Re: [Vo]:Mizuno slides coming
For D, certainly He4 is detected, correlated with heat. For H, it's a harder issue, so we cannot know if its is fusion which is happening, that is, with correlation with heat. 2014-03-29 2:47 GMT-03:00 Axil Axil janap...@gmail.com: You said Certainly, the only proof for fusion is the generation of He4 Will you accept this since the is no He4 detected, then there is no fusion taking place. -- Daniel Rocha - RJ danieldi...@gmail.com
RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Part I The recent Mizuno (Yoshino) presentation at the MIT colloquium and the surprising implication of finding about twice the quantity of hydrogen appearing as ash from deuterium reactions (as the starting gas) after a month long run - has been the inspiration for the following early stage hypothesis. This is a revision to focus on nano-magnetics and the SPP contribution. In answer to those who say that such an analysis before this experiment has been replicated is premature, my answer is that rewards of finding an early helpful answer to the mystery outweighs the risk of adding more confusion to an already confusing field. Very simply, what is being proposed is a new version of the Oppenheimer-Phillips effect. The Oppenheimer-Phillips reaction is also known as deuterium stripping. Stripping typically removes a proton from the deuteron at a tiny fraction of the thermonuclear requirement. In the case of the Farnsworth Fusor the threshold is reduced from 2.2 MeV to around 50 keV, or a factor of 40:1. This revised version, which has been tailored to the Mizuno results could be called a bi-stripping or the BOP reaction (Bi-Oppenheimer-Philips). This is different from Passell's version of the O-P presented at ICCF18 in that the nickel host provides ferromagnetic containment, but does not participate as a reactant. In both cases this is a quantum mechanical reaction similar to nuclear tunneling, but with a magnetic near-field component. The deuteron has only one bound state in which the magnetic moment (+0.8574) is a function of the proton positive value (+2.7928) and the neutron negative value (-1.9130) at a rather large separation distance. In short, this isotope is not strongly bound to begin with, and the linear bond lacks flexibility in torsion, so that when there is a sudden magnetic torque at the nanoscale, the bond can be broken without thermodynamics. The strong-force is spin dependent with deuterium. Thus, a nuclear reaction that looks endothermic (from a thermonuclear POV) can be made exothermic via spin dynamics. There is not a violation of conservation of energy, since the gain is nuclear. This is the heart of the Oppenheimer Philips effect. However, a strong local magnetic field is required. Polaritons and the SPP can provide a multi-Tesla local field to provide magnetic torque. This sets the stage for exotherm following nuclear shear from spin-coupling the deuteron to a circularly polarized magnetic field in a ferromagnetic surface feature- to with which to break the bond without thermonuclear brute force. The field is a function of SPP - surface plasmon-polaritons. When the reaction happens with two deuterons at low temperature, no fusion is possible, but nanocavity surface pitting provides near-field magnetic polarization via SPP. Thus the net reaction looks more like fission than fusion. Deuterons are effectively slit with instantaneous neutron decay such that D2 converts to 2H2 with about 400 keV of net mass-energy and no neutrino. To be continued in Part II Jones _ attachment: winmail.dat
RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Part II When a free neutron decays to a proton, substantial energy is released as well as a neutrino - which carries away about 40% of the net energy undetected. That is the main problem to overcome in framing a putative exothermic deuterium reaction in place of the endotherm which would normally appear. There is a valid QM rationalization for this, but the probability of it happening is unknown. Outside the nucleus, free neutrons are unstable and have a mean lifetime of about 15 minutes. They beta decay with the emission of an electron and electron antineutrino leaving a fairly cold proton. The decay energy for this process is up to 0.78 MeV for the electron, but is highly variable- unlike almost any other nuclear reaction. The energy of the unseen neutrino which is emitted is about 500 keV on average - which explanation resolves problems of conservation of spin and the lower net energy which is sometimes seen in experiment. The variability of energy release is hard to reconcile without a kludge of some kind - which is the neutrino. The reality of the neutrino in general is not in question here, but its application to a related reaction is in question, since it may not be required when the need is obviated. The free neutron mass is larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. The difference is ~1.3 MeV. Since the apparent energy release from neutron decay is occasionally nearly the entire value of the theoretical mass difference, we must ask: is the neutrino really necessary in a D+D collision, or any other without allowed spin problems, or is a relic of trying something else which has taken on a