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

  • Gunnar Lindberg

    Dear Andrea Rossi,
    I understand the ongoing test is an important step in the development of the e-cat. However, what make me curious is you repeated statements tha the outcome may be negative. You already know from previous tests that the e-cat produce energy. Otherwise you would not Thus, what is outcome you suspect may be negative?
    Good luck and Best Regards
    Gunnar

  • Wladimir Guglinski

    Joe,
    there is an error in the calculation:

    ——————————
    7-a) The mass of 3Li7 is 7,0160

    7-b) The mass of He4+deuteron in 3Li7 is 6,0167 – 0,01 = 6,0067

    7-c) The mass of the neutron is 1,0087

    7-d) The mass of He4+deuteron+neutron in 3Li7 is 6,0067 + 1,0087 =7,0154
    ——————————-

    Because actually the mass of 2He4+deuteron+neutron in the item 7-d must be greater than 7,0160

    This means that the mass defect of the 2He4-deuteron in 3Li7 must be actually a little smaller than 0,01.

    .

    Let’s consider the mass defect of the 2He4-deuteron in 3Li7 is 0,007.
    So we have:

    7-b.1) The mass of He4+deuteron in 3Li7 is 6,0167 – 0,007 = 6,0097

    7-c.1) The mass of the neutron is 1,0087

    7-d.1) The mass of He4+deuteron+neutron in 3Li7 is 6,0097 + 1,0087 = 7,0184

    7-e.1) The difference 7,0184 – 7,0160 = 0,0024 is the mass defect due to the binding energy of the spin-interaction between the neutron and the deuteron.

    7-f.1) Therefore, in 3Li7 we have:

    Mass defect due to magnetic force 2He4+deuteron = 0,007

    Mass defect due to spin-interaction deuteron+neutron = 0,0024

    From the considerations above, in the 3Li7 the binding energy due to the magnetic force is 2,9 times stronger.

    But if we consider the mass defect due to magnetic force of the 2He4+deuteron smaller than 0,007 as considered here, the tendency is to have an equalization of the two binding energies.

    regards
    wlad

  • Joe

    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.

    All the best,
    Joe

  • Andrea Rossi

    Brian Ho:
    I have been in the Hawaii Islands four years ago: what a wonderful place to visit!
    Thank you for your kind words. I am honoured from the attention of people interested in what we are doing.
    I do not expect any triumph, I just hope to continue the work. There is an enormous work to do.
    Warm Regards,
    A.R.

  • Brian Ho

    Dear Andrea Rossi,

    It must be gratifying to know that many people from around the world are watching closely the independent testing of your E-Cat. We all await in anticipation for the final report and we all hope that this will finally be the dawn of a new age for humanity. And, for you, a major triumph after all your hard work and the many battles you had to fight.

    Wishing you the very best!

    Kind Regards,

    Brian Ho
    Honolulu, Hawaii

  • Andrea Rossi

    Frankk Acland:
    1- yes
    2- yes
    Warm Regards,
    A.R.

  • Andrea Rossi

    Paul:
    It is state of the art, so far; as for the future…”panta rei”.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe,
    I calculated the relation between the binding energy due to the magnetic force deuteron-2He4 and due to the spin-interaction between the deuteron-neutron, for the nucleus 3Li7, as I show ahead.

    1) We start by calculating the binding energy due to the magnetic force attraction between the deuteron and the central 2He4 in the nucleus 3Li6, as follows:

    1-a) The mass of 3Li6 is 6,0151

    1-b) The mass of the deuteron is 2,0141

    1-c) The mass of 2He4 4,0026

    1-d) The mass of 3Li6 formed by He4+deuteron is 6,0167

    1-g) Therefore the difference 6,0167 – 6,0151 = 0,0016 is the mass defect due to the binding enrgy between the deuteron and the 2He4, in the nucleus 3Li6

    .

    2) The magnetic moment of 3Li6 is 0,822

    3) According to my nuclear model, the distance between the deuteron and the 2He4 in the nucleus 3Li6 is 0,355fm (see Fig. 18 in the paper Stability of Light Nuclei).

    4) The magnetic moment of 3Li7 is 3,256

    5) According to Fig. 19 the distance between the deuteron and the 2He4 in the nucleus 3Li7 is 0,405

    6) Considering that the binding energy is directly proportional to the magnetic moment of the nucleus and inversely proportional to the square of the distance, then we calculate the mass defect due to the binding energy between the deuteron and the 2He4 in the 3Li7, as follows:

    0,0016 x (3,256/0,822) x (0,355/0,405)² = 0,01

    .

    7-a) The mass of 3Li7 is 7,0160

    7-b) The mass of He4+deuteron in 3Li7 is 6,0167 – 0,01 = 6,0067

    7-c) The mass of the neutron is 1,0087

    7-d) The mass of He4+deuteron+neutron in 3Li7 is 6,0067 + 1,0087 =7,0154

    7-e) The difference 7,0160 – 7,0154 = 0,006 is the mass defect due to the binding energy of the spin-interaction between the neutron and the deuteron.

