The Quantum Space Theory (QST) could explain the LENR

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by
F. Santandrea
R&D systems analyst – Labor s.r.l. Rome Italy
E-mail: f.santandrea@labor-roma.it
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U. Abundo
Physics teacher – Leopoldo Pirelli I.T.I.S. high school Rome Italy
E-mail: interprogetto@email.it
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The QST theory elaborated in 1994 by F. Santandrea, now under revision, contains some topics concerning the LENR recently submitted and appreciated from LENR researchers, QST could giving an unifying point of view on the whole Physics.
For further detailed please refer to the following link QST updated topics:
Ten years later the same basic ideas were independently approached by U. Abundo employing  the tools offered by the J.Von Neumann’s Cellular Automata from a point of view focused on information traveling, please refer to the following link:
The well known Widom-Larsen theory, basically focused on the cooperating behavior of the electrons in condensed matter (tuned with the theory of G. Preparata) may be regarded as a special case, under specific conditions, of what is predictable by the QST.
According with QST, it is naturally predictable the loss of identity of the electrons confined into condensed matter lattice, while the properties of space have priority and permit/control existence and behavior of electrons, so giving a natural coherence to the assumptions of Widom-Larsen.
Into the present new approach to space and particles structure, the latter become just expression of stable resonance frequencies of space; the same electron, particles and generally condensed matter are “electromagnetic objects” constituted of standing waves into the space quantum found by TSQ.
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366 comments to The Quantum Space Theory (QST) could explain the LENR

  • Antenna

    Will the home e-cat require FCC certification for RFI?

  • Andrea Rossi

    Dear Avi:
    We will be able to sell the domestic E-Cat only after the certification will have been completed, not before.
    Warm Regards,
    A.R.

  • Avi

    Dear Andrea Rossi
    When the certifiers demands will be done (the changes in the domestic ECAT), you can sell this?
    Warm Regards

  • Andrea Rossi

    Dear Avi:
    Yes,
    Warm Regards,
    A.R.

  • Avi

    Dear Andrea Rossi
    Did the certifiers want from you to do any changes in the domestic ECAT so far?

    Warm Regards

  • Andrea Rossi

    Dear Jaroslaw Huba:
    Leonardo Corp. has forwarded to the Licensee of your Territory your request.
    Warm Regards,
    A.R.

  • Jaroslaw Huba

    Dear Andrea Rossi.
    I wrote e-mail with my complete adress to info@leonardocorp1996.com, as you requested, but I didn’t get any answer. I send it at 25th October twice and 4th November once (all are different). I’m worried my e-mails was lost because some spam settings. Are they get to proper hands?
    Warm Regards,
    Jaroslaw Huba.

  • Andrea Rossi

    Dear Captain:
    Thank you!
    Warm Regards,
    A.R.

  • captain

    Dear Andrea,
    with your hard work 16/24, with all kinds of difficulties, but trusting in the help of our Lord, U’re giving the world a new hope: not words but facts, products, a virtually unlimited source of cheap energy for everybody, safe, clean.
    You’re currently the only true pioneer in the practical realization of an operative device, even too efficient, which will allow the phasing out of the current costly energy sources that pollute, are dangerous for humanity and negatively affect our life on the planet.
    Go on and keep on doing it with faith, our Lord will surely help U in your mission.
    In God we trust.

  • Andrea Rossi

    Dear Herb Gills:
    No, because are under NDA.
    Warm Regards,
    A.R.

  • Joe

    Daniel,

    Thank you for your response.

    But, I see 3 problems with your scenario:

    1. I doubt very much that a signal generator could create the kinds of pressure that would make the proton overcome the Coulomb barrier of the overall Ni nucleus.

    2. If the proton did overcome the Coulomb barrier, I doubt that the force of attraction between the proton and an outer neutron would be greater than the force of attraction between the central He nucleus and that very neutron. Thus, the proton could never steal the neutron away.

    3. The proton therefore runs the risk of being absorbed itself into the halo of the Ni nucleus by way of combining itself with the neutron. So, ironically, instead of liberating a neutron, a proton gets captured. But, of course, this implies a transmutation from Ni to Cu – a process which is essentially absent from the E-Cat.

    The answer to the E-Cat mystery must lie somewhere else.

