Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Yes, that is the assumption. The issue is whether that assumption is valid. Can a large number of sinks participate in what is a random process such that they can share mass-energy? Can this collection remain intact for the time required for the process to go to completion. You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. This concept is in conflict with the laws of thermodynamics. Ed Storms Eric
Re: [Vo]:Christopher H. Cooper
This idea of fractioning the energy is similar to the scheme that Takahashi described in his TSC theory. Be-8* splits into two He* which decay to ground state with a little kinetic energy and a lot of low energy photons taken up by the lattice. Takahashi make the point that the coupling of the He* during decay to the lattice needs more work. This was in 2010. I do not know whether he has finished the coupling mechanism. I too was surprised at Haglestein's obvious neglect of the spin issues in the presentation of his theory at his 5th day session just recently. I have a similar idea for the Ni-H system--a proton-electron pair is positioned near a Ni on the surface or in a crack and the combined virtual particle (Ni Nucleus and proton-electron pair) reacts to form a new nucleus. The following may be what happens: The new Ni daughter decays as it will to copper or whatever. Ni-59 gives off it positron and the positron-electron initiation occurs with its .51 mev gammas. However with the proper temperature and black body background radiation for the Ni system the system favors reactions that distribute energy to the lattice or other Ni nuclei via spin coupling, and the Ni-59 may not form with the reaction ending up with Cu-59 directly, avoiding the positron associated radiation. The black body background radiation, having an entire spectrum of oscillating electro-magnetic fields in all directions, interact with Ni nuclei via their magnetic moments at the resonant frequencies making the release of many quanta possible from each excited Ni nucleus during the fractionation required by the main transition with its loss of mass. Here again the virtual Ni* first exists in a high spin energy state and decays via spin coupling to the other activated nuclei in the local system. A local temperature increase changes the reaction probability so that no more than one reaction occurs at a time and the system does not destroy itself. Other surfaces and cracks act the same way with a frequency controlled by the temperature. A time constant is associated with the change in the black body radiation spectrum and does not allow coupling of too many Nuclei at the same time given the constant removal of the resonant photons needed to activate the spin states of the Ni nuclei. Axil probably can add some obvious steps that I have omitted.(:-) Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 10:10 PM Subject: Re: [Vo]:Christopher H. Cooper On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Eric
Re: [Vo]:Christopher H. Cooper
The TSC theory has such a kinetic energy for the alphas identified Bob. - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 10:18 PM Subject: Re: [Vo]:Christopher H. Cooper I wrote: If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. This was stated incorrectly. To the extent that there is binding between the [dd]* state and one or more nearby ion cores, I assume the daughter 4He would be imparted kinetic energy in corresponding measure. So if this system is anywhere near what is really going on, we have a parameter that we can play with and adjust to match the actual kinetic energies that are seen (not very much). The more there is interaction with the electronic structure, and the less there is interaction with the ion cores, the less kinetic energy imparted to the daughter 4He. Eric
Re: [Vo]:Christopher H. Cooper
Eric, if the photons were to be emitted in random directions by the excited He4, then little kinetic energy would be imparted upon the nucleus.I suspect this is what you are referring to. Dave -Original Message- From: Eric Walker eric.wal...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Mar 6, 2014 1:19 am Subject: Re: [Vo]:Christopher H. Cooper I wrote: If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. This was stated incorrectly. To the extent that there is binding between the [dd]* state and one or more nearby ion cores, I assume the daughter 4He would be imparted kinetic energy in corresponding measure. So if this system is anywhere near what is really going on, we have a parameter that we can play with and adjust to match the actual kinetic energies that are seen (not very much). The more there is interaction with the electronic structure, and the less there is interaction with the ion cores, the less kinetic energy imparted to the daughter 4He. Eric
Re: [Vo]:Christopher H. Cooper
Mark-- Its hard to keep track of who says what in these threads. Sorry, Thanks for the correction. Bob - Original Message - From: MarkI-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 10:52 PM Subject: RE: [Vo]:Christopher H. Cooper Bob: It wasn't I, Jones referenced that paper in a posting dated: Tue 3/4/2014 8:11 AM. Credit where credit is due... -mark iverson -Original Message- From: Bob Cook [mailto:frobertc...@hotmail.com] Sent: Wednesday, March 05, 2014 1:22 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Christopher H. Cooper Robin-- If carbon nano tubes are the quantum cavity you refer to their dimensions can be greater--maybe up to 14 to 16 manometers. A mixture of sizes may allow absorption at may varied frequencies depending upon the temperature. The following paper addresses CNT size effects: http://arxiv.org/ftp/arxiv/papers/1202/1202.1328.pdf It was identified by MarkI-zero point two days ago. Bob - Original Message - From: mix...@bigpond.com To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 12:37 PM Subject: Re: [Vo]:Christopher H. Cooper In reply to Bob Cook's message of Tue, 4 Mar 2014 21:58:10 -0800: Hi, [snip] These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. ...this is on the order of hundreds of eV, perhaps coincidentally the same energy range one might also expect from either Hydrino formation or IRH. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 6:00 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Yes, that is the assumption. The issue is whether that assumption is valid. Can a large number of sinks participate in what is a random process such that they can share mass-energy? Can this collection remain intact for the time required for the process to go to completion. You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. This concept is in conflict with the laws of thermodynamics. Ed Storms Eric
Re: [Vo]:Christopher H. Cooper
Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 6:00 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Yes, that is the assumption. The issue is whether that assumption is valid. Can a large number of sinks participate in what is a random process such that they can share mass-energy? Can this collection remain intact for the time required for the process to go to completion. You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. This concept is in conflict with the laws of thermodynamics. Ed Storms Eric
Re: [Vo]:Christopher H. Cooper
Ed- The differential energy states of a nucleus associated with different spin states are not all that big. They come in units of Plank's constant. (Check out the discussion of spin in Wikipedia, https://www.google.com/webhp#q=nuclear+spin+quantum+number The following abstract of an article addresses the coupling between a nucleus and the electrons in a molecule--the Coulomb barrier does not come into play since the interaction is via the magnetic fields. You keep arguing about the Coulomb barrier--think magnetic coupling and spin coupling as the operative phenomena. http://scitation.aip.org/content/aip/journal/jcp/30/1/10.1063/1.1729860 The valence-bond theory for the contact electron-spin coupling of nuclear magnetic moments is used to calculate the proton-proton, proton-fluorine, and fluorine-fluorine coupling constants in ethanic and ethylenic molecules. A considerable simplification is introduced into the theory by approximations which reduce the problem to one involving only a small number of electrons and canonical structures. The agreement between calculated and experimental values is such as to demonstrate that the mechanism considered is the one of primary importance for the nuclear coupling in the compounds studied. Of particular interest is the theoretical confirmation of the observation that in ethylenic compounds the trans coupling between nuclei (HH, HF, FF) is considerably larger than cis coupling. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 6:00 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Yes
Re: [Vo]:Christopher H. Cooper
Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 6:00 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Yes, that is the assumption. The issue is whether that assumption is valid. Can a large number of sinks participate in what is a random process such that they can share mass-energy? Can this collection remain intact for the time required for the process to go to completion. You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. This concept is in conflict with the laws of thermodynamics. Ed Storms Eric
