To Understand The Basics Of Black Hole Cosmology

.
by
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U.V.S. Seshavatharam
Honorary faculty, I-SERVE, Alakapuri,
Hyderabad-35, AP, India
Email: seshavatharam.uvs@gmail.com
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S. Lakshminarayana
Dept. of Nuclear Physics, Andhra University,
Visakhapatnam-03, AP, India
Email: lnsrirama@yahoo.com
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.
Introduction
In this paper by highlighting the following 28 major short comings of modern big bang cosmology the authors made an attempt to develop a possible model of Black hole cosmology in a constructive way [1-3].
From now onwards instead of focusing on ‘big bang cosmology’ it is better to concentrate on ‘black hole cosmology’.
Its validity can be well confirmed from a combined study of cosmological and microscopic physical phenomena.
It can be suggested that, there exists one variable physical quantity in the presently believed atomic and nuclear physical constants and “rate of change” in its magnitude can be considered as a “standard measure” of the present “cosmic rate of expansion”.
Michael E. McCulloch says [4]: For an observer in an expanding universe there is a maximum volume that can be observed, since beyond the Hubble distance the velocity of recession is greater than the speed of light and the redshift is infinite: this is the Hubble volume.
Its boundary is similar to the event horizon of a black hole because it marks a boundary to what can be observed.
This means that it is reasonable to assume that Hawking radiation is emitted at this boundary both outwards and inwards to conserve energy, and any wavelength that does not fit exactly within this size cannot be allowed for the inwards radiation, and therefore also for the outwards radiation.
According to Hawking, the mass of a black hole is linearly related to its temperature or inversely-linearly related to the wavelength of the Hawking radiation it emits.
Therefore, for a given size of the universe there is a maximum Hawking wavelength it can have and a minimum allowed gravitational mass it can have.
If its mass was less than this then the Hawking radiation would have a wavelength that is bigger than the size of the observed universe and would be disallowed.
The minimum mass it predicts is encouragingly close to the observed mass of the Hubble volume.
Thus it is possible to model the Hubble volume as a black hole that emits Hawking radiation inwards, disallowing wavelengths that do not fit exactly into the Hubble diameter, since partial waves would allow an inference of what lies outside the horizon.
According to Tinaxi Zhang [5-7], the universe originated from a hot star-like black hole with several solar masses and gradually grew up through a super massive black hole with billion solar masses to the present state with hundred billion-trillion solar masses by accreting ambient materials and merging with other black holes.
According to N. J. Poplawski [8-11], the Universe is the interior of an Einstein-Rosen black hole and began with the formation of the black hole from a supernova explosion in the center of a galaxy.
He theorizes that torsion manifests itself as a repulsive force which causes fermions to be spatially extended and prevents the formation of a gravitational singularity within the black hole’s event horizon.
Because of torsion, the collapsing matter on the other side of the horizon reaches an enormous but finite density, explodes and rebounds, forming an Einstein-Rosen bridge (wormhole) to a new, closed, expanding universe.
Analogously, the Big Bang is replaced by the Big Bounce before which the Universe was the interior of a black hole.
The rotation of a black hole would influence the space-time on the other side of its event horizon and results in a preferred direction in the new universe.
Most recently cosmologists Razieh Pourhasan, Niayesh Afshordi and Robert B. Manna have proposed [12] that the Universe formed from the debris ejected when a four-dimensional star collapsed into a black hole – a scenario that would help to explain why the cosmos seems to be so uniform in all directions.

