Advanced concepts in black hole cosmology

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by
U.V.S.Seshavatharam
Honorary faculty, I-SERVE, Alakapuri, Hyderabad-35, AP, India
QA-Spun division, LANCO Industries Ltd, Srikalahasti-517641, AP, India
E-mail: seshavatharam.uvs@gmail.com
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Abstract
Based on the big bang concepts- in the expanding universe, ‘rate of decrease in CMBR temperature’ is a measure of  the cosmic ‘rate of expansion’. Modern standard cosmology is based on two contradictory statements. They are – present CMBR temperature is isotropic and the present universe is accelerating. In particle physics also, till today laboratory evidence for the existence of ‘dark matter’ and ‘dark energy’ is very poor. Recent observations and thoughts supports the existence of the ‘cosmic axis of evil’. In this connection an attempt is made to study the universe with a closed and growing model of cosmology. If the primordial universe is a natural setting for the creation of black holes and other non-perturbative gravitational entities, it is also possible to assume that throughout its journey, the whole universe is a primordial (growing and rotating) cosmic black hole. Instead of the Planck scale, initial conditions can be represented with the Coulomb or Stoney scale. Obtained value of the present Hubble constant is close to 71 Km/sec/Mpc.
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333 comments to Advanced concepts in black hole cosmology

  • Wladimir Guglinski

    Dears Joe and Steven N. Karels

    The excited 6C12 is a good example of qualitative experiment which result is capable to disprove a theory.

    When you analyse if the current Nuclear Physics is able to be fit to the two nuclear properties i=2 and µ= 0 of the excited 6C12, the question to be responded is the following:

    QUESTION:
    Is it possible to explain i=2 and µ= 0 of excited 6C12 by considering the current Nuclear Physics?

    And only two answers are possible:

    YES or NO .

    If the answer is NO, the theory is WRONG, because:

    1- nobody can corrupt the answer NO with mathematical calculations.

    2- the answer NO is definitive, and it means that it`s IMPOSSIBLE to get i=2 and µ= 0 of excited 6C12 by any combination between:
    ———————————–
    * the spins 1/2 (of the 6 protons and 6 neutrons)

    * and the magnetic moments µ=+2,973 of the proton and µ=-1,913 of the neutron.
    ———————————–

    This is the virtue of the qualitative experiments, and this is the reason why the physicists hate them.

    Unlike, when you have a quantitative experiment, you can introduce several degrees of accuracy, and so often there is no way to discard definitivelly a theory by using quantitative experiments.
    Besides, you can change the equations, by improving new parameters, so that to fit the results to the experimental data.

    This is the difference between the qualitative and quantitative experimets.

    Of course quantitative experiments are very important, because they can point out if the theory is in the correct way.
    However often a quantitative experiment is NO definitive, because NUMBERS ARE SUBJECTED TO BE CORRUPTED by the manipulation of the mathematics.

    Other important qualitative experiment was that made by Michelson.
    With his experiment, Michelson made the following question to the Nature:

    Is there a luminiferous aether filling the space?

    The experiment had responded: NO

    That answer was definitive: the luminiferous aether, supposed by the theorists of the 19th Century, does not exist.

    regards
    wlad

  • Eric Ashworth

    With regards current discussions that interest me. Just a few of my own thoughts. With regards what Wladimir has to say. Photons you mention have orthogonal feature of motion, this I presume relates to a helical trajectory of motion. This I suspect is due to the integral make-up of the photon regarding quarks of aether and the medium through which it passes. Nothing can escape the positive and negative forces that exist in nature unless the item has no charge. This is because you are within a charge of the macro environment and a structure with a charge responds accordingly as it traverses within i.e. it rotates upon a helical trajectory to remain neutral. The macro environment has an axis composed of two radials, this being the major macro potential forces. The minor macro forces span off the axis and it is these two forces, I believe, that are responsible for helical trajectories with regards photon activity. A photon with regards an aether could be explained as such. A photon is comprised of particles that go towards the make-up of a photon. These particles are quarks and the make-up of the quark being aether. Photons and quarks are therefore structures but aether is not a structure it contains no particles. Aether is therefore a substance or it could be termed an essential substance. Your mention of ‘the sound of aether’ intrigues me because you mention a velocity faster than light. You also mention wave activity of photons as against longitudinal activity of aether. The photon has to travel on a wave which is a helical trajectory to remain neutral to the exterior forces whereas the aether being mass less requires no need to remain neutral having no charge and thereby is able to travel on a vector that shortens its travel time due to less distance. It could in theory travel at the same velocity as a photon. Aether as presviously stated responds to empty space. Heat is an interesting subject that I will put some thought into with regards aether,containers and velocity. I believe everything in nature is binary. If so, if linear flow is on the aether dimension then linear flow must be possible on the atomic curvature dimension and it is but only by integrating curvature forces will linear forces be produced. This phenomena has been demonstrated but I realise this is straying off the subjects in question and I mention it as a point of interest. Regards Eric Ashworth.

  • JR

    Dear Wladimir,

    Since you won’t point to anything that actually suggests that an excited state of carbon with i=2, mu=0 exists, and consider my request for any such evidence to be ‘bla-bla-bla’, I’ll have to stick with my original answer: As far as I know, no such state exists.

    FYI, the table is not THAT difficult to understand. Each line is one measurement, in almost all cases, a given measurement extracts EITHER mu or Q(b). The method used for each measurement is also quoted, and the only measurement for the 12C excited state used the “CER” (Coulomb Excitation Reorientation) technique which only provides Q(b) values for all cases I could find. You could, maybe, look up the original reference to see what they measured, or look elsewhere for a measurement of the magentic moment. Let me know if you find anything :)

    So since both a cursory and detailed reading of the table suggests that a blank space is not equivalent to a measurement of zero (which, by the way, is also just obvious), then you are wrong in yet another assertion that this is a phenomenon that cannot be explained in normal models. So at this point, I believe that I have answered (or pointed to answers) all of your questions, and you have yet to show that any of your assertions about the failing of conventional nuclear physics hold water.

