Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Bob, these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear process. Ed Storms Sent from my iPad On Feb 12, 2014, at 3:00 PM, Bob Cook frobertc...@hotmail.com wrote: Jones--Bob Cook Here-- Can you show how the p-e-p reaction as you understand it conserves spin? I would think that the newly fused particle, whatever it is, would have 1/2 or 3/2 spin--I do not know. If a positron is emitted, its spin would be -1/2 I think. That would make the new particle have 0 or 1 spin. The reaction of the positron and electron give photons with 0 spin. Bob . -Original Message- From: Jones Beene tt Sent: Wednesday, February 12, 2014 1:10 PM To: vortex-l@eskimo.com Subject: RE: [Vo]:a note from Dr. Stoyan Sargoytchev -Original Message- From: mix...@bigpond.com The most elegant answer begins with the obvious assertion that there are no gammas ab initio, which means that no reaction of the kind which your theory proposes can be valid because gammas are expected. Actually not only would I not expect to detect any gammas from a p-e-p reaction, I wouldn't expect to detect any energy at all. That's because the energy of the p-e-p reaction is normally carried away by the neutrino, which is almost undetectable. Hi, Not so - the reaction produces a positron, which annihilates with an electron producing 2 gammas. They net energy is over 1 MeV and easily detectable. Jones
RE: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
-Original Message- From: Bob Cook Jones--Bob Cook Here-- Can you show how the p-e-p reaction as you understand it conserves spin? Hi Bob, No one can adequately explain the many inconsistencies of this proposed route to gain in LENR. Were it not for the reputation of the proponents, it would be ignored. It is obviously a ploy to shoehorn a known fusion reaction into experimental results where it does not fit - and there is no way to do this without resort to fiction. In short, the proposed proton fusion to deuterium in LENR - cannot be the known version which eventually powers our sun. It could happen rarely but that is the best that can be said, without some minimum level of proof. BC: I would think that the newly fused particle, whatever it is, would have 1/2 or 3/2 spin--I do not know. If a positron is emitted, its spin would be -1/2 I think. That would make the new particle have 0 or 1 spin. This depends on what details, precisely, have been invented in place of the known reaction. The known proton reaction on the sun conserves spin with an extraordinarily rare beta decay, but that would be problematic for LENR in its rarity and also it would be obvious, and we know there is no beta decay, as there is no signature - thus an new kind of proton reaction has to be invented in order to match the experimental results. As you might guess, I am not enthusiastic about this route, but it may happen on occasion. Jones
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
I say in LENR that the double proton(spin 0) fusion happens then after this fusion occurs an electron(spin -1/2) capture comes next (reverse beta decay) the spin of 1/2 is removed by an electron neutrino.
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Axil, if the reaction involves the capture of an electron, and there are many available nearby, the positron - electron annihilation would not occur. This would explain why no 511 keV radiation is seen. Of course the energy escaping via the neutrino would be significant. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Feb 13, 2014 10:08 am Subject: Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev I say in LENR that the double proton(spin 0) fusion happens then after this fusion occurs an electron(spin -1/2) capture comes next (reverse beta decay) the spin of 1/2 is removed by an electron neutrino.