637 comments to To Understand The Basics Of Black Hole Cosmology

  • Robert Curto

    DTravchenko,
    They are talking about Dr. Rossi all over the World. Even Bill Gates is getting
    involved.
    I am pleased to see the two men in the World, that I admire the most.
    Drs. Rossi and Matsumura, after an uphill Battle getting the recognition they deserve for their outstanding achievements.
    Dr. Rossi in the energy World.
    Dr. Matsumura in the cancer World.
    I am proud to say they are both my email buddies. We have exchanged
    2,859 emails.
    How many people can say that BOTH of these men are their email buddies ?
    How lucky can you get ?
    If you are interested in cancer or cancer research, you may email me at:
    BOBBYCURTO@WEBTV.NET

    Robert Curto
    Ft. Lauderdale, Florida
    USA

  • Andrea Rossi

    D. Travhenko:
    Unbilievable.
    Warm Regards,
    A.R.

  • Curiosone

    Dear Dr Rossi:
    In the paper you wrote with Prof. Cook is written that your effect can be based upon the Mossbauer effect. Can you explain, with your usual ability to make easy difficult things ? I tried to understand what is the Mossbauer effect for, but understood nothing.
    Thank you for your patience,
    W.G.

  • DTravchenko

    Dr Andrea Rossi:
    Do you know that he paper Cook-Rossi has been translated and published in Russian, Chinese, Indi, French, Spanish, French… has been read all over the World, you know?
    Warm Regards,
    D.T.

  • DTravchenko

    Andrea Rossi:
    I have seen that in the Cook-Rossi article is dedicated strong citation to our Russian scientist Alexander Parkhomov: thank you!
    DT

  • Andrea Rossi

    Andrea Calaon:
    I cannot give further information after what we wrote, so far. In next papers new issues will be taken into consideration.
    Information regarding the 1 MW E-Cat will be given after the test and the R&D on course will have been completed.
    Warm Regards,
    A.R.

  • Andrea Calaon

    Dear Andrea Rossi,
    I have read the recent article you and Prof. Cook wrote.
    It considers in detail only the energy contribution of Li7. I remain convinced that the primary role of Li is in increasing the number of superficial vacancies in the Ni grains and in lowering the hopping energy of these vacancies (NAE) by working as interstitial at the surface.

    The reaction you propose, namely:
    Li7+p -> Be8 -> 2He4 + 17.26 [MeV],
    is similar to reaction 11 of the theory I am proposing:
    11 :Li7+ep -> Be8 + e + 16.74 [MeV] – Gp
    Be8 -> 2He4 + 0.09184 [MeV]
    With Gp probably equal to 1.754 [MeV].

    Since the elements are the same the energies should be the same. I quickly checked the contribution of the first part (from Li7 to Be8) and I do not get your 17.17 [MeV]. I use data from the iaea database:
    https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

    It would be interesting to know if you have ever measured the 1.745 [MeV] gammas from an unshielded charge, similarly to what Iwamura does. It would confirm the presence of at least one emission mechanism that is independent from the metal matrix used (he uses Pd).

    I have also a question about the COP. Are you now tapping hot vapour from the collector and using it to heat the reactors back?

    Congratulations for your high COP!
    Hot regards,
    Andrea Calaon

  • Andrea Rossi

    Hank Mills:
    Thank you for your insight,
    Warm Regards
    A.R.

  • Andrea Rossi

    Steven N. Karels:
    Please read better the Lugano Report and the Cook-Rossi paper.
    I cannot add information.
    Warm Regards
    A.R.

  • Andrea Rossi

    Magnus Holm:
    Thank you, very interesting.
    Warm Regards,
    A.R.

  • Dear Andrea,

    After reading your article with Cook, On the Nuclear Mechanisms Underlying the Heat Production by the E-Cat, I just had a reflection on the Lithium 7 problem in Cosmology.

    I do not know if you are familiar with the Lithium 7 problem in cosmology but the problem is that from all known processes of nucleosynthesis there should be 2-3 times higher concentrations of Lithium 7 in the universe than what is observed.

    Mystery over apparent dearth of lithium 7 in universe deepens
    Measurement at Big Bang conditions confirms lithium problem

    Cook-Rossi theory might be able to shed new light on this problem if one can explain how the Cook-Rossi(LENR) Li+p->Be->2He process might appear in stars as well at higher rates than expected by classical nucleosytnthesis models.

    Very interesting theory.

