[Vo]:Converting hydrogen to work better in the Ni/H reactor
Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% . Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive. This is a way to increase parahydrogen hydrogen by using a noble metal catalyst. see Catalytic process for ortho-para hydrogen conversion http://www.google.com/patents/US3383176 Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?
[Vo]:new taxonomy of our field
Dear Friends, I have just published a paper about an important subject. http://egooutpeters.blogspot.ro/2014/09/new-lenr-taxonomy.html I think we have to focus more on the NiD way that is entirely new; we cannot be for ever angry with our colleague Mizuno just because he perfidiously converts Deuterium in other things, not in Heluim as all good LENR people do. Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
RE:[Vo]:Converting hydrogen to work better in the Ni/H reactor
Good insight! I know Jones has often mentioned these percentages of ortho to para but I don’t recall if he suggested methods to alter these numbers in favor LENR.. you know my attraction toward ZPE and I could even see ruthenium’s ability to disrupt these percentages as a type of demon sorting based on the geometry and quantum properties of the element. That energy being the bootstrap energy allowing hydrogen to do that which it obviously it can not do outside of the lattice wrt LENR. Fran From: Axil Axil [mailto:janap...@gmail.com] Sent: Monday, September 22, 2014 2:03 AM To: vortex-l Subject: EXTERNAL: [Vo]:Converting hydrogen to work better in the Ni/H reactor Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% . Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive. This is a way to increase parahydrogen hydrogen by using a noble metal catalyst. see Catalytic process for ortho-para hydrogen conversion http://www.google.com/patents/US3383176 Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?
RE: [Vo]:Converting hydrogen to work better in the Ni/H reactor
Hydrogen molecules are shown to be slightly diamagnetic no matter which alignment they are in. Since protons are fermions, the anti-symmetry of the wavefunction imposes restrictions on the rotational states - with the result that the molecule is always diamagnetic. Consequently, ortho/para alignment would be irrelevant or counter-productive to a fusion process with a heavy metal like nickel, since it implies three body. Three-body processes are extremely rare, even in plasmas, and H2 as a molecule would not be involved as a reactant in LENR, unless one is talking about Storms’ reaction of two protons fusing to deuterium, for which there is no physical evidence. If this reaction was happening, tritium would be expected, since its formation has a much higher cross-section than two protons. As a monatomic species, there would be no relic of the prior spin alignment of hydrogen, in fusion with nickel. However, some or most of the energy of LENR could be non-fusion related. In that case, ortho/para cycling could be relevant From: Axil Axil Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% . Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive. This is a way to increase parahydrogen hydrogen by using a noble metal catalyst. see Catalytic process for ortho-para hydrogen conversion http://www.google.com/patents/US3383176 Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?
RE: [Vo]:Mizuno, Rossi copper transmutation
I've looked through the isotope charts again - searching for reactions that rapidly decay back to the starting element or to any stable isotope which has already been reported to be there, and have not found any other possibility... ...other than Ni58 (d,Cu59) - Ni60 which happens by EC or positron emission, with a half-life of 20 minutes or so, and which fits the facts as reported in the most robust experiments (Rossi, DGT, Thermacore, Mills). 1) No or few gamma 2) No or little radioactive ash 3) No tritium, helium or positron annihilation 4) No or little bremsstrahlung 5) Excess energy which is at least 1000 times more than chemical Since nickel absorbs a deuteron and decays back to nickel in minutes, with low energy release, this reaction fits the bill. You may be thinking - what about the positron (beta positive) decay? No problem there, since nuclei which decay by positron emission also decay by electron capture in a known branching ratio which is dependant on the net energy of reaction. According to wiki-the-wonderful, in low-energy decays, electron capture is energetically favored by reactions below 1.022 MeV. The final state will have an electron added or a positron removed - and so the energy released is determinative of what can happen in the branching. As the energy of the decay goes up, so does the branching ratio towards positron emission. However, if the energy difference is low, then positron emission cannot occur, and electron capture is the sole decay mode. This would seem to be ready-made for the DDDL or deuteron-deep-Dirac-level species, which uses its tight electron for more than one purpose and probably reduces the net energy of the reaction as well. This still leaves spin conservation as the major problem. The end products of this reaction would be Ni60, and the starting nickel would be Ni58, so that is no problem. Both are spin 0. But the intermediary isotope, with short half-life would be Cu60 which is spin 2+ and the deuterium can only add is 1+ spin, and the EC electron another ½ spin. This over-simplification of spin issues - probably means that the reaction can only happen if a neutrino is captured, or else the inherent spin deficit decreases the half-life even more than its short nature. Probably the neutrino. Best of all - as a general working hypothesis which would make this relevant to LENR but is not expected to be seen anywhere else (which explains why it is not documented in the physics literature, as of now) there is NO other isotope in the periodic table (other than Ni58) - which is both a proton conductor and demonstrably neutron-deficient ! (the proof of that being that Ni-58 is lower amu than the preceding lower Z element (cobalt-59). That's right it is a perfect storm scenario. If this evolving explanation is correct, it will be seen nowhere else in the periodic table, since it demands conditions which do not exist anywhere else. This means, anthropomorphically speaking - that Ni58 desperately wants two more neutrons, and to get them, it essentially steals from its surroundings, whenever a deuteron comes too close... especially a DDDL. Falsifiability? Yes, this is falsifiable in three different way, which is a big advantage. Give me a working Rossi reactor :-) and a few months: if the [Ni-Ni] explanation is true, if will be proved beyond all reasonable doubt. P.S. do I get to keep the reactor? _ One more thing to add ... wrt the overdue suggestion (Doh, slaps forehead) that Rossi's secret sauce is looking like it is deuterium. Thank you, Clean Planet. The reaction would probably work best if it is started with regular hydrogen, and then deuterium is added later. This is because the exchange reaction between hydrogen and deuterium itself is so robust. In fact, many of the early critics of LENR thought that the entire phenomenon could be related to deuterium exchange. It is that energetic. As we know, Rossi has this mysterious system - which he calls cat-and-mouse. He has been intentionally vague on how it functions. Yet in reappraisal, this system is fully consistent with having two chambers, the main one containing hydrogen and the nickel reactant - and the smaller one deuterium (or a mix of H and D). The metering response can be simply by voltage to a window, since deuterium will diffuse through many proton conductors in direct proportion to negative charge. Positive charge stops the diffusion, which is easily controllable by a sensor. The purpose of the small chamber (mouse) is to meter D into the main chamber at a controlled rate, to avoid a runaway. If Rossi can be believed, he suffered several runaways with the HotCat which we can imagine did not have this kind of metering device. This seems to fit into everything we know, so long as one ignores Rossi's own
Re: [Vo]:Converting hydrogen to work better in the Ni/H reactor
The first step in the hydrogen doublet fusion process is the formation of one or more atoms of 2He. Helium-2 or 2He, also known as a diproton, is an extremely unstable isotope of helium that consists of two protons without any neutrons. According to theoretical calculations it would have been much more stable (although still beta decaying to deuterium) had the strong force been 2% greater. Its instability is due to spin-spin interactions in the nuclear force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the diproton a negative binding energy. By the way, the ash produced by the LENR reaction will have a non zero nuclear spin such as lithium, boron, and beryllium. This is due to the fact that the ash is at the end of the LENR reaction chain. By the way, all the isotopes of copper have a non zero nuclear spin. On Mon, Sep 22, 2014 at 10:11 AM, Jones Beene jone...@pacbell.net wrote: Hydrogen molecules are shown to be slightly diamagnetic no matter which alignment they are in. Since protons are fermions, the anti-symmetry of the wavefunction imposes restrictions on the rotational states - with the result that the molecule is always diamagnetic. Consequently, ortho/para alignment would be irrelevant or counter-productive to a fusion process with a heavy metal like nickel, since it implies three body. Three-body processes are extremely rare, even in plasmas, and H2 as a molecule would not be involved as a reactant in LENR, unless one is talking about Storms’ reaction of two protons fusing to deuterium, for which there is no physical evidence. If this reaction was happening, tritium would be expected, since its formation has a much higher cross-section than two protons. As a monatomic species, there would be no relic of the prior spin alignment of hydrogen, in fusion with nickel. However, some or most of the energy of LENR could be non-fusion related. In that case, ortho/para cycling could be relevant *From:* Axil Axil Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% . Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive. This is a way to increase parahydrogen hydrogen by using a noble metal catalyst. see Catalytic process for ortho-para hydrogen conversion http://www.google.com/patents/US3383176 Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?
RE: [Vo]:Mizuno, Rossi copper transmutation
Jones, Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a spin 0. _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: lundi 22 septembre 2014 17:34 To: vortex-l@eskimo.com Subject: RE: [Vo]:Mizuno, Rossi copper transmutation Typo- the suggested reaction is Ni58 + D - Cu60 - Ni60 _ I've looked through the isotope charts again - searching for reactions that rapidly decay back to the starting element or to any stable isotope which has already been reported to be there, and have not found any other possibility... ...other than Ni58 (d,Cu59) - Ni60 which happens by EC or positron emission, with a half-life of 20 minutes or so, and which fits the facts as reported in the most robust experiments (Rossi, DGT, Thermacore, Mills). 