    -John

    p.s. Since I’m confident that you’re not going to bother trying to support your assertion, I went ahead and looked up the reference for the original measurement; no mention of any measurement of a magnetic moment is given. No indication searching more broadly for such information either, but there’s only so much time I want to waste on this.

  • Andrea Rossi

    Giancarlo:
    We will donate it to the family of Prof. Sergio Focardi.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    JR wrote in July 10th, 2013 at 5:03 PM
    Dear Wladimir,

    I note that only one of the nuclei you list (other than 12C, which we were discussing) appears in the tables you pointed me to. 20Ne has two excited states listed, both of which have non-zero magnetic moments. Sorry, I made a mistake: 16O is also listed, and also has a non-zero magnetic moment. So I believe that my interpretation of “no number given means the value is unknown” is a reasonable interpretation.

    COMMENT

    JR, as I said, I will not waste my time with yours bla-bla-bla-bla…bla…

    All NON-EXCITED nuclei with Z=N=pair have null magnetic moment:
    2He4
    4Be8
    6C12
    8O16
    10Ne20
    12Mg24
    14Si28
    16S32
    18Ar36
    20Ca40
    etc.
    etc.
    etc.

    However, all those nuclei when EXCITED exhibit NON-NULL magnetic moment

    For instance:
    10Ne20 has two excited states, one has µ= 1,08 and the other has µ=1,5.

    But the excited 6C12 is an exception: it has spin i=2 and µ=0.

    Therefore, it`s impossible to explain the null magnetic moment of 6C12 by considering the current theories of nuclear physics

    If you dont know, Einstein had said:
    all experiments cannot prove a theory to be correct, but one unique experiment is able to prove it is wrong.

    And the excited 6C12 proves that current nuclear theories are wrong, since it is IMPOSSIBLE to explain its null magnetic moment by considering the current theories.

    Dear JR,
    your claim that “no number given means the value is unknown” is NOT a reasonable interpretation, because your interpretation actually shows that you are not able even to interpret correctly a nuclear table.

    I understand that you want to save the current theories, by such a dishonest way of argumentation.

    But I cannot waste my time with a dishonest person who even does not know to interpret correctly a nuclar table.

    regards
    wlad

  • Wladimir Guglinski

    Dears Joe and Steven N. Karels

    I would like to hear your oppinions on the problem concerning the halo neutron of 4Be11, with radius 7fm.

    JR had exhibited the following argument, so that to explain the radius 7fm detected in the experiment published in 2009:

    =================================================
    There is no reason that the size of a bound state must be less than the ‘approximate range’ of the relevant force. You can solve a system with square well potential with a well defined maximum range, and the particles can be further from the center than this range. In fact, even a point interaction can yield a bound state with finite size. So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.
    =================================================

    Pay attention to his argument:

    So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.

    And my reply:

    =================================================
    I would like to emphasize that you explanation is 100% INCORRECT and UNACCEPTABLE

    Because:

    In 97,1% of the decays of 11Be, it transmutes to 11B.

    Therefore, the halo neutron has a decay and it transmutes to a proton.

    Well, earlier the decay of 11Be the halo neutron was in a distance of 7fm from the rest of the nucleus.
    You claim that the halo neutron is very weakly bound to the rest of the nucleus.

    However, when the neutron transmutes to a proton, a very strong Coulomb force tries to expel the proton from the rest of the nucleus.
    Remember that the proton is far away 7fm from the nucleus, and so it is very weakly bound.

    Therefore the proton cannot go back to the nucleus, so that to form the 11B, since the proton is quickly expelled because of the strong Coulomb repulsion.

    CONCLUSION:
    from the principles of current Nuclear Physics, the decay 11Be -> 11B is IMPOSSIBLE !!!!
    but experiments show that it occurs in 97,1% of 11Be decays.
    =================================================

    Dears Joe and Steven,
    as you know, 5B11 is stable, and it has NOT a halo neutron.

    Therefore my interpretation is the unique way to explain what occurs in the 4Be11 decay:

    1- the neutron decays to a proton
    2- the proton goes back to the nucleus (which is IMPOSSIBLE by considering the current nuclear theories, since the proton has a big repulsion with the rest of the nucleus 5B11, and so it cannot go back to the nucleus, since it is very weakly bound (as the own JR had claimed).

    So, dears Joe and Steven,
    please tell me your honest opinion:
    do you think is it possible to explain the halo neutron with radius 7fm of 4Be11, and its decay in 5B11, by considering the current nuclear theories ?

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in July 10th, 2013 at 6:05 PM

    Wladimir,

    1- You say that a wave is not caused by the interaction of particles.
    Yet, you have compared the helical trajectory of a particle to a non-symmetrical parachute rushing against air. This implies the collision (interaction) of particles.

    COMMENT
    This imply that the helical trajectory of the non-symmentrical parachute is caused by the interaction of the matter (parachute) with the medium (air).

    This imply that the helical trajectory of the electron with spin is caused by the interaction of matter (electron) with a medium (aether).

    2- You also say that a wave is a disturbance in a medium.
    But, if a disturbance in a medium does not involve interacting particles, how can a wave transfer its energy and momentum except that it is an entity that is INDEPENDENT of the medium in which it finds itself.

    COMMENT
    Wave is interaction of particles, but not in the sense you claim.
    Wave is a disturbance caused by statistical interaction of matter (molecules of the air, molecules of the water, particles of the aether).

    Wave is not a property of ONE particle (molecule, atom, or particle of the aether).

    3- So, is wave nature dependent or independent of particle nature?

    COMMENT
    No, wave is a statistical disturbance of the particles of a medium

    The SO CALLED wave feature of a particle is not a statistical propagation of a medium. Because a particle moving with helical trajectory yields diffraction (characteristic of waves) thanks to the helical trajectory.

    regards
    wlad

  • Hoping to donate it to your staff, in memory of Sergio Focardi, when a public device will be born!

    Best Regards

    Giancarlo Di Grigoli

  • Andrea Rossi

    Giancarlo:
    Very nice, you made a professional job!
    Warm Regards,
    A.R.