life of its own? When two neutrons decay together immediately on the impact of two deuterons which do not have enough momentum to fuse, the collision can be a mini QCD version of quark soup that seldom overcomes the barrier for fusion to helium, but is nevertheless energetic. Moreover there is no allowed spin problem. Consider the spins of the electron and antineutrino with a net spin of zero. This is a Fermi decay since the electron and antineutrino take no spin away, and the nuclear spin cannot change. The other possibility allowed by QM is that spins combine into a net spin of one: Gamow-Teller decay. The angular momentum can change by up to one unit in an allowed double beta decay, which is the closest analogy. Consequently, there is a distinct possibility for spin issues to be resolved in the context of two inseparable reactions involving deuterons, but without neutrino emission. There is another issue - the extended half-life of free neutrons - which means that decay energy is not normally available instantaneously, to lend in the sense of quantum mechanics. This is where QM enters the picture in two different ways. The mass of the deuteron is 1875.613 MeV. The mass of a free neutron plus a free proton is 1877.8374 - thus about 2.2 MeV would be required (to be supplied via kinetic energy) in order to split the deuteron - without QM. The net deficit of this reaction is somewhere around ~900 keV if the neutrino is avoided. So far, even assuming a time reversed borrowing, we are still at endotherm unless the same initial kinetic energy provides two identical reactions. Voila! ... then there is net gain to the extent neutrino release is avoided. An apparent endotherm is the only reason why no one ever imagined Oppenheimer Philips as being relevant before now. It looks endothermic, without Heisenberg uncertainty - and even more so without neutrino suppression. However, one can surmise that when two deuterons approach each other so that both undergo the OP splitting reaction instantaneously as a result of the single impact, then the same 2.2 MeV of kinetic energy results in both reactions. This is an implication of Heisenberg. A net energy release of 2.6 MeV is then seen (from two instantaneous neutron decays without neutrinos). Most of the threshold energy can be borrowed. The two neutrons have decayed to protons instantly, instead of with an extended half-life and we have an allowed spin state without neutrino release. Thus the net reaction gain is 300-400 keV imparted to two electrons. The stretch of the imagination is that the same kinetic energy can split both atoms at exactly the same time, invoking quantum uncertainty. Thus, using borrowed energy from the net reaction - with neutrino emission suppressed we now have a net gainful reaction. Admittedly, this is a stretch, but isn't everything in QM, especially when first invoked ? The reality of this or any such QM explanation for an experimental result is dependent on the accuracy of Mizuno's mass spectroscopy. If Mizuno is correct, this is a defensible first step to consider towards a viable answer to the finding (of twice the quantity of gas in the ash of the reaction). Can anyone propose another defensible hypothesis for gain, giving benefit of doubt to Mizuno, which can support these findings? There is the
[Vo]:2014 CF/LANR Colloquium all files page
We are assembling a page for all 2014 CF/LANR Colloquium at MIT audio, video, .pdfs, and links to affiliated institutions: http://coldfusionnow.org/interviews/2014-cflanr-colloquium-at-mit-full-coverage/ New material will be added here as they are available throughout the week (or two!) Your comments and suggestions are welcome. Ruby -- Ruby Carat r...@coldfusionnow.org Skype ruby-carat www.coldfusionnow.org
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
I favor the idea that a strong magnetic field can catalyze pion virtual particles from the energy borrowed from the vacuum that can disrupt the nucleus of the atom along the path of that magnetic field. If a plus pion transmutes a neutron into a proton using borrowed energy from the vacuum (135 MeV), no angular momentum is carried into the reaction. Furthermore, all energy issues would be resolved through instantaneous decay of the neutron as transformed at that instant by the plus pion. That is, no 14 minute neutron decay time would occur. No electron and neutrino would be produced in the reaction since the neutron to proton transmutation process is a run of the mill nuclear one and not a radioactive one. The energy of the reaction being generated is simply the energy equivalent of the mass difference between the deuterium atom and 2 protons. That energy would be transferred by the magnetic field to the origin of the magnetic field and thermalized without loss in a dark mode within a positive feedback loop. The spin of the D being one would be broken into 2 spin ½ protons thus with spin having been conserved. On Sat, Mar 29, 2014 at 11:20 AM, Jones Beene jone...