to refill or recycle the used reactors.
If we consider the amount of one-way and single-use stuff we use nowadays, even for materials that could be refurbished with some effort e.g. bottles and packaging of foodstuffs, construction materials and some complex technological products and their peripherals (adapters, connectors,…) and how much of the total energy in the world dat is consumed to replace or recycle this stuff, it will be very important not to make the same mistake with the birth of this energy revolution called E-cat.
It may even be very “educative” or “healing” that mankind has to suffer for a while, having less energy and less income to spend for “stuff” in order to learn to re-use again the materials already produced, and come familiar again with an atitude of respect for what nature and efficient production delivers to us.
Thereby comes the next question to Andrea: do you think, from the first experiments, that the mouse of the mouse-and-cat-cat will be the only part that has to be serviced every six months ? And that the higher stages can last for much longer periods since they are of different concept ?
Also very interesting topic of dark lightning, introduced by Enrico Billy. Andrea, do you think that the use of “informatic” to understand and get grip on this effect is possible ? There might be much to do with micro-sized, well organised e-cats, and supercapacitors. Maybe an LENR electric battery. Must be feasible in mid-term range.
Best Regards,
Koen, fighting the spam-robot.
Dear ing. Rossi,
is the EMF (electromotive force) (you spoke in past) caused by a lot of electrons coming from beta minus decay?
Best Regards
Dear Paul:
The tests have involved 15 persons between Professors and Researchers from 4 Universities and I do not know where the publication will be made. I cannot know if between them there was also a reviewer of a scientific magazine. I have not been told anything about this issue.
Warm Regards,
A.R.
Dear DB:
Thank you very much, very interesting.
Warm Regards,
A.R.
@Enrico @Andrea
Maybe this is then also of interest for you, its related to the same claim.
http://www.agu.org/news/press/pr_archives/2013/2013-15.shtml
Andrea,
Where the last set of tests by the University representatives observed by the reviewers of the journal?
If that was the case then you had independent fourth party observers monitoring independent third party testers. If that does not satisfy Carl Sagan’s “Extraordinary claims require extraordinary evidence”, I do not know what will.
Paul
Dear Enrico Billi,
Very interesting link: the study of the dark lighting is an important source of information regarding the LENR. I am glad to see that you, a Physic from the University of Bologna that has made his way in China, are so smart.
Lavolale, lavolale!
Warm Regards,
A.R.
Dear Readers,
i saw the first pubblication about DARK-LIGHTNING, in other words Gamma Ray Burst from coulds. The interesting thing is the source of the gamma burst is not on the top of the cloud but inside the cloud and it is connected with the lightning and radio emissions from the clould.
http://web.ift.uib.no/~nikost/papers/Lis_Rhessi_rev_bold.pdf
LavoLaLe lavoLaLe
Enrico Billi
Dear Gherardo:
1- it depends on specific situations. Hot Cats can be well integrated in big power plants, but what you say can be correct in other situations
2- generally speaking, yes, but in specific situations maybe different
3- you are right, but still we can integrate very well existing big power plant.
I know your feeling, I am very anxious too to read the report of the Professors of the Third Indipendent Party.
Warm Regards,
A.R.
Dott.Rossi,
the 1GW electrical plant is not the right way to use the hot-cat and you know it.
There is no scaling efficency in your cats and so centralizing production is not an efficent solution since it would still get network distribution losses without benefits.
IMHO smaller cat-production plants should be build at electrical intermediate stations to get better average electrical loads while beeing close to users. Any remaining hot water could be punmped nearby to heat houses.
Same logic could be applied to neighbourhood plants or even condominiums.
1) do you generally agree?
2) Going back to my hypotesis, can you confirm that all cat types do not benefit from scaling above the 1MW assembly? I do remember that the 1MW allows better reactors control.
3) let’s not forget that the cop with cat and mouse will likely improve otherwise why make it… Isn’t it?