Re: [Vo]:Christopher H. Cooper
Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 6:00 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 11:10 PM, Eric Walker wrote: On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter
Re: [Vo]:Christopher H. Cooper
I do not see how the concept of spin has any relevance to the discussion. Both Rossi and DGT state that nickel isotopes of zero spin will react and nickel isotopes with non zero spins do not. This is both experimental data and an engineering requirement. The theory that purports to describe LENR must account for this spin based characterization. I will not accept a theory that does not explain spin as a factor in the LENR reaction. On Thu, Mar 6, 2014 at 2:38 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Thursday, March 06, 2014 10:49 AM *Subject:* Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although not of any practical intensity. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Thursday, March 06, 2014 6:00 AM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5
Re: [Vo]:Christopher H. Cooper
OK, Axil. We have an impasse. I will not accept any claim made by DGT unless the study is described in detail and can be evaluated. People seem to accept their statements without question. Where is the basic skepticism typical of all good science? In addition, I do not believe the Ni has any direct role in the nuclear process. The heat is only generated by fusion of H as I have described. So we have no more to discuss. Ed Storms On Mar 6, 2014, at 12:49 PM, Axil Axil wrote: I do not see how the concept of spin has any relevance to the discussion. Both Rossi and DGT state that nickel isotopes of zero spin will react and nickel isotopes with non zero spins do not. This is both experimental data and an engineering requirement. The theory that purports to describe LENR must account for this spin based characterization. I will not accept a theory that does not explain spin as a factor in the LENR reaction. On Thu, Mar 6, 2014 at 2:38 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple
Re: [Vo]:Christopher H. Cooper
- Original Message - From: Axil Axil To: vortex-l Sent: Thursday, March 06, 2014 11:49 AM Subject: Re: [Vo]:Christopher H. Cooper I do not see how the concept of spin has any relevance to the discussion. Both Rossi and DGT state that nickel isotopes of zero spin will react and nickel isotopes with non zero spins do not. This is both experimental data and an engineering requirement. The theory that purports to describe LENR must account for this spin based characterization. I will not accept a theory that does not explain spin as a factor in the LENR reaction. On Thu, Mar 6, 2014 at 2:38 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system. Nano particles, although not as large as a crystals, are also probably a QM system with many atoms. All molecules are QM systems and when close together may have various coupling mechanisms although
Re: [Vo]:Christopher H. Cooper
Ed -- I find that I agree with Axil this time. The Pauli Exclusion Principle is a key theory of physics and chemistry. Without it matter would be unstable. Electrons would collapse to the attraction of the protons and there would be no electronic structure of a molecule--no molecules period. Spin is a characteristic of primary particles and the quarks that make up compound particles. Particles with zero spin seem to have a certain characteristic--they have no potential energy associated with angular momentum. Photons have positive spin and angular momentum pointing in the direction of their motion. Reactions of photons with other particles must conserve angular momentum. I think this is a key restriction on various chemical reactions that are light sensitive. I do not agree that it is warranted to disregard a key parameter of particles in the consideration of LENR unless of course there are experiments that indicate the parameter is not real. It is like saying electrons do not have a charge or mass or that electrons are not real even though much evidence supports their reality. Bob - Original Message - From: Axil Axil To: vortex-l Sent: Thursday, March 06, 2014 11:49 AM Subject: Re: [Vo]:Christopher H. Cooper I do not see how the concept of spin has any relevance to the discussion. Both Rossi and DGT state that nickel isotopes of zero spin will react and nickel isotopes with non zero spins do not. This is both experimental data and an engineering requirement. The theory that purports to describe LENR must account for this spin based characterization. I will not accept a theory that does not explain spin as a factor in the LENR reaction. On Thu, Mar 6, 2014 at 2:38 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry
Re: [Vo]:Christopher H. Cooper
Ed, Axil etak-- The following link is a good tutorial on nuclear/electronic spin coupling and work to understand the mechanism. http://gabriel.physics.ucsb.edu/~balents/projects/Central-spin.html Bob - Original Message - From: Axil Axil To: vortex-l Sent: Thursday, March 06, 2014 11:49 AM Subject: Re: [Vo]:Christopher H. Cooper I do not see how the concept of spin has any relevance to the discussion. Both Rossi and DGT state that nickel isotopes of zero spin will react and nickel isotopes with non zero spins do not. This is both experimental data and an engineering requirement. The theory that purports to describe LENR must account for this spin based characterization. I will not accept a theory that does not explain spin as a factor in the LENR reaction. On Thu, Mar 6, 2014 at 2:38 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, let me see if I can simplify the issue. For fusion to occur, two D must get close enough for the two nuclei to combine. This process is prevented by the Coulomb barrier, which requires energy to overcome. A static magnetic field does not supply energy. Once the two nuclei combine, the mass-energy must be dissipated. This can be done by fragmentation of the resulting nucleus, i.e. hot fusion, or by release of energy as many photons. Observation places a limit on the energy the photons can have. You bring spin into the discussion. The spin state has a limit to how much energy it can hold. In addition, if spin is accepted as an actual rotation about an axis, creating this spin requires the law of conservation of momentum be considered and a process needs to be identified that can apply a force to the particle such that it spins rather than moves in a line. I see no way for this to happen in your description. If spin is viewed only as another variable in equations to allow them to fit data, then I do not know how to evaluate your claim. We know that all energy that is emitted with the alpha particle eventually appears as heat and the helium ends up with its normal spin state. Therefore, energy imagined to exist as spin acts exactly like translational energy in the real world. Therefore, I do not see how the concept of spin has any relevance to the discussion. Ed Storms On Mar 6, 2014, at 12:19 PM, Bob Cook wrote: Ed--The ionic bonds of a host lattice are not the issue when it comes to the transfer of energy in small bits. Its whether or not the small bits can find a host in another nucleus of the QM system or in the spin state of an electron in that lattice. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Thursday, March 06, 2014 10:49 AM Subject: Re: [Vo]:Christopher H. Cooper Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV, yet you propose to require this bonding to share and dissipate energy at the MeV level within a cluster of atoms. Only in the nucleus itself is this level of bonding and interaction available. Atoms are not attached to each other with the necessary force to share and transmit this level of energy. In addition, for nuclear interaction to take place, the Coulomb barrier must be overcome. This barrier is real and its magnitude is well known and far in excess of any source of energy available in a chemical system. LENR requires a new and so far unknown process to do this. I see no effort to effectively identify this process. Simply applying IF statements is not a solution. Simply applying QM using equations containing arbitrary assumptions does not change how chemical systems are known to behave. The people discussing these issues on Vortex seem to be in a different reality than the one I have occupied for over 60 years of scientific study of LENR, chemistry, and physics. Any imagined or assumed process described in the modern literature seems to be as important as what has been observed and accepted in science for the last 100 years. Any new observation in physics seems to be fair game as an explanation of LENR whether it has any real world support of not. In fact, many of the papers used as justification for the proposals are simply based on more theory and assumptions. Ed Storms On Mar 6, 2014, at 8:54 AM, Bob Cook wrote: Ed You said: You must assume that a nuclear energy state can form between a large number of atoms in a chemical system. Yes I do assume that. Crystals like in Pd metal I would consider to be one QM system as long as long as the ionic chemical bonds hold the atoms together. The nuclear magnetic moments of a crystal clearly couple with the electrons in the system
Re: [Vo]:Christopher H. Cooper