1) No or few gamma 2) No or little radioactive ash 3) No tritium, helium or positron annihilation 4) No or little bremsstrahlung 5) Excess energy which is at least 1000 times more than chemical Since nickel absorbs a deuteron and decays back to nickel in minutes, with low energy release, this reaction fits the bill. You may be thinking - what about the positron (beta positive) decay? No problem there, since nuclei which decay by positron emission also decay by electron capture in a known branching ratio which is dependant on the net energy of reaction. According to wiki-the-wonderful, in low-energy decays, electron capture is energetically favored by reactions below 1.022 MeV. The final state will have an electron added or a positron removed - and so the energy released is determinative of what can happen in the branching. As the energy of the decay goes up, so does the branching ratio towards positron emission. However, if the energy difference is low, then positron emission cannot occur, and electron capture is the sole decay mode. This would seem to be ready-made for the DDDL or deuteron-deep-Dirac-level species, which uses its tight electron for more than one purpose and probably reduces the net energy of the reaction as well. This still leaves spin conservation as the major problem. The end products of this reaction would be Ni60, and the starting nickel would be Ni58, so that is no problem. Both are spin 0. But the intermediary isotope, with short half-life would be Cu60 which is spin 2+ and the deuterium can only add is 1+ spin, and the EC electron another ½ spin. This over-simplification of spin issues - probably means that the reaction can only happen if a neutrino is captured, or else the inherent spin deficit decreases the half-life even more than its short nature. Probably the neutrino. Best of all - as a general working hypothesis which would make this relevant to LENR but is not expected to be seen anywhere else (which explains why it is not documented in the physics literature, as of now) there is NO other isotope in the periodic table (other than Ni58) - which is both a proton conductor and demonstrably neutron-deficient ! (the proof of that being that Ni-58 is lower amu than the preceding lower Z element (cobalt-59). That's right it is a perfect storm scenario. If this evolving explanation is correct, it will be seen nowhere else in the periodic table, since it demands conditions which do not exist anywhere else. This means, anthropomorphically speaking - that Ni58 desperately wants two more neutrons, and to get them, it essentially steals from its surroundings, whenever a deuteron comes too close... especially a DDDL. Falsifiability? Yes, this is falsifiable in three different way, which is a big advantage. Give me a working Rossi reactor :-) and a few months: if the [Ni-Ni] explanation is true, if will be proved beyond all reasonable doubt. P.S. do I get to keep the reactor? _ One more thing to add ... wrt the overdue suggestion (Doh, slaps forehead) that Rossi's secret sauce is looking like it is deuterium. Thank you, Clean Planet. The reaction would probably work best if it is started with regular hydrogen, and then deuterium is added later. This is because the exchange reaction between hydrogen and deuterium itself is so robust. In fact, many of the early critics of LENR thought that the entire phenomenon could be related to deuterium exchange. It is that energetic. As we know, Rossi has this mysterious system - which he calls cat-and-mouse. He has been intentionally vague on
Re: [Vo]:Mizuno, Rossi copper transmutation
If you look at the ICCF-18 transmutation study of nickel and palladium study by Cook, you will see that Mizuno shows the same isotopic shifts in nickel that DGT shows. Ni61 does not participate in the reaction but all other isotopes of nickel do. Sorry, that link to this reference is broken. On Mon, Sep 22, 2014 at 12:01 PM, Arnaud Kodeck arnaud.kod...@lakoco.be wrote: Jones, Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a spin 0. _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: lundi 22 septembre 2014 17:34 To: vortex-l@eskimo.com Subject: RE: [Vo]:Mizuno, Rossi copper transmutation Typo- the suggested reaction is Ni58 + D - Cu60 - Ni60 _ I've looked through the isotope charts again - searching for reactions that rapidly decay back to the starting element or to any stable isotope which has already been reported to be there, and have not found any other possibility... ...other than Ni58 (d,Cu59) - Ni60 which happens by EC or positron emission, with a half-life of 20 minutes or so, and which fits the facts as reported in the most robust experiments (Rossi, DGT, Thermacore, Mills). 1) No or few gamma 2) No or little radioactive ash 3) No tritium, helium or positron annihilation 4) No or little bremsstrahlung 5) Excess energy which is at least 1000 times more than chemical Since nickel absorbs a deuteron and decays back to nickel in minutes, with low energy release, this reaction fits the bill. You may be thinking - what about the positron (beta positive) decay? No problem there, since nuclei which decay by positron emission also decay by electron capture in a known branching ratio which is dependant on the net energy of reaction. According to wiki-the-wonderful, in low-energy decays, electron capture is energetically favored by reactions below 1.022 MeV. The final state will have an electron added or a positron removed - and so the energy released is determinative of what can happen in the branching. As the energy of the decay goes up, so does the branching ratio towards positron emission. However, if the energy difference is low, then positron emission cannot occur, and electron capture is the sole decay mode. This would seem to be ready-made for the DDDL or deuteron-deep-Dirac-level species, which uses its tight electron for more than one purpose and probably reduces the net energy of the reaction as well. This still leaves spin conservation as the major problem. The end products of this reaction would be Ni60, and the starting nickel would be Ni58, so that is no problem. Both are spin 0. But the intermediary isotope, with short half-life would be Cu60 which is spin 2+ and the deuterium can only add is 1+ spin, and the EC electron another ½ spin. This over-simplification of spin issues - probably means that the reaction can only happen if a neutrino is captured, or else the inherent spin deficit decreases the half-life even more than its short nature. Probably the neutrino. Best of all - as