@pacbell.net wrote: Part II When a free neutron decays to a proton, substantial energy is released as well as a neutrino - which carries away about 40% of the net energy undetected. That is the main problem to overcome in framing a putative exothermic deuterium reaction in place of the endotherm which would normally appear. There is a valid QM rationalization for this, but the probability of it happening is unknown. Outside the nucleus, free neutrons are unstable and have a mean lifetime of about 15 minutes. They beta decay with the emission of an electron and electron antineutrino leaving a fairly cold proton. The decay energy for this process is up to 0.78 MeV for the electron, but is highly variable- unlike almost any other nuclear reaction. The energy of the unseen neutrino which is emitted is about 500 keV on average - which explanation resolves problems of conservation of spin and the lower net energy which is sometimes seen in experiment. The variability of energy release is hard to reconcile without a kludge of some kind - which is the neutrino. The reality of the neutrino in general is not in question here, but its application to a related reaction is in question, since it may not be required when the need is obviated. The free neutron mass is larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. The difference is ~1.3 MeV. Since the apparent energy release from neutron decay is occasionally nearly the entire value of the theoretical mass difference, we must ask: is the neutrino really necessary in a D+D collision, or any other without allowed spin problems, or is a relic of trying something else which has taken on a life of its own? When two neutrons decay together immediately on the impact of two deuterons which do not have enough momentum to fuse, the collision can be a mini QCD version of quark soup that seldom overcomes the barrier for fusion to helium, but is nevertheless energetic. Moreover there is no allowed spin problem. Consider the spins of the electron and antineutrino with a net spin of zero. This is a Fermi decay since the electron and antineutrino take no spin away, and the nuclear spin cannot change. The other possibility allowed by QM is that spins combine into a net spin of one: Gamow-Teller decay. The angular momentum can change by up to one unit in an allowed double beta decay, which is the closest analogy. Consequently, there is a distinct possibility for spin issues to be resolved in the context of two inseparable reactions involving deuterons, but without neutrino emission. There is another issue - the extended half-life of free neutrons - which means that decay energy is not normally available instantaneously, to lend in the sense of quantum mechanics. This is where QM enters the picture in two different ways. The mass of the deuteron is 1875.613 MeV. The mass of a free neutron plus a free proton is 1877.8374 - thus about 2.2 MeV would be required (to be supplied via kinetic energy) in order to split the deuteron - without QM. The net deficit of this reaction is somewhere around ~900 keV if the neutrino is avoided. So far, even assuming a time reversed borrowing, we are still at endotherm unless the same initial kinetic energy provides two identical reactions. Voila! ... then there is net gain to the extent neutrino release is avoided. An apparent endotherm is the only reason why no one ever imagined Oppenheimer Philips as being relevant before now. It looks endothermic, without Heisenberg uncertainty - and even more so without neutrino suppression. However, one can surmise that when two deuterons approach each other so that both undergo the OP splitting reaction instantaneously as a result of the single impact,
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Come to think of it, this pion reaction is a mind bender. This Mizuno reaction is the exact opposite of the proton-proton chain reaction. That is the fusion reaction by which stars convert hydrogen to helium. If the proton-proton chain reaction produces positive energy, then its opposite fission reaction must need the same energy supplement to occur. If the Mizuno reaction is occurring, the energy needed for it to occur must be coming from the vacuum, fed by the magnetic field that connects the location of vacuum activity to the nucleus. This Mizuno experiment is such a pure, seemingly uncomplicated, and straightforward experiment, the basic reaction is self-evident, yet it is unbelievable by all conventional standards. They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum. On Sat, Mar 29, 2014 at 1:14 PM, Axil Axil janap...@gmail.com wrote: I favor the idea that a strong magnetic field can catalyze pion virtual particles from the energy borrowed from the vacuum that can disrupt the nucleus of the atom along the path of that magnetic field. If a plus pion transmutes a neutron into a proton using borrowed energy from the vacuum (135 MeV), no angular momentum is carried into the reaction. Furthermore, all energy issues would be resolved through instantaneous decay of the neutron as transformed at that instant by the plus pion. That is, no 14 minute neutron decay time would occur. No electron and neutrino would be produced in the reaction since the neutron to proton transmutation process is a run of the mill nuclear one and not a radioactive one. The energy of the reaction being generated is simply the energy equivalent of the mass difference between the deuterium atom and 2 protons. That energy would be transferred by the magnetic field to the origin of the magnetic field and thermalized without loss in a dark mode within a positive feedback loop. The spin of the D being one would be broken into 2 spin 1/2 protons thus with spin having been conserved. On Sat, Mar 29, 2014 at 11:20 AM, Jones Beene jone...