    Kind regards,
    Magnus

  • Steven N. Karels

    Dear Andrea Rossi,

    As I study your latest publication (“On the Nuclear Mechanisms…), it appears the hypothesis is that the energy comes from conversion of 7Li4 into helium and the conversion of 61Ni33 into 62Ni34. A quick energy calculation shows that there is insufficient 7Li4 and 61Ni33 in the one gram sample to release the energy observed and reported in the Lugano test. This is based on the amount of 7Li4 produced energy suggested in the Lugano report (0.72MWh), the total excess energy produced (1.5MWh) and the small natural abundance of 61Ni34 (1.14%). Total nickel content in the one gram fuel sample is assumed to be between 0.55 and 0.989 grams. I suggest there must be another energy producing reaction to account for the total excess energy observed.

  • Hank Mills

    Dear Andrea,

    Your theory paper specifies that the Ni + LiAlH4 combination is the essence of the hot cat. From what we are learning from replicators – in addition to the Lugano results and statements from yourself – the properties of this nuclear technology seem to be overwhelmingly favorable. The power density is extremely high (at a minimum over 1 kilowatt of nuclear heat production per gram of fuel), the radiological profile is favorable (no radioactive fuel, no emission of radioactivity into the environment, no nuclear waste), fuel cost is practically zero (pennies per gram), temperature range is high (1400C even allows for thermal-photo-voltaic systems to be considered), and the energy content is enormous (who knows how many months the Lugano device could have continued operating).

    Of course there is a gigantic difference between replicating a basic test setup using this combination of fuel and the massive effort and expense of bringing a product to the market. But the fact the essence of this technology is now loose for professional scientists and engineers in laboratory settings to test and measure for themselves is thrilling. The “Rossi Effect” that can be produced by Ni+LiAlH4 is not simply another in a long line of setups with puny, miniscule output, but a massive leap forward. Literally speaking, it’s like a technology from a few hundred years in the future has been dropped into the present.

    Very often I read about the efforts of nations to build and test hugely expensive (sometimes planning to spend hundreds of millions or billions of dollars) nuclear reactors – both fission and hot fusion based. These highly experimental systems almost all have several things in common: the use of radioactive fuel, a large size, and an unknown completion date. In many instances, they are not even expected to be constructed for decades. Then there could be years of testing. Actual useful power production could take even longer. So in the best case considering many of these systems, infants born this year may see a benefit when they are in their twenties, thirties, or maybe never.

    The E-Cat is producing useful low temperature steam for a manufacturer producing a product today. The usefulness of this technology has already begun – we don’t have to look decades into the future. Soon, I expect the hot cat will be used to build a plant suitable to produce electricity using higher temperature steam. From there, once thousands of plants start being sold, the sky is the limit. Home heaters, portable electrical generators, batteries, vehicles, aircraft, spaceships: they are all coming. There is an enormous amount of work that must be performed to make this happen; fortunantly, I think the significance of this technology will push it forward.

    Photovoltaic panels are being further developed by dozens of companies, and have improved dramatically over the years in reliability, efficency, and cost. But even an impossibly 100% efficent solar panel at a fraction of current prices will pale in comparison to what the E-Cat will evolve into. The inherent properties of the E-Cat , the communication tool called the internet, and the level of scientific knowledge we have today (far greater than in the early days of solar panels in the 1970s for example) will combine to boost the evolution of the E-Cat into overdrive.

    All of this is apparent to some of us who are closely following this saga. I’m ready to watch – sipping on a chocolate Atkins lowcarb weight loss shake instead of munching on high carb popcorn – the waves of awareness about the reality of the E-Cat start crashing into the shores of the “mainstream” media and scientific community in the near future. The show should be exciting as so many minds realize a near limitless source of dense, cheap, clean, and safe energy exists.

    As I sit here in my chair, unable to sleep, I can’t help but wonder how everything is going to start changing, moving forward, and accelerating in the months to come. I know it will be worth following – the E-Cat is a thousand times more significant than the next iphone or electronic gizmo.