a general working hypothesis which would make this relevant to LENR but is not expected to be seen anywhere else (which explains why it is not documented in the physics literature, as of now) there is NO other isotope in the periodic table (other than Ni58) - which is both a proton conductor and demonstrably neutron-deficient ! (the proof of that being that Ni-58 is lower amu than the preceding lower Z element (cobalt-59). That's right it is a perfect storm scenario. If this evolving explanation is correct, it will be seen nowhere else in the periodic table, since it demands conditions which do not exist anywhere else. This means, anthropomorphically speaking - that Ni58 desperately wants two more neutrons, and to get them, it essentially steals from its surroundings, whenever a deuteron comes too close... especially a DDDL. Falsifiability? Yes, this is falsifiable in three different way, which is a big advantage. Give me a working Rossi reactor :-) and a few months: if the [Ni-Ni] explanation is true, if will be proved beyond all reasonable doubt. P.S. do I get to keep the reactor? _ One more thing to add ... wrt the overdue suggestion (Doh, slaps forehead) that Rossi's secret sauce is looking like it is deuterium. Thank you, Clean Planet. The reaction would probably work best if it is started with regular hydrogen, and then deuterium is added later. This is
RE: [Vo]:Mizuno, Rossi copper transmutation
From: Arnaud Kodeck Jones, Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a spin 0. _ the suggested reaction is Ni58 + D - Cu60 - Ni60 Arnaud, This would be a three body reaction, no? You may be suggesting this reaction - in the event that Rossi does not use deuterium. That is wise to consider - since he professes not to, despite a tank of it being seen in his lab, early on. There is an even better possibility when two protons densified as a DDL molecule, and would act like the two needed neutrons, to make this reaction work. If my understanding is correct, nickel-58 is active ONLY because it is neutron-deficient, and the two protons do not help the immediate situation, at least not on the surface - even if both protons decay to neutrons, eventually. However, all bets are off with the DDL, since it allows the protons to look like virtual neutrons. There is nothing out there, which fits all of the parameters seamlessly, so in the end - we need reliable data. But it looks like we are framing a workable situation with enough variable to accommodate either D, H or H+D as the active gases. In short, your suggestion may work well - an especially if Rossi uses hydrogen only, and even more so - if the signature of the DDL formation (soft x-ray) is documented. Jones attachment: winmail.dat
Re: [Vo]:Converting hydrogen to work better in the Ni/H reactor
It seems that the popular LENR catalyst acts like a superconductor for protons where protons pair up into a cooper pair. See *http://arxiv.org/pdf/0807.1386.pdf* http://arxiv.org/pdf/0807.1386.pdf This work emphasizes that atoms in the crystal-field of KHCO3 are not individual particles possessing properties on their own right. They merge into macroscopic states and exhibit all features of quantum mechanics: nonlocality, entanglement, spin- symmetry, superposition and interferences. There is every reason to suppose that similar quantum effects should occur in many hydrogen bonded crystals undergoing structural phase transitions. I understand spin- symmetry is a zero spin. This catalyst provide a proton dimer of zero spin to the reaction. This is the reason why this catalyst enhances electrolytic LENR. On Mon, Sep 22, 2014 at 11:28 AM, Axil Axil janap...@gmail.com wrote: The first step in the hydrogen doublet fusion process is the formation of one or more atoms of 2He. Helium-2 or 2He, also known as a diproton, is an extremely unstable isotope of helium that consists of two protons without any neutrons. According to theoretical calculations it would have been much more stable (although still beta decaying to deuterium) had the strong force been 2% greater. Its instability is due to spin-spin interactions in the nuclear force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the diproton a negative binding energy. By the way, the ash produced by the LENR reaction will have a non zero nuclear spin such as lithium, boron, and beryllium. This is due to the fact that the ash is at the end of the LENR reaction chain. By the way, all the isotopes of copper have a non zero nuclear spin. On Mon, Sep 22, 2014 at 10:11 AM, Jones Beene jone...@pacbell.net wrote: Hydrogen molecules are shown to be slightly diamagnetic no matter which alignment they are in. Since protons are fermions, the anti-symmetry of the wavefunction imposes restrictions on the rotational states - with the result that the molecule is always diamagnetic. Consequently, ortho/para alignment would be irrelevant or counter-productive to a fusion process with a heavy metal like nickel, since it implies three body. Three-body processes are extremely rare, even in plasmas, and H2 as a molecule would not be involved as a reactant in LENR, unless one is talking about Storms’ reaction of two protons fusing to deuterium, for which there is no physical evidence. If this reaction was happening, tritium would be expected, since its formation has a much higher cross-section than two protons. As a monatomic species, there would be no relic of the prior spin alignment of hydrogen, in fusion with nickel. However, some or most of the energy of LENR could be non-fusion related. In that case, ortho/para cycling could be relevant *From:* Axil Axil Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% . Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive. This is a way to increase parahydrogen hydrogen by using a noble metal catalyst. see Catalytic process for ortho-para hydrogen conversion http://www.google.com/patents/US3383176 Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?