  • Dear Andrea,

    When I painted this reproduction of the work of Giambattista Tiepolo “The triumph of fortitude and wisdom”,
    https://picasaweb.google.com/117535862209080486614/Mu#5899326774588294114

    I often thought at your crazy and brilliant attempt to “donate” to humanity the possibility of turn on a new type of “fire.”
    In the allegory (the original is at the Civic Museums of Udine – Italy), the figure depicts the Wisdom holding in his left hand the lumen of truth (… the new “fire” :-)) and she is flanked by the Fortitude, armed with a spear.
    Their triumph over Ignorance, the figure falling in the darkness covering his face in shame, is symbolized by the ringing of trumpets in the high part of the composition.

    Best regards

    Giancarlo Di Grigoli

  • Andrea Rossi

    Svein Utne:
    1- yes
    2- no
    I am extremely engafed now here because our technological development requires a long process, involving many changes as the technology moves forward. E-Cat is undergoing that process now.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Sterling Allen:
    We have great hopes for the E-Cat and what it can accomplish, and we are pleased about the findings of the other scientists and persons who have participated in evaluating so far. As this technology is still also in the phase of R&D and undergoing rigorous review, we will continue the testing process to determine the potential and uses of the same. We are pleased with our progress to date and I will share more as our work continues.
    Thank you for your continue attention,
    Warm Regards,
    A.R.

  • Dear Andrea Rossi,
    On July 5th you confirmed that you are working full time for the “USA Concern”. Is this the
    USA Partner and world Licensee for the manufacturing of the E-cat?
    Is this really the same non profit organization with this web page http://www.concernusa.org
    Regards
    Svein

  • LENR-to-Market Weekly — July 11, 2013 – Highlights this week include: US E-Cat manufacturing plant allegedly successfully builds working E-Cat; NAVY LENR Patent Granted – Transmutes Radioactive Waste; Synopsis of recent Finnish patent; John O’ Mara Bockris passes (PESN; July 11, 2013)

  • Joseph Fine

    Andrea,

    I know you and your readers are busy but there may be interest in the following

    (Below text was copied from cover sheet)

    http://www.sstp.org/showcase

    “You Are Cordially Invited to Attend the
    Sandia Research & Technology Showcase.”

    Learn about some of the cutting edge research and technology
    development taking place at Sandia National Laboratories.

    The event will also provide information on doing business with
    Sandia National Laboratories through licensing, partnerships,
    procurement, and economic development programs.

    Date and Time
    Tuesday, September 10, 2013
    8:30 am – 4:00 pm
    (Registration and check-in begins at 8:00 am)

    Location
    Embassy Suites
    1000 Woodward Place NE
    Albuquerque, New Mexico 87102

    Event is free and open to the public, but online registration is required.
    For more information and registration, visit http://www.sstp.org/showcase

    http://www.sstp.org/showcase

    (FYI: Sandia means ‘Watermelon’ in Spanish.)

    Best regards,

    Joseph Fine

  • Joe

    Wladimir,

    You say that a wave is not caused by the interaction of particles.
    Yet, you have compared the helical trajectory of a particle to a non-symmetrical parachute rushing against air. This implies the collision (interaction) of particles.

    You also say that a wave is a disturbance in a medium.
    But, if a disturbance in a medium does not involve interacting particles, how can a wave transfer its energy and momentum except that it is an entity that is INDEPENDENT of the medium in which it finds itself.

    So, is wave nature dependent or independent of particle nature?

    All the best,
    Joe

  • JR

    Dear Wladimir,

    You said: From your argument, we infer that all the nuclei with Z=N= pair do not have null magnetic moment, since the value of their magnetic moments do not enter in the nuclear tables. and then listed several even-even nuclei.

    I note that only one of the nuclei you list (other than 12C, which we were discussing) appears in the tables you pointed me to. 20Ne has two excited states listed, both of which have non-zero magnetic moments. Sorry, I made a mistake: 16O is also listed, and also has a non-zero magnetic moment. So I believe that my interpretation of “no number given means the value is unknown” is a reasonable interpretation.

    I realize that you can’t be bothered to read other people’s responses or the references they point you to in trying to make honest arguments. But you could at least glance at the articles that YOU are using to justfy your insulting rants.

    As for your follow up comment, you say “Two neutrons interact via strong force. It does no matter if neutron-neutron interaction is different of the proton-neutron interaction.” I’m very curious to hear your explanation as to why the force between two objects is unrelated to whether or not they form a bound state. By the way, just in case it was unclear, when I say that they have different interactions I mean that the strong force yields different interactions between neutron-neutron (or proton-proton) and neutron-proton configurations. So again, I would say that I DID answer your question, you just don’t like the answer.

    And, as noted before, I’m not arguing that everything is fully and completely understood or that there’s no room for new ideas. I’m arguing that you have an extremely poor understanding of conventional nuclear theory, and so your attempts to describe it as a dismal failure are, to an very good approximation, information free. This, coupled with your clear lack of interest in gaining a better understanding of the physics involved makes me quite fine with the idea of not discussing this any more.

  • Wladimir Guglinski

    JR wrote in July 10th, 2013 at 8:44 AM

    Dear Wladimiar,
    Why is there no di-neutron?: because the interaction between two neutrons is different from the interaction between a neutron and a proton.

    COMMENT
    Dear Jr,
    again your explanation sounds like a joke.

    Two neutrons interact via strong force.
    It does no matter if neutron-neutron interaction is different of the proton-neutron interaction

    What is important is the following:

    two neutrons interact via strong force, as the experiments have detected, and the magnitude of strong force is 100 times stronger than that of the electromagnetism.

    So, as the interaction neutron-neutron is very strong (100 times stronger than electromagnetism), then two neutrons would have to be bound in the dineutron, according to Nuclear Physics.

    So, dear JR,
    YOU DID NOT ANSWER WHAT IS THE FORCE OF REPULSION WHICH DO NOT PERMIT TWO NEUTRONS TO FORM A DINEUTRON

    But how did Heinsenberg solve this problem?

    Heinseberg had solved it by a very confortable way:

    He solved it via mathematics.