    7-f) Therefore, in 3Li7 we have:

    Mass defect due to magnetic force 2He4+deuteron = 0,01

    Mass defect due to spin-interaction deuteron+neutron = 0,006

    As we realize, in the 3Li7 the binding energy due to the magnetic force is 1,7 times stronger.

    But obviously such relation changes in other nuclei, because of the changing in the magnetic moment of the nucleus and the distance deuteron-2He4.

    Regards
    wlad

  • Wladimir Guglinski

    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. That is what we know empirically. The other variables (a, c, d) that you mention in your list have a dipolar nature that can not possibly logically have an effect to change a property of a monopolar nature. Switching spins, or reversing fluxes n(o), will not affect the electric charge of a particle. It could possibly logically only affect the dipolar properties of that particle. The way that you would have it, the deuteron would be treated no different from the neutron, even though the deuteron has something extra: an electric charge. What you are actually doing is ignoring the electric charge when considering induced magnetic moments from rotation of nucleons.
    ——————————————–

    COMMENT

    Joe,
    I dont see any problem, and let me tell you why, as follows.

    1) Consider a nucleus 2He4.
    As its two deuterons have contrary spins, the 2He4 has a total magnetic moment zero. And therefore the two charges of the two deuterons cannot induce a magnetic field, since the two charges are gyrating within a space without any magnetic field.

    2) Now let us consider the 3Li6 as example for explaining how occurs the magnetic force Fm between the deuteron and the central 2He4, as follows.

    3) The deuteron of the 3Li6 is captured by the flux n(o) of the 2He4, and it is dragged by the rotation of the flux n(o), in order that the rotation of the deuteron induces a magnetic moment in the 3Li6.

    4) The magnetic moment of the 3Li6 is +0,822, measured in experiments.

    5) Suppose that the rotation of the deuteron induces a magnetic moment +0,820

    6) The two deuterons of the 2He4 have a very short orbit radius of rotation about the center of the 3Li6 nucleus. As the two deuterons are gyrating wihin a magnetic moment +0,820 induced by the deuteron captured by the flux n(o), then the two positives charges of the 2He4 (gyrating within a magnetic moment +0,820)_ induce an additional very weak magnetic moment (because the radius of their orbits is very short).

    7) Suppose that the additional magnetic moment induced by the two deuterons of the central 2He4 is +0,002.

    8) Then the total magnetic moment of the 3Li6 becomes +0,820 + 0,002 = +0,822

    Note that the two charges of the deuterons of the central 2He4 induce a weak magnetic moment only when a particle with charge is captured by the flux n(o) of the 2He4, as occurs with a proton or a deuteron.
    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.

    As the neutron is not captured by the flux n(o), the phenomenon does not occur with the neutron.
    A neutron influences the magnetic moment of a nucleus (for instance the 3Li7) only when the neutron is captured by a deuteron via spin-interaction, and then the neutron also contributes for the magnetic moment of the 3Li7.

    I hope to have responded your question

    regards
    wlad

  • Paul

    Andrea,

    Is the e-cat tested by the independent third party for the last months still “state of the art”?

    Or has it become obsolete due to further development?

    Paul

  • Frank Acland

    Dear Andrea,

    Are the professors doing the testing at liberty to make adjustments to the E-Cat setup? In other words are they allowed to change performance parameters during the testing period?

    Many thanks,

    Frank Acland

  • Andrea Rossi

    Michael Schneider:
    I share the effort of cross fingers.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Orsobubu:
    Whom to you believe to: me or her?
    A.R.

  • Andrea Rossi

    Steven N. Karels:
    The Hot Cat under examination from the third indipendent party has never been stopped and it run continuously with no interruptions of sort.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    Can you share if the scientists continuously ran one eCat or was it stopped and started multiple times?

  • orsobubu

    “The reliability has been reached”. This sentence sounds to me a little like “The Eagle has landed” by Neil Armstrong, or another historical one…

    … After twenty minutes Fermi did reinsert the bars and everyone felt more relieved because the counters stopped ticking. Then there were handshakes and was told that the experiment was over. Professor Eugene Wigner, who came from Princeton, took out a bottle of Chianti wine and some paper cups: all drank an inch of wine and signed the fiasco. Soon after Arthur Compton, director of the Chicago laboratories, phoned Conant at Harvard University. He said: “The Italian navigator has landed in the New World”. Although the code was improvised offhand by Compton, Conant understood perfectly and anxiously asked: “How were the natives?” Compton: “Very friendly”.

  • orsobubu

    >… in that case all you get is an annichilation, but also a lot of energy, which becomes a second order serendipity.