RE: [Vo]:Mizuno, Rossi copper transmutation
Yes, in my view, the DDL state diatomic hydrogen (shrunken hydrogen) reacts with Ni58. Should both atoms be in shrunken state? Is the DDL atoms small enough to go in the lattice? We can consider as well with pD or DD DDL state if natural hydrogen is used. _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: lundi 22 septembre 2014 18:28 To: vortex-l@eskimo.com Subject: RE: [Vo]:Mizuno, Rossi copper transmutation From: Arnaud Kodeck Jones, Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a spin 0. _ the suggested reaction is Ni58 + D - Cu60 - Ni60 Arnaud, This would be a three body reaction, no? You may be suggesting this reaction - in the event that Rossi does not use deuterium. That is wise to consider - since he professes not to, despite a tank of it being seen in his lab, early on. There is an even better possibility when two protons densified as a DDL molecule, and would act like the two needed neutrons, to make this reaction work. If my understanding is correct, nickel-58 is active ONLY because it is neutron-deficient, and the two protons do not help the immediate situation, at least not on the surface - even if both protons decay to neutrons, eventually. However, all bets are off with the DDL, since it allows the protons to look like virtual neutrons. There is nothing out there, which fits all of the parameters seamlessly, so in the end - we need reliable data. But it looks like we are framing a workable situation with enough variable to accommodate either D, H or H+D as the active gases. In short, your suggestion may work well - an especially if Rossi uses hydrogen only, and even more so - if the signature of the DDL formation (soft x-ray) is documented. Jones attachment: winmail.dat
Re: [Vo]:new taxonomy of our field
*Dear Peter, * *This comment too long to put into your last blog post: *NEW LENR TAXONOMY *There is good reason to believe that magnetism is the prime mover in LENR. Under this speculative paradigm, it is interesting to consider the options and consequences of this conjecture. In such a paradigm, any technology that is friendly to magnetism would be good for LENR, and conversely, a technology that undercuts the strength of magnetism is bad.* *The Pd/D wet technology is more unfriendly to magnetism than nickel because it makes magnetism more difficult to maintain. Firstly as a general technological principle, an isotope must have a nuclear spin of zero to enable the LENR reaction. There is much experimental evidence to support this conjecture. For an explanation see below. In this respect, palladium has a nuclear spin profile that is about 78% effective. 105Pd has a non-zero spin and is 22% of the isotopic contents of run of the mill palladium. * *On the other hand, Nickel is much more efficient in terms of supporting magnetism. 61Ni has a non-zero nuclear spin, but that isotope is only 1.14% of the isotopic content of Nickel.* *Palladium is paramagnetic and Nickel is ferromagnetic. So nickel is more desirable than palladium as a magnetic reaction catalyst.* *In more detail, this thinking is underpinned by a speculative LENR reaction rule that is interesting to explore. That rule is that the LENR reaction must occur among atomic ions that have zero nuclear spin.* *In explanation, Nuclear magnetic resonance (NMR) is a physical phenomenon in which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. This energy is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms; in practical applications, the frequency is similar to old style VHF and UHF television broadcasts (60–1000 MHz). NMR allows the observation of specific quantum mechanical magnetic properties of the atomic nucleus. * *All isotopes that contain an odd number of protons and/or of neutrons have an intrinsic magnetic moment and angular momentum, in other words a nonzero spin, while all nuclides with even numbers of both have a total spin of zero. The most commonly studied NMR active nuclei are 1H and 13C, although nuclei from isotopes of many other elements (e.g. 2H, 6Li, 10B, 11B, 14N, 15N, 17O, 19F, 23Na, 29Si, 31P, 35Cl, 113Cd, 129Xe, 195Pt) have been studied by high-field NMR spectroscopy as well.* *It is now known that Ni61 does not participate in the LENR reaction. Ni61 is a NMR active isotope. When a magnetic field is applied to an NMR active isotope, the magnetic energy imparted to the nucleus is dissipated by induced nuclear vibrational energy which is radiated away as rf energy. The non-zero spin of the the nucleus shields the nucleus from the external magnetic field not allowing that field to penetrate into it. External magnetic fields catalyze changes in the protons and neutrons in the nucleus as well as enabling accelerated quantum mechanical tunneling. If this external magnetic field is shielded by NMR activity, LENR transmutation of the protons and neutrons in the nucleus is made more difficult.* *Therefore, during the course of an extended LENR reaction cycle, isotope depletion will tend to favor the enrichment and buildup of NMR active elements.* *Hydrogen with non-zero spin will not participate in the LENR reaction whereas cooper pairs of protons will. Expect LENR reactions centered on pairs of protons with zero spin.* *Also, as the LERN reaction matures and more NMR active isotopes accumulate, the LENR reactor will put out increasing levels or rf radiation derived from the nuclear vibrations of the NMR isotope.* *This NMR thinking also applies to the nature of the various isotopes of hydrogen.* *Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% parahydrogen.* *Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive.* *This is a way to increase parahydrogen hydrogen by using a noble metal catalyst.* *see* *Catalytic process for ortho-para hydrogen conversion* *http://www.google.com/patents/US3383176 http://www.google.com/patents/US3383176* *Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?* *The first step in the hydrogen doublet fusion process is the formation of one or more atoms of 2He.* *Helium-2 or 2He, also known as a diproton, is an extremely unstable isotope of helium that consists
