Physics of rotating and expanding black hole universe

by
U.V.S.Seshavatharam
Honorary Faculty, Institute of Scientific Research on Vedas(I-SERVE)
Hyderabad-35, India
Email: seshavatharam.uvs@gmail.com
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Introduction
Now as recently reported at the American Astronomical Society a study using the Very Large Array radio telescope in New Mexico and the French Plateau de Bure Interferometer has enabled astronomers to peer within a billion years of the Big Bang and found evidence that black holes were the first that leads galaxy growth. The implication is that the black holes started growing first. Initially astrophysicists attempted to explain the presence of these black holes by describing the evolution of galaxies as gathering mass until black holes format their center but further observation demanded that the galactic central black hole co-evolved with the galactic bulge plasma dynamics and the galactic arms. This is a fundamental confirmation of N. Haramein’s theory described in his papers as a universe composed of “different scale black holes from universal size to atomic size”.
This clearly suggests that: galaxy constitutes a central black hole; the central black hole grows first; Star and galaxy growth goes parallel or later to the central black holes growth. The fundamental questions are: If “black hole” is the result of a collapsing star, how and why a stable galaxy contains a black hole at its center? Where does the central black hole comes from? How the galaxy center will grow like a black hole? How its event horizon exists with growing? If these are the observed and believed facts — not only for the author — this is a big problem for the whole science community to be understood.
Any how, the important point to be noted here is that “due to some unknown reason galactic central black holes are growing”! This is the key point for the beginning of the proposed expanding or growing cosmic black hole! See this latest published reference for the “black hole universe”. In our daily life generally it is observed that any animal or fruit or human beings (from birth to death) grows with closed boundaries (irregular shapes also can have a closed boundary). An apple grows like an apple. An elephant grows like an elephant. A plant grows like a plant. A human grows like a human. Through out their lifetime they won’t change their respective identities. These are observed facts. From these observed facts it can be suggested that “growth” or “expansion” can be possible with a closed boundary. By any reason if the closed boundary is opened it leads to “destruction” rather than “growth or expansion”. Thinking that nature loves symmetry, in a heuristic approach in this paper author assumes that“ through out its lifetime universe is a black hole”. Even though it is growing, at any time it is having an event horizon with a closed boundary and thus it retains her identity as a black hole forever. Note that universe is an independent body. It may have its own set of laws. At any time if universe maintains a closed boundary to have its size minimum at that time it must follow “strong gravity” at that time.
If universe is having no black hole structure any massive body(which is bound to the universe) may not show a black hole structure. That is black hole structure may be a subset of cosmic structure. This idea may be given a chance.
Rotation is a universal phenomenon. We know that black holes are having rotation and are not stationary. Recent observations indicates that black holes are spinning close to speed of light.
In this paper author made an attempt to give an outline of “expanding and light speed rotating black hole universe” that follows strong gravity from its birth to end of expansion.
Stephen Hawking in his famous book A Brief History of Time, in Chapter 3 which is entitled The Expanding Universe, says: “Friedmann made two very simple assumptions about the universe: that the universe looks identical in which ever direction we look, and that this would also be true if we were observing the universe from anywhere else. From these two ideas alone, Friedmann showed that we should not expect the universe to be static. In fact, in 1922, several years before Edwin Hubble’s discovery, Friedmann predicted exactly what Hubble found… We have no scientific evidence for, or against, the Friedmann’s second assumption. We believe it only on grounds of modesty: it would be most remarkable if the universe looked the same in every direction around us, but not around other points in the universe”.
From this statement it is very clear and can be suggested that, the possibility for a “closed universe” and a “flat universe” is 50–50 per cent and one cannot completely avoid the concept of a “closed universe”.
Clearly speaking, from Hubble’s observations and interpretations in 1929, the possibility of “galaxy receding” and “galaxy revolution” is 50–50 per cent and one cannot completely avoid the concept of “rotating universe”.
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437 comments to Physics of rotating and expanding black hole universe

  • Koen Vandewalle

    Dear Andrea,
    Do you already have a – maybe theoretical, draft or imaginary – reactor design that produces and exchanges enough heat to have the abilities to power an aircraft on LENR only ?
    I know this depends on engine design, but since you are working very well with a very good team, and we do not have any idea how the latest versions of your inventions look like, I dare to ask.
    Please consider the theoretical or imaginary option, based on your daily experiments and experience.
    Maybe a wing-shaped thruster with very few moving parts, e.g. pulsejet or ramjet like device, but flat.
    I do not know if it is possible to create a flat hot cat of huge dimensions.
    I hope you have enough time to dream about new things.
    Kind Regards,
    Koen

  • Andrea Rossi

    Eernie1:
    I can say nothing until the report of the Third Indipendent Party is published. I also must remind you that to talk of COP is premature, because at the moment I must say that the results could be positive or negative.
    Warm Regards,
    A.R.

  • eernie1

    Dear Andrea,
    I know you cannot talk about the work of the Third Indipendent Party because of the NDA. Does this NDA also include your present test and development efforts? Can you tell us if you are achieving greater than 6 COP? Are your efforts more positive than you expected?
    Thank you for any provided information you are allowed to disclose.

  • Andrea Rossi

    Steven N. Karels:
    You got the point!
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    Compliments to you for your burning desire to see LENR work. I would never have taken the risk. It must be extremely dis-heartening and discouraging when you are widely criticized for only trying to protect your intellectual property. You must face a daily struggle on how much information to release. Yet, I will continue to inquire and ask that which you may not wish to disclose. The game goes on, my friend.

