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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439 comments to Physics of rotating and expanding black hole universe

  • Frank Acland

    Dear Andrea,

    As Steven Karels mentions, the amount of input power required to run an E-Cat is a significant issue when looking at the economic viability of the E-Cat compared to other energy technologies.

    I am wondering:

    1. Have you been able to achieve satisfactory E-Cat performance with natural gas as an input power source?

    2. Have you found any new ways (since discovering the cat and mouse configuration)to maintain optimum E-Cat output performance while reducing the amount of input energy?

    3. Is it possible to control the E-Cat using methods other than applying heat?

    4. If yes (question 3), are you working on developing those systems?

    Thank you very much!

    Frank Acland

  • Steven N. Karels

    Wladimir,

    While LENR technology looks promising, nuclear is present technology. Its ugly parts are known. I would contend we don’t know yet if LENR has “ugly” parts.

    My largest concern for eCat technology is the performance issue, aka COP. Nuclear, like hydro-electric, requires very little input power. If the eCat technology, for whatever reason, is limited to an effective COP of three, then it probably won’t be used for commercial electric power generation. I would guess an effective COP of 12 or higher — then it is a viable candidate.

    You are right about a blend of technologies existing in the near-term. I hope for LENR — but I still have my concerns and doubts. We will see.

  • Wladimir Guglinski

    Mark wrote in April 15th, 2014 at 8:28 AM

    I’m not here to advocate Liquid-Fluoride Thorium reactor technology but my points are to compare the operational costs of LENR and LFTR technologies. The real question is: apart from LFTR and LENR, what else do you have on the “table” for humanity? (please don’t mention renewable!). By the way, you should read more about the safety features of a LFTR reactor. I pray that Andrea will make it but he has many battles to win.
    ——————————-

    Dear Mark,
    as Andrea Rossi told here several times, the production of various energy technologies can coexist peacefully for decades.

    However, if two technologies are competitive, and one of them is perfectly ecologically clean, and it is entirely safe, while the other is potentially threatened by catastrophic disasters caused by tsunamis, earthquakes, terrorism, sabotage, operational errors, etc., then obviously after some decades the potentially dangerous technology will not survive.

    In my case, I am not intested in LFTR technology, but because of other reason. My miss of interest is because it is a technology based on the current old paradigm of Physics.

    While the Rossi’s eCAt is a new revolutionary technology, which is inaugurating a New Era for the Humanity, openning the door for many other discoveries beyond the search for energy alternative technologies.

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in April 15th, 2014 at 7:08 AM

    But the problem that remains in QRT is that the spins are VERTICAL at all times, which is perpendicular to the direction of displacement. This means no helical trajectory ever possible and therefore no induced magnetic moments either. Only intrinsic magnetic moments would be allowed. This, of course, would eviscerate QRT.
    —————————-

    COMMENT

    Joe,
    the problem remains also for all the other even-even nuclei with Z=N, as 4Be8, 6C12, 8O16, 10Ne20, etc., where the problem is worst, because their positive charges due to the proton have a larger radius of rotation than it happens in the case of the 2He4.

    As they have several protons gyrating in the same direction, due to the monopolar nature of the charge all those nuclei could not have magnetic moment zero.

    But consider the following:

    The magnetic moment induced by the charge is due to the direct interaction between the charged particle and the particles of the aether where the particle is moving.

    In the even-even nuclei with Z=N as 4Be8, 6C12, 8O16, etc, all they are formed by deuterons captured by the central 2He4.

    Suppose that in each one of all those deuterons, the neutron of that deuteron gyrates about the proton.
    As the neutron has charge zero, it is possible the neutron creates a neutral blindage (armour) about the proton, and such blindage avoid the interaction of the proton’s charge with the particles of the aether.

    By this way, all the positive charges of the deuterons in the nuclei 4Be8, 6C12, 8O16, etc do not induce magnetic moment due to the rotation of the nucleus, because the blindage of the neutron do not allow them to interact with the aether.

    The same can happens in the case of the nucleus 2He4.

    I cant see any other explanation, Joe.
    And as you may realize, by considering such hypothesis we keep the monopolar nature of the charge.
    So, the monopolar nature of the charge does not work in any nucleus in which a neutron gyrates about the proton (in the structure of the deuteron).
    Such “particular micro-law” (not followed by the monopolar nature of the charge) is valid for any nucleus, and not only the even-even nuclei with Z=N.