@pacbell.net wrote: Part II When a free neutron decays to a proton, substantial energy is released as well as a neutrino - which carries away about 40% of the net energy undetected. That is the main problem to overcome in framing a putative exothermic deuterium reaction in place of the endotherm which would normally appear. There is a valid QM rationalization for this, but the probability of it happening is unknown. Outside the nucleus, free neutrons are unstable and have a mean lifetime of about 15 minutes. They beta decay with the emission of an electron and electron antineutrino leaving a fairly cold proton. The decay energy for this process is up to 0.78 MeV for the electron, but is highly variable- unlike almost any other nuclear reaction. The energy of the unseen neutrino which is emitted is about 500 keV on average - which explanation resolves problems of conservation of spin and the lower net energy which is sometimes seen in experiment. The variability of energy release is hard to reconcile without a kludge of some kind - which is the neutrino. The reality of the neutrino in general is not in question here, but its application to a related reaction is in question, since it may not be required when the need is obviated. The free neutron mass is larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. The difference is ~1.3 MeV. Since the apparent energy release from neutron decay is occasionally nearly the entire value of the theoretical mass difference, we must ask: is the neutrino really necessary in a D+D collision, or any other without allowed spin problems, or is a relic of trying something else which has taken on a life of its own? When two neutrons decay together immediately on the impact of two deuterons which do not have enough momentum to fuse, the collision can be a mini QCD version of quark soup that seldom overcomes the barrier for fusion to helium, but is nevertheless energetic. Moreover there is no allowed spin problem. Consider the spins of the electron and antineutrino with a net spin of zero. This is a Fermi decay since the electron and antineutrino take no spin away, and the nuclear spin cannot change. The other possibility allowed by QM is that spins combine into a net spin of one: Gamow-Teller decay. The angular momentum can change by up to one unit in an allowed double beta decay, which is the closest analogy. Consequently, there is a distinct possibility for spin issues to be resolved in the context of two inseparable reactions involving deuterons, but without neutrino emission. There is another issue - the extended half-life of free neutrons - which means that decay energy is not normally available instantaneously, to lend in the sense of quantum mechanics. This is where QM enters the picture in two different
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
On Sat, Mar 29, 2014 at 3:03 PM, Axil Axil janap...@gmail.com wrote: They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum. I have always thought so. But, then, I have been a Puthoff fan-boy for ages. :-)
RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Yes - even if plausible way exists in QM for converting deuterium to hydrogen with gain, that gain obviously does not derive from the mass of the deuterium, per se. This leaves these main possibilities, and a few others 1) vacuum energy (ZPE) 2) nickel mass via spin coupling 3) Mills version of redundant ground states It could be possible that all of these are entwined. -Original Message- From: Terry Blanton Axil wrote: They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum? I have always thought so. But, then, I have been a Puthoff fan-boy for ages. :-) attachment: winmail.dat
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Entwined as in a tree with vacuum energy (ZPE) as the tap root, and the others (if real) as emergent pathways of energy flow. On Sat, Mar 29, 2014 at 3:39 PM, Jones Beene jone...@pacbell.net wrote: Yes - even if plausible way exists in QM for converting deuterium to hydrogen with gain, that gain obviously does not derive from the mass of the deuterium, per se. This leaves these main possibilities, and a few others 1) vacuum energy (ZPE) 2) nickel mass via spin coupling 3) Mills version of redundant ground states It could be possible that all of these are entwined. -Original Message- From: Terry Blanton Axil wrote: They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum? I have always thought so. But, then, I have been a Puthoff fan-boy for ages. :-)
Re: [Vo]:Mizuno slides coming