  • Andrea Rossi

    Mark:
    Try to merit what God gave me: my life.
    Warm Regards,
    A.R.

  • Mark

    Hi Andrea,

    I guess that your motivations are not all based on money and returns. What else is motivating your hard work and dedication?

    Mark

  • Andrea Rossi

    Andrew:
    My personal investments in this technology began in 1997 in the USA, when I started to spend real money after some year of theoretical studies on the issue. Since then I spent for this barely all my available money, whose source was mainly the manufacturing of energy generators working with vegetable derivated fuels ( a very difficult art too!), a field in which I had strong bases from my past ( see http://www.ingandrearossi.com). To the R&D for the E-Cat have been spent also all the proceeds I got when I decided to leave any other work and focus exclusively on the LENR, without possibility of return: sort of “make this or die”. Without this decision I could not crop the ferocity necessary to get real results. It is impossible to calculate how much I invested in about 16 years: I didn’t hold the record of the worth , but we are in the range of millions.
    Warm Regards,
    A.R.

  • Andrew

    Dear Andrea Rossi,
    Is it possibile to quantificate the amount of money that has been invested from you in your technology from the very beginning?

  • Wladimir Guglinski

    Joe wrote in April 2nd, 2014 at 2:39 AM

    Wladimir,

    1. In QRT, why do the deuterons of 2He4 have no spin? Are they not centred on each others’ gravitational fluxes n(o) and have spin interaction (SI) between them?
    ———————————————
    COMMENT

    The deuterons of the central 2He4 have spin. The induction of the flux n(o) of the nuclei is just due to the spin of the deuterons of the 2He4.

    .

    ———————————–
    Are they not centred on each others’ gravitational fluxes n(o) and have spin interaction (SI) between them?
    ———————————-

    COMMENT
    They are centred on each other.
    The Fig. 4 does NOT show them centered each other because it is only an artistic picture, so that to make easier the understanding, because if I did not separate the two deuterons in the figure, there would be an overlap of the flux n(o)-up and the flux n(o)-down, and therefore would be difficult for the reader to understand how the two fluxes work.

    The real position of the deuterons in the 2He4 is shown in the Fig. 6, and many other figures along the paper, showing the two deuterons always centered.

    .

    —————————————-
    2. You state the following:
    “Therefore the proton and the side ANA of the 2He4 induce a collective field of magnetons m(+), responsible for a magnetic force of attraction between the proton and the 2He4.”
    (I presume that conversely, on the side of DOUGLAS, the field would be m(-).)
    —————————————–

    COMMENT

    No, in the side DOUGLAS the field is not m(-). It is also m(+), since it is generated by the proton.
    However, as the spin of the magnetons m(+) in the outer side of ANA is contrary of the spin of magnetons m(+) in the outer side of DOUGLAS, then the total magnetic moment of the 2He4 is zero.
    —————————————-

    .

    ————————————-
    My question is, should not the proton and 2He4 be repelled rather than attracted to each other if they have the same magnetic sign?
    ———————————–

    COMMENT

    Look at the side ANA. It has two sides: OUTER side and INNER side.
    In the outher side the flux n(o)-up is moving up, while in the inner side the flux n(0) is going down.
    A proton situated in the OUTER side of ANA induces magnetons m(+) with spin-up, because the flux n(0) is moving in the up direction when it crosses the proton.

    Unlike, the deuteron of the 2He4 of the side ANA induces magnetons m(+) with spin-down, because the flux n(o) moves along down direction when it crosses the proton (of the deuteron of 2He4).

    Therefore the magnetic field induced by the proton in the OUTER side of ANA has ATTRACTION with the magnetic field induced by the deuteron of the 2He4, because:

    a) The field induced by the proton in the OUTER side is produced by magnetons m(+) with spin-up.

    b) The field induced by the deuteron of the 2He4 is produced by magnetons m(+) with spin-down.

    c) Therefore the proton-deuteron works as it should be a micro-magnetic-loadswtone with poles N and S, with the poles N and S in a position along the horizontal line.
    ——————————-

    .

    ——————————–
    3. Even if the neutron has effectively no gravitational flux n(o) to excite permeabilitons to create a magnetic field for it and have it interact with the magnetic field of 2He4 in order to create a supposed attraction between the two of them, the neutron still has an intrinsic magnetic moment that should interact with the magnetic field of 2He4.
    —————————-

    COMMENT
    No, the poles N and S of the neutron are situated along a vertical line, because of the spin of the neutron (see a neutron shown as yellow in Fig. 3, with its two poles (+) and (-) in the vertical line).

    And remember that the loadstone formed by the proton-deuteron has the poles N and S situated along an horizontal line.

    Therefore the attraction of the pole (-) of the neutron with the loadstone proton-He4 is cancelled by the repulsion of its pole (+) with the loadstone proton-2He4.
    ————————————–

    .

    —————————————
    4. You state that the neutron is captured by 2He4, even if only momentarily. But how is this done if the neutron has no effective gravitational flux n(o) with which to interact with the gravitational flux n(o) of 2He4? Why are neutrons that are in stable orbit about 2He4 constrained to the gravitational fluxes n(o) of 2He4 if neutrons do not even have effective gravitational fluxes n(o) of their own that would help to accomplish this?
    —————————————-

    COMMENT
    Neutrons are always kept in the nuclei via their spin-interaction with deuterons.