RE: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
-Original Message- From: Edmund Storms Bob, ES: these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear process. Ed, Bob I fully agree with this summary to the extent that when Ed's proposal is slanted slightly more into the f/H (fractional hydrogen) version, which is another way to look at a variation of Mills' CQM - and the deuteron loses energy as the ground state collapses, then it makes far more sense to imagine gammaless fusion as the QM result of the lowest state of two deuterons in the deep Dirac layer (DDL). This is the state that then goes to helium - via QM time reversal and recaptures the energy already expended in the prior shrinkage where UV photons have been shed all the way down. Very elegant ... at least with deuterium, this is very elegant. However, it is probably wise to acknowledge Mills' contribution and notably the argument does not explain more than necessary. The He nucleus is essentially paying back energy already shed so the lack of gamma can be explained away. This makes sense with bosons, but Pauli prevents this from happening with protons. There is no spin problem with deuterium going to helium. That would be another way of looking at Bob Cook's spin objection. However, one cannot transfer this lovely deuteron vehicle over to protons, without getting a speeding ticket. Instead something else must happen for energy to occur without the problems of spin and other major difficulties. The solution is obvious to me - and it is also unproved but suffers fewer objections when looking at experimental results. That something else, can indeed be based on the solar model and without a problem with spin. It is RPF which is also known as the diproton reaction. P+P - 2He There is no permanent fusion in RPF and no gamma. The two protons are ultimately unbound but this is not elastic collision and they do fuse for a tiny amount of time. QCD (the strong nuclear force) replaces the energy which has been shed on the atomic shrinkage, and that energy comes from excess proton mass. Again there is no proof of RPF either - but it comes with less baggage than P-e-P. That is the crux of the argument which Ed and I have several times per week - to the annoyance of others, no doubt. However, the ongoing argument has allowed both of us to hone our approaches- by focusing on the obvious weaknesses of the alternative. Jones
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
There is one complication that does not fall out of these various single track theories of LENR fusion. That complication is the Fission/Fusion reaction. What causes many protons to fuse with a high Z element like nickel? This process results in many and various secondary reaction trees producing one or more of the light elements including helium, boron, beryllium, and lithium to form coming out of one LENR reaction. Such a fission/fusion reaction can be explained by a complete suspension of the coulomb barrier in a volume of neighboring nuclei would allow a single nucleus to form with a very large and unsustainable amount of positive charge to be accumulated in one unstable large nucleus. When the Coulomb barrier is switched back on, the unstable nucleus with many excess protons would fission into many light atoms and one or two heaver atoms. In a series of independent secondary reactions, excess protons within the nuclei of this collection of multiple fission reaction products would then be converted to neutrons through electron capture after the primary fusion event has occurred. On Thu, Feb 13, 2014 at 10:07 AM, Axil Axil janap...@gmail.com wrote: I say in LENR that the double proton(spin 0) fusion happens then after this fusion occurs an electron(spin -1/2) capture comes next (reverse beta decay) the spin of 1/2 is removed by an electron neutrino.
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
In reply to Bob Cook's message of Thu, 13 Feb 2014 06:52:29 -0800: Hi, [snip] Ed--Bob here-- The protons are fermi particles with a spin of 1/2, so 2 protons would create a new particle spin of 1 plus the 1/2 from the electron for a total of +1-1/2. I think deuteron's are Bose particles with a spin of 0. Correct me if I am wrong. What happens to the excess spin? I think a spin of 1/2 actually means +- 1/2, i.e. in any given instance it may be either +1/2 or -1/2. D has a spin of 1. P - +1/2 P - +1/2 e - +1/2 neutrino -1/2 - D1 = Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
When the coulomb repulsion is removed from a pile of protons, they will attract each other and minimize energy by forming a pair with zero spin. In other words, large scale wide area charge screening is required to get protons to pair up based on opposite spins. This happens with electrons in superconductivity. On Thu, Feb 13, 2014 at 5:39 PM, Bob Cook frobertc...@hotmail.com wrote: Axil--Bob Cook here- That sounds possible from the spin part. How does the double proton form? I think the electrons and the two protons may all react in the QM Ni system at the same time ( 10 x e-18 sec.) Bob *From:* Axil Axil janap...@gmail.com *Sent:* Thursday, February 13, 2014 9:54 AM *To:* vortex-l vortex-l@eskimo.com *Subject:* Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev There is one complication that does not fall out of these various single track theories of LENR fusion. That complication is the Fission/Fusion reaction. What causes many protons to fuse with a high Z element like nickel? This process results in many and various secondary reaction trees producing one or more of the light elements including helium, boron, beryllium, and lithium to form coming out of one LENR reaction. Such a fission/fusion reaction can be explained by a complete suspension of the coulomb barrier in a volume of neighboring nuclei would allow a single nucleus to form with a very large and unsustainable amount of positive charge to be accumulated in one unstable large nucleus. When the Coulomb barrier is switched back on, the unstable nucleus with many excess protons would fission into many light atoms and one or two heaver atoms. In a series of independent secondary reactions, excess protons within the nuclei of this collection of multiple fission reaction products would then be converted to neutrons through electron capture after the primary fusion event has occurred. On Thu, Feb 13, 2014 at 10:07 AM, Axil Axil janap...@gmail.com wrote: I say in LENR that the double proton(spin 0) fusion happens then after this fusion occurs an electron(spin -1/2) capture comes next (reverse beta decay) the spin of 1/2 is removed by an electron neutrino.