    All the best,
    Joe

  • Herb Gillis

    Andrea Rossi:
    In regard to the certification process. If I understand correctly; you had said that more extensive data will need to be collected from the industrial (1MW) Ecats in operation in order to convince the organizations responsible for certification that the domestic Ecats are safe (ie. can be certified for sale). Does data collected by the un-named customer (of October-28-2011) count in this process? Does data collected internally within your company count in this process?
    Regards; HRG.

  • Bernie Koppenhofer

    Patience please…….. I know it is simplistic to compare the E-Cat to the Microwave Oven but a little history: This from Wikipedia:

    “Percy Spencer invented the first microwave oven after World War II from radar technology developed during the war. Named the “Radarange”, it was first sold in 1947. Raytheon later licensed its patents for a home-use microwave oven that was first introduced by Tappan in 1955, but these units were still too large and expensive for general home use. The countertop microwave oven was first introduced in 1967……”

  • Bernie Koppenhofer

    Pekka Janhunen “certificators” Thank you, I was looking for that word for 15 minutes instead of my
    “certifiers”. (:

  • Steven Karels

    Dear Man,

    Thank you for the more civil tone. What I said was
    “Given the statement that there are no releases of radiation outside the eCat, then everything released must logically have been absorbed within the eCat.”

    I personally have never seen an eCat in operation. We have been told there is not outside occurance of radiation. Logically, it therefore must have been absorbed within the eCat if it ever existed. Agreed?

    We specualate on many things in life. When we lack complete knowledge, we guess, we speculate. This is still information although we must take it with a “grain of salt” – to be wary.

    I do not understand how the eCat contains the gamma or xray energy without large amounts of material shielding. There have been some suggestions that it is “trapped” in the surrounding material atomic structure but I have not investigated this.

    My posting was an attempt to bring together several seemingly contradicting reports – no significant generation of Copper, no detectable generation of 59Ni, the need for isotope enhancement of 62Ni and 64Ni, the lack of any Helium or Deuterium formation. With the restricted information being released from Andrea Rossi (to protect his intellectual property), I see energy being produced without an apparent “ash” — no by-product. So I did not concern myself with positron generation and its decay and energy capture. I looked to see what were the essential reactions and how this might work. You all are my peer review and maybe, just maybe, an idea might spring forth that was useful. Okay?

  • @Joe,

    Pure speculative, and based on QRT, where the excess neutrons of the heavy isotopes (of 62Ni and 64Ni) are hypothised on the edge (and not in the center 😉 of the nucleus, and that they tend to pair with protons (also according to QRT), and combine this with an oscillating pressure wave (cfr. ‘frequencies’) of the loaded hydrogen… But just consider it as an idea coming out of brainstorming, and not as official science… 😉

    Because, if no significant amount of copper is formed, something else must happen, isn’t it? 😉

  • Man

    @ Steven N. Karels

    first example, You wrote:

    62Ni + H -> 63Cu (stable)
    64Ni + H -> 65Cu (stable)

    forgetting that the each above nuclear reaction produces about 6-7 MeV of Gamma Rays
    http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf

    So try to explain to me how You are able to transform in heat and to fully shield 6 MeV energy of Gamma Rays.

  • Man

    @ Steven N. Karels

    You aren’t adding information, only speculations.
    Okay let’s try to discuss the matter, you say that there is no radiation outside, do you have idea of the energy of radiation produced according to your assumptions and what would be needed to stop them so they don’t come out?

  • Steven N. Karels

    Dear Man,

    Thank you for your wonderful, supporting comments. I post my analyses to provoke thought and try to include enough information that others can follow, duplicate and expand on my posts.

    I encourage you to modify the analyses I make, adding the full nuclear equations with their individual products and energies and see if it makes any difference in the end result. Given the statement that there are no releases of radiation outside the eCat, then everything released must logically have been absorbed within the eCat. If this is true, then an analysis using only mass defect converted to energy is a valid approach.

    As Einstein said “Everything should be made as simple as possible, but not simpler.”

    Meanwhile, my friend, lose the attitude. It serves you not.

  • Joe

    Daniel De Caluwe,

    How do you imagine that a proton could be a catalyst for neutron emission?