All those writing while waiting for the official test results…
Thanks a lot for your always kind responses, Gherardo
Dear Andre Blum:
1- still working on this promising issue
2- yes
3- when we will have certain results and explications of phenomenons we are observing
4- nothing more: we are in a R&D phase, too premature.
Warm Regards,
A.R.
Dear Mr Rossi,
It has been almost four months since you answered to questions by Steven Karels about EMF (electromotive force) directly from the reactor core.
1) Is EMF still looking possible, or have results been discouraging?
2) Who is working on this? Is a separate team working on it (like previously you talked about a separate hot-Cat team)?
3) When do you expect you can tell us more about it?
4) The key question: Steven asked whether you thought it was possible to have EMF at decreased temperature, or even without heat production. At the time, your answer was that it was too soon to tell, but you hoped that it would. What can you tell us now?
Many thanks and wishing you all the best
Andre Blum
Dear k.d.
thanks ! I’m learnig so much from you, about italy, my homeland.
thanks again !
warm regards
gio
Dear gio
You did not make mistake with your English. It might better than mine.
In the past I worked with Italian people, with very friendly atmosphere.
But Italian-Americans are American first. To expand Mr. Rossi business involvement in Italy, it is necessary a more friendly atmosphere.
I am talking about past negative experiences of Mr. Rossi in Italy. And this bureaucratic condition exists in many European countries.
But it looks like Sweden is a right place to start.
The problem with traffic jam? Mr. Rossi have solution too.:)) Bicycle
Joe wrote in April 24th, 2013 at 1:18 AM
Wladimir,
1. You state that a bound particle loses its helical trajectory because of the “interaction” between the particle and its central partner. What is the nature of this interaction? (The great speed alone can not account for the loss of radius in the helical trajectory since the speed does not reach that of light c.)
RESPONSE:
I dont know.
The electron can interact with the proton through severa interactions
1- Coulomb attraction
2- Magnetic attraction
3- Spin-interaction
2. How are quarks created? How are they annihilated?
RESPONSE:
Ask it to God
3. 3Li5 should be able to exist as long as the orbiting proton has angular velocity equal to (1.6*10^-19)*((mass of the proton)^-1).
Derivation:
F(M) = F(C)
(1.6*10^-19)*(ang vel)*R = (mass of proton)*((ang vel)^2)*R
ang vel = (1.6*10^-19)*((mass of proton)^-1)
RESPONSE:
Joe, you are forgetting to consider the influence of the flux n(o).
3Li5 must be formed by the capture of a proton by the flux n(o) of a 2He4.
The flux n(o) of the 2He4 is able to capture a deuteron, because the flux n(o) crosses two rings: 1proton+1neutron.
They reinforce the flux n(o).
The deuteron gets its angular velocity thanks to its capture by the flux n(o).
So, when the flux n(o) 2He4 captures a deuteron, and it gets the same angular velocity of the 2He4, the deuteron starts to increase the radius of its orbit, and takes place the mechanism of equilibrium between F(C) and F(M).
The flux n(o) is not able to capture one proton, so that to force the proton to get angular velocity together with the spin of the 2He4.
The proton is captured by the 2He4 with a radius orbit practically null.
Then the mechanism of equilibrium between F(C) and F(M) do not occurs, because the proton does not get angular velocity.
4. The deuterons in 4Be8 are settled in different fluxes n(o), in different halves of the 2He4 space (Ana versus Douglas), approaching an electrically repulsive 2He4, from opposite sides.
a) Why would they even be allowed close to each other, much less to form a 2He4?
RESPONSE:
Look at the Fig. 2.8 of the page 198 of my book (shown in the page 13 the present article).
The Coulomb repulsion existing within the nuclei is due to the secondary Coulomb interactions only. They are very waker than the known principal Coulomb interactions known in Classical Nuclear Physics.
Therefore, the spin-interaction between two deuterons can win the secondary Coulomb repulsion witht the central 2He4.