On Thu, Mar 6, 2014 at 7:24 AM, David Roberson dlrober...@aol.com wrote: Eric, if the photons were to be emitted in random directions by the excited He4, then little kinetic energy would be imparted upon the nucleus.I suspect this is what you are referring to. Perhaps; I'm not sure. I had in mind something like this: an excited [dd]* or [pNi]* state is like a capacitor that will discharge. In a vacuum it will discharge either by emitting a gamma, which takes a while, or by breaking apart, which happens more quickly. But at the surface of or within a few layers of a metal like nickel, there is an environment rich in electrostatic charge, provided by the electrons and the lattice sites (sometimes called ion cores, since they're positively charged). If the [pNi]* excited state discharges like a capacitor within this environment with all of the electrostatic charge, I'm assuming there will be electromagnetic coupling between the excited state and the electrostatic sources, in the sense that they will form a system and interact. There will be a strong repulsive force given off by the [pNi]* state as it decays to whatever it decays to (for example, 63Cu), and this repulsive force will push away the nearby electrons and ion cores. The more it pushes away the electrons, the more you'll get a bath of photons. The more it pushes away the ion cores, the more kinetic energy will be imparted to the daughter of the decay. This is because electrons are nearly massless, and so receive the majority of the impulse, while the ion cores have a mass nearly equal to the daughter, and so push back on the resulting daughter much more than the electrons. I am not yet sure how the electromagnetic interaction relates to spin coupling, although I think Bob sees something in this. Eric
Re: [Vo]:Christopher H. Cooper
On Thu, Mar 6, 2014 at 10:49 AM, Edmund Storms stor...@ix.netcom.comwrote: Bob, you fail to take into account the known and well documented bonding energy that can exist in a chemical system. This bonding is limited to no more than about 10 eV ... Is this the energy required for a dislocation? Wouldn't it be higher? Eric
RE: [Vo]:Christopher H. Cooper
From: Eric Walker * This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. Sure… My working assumption is that both NiH and PdD (as well as W, Ti, etc.) involve fusion in some way. Both are without gammas This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. And the willingness of LENR proponents to jump on an unjustified assumption at the very core of the phenomenon - is most of the problem, since after many years or holding onto this assumption (which was never more than a placeholder for real understanding) they fail to realize that the assumption was never valid in the first place. Known nuclear reactions produce gammas or at least bremsstrahlung. That is fact. Since neither are observed, in a significant fashion, in LENR, the best assumption is that no known nuclear reaction can be involved. That should be obvious, but somehow it is glossed over. Of course, there are those who will say that they intended all along that the LENR reaction was not the same as the thermonuclear variety. If we start with that premise of a novel nuclear reaction – which is almost unassailable as the most logical premise on which to build, then it is much easier to understand that IF there are novel nuclear reactions at play, and especially QM versions and variants of known thermonuclear reactions, then there could be many reaction in LENR since there are at least a dozen types of nuclear reactions (over and above fission and fusion, some are included at the end since many vorticians are unaware of this important detail) which could be amenable to QM. A QM version of a rare reaction, which has been either unknown in the past or underappreciated (the diproton reaction is known, but underappreciated), sets the groundwork for my belief that then there is no logical reason to suggest there is only one novel reaction in LENR Jones In addition to fission and fusion, there are other possible rare nuclear reactions which could show up in LENR as QM variants of the already rare reactions. These are listed in order of relative ease (perceived ease) of having QM variants based on nuclear tunneling. There are surely more so please add to the list. 1) Spallation — a nucleus is hit by a particle with sufficient energy to knock out two or more smaller fragments. This could be called partial fission. A) Neutron spallation (as opposed to neutron induced fission) B) Proton spallation C) Electron spallation D) Photon spallation 2) Induced photon emission (often call induced gamma emission or IGE which is a subset of IPE). This is the typical “halo” nucleus reaction. 3) Spontaneous fission (placeholder for an unknown type of fission which simply happens). This may also be a halo nucleus reaction but it has been too rate to quantify. 4) Double or triple alpha emission, or decay. This is a stimulated decay. 5) Ternary fission (can be induced or not) and similar to 4) 6) Rare types of beta or positron decay (there are many types of beta decay that involve a different route or variation than a hot electron emission BTW once we agree that LENR is a QM version of a known nuclear reaction which is not normal fission or fusion, then the rarity of the known version is immaterial when we move over to quantum mechanics. PLUS the QM version could be related to a completely unknown prior reaction. attachment: winmail.dat
Re: [Vo]:Christopher H. Cooper
On Wed, Mar 5, 2014 at 6:15 AM, Jones Beene jone...@pacbell.net wrote: From: Eric Walker * This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. Sure… My working assumption is that both NiH and PdD (as well as W, Ti, etc.) involve fusion in some way. Both are without gammas This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. When we come across an anomaly whose possible explanation is equivocal (i.e., we don't have enough data to say one way or the other), we have the option of adopting a working assumption vis-a-vis that anomaly. By working assumption I'm thinking of a placeholder of some kind to stand in for whatever the explanation ends up being when we have sufficient experimental data to remove the ambiguity in the data. Working assumptions are something we can throw away later when more evidence comes to light. In this sense they're not a blind assumptions, implicitly adopted. They're adopted consciously and tentatively. In this case I'm working from these details: - Skillful experimentalists have observed in the PdD system a correlation between 4He levels and excess heat that strongly suggests that there is d+d fusion going on, somehow. - The Elforsk team saw what they believe to be heat above and beyond what can be produced by a chemical reaction in Rossi's NiH system. - Other experimentalists looking at the NiH system have also seen what they believe to be heat above what can be produced by a chemical reaction. Now here are my working assumptions: - There's only two ways to get energy out of a system above a chemical reaction, and that's through fission or fusion. There is no other supra-chemical means of getting energy out of a system. - There's no reason to go for two different sets of explanations to explain an excess heat anomaly when the evidence is equivocal on what's going on. My own bias is towards one explanation, so I go with my bias. - There is a mechanism that has not yet been carefully characterized in which fusion can proceed without penetrating radiation. From these observations and working assumptions taken together I infer, consciously, aware of the implications, that what's going on in the NiH system is some kind of fusion. A conclusion I am quite happy with for the moment given my working assumptions and starting point. Eric
RE: [Vo]:Christopher H. Cooper
From: Eric Walker This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. Sorry to have made this blanket statement in regard to your prior post specifically, Eric, since it is a generic criticism to many of the posts on Vortex and not personal - but… No, it’s not opinion when 100% of the available proof is on your side. It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. Since 1989, there have been assertions and claims, but they are only assertions, that LENR is proof of a gammaless nuclear reaction, but that is circular logic. LENR is proof of a thermal anomaly, and helium is seen in the ash, but that is all that can be said logically. Even if helium is seen in proportion to the excess heat, which is in dispute, that does not raise LENR to the level of a known fusion reaction which is gammaless, at least not so long as there are other valid explanations. To be raised to this level the claimant must also demonstrate in an experiment not involving LENR that 24 MeV gammas can be completely suppressed by any mechanism. Any mechanism will suffice. This has not been done, even with 1 MeV gammas since there is always leakage – even with lead shielding. By definition, cold fusion cannot be the same known reaction as deuterium fusion to helium, which was known prior to 1989 - if it is gammaless – unless and until it can be shown that there is a real physical mechanism for not only for suppressing gammas, but for suppressing 100% of them without exception. How is that opinion? Jones
Re: [Vo]:Christopher H. Cooper