    Maybe you’ve been lucky and you got the energy and in the end also her serendipity, dear Andrea; but she keeps telling me that the contact is impossible

  • Michael Schneider

    Dear Andrea Rossi,

    Today would be a perfect day for a COMING OUT 😉
    G20 in Washington & Russia threatening to stop Gas furniture…

    The next weeks should be interesting !

    I cross fingers too, as certainly most followers of this extraordinary blog.

    Kind regards,

    Michael S

  • Andrea Rossi

    Frank Acland:
    Yes, the reliability has been reached.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Orsobubu:
    … in that case all you get is an annichilation, but also a lot of energy, which becomes a second order serendipity.
    Warm Regards,
    A.R.

  • orsobubu

    If the beautiful daughter of the farmer, with quantic state +1 and positive serendipity, rejects you, one has to switch to the realm of antimatter physics, because the contact is impossible.

  • Frank Acland

    Dear Andrea,

    I’m glad to hear that the first reactor has worked as intended. You must be pleased about that.

    Best wishes,

    Frank

  • Andrea Rossi

    Italo R.:
    “Serentipity” means the discovery of something you were not searching for, while trying to discover something else: for example, the discovery of pennycillin by Fleming has been a serendipity. In this sense you can say that also to find the daughter of the farmer while searching a needle in a haystack is a serendipity. If the daughter of the farmer id beautiful, the serendipity is positive ( +1), if ugly, the serentipity is negative (-1) if so and so is neutral (0). Therefore the serendipity has three possible quantic statuses .
    Warm Regards,
    A.R.

  • Italo R.

    Dear Dr, Rossi, I really hope that the Team is discovering many farmer’s daughters..:-)

    « Serendipity is looking in a haystack for a needle and discovering a farmer’s daughter. » (IT)
    « La serendipità è cercare un ago in un pagliaio e trovarci la figlia del contadino. »
    (Julius Comroe Jr., 1976)

    Kind Regards,
    Italo R.

  • Andrea Rossi

    Koen Vandewalle:
    No, the total endurance , or the life-span of the charge and the E-Cat are not in the protocol of the experiment. Obviously, if the exhaustion will happen during the experiment, that would be a serendipity.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Frank Acland:
    We sent 3 of them, as spare parts, but ( this I can say) we did not have breaks or malfunctions, so far, so the spare parts are intact.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    You mentioned you sent your E-Cat to the neutral location when you learned about it. Did you send multiple reactors, or only one?

    Many thanks,

    Frank Acland

  • Koen Vandewalle

    Dear Andrea,
    Supposed that at the beginning of the test, the Professors did discover that the E-Cat does what is is meant to do. One question that should be asked: how much energy can produce one E-Cat ? Did you make agreements on this item ?
    And when the answer is that it continues in eternity to “catalyze energy”, then the test will never be over. A catalyzer is not consumed….
    Did you agree on limits in time or amount of energy ?
    Kind Regards,
    Koen.
    P.S. : the congratulations were not for the results of these tests, but for the respect that is finally given to your work.

  • Andrea Rossi

    Andrew:
    1- If the results will be negative, we will study the results and work to resolve the problems, as always
    2- My engagement with this technology will continue until I will be alive.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Giovanni Guerrini:
    Ssssswwishhhhh !!!
    Warm Regards,
    A.R.

  • Andrea Rossi

    Koen Vandewalle:
    Wait for the results, before congratulate.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Koen Vandewalle:
    Wait the results before congratulate…
    Warm Regards,
    A.R.

  • Andrea Rossi

    Teemu:
    The Professors have worked and are working in a laboratory that is not owned by us, is totally out of our premises and that we never used before. We knew of it few days before the beginning of the test and sent there the E-Cat. It is located in a Country that is not Italy and is not USA. I cannot give further information, but, obviously, the precise location where the test will have been completed will be described in the Report that will be written by the Third Indipendent Party. When we arrived there for the assembling of the reactor, some of the components of the t.i.p. were already there for the set up.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Don Witcher:
    1- we are working on it
    2- is a strategic core
    Warm Regards,
    A.R.

  • Andrea Rossi

    Gherardo:
    Interesting,
    Warm Regards,
    A.R.

  • Koen Vandewalle

    Dear Andrea,
    Report or Not Report, that is not the question.
    They have heard of it.
    The data are recorded.
    Congratulations,
    Koen

  • Gherardo

    Dott.Rossi,
    this invention would be great if could use electric power from renewable sources.
    Do you know any? 😉

    http://www.dailymail.co.uk/sciencetech/article-2599036/The-plane-powered-WATER-US-Navy-reveals-radical-new-game-changing-process-power-jets-boats-seawater.html

    Gherardo

  • Don Witcher

    Dear Andrea Rossi
    It has been a year now since you told us that you were very close to producing electric power with Hot Cat prototypes.http://www.journal-of-nuclear-physics.com/?p=802&cpage=5#comment-686269“.
    Can you tell us now whether (1) you have achieved this very important goal in laboratory tests or (2) if not yet when you might expect to achieve this goal?
    Warm regards
    Don Witcher

  • Giovanni Guerrini

    ….We all need SuperRossi! With a flyng E-Cat suit,of course!! eh,eh..