    All the best,
    Joe

  • Dear Andrea,
    Still about home unit certification,
    1) Is it correct to say that the certificators are not concerned about anything in particular, but they are just worried about possible “unknown unknowns”?
    2) How many testing hours do they want to see, ten million or more?
    regards,/pekka

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    The difficulty is objective, not subjective. An apparatus working with LENR is difficult to digest for a certifier that has to get the liability to certify such an apparatus to go in apartments where no safety rules will be respected, in many cases. The obstacle will be overcome in time, with good statistics from the industrial plants.
    Warm Regards,
    A.R.

  • Bernie Koppenhofer

    Mr. Rossi: Thanks in advance for answering, could you tell us what is holding up the home E-Cat. What specifically are the certifiers not accepting?

  • Man

    @ Steven N. Karels

    It’s useless to talk of isotopes of Nickel and Copper without writing down nuclear equations that involve energies and radiation released. Are You able to do this?
    Enough with the simplistic assumptions.

  • @Herb Gillis,

    Yes, I agree, if the heavy isotopes lose neutrons, then that should show up in the isotope analysis of the spent fuel, that should be different from the fuel they started with. But as the fuel always has been enriched (to obtain more 62Ni en 64Ni than naturally occuring), and as some measurements indicated that the spent fuel showed more or less the natural concentrations of the isotopes, I presume that the heavy isotopes lose neutrons in the process. But this doesn’t mean that there’s a strong neutron-radiation/flux (like in ordinary fission reactions), because I think here it happens in a smooth way. And the protons of the loaded hydrogen then are, in a certain way, a catalysator for this…

    But I don’t think that Dr. Rossi likes that we communicate a lot or in detail about this, because this probably or gradually/eventually could reveal part of the working of the E-cat.

    Kind Regards.

  • Herb Gillis

    Daniel:
    If the heavy isotopes lose neutrons then that would show up in the isotope analysis of the spent fuel. Unless I misunderstood something I think A.R said that he did not see significant changes in the nickel isotope distribution. Is my understanding correct?
    Regards; HRG.

  • @Steven N. Karels,

    I think another mechanism is involved. I think the heavy isotopes loose neutrons.

    Kind Regards,

  • Andrea Rossi

    WARNING:
    A PERSON WHOSE NAME IS ANTONIO MARCHIONNI HAS PUBLISHED ON A BLOG THAT A COMPANY RELATED TO HIM HAS BOUGHT SOMEWHERE DOMESTIC E- CATS, AND THAT HE OR A PERSON TO HIM RELATED HAS EVENTUALLY RECEIVED MATERIALS THAT GAVE HIM TROUBLES: ATTENTION, THIS IS A FRAUD. WE NEVER SOLD DOMESTIC ECATS TO ANYBODY, ANYWHERE IN THE WORLD, BECAUSE THEY HAVE NOT YET OBTAINED THE NECESSARY CERTIFICATION. REMEMBER: IF ANYBODY OFFERS FOR SALE OUR DOMESTIC E-CATS IS FRAUDOLENT, BECAUSE NO DOMESTIC E-CATS ARE FOR SALE, SO FAR. IF ANYBODY OFFERS YOU OUR PRODUCTS, PLEASE CONTACT DIRECTLY US AT THE ADDRESS BELOW:
    info@leonardocorp1996.com
    GIVING US THE NAME OF THE PERSON WHO IS OFFERING YOU OUR PRODUCTS: WE WILL CHECK IF THEY ARE OUR LICENSEES OR IF THEY ARE JUST BOGUS SELLERS.
    AT THE MOMENT WE ARE SELLING ONLY THE 1 THERMAL MW INDUSTRIAL PLANTS.
    WARM REGARDS,
    ANDREA ROSSI, PRESIDENT OF LEONARDO CORPORATION

  • Andrea Rossi

    Dear Steven N. Karels,
    Thank you for your insight, as I said I cannot give further information on this issue.
    Warm Regards,
    A.R.

  • Steven N. Karels

    High Grade isotopes of Nickel are commercailly obtained from several companies.

    Let us assume the pre-operation mixture of the “fuel” is 80% 60Ni and 5% 62Ni and 15% 64Ni.

    The primary reaction will be:

    60Ni + H -> 61Cu (half-life of 3.33 hrs) -> 61Ni
    61Ni + H -> 62Cu (half-life 9.67 sec) -> 62Ni
    62Ni + H -> 63Cu (stable)
    64Ni + H -> 65Cu (stable)

    As the continuous eCat operation occurs, under this assumed scenario, the 60Ni will become 61Ni which will become 62Ni which stops with 63Cu.