You have to consider also that the central 2He4 is formed by two deuterons, and they have spin. So, the central 2He4 can also attract the two deuterons in 4Be8, because:
1- as 4Be8 has null magnetic moment, the two deuterons are captured along the z-axis (the radius of their orbits about the z-axis is zero)
2- as the two deuterons are captured along the z-axis, they are not captured by the flux n(o) of the 2He4
3- in order a deuteron to be captured by the flux n(o), the newborn nucleus to be formed must have no null magnetic moment. By this way the deuteron is not captured along the z-axis, and so it can be captured by one of the two sides Ana or Douglas.
b) Even if they would be allowed to interact, why would they not be as easily emissible BEFORE forming a 2He4 as after?
RESPONSE:
One deuteron interacts with one deuteron of the central 2He4, and a new 2He4 is formed
The other deuteron interacts with the other deuteron of the central 2He4, and a new 2He4 is formed.
The reason why it occurs had already been explained in your previous item “a”.
regards
wlad
Dear gio:
he,he,he,he… I too like Benigni.
Warm Regards,
A.R.
Dear k.d.
i made a mistake with my english!!!
I would have to write:”I bet on an italian-american partner” and not hope.
What a mistake!
Warm regards
gio
P.s.:you are right when say that bureaucracy is a problem in italy, but we have a problem worse than bureaucracy…..maybe the worst problem in itay: the traffic jam !
Dear Steven N. Karels:
Yes, you are right.
Thanks,
Warm Regards,
A.R.
Dear Andrea Rossi,
Thank you for your response. Based on the roughly 500,000 modules per year (and I assume a module can output 10kW of Hot eCat power), there should be no problem of supplying the eCat modules for such a theoretical 1GW electric power plant. A single plant does not appear to affect your total possible production capacity. If you reach that point, you will be successful beyond your wildest dreams. Keep up the good work.
gio
April 24th, 2013 at 5:19 AM
Dear Ing. Rossi
well, i hope for an ” Italian-American” partner.
Warm regards
gio!
Sympathizers from every Country like to be Mr. Rossi – Partner.
The problem is the bureaucracy in many countries.
Including Italy
Dear Steven N. Karels:
This question needs a study that cannot be made in minutes. We will face this problem when and if we will foresee that such a problem is in the horizon. In any case, the robotized line we already have designed is able to make half million modules per year, and to double this capacity is not a big problem; besides, much work can be outsourced, and we have already organized a network of manufacturers we can outsource our production to ( for the non confidential parts).
Warm Regards,
A.R.
Dear Ing. Rossi
well, i hope for an ” Italian-American” partner.
Warm regards
gio
Dear Andrea Rossi,
Regarding the question on a 1GW electric power plant using Hot eCat technology.
To produce 1 MegaWatt of electrical power at 47% efficiency requires a thermal source of 2.128 MegaWatts or about 213 Hot eCats each producing 10kW of thermal energy. Or, each 1 GigaWatt of electrical output will require 213,000 Hot eCat 10 kW units. (I ignore any COP issues, assuming the plant would tap some of its output to control the eCats.)
Question: Assuming you and your partner agreed to provide the eCats for such an activity (assuming another corporate entity would actually build the plant), how long would it take to ramp up production to produce the required number (213,000) of Hot eCat units (assuming certifications, etc had been approved)? Months, years?
Dear gio:
Of course there will be this chance.
Warm Regards,
A.R.
Dear Ing. Rossi
you always speak about Usa partner, but there was never the chance to have an Italian partner…….or maybe there will be this chance?
Warm regards
gio
Wladimir,
1. You state that a bound particle loses its helical trajectory because of the “interaction” between the particle and its central partner. What is the nature of this interaction? (The great speed alone can not account for the loss of radius in the helical trajectory since the speed does not reach that of light c.)
2. How are quarks created? How are they annihilated?
3. 3Li5 should be able to exist as long as the orbiting proton has angular velocity equal to (1.6*10^-19)*((mass of the proton)^-1).