On Fri, Mar 28, 2014 at 8:06 AM, Jones Beene jone...@pacbell.net wrote: There would be a net decrease in gas quantity under any scenario in which D2 reacts with nickel – never wound an increase be expected, even small - much less a ~2:1 increase in gas quantity. Amazing. I think the lead that you and Axil are pursuing on the possibility of some way of splitting the deuterons in a special near-field magnetic field in the environment provided by nickel cavities is a thought-provoking one, and I'll be interested to see where the thought experiment goes. For the moment, I figure we each of us gets to egregiously ignore at least one major claim or implication of any item of news until one has lost enthusiasm for what one gets in return (in doing this, I'm just formalizing the existing practice on this list). The claim I will egregiously ignore for the moment as either being artifact or something that is different from what we currently understand it to be is the idea that there were twice as many gas molecules after the experiment had run than at the time it had started. (Because I'm *egregiously* ignoring the detail, I make no claims as to the plausibility that something is wrong with it.) What this gets me in return: - The p+Ni lead appears to align with the thoughts of the experimenters themselves, who included graphs of the neutron capture cross sections for nickel in their slides. - The p+Ni lead takes on a similar shape to earlier speculations about proton capture in the NiH system and to 4He generation in the PdD system (e.g., involving electric arcs between insulated grains). - The authors mention that if you calculate the amount of energy that would be expected of reactions generating between 3-4 MeV each, you would get less energy than they observed. This is a detail you have to egregiously ignore to put forward a reaction that produces on the order of 400 keV apiece. (Another detail *I'm* egregiously ignoring at the moment is the expected Bremsstrahlung radiation from fast protons; I'm starting to wonder whether whatever is going on can diffuse even kinetic energy into the electronic structure.) But, again, I like where you're going with the deuteron splitting and the neutron either decaying over a period of minutes, or instantaneously changing to a proton due to the weird way the process unfolds. Can we agree on this -- your argument will not be expected to predict beta+ or beta- decay signatures in any significant amount, whereas mine will? Eric
Re: [Vo]:Mizuno slides coming
I wrote: - The p+Ni lead appears to align with the thoughts of the experimenters themselves, who included graphs of the neutron capture cross sections for nickel in their slides. I wrote p+Ni, but I meant d+Ni. Eric
RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
One other exotic possibility comes to mind, thinking about Ni-64. This nickel isotope appears to be unique in the periodic table, being the highest ratio of excess neutrons in a stable isotope, compared to the most common isotope of that element (36/30 = 12%) in nature. (hydrogen-deuterium does not qualify since H has no neutron) Not sure what that means, but when one finds an anomaly juxtaposed with a singularity - the two are unlikely to be the result of coincidence. There is not much of this isotope available ... OTOH there is more of it than there is U235 in natural U. _ Yes - even if plausible way exists in QM for converting deuterium to hydrogen with gain, that gain obviously does not derive from the mass of the deuterium, per se. This leaves these main possibilities, and a few others 1) vacuum energy (ZPE) 2) nickel mass via spin coupling 3) Mills version of redundant ground states It could be possible that all of these are entwined. -Original Message- From: Terry Blanton Axil wrote: They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum? I have always thought so. But, then, I have been a Puthoff fan-boy for ages. :-) attachment: winmail.dat
RE: [Vo]:Mizuno slides coming
From: Eric Walker The p+Ni lead appears to align with the thoughts of the experimenters themselves, who included graphs of the neutron capture cross sections for nickel in their slides. I wrote p+Ni, but I meant d+Ni. The d+Ni reaction would have to be the Oppenheimer-Phillips version, to be statistically relevant. Here is a blip on Passell’s O-P theroy. I have not found it as a separate file. http://coldfusionnow.org/iccf-18-day-5-presentations-and-awards/ attachment: winmail.dat
Re: [Vo]:Mizuno slides coming
On Sat, Mar 29, 2014 at 1:31 PM, Jones Beene jone...@pacbell.net wrote: I wrote p+Ni, but I meant d+Ni. The d+Ni reaction would have to be the Oppenheimer-Phillips version, to be statistically relevant. Here is a blip on Passell’s O-P theroy. I have not found it as a separate file. http://coldfusionnow.org/iccf-18-day-5-presentations-and-awards/ Yes! Yesterday I borrowed (stole) the OP idea from you. (I didn't know that I also borrowed it from Thomas Passel, although he seems to be looking at palladium.) http://www.mail-archive.com/vortex-l@eskimo.com/msg92381.html I just discovered that you wrote concerning the OP angle back in 2010 (and Abd Lomax replied): https://www.mail-archive.com/vortex-l@eskimo.com In answer to Abd Lomax's question about how to accelerate the deuteron, I'm venturing that this would be done by way of an electric arc between two insulated nickel grains. Although one expects a fast proton as a result, I think we have to posit something that short-circuits the resulting kinetic energy in order to avoid a situation where there is a Chernobyl's worth of Bremsstrahlung coming out of the system. Eric
[Vo]:Modern technology for every time in ones life.