On Wed, Mar 5, 2014 at 8:21 AM, Jones Beene jone...@pacbell.net wrote: No, it’s not opinion when 100% of the available proof is on your side. That's a pretty strong assessment of the merits of your position. :) It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. ... By definition, cold fusion cannot be the same known reaction as deuterium fusion to helium, which was known prior to 1989 - if it is gammaless – unless and until it can be shown that there is a real physical mechanism for not only for suppressing gammas, but for suppressing 100% of them without exception. Does either of these statements contradict anything I've said or assumed? I hope my outlining of my assumptions demonstrates that I do not have the typical fusion branches in mind. I have the general notion of two nucleons combining to create a larger nucleon with less mass and a release of energy. The branches would need to be different. Eric
Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 9:21 AM, Jones Beene wrote: From: Eric Walker This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. Sorry to have made this blanket statement in regard to your prior post specifically, Eric, since it is a generic criticism to many of the posts on Vortex and not personal - but… No, it’s not opinion when 100% of the available proof is on your side. It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. Jones, this statement is not correct. LENR emits photons. These photons are not as energetic as those produced by many normal nuclear reactions, hence most do not escape the apparatus. Nevertheless, the mass-energy is released as photons as is normal and is required of a nuclear reaction. The only unknown is the mechanism causing this process. Obviously, a process is required that does not operate during hot fusion. Nevertheless, nuclear products are formed that can only result from a nuclear reaction having the known and well understood consequences. Ed Storms
RE: [Vo]:Christopher H. Cooper
Eric, Again, I apologize for any inference that this is personal or related precisely to your prior post. My comment was intended to show only that: 1)LENR is NOT a known nuclear fusion reaction since all known fusion reactions produce gamma radiation. 2)Since there is a novel reaction at play, then there is no valid reason to suggest that there is only one novel reaction which is possible in LENR - other than some vague notion of parsimony which is always wrong when it comes to QM. 3)If we admit that QM should in principle allow for novel variants of all rare nuclear reactions, and there are a dozen or more of these novel reactions, then there are at least that many different types of LENR reactions which are possible, and certainly not limited to any constraint based on historical precedence- i.e. Pd-D coming along first historically in the progression towards Ni-H. In short – all of this goes back to the ongoing debate on Vortex: on the question that there is only one basic type of reaction, and that Ni-H is a type of Pd-D. This is complete nonsense IMO especially when we reduce the argument to the fact that even Pd-D is NOT THE KNOWN fusion reaction of nuclear science prior to 1989. With QM in the picture – all bets are off, and instead of parsimony we expect added complexity. QM is anti-Ockham. Most disturbing for the theorist is that in any experiment there could be several if not many types of gainful QM reactions taking place at the same time. From: Eric Walker Does either of these statements contradict anything I've said or assumed? I hope my outlining of my assumptions demonstrates that I do not have the typical fusion branches in mind. I have the general notion of two nucleons combining to create a larger nucleon with less mass and a release of energy. The branches would need to be different. Eric
RE: [Vo]:Christopher H. Cooper
From: Edmund Storms LENR emits photons. These photons are not as energetic as those produced by many normal nuclear reactions, hence most do not escape the apparatus. Where is the documented proof and spectra of these photons?
Re: [Vo]:Christopher H. Cooper
Ed and Jones-- The definition of gamma emission is cropping up again. Jones I assume you mean any electromagnetic radiation that stems from a nuclear transition of some sort. This would include low energy photons that Ed describes as well as nuclear magnetic resonance transitions to higher spin states activated by external oscillating magnetic fields and subsequent radiation emitted by the excited nucleus. However, I assume gamma radiation would NOT include small amounts of electromagnetic radiation released by electrons in a metal lattice that share spin energy/angular momentum with near by nuclei, but are not part of the nuclei. Bob From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 8:34 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 9:21 AM, Jones Beene wrote: From: Eric Walker This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. Sorry to have made this blanket statement in regard to your prior post specifically, Eric, since it is a generic criticism to many of the posts on Vortex and not personal - but… No, it’s not opinion when 100% of the available proof is on your side. It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. Jones, this statement is not correct. LENR emits photons. These photons are not as energetic as those produced by many normal nuclear reactions, hence most do not escape the apparatus. Nevertheless, the mass-energy is released as photons as is normal and is required of a nuclear reaction. The only unknown is the mechanism causing this process. Obviously, a process is required that does not operate during hot fusion. Nevertheless, nuclear products are formed that can only result from a nuclear reaction having the known and well understood consequences. Ed Storms
RE: [Vo]:Christopher H. Cooper
From: Bob Cook The definition of gamma emission is cropping up again. Jones I assume you mean any electromagnetic radiation that stems from a nuclear transition of some sort. The trend in science, and even in physics, is to avoid the origin, since it cannot always be known, and to use the frequency/wavelength only. Wiki sez X-radiation is in the range of 0.01 to 10 nanometers, (corresponding to frequencies in the range 30 petaHertz to 30 exaHertz) and energies in the range 100 eV to 100 keV. Personally, the range of 1000 eV to 200 keV seems more logical to me for x-rays, since EUV better describes the range of 100 eV to 1000 eV. But anyway - if there is decent proof of copious LENR photons of something greater than 100 eV, I would like to see that data. I suspect that there is not good data, since it would mean that a quartz LENR cell would be emitting so much UV that the experimenter would have a severe sunburn in a few hours. Even if it favors Mills' theory more than LENR - photons in the EUV or x-ray spectrum would be an important detail to mix into the picture - if they were substantial and with reliable proof and especially if they can be differentiated from Mills. Jones
Re: [Vo]:Christopher H. Cooper
Eric-- You wrote: I have the general notion of two nucleons combining to create a larger nucleon with less mass and a release of energy. The branches would need to be different. I have had a similar notion relative to the Pd-D system. Specifically two D come together to form a virtual excited He particle with high spin energy that fractionates its high spin energy to electrons and other coupled particles to attain the desired low energy associated with the stable He particle. Only many low energy photons are involved. to balance the lower mass of the He compared to the starting material. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 8:26 AM Subject: Re: [Vo]:Christopher H. Cooper On Wed, Mar 5, 2014 at 8:21 AM, Jones Beene jone...@pacbell.net wrote: No, it’s not opinion when 100% of the available proof is on your side. That's a pretty strong assessment of the merits of your position. :) It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. ... By definition, cold fusion cannot be the same known reaction as deuterium fusion to helium, which was known prior to 1989 - if it is gammaless – unless and until it can be shown that there is a real physical mechanism for not only for suppressing gammas, but for suppressing 100% of them without exception. Does either of these statements contradict anything I've said or assumed? I hope my outlining of my assumptions demonstrates that I do not have the typical fusion branches in mind. I have the general notion of two nucleons combining to create a larger nucleon with less mass and a release of energy. The branches would need to be different. Eric
Re: [Vo]:Christopher H. Cooper