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
In reply to Axil Axil's message of Thu, 13 Feb 2014 12:54:11 -0500: Hi, [snip] In a series of independent secondary reactions, excess protons within the nuclei of this collection of multiple fission reaction products would then be converted to neutrons through electron capture after the primary fusion event has occurred. You don't really need any weak force conversions in this scenario, because the initial heavy nucleus (e.g. Ni) already has a higher neutron:proton ratio than required by many light elements. (e.g. 60 Ni = 32N:28P). Many light elements have a 1:1 ratio, so there are 4 neutrons to spare from the Ni that can combine with 4 free protons to produce low mass elements with a 1:1 ratio (e.g. 24Mg, 28Si, 32S, 36Ar etc., all of which are stable isotopes.) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
With fusion/fission in mind and its need for neutrons, it might make things go smother in the reaction if the nickel's isotope neutron profile was increased say to Ni62 or Ni64. That will supply 34 or 36 extra neutrons to form light elements more readily. On Thu, Feb 13, 2014 at 9:04 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Thu, 13 Feb 2014 12:54:11 -0500: Hi, [snip] In a series of independent secondary reactions, excess protons within the nuclei of this collection of multiple fission reaction products would then be converted to neutrons through electron capture after the primary fusion event has occurred. You don't really need any weak force conversions in this scenario, because the initial heavy nucleus (e.g. Ni) already has a higher neutron:proton ratio than required by many light elements. (e.g. 60 Ni = 32N:28P). Many light elements have a 1:1 ratio, so there are 4 neutrons to spare from the Ni that can combine with 4 free protons to produce low mass elements with a 1:1 ratio (e.g. 24Mg, 28Si, 32S, 36Ar etc., all of which are stable isotopes.) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
On Thu, Feb 13, 2014 at 12:54 PM, Axil Axil janap...@gmail.com wrote: There is one complication that does not fall out of these various single track theories of LENR fusion. That complication is the Fission/Fusion reaction. What causes many protons to fuse with a high Z element like nickel? This process results in many and various secondary reaction trees producing one or more of the light elements including helium, boron, beryllium, and lithium to form coming out of one LENR reaction. Such a fission/fusion reaction can be explained by a complete suspension of the coulomb barrier in a volume of neighboring nuclei would allow a single nucleus to form with a very large and unsustainable amount of positive charge to be accumulated in one unstable large nucleus. When the Coulomb barrier is switched back on, the unstable nucleus with many excess protons would fission into many light atoms and one or two heaver atoms. Barriers in general perform two functions. They keep something out as well keeping something in. The coulomb barrier keeps protons apart, but could it be argued that they keep mass-energy locked inside. If so, if it were possible to switch the barrier off, that would be like opening a floodgate, so to speak. Fusion is likely to follow but it would not be a necessity for the release of energy. Harry
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Yes, 1/2 and the electron neutrino has that same spin. On Thu, Feb 13, 2014 at 9:59 PM, Bob Cook frobertc...@hotmail.com wrote: Axil--Bob Cook Here-- On Thu, Feb 13, 2014 at 10:07 AM, Axil Axil janap...@gmail.com wrote: I say in LENR that the double proton(spin 0) fusion happens then after this fusion occurs an electron (spin -1/2) capture comes next (reverse beta decay) the spin of 1/2 is removed by an electron neutrino. Axil I believe electrons have a spin of +1/2, not -1/2. The outgoing neutrino would have to have a -1/2 spin--maybe an electron anti neutrino or whatever its called--a positron neutrino. Bob