    But no matter what is the explanation, dear Joe, we realize the following: from the Standard Physics it is impossible to explain why even-even nuyclei with Z=N have magnetic moment zero.

    regards
    wlad

  • Andrea Rossi

    Teemu:
    ( second answer)
    I have a great number of ideas for how to maximize the E-Cat, we are testing many of them. Every time we have an idea it is a rigorous process to evaluate and test it- which means we must have some flexibility. This process is experimental in nature and reminds me of the value of strong and trustworthy business and scientific Partners.
    Warm Regards,
    A.R.

  • Mark

    Hi Wlad,

    “and in terms of nuclear fuel leak caused by tsunamis or operational errors, and thousands of people infected getting cancer, the LFTR will be competitive with Fukushima or Chernobyl”

    nice sarc..!!How many people died and are dying from pollutions from coal fired power stations? And diseases? And lack of clean water? Much more that Fukushima and Chernobyl combined

    I’m not here to advocate Liquid-Fluoride Thorium reactor technology but my points are to compare the operational costs of LENR and LFTR technologies. The real question is: apart from LFTR and LENR, what else do you have on the “table” for humanity? (please don’t mention renewable!). By the way, you should read more about the safety features of a LFTR reactor. I pray that Andrea will make it but he has many battles to win.

    Mark

  • Joe

    Wladimir,

    The following is my explanation for the direct relationship between the direction of the induced magnetic dipole moment and the direction of rotation of electric charge.

    Part 1: Monopolar versus Scalar
    A neutron has a magnetic moment which manifests itself as dipolar.
    An electron has the same type of magnetic moment as the neutron, PLUS another type of magnetic dipole moment which includes electric charge as ONLY ONE of its many variables.
    What we see as a monopolar property in the case of electric charge is really just a SCALAR property within the larger context of a dipolar property.
    So now it is logical for this “monopolar” property (electric charge) to induce a dipolar property (magnetic moment).
    It is not really charge that is moving but rather an electron that is moving and carrying the charge.

    Part 2: Intrinsic Spin versus Helicity
    Since we now know from Part 1 above that it is not electric charge that is responsible for inducing magnetic moments, we must discover the true cause of directionality of magnetic moments induced by the rotation of charge.
    It is obvious that the intrinsic spin of the electron retains the same sign since only the direction of rotation of the electron causes a change in the direction of the induced magnetic moment. Otherwise, directionality would be chaotic. Therefore, the intrinsic spin of the electron can not be the cause since it is constant between opposite rotations.
    Logically the only property left is helicity which is directly related to intrinsic spin. The handedness changes between opposite rotations with the sign of the intrinsic spin of the electron remaining constant.

    Part 3: Standard Physics versus QRT
    For two similar charges rotating in the same direction, a null result for the induced magnetic moment could only be achieved by having the sign of the intrinsic spin of one of the two particles changed.
    This means that for a rotating 2He4, standard physics can account for mu = 0.
    QRT can also seemingly account for mu = 0 with the presence of gravitational fluxes n(o) penetrating the rings of the nucleons in opposite directions and causing opposite spins. But the problem that remains in QRT is that the spins are VERTICAL at all times, which is perpendicular to the direction of displacement. This means no helical trajectory ever possible and therefore no induced magnetic moments either. Only intrinsic magnetic moments would be allowed. This, of course, would eviscerate QRT.

    All the best,
    Joe

  • Wladimir Guglinski

    Mark wrote in April 14th, 2014 at 9:15 PM

    Hi Andrea,
    My points are, if LENR is going to make to the commercial stage within 5-10 years time, its main technological competitor, in term of CO2 emission and cost, will be LFTR.
    ——————————————-

    Dear Mark
    and in terms of nuclear fuel leak caused by tsunamis or operational errors, and thousands of people infected getting cancer, the LFTR will be competitive with Fukushima or Chernobyl?

    regards
    wlad

  • orsobubu

    This is a negative article about thorium, but remember to read the more balanced comments written by interested supporters.

    http://www.fukuleaks.org/web/?p=3101

    The technology seems promising, above all for positive applications in plutonium wastes recycling. More critical points are possible dangerous thorium leaks (lungs, pancreas, kidneys, liver, blood cancer)and obviously the immense risk connected to transportation of plutonium wastes. But, as always, the real problem is economics. In 1993 nuclear plants produced 17% of world energy, today, 10%. They are at loss and many are closing down. Moreover, uranium extracting price is 70 dollars/pound, while it is currently sold at 35 dollars. So, even the extractive industry is at risk. Remember that nuclear plants are very costly to maintain and securing wastes. This impacts on energy competitivity, and this is also true for thorium, a very complex technology, still not fully addressed. Today, thorium is much more costly to extract and absolutely not competititve with uranium. The price could fall well under uranium with extensive usage, but the problem would probably represent in non-profitable market prices of both the metal and the energy. In capitalism, the tragedy is that we’ll have too much energy, not the contrary.