  • Andrea Rossi

    Peter Metz:
    Now I have more time, can answer better:
    Yes, but not in the same measure, because in small units we have not the synergies we get with the 1MW E-Cat.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Valery Tarasov:
    Thank you for your insight.
    Warm Regards
    A.R.

  • Dear Andrea Rossi,
    I would like to share some information, which can potentially explain disappearance of 58Ni (and the observed Ni isotopes ratio change in Lugano report) by appearance of Fe in the ash, hided by the presence of Fe already in the initial fuel (from Lugano report: “From all combined analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and H in addition to Ni.”).

    First, the interaction of 58Ni with alpha particles (generated from the reaction of protons with 7Li) can result in formation of 56Ni, 57Ni, 55Co, 56Co, 57Co, and 58Co (http://journals.aps.org/pr/abstract/10.1103/PhysRev.134.B783). Then, second reaction, the electron capture for 56Ni, 56Co, 57Co, and 57Co will result in the formation of stable isotopes Fe56, Fe57, Fe58. Thus, 58Ni will disappear and Ni isotopes ratio will be changed.

    Best wishes,
    Valeriy Tarasov

  • Andrea Rossi

    Peter Metz:
    Thank you for kind words.
    The answer to your question is:
    Yes.
    Warm Regards,
    A.R.

  • Peter Metz

    Dear Andrea Rossi,

    Are the improvements you are currently seeing in the COP of the 1MW plant (not known with certainty until the end of the trial period) possibly transferable in whole or in part to the Home E-Cat?

    It must be very difficult to be quietly working in your shipping container office whilst witnessing these potentially world shaking results. I would like to add I really appreciate the news you are giving us. Thanks!

    Sincerely,
    Peter Metz

  • claudio ciani

    Caro dr. Rossi,

    ho appena visto ed ascoltato con attenzione la conferenza della
    dott.ssa Judy Wood [1]

    L’evidenza scientifica (non ipotetica) che le due torri siano crollate
    per “fusione fredda”, vale a dire con un processo implicante la
    “fusione fredda”, che la dott.ssa Wood definisce “polverizzazione”,
    credo debba prendersi in considerazione.

    Lei che ha studiato la “fusione fredda” nell’applicazione al concreto,
    cosa ne pensa? Cosa pensa di ciò che la dott.ssa Wood afferma?
    Il processo di “fusione fredda” è in grado di “polverizzare” la
    materia e dunque anche le torri gemelle?

    La ringrazio per la disponibilità.

    Buon lavoro per le sue ricerche.

    Claudio Ciani

    Links:
    ——
    [1] http://www.drjudywood.com/

  • Andrea Rossi

    Claudio Ciani:
    Your comment in Italian substantially asks what I think about the assumption made by Dr Judy Woods that the Twin Towers could have fallen for LENR she calls “pulverization process” started by the energy released by the airoplanes.
    I am not able to answer, because I have not found the necessary data. Without precise calculations, based upon precise equations, you can say everything and the contrary of everything.
    This reminds me the Fermat theorem.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Hank Mills:
    Sorry, but your theoretical question is based upon assumptions fundamentally wrong and I cannot answer, maybe some Reader will answer you.
    You also can try this:
    FF Chen, MD Smith 1984, Wiley Online Library.
    About the use of the so called Rossi Effect as a source for the production of charged particles, the equations you imply are barred.
    Thank you anyway for your very, very personal insights, perhaps somebody will be inspired by them: it sometimes happens that wrong assumptions have generated right consequences…
    Warm Regards,
    A.R.

  • Andrea Rossi

    Oeystein Lande:
    This issue will be the object of the continuation of the theoretical work I am making with Prof. Cook. I prefer not to give any information before complete publication on this part of the R&D: probably you have noticed that I prefer not to talk about an issue until the related work is completed. Inside the computers container of the 1 MW E-Cat we are continuing the experiments and the R&D regarding the Hot Cat and its future applications also to domestic appliances and our theoretical work grows up in parallel with a synergetic interdependence.
    Warm Regards,
    A.R.