I attended to a funeral two weeks ago. He was a veteran of WW2. The bugle player was excellent. He played taps. Many cried. I went to another funeral today. Taps were played again,the bugle player sneezed and, the bugle kept on playing. I asked about it, They are now using digital bugles. They play with the touch of a button. What next? Frank.
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
Just wanted to add one minor thought to the discussion. Could it be that the breaking up of the D into pieces might actually take energy from the system that is then added back by a relatively minor amount of more or less standard H reaction with nickel? The implications of such a process are strange indeed, but at least the energy output tracks with what Rossi has been producing. I suppose that someone might point out that these guys favor using D instead of H as reactant gases. Perhaps the test cells varied in efficiency and the D cells won in this special case. This whole concept makes me wonder about the ability of nano particle heavy nickel to act as a catalyst to split apart D atoms. AFAIK, no one has reported such behavior in previous experiments. Is there reason to consider this particular experiment as being substantially different from previous ones where nickel was exposed to D? In other words, why now? I admit that I need the same break that Eric requests. Sometimes you must wait on the sidelines a bit longer than others before you get the call to enter the game. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Sat, Mar 29, 2014 4:25 pm Subject: RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough One other exotic possibility comes to mind, thinking about Ni-64. This nickel isotope appears to be unique in the periodic table, being the highest ratio of excess neutrons in a stable isotope, compared to the most common isotope of that element (36/30 = 12%) in nature. (hydrogen-deuterium does not qualify since H has no neutron) Not sure what that means, but when one finds an anomaly juxtaposed with a singularity - the two are unlikely to be the result of coincidence. There is not much of this isotope available ... OTOH there is more of it than there is U235 in natural U. _ Yes - even if plausible way exists in QM for converting deuterium to hydrogen with gain, that gain obviously does not derive from the mass of the deuterium, per se. This leaves these main possibilities, and a few others 1) vacuum energy (ZPE) 2) nickel mass via spin coupling 3) Mills version of redundant ground states It could be possible that all of these are entwined. -Original Message- From: Terry Blanton Axil wrote: They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum? I have always thought so. But, then, I have been a Puthoff fan-boy for ages. :-)
Re: [Vo]:Mizuno slides coming
On Sat, Mar 29, 2014 at 2:15 PM, Eric Walker eric.wal...@gmail.com wrote: I just discovered that you wrote concerning the OP angle back in 2010 (and Abd Lomax replied): https://www.mail-archive.com/vortex-l@eskimo.com That second link above, where Jones and Abd Lomax discuss the Oppenheimer-Phillips process in the context of d+Ni, was a little too general. The link was supposed to be: https://www.mail-archive.com/vortex-l@eskimo.com/msg39383.html Eric
[Vo]:1-minute video The Frequency of Humanity
A video about what happens in one minute. Mostly off topic but the amount of garbage produced and CO2 released are awesome. http://www.theatlantic.com/video/index/359066/the-frequency-of-humanity/
Re: [Vo]:Mizuno slides coming
I wrote: The claim I will egregiously ignore for the moment as either being artifact or something that is different from what we currently understand it to be is the idea that there were twice as many gas molecules after the experiment had run than at the time it had started. I think I found a way out of this difficulty. There might be a straightforward way to explain the increase in the number of gas molecules after the runs by Yoshida et al. If we're seeing neutron capture after a deuteron has been forced to approach a nickel lattice site, with a corresponding expelling of a 5-7 MeV proton, we can expect there to be a lot of spallation. Here is an image of what I have in mind: http://i.imgur.com/cATIdcT.png