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Theory has to account for the DGT ash assay released in ICCF-17. http://cdn.coldfusionnow.org/wp-content/uploads/2013/04/2012-08-13-ICCF-17__Paper_DGTGx.pdf See Table 3. We learned from the MIT lecture that the coupling between the nuclear reaction site and the gamma/energy receiver must be strong. That connection must be an EMF coupling because gamma rays are EMF. It can't be charge, it might be photons, but it is most likely magnetic. On Thu, Feb 13, 2014 at 9:39 PM, H Veeder hveeder...@gmail.com wrote: On Thu, Feb 13, 2014 at 12:54 PM, Axil Axil janap...@gmail.com wrote: There is one complication that does not fall out of these various single track theories of LENR fusion. That complication is the Fission/Fusion reaction. What causes many protons to fuse with a high Z element like nickel? This process results in many and various secondary reaction trees producing one or more of the light elements including helium, boron, beryllium, and lithium to form coming out of one LENR reaction. Such a fission/fusion reaction can be explained by a complete suspension of the coulomb barrier in a volume of neighboring nuclei would allow a single nucleus to form with a very large and unsustainable amount of positive charge to be accumulated in one unstable large nucleus. When the Coulomb barrier is switched back on, the unstable nucleus with many excess protons would fission into many light atoms and one or two heaver atoms. Barriers in general perform two functions. They keep something out as well keeping something in. The coulomb barrier keeps protons apart, but could it be argued that they keep mass-energy locked inside. If so, if it were possible to switch the barrier off, that would be like opening a floodgate, so to speak. Fusion is likely to follow but it would not be a necessity for the release of energy. Harry
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
On Thu, Feb 13, 2014 at 6:39 PM, H Veeder hveeder...@gmail.com wrote: If so, if it were possible to switch the barrier off, that would be like opening a floodgate, so to speak. Fusion is likely to follow but it would not be a necessity for the release of energy. It seems to me that in order for Coulomb barrier to be switched off in this manner, it's not going to come for free. You're going to have to feed a ghastly amount of energy into the system somehow in order to effect this hypothetical flipping of the light switch. In other words, the switch is probably going to be unlike a normal light switch and instead will be very difficult to flip on and off. The energy has to come from somewhere in order to keep the books of the Bank of Heisenberg balanced. Eric
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
The energy for the switch comes from infrared heat concentrated by 20 orders of magnitude and positive feed back of nuclear energy from the nuclear reaction. The magnitude of the magnetic field that does the screening is between 10^5 and10^12 tesla. On Thu, Feb 13, 2014 at 10:19 PM, Eric Walker eric.wal...@gmail.com wrote: On Thu, Feb 13, 2014 at 6:39 PM, H Veeder hveeder...@gmail.com wrote: If so, if it were possible to switch the barrier off, that would be like opening a floodgate, so to speak. Fusion is likely to follow but it would not be a necessity for the release of energy. It seems to me that in order for Coulomb barrier to be switched off in this manner, it's not going to come for free. You're going to have to feed a ghastly amount of energy into the system somehow in order to effect this hypothetical flipping of the light switch. In other words, the switch is probably going to be unlike a normal light switch and instead will be very difficult to flip on and off. The energy has to come from somewhere in order to keep the books of the Bank of Heisenberg balanced. Eric
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Ed --Bob Here- I have assumed spin--angular momentum--is conserved. Are you saying forget about that conventional thinking--that angular momentum is not conserved in the lenr new nuclear process? Bob -Original Message- From: Bob Cook Sent: Thursday, February 13, 2014 6:52 AM To: vortex-l@eskimo.com Subject: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Ed--Bob here-- The protons are fermi particles with a spin of 1/2, so 2 protons would create a new particle spin of 1 plus the 1/2 from the electron for a total of +1-1/2. I think deuteron's are Bose particles with a spin of 0. Correct me if I am wrong. What happens to the excess spin? Bob -Original Message- From: Edmund Storms Sent: Thursday, February 13, 2014 6:30 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Bob, these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear process. Ed Storms Sent from my iPad On Feb 12, 2014, at 3:00 PM, Bob Cook frobertc...