    Of course his solution is ABSURD, because the mathematics cannot create a physical force of repulsion, able to separate two neutrons tied by the strong force.

    Now let`s go to the point:

    1- Nuclei with Z=N=pair have to have a spherical shape, because the nucleons within a nucleus behave like a gas of molecules within a spherical vessel, according to Nuclear Physics

    2- There is not any force (according to Nuclear Physics) capable to apply a distortion on those nuclei.

    3- However the expeeriments show that such nuclei do not have a spherical shape.

    How to solve such problem?

    It`s very easy, and the nuclear theorists have done it: they simply apply the ABSURD Heisenberg method

    As the theorists cannot find any PHYSICAL force which causes the distortion of those nuclei, then they simply invent some equations with the aim to describe the distortion.

    But they do not explain what is the physical force with causes the distortion

    THIS IS NOT PHYSICS, dear JR.
    The solution presented by nuclear theorist is not Physics.
    It`s Mathematics, without connection with the physical reality.

    Dear JR,
    you are in the wrong place.

    There is a lot of magazines, like Scientific American, where physicists like you can publish articles, so that to mislead the lay public into believing that everything is 100% OK with current theories of physics.

    Go publish your lies in those sort of magazines. You will have a big audience in there.

    Here we are not interested in lies.
    Here we are interested in the true.

    If we had interested in lies, we would not be here, reading about Rossi`s eCat, which working is absurd according to the claim of physicists like you who claimed along 20 years that cold fusion is a fraud and Fleischmann was a charlatan.

    As you are not honest in your replies, because you try to find dishonest subterfuges so that to try to hide the true, I will not waste my time in discussing with you anymore.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 10th, 2013 at 9:14 AM

    Dear Wladimir,
    Also, thank you for the reference, but I should note that the fact that a value is not entered for the magnetic moment usually means that it’s unknown, not that it’s zero

    COMMENT

    Dear JR
    you have made a mystake when you decided to become a physicist. You actually had to become a comic artist.

    From your argument, we infer that all the nuclei with Z=N= pair do not have null magnetic moment, since the value of their magnetic moments do not enter in the nuclear tables.
    For instance:
    2He4
    4Be8
    6C12
    8O16
    10Ne20
    12Mg24
    etc
    etc
    etc

    Nuclear tables do not give information on their magnetic moments.

    So, according to your argument, it means that their magnetic moments is unknown, not that they are zero… ha, ha, ha, ha…

    Dear JR,
    you are very funny.
    Your attempt to defend the nuclear theory is pathetic.

    Regards
    wlad

  • JR

    Dear Wladimir,

    On the 11Be decay: There is nothing exotic or unusual about it’s decay. The neutron becomes a proton, the coulomb interaction is very weak and the proton is in a lower energy state (since there are fewer protons and it can fill a lower shell state), yielding a more tightly bound 11B nucleus. You don’t have to like the explanation, but it’s completely conventional.

    Note that nothing says that the neutron that’s further out has to be the one that changes to a proton. Since it’s energetically favorable for it to end up in the 11B final state, so I would think that it would be more likely that one of the many neutrons in the ‘core’ would be the one to decay, just on energy/phase-space arguments.

    p.s. accidentally put the end of my last post in bold, feel free to ignore that fact – it meant nothing

  • JR

    Dear Wladimir,

    First, I think it’s great that lots of people are thinking about all of these issues, and I have no problem with people who aren’t active physicists thinking about these problems. But if you’re going to make statements about things that do/don’t work in conventional theory, you really need to understand the theory at some level. I know next to nothing about QRT, and so I don’t spend time arguing that it’s wrong. If I were to try and say QRT was wrong by misrepresenting what you’re said, you should certainly call me on it. That’s what I’m doing when you misrepresent conventional nuclear theory, or when you misrepresent various experiments in nuclear structure of quantum mechanics it order to claim that they support your theory.

    On to the previous questions:

    1) Again, you totally missed the point. Your argument is that no even-even nuclei can be non-spherical in conventional nuclear physics, and I pointed out that this is simply not true and that many calculations exist for deformed nuclei. Whether 16O is or isn’t spherical has nothing to do with whether your statement is incorrect (it is). As a side note, I suspect you are mistaken when you say that Freer changed his mind.

    2A) First, don’t pretend that my not being able to answer the question means that it isn’t understood in conventional physics. Second, I pointed out that calculations of this exist and showed you where they are. While you left this part out of your follow up question, it demonstrates that conventional nuclear physics can explain it. If you or I don’t intuitively understand why it happens, it may not be satisfying, but it does take care of your argument that conventional physics does not allow for such a phenomenon.

    [I didn’t answer the neutron question at the time because it hadn’t showed up on the board; but it was a simple question, answered in my last reply]

    2B) 1- Yes, nuclei are bound by the strong force

    2- No, it does not have a sharp cutoff at 3fm, although it is very weak at large distances

    3,4 – Incorrect, both because of 2 above and because the range of the force does not yield an identical limit on the size of the bound state. If we were talking balls on strings or sticks, you would be correct. If you want to dismiss quantum mechanics as a meaningless mathematical construct, you can feel free to be unsatisfied. But you can’t just ignore quantum mechanics if you’re going to argue the conventional models (which include QM) don’t work.

    2C) Once again, I’m not arguing that I understand everything, simply demonstrating that you don’t. It’s fine if you don’t understand some things, but when you don’t understand the things you’re “proving” are wrong, it’s an issue.

    Also, thank you for the reference, but I should note that the fact that a value is not entered for the magnetic moment usually means that it’s unknown, not that it’s zero. So I believe you are incorrect in stating that such a state exists (but feel free to check on the original reference if you want to be certain). Once again, as a reminder, even if it did have mu=0 and both you and I were unable to combine the contributions to get zero, that doesn’t mean that it’s beyond the range of the nuclear physics models.

    So try and come up with some actual questions that current theories actually can’t explain. If I still insist that there are not problems, then you can feel free to call me a betrayer.