  • Andrea Rossi

    TO ALL THE READERS OF THE JoNP:
    HAPPY EASTER TO YOU ALL AND YOUR FAMILIES FROM THE TEAM I WORK WITH AND MYSELF.
    ANDREA ROSSI

  • Andrea Rossi

    Mark:
    If you say so…
    Warm Regards,
    A.R.

  • Mark

    Hi Andrea,

    First developed by ORNL in the 60s-70s by Dr. Weinberg, we understand LFTR technology well and that its waste components are about 10,000X less than existing light water nuclear reactor (most can be recycled). My points are, if LENR is going to make to the commercial stage within 5-10 years time, its main technological competitor, in term of CO2 emission and cost, will be LFTR.

  • Andrea Rossi

    Mark:
    LFTR is a nuclear plant, it works with radioactive fuel with high level of radioactivity. Nothing to do with us, we do not use radioactive materials and we do not produce radioactive wastes. About the 2-3 cts/kWe: did you calculate the cost of nuclear wastes disposal ?
    Warm Regards,
    A.R.

  • Mark

    Hi Andrea,

    Will your E-Cat technology be able to compete with LFTR (liquid fluoride Thorium reactor) technology, which could be as low as 2-3cents per KWh (electric)? China has planned to produce 200MW distributed modular units within 10 years time.

    https://www.youtube.com/watch?v=ayIyiVua8cY

    Mark

  • Wladimir Guglinski

    JR wrote in April 14th, 2014 at 8:40 AM

    Wladimir,

    1)——————————
    The magnetic moment of 4He is zero because the magnetic moment is defined in such a way that it is zero for spin-0 particles; nothing else is needed.
    ——————————–

    COMMENT

    Show us here where did you find such definition of the magnetic moment

    2) —————————-
    If you wish to define some new quantity that is similar to the magnetic moment but different, that’s fine. But you should (1) provide a definition (2) call it something different (3) not confuse it with the already-defined magnetic moment and (4) not assume it’s zero because the magnetic moment is zero.
    ———————————–

    COMMENT:

    The definition of magnetic moment is independent of the spin of the nuclei.

    Definition by wikipedia:
    http://en.wikipedia.org/wiki/Magnetic_moment
    “The magnetic moment of a magnet is a quantity that determines the torque it will experience in an external magnetic field. A loop of electric current, a bar magnet, an electron, a molecule, and a planet all have magnetic moments“.

    And magnetic moment of a nucleus:
    “Since the electromagnetic moments of the nucleus depend on the spin of the individual nucleons, one can look at these properties with measurements of nuclear moments, and more specifically the nuclear magnetic dipole moment.”

    Therefore, although the magnetic moment of a nucleus depends on the spin of individual nucleons, however it does not means that it depeneds ONLY on the spin of the INDIVIDUAL nucleons.

    The nuclei have also rotation:
    http://journals.aps.org/pr/abstract/10.1103/PhysRev.53.778

    And therefore the magnetic moment depends also on the rotation of the nucleus.

    The total spin due to the individual nucleons in the 2He4 is zero.
    However, as the two protons of the 2He4 gyrate in the same direction, then (by considering the monopolar nature of the charge) the two protons have to induce a magnetic field, which will be responsible for a magnetic moment for the 2He4.

    But as the experiments detect that 2He4 has no magnetic moment, this means that 2He4 violates the monopolar nature of the charge, by considering the current nuclear models.

    There is no need to be a genius to understand it.
    And you, Mr. JR, you are actually showing to everybody that your understanding of Physics is very poor.

    regards
    wlad

  • Andrea Rossi

    Teemu:
    When we will be ready for the market with that.
    Warm Regards,
    A.R.