  • Øystein Lande

    Dear Rossi,

    Can you reveal anything more on the important issue of He measurement?
    Equipment used? He isotopes?
    How much He was measured (ppm?) related to what level of energy production?

    With Regards

  • Hank Mills

    Dear Andrea,

    I have a physics question for you that I am perplexed about. The question is laid out in more detail at the following page, but I will give a short
    version here.

    http://pesn.com/2014/10/24/9602555_Late-Night-Speculative-Raving_From_Hank-Mills/

    I’ve done a significant amount of reading online, and I cannot find a reference to answer the following question: what is the source of energy that creates the magnetic field of a changed particle – such as an alpha particle?

    Many references state that the amount of kinetic energy of a changed particle is exactly equivalent to the binding energy released in the nuclear reaction. So if you were to run the reaction backwards, you could transform 100% of the kinetic energy into binding energy. But no reference discusses the source of the magnetic field which seems to just appear. I may be totally wrong, but it seems the magnetic field seems to be something extra that cannot be thermodynamically accounted for.

    At the link above there are links to a resonant nuclear battery that was claimed to produce huge amounts of electrical power by harnessing not only the kinetic energy of charged particles emited by decaying radioactive isotopes, but also the magnetic energy. By absorbing the magnetic energy as well, the electrical output of the battery was increased dramatically.

    I’m curious if the magnetic field of a charged particle is indeed seperate from the kinetic energy, and, if so, can it be extracted. If this is possible then your technology could be an ideal alpha particle source emitting a thousand times as many charged particles as Strontium. Maybe not only their kinetic energy but magnetic energy could be harnessed and converted to electricity.

    Thank you.

  • Andrea Rossi

    WaltC:
    Also your comment, as many others, have been unspammed by me from the spam where our robot has sent them, for reasons I do not know. Randomly I check some spam page and fish out good comments I find unproperly spammed.
    As I already said, the Readers who send a comment and find it spammed, can email their comment to
    info@journal-of-nuclear-physics.com
    It will be recovered. It is rare that I spam comments, unless they contain insults of any kind against any person or blatant falsities.
    Warm Regards,
    A.R.

  • WaltC

    Steven N. Karels:
    I was wondering the same thing. According to their paper, you’d need some sort of lattice.

    Speculating: the lattice might need to act somewhat like a “hydrogen sponge”. They also talk about NAE (nuclear active environment– a term used by Edmund Storms (his LENR book is interesting)) which could imply certain lattice characteristics, like free electron availability.

  • Andrea Rossi

    Steven N. Karels:
    I cannot add anything to what has been published.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    After a cursory reading of your paper (very interesting I might add), it seems like it is possible to have the “Rossi Effect” with only lithium (no nickel). Can you comment on your paper in this regard?

  • Andrea Rossi

    Steven N. Karels:
    Yes.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    You posted “Now we are working on the calculation of the He produced, that we have already found.” So you have experimentally verified the production of helium in the eCat operation?

  • Andrea Rossi

    Hugh De Vries:
    Maybe in future
    Warm Regards
    A.R.

  • Hugh DeVries

    Dear Andrea

    The LENR field continues to get more and more interesting. Now if we can select the right fuel maybe we can produce energy and element ash that is of high value as well.

    Best regards,
    Hugh

  • Andrea Rossi

    Daniel De Caluwé:
    We asked due corrections of all the typos we found: should be made asap.
    Warm Regards,
    A.R.

  • Daniel De Caluwé

    Dear Dr. Rossi,

    Just some small remarks/typo’s:

    i) About my previous remark: Concerning the description of isotopes, I was used to give the number of protons (= Z = Atomic number) and the number of nucleons (= protons + neutrons), so I write 3Li7 in stead of ‘7Li4’, but the notation used in the paper is right when we take into account that the first number is the total number of nucleons, and the second is the number of neutrons. Then 3Li7 is 7Li4, because 3Li7 has, indeed, 4 neutrons (and 3 protons).