    In the meantime, the following principles are worth remembering:

    “I can’t figure out how to do it” does not prove “it can’t be done”

    “I’m right” is not an argument to establish “you’re wrong”

  • JR

    Dear Wladimiar,

    Why is there no di-neutron?: because the interaction between two neutrons is different from the interaction between a neutron and a proton. That is the answer according to conventional nuclear physics, and detailed models of the two-nucleon interaction do an extremely good job of calculating detailed masses and excited state energy levels for many many states in a variety of light nuclei.

    One simple example to illustrate, since you don’t favor detailed mathematical descriptions: we often discuss the long-range interaction in terms of pion exchange: a neutral pion can be exchanged between any two nucleons, but positive and negative pions can only be exchanged between a proton and neutron (the proton emits a positive pion and turns into a neutron, the other neutron can absorb that pion and become a proton, while a second proton would not be able to do so as it would have charge=2). Thus, n-p is very different from n-n or p-p.

    More specifically, the tensor part of the interaction is important to binding and is applies only to the interaction between a proton and a neutron.

    So no, there is no repulsive force that overcomes the attraction. This is a long answered and well understood question in conventional nuclear physics. If you don’t know the theory well enough to know the answer then it’s not surprising that you have to dismiss it as a meaningless mathematical concept. If you’re not satisfied with that, then that’s OK. But clearly you’re not in a position to be bashing the theory as not being able to explain simple questions given your level of understanding, or to be criticizing people who do understand the theory for thinking it does a pretty good job.

    p.s. You told Andreas Moraitis that the link provided did not have to do with free tetra neutrons. However, if you read the paper (or even the abstract), you’ll see that they are talking about the formation of free tetraneutrons in the breakup of light, neutron-rich nuclei.

    Andreas – I should note that the original signal was not definitively connected to tetraneutron states. While it got many people excited (both those who believed it was real and those who did not), I do not think that further experimental investigations showed any support for the idea that what they observed were bound tetraneutron states. But if such a state exists, it would certainly pose problems for modern models of neutron-neutron interactions.

  • Wladimir Guglinski

    JR wrote in July 9th, 2013 at 3:38 PM

    Wladimir,
    So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.

    COMMENT
    I would like to emphasize that you explanation is 100% INCORRECT and UNACCEPTABLE

    Because:

    In 97,1% of the decays of 11Be, it transmutes to 11B.

    Therefore, the halo neutron has a decay and it transmutes to a proton.

    Well, earlier the decay of 11Be the halo neutron was in a distance of 7fm from the rest of the nucleus.
    You claim that the halo neutron is very weakly bound to the rest of the nucleus.

    However, when the neutron transmutes to a proton, a very strong Coulomb force tries to expel the proton from the rest of the nucleus.
    Remember that the proton is far away 7fm from the nucleus, and so it is very weakly bound.

    Therefore the proton cannot go back to the nucleus, so that to form the 11B, since the proton is quickly expelled because of the strong Coulomb repulsion.

    CONCLUSION:
    from the principles of current Nuclear Physics, the decay 11Be -> 11B is IMPOSSIBLE !!!!
    but experiments show that it occurs in 97,1% of 11Be decays.

    So, dear JR,
    your explanation is 100% INCORRECT and UNACCEPTABLE.

    regards
    wlad

  • Andrea Rossi

    Riccardo:
    1- the production of all the E-Cats will be made in the USA
    2- I know the theories of Todeschini, and they have nothing to do with our effect
    Next time, please translate in English, because the 90% of our readers can speak English and only the 10% can speak Italian. Just write in Italian and in English ( can use Google translator: better bad English than no English).
    Warm Regards,
    A.R.

  • Riccardo

    Il fatto che il partner americano è in grado di proseguire senza la tua assistenza è un ottima notizia ed un altro risultato miliare, svincola la sua persona dalla replicabilità dell’Ecat, e non è una notizia da poco !! La produzione quindi avverrà esclusivamente negli states? Per i partner degli altri paesi rimarrà tutto uguale o il partner americano ha o avrà l’esclusiva mondiale oltre che della produzione anche della commercializzazione??
    ps A proposito, ha considerato qualche dato di ricerca contenuto nelle teorie dell’universo superfluido, o i lavori di Todeschini?

  • Steven N. Karels

    Dear Andrea Rossi,

    I recall in my many days (now years) of engineering during the days of film photography, before really high speed film was invented, intelligent services would place a small light bulb within a camera to pre-expose the film for dark collection conditions. This resulted in not great imagery but allowed their operatives to collect photos without flashbulbs, etc.

    It occurs to me that something similar might be of benefit to eCat technology. If I assume the power mechanism involves the conversion of Hydrogen molecules to Hydrogen atoms, then some input energy is required to make them dis-associate. Perhaps a low level radioactive source outputting the correct energy level to make the dis-association(or whatever the process is) but in small enough quantity to not sustain a reaction, could be placed to “pre-charge” the environment.

    Then your Control system only needs to add the additional energy to raise or lower the reaction rate.

    If your container that holds the pressurized Hydrogen gas and the other eCat components can also absorb and contain the radiation products, there should be no certification issue. The amount of radiation within the eCat container would be so small as to not be noticeable in terms of heat generation.

    Anyways, as I have thoughts on eCat technology that I think might be useful to you, I will continue to suggest them and you can ignore, reject or use them as you wish.

    P.S.

    Really great news on the independent eCat production. Not many people realize how important that is. A Major Milestone.

  • Wladimir Guglinski

    Andreas Moraitis wrote in July 10th, 2013 at 1:15 AM
    Wladimir,

    In 2002, Marques et al. from GANIL at Caen in France found some hints for the possible existence of a tetraneutron:

    http://de.arxiv.org/abs/nucl-ex/0111001

    Dear Andreas,
    the link above is concerning neutron confined within nuclei.

    My discussion with JR is about free neutrons.

    regards
    wlad

  • Wladimir Guglinski

    Joe wrote in July 9th, 2013 at 8:37 PM
    Wladimir,

    1. One qualitative aspect of QRT is its inability to explain the missing QED baryons (composed of the QED quarks from QRT) in Nature. Therefore, how can QRT ever be considered a correct theory?

