RE: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
The vacuum according to experts, including Dirac, is not empty. If the vacuum reacts with any kind of matter, it reacts with hydrogen. The prior post was based on the proton mass of 938.27231 MeV or 1876.545 for two protons, and the mass of the deuteron being 1875.613 MeV. Thus splitting deuterium, with subsequent neutron decay (even when instantaneous and with no neutrino emission) cannot be net exothermic without something else, since the difference is -0.932 MeV. Curiously the mass-energy of electron/positron annihilation is 1.022 MeV. This ties into Dirac and Hotson's epo field interpretation. The epo field, now called the BEC, is the superset of ZPE. Dirac's sea of negative energy is also approximately the same background entity as the zero point field, or simply the vacuum. If you merge Puthoff and Hotson, (two of Terry's favorites) then deuterium fission into protons will be net energetic if virtual Ps enters into the reaction. In fact the energy release will be strong for chemical or weak for nuclear, matching experimental finding - and in the zone of non-detectability by gamma detectors, since the reactor walls will absorb this level, but not much higher. D + Ps - 2H + 90 keV ... which is about 100,000 times more energy release than burning hydrogen. ( Ps is positronium). The next problem is conservation of charge... which means you must annihilate the electron from neutron decay with a positron from the epo field. Falsifiability - look for radiation inside the reactor at the approximate level of 90 or 45 keV. _ 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
On Sun, Mar 30, 2014 at 10:53 AM, Jones Beene jone...@pacbell.net wrote: Falsifiability - look for radiation inside the reactor at the approximate level of 90 or 45 keV. And, while you're looking, check for spontaneous emission, ie coherence.
Re: [Vo]:The DD-BOP reaction in the context of Mizuno's new breakthrough
When the understanding and credibility of LENR is well entrenched is main stream science, the Mizuno (Yoshino) experiment and others like it will be a first of the primary and indispensable tools used to explore and quantify dark energy and its place in the universe. On Sun, Mar 30, 2014 at 10:53 AM, Jones Beene jone...@pacbell.net wrote: The vacuum according to experts, including Dirac, is not empty. If the vacuum reacts with any kind of matter, it reacts with hydrogen. The prior post was based on the proton mass of 938.27231 MeV or 1876.545 for two protons, and the mass of the deuteron being 1875.613 MeV. Thus splitting deuterium, with subsequent neutron decay (even when instantaneous and with no neutrino emission) cannot be net exothermic without something else, since the difference is -0.932 MeV. Curiously the mass-energy of electron/positron annihilation is 1.022 MeV. This ties into Dirac and Hotson's epo field interpretation. The epo field, now called the BEC, is the superset of ZPE. Dirac's sea of negative energy is also approximately the same background entity as the zero point field, or simply the vacuum. If you merge Puthoff and Hotson, (two of Terry's favorites) then deuterium fission into protons will be net energetic if virtual Ps enters into the reaction. In fact the energy release will be strong for chemical or weak for nuclear, matching experimental finding - and in the zone of non-detectability by gamma detectors, since the reactor walls will absorb this level, but not much higher. D + Ps - 2H + 90 keV ... which is about 100,000 times more energy release than burning hydrogen. ( Ps is positronium). The next problem is conservation of charge... which means you must annihilate the electron from neutron decay with a positron from the epo field. Falsifiability - look for radiation inside the reactor at the approximate level of 90 or 45 keV. _ 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]: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
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]: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]: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. :-)