  • teemu

    Dear Andrea Rossi,

    When will we hear more about the electric power generation that you have supposedly managed to achieve?

    Best Regards,
    Teemu

  • JR

    Wladimir,

    I’m not sure how you propose to explain the properties of nuclei if you aren’t willing to define what those properties are and then stick to those definitions. What you asked is how conventional models explain the fact that 4He has a zero magnetic moment. The magnetic moment of 4He is zero because the magnetic moment is defined in such a way that it is zero for spin-0 particles; nothing else is needed.

    If you wish to define some new quantity that is similar to the magnetic moment but different, that’s fine. But you should (1) provide a definition (2) call it something different (3) not confuse it with the already-defined magnetic moment and (4) not assume it’s zero because the magnetic moment is zero.

  • Wladimir Guglinski

    Joe wrote in April 14th, 2014 at 12:20 AM

    Wladimir,

    It may be that the ultimate test of veracity in QRT is in explaining how 4Be8 is unstable while 6C12 and other nuclei with a symmetrical distribution of only deuterons, are stable.
    —————————-

    Joe,
    the reason why 4Be8 is not stable is shown in the page page 17, item 3.13.5, Fig. 14, of the paper Stability of Light Nuclei
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    As we see in the Fig. 14, the 4Be8 is the unique nucleus in which there are two deuterons occupying opposite perfectly symmetric positions regarding to the central 2He4.

    A perfect symmetry occurs when:

    1- one deuteron is in the side ANA, and the other in the side DOUGLAS

    2- one deuteron is in the inferior part of ANA, and the other in the superior part of DOUGLAS (or vice-versa).

    A partial symmetry occurs when one deuteron is in the superior part of ANA and the other deuteron is also in the superior part, but in the side of DOUGLAS.
    A partial symmetry between a deuteron and two neutrons can be seen in the superior part of the Fig. 7.

    Looking at the Fig. 14 you realize that the spin-interaction Fsi(green arrows) promotes a force of attraction between the two deuterons (the red arrows show only the direction of their spins)

    In the Fig. 14 I supposed that the spin-interaction Fsi is stronger than the repulsion force Fr (pink arrow) because the 4Be8 decays in two alpha particles , and so such sort of decay requires to suppose that the two deuterons are captured by the central 2He4, and the 4Be8 decays emitting two nucleons 2He4:
    4Be8 -> 2He4 + 2He4

    If in the Fig. 14 the repulsion force Fr should be stronger than Fsi, then the two deuterons would be expelled in contrary direction, and the decay of 4Be8 would be:
    4Be8 -> 2He4 + D + D

    .

    The reason why 6C12 is stable is shown in the item 3.13.6, Fig. 15 and 16.

    .

    With the 8O16 the first hexagonal floor is complete, and so the nuclei with Z > 8 (as 10Ne20, 12Mg24, 14Si28, etc) are stable thanks to the spin-interactions between their deuterons.

    regards
    wlad

  • Joe

    Wladimir,

    It may be that the ultimate test of veracity in QRT is in explaining how 4Be8 is unstable while 6C12 and other nuclei with a symmetrical distribution of only deuterons, are stable.

    All the best,
    Joe

  • Andrea Rossi

    Frank Acland:
    No role in the measurements, just check that the E-Cat was working properly, checking the control system, looking at the surface, to be sure no cracks emerged, things like these.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    You mention attending the testing for about 30 per cent of its duration. What has been your role as you have been there?

    Many thanks,

    Frank Acland

  • Andrea Rossi

    Eernie1:
    For the industrial plants we already have obtained the safety certification. For the domestic the situation is totally different, as I have explained many times.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Ecco Liberation:
    I do not know, if yes we will see in the report. I have attended now and again, maybe 30% of the time od the test.
    Warm Regards,
    A.R.

  • Ecco Liberation

    Dr. Andrea Rossi,
    Did the team of professors involved with the third party testing compare the E-Cat’s thermal output with an equivalent blank/inert/dummy load? E.L.

  • Wladimir Guglinski

    Eric Ashworth wrote in April 13th, 2014 at 11:35 AM

    Dear Wladimir,

    Wladimir do you wonder why your QR theory is rejected so strongly by certain individuals or have you a good idea why? because there is one.
    ———————————–

    COMMENT

    DEar Eric,
    by looking at the explanation given by Mr. JR we realize why.