    COMMENT
    Joe,
    as I already had said, my theory is no complete.
    When I had started the development of my theory, I had the following thought:
    before to develop a Particle Physics, there is need to discover a correct Nuclear Physics, because the fundamental laws to be discovered in the level of Nuclear Physics will work in the level of Particle Physics.

    So, if we do not find the correct laws for Particle Physics, it`s impossible to develop a correct Particle Physics.

    2. Since you believe that a wave is a behavior that is caused by the interaction of particles, and not a nature that exists independently of particles, how do you define particles themselves? Are they a nature unto themselves? Or are they simply, like a wave, a behavior caused by the interaction of a deeper level of phenomenon?

    COMMENT
    I dont believe that a wave is a behavior caused by interaction of particles.

    Waves exist in Nature.
    Wave is a disturbance of a medium (it can be air, water, aether, etc.)

    For instance, in my book Quantum Ring Theory I show that there is an electromagnetic disturbance of the aether which I named the Sound of Aether.

    They were detected experimentally, and move faster than light.

    The difference between light and the sound of aether is basically because the photon is a particle with orthogonal feature of motion, while the sound of aether moves with longitudinal feature.

    One of the properties of the wave is the diffraction.
    The particles have not such a property.
    But a particle with helical trajectory exhibits the property of diffraction.

    Difference between particles and waves:

    1- Electromagnetic waves are a propagation of a longitudinal disturbance of the elementary particles of he aether.

    2- Particles are composed by quarks formed by the elementary particles of the aether.

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in July 9th, 2013 at 3:38 PM
    Wladimir,

    1) I said that even-even nuclei are not required to be spherical in conventional models. I did not say that no nuclei were allowed to be spherical. So the fact that 16O is spherical does not contradict what I said in any way.

    COMMENT
    So, you prove that you dont konw what you are saying, because Martin Freer publication in 2010 is wrong, since 8O16 is non-spherical.
    In 2012 he changed his mind.

    As a non-spherical nucleus must have non-null electric quadrupole moment, in 2012 the journal Nature had published a plagiarism of mine published in my book Quantum Ring Theory, in 2006, where I explain why 8O16 has null electric quadrupole moment, in spite of it is non-spherical.

    The argument published in the journal Nature in 2012 is the same argument published in my book in 2006.

    2A) I personally can’t explain why 224Ra and 220Rn are not spherical off the top of my head. But you’re the one claiming that this is absolutely impossible within conventional models of nuclear physics, so I would claim that the burden of proof is on you.

    COMMENT
    Yes, I can prove if you give us answer on my question: what is the repulsion force which wins the strong force of two neutrons and do not allow them to form a dineutron
    How can I prove you are wrong if you dont give us the answers I ask ??????

    2B) There is no reason that the size of a bound state must be less than the ‘approximate range’ of the relevant force. You can solve a system with square well potential with a well defined maximum range, and the particles can be further from the center than this range. In fact, even a point interaction can yield a bound state with finite size. So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.

    COMMENT
    So, you did explain NOTHING, because:

    1- According to current Nuclear Physics the nucleons within nuclei are bond by the strong nuclear force

    2- The strong force actuates in a maximum distance of 3fm.

    3- Therefore the neutron halo in 11Be cannot be tied to the rest of the nucleus.

    4- Conclusion: according to current Nuclear Physics the 11Be cannot exist

    2C) Again, you are asserting that it is not possible, so I would say that it would be more helpful if you found a work that claimed that it cannot be explained by conventional models.

    COMMENT
    Dear JR
    I dont want bla-bla-bla
    Show us the solution according to current Nuclear Physics.

    You are like all the betrayers of the scientific method.
    All you claim do not know any problem with the current Nuclear Physics.
    But when we ask solutions for some question that current theories cannot explain, all you start up with bla-bla-bla.
    I am tired of bla-bla-bla.

    But before that, let me ask why you say that there is a spin-2 excited state of carbon with zero magnetic moment.

    COMMENT

    Nuc….E(level)…s1/2….I……..µ(nm)………Q(b)

    12C…..4438……45 fs…2+……………….+0.06(3)

    http://faculty.missouri.edu/~glaserr/8160f09/STONE_Tables.pdf

    It may be the case, but to the best of my knowledge, we do not have evidence for such a state.

    COMMENT
    Then please go to change the nuclear tables, since they exhibit nuclear properties that descredit your current Nuclear Physics.
    After you change the nuclear tables, come back to discuss Nuclear Physics with me.

    regards
    wlad

  • Andreas Moraitis

    Wladimir,

    In 2002, Marques et al. from GANIL at Caen in France found some hints for the possible existence of a tetraneutron:

    http://de.arxiv.org/abs/nucl-ex/0111001

    Nevertheless, their results – which were inconsistent with “state-of-the-art”-physics – have been called into question. Would tetraneutrons be possible according to QRT? (I’m not a physicist, sorry…)

    Regards,
    Andreas

  • Joe

    Wladimir,

    1. One qualitative aspect of QRT is its inability to explain the missing QED baryons (composed of the QED quarks from QRT) in Nature. Therefore, how can QRT ever be considered a correct theory?

    2. Since you believe that a wave is a behavior that is caused by the interaction of particles, and not a nature that exists independently of particles, how do you define particles themselves? Are they a nature unto themselves? Or are they simply, like a wave, a behavior caused by the interaction of a deeper level of phenomenon?

    All the best,
    Joe

  • Andrea Rossi

    Giuseppe:
    Thank you for your kind words.
    Warm Regards,
    A.R.

  • Dear Andrea,
    you are italian and we all are proud for this, but as Meucci, Fermi, Marconi and many others did ,your success is far from here and this breaks my heart.
    Sad regards

  • JR

    Wladimir,

    1) I said that even-even nuclei are not required to be spherical in conventional models. I did not say that no nuclei were allowed to be spherical. So the fact that 16O is spherical does not contradict what I said in any way.