    The neutrons do not stimulate the permeabilitons, since the flux n(0) of neutrons is formed by gravitons g(+) and g(-), and they cancell each other. Therefore neutrons cannot interact with the central 2He4 as happens with protons and deuterons.

    However, I suppose that a flux n(o) of gravitons g(+) can interact with another flux n(o) of gravitons g(+) when they have relative motion between them.
    I suppose this is the origin of the strong nuclear force, detected in experiments of scaterring proton-proton. The magnitude of the interaction depends on the speed of their relative motion.

    Two gravitons g(+) do not interact if one is at rest regarding to the other (without relative motion).

    But they can interact if they have relative motion.

    So, the flux of graviitons g(+) of the proton in the structure of the neutron can interact with the flux n(o) of the 2he4.
    But obviously it is a weak interaction.
    ———————————-

    regards
    wlad

  • Andrea Rossi

    TO THE READERS OF THE JoNP:
    Today has been presented the book of the scientific journalist Mats Lewan ” An Impossible Invention”: it is professional, honest and sincere as his Author. For those interested, the contact is
    http://www.animpossibleinvention.com
    Andrea Rossi

  • Joe

    Wladimir,

    1. In QRT, why do the deuterons of 2He4 have no spin? Are they not centred on each others’ gravitational fluxes n(o) and have spin interaction (SI) between them?

    2. You state the following:
    “Therefore the proton and the side ANA of the 2He4 induce a collective field of magnetons m(+), responsible for a magnetic force of attraction between the proton and the 2He4.”
    (I presume that conversely, on the side of DOUGLAS, the field would be m(-).)
    My question is, should not the proton and 2He4 be repelled rather than attracted to each other if they have the same magnetic sign?

    3. Even if the neutron has effectively no gravitational flux n(o) to excite permeabilitons to create a magnetic field for it and have it interact with the magnetic field of 2He4 in order to create a supposed attraction between the two of them, the neutron still has an intrinsic magnetic moment that should interact with the magnetic field of 2He4.

    4. You state that the neutron is captured by 2He4, even if only momentarily. But how is this done if the neutron has no effective gravitational flux n(o) with which to interact with the gravitational flux n(o) of 2He4? Why are neutrons that are in stable orbit about 2He4 constrained to the gravitational fluxes n(o) of 2He4 if neutrons do not even have effective gravitational fluxes n(o) of their own that would help to accomplish this?

    All the best,
    Joe

  • Andrea Rossi

    Mark:
    They are in the neutral laboratory where the test is done in turns or all together, depending on the work they have to do, they have faculty of access to the laboratory 24 hours per day, at any time and they have access 24 hours per day to the data registration of temperatures and energy, which is permanent 24 hours per day, every day with no interruption of sort. They told me two days ago that they have elaborated more than one million of data already. I have no data at all, being the registration made by their instrumentation and being restricted to them the consultation of the same. In the lab there are four videocameras that make the registration of any leaf that can fly in any part of the laboratory.
    Warm Regards,
    A.R.

  • Mark

    Hi Andrea
    Are you telling us that the professors “live” with the E-Cat in the same building?

    Mark

  • Andrea Rossi

    Giuliano Bettini:
    Believe it or not, I have no idea of the measurements made by the Professors, their methodology and their calculations, therefore at the moment I do not know if their results will be positive or negative. They are on the piece, our electronic engineer can only give assistance in case of malfunctions ( no one happened so far), I am distant from the Hot Cat; besides, I have to make my normal work in other places most of the time, so I do not know the numbers obtained in most of the test; I have no access to the registrations too, that the Professors take 24 hours per day regarding the temperatures and the consume of energy. Sincerely, I do not know if the results will be positive or negative. Obviously, to be totally sincere, I have my impression about how things are going, but I signed an NDA that forbids to all of us to give any bit of information to the public before the report is published.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Frank Acland:
    Your assessment is relevant.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear readers of the JoNP
    God would be very stupid if He had created the Universe from an empty space (with no structure formed by particles and antiparticles), as considered in Modern Physics, and also if He had conceived the structure of the matter with the ingenuous model proposed in the Particle Physics and Nuclear Physics.

    Happily, God did not use the Heisenberg’s phantasmagoric scientific method when He decided to create the Universe. Otherwise His attempt would not succeed, as occurred with the theoretical attempt made by the physicists.

    regards
    wlad

  • Frank Acland

    Dear Andrea,

    I believe that it would be a great service to the general public if the publication of the report was accompanied by a press conference. I hope you and the Industrial Heat team will hold one, whatever the results of the report are.

    Many thanks,

    Frank Acland

  • Wladimir Guglinski

    Dear Joe
    you are doing an accurate analysis of my nuclear model and the structure of the aether.

    And we are realizing that the there is logic in the model.

    But, beyond the logic intrinsic in the theory, we are realizing that it is able to explain the nuclear properties of nuclei, and many puzzles of Modern Physics, as for instance that question risen by you: “preference for matter over antimatter in the Universe“, not responded yet by any theory.

    So, there is an interesting overlap, indispensable for any acceptable theory:
    1- The theory is logic
    2- The theory is able explain to the nuclear properties and the experimental findings.