    Look at his “brilliant” solution:

    “4He is a spin-0 nucleus and so, by definition, has no magnetic moment”

    The reason is because the theorists have not respect to the fundamental questions in Physics.

    When the experimental data prove that current nuclear models violate some fundamental law and they cannot give a satisfactory explanation (satisfactory for a honest scientist), they give explanations which disagree to what we expect from a honest scientist:

    they give orders to the nucleus, telling him the rules he has to follow, in order do not defy their nuclear models.

    regards
    wlad

  • eernie1

    Dear Andrea,
    In pharmaceutical testing procedures, a drug that shows positive results in mitigating physical problems is immediately released for patient use. Do you think that very positive results from the TPI tests will result in quick approval of your devices for commercial and public use in the USA?

  • Wladimir Guglinski

    Dears Joe and Mr. JR

    there is a 4th sort of solution which you may propose, for the violation of the monopolar nature of the electric charges by the nuclei.

    The 4th solution you can propose I explain ahead.

    As we know, when Bohr discovered his hydrogen model, he realized that his model violates some fundamental known Laws of Physics.

    Then Bohr proposed a solution based on postulates, by claiming that the atom is able to violate some fundamental laws.

    So, dears Joe and Mr. JR, you can propose the following postulate:

    The light even-even nuclei with Z=N=pair can violate the monopolar nature of the electric charges.

    .

    Therefore, my dears friends, you have now 4 alternatives for chosing what is the best solution.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in April 13th, 2014 at 11:11 AM

    4He is a spin-0 nucleus and so, by definition, has no magnetic moment.
    ———————————–

    COMMENT

    Dear JR
    nuclear properties of nuclei cannot be established by definition

    The nuclei have nuclear properties, which have to be explained by any nuclear theory taking in consideration the known Laws of Physics

    If a nuclear model violates a known Law of Physics, as the current nuclear models are violating the monopolar nature of the electric charges, the theorists have two alternatives:

    1- To reject the nuclear model, and to look for another model able to be suit to the law (in the present case the monopolar nature of the electric charges).

    2- To discover the reason why the Law is being seemingly violated

    .

    Dear JR
    such a solution of yours (claiming that “4He is a spin-0 nucleus and so, by definition, has no magnetic moment”) is actually proposed according to the phantasmagoric Heisenberg’s method.

    The physicists who do not have interest to solve the questions regarding the Fundamental Physics, as happens to you, may be satisfied with your solution.

    But any sincere physicists who has respect to his own honesty cannot accept such sort of explanation.
    Because your explanation is actually a way of running away of the theoretical puzzles which defy the foundations of Modern Physics.

    regards
    wlad

  • Andrea Rossi

    Andreas Moraitis:
    The generated heat is withdrawn by means of heat exchangers, based on well known technologies. The reliability doesn’t depend on the heat exchangers. Obviously, the temperature of the secondary depends on the heat exchanger, while the primary temperature depends only by the E-Cat. The lower the flow of the secondary, the higher the secondary temperature, and vice versa.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Herb Gillis:
    Sugars and cellulose can yield 20% of oil, 40% of charcoal, 40% of gas.
    From 2003 through 2007 I manufactured power plants working with oil made by means of not utilized cascaded substances derivated from the production of food vegetable oil. That’s how I made the money to work on the LENR.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Bernie Koppenhofer:
    I never wrote that power generation eludes me.
    Warm Regards,
    A.R.

  • Bernie Koppenhofer

    Dr. Rossi: In one of your answers you said power generation still eludes you. Could you give us the reasons in general terms? Maybe this group could give some ideas and maybe “serendipity” will occur. (:

  • Eric Ashworth

    Dear Wladimir,

    I do not profess to know nuclear physics as you do and I do not think many other people do also, other than to days modern day physicists. What you wrote April 13 at 5.58 am are, I beleive, some of the clearest statements you have made so far, as far as I am concerned, with regards an explanation of this anomaly. From what I gather it is to do with a monopolar nature and a zero reading. From my own understanding of nature and investigations of natural phenomena you cannat have a mono situation, it has to be binary i.e. a duality to create,exist and destroy at its most primary principle. Having spent many years developing a technology based upon this binary understanding of nature and considerable outside pressure in specific areas against its development I have discovered that some mysteries must remain mysteries for a specific reason. Wladimir do you wonder why your QR theory is rejected so strongly by certain individuals or have you a good idea why? because there is one. The reason is well worth finding out. All the best Eric Ashworth.