Derivation:
F(M) = F(C)
(1.6*10^-19)*(ang vel)*R = (mass of proton)*((ang vel)^2)*R
ang vel = (1.6*10^-19)*((mass of proton)^-1)
4. The deuterons in 4Be8 are settled in different fluxes n(o), in different halves of the 2He4 space (Ana versus Douglas), approaching an electrically repulsive 2He4, from opposite sides.
a) Why would they even be allowed close to each other, much less to form a 2He4?
b) Even if they would be allowed to interact, why would they not be as easily emissible BEFORE forming a 2He4 as after?
All the best,
Joe
Dear frank:
1- we would have to study very carefully the request together with our US Partner.
2- You are so young, that it is impossible to answer: the world changes so fast…
Warm Regards,
A.R.
Dear Dr. Rossi,
just a couple questions, I hope this doesn’t bother you.
if tomorrow morning the government of a western nation would ask you to build (or perhaps would ask for the license of your tech) the thermal apparatuses for a new 1GW thermo-electric plant, would you accept?
I’m about 30 now, do you think I’ll be able to take a trip to low earth orbit with a lenr-propelled vessel, in my lifetime?
Thanks
Dear Brian:
Thank you for your interesting question-information. We do not have contacts with this Laboratory and consequently I have not information about their work, nor I am making any work with them, nor I gave them any information. In any case, I am honoured to read that there are Laboratories that are studying our work.
Warm Regards,
A.R.
Mr. Rossi
I hope this finds you well.
It is being reported here that Elforsk, a Swedish research organization, has designated money for the study of the E-Cat. Are you able to discuss anything about your work with Elforsk? Were the third party testers working for Elforsk?
http://www.e-catworld.com/2013/04/swedish-utilities-rd-organization-elforsk-fund-study-of-e-cat/
Thank you for taking the time to answer my question.
Brian
Dear Robert Curto:
Thank you for your homage to Ferrara: it is a jewel town, dense of History and Art, home of an important University, where we have chosen to put our R&D Center, and where have been made series of tests, among which the Third Party Indipendent test that probably will pass to the History, whatever its content, because it has been the first third party indipendent test run for 120 straight hours.
Warm Regards,
A.R.
Dear Jan.Gustavsson:
In Sweden with our partners of Hydrofusion we are organizing a local production of plants to sell heat to centralized distributors of thermal energy. The particular economic and metheorologic conditions of Sweden makes this Nation particularly fit for our technology.
Warm Regards,
A.R.
Dear Dr Rossi,
you wrote that you are putting strong bases in Sweden: can you explain?
I have read http://www.ingandrearossi.com and I saw your Calvary; now I understand why you don’t care of the snakes: they are like mosquitos biting an elephant.
Continue the good job,
Jan Gustavsson
Dr. Rossi, on Google please enter:
FERRARA ITALY
Click on:
Ferrara-Wikipedia, the free encyclopedia.
Robert Curto
Ft. Lauderdale, Florida
USA
Joe wrote in April 22nd, 2013 at 1:10 AM
Wladimir,
1. Does the fact that a neutrino’s positron orbits the electron in a classical trajectory mean that there exists no aether atmosphere around the electron, since it is the movement of a particle through aether that generates a helical trajectory?
RESPONSE:
No.
The electron within the structure of the neutron also orbits the proton in a classical trajectory, but that does not means that there is no aether between the proton and the electron.
When the speed of a free electron tends to zero, the radius of its helical trajectory tends to infinite.
When the speed of a free electron tends to velocity c of light, the radius of its helical trajectory tends to zero.
Within the neutron the speed of electron is 0,92.c , but as the electron is very close to the proton, due to their interaction the electron loses the helical trajectory (also because its speed is close to speed c of light).
Unlike, a free electron with speed 0,92.c has not a classical trajectory (but the radius of its helical trajectory is very small, because its speed is near to speed c of light).
Only when the electron with speed 0,92.c is captured by the proton (and they form the neutron) the electron loses its helical trajectory.
The same happens within the neutrino, with the electron orbiting the positron.