    And as you know, the indispensable overlap does not occur in the current Classical Nuclear Physics, because there are many unsolved puzzles, and the new experiments made between 2009 and 2013 with light nuclei are bringing more and more unsolved puzzles for the Classical Nuclear Physics, while the nuclear theorists are trying uselessly desperated attempts so that to solve the puzzles, via the phantasmagoric Heisenberg method.

    When in 1990 I had undertaken to discover a new nuclear model, I decided to face that challenge because I had realized that current Nuclear Physics is unable to satisfy the indispensable overlap between logic and the nuclear properties of nuclei.

    The experimental findings concerning the light nuclei published between 2009 and 2013 are reinforcing that my initial certainty of 1990: some fundamental principles are missing in the current Nuclear Physics.

    regards
    wlad

  • Giuliano Bettini

    Dear Andrea,
    you wrote for the nth time on March 31st, 2014 at 1:52 PM:
    (…) Now it is very important to see the results of the third indipendent party long run test. (…) I have still to say that the results can be positive or negative (…).
    That’s very interesting but, since you’re an Italian, it is obviously not credible that at this point you don’t know anything about the “positive or negative”.
    I mean, you know nothing of what professors will write, but obviously you know very well how things went.
    Your personal feeling, about which you obviously can not speak, what is it?
    :)
    Indiscreet regards,
    Giuliano Bettini.

  • Andrea Rossi

    Mark:
    Only if there will be a demand for it. In this case, it will be held either the results will be positive or negative for us, for intellectual honesty.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe wrote in March 30th, 2014 at 9:00 PM
    Wladimir,
    1. If QRT can allow a weak spin interaction (SI), which is also necessarily an interaction of magnetic moments, between the halo neutron and part of the non-magnetic nuclear core 4Be10 (a deuteron) of 4Be11, why could QRT not also allow the same in the case of 2He5 (one of the deuterons in 2He4) which would increase the half-life of 2He5 to a level similar to 4Be11?
    ———————————————————————————-

    COMMENT
    There is a difference between the deuteron of 4Be10 and the deuterons of the 2He4, as follows:

    a) The deuterons of 2He4 produce the flux n(o) formed by gravitons g(+), and this flux captures other deuterons. The deuterons of 4He4 do not gyrate individually, so that they could get spin-interaction with other nucleons, because those other nucleons are just captured by the flux n(o) produced by the deuterons of the 2He4.

    b) Each of the two deuterons of 4Be10 are captured by the flux n(o) of the 2He4. Those two deuterons gyrate (spin) having as center of rotation the flux n(o) of the 2He4 which crosses the two deuterons, and therefore those two deuterons of 4Be10 have each one an individual spin, able to interact with other nucleons, also captured by the flux n(o) of the 2He4.

    .

    —————————————
    2. You say the following:
    •There is no magnetic interaction when the halo neutron is close to the cluster, within the structure of 4Be11
    •There is magnetic attraction between the halo neutron and the cluster when they are separated by a distance of 7fm
    i) How can distance dictate whether or not a magnetic interaction occurs?
    ii) How can there be a magnetic interaction between the halo neutron and the core since the core (4Be10) has mu = 0? (Unless, of course, you mean a weak magnetic interaction between the halo neutron and a deuteron in the core as mentioned in (1) above.)
    —————————————————————

    COMMENT
    Joe,
    you are right.
    Consider a core with magnetic moment. A neutron very near to the core is situated within the field Sp of the core. And a halo-neutron in a distance of 7fm of the core is penetrating in the field Sn (responsible for electromagnetic fields). Therefore, for a core with magnetic moment we can explain how distance can dictate whether or not a magnetic interaction occurs.

    But the 4Be10 has no magnetic moment, and so its field Sn has no magnetic moment too, and there is no way to explain how can distance dictate whether or not a magnetic interaction occurs (in the case of the halo-neutron of 4Be11).

    So, it think we have to consider your suggestion: “a weak magnetic spin-interaction between the halo neutron and a deuteron in the core as mentioned in (1) above.)

    Therefore, the radius of the orbit of the neutron in the 4Be11 grows slowly, and when the neutron arrives to the point with radius 7fm it decays, emitting an electron.

    .

    —————————————-
    3. What classical electromagnetic law are you applying to justify a charged particle being attracted magnetically centripetally toward (the core of) a nucleus which has no magnetic moment? (Example, 3Li5 with its free proton spiraling toward 2He4.)
    ——————————————————————

    COMMENT
    Joe,
    Faraday made several experiments, so that to discover the laws of the electromagnetism. But I cannot make experiments so that to discover the laws within the nucleus. I have to use the logic.

    If the charged particle was not be attracted centripetally toward the core (as a proton moving about the 2He4 in the isotope 3Li5), then it would be impossible to get the stability of the nucleus.

    There is not a similar “classical electromagnetic” macroscopic Law in the sense of those discovered by Faraday. He discovered macroscopic laws. Within the nucleus we have to discover microscopic laws.

    But let us try to understand the mechanism of such a new microscopic Law.

    a) The proton has a flux n(o) formed by gravitons g(+).

    b) The 2He4 has a flux n(o) formed by gravitons g(+)

    c) The proton is captured by the flux n(o) of the 2He4, and so the proton has two motions:
    c-1) About the 2He4
    c-2) A spin, which center of rotation is the flux n(o) of the 2He4

    Let us remember the Law 3 concerning the permeabilitons:

    Law 3- The permeabilitons p(+) excited by the flux n(o) of the proton capture magnetons m(+), and so a magnetic field is formed by magnetons m(+) in the proton.