  • JR

    Wladimir said:

    The readers of the JoNP will feel themselves very happy if you come here to explain us how the nucleus 2He4 can have magnetic moment zero, according to the current models of the Standard Nuclear Physics.

    All of us will be very thankfull to your explanation

    I doubt that you will be thankful, since you didn’t like the explanation very much the last time I answered this question (for 12Be), but here it goes:

    4He is a spin-0 nucleus and so, by definition, has no magnetic moment.

    In conclusion, I don’t believe that your misunderstanding of how magnetic moments are defined is enough to throw out all of classical electromagnetism or quantum electrodynamics. But keep trying.

    -John

  • Wladimir Guglinski

    Joe wrote in April 12th, 2014 at 11:46 PM

    Wladimir,

    1- I do not know of any experiment that has switched the intrinsic spin to observe if the same electric charge with the same direction of rotation would produce a magnetic moment in the opposite direction.
    Logically there should be no difference since electric charge and intrinsic spin are independent of one another. For example, a neutrino has no electric charge but does have an intrinsic spin.
    ———————————

    COMMENT

    Consequence:

    Therefore it is IMPOSSIBLE the existence of the nucleus 2He4, because the experiments measured that it has magnetic moment zero, however due to fact that charge is monopolar, it is IMPOSSIBLE to exist the nucleus 2He4 with magnetic moment zero.

    ——————————————
    To have the behavior that you want, a switched intrinsic spin would have to effect a corresponding switch in the electric charge. But the electric charge is monopolar: it can not be flipped to an alternate state that does not exist.
    ——————————————

    COMMENT

    I dont want nothing.

    As the electric charge is monopolar, therefore the 2He4 nucleus cannot exist, since it has magnetic moment zero as measured by experiments, but due to its rotation and the fact that electric charge is monopolar the 2He4 cannot have magnetic moment zero.

    The same we can say about all the other light even-even nuclei with Z=N pairs: 4Be8, 6C12, 8O16, 10Ne20, etc.

    Therefore we have to eliminate all those nuclei from the Periodic Table of Elements, because from the known principles of Physics they cannot exist.

    So, I propose a New Periodic Table, where the elements with nuclei 2He4, 4Be8, 6C12, 8O16, 10Ne30, 12Mg24, etc. will be eliminated, because it is IMPOSSIBLE to explain their magnetic moment zero from any nuclear model based on the monopolar nature of the electric charge.
    And the sequence of the elements will become the following:

    1H

    2He2, 2He3, ….(eliminated 2He4)… , 2He5, 2He6 , etc

    3Li

    4Be5, 4Be6, 4Be7, …(eliminated 4Be8)… , 4Be9 , 4Be10, 4Be11, etc

    5B

    6C8, 6C9, 6C10, 6C11, ….(eliminated 6C12)…. , 6C13, 6C14 , etc

    7N

    8O13, 8O14, 8O15, …. (eliminated 8O16)… , 8O17, 8O18, etc

    9F

    etc, etc.

    Unless you propose, dear Joe, that the rotation of the nuclei does not exist.
    But unfortunatelly, by this sort of solution is impossible to explain the magnetic moments of the other nuclei.

    Or perhaps you will claim that only the light even-even nuclei with Z=N pairs do not have rotation.

    Dear Joe,
    please ask the help of Mr. JR so that to decide what is the best solution:

    1- To eliminate the nuclei 2He4, 4Be8, 6C12, 8O16, etc. from the Periodic Table

    2- To reject the hipothesis of rotation of the nuclei

    3- To propose that only the nuclei 2He4, 4Be8, 6C12, etc. do not have rotation

    I will be waiting the decision of yours and Mr. JR, telling me what is the best solution.

    regards
    wlad

  • Herb Gillis

    Andrea Rossi:
    I just read Mats Lewan’s book [“An Impossible Invention”] about you and your work. It was an inspiring story of rugged individualism. I think everyone should read it.
    In regard to your earlier work on the Petroldragon technology: Can you tell me if the technology can convert certain clean organic substances (not waste) into a useful liquid fuel. Specifically I am interested in:
    a) sugars, such as sucrose?; and
    b) cellulose?
    Your comments are much appreciated.
    Regards; HRG.