    64Ni becomes 65Cu.

    If I assume a certain amount of 62Ni and 64Ni is needed, for some unknown to me reason, to be necessary to act as a catalyst for the reaction, the 62Ni is continuously replenished by the other Nickel reactions. But the 64Ni can become exhausted by its conversion to 65Cu. Perhaps that is why the eCat is limited to 6 months of operation?

    Probably wrong, but I see it as a possible candidate consistent with the lack of 59Ni generation.

  • Andrea Rossi

    Dear Steven N. Karels:
    Further information about this issue is confidential.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Herb Gillis:
    I already answered to the ratio between energy necessary to make the charge and energy produced: the energy consumed to make the charge is irrilevant, due to the fact that we consume electrochemical energy and produce nuclear energy.
    Further information about this issue is confidential.
    Warm Regards,
    A.R.

  • Herb Gillis

    Andrea Rossi:
    1) If the heavy nickel isotopes (62 and 64) are not being consumed (since you said you have no evidence they are being consumed) and you think these isotopes are involved in the primary energy production mechanism, why is it necessary to change the fuel every 6 mos.?
    2) I am confused by your response on overall energy balance. Even if you are making nuclear energy from electromechanical energy, you have a limited COP (of 6) and you most likely have to convert heat to electricity to get the electrical energy for the isotope separation. The efficiency of that coversion is seldom greater than 40%, often less. If you have to convert nickel into a volatile form (a compound) to facilitate isotope separation, then do the separation, then convert the volatile nickel compound back to the metal, then make the fuel particles of optimum size; it seems to me that processing could still make a significant dent in the overall amount of useful energy you can get out of the process. Have you looked at the overall energy balance (including isotope separation, the limited COP, and the thermal to electric conversion efficiency)?

  • Steven N. Karels

    Dear Andrea Rossi,

    I am searching for an understanding of Nickel enrichment, no production of 59Ni and Copper production as a secondary energy process. Isotope enrichment is really a depletion of an unwanted isotope leaving a larger percentage of the desired isotope(s).

    a. Can you reveal what isotope distribution is present in your “fuel” before the start of the eCat energy production process?

    b. Are substantial quantities of 60Ni and 61Ni still present?

    c. Do you think the significant reactions for energy production are based on 60Ni and 61Ni? (62Ni -> 63Cu and 64Ni -> 65Cu)

    d. Does the elevated isotopic presence of 62Ni and 64Ni act as a catalyst to enable the 60Ni and/o 61Ni reactions?

  • Andrea Rossi

    Dear Daniel De Caluwe’:
    1- No
    2- No
    3- No
    4- Confidential
    Warm Regards,
    A.R.

  • Dear Dr. Rossi,

    Although I don’t want to take part in a Q&A game, where you gradually (step by step) and eventually are forced to reveal the secret working of the E-cat (even when that maybe is against your initial will), the questions of some participants and your answers also make me curious, and therefore these questions: (but of course, if some of these questions would reveal too much of your secrets, I accept that you don’t answer them or even remove some of them)

    1) Did you initially enrich Ni (to get more 62Ni and 64Ni) because you initially thought/presumed that the energy maybe was coming from the transmutation tot Cu (62Ni + proton -> stable 63Cu; and 64Ni + proton -> stable 65Cu)?

    2) But, from the moment you knew that the transmutation to copper (Cu) only was a side-effect, did you try normal (not enriched) Ni back again?

    3) And if yes (to question 2), did normal (not enriched) Ni also work and produces the same amount of energy?

    4) If the enrichment of Ni (to get more 62Ni and 64Ni than naturally occuring) really is necessary to get the E-cat work (or to get the necessary amount of energy), could the fact that 62Ni and 64Ni, who are the most heavy of the stable isotopes of Ni, have most free (excess-) neutrons (6 neutrons in excess for 62Ni and 8 neutrons in excess for 64Ni), and that, thinking on QRT of Wladimir Guglinsky (where the structure of the nucleus is supposed/hypothised by a He-core, surrounded by deuterions (proton-neutron pairs) in hexagonal layers (with max 6 deuterons per layer)), so that more excess neutrons tend to enhance ‘proton-capture’ (to form deuterons in the outside hexagonal layers)?