@hotmail.com wrote: Jones--Bob Cook Here-- Can you show how the p-e-p reaction as you understand it conserves spin? I would think that the newly fused particle, whatever it is, would have 1/2 or 3/2 spin--I do not know. If a positron is emitted, its spin would be -1/2 I think. That would make the new particle have 0 or 1 spin. The reaction of the positron and electron give photons with 0 spin. Bob . -Original Message- From: Jones Beene tt Sent: Wednesday, February 12, 2014 1:10 PM To: vortex-l@eskimo.com Subject: RE: [Vo]:a note from Dr. Stoyan Sargoytchev -Original Message- From: mix...@bigpond.com The most elegant answer begins with the obvious assertion that there are no gammas ab initio, which means that no reaction of the kind which your theory proposes can be valid because gammas are expected. Actually not only would I not expect to detect any gammas from a p-e-p reaction, I wouldn't expect to detect any energy at all. That's because the energy of the p-e-p reaction is normally carried away by the neutrino, which is almost undetectable. Hi, Not so - the reaction produces a positron, which annihilates with an electron producing 2 gammas. They net energy is over 1 MeV and easily detectable. Jones
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
One big limitation of gamma decay is for nuclear states of zero spin. This is the usual case in LENR. A state of zero spin cannot transition to another state of zero spin by emitting a photon. As discussed in chapter this violates conservation of angular momentum. But there are other ways that a nucleus can adjust energy besides emitting an electromagnetic photon. One way is by kicking an atomic electron out of the surrounding atom. This process is called internal conversion because the electron is outside the nucleus. It allows transitions between states of zero spin. For atoms, two-photon emission is a common way to achieve decays between states of zero angular momentum. However, for nuclei this process is less important because internal conversion usually works so well. Internal conversion is also important for other transitions. Gamma decay is slow between states that have little difference in energy and/or a big difference in spin. For such decays, internal conversion can provide a faster alternative. Internal conversion may be where the excess elections come from in LENR systems. Electrons could be carrying away spin in a zero spin nuclear reaction. I am sure that in a complex cluster fusion/fission reaction nature will balance the spin books correctly. On Thu, Feb 13, 2014 at 10:28 PM, Bob Cook frobertc...@hotmail.com wrote: Ed --Bob Here- I have assumed spin--angular momentum--is conserved. Are you saying forget about that conventional thinking--that angular momentum is not conserved in the lenr new nuclear process? Bob -Original Message- From: Bob Cook Sent: Thursday, February 13, 2014 6:52 AM To: vortex-l@eskimo.com Subject: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Ed--Bob here-- The protons are fermi particles with a spin of 1/2, so 2 protons would create a new particle spin of 1 plus the 1/2 from the electron for a total of +1-1/2. I think deuteron's are Bose particles with a spin of 0. Correct me if I am wrong. What happens to the excess spin? Bob -Original Message- From: Edmund Storms Sent: Thursday, February 13, 2014 6:30 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Bob, these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear process. Ed Storms Sent from my iPad On Feb 12, 2014, at 3:00 PM, Bob Cook frobertc...@hotmail.com wrote: Jones--Bob Cook Here-- Can you show how the p-e-p reaction as you understand it conserves spin? I would think that the newly fused particle, whatever it is, would have 1/2 or 3/2 spin--I do not know. If a positron is emitted, its spin would be -1/2 I think. That would make the new particle have 0 or 1 spin. The reaction of the positron and electron give photons with 0 spin. Bob . -Original Message- From: Jones Beene tt Sent: Wednesday, February 12, 2014 1:10 PM To: vortex-l@eskimo.com Subject: RE: [Vo]:a note from Dr. Stoyan Sargoytchev -Original Message- From: mix...@bigpond.com The most elegant answer begins with the obvious assertion that there are no gammas ab initio, which means that no reaction of the kind which your theory proposes can be valid because gammas are expected. Actually not only would I not expect to detect any gammas from a p-e-p reaction, I wouldn't expect to detect any energy at all. That's because the energy of the p-e-p reaction is normally carried away by the neutrino, which is almost undetectable. Hi, Not so - the reaction produces a positron, which annihilates with an electron producing 2 gammas. They net energy is over 1 MeV and easily detectable. Jones