  • Andreas Moraitis

    Dear Andrea Rossi,

    It’s good news that there were no malfunctions of the Hot-Cat during the test. Obviously, you have reached by now a high level of reliability – which is everything else but natural for a completely new invention. Congratulations for that!
    My question is if you are already at a point where the process works as reliable if significant amounts of the generated heat are withdrawn systematically from the system. Evidently, that would be a precondition for the practical utilization of the generated energy.

    Best regards,
    Andreas Moraitis

  • Andrea Rossi

    orsobubu:
    1- yes, but evolution is permanent ( an adjective you should like)
    2- yes
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear Joe,
    you and me have proven here, together, the following:

    1- It is impossible to explain the magnetic moment zero of the 2He4 by keeping the current dogma of the monopolar nature of the electric charge.

    And as consequence:

    2- Therefore it is impossible to explain the magnetic moment zero of the 2He4 by taking as starting point any nuclear model which do not work by considering a structure of the aether formed by particles and antiparticles as gravitons, magnetons, electricitons, etc.

    3- Any nuclear model developed by no considering the participation of the aether within its structure is according to the monopolar nature of the electric charge, and therefore cannot be correct.

    4- All the current nuclear models of the Standard Nuclear Physics are wrong, because they do not consider the aether within their structure, and so they cannot explain the magnetic moment zero of the 2He4.

    5- Nowadays all the nuclear theories of the authors trying to explain the cold fusion cannot be entirely successful, because they are developed by considering wrong nuclear models, since all they are based on the monopolar nature of the electric charge.

    .

    I and the readers of the JoNP hope you have the honesty to admit it.

    regards
    wlad

  • orsobubu

    Andrea,

    I’m delighted with the idea that testers could limit or push the experimental set at their free choice without external conditions. When you say that the e-cat tested by the independent third party for the last 6 months is still “state of the art”, does it mean:

    1- the core of the tested hot cat completely satisfies you and currently no other work is necessary to improve its performance and stabilty, so the research is directed on equipment around the beast

    2- the core tested is the best stable version you can provide but you are currently working both on the equipment around the cat and on new ideas about the beast itself

  • Joe

    Wladimir,

    I do not know of any experiment that has switched the intrinsic spin to observe if the same electric charge with the same direction of rotation would produce a magnetic moment in the opposite direction. Logically there should be no difference since electric charge and intrinsic spin are independent of one another. For example, a neutrino has no electric charge but does have an intrinsic spin.

    To have the behavior that you want, a switched intrinsic spin would have to effect a corresponding switch in the electric charge. But the electric charge is monopolar: it can not be flipped to an alternate state that does not exist. And flipping the spin is not going to change the sign of the charge (which would give you the behavior that you want). For example, there is no relationship such as, electron implies negative spin, and positron implies positive spin. Proof of this is in the fact that a maximum of two electrons (same sign) can occupy one orbital, and each electron must have a spin that is the opposite of the other. This is the Pauli Exclusion Principle.

    All the best,
    Joe

  • Wladimir Guglinski

    Dear Mr JR

    The readers of the JoNP will feel themselves very happy if you come here to explain us how the nucleus 2He4 can have magnetic moment zero, according to the current models of the Standard Nuclear Physics.

    All of us will be very thankfull to your explanation

    regards
    wlad

  • Wladimir Guglinski

    Wladimir Guglinski
    April 11th, 2014 at 6:01 PM

    Joe wrote in April 10th, 2014 at 1:59 AM

    Wladimir,

    I repeat what I answered in (1). Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation.
    ———————————-

    Joe,
    we had a good discussion here.
    And we arrived to a very interesting conclusion:

    It’s IMPOSSIBLE to explain the magnetic moment zero of the 2He4 by considering the current nuclear models of the Standard Nuclear Physics

    Indeed, let us see why:

    1- All the nuclei have rotation. The Nobel Laureate Hans Bette estimated that 10% of the magnetic moment of the nuclei is due to their rotation.
    For instance, consider the 3Li6. Suppose that 2 protons and 2 neutrons cancell each other their magnetic moment. The third proton has magnetic moment +2,793, while the third neutron has magnetic moment -1,913.
    If the 3Li6 had no rotation, its magnetic moment would have to be +2,793-1,913 = +0,880. But the magnetic moment of 3Li6 is +0,835 , and the difference +0,045 is due to the rotation of the 3Li6.