Re: [Vo]:new taxonomy of our field
thank you I will read it tomorrow Peter On Mon, Sep 22, 2014 at 9:57 PM, Axil Axil janap...@gmail.com wrote: *Dear Peter, * *This comment too long to put into your last blog post: *NEW LENR TAXONOMY *There is good reason to believe that magnetism is the prime mover in LENR. Under this speculative paradigm, it is interesting to consider the options and consequences of this conjecture. In such a paradigm, any technology that is friendly to magnetism would be good for LENR, and conversely, a technology that undercuts the strength of magnetism is bad.* *The Pd/D wet technology is more unfriendly to magnetism than nickel because it makes magnetism more difficult to maintain. Firstly as a general technological principle, an isotope must have a nuclear spin of zero to enable the LENR reaction. There is much experimental evidence to support this conjecture. For an explanation see below. In this respect, palladium has a nuclear spin profile that is about 78% effective. 105Pd has a non-zero spin and is 22% of the isotopic contents of run of the mill palladium. * *On the other hand, Nickel is much more efficient in terms of supporting magnetism. 61Ni has a non-zero nuclear spin, but that isotope is only 1.14% of the isotopic content of Nickel.* *Palladium is paramagnetic and Nickel is ferromagnetic. So nickel is more desirable than palladium as a magnetic reaction catalyst.* *In more detail, this thinking is underpinned by a speculative LENR reaction rule that is interesting to explore. That rule is that the LENR reaction must occur among atomic ions that have zero nuclear spin.* *In explanation, Nuclear magnetic resonance (NMR) is a physical phenomenon in which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. This energy is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms; in practical applications, the frequency is similar to old style VHF and UHF television broadcasts (60–1000 MHz). NMR allows the observation of specific quantum mechanical magnetic properties of the atomic nucleus. * *All isotopes that contain an odd number of protons and/or of neutrons have an intrinsic magnetic moment and angular momentum, in other words a nonzero spin, while all nuclides with even numbers of both have a total spin of zero. The most commonly studied NMR active nuclei are 1H and 13C, although nuclei from isotopes of many other elements (e.g. 2H, 6Li, 10B, 11B, 14N, 15N, 17O, 19F, 23Na, 29Si, 31P, 35Cl, 113Cd, 129Xe, 195Pt) have been studied by high-field NMR spectroscopy as well.* *It is now known that Ni61 does not participate in the LENR reaction. Ni61 is a NMR active isotope. When a magnetic field is applied to an NMR active isotope, the magnetic energy imparted to the nucleus is dissipated by induced nuclear vibrational energy which is radiated away as rf energy. The non-zero spin of the the nucleus shields the nucleus from the external magnetic field not allowing that field to penetrate into it. External magnetic fields catalyze changes in the protons and neutrons in the nucleus as well as enabling accelerated quantum mechanical tunneling. If this external magnetic field is shielded by NMR activity, LENR transmutation of the protons and neutrons in the nucleus is made more difficult.* *Therefore, during the course of an extended LENR reaction cycle, isotope depletion will tend to favor the enrichment and buildup of NMR active elements.* *Hydrogen with non-zero spin will not participate in the LENR reaction whereas cooper pairs of protons will. Expect LENR reactions centered on pairs of protons with zero spin.* *Also, as the LERN reaction matures and more NMR active isotopes accumulate, the LENR reactor will put out increasing levels or rf radiation derived from the nuclear vibrations of the NMR isotope.* *This NMR thinking also applies to the nature of the various isotopes of hydrogen.* *Molecular hydrogen occurs in two isomeric forms, one with its two proton spins aligned parallel (orthohydrogen), the other with its two proton spins aligned antiparallel (parahydrogen). At room temperature and thermal equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25% parahydrogen.* *Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because the non zero spin wastes magnetic energy by producing RF radiation. Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this type of hydrogen is magnetically inactive.* *This is a way to increase parahydrogen hydrogen by using a noble metal catalyst.* *see* *Catalytic process for ortho-para hydrogen conversion* *http://www.google.com/patents/US3383176 http://www.google.com/patents/US3383176* *Could this metallic ruthenium and certain ruthenium alloys be Rossi's secret sauce?* *The first step
RE: [Vo]:Mizuno, Rossi copper transmutation