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
http://en.wikipedia.org/wiki/Internal_conversion *Internal conversion* is a radioactive decayhttp://en.wikipedia.org/wiki/Radioactive_decayprocess where an excited nucleus http://en.wikipedia.org/wiki/Atomic_nucleus interacts electromagnetically http://en.wikipedia.org/wiki/Electromagnetism with an electron http://en.wikipedia.org/wiki/Electron in one of the lower atomic orbitals http://en.wikipedia.org/wiki/Atomic_orbital, causing the electron to be emitted (ejected) from the atom.[1]http://en.wikipedia.org/wiki/Internal_conversion#cite_note-Loveland-1Thus, in an internal conversion process, a high-energy electron is emitted from the radioactive atom, but not from a nucleon in the nucleus. Instead, the electron is ejected as a result of an interaction between the entire nucleus and an outside electron that interacts with it. For this reason, the high-speed electrons from internal conversion are not beta particleshttp://en.wikipedia.org/wiki/Beta_particle, since the latter come from beta decayhttp://en.wikipedia.org/wiki/Beta_decay, where they are newly created in the process. Since no beta decay takes place during internal conversion, the element atomic number does not change, and thus (as is the case with gamma decayhttp://en.wikipedia.org/wiki/Gamma_decay) no transmutation of one element to another is seen. However, since an electron is lost, an otherwise neutral atom becomes ionizedhttp://en.wikipedia.org/wiki/Ionization. Also, no neutrino is emitted during internal conversion. Internally converted electrons do not have the characteristic energetically spread spectrum of beta particles, which results from varying amounts of decay energy being carried off by the neutrino (or antineutrino) in beta decay. Internally converted electrons, which carry a fixed fraction of the characteristic decay energy, have a well-specified discrete energy. The energy spectrum of a beta particle is thus a broad hump, extending to a maximum decay energy value, while the spectrum of internally converted electrons has a sharp peak. If these internal conversion electrons are dipole electrons and have been absorbed inside the solation, the nuclear decay energy would be transferred directly into the solation and be thermalized by resonance. On Thu, Feb 13, 2014 at 10:44 PM, Axil Axil janap...@gmail.com wrote: One big limitation of gamma decay is for nuclear states of zero spin. This is the usual case in LENR. A state of zero spin cannot transition to another state of zero spin by emitting a photon. As discussed in chapter this violates conservation of angular momentum. But there are other ways that a nucleus can adjust energy besides emitting an electromagnetic photon. One way is by kicking an atomic electron out of the surrounding atom. This process is called internal conversion because the electron is outside the nucleus. It allows transitions between states of zero spin. For atoms, two-photon emission is a common way to achieve decays between states of zero angular momentum. However, for nuclei this process is less important because internal conversion usually works so well. Internal conversion is also important for other transitions. Gamma decay is slow between states that have little difference in energy and/or a big difference in spin. For such decays, internal conversion can provide a faster alternative. Internal conversion may be where the excess elections come from in LENR systems. Electrons could be carrying away spin in a zero spin nuclear reaction. I am sure that in a complex cluster fusion/fission reaction nature will balance the spin books correctly. On Thu, Feb 13, 2014 at 10:28 PM, Bob Cook frobertc...