Jones-- There are nuclear events that occur without emission of gammas. The decay of Ni-59 is an example. What's different in Ni-59 with respect to most other radioactive decay? Bob - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 8:21 AM Subject: RE: [Vo]:Christopher H. Cooper From: Eric Walker This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. Sorry to have made this blanket statement in regard to your prior post specifically, Eric, since it is a generic criticism to many of the posts on Vortex and not personal - but… No, it’s not opinion when 100% of the available proof is on your side. It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. Since 1989, there have been assertions and claims, but they are only assertions, that LENR is proof of a gammaless nuclear reaction, but that is circular logic. LENR is proof of a thermal anomaly, and helium is seen in the ash, but that is all that can be said logically. Even if helium is seen in proportion to the excess heat, which is in dispute, that does not raise LENR to the level of a known fusion reaction which is gammaless, at least not so long as there are other valid explanations. To be raised to this level the claimant must also demonstrate in an experiment not involving LENR that 24 MeV gammas can be completely suppressed by any mechanism. Any mechanism will suffice. This has not been done, even with 1 MeV gammas since there is always leakage – even with lead shielding. By definition, cold fusion cannot be the same known reaction as deuterium fusion to helium, which was known prior to 1989 - if it is gammaless – unless and until it can be shown that there is a real physical mechanism for not only for suppressing gammas, but for suppressing 100% of them without exception. How is that opinion? Jones
Re: [Vo]:Christopher H. Cooper
Eric- add this to my previous email. The D particles arrive together with antiparallel spins (a Bose particle) and in the process of forming a He, each assumes a high spin state--one negative and one positive--conserving angular momentum and producing zero angular momentum in the virtual He particle. The spin coupling to the electronic structure is the unknown sauce. Bob - Original Message - From: Bob Cook To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 9:39 AM Subject: Re: [Vo]:Christopher H. Cooper Eric-- You wrote: I have the general notion of two nucleons combining to create a larger nucleon with less mass and a release of energy. The branches would need to be different. I have had a similar notion relative to the Pd-D system. Specifically two D come together to form a virtual excited He particle with high spin energy that fractionates its high spin energy to electrons and other coupled particles to attain the desired low energy associated with the stable He particle. Only many low energy photons are involved. to balance the lower mass of the He compared to the starting material. Bob
RE: [Vo]:Christopher H. Cooper
From: Bob Cook There are nuclear events that occur without emission of gammas. The decay of Ni-59 is an example. What's different in Ni-59 with respect to most other radioactive decay? Bob - It is not gammas alone which are absent in LENR - but gammas and bremsstrahlung… which of course is lower energy - x-ray level and EUV but still measurable. In these posts - we do not always type in both words in every post - since the latter is so damn hard to spell, but when you have one MeV in excess energy - as does Ni-59, you should have measurable radiation and especially when the reactor is opened, it will be noticed due to the rather long half-life. However, of all the possible novel Ni-H reactions which could be proposed – a QM variation on this one would be a decent fit – as EC would be easier to hide. Substantial cobalt in the ash – instead of copper - would be proof. One could imagine a DDL of the H atom using its reduced electron orbital to tunnel into Ni-58, taking the nucleus to Ni-59 in an energy-deficient way if the spin problem can be dealt with, as if it were an energy-deficient neutron, and having only about 100-200 keV of excess energy which would almost fit the Rossi evidence if the half-life was reduced. The amount of cobalt which should be in the ash is predictable. Is it there?
Re: [Vo]:Christopher H. Cooper
Jones, bremsstrahlung or slowing down radiation is not produced by photons. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. Ed Storms On Mar 5, 2014, at 12:04 PM, Jones Beene wrote: From: Bob Cook There are nuclear events that occur without emission of gammas. The decay of Ni-59 is an example. What's different in Ni-59 with respect to most other radioactive decay? Bob - It is not gammas alone which are absent in LENR - but gammas and bremsstrahlung… which of course is lower energy - x-ray level and EUV but still measurable. In these posts - we do not always type in both words in every post - since the latter is so damn hard to spell, but when you have one MeV in excess energy - as does Ni-59, you should have measurable radiation and especially when the reactor is opened, it will be noticed due to the rather long half-life. However, of all the possible novel Ni-H reactions which could be proposed – a QM variation on this one would be a decent fit – as EC would be easier to hide. Substantial cobalt in the ash – instead of copper - would be proof. One could imagine a DDL of the H atom using its reduced electron orbital to tunnel into Ni-58, taking the nucleus to Ni-59 in an energy-deficient way if the spin problem can be dealt with, as if it were an energy-deficient neutron, and having only about 100-200 keV of excess energy which would almost fit the Rossi evidence if the half-life was reduced. The amount of cobalt which should be in the ash is predictable. Is it there?
Re: [Vo]:Christopher H. Cooper
Jones-- The fact that Rossi has not, to my knowledge, reported anything except Cu. I have not heard of Co ash from anyone. They may be holding back key information. Bob - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 11:04 AM Subject: RE: [Vo]:Christopher H. Cooper From: Bob Cook There are nuclear events that occur without emission of gammas. The decay of Ni-59 is an example. What's different in Ni-59 with respect to most other radioactive decay? Bob - It is not gammas alone which are absent in LENR - but gammas and bremsstrahlung… which of course is lower energy - x-ray level and EUV but still measurable. In these posts - we do not always type in both words in every post - since the latter is so damn hard to spell, but when you have one MeV in excess energy - as does Ni-59, you should have measurable radiation and especially when the reactor is opened, it will be noticed due to the rather long half-life. However, of all the possible novel Ni-H reactions which could be proposed – a QM variation on this one would be a decent fit – as EC would be easier to hide. Substantial cobalt in the ash – instead of copper - would be proof. One could imagine a DDL of the H atom using its reduced electron orbital to tunnel into Ni-58, taking the nucleus to Ni-59 in an energy-deficient way if the spin problem can be dealt with, as if it were an energy-deficient neutron, and having only about 100-200 keV of excess energy which would almost fit the Rossi evidence if the half-life was reduced. The amount of cobalt which should be in the ash is predictable. Is it there?
RE: [Vo]:Christopher H. Cooper
From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. Where is the documentation? Jones attachment: winmail.dat
Re: [Vo]:Christopher H. Cooper
Jones-- Alphas would not produce Bremstrallung, if they gain no kinetic energy in being produced. Energy in the form of angular momentum would not produce the B word. Bob Bob - Original Message - From: Jones Beene jone...@pacbell.net To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 11:28 AM Subject: RE: [Vo]:Christopher H. Cooper From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. Where is the documentation? Jones
Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
In reply to Bob Cook's message of Tue, 4 Mar 2014 21:58:10 -0800: Hi, [snip] These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. ...this is on the order of hundreds of eV, perhaps coincidentally the same energy range one might also expect from either Hydrino formation or IRH. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Christopher H. Cooper
-Original Message- From: Bob Cook Jones-- Alphas would not produce Bremstrallung, if they gain no kinetic energy in being produced. Energy in the form of angular momentum would not produce the B word. Bob- That much is almost true, but you overlook the 800 pound gorilla in the corner - TSC. Maybe you are unfamiliar with it. This happens to be one of the more credible versions of Pd-D in my opinion since no gamma is expected - yet the kinetics of the reaction produce bremsstrahlung. Apparently many in Japan think that Takahashi's version makes the most sense also but AFAIK he has not documented the spectra of the B word. He postulates that a BEC of deuterons is more likely to produce 2 energetic alphas from 4 deuterons than the alternative situation. TSC produces 8Be first which decays into 4He + 4He liberating up to 47.6 MeV of kinetic energy, no gamma and the reaction is known in cosmology from supernova - so it is not an invention. As an alternative of D+D - He, which has to overcome the huge problem of least-favored-channel, TSC is a superior Point of View to many. It is a very strong argument since no gammas are expected - all kinetic. All B-word. Jones
Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 1:45 PM, Jones Beene wrote: -Original Message- From: Bob Cook Jones-- Alphas would not produce Bremstrallung, if they gain no kinetic energy in being produced. Energy in the form of angular momentum would not produce the B word. Bob- That much is almost true, but you overlook the 800 pound gorilla in the corner - TSC. Maybe you are unfamiliar with it. This happens to be one of the more credible versions of Pd-D in my opinion since no gamma is expected - yet the kinetics of the reaction produce bremsstrahlung. Apparently many in Japan think that Takahashi's version makes the most sense also but AFAIK he has not documented the spectra of the B word. He postulates that a BEC of deuterons is more likely to produce 2 energetic alphas from 4 deuterons than the alternative situation. TSC produces 8Be first which decays into 4He + 4He liberating up to 47.6 MeV of kinetic energy, no gamma and the reaction is known in cosmology from supernova - so it is not an invention. As an alternative of D+D - He, which has to overcome the huge problem of least-favored-channel, TSC is a superior Point of View to many. Jones, Hagelstein showed that this proposed reaction was not consistent with what is observed. As a result, Takahashi changed his explanation to claim that Be8 formed and dissipated most of the mass-energy as photons before it split into two alpha. Unfortunately, this additional required feature subtracts from the plausibility of the basic idea. Ed Storms 1.Hagelstein, P.I., Secondary Neutron Yield in the Presence of Energetic Alpha Particles in PdD. J. Cond. Matter Nucl. Sci., 2010. 3: p. 41-49. 2.Hagelstein, P.I., On the connection between Ka X-rays and energetic alpha particles in Fleischmann–Pons experiments. J. Cond. Matter Nucl. Sci., 2010. 3: p. 50-58. 3.Hagelstein, P.L., Simple Parameterizations of the Deuteron–Deuteron Fusion Cross Sections. J. Cond. Matter Nucl. Sci., 2010. 3: p. 31-34. 4.Hagelstein, P.L., Neutron Yield for Energetic Deuterons in PdD and in D2O. J. Cond. Matter Nucl. Sci., 2010. 3: p. 35-40. It is a very strong argument since no gammas are expected - all kinetic. All B-word. Jones
Re: [Vo]:Christopher H. Cooper
Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Ed: Things in LENR are more complicated than you are stating. Sometimes gammas are produced in LENR and most times it isn't. The cause of Gamma thermalization is connected with a nuclear based positive feedback loop in the energy conversion/thermalization mechanism. But LENR can happen even when only gammas are produced. On Wed, Mar 5, 2014 at 4:09 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 12:06 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
There is more than enough evidence to zero in on the prime cause of LENR both in orthodox science and LENR data. You have not put the work into utilizing all the data that is available. On Wed, Mar 5, 2014 at 10:50 AM, Eric Walker eric.wal...@gmail.com wrote: On Wed, Mar 5, 2014 at 6:15 AM, Jones Beene jone...@pacbell.net wrote: From: Eric Walker * This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. Sure... My working assumption is that both NiH and PdD (as well as W, Ti, etc.) involve fusion in some way. Both are without gammas This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. When we come across an anomaly whose possible explanation is equivocal (i.e., we don't have enough data to say one way or the other), we have the option of adopting a working assumption vis-a-vis that anomaly. By working assumption I'm thinking of a placeholder of some kind to stand in for whatever the explanation ends up being when we have sufficient experimental data to remove the ambiguity in the data. Working assumptions are something we can throw away later when more evidence comes to light. In this sense they're not a blind assumptions, implicitly adopted. They're adopted consciously and tentatively. In this case I'm working from these details: - Skillful experimentalists have observed in the PdD system a correlation between 4He levels and excess heat that strongly suggests that there is d+d fusion going on, somehow. - The Elforsk team saw what they believe to be heat above and beyond what can be produced by a chemical reaction in Rossi's NiH system. - Other experimentalists looking at the NiH system have also seen what they believe to be heat above what can be produced by a chemical reaction. Now here are my working assumptions: - There's only two ways to get energy out of a system above a chemical reaction, and that's through fission or fusion. There is no other supra-chemical means of getting energy out of a system. - There's no reason to go for two different sets of explanations to explain an excess heat anomaly when the evidence is equivocal on what's going on. My own bias is towards one explanation, so I go with my bias. - There is a mechanism that has not yet been carefully characterized in which fusion can proceed without penetrating radiation. From these observations and working assumptions taken together I infer, consciously, aware of the implications, that what's going on in the NiH system is some kind of fusion. A conclusion I am quite happy with for the moment given my working assumptions and starting point. Eric
Re: [Vo]:Christopher H. Cooper
Bob, I agree with you that two particles are not required to conserve linear momentum. I have difficulty accepting the notion that potential energy can be converted into angular momentum. Angular momentum can not be generated in a closed system IIRC unless an equal amount of the opposite sign is co generated. The net system AM remains constant. If your assumed reaction includes a larger system of particles than the two initial particles then energy and momentum can be traded among the larger number. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 4:01 pm Subject: Re: [Vo]:Christopher H. Cooper Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
In reply to Bob Cook's message of Wed, 5 Mar 2014 13:22:16 -0800: Hi, I think the 1-2 nm size came from Axil, not from me. Robin-- If carbon nano tubes are the quantum cavity you refer to their dimensions can be greater--maybe up to 14 to 16 manometers. A mixture of sizes may allow absorption at may varied frequencies depending upon the temperature. The following paper addresses CNT size effects: http://arxiv.org/ftp/arxiv/papers/1202/1202.1328.pdf It was identified by MarkI-zero point two days ago. Bob - Original Message - From: mix...@bigpond.com To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 12:37 PM Subject: Re: [Vo]:Christopher H. Cooper In reply to Bob Cook's message of Tue, 4 Mar 2014 21:58:10 -0800: Hi, [snip] These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. ...this is on the order of hundreds of eV, perhaps coincidentally the same energy range one might also expect from either Hydrino formation or IRH. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
One of the mistakes that LENR duplicators almost always make is not including 5 micron particles in the particle mix. This particle size is the black body resonance size for 400C to 600C. And they do not coat the particle with nano hairs to get the dipole energy to the smaller nano particles. On Wed, Mar 5, 2014 at 4:22 PM, Bob Cook frobertc...@hotmail.com wrote: Robin-- If carbon nano tubes are the quantum cavity you refer to their dimensions can be greater--maybe up to 14 to 16 manometers. A mixture of sizes may allow absorption at may varied frequencies depending upon the temperature. The following paper addresses CNT size effects: http://arxiv.org/ftp/arxiv/papers/1202/1202.1328.pdf It was identified by MarkI-zero point two days ago. Bob - Original Message - From: mix...@bigpond.com To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 12:37 PM Subject: Re: [Vo]:Christopher H. Cooper In reply to Bob Cook's message of Tue, 4 Mar 2014 21:58:10 -0800: Hi, [snip] These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. ...this is on the order of hundreds of eV, perhaps coincidentally the same energy range one might also expect from either Hydrino formation or IRH. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
Axil, I would be interested in your statements of absolute certainty if I had not studied LENR in great depth. Nothing personal, but you do not know what you are talking about. Ed Storms On Mar 5, 2014, at 2:17 PM, Axil Axil wrote: Ed: Things in LENR are more complicated than you are stating. Sometimes gammas are produced in LENR and most times it isn't. The cause of Gamma thermalization is connected with a nuclear based positive feedback loop in the energy conversion/thermalization mechanism. But LENR can happen even when only gammas are produced. On Wed, Mar 5, 2014 at 4:09 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Ed, You haven't considered the work done by Piantelli on gamma radiation, Rossi's statement about pair production, and radiation seen in Rossi's early reactors (his first demo in January when gammas were seen at startup). On Wed, Mar 5, 2014 at 4:29 PM, Edmund Storms stor...@ix.netcom.com wrote: Axil, I would be interested in your statements of absolute certainty if I had not studied LENR in great depth. Nothing personal, but you do not know what you are talking about. Ed Storms On Mar 5, 2014, at 2:17 PM, Axil Axil wrote: Ed: Things in LENR are more complicated than you are stating. Sometimes gammas are produced in LENR and most times it isn't. The cause of Gamma thermalization is connected with a nuclear based positive feedback loop in the energy conversion/thermalization mechanism. But LENR can happen even when only gammas are produced. On Wed, Mar 5, 2014 at 4:09 PM, Edmund Storms stor...@ix.netcom.comwrote: Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 12:06 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Piantelli has seen a 6 MeV proton in a cloud chamber. On Wed, Mar 5, 2014 at 4:34 PM, David Roberson dlrober...@aol.com wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 12:06 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