2. You state that a particle is an agglutination of aether particles which move along a ring-like trajectory, inducing gravitational fluxes n(o) which, in turn, form electric and magnetic fields. But since a magnetic field is initially responsible for circular motion of any charged particles, including aether particles, does this imply that matter can only be created in the magnetic field of existent matter?
RESPONSE:
Magnetic field is NOT initially responsibe for circular motion of any charged particles. The laws within the elementary particles are not the same of the known electromagnetism.
When a particle is created, at the same time it’s created its ring formed by 3 quarks having spin and the flux n(o) crossing within the ring.
They cannot be created separatelly. They are created together. They cannot exist alone.
3. You state that an isotope with greater angular velocity gives greater centripetal force to a nucleon. So in the case of 3Li5, why is it impossible to achieve a stable equilibrium of forces with the lone proton orbiting the 2He4 with a greater angular velocity in order to avoid collapse toward the 2He4?
RESPONSE:
As I said, there is no STABLE equilibrium in nuclei with excess protons.
Read the item 3.13.3 in the page 15:
So, the magnetic force Fm= q.v.B on the proton in 3Li5 is very stronger than the
magnetic force on the deuteron in 3Li6. But the deuteron’s mass is twice of the proton.
Therefore, while the centripetal force on the deuteron succeeds to keep it in equilibrium
in 3Li6, however it is no able to keep the proton in equilibrium in 3Li5, and the proton
is pulled toward the central 2He4 direction by a strong magnetic force, and so it enters
within the structure of the central 2He4, as shown in the inferior side of the Fig. 12, it
breaks up the helium structure, they form a 3Li5 with spin 3/2, and it emits a proton:
3Li5 −> 2He4 + p.
4. You state that similarity of structure gives similarity of decay. And the example that you chose is between 6C8 and 4Be8 (Fig.14), stating, “Like in the case of 4Be8, the daughter isotopes are two 2He4″ (comment from April 14th, 2013 at 12:18 PM). While it is true that 4Be8 decays into two 2He4, 6C8 seems to decay into only one 2He4:
6C8 —> 4Be6 + two protons
4Be6 —> 2He4 + two protons.
Hence, there does not seem to be a similarity of decay.
RESPONSE:
I made a mistake:
4Be8 -> 2He4 + ( 2He4 )
6C8 -> 4Be6 + 2p -> 2He4 + ( 2p + 2p )
But their decay is similar (because the mechanism of decay is similar). Look:
In 4Be8 the two deuterons are captured along the line center of the central 2He4, and therefore there is no centripetal force on them, because their orbit radius is zero. So, the interact with the central 2He4, and it is formed two nucleons 2He4.
In the 6C8 the same mechanism happens with the four protons.
But as they cannot form a 2He4 when they interact with the central 2He4, then the four protons are emitted.
regards
wlad
Dear Kari:
he,he,he…OK, we will publish the photos of the delivery we will make on the 30st of April, all right?
Warm Regards,
A.R.
Dear Dott. Rossi,
Things are quiet and I’m getting bored..why don’t you get us something to chew on, like a picture of the soon to be shipped 1MW plant.
With best regards,
Kari
Wladimir,
1. Does the fact that a neutrino’s positron orbits the electron in a classical trajectory mean that there exists no aether atmosphere around the electron, since it is the movement of a particle through aether that generates a helical trajectory?
2. You state that a particle is an agglutination of aether particles which move along a ring-like trajectory, inducing gravitational fluxes n(o) which, in turn, form electric and magnetic fields. But since a magnetic field is initially responsible for circular motion of any charged particles, including aether particles, does this imply that matter can only be created in the magnetic field of existent matter?
3. You state that an isotope with greater angular velocity gives greater centripetal force to a nucleon. So in the case of 3Li5, why is it impossible to achieve a stable equilibrium of forces with the lone proton orbiting the 2He4 with a greater angular velocity in order to avoid collapse toward the 2He4?