    The 2He4 has two fluxes, n(o)-up and n(o)-down (see Fig. 4. In the paper Stability of Light Nuclei).

    Suppose that the proton is captured by the flux n(o)-up in the side ANA of the 2He4.

    In spite of it is zero the TOTAL magnetic moment produced by n(o)-up and n(o)-down (and that’s why 2He4 has null magnetic moment), however the flux n(o)-up captures permeabilitons with spins in the same direction of the spins of permeabilitons captured by the flux n(o) of the proton. Therefore the proton and the side ANA of the 2He4 induce a collective field of magnétons m(+), reponsible for a magnetic force of attraction between the proton and the 2He4.

    If the proton is captured by the flux n(o)-down of the side DOUGLAS, there is changing in the direction of the spin of permeabilitons captured by the 2He4, but the proton also changes its spin in the side DOUGLAS (see Fig. 4), and therefore the permeabilitons captured by the proton have again the same direction of the spins of permeabilitons captured by the flux n(o)-down of 2He4.

    In the case of the neutron (formed by proton+electron), as the proton has a flux n(o) formed by g(+), and the electron has a flux n(o) formed by g(-), the capture of the neutron by the 2He4 does not generate a magnetic force of attraction, because the flux n(o) of the electron cancells the action of the flux n(o) of the proton.

    Regards
    wlad

  • Mark

    Hi Andrea
    Do you think a press conference will be held when the report is released?

    Mark

  • Andrea Rossi

    Bernie Koppenhofer:
    We have all the resources we need. Our work is going on as scheduled. Now it is very important to see the results of the third indipendent party long run test. In the meantime we are carrying on an intensive work of R&D. I have still to say that the results can be positive or negative and so far we have to work hard and be patient.
    Warm Regards,
    A.R.

  • Bernie Koppenhofer

    Dr. Rossi: Would adding additional resources to your research/production efforts speed up the introduction and use of your new technology?

  • Joe

    Wladimir,

    1. If QRT can allow a weak spin interaction (SI), which is also necessarily an interaction of magnetic moments, between the halo neutron and part of the non-magnetic nuclear core 4Be10 (a deuteron) of 4Be11, why could QRT not also allow the same in the case of 2He5 (one of the deuterons in 2He4) which would increase the half-life of 2He5 to a level similar to 4Be11?

    2. You say the following:

    •There is no magnetic interaction when the halo neutron is close to the cluster, within the structure of 4Be11

    •There is magnetic attraction between the halo neutron and the cluster when they are separated by a distance of 7fm

    i) How can distance dictate whether or not a magnetic interaction occurs?

    ii) How can there be a magnetic interaction between the halo neutron and the core since the core (4Be10) has mu = 0? (Unless, of course, you mean a weak magnetic interaction between the halo neutron and a deuteron in the core as mentioned in (1) above.)

    3. What classical electromagnetic law are you applying to justify a charged particle being attracted magnetically centripetally toward (the core of) a nucleus which has no magnetic moment? (Example, 3Li5 with its free proton spiraling toward 2He4.)

    All the best,
    Joe

  • Andrea Rossi

    Herb Gillis:
    I am not able to comment the issue you proposed, because I did not observe it so far.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe,
    I made a mistake in my previous explanation of the half-life of 2He5:
    ————————————————–
    So, the time-life of 2He5 cannot be compared with that of 4Be11, because in 2He5 the proton has a strong attractive magnetic force with the core, and also because the proton is very near to the core.
    ————————————————–
    because actually there is an excess neutron in 2He5, and NOT an excess proton.

    So, let me propose another explanation for your question:
    —————————————–
    1. In QRT, the halo neutron of 4Be11 has a repulsive centripetal force.
    Therefore, 4Be11 can not possibly have a half-life anywhere near 13.8s by the laws of QRT. The precedent is that of 2He5 which has a half-life of only 7.00*10^-22s because the gravitational flux n(o) can not counter the repulsive centripetal force acting on the neutron orbiting 2He4 (mu = 0).
    —————————————–

    Joe,
    look at the Fig. 56 (page 74) of the paper Stability of Light Nuclei, showing the halo-neutron of the 4Be11:

    we see that the halo-neutron n-2 in that figure has a spin-interaction with the deuteron D-2.
    So, I suppose that such spin-interaction is weak, and it cannot keep the neutron n-2 in the structure of 4Be11, however the spin-interaction is enough to increase the time of the halo-neutron going around the core.

    Unlike, in the case of the 2He5, the excess neutron has not any deuteron available, so that to get a spin-interaction with it. That’s why the neutron goes very fast moving away of the core.

    It seems the explanation now is very coherent

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in March 29th, 2014 at 4:41 AM

    2. In QRT, 5B11 (halo proton in passing only) is initially similar to 3Li5 which is unstable (half-life = 3.70*10^-22s) due to the gravitational flux n(o) not being able to counter the sum of the repulsive centripetal and electrostatic forces, and an ODD attractive magnetic force induced by the initial momentary orbit of the proton’s charge about the 2He4 (mu = 0) caused by the gravitational flux n(o), acting on the orbiting proton. Stability is gained when the proton that heads toward 4Be10 (mu = 0) couples to a neutron along the way to form a deuteron.
    ———————————————–

    COMMENT

    Joe
    you cannot compare 3Li5 with 5B11.