RE: [Vo]:Christopher H. Cooper
From: Edmund Storms Jones, Hagelstein showed that this proposed reaction was not consistent with what is observed. We must also realize that Hagelstein is promoting his own theory which is not consistent with the rest of nuclear physics. As a result, Takahashi changed his explanation to claim that Be8 formed and dissipated most of the mass-energy as photons before it split into two alpha. Unfortunately, this additional required feature subtracts from the plausibility of the basic idea. Or else it improves it. Well in the end, it is no less plausible than your contention that a strong gamma can be dissipated completely in the form of low energy photons, and in fact TSC still does not have to deal with the strong disproportion of huge gamma, since the alphas are so massive compared to any photon. It's all about disproportion. Jones
Re: [Vo]:Christopher H. Cooper
Jones-- I am not familiar with TSC. Can you give a reference? Bob - Original Message - From: Jones Beene jone...@pacbell.net To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 12:45 PM Subject: RE: [Vo]:Christopher H. Cooper -Original Message- From: Bob Cook Jones-- Alphas would not produce Bremstrallung, if they gain no kinetic energy in being produced. Energy in the form of angular momentum would not produce the B word. Bob- That much is almost true, but you overlook the 800 pound gorilla in the corner - TSC. Maybe you are unfamiliar with it. This happens to be one of the more credible versions of Pd-D in my opinion since no gamma is expected - yet the kinetics of the reaction produce bremsstrahlung. Apparently many in Japan think that Takahashi's version makes the most sense also but AFAIK he has not documented the spectra of the B word. He postulates that a BEC of deuterons is more likely to produce 2 energetic alphas from 4 deuterons than the alternative situation. TSC produces 8Be first which decays into 4He + 4He liberating up to 47.6 MeV of kinetic energy, no gamma and the reaction is known in cosmology from supernova - so it is not an invention. As an alternative of D+D - He, which has to overcome the huge problem of least-favored-channel, TSC is a superior Point of View to many. It is a very strong argument since no gammas are expected - all kinetic. All B-word. Jones
Re: [Vo]:Christopher H. Cooper
Axil, if the reaction you find interesting emits a high energy proton then conservation laws can readily be observed. The best way I have found to analyze two particle collisions is to become an observer located at a point where the linear momentum of the two incoming particles sums to zero. The result of the collision must then remain stationary until the energy is released. At that time at least two things must exit the reaction to conserve both linear momentum and energy. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 4:37 pm Subject: Re: [Vo]:Christopher H. Cooper Piantelli has seen a 6 MeV proton in a cloud chamber. On Wed, Mar 5, 2014 at 4:34 PM, David Roberson dlrober...@aol.com wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
RE: [Vo]:Christopher H. Cooper
-Original Message- From: Bob Cook I am not familiar with TSC. Can you give a reference? Bob Go to this page and type TSC in the search box. Many good papers http://lenr-canr.org/
Re: [Vo]:Christopher H. Cooper
Dave-- I think there is a large number of particles involved in the fractionation of energy resulting from LENR. Otherwise the structure would be damaged so as not to produce LENR anymore. I agree that angular momentum can not be generated, however, if two particles with equal but opposite spin--angular momentum--in the same system come together the net angular momentum is zero. How the spin energy for a system couples and excanges with potential energy is where better understanding is required. You noted the following: I have difficulty accepting the notion that potential energy can be converted into angular momentum. What is the basis for this lack of acceptance? Bob - Original Message - From: David Roberson To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 1:27 PM Subject: Re: [Vo]:Christopher H. Cooper Bob, I agree with you that two particles are not required to conserve linear momentum. I have difficulty accepting the notion that potential energy can be converted into angular momentum. Angular momentum can not be generated in a closed system IIRC unless an equal amount of the opposite sign is co generated. The net system AM remains constant. If your assumed reaction includes a larger system of particles than the two initial particles then energy and momentum can be traded among the larger number. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 4:01 pm Subject: Re: [Vo]:Christopher H. Cooper Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
Jones-- Got it, thanks. Bob - Original Message - From: Jones Beene jone...@pacbell.net To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 1:51 PM Subject: RE: [Vo]:Christopher H. Cooper -Original Message- From: Bob Cook I am not familiar with TSC. Can you give a reference? Bob Go to this page and type TSC in the search box. Many good papers http://lenr-canr.org/
Re: [Vo]:Christopher H. Cooper
Bob, Momentum in a linear product of mass and velocity. Energy is a non linear product with velocity being squared in the equations. The two are not compatible. There should be no problem taking two non spinning particles and ending up with opposite spins due to internal forces. These could independently interact with other particles to transmit the energy. Of course the initial spin energy of the two static particles must be derived from some other potential source of energy. It is important to keep the concept of angular energy and angular momentum separate just as with linear momentum and kinetic energy. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 5:01 pm Subject: Re: [Vo]:Christopher H. Cooper Dave-- I think there is a large number of particles involved in the fractionation of energy resulting from LENR. Otherwise the structure would be damaged so as not to produce LENR anymore. I agree that angular momentum can not be generated, however, if two particles with equal but opposite spin--angular momentum--in the same system come together the net angular momentum is zero. How the spin energy for a system couples and excanges with potential energy is where better understanding is required. You noted the following: I have difficulty accepting the notion that potential energy can be converted into angular momentum. What is the basis for this lack of acceptance? Bob - Original Message - From: David Roberson To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 1:27 PM Subject: Re: [Vo]:Christopher H. Cooper Bob, I agree with you that two particles are not required to conserve linear momentum. I have difficulty accepting the notion that potential energy can be converted into angular momentum. Angular momentum can not be generated in a closed system IIRC unless an equal amount of the opposite sign is co generated. The net system AM remains constant. If your assumed reaction includes a larger system of particles than the two initial particles then energy and momentum can be traded among the larger number. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 4:01 pm Subject: Re: [Vo]:Christopher H. Cooper Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism
Re: [Vo]:Christopher H. Cooper
It only takes one good experiment to disprove a generalization. Piantelli, like the great scientist that he is has done good work on the nature of the LENR reaction. In the Piantelli reactor, it is certain that nickel transmutation to copper is just one of many possible reactions that occur in the Ni/H reaction. Let us focus on this one particular copper based reaction for a moment. Some important insight can be gleaned from the Piantelli cloud chamber experiment about some of the quantum mechanics of the LENR transmutation of nickel into copper. While the Piantilli reactor is in operation, no large photons or particle are seen. But in this cloud chamber experiment, Piantelli removes one of his nickel rods from his reactor and places into in a cloud chamber. This move of the bar into the chamber must have had to take an extended period of time assuming the reactor is cooled down enough to be disassembled. This means that the release of 6 MeV of LENR reaction energy derived from the binding force of nickel after it is transmuted into copper of a high energy proton takes a macroscopic amount of time: taking from minutes to hours. What supports this delay? This long delay in the relaxation of the excited nickel nucleus means that after the double proton pair (proton dimer) has entered into the nickel nucleus, the nickel retains its original atomic number, that is, it is still nickel; it remains nickel even though there is many additional protons resident inside the nickel nucleus. Let us consider Quantum superposition. Quantum superposition is a fundamental principle of quantum mechanics. It holds that a physical system -- such as a nucleus--can exist partly