4. You state that similarity of structure gives similarity of decay. And the example that you chose is between 6C8 and 4Be8 (Fig.14), stating, “Like in the case of 4Be8, the daughter isotopes are two 2He4” (comment from April 14th, 2013 at 12:18 PM). While it is true that 4Be8 decays into two 2He4, 6C8 seems to decay into only one 2He4:
6C8 —> 4Be6 + two protons
4Be6 —> 2He4 + two protons.
Hence, there does not seem to be a similarity of decay.
All the best,
Joe
Wladimir, Just read your communication with ‘Journal Science’ 11th March inst. Last question with regards developing the science. They obviously do because they do not want to develop. My reply is that science can reveal a secret that they are paid to keep. Regards Eric Ashworth.
Wladimir, This is a small piece of information that may help explain some anomalies. This is a second part to my earlier theory regarding quarks as constructing containers. Aether is mass less because it contains no gravity i.e. empty space. There are only two primary states from which structured aether is formed. These being mass less aether and empty space, both of which although being integral to structure constitute as being nothing when out of geometric format because aether has no charge. It is the interaction of these two particulars that create the material world. The first force in this interaction comes from an outside source that reduces a volume of empty space that aether has been able to freely roam within. With a decrees in volume the aethers achieve similarity of vibration resulting in aether rejection from the central mass in an outward direction. Aether consequently travels outwards on a vector which is a radial dimension of the environmental critical mass. The radial dimension comprised of aether by its activity creates an empty space behind it i.e. from the inner most position. Because aether has an affinity to empty space its vector is changed into a tangential curvature circumferential dimension. This activity is with regards an aether closest to the central position of the mass and thereby because the aether in front detects an empty space behind because of the previous activity of an aether it too goes through the same motion but it fails to get into the empty space due to the approach of the first aethers arrival and therefore it continues downward to fill the newly formed empty space from which the first aether has now departed. Consequently, the third aether above senses the empty space created by the activity of the second aether behind and it too reacts in a similar fashion. The first aether to turn forms a tight loop being most inner most. The second aether to turn forms a loop less tight and the third aether a less tight loop than the second. Each aether has to circumnavigate a 180 degree trajectory in order to satisfy a dimension of empty space and to do this the structure has to spin on its diameter. Thereby the structure that is now formed is comprised of two aethers trvelling up on a diametrical dimension and one aether travelling down on a circumferential dimension. What has been formed is an aether energy knot comprised of three gravity values that cannot be undone because this knot has been formed by the primary environmental creating force. This aether knot is a kinetic neutral that could be called a quark or a photon depending whether it is a part of a triplicity or whether it is out of its triple interaction. In the next phase quarks traveling on a vector go through the same motions as the aether and this structure is now called proton, neutron and electron that by consequence make an atom. Why should a neutron equal a proton and an electron?. Each of these structures is with regards to the environmental creating force with regards position upon a radial dimension or you could say time dimensional units of force that are tied into a knot. Consequently the neutron does not have to equal the proton and electron. The proton is u.u.d. with regards quark behaviour and its knot. The neutron is d.d.u. with regards quark behaviour and its knot. These knots are comprised of a spinning 180 degree of their quarks (all aethers and quarks travel on helical trajectories) and each spin upon their diametrical dimension. To unify the proton and neutron the neutron flips through 180 degrees because of polar attraction between both units and in this configuration the proton and neutron oscillate towards and away from each other due to the 180 degree interaction of the circumferential quarks. The electron I guess has the same u.u.d. configuration as the proton but because of its massive exterior trajectory it lacks detectable structural density.
My previous reference in my last message to exterior radials was with regards to end systems/structures that have completed their rounds with regards exterior environmental force and thereby by polarization, an exteriuor vector force activity is set in motion due to none inclusiveness within the structure of its aether. This information I believe is a tiny part of a massive riddle that will eventually unfold itself as more people become involved. It is as you have said Wladimir it will take more than one mind to solve the puzzle but as we all know from a realization will come a revalation. I shall continue to study your QRT material and keep you informed of my thoughts. All the best to both you and Andrea Rossi Eric Ashworth.