    The 3Li5 instability is analysed in the Figure 12 at the page 15 of the paper Stability of Light Nuclei
    As there is not a neutron able to capture the proton, and so the proton is captured by the central 2He4

    Unlike, in the core of 5B11 there is a neutron able to capture the helo-proton

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in March 29th, 2014 at 4:41 AM

    1. In QRT, the halo neutron of 4Be11 has a repulsive centripetal force.
    Therefore, 4Be11 can not possibly have a half-life anywhere near 13.8s by the laws of QRT. The precedent is that of 2He5 which has a half-life of only 7.00*10^-22s because the gravitational flux n(o) can not counter the repulsive centripetal force acting on the neutron orbiting 2He4 (mu = 0).
    ————————————————–

    COMMENT

    Joe,
    in the item 3.8- Equilibrium via strong spin-interaction and via magnetic force at the page 7 of the paper Stability of Light Nuclei it is written:

    ========================================
    A deuteron alone can be captured by the flux n(o) and a stable isotope can be formed, as
    for instance in the case of the isotope 3Li6. As there is not another nucleon to have spin-interaction with that deuteron, then it is kept in 3Li6 via magnetic interaction, as
    follows: the magnetic force on the deuteron is equilibrated by the centripetal force on it,
    as we will see in the calculations.
    Unlike, an alone neutron or proton captured by the flux n(o) cannot form a stable isotope because of the following reasons:
    a) neutron: it is no able to yield a magnetic force within the magnetic field of a nucleus, because it has no charge.
    b) proton: the magnetic field on the proton is stronger than the centripetal force on
    it, because its mass is half of the deuteron’s mass. So the proton is captured by
    the central 2He4, and the nucleus decays.
    c) That’ why 2He5 and 3Li5 cannot be stable.
    ========================================

    So, the time-life of 2He5 cannot be compared with that of 4Be11, because in 2He5 the proton has a strong attractive magnetic force with the core, and also because the proton is very near to the core.

    The time-life of 4Be11 is explained in the page 72:

    =========================================
    NOTE: The magnetic interaction between two loadstones A and B depend on the relative position of their south and north poles.
    If the north pole of A is close to the north pole of B there is repulsion. If the south pole of A is close to the north pole of B, there is attraction. The same happens with the nu
    cleons. But when they are within the structure of an isotope, they do not experience magnetic interaction similar to that between the loadstones A and B, because the nucleons are between the two poles north and south of the isotope, since they are within its. Fig. 55 shows the difference:

    •There is no magnetic interaction when the halo neutron is close to the custer, within the structure of 4Be11

    •There is magnetic attraction between the halo neutron and the cluster when they are separated by a distance of 7fm

    4. As the angular velocity is decreasing, also decreases the centripetal force. But as the magnetic moment of the neutron do not decrease, when it arrives to the orbit with radius R= 7fm the orbit radius stops to grow, because the centripetal force Fc= m.ω2.R decreases with the square of the angular velocity (which decreases up to the instant when Fm and Fc get equi
    librium in the orbit with R=7fm)

    5. Finally the neutron n-2 stays orbiting the cluster with radius 7fm along 13,81
    seconds, as shwon in Fig. 56. Wasted that time, the halo neutron n-2 decays emitting an electron.
    =========================================

    regards
    wlad

  • Herb Gillis

    Dr. Rossi:
    Do you think it is possible that endothermic nuclear reactions might be happening (in addition to exothermic reactions you know about) in LENR devices like the Ecat? Could it be that a portion of the energy released in the exothermic reactions is being consumed in other endothermic reactions? I cannot think of any reasons why endothermic reactions could not be driven by whatever lattice phenomena are responsible for exothermic reactions. Unlike “hot” fusion, LENR is probably not a thermo-nuclear process. Can you comment on this issue?
    Kind Regards; HRG.

  • Andrea Rossi

    Eric Ashworth:
    I appreciate your help.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Ville Kanninen:
    Thank you for the information,
    Warm Regards,
    A.R.

  • Dear Andrea Rossi,

    There might be something that interests you in this article

    – Popularized version in phys.org
    Physicists propose explanation for metals behaving badly, http://phys.org/news/2014-03-physicists-explanation-metals-badly.html

    – Original article in Nature Physics
    Origins of bad-metal conductivity and the insulator–metal transition in the rare-earth nickelates, http://www.nature.com/nphys/journal/vaop/ncurrent/full/nphys2907.html

    kind regards

    Ville Kanninen

  • Eric Ashworth

    Dear Andrea, As I understand, the e-cat works upon a natural process of LENR. It runs in self sustain mode for so long after which it requires an injection of energy to bring it back into self sustain mode for another duration of time. Have you considered that macro effects could be totally different than micro effects of the same process. What I am getting at is maybe if you could build a macro e-cat say 20 times bigger than what you have then maybe the effect produced would be that much different, a sustainable source of heat. What I am thinking is that at a macro size the major hurdle could be to control the overal output from running away, if so perhaps by reducing the hydrogen to maintain a required level. If this suggestion is too simplistic or irrelevant. You do not need to reply. All the best Eric Ashworth.

  • Joe

    Wladimir,

    (In my last posts, it was wrong of me to say “nucleus”. Rather, it should read “core of the nucleus”.)