    2A) I personally can’t explain why 224Ra and 220Rn are not spherical off the top of my head. But you’re the one claiming that this is absolutely impossible within conventional models of nuclear physics, so I would claim that the burden of proof is on you. But you can also look at the nature article you cited and see predictions of the E3 transition, related to the octupole deformation, and see that models did predict the approximate scale of these transitions, although the exact values are somewhat off. While it’s possible that I’m misunderstanding what the shows, it looks to me like there are multiple calculations of octupole deformation in these nuclei, and you can look at the papers describing those calculations if you want to know why they are pear-shaped.

    2B) There is no reason that the size of a bound state must be less than the ‘approximate range’ of the relevant force. You can solve a system with square well potential with a well defined maximum range, and the particles can be further from the center than this range. In fact, even a point interaction can yield a bound state with finite size. So there’s nothing to explain, other than the fact that 11Be must be very weakly bound.

    2C) Again, you are asserting that it is not possible, so I would say that it would be more helpful if you found a work that claimed that it cannot be explained by conventional models. But before that, let me ask why you say that there is a spin-2 excited state of carbon with zero magnetic moment. It may be the case, but to the best of my knowledge, we do not have evidence for such a state.

  • Wladimir Guglinski

    Dear JR,
    there is other fundamental question I would like you explain to us.

    There is no repulsion between two neutrons. But two neutrons interact through the strong force.
    Therefore two neutrons would have to form the dineutron.

    But the dineutron does not exist in Nature.

    How do you explain why dineutrons do not exist?
    Please explain us the PHYSICAL cause which do not allow two netrons to bind via strong force.

    And please do not mention Heisenberg Isospin, because it is a mere mathematic concept, and it cannot explain the reason why dineutrons do not exist.

    Because as there is a physical force of ATTRACTION tying two neutrons via strong force, then only another physical force of REPULSION can be able to separate them, in order do not allow that a dineutron can be formed.

    As a mathematic concept cannot create a PHYSICAL force of repulsion, then obviously the isospin cannot be the physical cause that prevent two neutrons do not be linked by the strong force.

    So, dear JR,
    please explain us what is the origin of the force of repulsion which do not allow the dineutron to be formed.

    regards
    wlad

  • Andrea Rossi

    Steven N Karels:
    I never used radioactive materials, so I do not know.
    Warm Regards,
    A.R.

  • Eric Ashworth

    To the readers of JONP. With regards the present interest in theoretical and applied physics I must say I enjoy reading all the posts (Joe and Wladimir are very interesting although I cannot understand everything not being an academic). For me the subjects seem to be becoming more esoteric due to the more recent subject in question, this being reality that covers a whole spectrum. Anyway, one issue I have read about was structure, whether it is pear shape or spherical. For me it can be a minimum of three shapes, 1. None spherical,in a phase of becoming, 2. Spherical, 3. Pear shaped in a phase of going. During which time every structure rotates upon an axis to remain neutral within its body of creation (BOC) i.e. the macro creating environment. Structure goes through three stages over three distinct dimensions in its development stage from birth to death and consequential transmutation. It travels with its axis comprised of a helical trajectory and upon an axis of the BOC that is also of a helical trajectory. Every axis is comprised of two radial dimensions that constitute a diametrical dimension of the whole. In the lower radial the BOC of formation the structure is none spherical. When the structure reaches the midway position of the BOC it becomes spherical because the radials square in this position. When the structure crosses the midway position the structure becomes pear shape due to entering a more positive field of expansion with a directional pull. The creating environment is pear shape being within a greater body and being part of a systemic system. Spherical shape equals the field at the central position because it satisfies both positional shapes of the BOC. The lower radial of any field i.e. half of its axis, represents the fusion dimension which supports intelligence due to the BOC force. The mid position represents the position of conflict between the lower field of fusion and the higher field of fission that supports the higher intelligence. Intelligence is that which surrounds the field of any structure. Minerals/vegetables do not have intelligence as per se. but they do have response mechanisms or r.ms. intelligence at their own level. This three stage structural development occurs at any point in time along the axis of the BOC. The Earth is known to be pear shape. This is because it is now in the intelligent zone of its creating environment. What is the connection between pear shape and sphere?. There are only two dimensions, fission and fusion. Fusion is the forming of structure from aether. Fission is the disintegration of that which is formed between which for a brief period of time is that which is formed/become. The BOC can be considered as a static environment although it is not (not necessary to explain). Physics envelopes three specific subjects of which each have their own identity and purpose. Consequently, to mix them together and then try to incorporate all three into one mathematical formulae is as impossible as putting a fetus, teenager and an old person in the same room and expecting them to all act the same way. The life of every structure depends upon its own ‘internal economy flow system’. The more economy a structure posses the better the quality and the longer the life. The understanding of ‘the static and mobile mechanics of energy interaction’ helps to throw some light on this complex subject involving the BOC and the created structure. I believe one mathematical formulae is unable to encompass all aspects, only geometry and maths will achieve it. In a nutshell the subject is able to explain how curvature force energy/aether structure of interacting flows between two identies are able to form a linear flow with direction and purpose. Identies are within the atom (lazers are formed from curvature force energy). Hydrogen has very little economy. This is why it is easily broken down or captured. From personal experience I have found that the subject of physics has always been to certain people treated as a noxious gas within a bottle of which the cork must never be removed completely but only very, very slowly which is to some people an annoyance. I realise this is none technical but an overview as I see it. Regards Eric Ashworth.

  • Andrea Rossi

    Italo Caproni:
    It is not mine… Very likely it will remain in the USA.
    Warm Regards,
    A.R.

  • italo caproni

    Dear Andrea Rossi
    As you wrote ” the first is always special” soo please do not use it anymore but keep it for history.
    Many museum WILL need it
    Best regards
    Italo

  • Wladimir Guglinski

    JR wrote in July 9th, 2013 at 8:44 AM

    Wladimir,

    1-
    While you assert that current nuclear theories of nuclear physics require that even-even nuclei be spherical, this is just not true.

    So for more than 50 years, it’s been know that doubly-magic nuclei (all of which are even-even) tend to prefer spherical shape but do not need to be spherical.