The idea is that the current from an arc between two grains is causing great downward pressure on deuteron ions, forcing them into a recess in one of the grains. (They're ionized because they're in the midst of an electric arc.) That pressure forces a deuteron at the bottom of the recess to approach close to one of the lattice sites. At some point the Oppenheimer-Phillips process takes over and strips the neutron from the deuteron, yielding a high-energy proton. While the lattice site barely moves, the proton flies with great force into the ions above it. As happens when a bullet is fired into water or sand, the momentum of the proton is quickly dampened. In the process you can expect a spallation, in which some of the other deuterons are broken apart into protons and neutrons. The neutrons will have a half-life of 14 minutes and will decay into protons. Outside of the electric arc the protons will combine to form some multiple of H2 molecules of the original number of D2/DH molecules that were fed into the system. Since the high-energy proton is colliding primarily with other ionized protons and deuterons, I'm guessing there will be little high-energy Bremsstrahlung from collisions with lattice site electrons. Presumably all of this happens before a dislocation occurs at the bottom of the recess and relieves some of the pressure. Eric
Re: [Vo]:Mizuno slides coming
The decay of the neutron must be instansious because no indications of neutron absorption into a deuterium nucleus to produce tritium. There is no indication that any atom larger in mass than deuterium had been generated. You must assume instansious completion of the reaction exclusive of any side reactions. On Sat, Mar 29, 2014 at 11:54 PM, Eric Walker eric.wal...@gmail.com wrote: I wrote: The claim I will egregiously ignore for the moment as either being artifact or something that is different from what we currently understand it to be is the idea that there were twice as many gas molecules after the experiment had run than at the time it had started. I think I found a way out of this difficulty. There might be a straightforward way to explain the increase in the number of gas molecules after the runs by Yoshida et al. If we're seeing neutron capture after a deuteron has been forced to approach a nickel lattice site, with a corresponding expelling of a 5-7 MeV proton, we can expect there to be a lot of spallation. Here is an image of what I have in mind: http://i.imgur.com/cATIdcT.png The idea is that the current from an arc between two grains is causing great downward pressure on deuteron ions, forcing them into a recess in one of the grains. (They're ionized because they're in the midst of an electric arc.) That pressure forces a deuteron at the bottom of the recess to approach close to one of the lattice sites. At some point the Oppenheimer-Phillips process takes over and strips the neutron from the deuteron, yielding a high-energy proton. While the lattice site barely moves, the proton flies with great force into the ions above it. As happens when a bullet is fired into water or sand, the momentum of the proton is quickly dampened. In the process you can expect a spallation, in which some of the other deuterons are broken apart into protons and neutrons. The neutrons will have a half-life of 14 minutes and will decay into protons. Outside of the electric arc the protons will combine to form some multiple of H2 molecules of the original number of D2/DH molecules that were fed into the system. Since the high-energy proton is colliding primarily with other ionized protons and deuterons, I'm guessing there will be little high-energy Bremsstrahlung from collisions with lattice site electrons. Presumably all of this happens before a dislocation occurs at the bottom of the recess and relieves some of the pressure. Eric
Re: [Vo]:Mizuno slides coming
On Sat, Mar 29, 2014 at 9:11 PM, Axil Axil janap...@gmail.com wrote: There is no indication that any atom larger in mass than deuterium had been generated. See the yellow arrow for species of mass 3 on pp. 38, 39, 41 and 42 of the slides (according to Chrome): http://lenr-canr.org/acrobat/YoshinoHreplicable.pdf At first the m=3 species go up. Only after some time do they go down. Eric