    In QRT, the nuclear core of both 4Be11 and 5B11 is 4Be10 (mu = 0).
    (Since both the neutron and the proton interact with 4Be10 in its entirety, it is not necessary to analyze their interaction with 4Be10 in its various parts.)

    1. In QRT, the halo neutron of 4Be11 has a repulsive centripetal force.
    Therefore, 4Be11 can not possibly have a half-life anywhere near 13.8s by the laws of QRT. The precedent is that of 2He5 which has a half-life of only 7.00*10^-22s because the gravitational flux n(o) can not counter the repulsive centripetal force acting on the neutron orbiting 2He4 (mu = 0).

    2. In QRT, 5B11 (halo proton in passing only) is initially similar to 3Li5 which is unstable (half-life = 3.70*10^-22s) due to the gravitational flux n(o) not being able to counter the sum of the repulsive centripetal and electrostatic forces, and an ODD attractive magnetic force induced by the initial momentary orbit of the proton’s charge about the 2He4 (mu = 0) caused by the gravitational flux n(o), acting on the orbiting proton. Stability is gained when the proton that heads toward 4Be10 (mu = 0) couples to a neutron along the way to form a deuteron.

    All the best,
    Joe

  • Andrea Rossi

    Steven N. Karels:
    I cannot give any information about this issue before the publication of the third indipendent party . The results could also be negative. Let’s be patient, work and wait the results. After the results, we will work consequently.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Hank,

    My understanding of eCat is a Cat and Mouse configuration.

    After the eCat has been operating for some time, the Cat is usually in a self-sustaining mode of operation. It outputs nominally 10kW of thermal power and does this for about 75% of the time, when averaged over a long period.

    The Mouse is on only about 25% to 33% of the time and when it is on it is “tickling” the Cat in some manner – heat, other? The CAT responds by going back into a self-sustaining mode of operation and the Mouse shuts off.

    With this understanding, the combined effect of the Mouse and Cat is an averaged COP of about 3 or 4. Clearly in an electricity production mode using an oveall Carnot efficiency of 40%, this is barely enough to output slightly more electricity than what is consumed on average by the Mouse (1/40% = 2.5 versus a COP of 3 or 4).

    Andrea Rossi – is this closer to how your system is envisioned to work or works?

  • Andrea Rossi

    Tom Conover:
    I prefer not to talk about this issue before the third indipendent party report is published.
    Warm Regards,
    A.R.

  • Tom Conover

    Dear Andrea,
    COP is not relevant to scientific experiments, but to production environments. However, (1) have the results from the current model diminished (for any reason) from the previous results earlier discussed on this blog that Hank Mills referred to, or (2) have they increased? Could you reply with (- or +) or ((1) or (2)) instead of (yes or no) perhaps, please? Many of us patiently await the end of the interim period. :-)

    Warm heart, cold weather …
    Tom Conover

  • Andrea Rossi

    Hank Mills:
    Dear Friend: no.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Maciek:
    Luckily, we have not war wars in the horizon.
    For what concerns our E-Cat, we are working at the maximum of our possibilities.
    Warm Regards,
    A.R.

  • Maciek

    Dear Andrea,

    I follow all are working on, and I wish you all the best.
    In Europe we have a world war again, please consider or not allowed to share the secret of the E-Cat-and to give people a new energy source discharging conflicts between nations, relating to the case of extraction of natural resources.

    Warm Regards,

    Maciej Z.
    Poland

  • Wladimir Guglinski

    Joe,
    there is other question to be considered.

    The magnetic moment +2,69 of the 5B11 must be actually considered only when the halo-proton is captured by the neutron, and the 5B11 becomes stable.

    So, we have to consider the newborn 5B11 without considering the contribution of +2,79 due to the halo-proton, and therefore the rest of the 5B11 (without the halo-proton) has actually a magnetic moment near to zero.

    Therefore there is only attraction proton-neutron

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in March 26th, 2014 at 4:14 PM

    Wladimir,

    1. In QRT, all the fields within a nucleus are of the same type: Sp. It is irrelevant that Sp < Sn. The classical laws involving Coulomb force and magnetic moments apply to Sp as much as Sn. Therefore, the magnetic moment of the free proton would be drawn to the nuclear magnetic moment of 5B11 if the latter were greater than the magnetic moment of one of its constituent neutrons. The free proton would not interact with a neutron. The free proton would head to the 2He4 and destroy it, either directly from the proton's last point of orbit, or indirectly by traveling along the z axis. There would be neither 5B11 formed, nor 3Li7 + 2He4, unless there were some sort of very quick reconfiguration of the constituent nucleons which has never been witnessed.

    2. I repeat what I wrote earlier. In QRT, if the nuclear magnetic moment of 5B11 were not only greater than the magnetic moment of a neutron but also had the same sign as the magnetic moment of the free proton, the free proton would be repelled by the nucleus.
    —————————————————

    Dear Joe,
    the magnetic moment of 5B11 is +2,69, of the neutron is -1,91 , and of the proton is +2,79
    http://faculty.missouri.edu/~glaserr/8160f09/STONE_Tables.pdf

    The neutron is nearest the proton, because the neutron is orbiting the center of the 5B11.
    In spite of the magnetic moment of the neutron is weaker that that of the 5B11, however there is a compensation, because of two facts:

    1) Force decreases with the square of the distance, so the force proton-neutron has a little growth regarding to the force proton-5B11

    2) The neutron has a radius orbit. Therefore it has an additinal magnetic moment due to its rotation about the 5B11 (due to the g-factor of the neutron).