_ From: Arnaud Kodeck Yes, in my view, the DDL state diatomic hydrogen (shrunken hydrogen) reacts with Ni58. Should both atoms be in shrunken state? Yes, that would seem to be highly beneficial. The reaction looks less like a three-body reaction if it happens with a tight DDL molecule, which is possibly in the 10s of Fermi size range. Is the DDL small enough to go in the lattice? Easily. The beauty of nickel as the host - from the Rydberg orbital viewpoint, is that it has two orbitals which are located at very good match in energy level for the correct hole, both of which are in its valence band at IP5 and IP6 ! This would essentially permit the DDL molecule, which has two electrons of a set Rydberg value, to find stability inside the shell - by replacing two normal electrons of nickel at a moderately deep level. Very few proton conductors have two deep orbitals which are adjoining in Rydberg values. Curiously, cobalt and iron are the others. This means any ferromagnetic material could be substituted for some of the nickel and host the DDL. From there, the excursion of the DDL to the nucleus on an occasional basis would seem to be highly favored. Only nickel has the neutron deficient isotope, however. As you can see, this is a mix of Mills CQM, the DDL version of other theorists and a few new additions. It would not require that excess energy is given up in shrinkage, as does Mills theory, since the progression goes to fusion eventually, which never happens according to CQM. This version does borrow the idea that the orbital electrons must have Rydberg values if they are to give up a proper hole for substitution (with the two electrons of the DDL). But AFAIK - Mills has not recognized the novelty of this suggestion, which is that adjoining holes in the valence shell of a ferromagnetic element like nickel, is the special parameter for LENR. Why would he? ...since he denies LENR is real, it has not occurred to him. In fact he uses other metals besides nickel these days instead in his own experiments. Maybe he intentionally avoids nickel :-) Jones attachment: winmail.dat
[Vo]:Chase Peterson dies
Chase Peterson, who was the President of University of Utah in 1989, died on September 14, 2014. See: http://infinite-energy.com/iemagazine/issue118/chase.html Here is most of chapter 12 of his book, which is the chapter about cold fusion: http://lenr-canr.org/acrobat/PetersonCtheguardia.pdf - Jed
Re: [Vo]:Mizuno, Rossi copper transmutation
In reply to Jones Beene's message of Mon, 22 Sep 2014 08:33:43 -0700: Hi, Typo- the suggested reaction is - Ni60 [snip] Ni58 + D - Cu60 + 11.252 MeV Normally one would expect prompt gammas from this reaction totaling 11.25 MeV. If no gammas are detected, what do you propose happens to the energy? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Re: [Vo]:Chase Peterson dies
Was he instrumental in releasing FP finding to the Press? Sent from my Verizon Wireless 4G LTE Smartphone - Reply message - From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com vortex-l@eskimo.com Subject: [Vo]:Chase Peterson dies Date: Mon, Sep 22, 2014 4:57 PM Chase Peterson, who was the President of University of Utah in 1989, died on September 14, 2014. See: http://infinite-energy.com/iemagazine/issue118/chase.html Here is most of chapter 12 of his book, which is the chapter about cold fusion: http://lenr-canr.org/acrobat/PetersonCtheguardia.pdf - Jed
RE: [Vo]:Mizuno, Rossi copper transmutation
The usual lame rationalizations we have used is that the energy was borrowed in advance to overcome the Coulomb barrier or shed in advance to achieve the redundancy ... But you're right - fusion numbers simply don't work well for the reality of a Rossi type reaction, as there is too much excess energy to hide... and this rationalization is no better than the factionalized gamma. So there you have it... back to no-fusion in LENR it is... -Original Message- From: mix...@bigpond.com Hi, Typo- the suggested reaction is - Ni60 [snip] Ni58 + D - Cu60 + 11.252 MeV Normally one would expect prompt gammas from this reaction totaling 11.25 MeV. If no gammas are detected, what do you propose happens to the energy? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Re: [Vo]:Chase Peterson dies
hohlr...@gmail.com hohlr...@gmail.com wrote: Was he instrumental in releasing FP finding to the Press? In the chapter I uploaded, he said no: Fleischmann reportedly said (for reasons never clear) that the University of Utah had required the two investigators to go public when they did. When I subsequently asked for clarification from the relevant university office, people there clearly stated that their policy was to honor all faculty requests with respect to publication and announcement, not initiate them. - Jed
Re: [Vo]:Re: [Vo]:Chase Peterson dies
On Mon, Sep 22, 2014 at 8:10 PM, Jed Rothwell jedrothw...@gmail.com wrote: hohlr...@gmail.com hohlr...@gmail.com wrote: Was he instrumental in releasing FP finding to the Press? In the chapter I uploaded, he said no: Fleischmann reportedly said (for reasons never clear) that the University of Utah had required the two investigators to go public when they did. When I subsequently asked for clarification from the relevant university office, people there clearly stated that their policy was to honor all faculty requests with respect to publication and announcement, not initiate them. It meant a lot to the university to be the first to announce. From: http://archive.wired.com/wired/archive/6.11/coldfusion_pr.html In their defense, Pons and Fleischmann explained that they couldn't reveal all the details because the University of Utah's patent had not yet been approved. They admitted that the press conference had been premature, but claimed the University had urged them to go public when another scientist - a physicist named Steve Jones - turned out to be pursuing similar work. Jones later became one of FP's greatest antagonists. The whole bloody fiasco probably set back CF 30 years. Sour grapes indeed.
[Vo]:John Farrell vs John Farrell
Obviously this John Farrell http://cosmosmagazine.com/features/in-wikipedia-we-trust/ (writing for Cosmos Magazine) is not the same as this John Farrell https://web.archive.org/web/20050128120420/http://www.blacklightpower.com/pdf/GUT/Review%20by%20John%20J.%20Farrell%20021004.pdf (who is obviously partial to Mills). Who is the John Farrell that writes for Cosmos Magazine?