    But you have chosen one other way to save the world…

    Superegards G G

  • Teemu

    Dear Andrea Rossi,

    When you say that the Professors are conducting their work “in a neutral laboratory”, do you mean they are still in your premises, or instead situated in a laboratory that is not owned by you?

    Best Regards,

    Teemu

  • Andrew

    Dear Andrea Rossi,
    As you have repeatadly stated, regarding the tests running on the e-cat, there is the possibility of getting negative results.
    I was wondering what was going to happen in this case?
    Would that be the end of it at that point ?
    I don’t want to seem too malicious with this question, but the thing is that, if I got it correctly, you don’t have any economic interest in the success of your invention anymore since you sold it.
    what would your position be in case of negative results?
    Despite my question I want to say that I believe in your hard work and wish the best luck.
    Regards
    Andrew

  • Andrea Rossi

    Mark Saker:
    I am not really able to answer to this fantascientific issue. Sorry for that.
    Warm Regards,
    A.R.

  • Mark Saker

    Dear Andrea,

    Care to step into fantasy land for a moment :)

    In the film Iron Man, the metal suit gives Tony Stark the power of flight. This is achieved via ion thrust (I think)

    Do you think there is the slightest possibility that the scale reduction of reactors or complete changes in design could lead to a propulsive flight suit that uses the ‘Rossi Effect’ or a derivative of?

    I do not know the calculations behind how much power in how small a space would be required to lift a human off the ground but is there any reason why this could never be done?

    Thanks for answering my “silly” question.

    Mark

  • Joe

    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. That is what we know empirically. The other variables (a, c, d) that you mention in your list have a dipolar nature that can not possibly logically have an effect to change a property of a monopolar nature. Switching spins, or reversing fluxes n(o), will not affect the electric charge of a particle. It could possibly logically only affect the dipolar properties of that particle. The way that you would have it, the deuteron would be treated no different from the neutron, even though the deuteron has something extra: an electric charge. What you are actually doing is ignoring the electric charge when considering induced magnetic moments from rotation of nucleons.

    All the best,
    Joe

  • Andrea Rossi

    Giuliano Bettini:
    You understood correctly.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Frank Acland:
    No.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear Joe,
    I think that I finally understood in detail the stability of the light nuclei.

    Let me tell you my conclusion.

    Look at the Figure 3 of the paper Stability of Light Nuclei:

    a) You see that the neutron in the outer side of the side DOUGLAS has positive magnetic moment +1,913

    b) You see that the deuteron in the inner side of the side DOUGLAS has positive magnetic moment +0,857

    Now look at the Figure 13 showing the 3Li7, with the neutron and the deuteron in the positions described in a) and b) above:

    c) The magnetic moment of the 3Li7 is +3,256

    d) Therefore the deuteron and the neutron would be submitted to a magnetic force of repulsion, if they were at rest with regard to the center of the 3Li7.

    e) But the deuteron is not at rest. Actally it gyrates about the central 2He4. Therefore the rotation of its magnetic moment +0,857 induces a magnetic force of attraction with the center of the 3Li7, in order that the force of repulsion is cancelled by the force of attraction

    f) The same happens with the neutron. As it gyrates about the central 2He4, it induces a magnetic force of attraction, which cancells the force of repulsion.

    g) Therefore the magnetic moments +0,857 of the deuteron and +1,913 of the neutron do not have influence in the equilibrium of the nucleus.

    h) It remains the magnetic force Fa of attraction due to the rotation of the positive charge of the deuteron gyrating about the magnetic moment +3,256 of the 3Li7.

    i) The magnetic force Fa of attraction is equilibrated by the centripetal force Fc on the deuteron-neutron

    regards
    wlad

  • Frank Acland

    Dear Andrea,

    Very interesting information about the authors and reviewers. From what you described it looks like the writing and reviewing process could take quite a long time — many months perhaps. Do you have any expectations on the approximate amount of time it may take?

    Many thanks,

    Frank Acland

  • Giuliano Bettini

    Dear Andrea,
    you wrote on April 8th, 2014 at 9:02 PM:
    “….7 Universities and Nuclear Physics Institutes of Europe, Asia, America …”
    They seem to be … 7 DIFFERENT institutes and universities, which is a big number …
    Did I understand correctly?
    Thank you, I wish you a good work,
    Giuliano Bettini.

  • Andrea Rossi

    Steven N.Karels:
    Fingers crossed.
    Warm Regards,
    A.R.

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