    Therefore there is a tendency of to cancell one each other.

    .

    We have also to consider that the proton is submitted to the following forces:

    a) Trying to expell it there is a centripetal force (half of the centripetal force on the deuterons, which are in equilibrium in the nucleus)

    b) Pushing it toward the center there is the magnetic force due to the charge of the proton (induced by its rotation about the 5B11). Such magnetic force has the same value of that on the deuterons.

    So,
    the proton is pushed to the center of the 5B11, but not due to the magnetic moments, but because the magnetic force is stronger than the centripetal force.
    And as the there is attraction proton-neutron, with the proton going to the center of the 5B11 it is captured by the neutron.

    regards
    wlad

  • Hank Mills

    Dear Andrea,

    The following is my understanding of how a hot cat should currently work using the information you have shared. Please correct me if I am in error or incorrect about anything.

    C – Hot Cat (Larger part of reactor.)
    M – Mouse (Smaller part of reactor inside the cat.)

    (The following assumes the module has fully warmed up and has been operating for some time.)

    1) The hot cat is self sustaining for three units of time maintaining the same temperature – NOT DECLINING – with no input power.

    2) If the module is rated for 10kW then 30 units of power in the form of heat will have been generated over these three time periods.

    3) With the temperature of the reactor the same or above that when self sustain mode began, the mouse activates consuming 3kW for one unit of time. In other words, the mouse is on 25% of the time.

    4) The mouse by itself has a COP of over 1, so it produces more than 3 units of heat. However, to be conservative, lets assume it generates the same ammount of power as it consumes.

    5) The function of the mouse, after stimulating the cat when the device is first turned on, is to stabilize the cat so the temperature of the module will not run away. So lets assume for the period of time the mouse is on the reactions in the cat are reduced so only 7kW are produced. You can then add the 3kW of heat emitted by the mouse.

    6) Some of the input power from the mouse will be converted back to electricity if the module is being used to produce steam for a generator. So it is not fair to say it is really all consumed or lost.

    7) So for this example, over the four units of time, we have 40 units of output (37 produced by the cat and 3 from the mouse) and 3 units of input used to control the reaction.

    8) 40 divided by three is 13.333.

    So, in conclusion, if an E-Cat works according to these principles and utilizrs TRUE self sistain mode – in which the temperature does not drop until power is applied again – the COP is at least 13.

    That sounds awesome to me.

    Not at all referring to the devices in the third party test, does this sound like how the E-Cats in your lab function? Are the results you personally are obtaining close to this?

    Thanks!

  • Andrea Rossi

    Tom Conover:
    Thank you very much, we are working on this. Very interesting.
    Warm Regards,
    A.R.

  • Tom Conover

    Dear Andrea,

    Thank you for your continued commitment is to contribute to the generation of cleaner energy and to help meet a world wide surge in demand for power during the upcoming decade.

    For your convienience here is the contact information for Toshiba JSW:

    http://www.toshiba-tjps.in/files/contactus.htm

    I too think that Toshiba JSW might be another “Partner” to consider for Industrial Heat, LLC … Some more info about them is shown below for you:

    Toshiba JSW Power Systems PVT. LTD.
    Chennai – Regd. Office Cum Factory
    S.NO. 74 – 95 , Vaikkadu Village,
    Andarkuppam Check Post,
    Manali New Town,
    Chennai- 600103.
    Phone No: 0091 44 30666000
    0091 44 30666500
    Email ID: TJPS-contactus@toshiba-tjps.in

    The About US page states:
    Toshiba JSW Power Systems Private Limited is a Joint Venture between Toshiba Corporation, Japan (Toshiba) and JSW Group, India (JSW) was initially incorporated to manufacture and market super-critical Steam turbines and Generators for Thermal Power Plants in India. It was established in September 2008, and started its commercial operation in January 2011.

    It has been merged with Toshiba Thermal & Hydro Power Systems Company, a division of Toshiba India Private Limited to broaden the gamut from Manufacturing to One-stop Solution for Engineering, Manufacturing, Procurement, Construction and Services (EMPCS) of Thermal Power Plant.

    The facility is located at Manali, about 18 kilometers north of central Chennai, Tamil Nadu with a ground area of approximately 400,000 square meters.
    Toshiba JSW will manufacture and market mid- to large-sized steam turbines and generators ranging from 500-megawatts (MW) to 1,000MW, for highly efficient super-critical thermal power plants in India.

    Toshiba’s power equipment production facility (Keihin Product Operations), in Yokohama, Japan, is supporting Toshiba JSW in ramping up manufacturing and in working toward establishing an independent production scale of 3,000MW a year by 2014.

    Our commitment is to contribute to the generation of cleaner energy and to help meet a surge in demand for power expected in India.

    We all hope the upcoming release of test results will warm our hearts!

    Sincerely
    Tom Conover

  • Andrea Rossi

    Andrea Calaon:
    Thank you very much, really interesting. I am studying carefully the issue.
    Warm Regards,
    Andrea Rossi

  • Andrea Rossi

    Neri B.:
    We must be patient. When the test will be condidered completed the third indipendent party’s Professors will notify it to me. The results will be communicated when their report will have been written.
    Warm Regards,
    A.R.

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