Re: [Vo]:Mizuno, Rossi copper transmutation
In reply to Jones Beene's message of Mon, 22 Sep 2014 16:32:02 -0700: Hi, [snip] The usual lame rationalizations we have used is that the energy was borrowed in advance to overcome the Coulomb barrier or shed in advance to achieve the redundancy ... But you're right - fusion numbers simply don't work well for the reality of a Rossi type reaction, as there is too much excess energy to hide... and this rationalization is no better than the factionalized gamma. So there you have it... back to no-fusion in LENR it is... I would still be inclined to consider reactions that produce heavy charged particles. The heavier and slower, the better. E.g. fusion/fission reactions. I think the secondary gammas from heavily charged slow moving daughter nuclei might have been shielded. One possibility is the p-B11 reaction which produces quite low energy alphas because there are three of them. Furthermore the double charge on the alpha particles means both rapid energy loss to electrons, and strong repulsion from other nuclei, thus strongly reducing the chances of creating secondary gammas. Also, if such a reaction were to occur within the mass of the Boron, then the short range of the alphas would mostly keep them in the Boron itself, and the lowest excited state of B11 is 2.1247 MeV. This is not much less than the energy of the alphas, so in order to produce any secondary gammas at all, such an alpha would need to collide directly with another B11 nucleus almost immediately after creation, i.e. before it lost too much energy through ionization. That thus works to reduce the intensity of any secondary radiation. (B10 OTOH has excited states at 718 740 keV, but then B10 is only 20% of natural Boron). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Mizuno, Rossi copper transmutation
On Mon, Sep 22, 2014 at 9:26 AM, Axil Axil janap...@gmail.com wrote: If you look at the ICCF-18 transmutation study of nickel and palladium study by Cook, you will see that Mizuno shows the same isotopic shifts in nickel that DGT shows. Ni61 does not participate in the reaction but all other isotopes of nickel do. I'm having trouble finding the transmutation study by Cook. I have found this: http://iccf18.research.missouri.edu/files/Poster/Cook.pdf Is Cook's study on transmutations in nickel and palladium available online? I take it that it is a summary of experiments and not a set of ab initio calculations? Eric
Re: [Vo]:Mizuno, Rossi copper transmutation
On Mon, Sep 22, 2014 at 8:01 PM, mix...@bigpond.com wrote: I would still be inclined to consider reactions that produce heavy charged particles. The heavier and slower, the better. E.g. fusion/fission reactions. The reactions I've been looking at recently have charged particles as daughters as well. But the daughters are generally protons in the 5-10 MeV range. The way I propose that gammas from excited nuclei are avoided is to suggest that the reactions occur at the surface and that the daughters fly out from the surface: + d +++ p p p +++ p p p d p p ++ p --- p d p +++ p p d p p p +++ p p p + p Here the (+)'s are nickel lattice sites. The p results from an Ni(d,p)Ni reaction. The arrow represents the momentum. Although the p is born with ~ 5-10 MeV of energy, it burrows into the other p's at the surface, quickly thermalizing to a much lower energy. Occasionally there is a d that is broken apart through spallation. This wouldn't happen very often with a normal hydrogen mix, because there are only ~ 1/6000 parts deuterium, and only a fraction of these would be encountered (and only a fraction of the neutrons resulting from such spallations would exit the system). I think the secondary gammas from heavily charged slow moving daughter nuclei might have been shielded. By this I take it you mean gammas from lattice sites excited through inelastic collisions? Eric
Re: [Vo]:Mizuno, Rossi copper transmutation
In reply to Eric Walker's message of Mon, 22 Sep 2014 21:08:59 -0700: Hi Eric, [snip] On Mon, Sep 22, 2014 at 8:01 PM, mix...@bigpond.com wrote: I would still be inclined to consider reactions that produce heavy charged particles. The heavier and slower, the better. E.g. fusion/fission reactions. The reactions I've been looking at recently have charged particles as daughters as well. But the daughters are generally protons in the 5-10 MeV range. The way I propose that gammas from excited nuclei are avoided is to suggest that the reactions occur at the surface and that the daughters fly out from the surface: ...but wouldn't you expect 1/2 to fly away from the surface, and half to fly into it? + d +++ p p p +++ p p p d p p ++ p --- p d p +++ p p d p p p +++ p p p + p Here the (+)'s are nickel lattice sites. The p results from an Ni(d,p)Ni reaction. The arrow represents the momentum. Although the p is born with ~ 5-10 MeV of energy, it burrows into the other p's at the surface, quickly thermalizing to a much lower energy. Occasionally there is a d that is broken apart through spallation. This wouldn't happen very often with a normal hydrogen mix, because there are only ~ 1/6000 parts deuterium, and only a fraction of these would be encountered (and only a fraction of the neutrons resulting from such spallations would exit the system). I think the secondary gammas from heavily charged slow moving daughter nuclei might have been shielded. By this I take it you mean gammas from lattice sites excited through inelastic collisions? Yes, that's what I meant. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Mizuno, Rossi copper transmutation
On Mon, Sep 22, 2014 at 10:33 PM, mix...@bigpond.com wrote: ...but wouldn't you expect 1/2 to fly away from the surface, and half to fly into it? I would expect there to be an anisotropy. As I envision it, there's an electric arc pulling a mass of protons into a recess. For a fraction of a moment, the pressure is astronomical. During this brief moment a deuteron (the smaller species are all ionized within the arc) is forced up against a lattice site, coming from the direction of the open area and the current towards the wall of the substrate. Unless there's some kind of rotation during the moment of contact, if the lattice site is on the left and the deuteron is coming from the right to the left, I would expect the daughter proton to push off of the daughter nickel and be expelled back out to the right, which is the open area. I assume this would all happen too quickly for any kind of rotation of the nickel/deuteron system. Eric