    COMMENT
    then I suggest you to go teaching nuclear physics to the nulcear physicist Martin Freer, since in 2010 he had published a paper in which he claims that oxygen 8O16 is spherical.
    http://www.scholarpedia.org/article/Clusters_in_nuclei

    2-
    You claim that your model does a better job because other models cannot predict certain phenomena. But so far, I have yet to see you give an correct example of such a case, as your arguments appear to be based mainly on your lack of understanding of modern nuclear theory or basic quantum mechanics

    COMMENT
    OK, dear JR,
    then give us explanations for the following questions:

    A)
    Why 224Ra and 220Rn are pear-shape, since they have Z=paiir, N=pair.

    B)
    As you know, the range of strong force is about 2fm. But an experiment published in 2009 had showed that Be-11 has a halo neutron in a distance of 7fm.
    http://phys.org/news154361753.html

    C)
    The excited carbon nucleus 6C12 has nuclear spin i=2 and magnetic moment µ=0.

    Please show to us (or quote to us any theory capable to explain it) , how is it possible to explain how can the excitd 6C12 to have i=2 and µ=0 , by considering the current Nuclear Physics.
    As you know, the magnetic moment of proton is µ=+2,973 , and the magnetic moment of neutron is µ=-1,913 , while both them have spin i=1/2.
    So, you have to show the combination of spins and magnetic moments capable to give a nucleus 6C12 with i=2 and µ=0.

    Please prove to us that my arguments appear to be based mainly on my lack of understanding of modern nuclear theory or basic quantum mechanics

    If you succeed to explain those three questions, I will ask you more questions after your reply.

    gook luck

    regards
    wlad

  • Steven N. Karels

    Dear Andrea Rossi,

    The suggestion for a radioactive supplement was a technical question, not a business recommendation. I understand there are all sorts of constraints on commercial items and the addition of a radioactive material has certain deteriments. But the question was whether the addition of a radioactive source would have any benefit to COP improvement or SSM duration? Even if it wonderfully improved these or other performances of eCat technology, there would be management and sales considerations that might override its usage.

  • Andrea Rossi

    Hans-Joachim Mueller:
    Thank you,
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe wrote in July 9th, 2013 at 12:54 AM
    Wladimir,

    You have suggested that, in the future, QRT will be improved by the work of other scientists. Are you implying that QRT will never be a finished theory? (This, of course, would contradict your philosophical outlook on such matters.)

    No, Joe
    I know that my theory is not complete yet. As I had explained earlier, in my book The Missed U-Turn I state that we are in the 4th stage of the development of Modern Physics.

    My modeles are getting qualitative correct predictions, as many recent experiments are showing.

    But there is need to improve the models, so that to get quantitative predictions.
    I already have obtained some quantitative results (as the calculation of magnetic moments, binding energy of light nuclei, etc.). But there is need much more.

    In the upcoming next 5th stage there is need to improve my models, so that to develop equations from which the theorists will be able to get quantitative predictions.
    Such process of equations development depends on some empirical constants (similar to what the nuclear physicists did for the development of the empirical equation of the Liquid Drop Model of the Nuclear Physics).

    Also, there is need to get a lot of answers for questions concerning the light nuclei (for instance, like that raised by you, with respect on how the nucleons take their positions in the excited light nuclei).

    regards
    wlad

  • JR

    Wladimir,

    I misunderstood your point about 238U, as I thought this was a direct comment on the recent nature article. However, that doesn’t change the main focus of my argument. While you assert that current nuclear theories of nuclear physics require that even-even nuclei be spherical, this is just not true.

    From the abstract of a 1959 Science article “Shape of the Nucleus”: “Nuclear shapes can vary rather widely, with doubly magic nuclei preferring spherical symmetry.” So for more than 50 years, it’s been know that doubly-magic nuclei (all of which are even-even) tend to prefer spherical shape but do not need to be spherical. The Nilsson model, a 1950s modification of the shell model that goes beyond spherical nuclei, yields deformations for even-even nuclei, and generates magic numbers for heavy nuclei which match data, while the purely spherical shell model does not.

    So you argument is fine if you want to exclude the purely spherical shell model. But that was demonstrated to be incomplete (or, if you prefer, “100% incorrect”) decades ago. So it is untrue to say that for the last 80 years, theorists have believed that even-even nuclei must be spherical. It’s probably even incorrect to say that it was believe 80 years ago, as the neutron had only just been discovered and nuclear structure models were somewhat incomplete.

    So we come back to comparing your model, which cannot make quantitative predictions for deformations, and modern nuclear theory which can. The quantitative models are more useful, are easier to test, AND give just as much qualitative insight into what’s actually going on. Your approach is much harder to evaluate or disprove, as you don’t actually make many significant predictions. You claim that your model does a better job because other models cannot predict certain phenomena. But so far, I have yet to see you give an correct example of such a case, as your arguments appear to be based mainly on your lack of understanding of modern nuclear theory or basic quantum mechanics (e.g. your total misunderstanding of the single/double slit interference experiments, which is introductory quantum mechanics level)

  • Andrea Rossi

    Greg Leonard:
    I agree.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Arthur B:
    Of course, as soon as we will make also electric power the Customers’ factory will be powered with the E-Cat.
    About my plans, please read the answer I gave to Eugenio Mieli.
    Warm Regards,
    A.R.

  • Hans-Joachim Müller

    Dear Mr. Rossi,
    I am very happy to learn about the progress in the fabrication of the E-Cat. We followers should not expect further progress in days or weeks, but we should work to spread the information about E-cat among the people.

    All the best for you and your team
    Hans-Joachim Müller

  • Andrea Rossi

    Eugenio Mieli:
    Thank you for your attention,
    Warm Regards,
    A.R.

  • Andrea Rossi

    Steven N Karels:
    Thank you, Iagree, the totally indipendent manufacturing by the Customer of the charge of the E-Cat and of the E-Cat itself has been an event to take record of.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Steven N Karels:
    To add radioactive material to the charge would annichilate all the certifications on the spot and would make unsellable all our production. Not really a good idea, Sir!
    Warm Regards,
    A.R.

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