    But if we follow the notation in the paper, I have a remark about the last line of page 8: where is written (between brackets): (58Ni30 + Alfa -> 62Zn32 -> 61Cu33 -> 62Ni34)

    In this line, 61Cu33 should be 62Cu33, because, in electron-capture, the total number of nucleons remains the same (a proton becomes a neutron), while the number of neutrons increase with one unit.

    ii) Your remark that Ni should be N on page 3: Yes, and there also is a reference to it on page 6

    iii) Some references (to the list in the last page) are wrongly indicated in the paper: For instance on page 5, where there is a reference to Parkomov: in the list at the last page it has reference number [2], but in the beginning of page 5 it is numbered [4], which is wrong.

    (I think this was the case with other references also, but I did not note them all).

    But, of course, this only are small details (typo’s).

    Kind Regards,

  • Andrea Rossi

    Pekka Janhunen:
    About the alternative reaction you suggest, I think they are unlikely and the reason is explained in the article: unlikely does not mean impossible, though. About the report of Dr Bianchini, there is one particular that noobody observed: if you do not consider the Sigma, neutron have been produced. The Sigma is a statistic datum, and there is no evidence that it has necessarily to bear a positive sign, statistically speaking. As a consequence of this consideration, should the Sigma be negative ( 50% of probabilities) we had a neutron emission in small measure. Speaking with Dr Bianchini, he told me it is a possibility and he said that more detailed measurements had to be done on this issue, but the aim of his measurements was related to the safety, so his work has been limited to verify that there was no danger regarding the radiations emitted from the Hot Cat in the environment of the laboratory the Professors were working in.
    When you toss a coin, the chance that it will be always head are pretty unlikely.
    Probabilistic Regards,
    A.R.

  • Andrea Rossi

    Pekka Janhunen:
    We found that calculating properly the Mossbauer effect the understandable dilemma you propose ( that obviously was clear to us from the beginning) could be resolved. I underline “could be”.
    We are still studying on this. I think the keys are 7Li3+p -> 8Be4 -> 2 Alpha -> Mossbauer Effect-> Heat + (2 Alpha + 2e)-> 2He
    Maybe we are wrong, but this is the reconciliation we propose. And this is the core of the Rossi Effect, we suppose. Now we are working on the calculation of the He produced, that we have already found.
    This will be the object of a next Cook-Rossi paper. The nuclear model of Prof. Norman Cook suits perfectly my effect.
    Warm Regards,
    A.R.

  • Dear Andrea,
    There exists an endothermic reaction Li7(alpha,n)B10 which has cross section 0.03 barn at 5 MeV and 0.25 barn at 7 MeV. I also read in wikipedia (article Neutron_source) that there exist americium-lithium neutron generators based on this reaction. If there are energetic alphas (energy about 17/2=8.5 MeV) in the reactor as per your paper’s suggestion, then there should also be some free neutrons generated from lithium-7 by this reaction. But Bianchini saw no neutrons. This looks like a dilemma.
    With mysterious regards, /pekka

  • Andrea Rossi

    Stan Lippman:
    Thank you for your attention.
    Your considerations are understandable.
    Warm Regards,
    A.R.

  • Stan Lippmann

    Dear Dr. Rossi,
    I enjoyed reading your paper. Although it doesn’t help with the mystery of start-up, I wonder if the alphas can transfer enough kinetic energy to the protons to sustain to reaction.

    it would seem easy enough to measure the He outgas with a mass spectrometer, this would get the attention of a lot of people if the saw proof that the right amount of helium were produced.

  • Andrea Rossi

    Giuliano Bettini:
    If you read carefully the paper, you will find what explains the reason of the limits regarding the disclosing of these issues.
    Warm Regards,
    A.R.

  • Giuliano Bettini

    Dear Andrea,
    in the ArXiv paper I read, among others, Li7+p=Be8, Ni61+p=Cu62.
    This suggests an old question.
    What lowers the Coulomb barrier, between the atomic nuclei of Hydrogen and Nickel? Or Lithium?
    I just read “If the high temperature, high-pressure conditions within the E-Cat provide sufficient energy to allow Hydrogen nuclei to overcome the Coulomb barrier and to approach Lithium nuclei, then….“
    Are there any idea on this issue?
    Nuclear regards,
    Giuliano Bettini.