    2- Consider the 2He4. Consider that, if the 2He4 had no rotation, its magnetic moment would be zero, since the two protons cancell each other their magnetic moment, and the two neutrons also cancell each other.

    3- However the 2He4 has rotation. And therefore the two protons move in the same direction.

    4- But “Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation”.
    And so by considering the current nuclear models the two protons of the 2He4 have to induce a magnetic moment.

    5- Therefore the 2He4 cannot have magnetic moment zero, according to the current nuclear models of the Standard Nuclear Physics.

    .

    I wonder if Mr. JR would come here to show us how such puzzle is solved in the Standard Nuclear Physics.

    regards
    wlad

  • Andrea Rossi

    Tom Conover:
    Good question: no, we sent 3 spare parts just to have safety boats in case of problems. Since if this test has a paramount importance for us, as everybody can understand, we have left nothing to chance, within the measure of possible.
    Warm Regards,
    A.R.

  • Tom Conover

    Hello Andrea,

    I was wondering if the reason that you sent 3 spare e-cats for the test might be that the test was being performed on an assembly of (more than one) e-cat units?

    CONGRATULATIONS on the stability of the units!

    Warm regards,

    Tom Conover

  • Andrea Rossi

    Hank Mills:
    Interesting question. We must wait fot the report of the Professors to know the answers.
    Warm regards,
    A.R.

  • Hank Mills

    Dear Rossi,

    You state that the professors are doing a very conservative work. If that means they are being conservative in their measurements and calculations that is a good thing. However, I hope they are not being overly conservative when it comes to pushing the E-Cat to obtain high performance. For example, in the previous report they only increased the temperature of the hot cat to around 500C. From what I have calculated, by increasing the temperature to 1000C the blackbody of the reactor would radiate much more power – actually, several times as much. Do you know if they are attempting to find the maximum safe performance of the module in addition to simply determing the reality of the technology? With six month or longer to test and spare parts (just in case pushing the reactor caused any leaks or damage) I would like to think they are trying to determine both. Thank you.

  • Andrea Rossi

    Gunnar Lindberg:
    You make your son well educated, well trained, and test him many times. Then he goes to the war: aren’t you afraid he can die?
    Beyond the metaphor: the test in course is a very long test, with a system of calculation very conservative, prepared after the test made by the same third indipendent party one year ago ( March 2013); during that test and the afterward discussions the third indipendent party has learnt all the possible further controls to make ; all the possible shortcomings have been experienced one year ago and now they are making a very conservative work. If you go to read the report of the March test published on Arxiv Physics, you will see that they had written that a long run test would have been necessary to make better measurements. Now you can understand the metaphor.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe
    April 12th, 2014 at 12:21 AM

    Wladimir,

    You state the following:

    “In the case of a free 2He4 the two charges of the deuterons do not induce magnetic field, since the two charges rotate in a space with no magnetic field.”

    This idea is false since rotating electric charges create their own magnetic field independent of any pre-existent magnetic field. Magnetic fields from various sources are simply summed. A null magnetic field from a combination of sources does not prevent another source from exhibiting its own magnetic field within that same space, and yielding a non-null net magnetic field for that space.
    —————————————–

    COMMENT

    Joe,
    consider the following:

    1- suppose the body of the electron gyrates in the clockwise direction (spin-up)

    2- consider one electron moving in an orbit with radius R about a vertical axis in the clockwise direction

    3- suppose a particle X with negative charge (equal to the charge of the electron) which body gyrates in the anti clockwise direction (spin-down, so contrary of the spin of the electron).

    4- consider the particle X moving in an orbit with radius R about a vertical axis in the clockwise direction, as happens with the electron.

    .

    Then I ask you:

    1- Will the electron and the particle X create the same magnetic field ?

    .

    You said:
    “I repeat what I answered in (1). Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation. That is what we know empirically.”

    However, all the experiments with a body electrically charged are made with a body which has its electric charge thanks to the electrons in its electrosphere. And the spin of the electron has always the same direction.

    But suppose we are be able to create a particle X with the same charge of the electron, but its spin being contrary of the spin of the electron.
    And suppose we create a body, which electrosphere is composed by the particle X.

    Then I ask you:

    2- Would electric charge to have a monopolar nature?

    3- Or, in another words:
    would a body formed by electrons in its electrosphere induce the same magnetic field of a body formed by particles X in its electrosphere?

    regards
    wlad

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