    Could that possibly be a mechanism? : ‘Proton-capture’ (62Ni/64Ni + proton -> (63Cu/65Cu)*, but not yet completely formed/finished, and therefore falling back (and/or via ‘pseudo electron-capture’ with beta and gamma radiaton that produces the heat) and going back and forth (because of the applied ‘frequencies’)?

    But of course, I accept if you don’t want to answer the last question.

    Kind Regards,

  • Andrea Rossi

    Dear Pwkka Janhunen:
    Considering a period of 6 mo, you are right, we consume mg.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Herb Gills:
    1- irrilevant, since we produce nuclear energy while consume electro-chemical energy
    2- no
    3- no
    Warm Regards,
    A.R.

  • Andrea Rossi

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

  • Andrea Rossi

    Dear Gian Luca:
    For now we have got the certification only for 1 MW industrial plants.
    Warm Regards,
    A.R.

  • GianLuca

    Ciao A.
    volevo chiedere se la taglia da 1 MW sia la sola prevista (sia per E che H CAT) oppure avete previsto anche la possibilità di investire su taglie minori. Volgio dire.
    All’arrivo sul mercato di HOT CAT, sarà possibile investire anche per chi magari non ha a disposizione 1ML di €?

  • Steven N. Karels

    Dear Andrea Rossi,

    I won’t even ask how you economically do isotope enhancement of 62Ni and 64Ni. I am sure it is proprietary.

    Should I therefore gather that you believe the primary energy production mechanism involves 62Ni and 64Ni?

  • Herb Gillis

    Andrea Rossi:
    You said that you enrich the fuel in the heavy (rare) isotoptes (62 and 64) of nickel. This sounds like an expensive and energy intensive pre-process. Can you tell us how this effects the overall energy balance of the reaction? In other words; what percentage of energy liberated must be used in the isotope separation? Can you tell us anything about what isotope separation process you use? If you are making large plants you must be doing a lot of isotope separation. Finally; do you have evidence that the heavy isotopes are being consumed selectively? If so; to what end product (since transmutation to copper does not account for the energy produced)?
    Regards; HRG.

  • Dear Andrea,
    I’m confused. You say correctly that 1 milligram is 23000 kWh. But there are 4400 hours in six months, so in 6 months a 10 kW device generates 44000 kWh, equivalent to 1.8 milligrams. One picogram is only worth of 100 joules of energy which is very little.
    In any case, I agree that even a milligram is so small amount that it’s probably very difficult to measure in comparison to the tens of grams of active mass.
    regards, /pekka

  • Andrea Rossi

    Dear Steven N. Karels:
    a- yes
    b- no
    c- not outside the natural range: remember that we enrich 62 and 64 Ni, whose changement in % is made before cherging
    d- yes, and the margin of error is superior to the range
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Pekka Janhunen:
    I mean picograms: i mg= 23 000 kWh!
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Julian Becker:
    I did say that as a side effect traces of Cu are formed, but that this is not the mail source of energy. I am not able to answer to your question, but our theory is quite different.
    Warm Regards,
    A.R.

  • julian_becker

    I don’t know why, but I cannot lose the feeling something is happening on a much smaller level inside the atom. Mr. Rossi, you said that only picograms of Nickel disappear and no copper is formed as a process out of it? I know maybe this sounds really foolish and I am also not an expert, but have you ever considered that singularities are created that evaporate immediately again? So the energy observed from the reaction is in fact Hawking Radiation? I know this sounds a bit Sci-Fi, but could it be a possibility?

    Kind regards,

    Julian

  • Dear Andrea,
    Just a note on terminology. When you say “picograms”, it seems that you actually mean milligrams. A picogram is 1e-12 grams. A 10 kW device consumes hundred picograms per second and 2 milligrams per 6 months.
    best regards, /pekka

  • Steven N. Karels

    Dear Andrea Rossi,

    I did not mean to imply a 1 gram conversion to energy as that would be more than the total energy output observed. But I do appreciate the information that the loss may have due to loss of humidity. Making mass change measurements when the eCat has 4,000 times the change in mass can be very challenging.

    a. Perhaps the conundrum could be explained in your instrumentation?

    b. Can you see a 1% change in isotope distributions with your SEM?

    c. For example, can you detect the natural Nickel isotope distributions of 61Ni and 64Ni?

    d. Do you have any information on the accuracy and precision of the instrumentation used?

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