  • Andrea Rossi

    Daniel De Caluwè:
    There are typos that we are correcting. Not just this, but also others, like Ni instead of N on pag. 3-4 and some other evident typos derived from a wrong trascription, like, for example, wrong numbers of the references, so that some reference is related to the wrong number…We are sorry for this inconvenience and we are making the corrections.
    Thank you for your attention and for your kind words.

  • Andrea Rossi

    Hank Mills:
    Thank you for your continue and kind interest to our work.
    Please read with attention the Cook-Rossi paper and the Lugano report: I am sure you will find the answers you need.
    I cannot add anything, so far, to what has been written already.
    Warm Regards,
    A.R.

  • Daniel De Caluwé

    Dear Dr. Rossi,

    I didn’t read your paper ( http://arxiv.org/abs/1504.01261 ) in detail yet, but in it and in its summary, is mentioned 7Li4, but shouldn’t that be 7Li3, as Z = 3 and not 4? So, 7Li3 + p -> 8Be4 -> 4He2 + 4He2?

    Anyhow, if I’m right, this is just a detail, but for the rest: april 7, 2015 as the date when you announced the presence of your paper, will be a historical date in science! The paper looks impressive and close to the truth, but as I was not following this forum for a while, I didn’t read it in detail yet.

    Kind Regards,

  • Hank Mills

    Dear Andrea,

    Please let me stress something from the start: your hot cat reactors – capable of exceeding 1400C – are completely safe devices for experts such as the employees at Industrial Heat to work with. They have proven to produce no gamma radiation, emit no neutrons, produce no nuclear waste, and simple cease functioning when pushed to rupture in torture tests. Most importantly, you have built and tested hundreds or thousands of them and are in good health.

    However, the recent paper published and statements about steel vaporizing in torture tests indicates that the nuclear reactions may continue when nickel is in the molten form. This may not have been the case in the low temp E-Cat, but it seems to be the case in the hot cat.

    There are currently arguments and chattering between individuals who insist such reactions when the nickel is in the molten form are impossible – due to your previous statements saying that all reactions cease when the nickel melts – and those who are more open to the idea that the reactions may continue.

    Could you please clarify this issue? I don’t think you would need to risk any IP to end this debate.

  • Andrea Rossi

    Bernie Koppenhofer:
    Negative advertising is very positive for products that give evidence of the falsity of the negative advertising itself: I call it ” compression”: when you make a negative compression of a product that is positive in itself, you produce the explosion of its diffusion. Snakes are useful. Snakes are precious. If you resist to their attack.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Valery Tarasov:
    Thank you for your attention.
    I will answer to all the issues regarding the paper in a next publication.
    Obviously I will not answer to questions that could lead to breach of IP
    Warm Regards
    A.R.

  • Dear Andrea Rossi,
    Congratulations with your paper!
    I have a related question. Since the interaction of 7Li with protons is the main LEN reaction producing two alpha particles in your Cats, I suppose even more effective can be to use LiBH instead of LiAlH. This is because, interaction of boron with proton (500 keV) results in generation of 3 alpha particles (in addition 2 alpha particles from 7Li + p): 1p + 11B → 3 4He + 8.7 MeV. Thus, it will be interesting to try Ni + LiBH in the same setup as it was for Ni + LiAlH. Have you tried this combination?
    Best wishes,
    Valeriy Tarasov

  • Bernie Koppenhofer

    Dr. Rossi: To follow up on my question about being explicitly able to show your customers savings. There are many historical examples of productive products being delayed or eliminated because of negative advertisement. It has been proven negative advertisement simply works. This negative advertisement can be overcome by you being able to show in a simple straight forward way your customers savings. Is your set up in your customer’s factory able to show a skeptical future customer who has been subjected to negative advertisement that your reactor can save him money?

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