@hotmail.comwrote: Ed --Bob Here- I have assumed spin--angular momentum--is conserved. Are you saying forget about that conventional thinking--that angular momentum is not conserved in the lenr new nuclear process? Bob -Original Message- From: Bob Cook Sent: Thursday, February 13, 2014 6:52 AM To: vortex-l@eskimo.com Subject: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Ed--Bob here-- The protons are fermi particles with a spin of 1/2, so 2 protons would create a new particle spin of 1 plus the 1/2 from the electron for a total of +1-1/2. I think deuteron's are Bose particles with a spin of 0. Correct me if I am wrong. What happens to the excess spin? Bob -Original Message- From: Edmund Storms Sent: Thursday, February 13, 2014 6:30 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev Bob, these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
Bob wrote: | Axil I believe electrons have a spin of +1/2, not –1/2. The outgoing neutrino would have to have | a –1/2 spin—maybe an electron anti neutrino or whatever its called—a positron neutrino. ... Sorry for barging into this convo here, but I thought it would be useful to briefly describe Spin, so we are all on the same footing... Spin (or more specifically, intrinsic spin) of electrons are either +1/2 or –1/2. Remember the Stern-Gerlach Experiment[1]? ... Electrons (which are Fermions) are spin-1/2 particles, meaning that they can possess a spin of +1/2 (Spin Up) or –1/2 (Spin Down), and that they exist in a superposition of the |+1/2 and |-1/2 states. The magnitude 1/2 is the value (in units of “h”) of the projection of the spin along the + or - z axis... The classical analogue of intrinsic spin, is to envision the electron as a spherical top that is spinning and there is a precession around the z axis, at an angle such that the value of the spin along the z axis is exactly +/- h/2 ... ... Just as a side note, photons are bosons that have a spin of +/- 1 . You can break down light (photons) as a superposition of right and left circularly polarized light; Spin +1 Photons are termed Right Circularly Polarized Photons and Spin –1 Photons are the Left Circularly Polarized Critters[2]... FYI: You won’t find the term “Critters” in the Wikipedia Notes, sorry! ... Spinless Particles are Particles of Spin 0 ... ... So the spin of the electronneutrino pair is a superposition of the possible spin states... Carry on! - Mark Jurich [1] http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment [2] http://en.wikipedia.org/wiki/Photon_polarization#Angular_momentum_and_spin
Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev
In reply to Bob Cook's message of Thu, 13 Feb 2014 18:59:06 -0800: Hi, [snip] Axil I believe electrons have a spin of +1/2, not 1/2. The outgoing neutrino would have to have a 1/2 spinmaybe an electron anti neutrino or whatever its calleda positron neutrino. Bob AFAIK The spin quantum number is just the absolute value of the allowed instantaneous values. I.e. a particle with spin=1/2 could have an actual instantaneous spin of either +1/2 or -1/2. This goes for all Fermions. The actual value being an indication of the direction of the angular momentum vector, which could be either up or down at any given instant. Obviously each combination of particles has a combination of spins that has a minimum energy. For D this means both the proton and the neutron have the same angular momentum vectors, and hence the deuteron has spin 1, when it's energy is minimal. When you are making a deuteron from its parts, you are free to choose what instantaneous spin values each of the constituents has, so you could choose e.g. P (- 1/2) + N (- 1/2) - D (-1) (absolute value 1), or P (+1/2) + N (+ 1/2) - D (+1) (absolute value 1). The only difference between the two D's is that one is upside down relative to the other. However given thermal and zero point motion, the relative orientation of the two nuclei would rapidly change (independently), so it doesn't usually make much sense to speak of signed spins. My point is that the individual building blocks (Fermions) can have either + or - spins at the time that they combine, so determining the final spin is not just a matter of adding 1/2's. However you need to be careful that if you start with an odd number of Fermions, then you also end up with an odd number of Fermions (ditto for even). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