in all its particular, theoretically possible states (or, configuration of its properties) simultaneously; but, when measured, it gives a result corresponding to only one of the possible configurations (as described in interpretation of quantum mechanics). The excited nickel nucleus is both nickel and copper at the same time. The principle of quantum superposition states that if a physical system: i.e. an arrangement of subatomic particles or fields may be in some configuration and if the system could also be in another configuration, then it is in a state which is a superposition of the two, where the probability of each configuration that is in the superposition is specified by a complex number. Erwin Schrödinger explained this concept through his famous thought experiment Schrödinger's cat The thought experiment is also often featured in theoretical discussions of the interpretation of quantum mechanics. In the course of developing this experiment, Schrödinger coined the term Verschränkung (entanglement). What would usually happen is that a 6 MeV photon would exit the nickel to copper nucleus. But when the LENR reaction cools, standard rules of physics take effect again. Because the proton is delayed in its exit from the nickel nucleus, this proves that an entangled proton pair enters into this nucleus because just like in the thought experiment Schrödinger's cat the proton pair infects the larger system: this nickel nucleus with proton entanglement. Only when energy is release does the superposition of the nuclear state resolve into decoherence. Decoherence always results from energy release from the nucleus. What a LENR theory must explain is what quantum mechanical conditions can produce an entangled proton dimer. There has to be an explanation of proton cooper pair formation. Cluster fusion of multiple nuclei happen and because the LENR positive feedback mechanism has been disabled when the Nickel bar is removed from the reactor( it cools). And the thermalization of gammas no longer occurs. On Wed, Mar 5, 2014 at 4:37 PM, Axil Axil janap...@gmail.com wrote: Piantelli has seen a 6 MeV proton in a cloud chamber. On Wed, Mar 5, 2014 at 4:34 PM, David Roberson dlrober...@aol.com wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission
Re: [Vo]:Christopher H. Cooper
Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms Where is the documentation? Jones winmail.dat
Re: [Vo]:Christopher H. Cooper
The incommensurability of momentum and energy plays tricks on people's intuition. A graphic example is the way the movie JFK used this in its climactic courtroom scene where the Zapruder film shows JFK's head going backwards giving the appearance of a second shooter coming from another direction than the Book Depository. If a bullet entered at high velocity from the back and dissipated its energy in JFK's brain in such a way as to pressurize it, then when it exited the forward side it would have exited at a lower velocity making a larger hole which would have been the preferred route of escape of the brain matter -- yielding a high mass flow in the forward direction. High mass flow at the same energy yields higher thrust. JFK's skull was a bit like a combustion chamber in a rocket and the larger hole at the front was the nozzle of the rocket engine. On Wed, Mar 5, 2014 at 4:18 PM, David Roberson dlrober...@aol.com wrote: Bob, Momentum in a linear product of mass and velocity. Energy is a non linear product with velocity being squared in the equations. The two are not compatible. There should be no problem taking two non spinning particles and ending up with opposite spins due to internal forces. These could independently interact with other particles to transmit the energy. Of course the initial spin energy of the two static particles must be derived from some other potential source of energy. It is important to keep the concept of angular energy and angular momentum separate just as with linear momentum and kinetic energy. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 5:01 pm Subject: Re: [Vo]:Christopher H. Cooper Dave-- I think there is a large number of particles involved in the fractionation of energy resulting from LENR. Otherwise the structure would be damaged so as not to produce LENR anymore. I agree that angular momentum can not be generated, however, if two particles with equal but opposite spin--angular momentum--in the same system come together the net angular momentum is zero. How the spin energy for a system couples and excanges with potential energy is where better understanding is required. You noted the following: I have difficulty accepting the notion that potential energy can be converted into angular momentum. What is the basis for this lack of acceptance? Bob - Original Message - *From:* David Roberson dlrober...@aol.com *To:* vortex-l@eskimo.com *Sent:* Wednesday, March 05, 2014 1:27 PM *Subject:* Re: [Vo]:Christopher H. Cooper Bob, I agree with you that two particles are not required to conserve linear momentum. I have difficulty accepting the notion that potential energy can be converted into angular momentum. Angular momentum can not be generated in a closed system IIRC unless an equal amount of the opposite sign is co generated. The net system AM remains constant. If your assumed reaction includes a larger system of particles than the two initial particles then energy and momentum can be traded among the larger number. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 4:01 pm Subject: Re: [Vo]:Christopher H. Cooper Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 12:06 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin
Re: [Vo]:Christopher H. Cooper
So your argument is that Hagelstein has generated incorrect arguments simply to support his own theory. And that no matter what is said about the Takahashi theory, it must be correct because it does not emit strong gamma and it must be better than my theory. You apparently do not acknowledge any fact of nature independent of personal motivation. Amazing. Ed Storms On Mar 5, 2014, at 2:38 PM, Jones Beene wrote: From: Edmund Storms Jones, Hagelstein showed that this proposed reaction was not consistent with what is observed. We must also realize that Hagelstein is promoting his own theory which is not consistent with the rest of nuclear physics. As a result, Takahashi changed his explanation to claim that Be8 formed and dissipated most of the mass-energy as photons before it split into two alpha. Unfortunately, this additional required feature subtracts from the plausibility of the basic idea. Or else it improves it. Well in the end, it is no less plausible than your contention that a strong gamma can be dissipated completely in the form of low energy photons, and in fact TSC still does not have to deal with the strong disproportion of huge gamma, since the alphas are so massive compared to any photon. It’s all about disproportion. Jones
RE: [Vo]:Christopher H. Cooper
From: Edmund Storms So your argument is that Hagelstein has generated incorrect arguments simply to support his own theory. They may or may not be incorrect, but they are definitely self-serving. And that no matter what is said about the Takahashi theory, it must be correct because it does not emit strong gamma and it must be better than my theory. Any theory of deuterium to 4He fusion is more likely to be correct to the degree that it does not wish away a strong gamma. You apparently do not acknowledge any fact of nature independent of personal motivation. Amazing. You have presented no fact of nature to consider, and no indisputable fact of any kind - so my personal motivation does not enter into the discussion. Jones
Re: [Vo]:Christopher H. Cooper
Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want a copy of a particular paper to read, ask and I will send what I have. Unfortunately, I can not send using Vortex and I can not send all the papers. This is generated by energetic electrons or particles such as alpha emission. LENR produces neither kind of radiation. What? Are you now saying that the helium you claim to see in Pd-D does not begin as an alpha particles? Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. Therefore, bremsstrahlung is not an issue because all the mass-energy is dissipated as photons. There is no proof of this. The proof is in the behavior. This is the only conclusion consistent with all behavior. Unfortunately, a book is required to present this information in a form and as complete as you require. I'm attempting to do this. Please be patient. The only question is how this happens. I have proposed a mechanism. The only issue is whether this mechanism is plausible and consistent will all the other observations. It is not plausible if you cannot document photons sufficient to account for the heat. I agree, the measurement of heat and radiation have not been done in a way to show a quantitative correlation. However, I suggest you apply this standard to the other explanations as well. If you do, I think you will have to agree that no explanation meeting this requirements presently exists, including your own. Ed Storms
Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 3:44 PM, Jones Beene wrote: From: Edmund Storms So your argument is that Hagelstein has generated incorrect arguments simply to support his own theory. They may or may not be incorrect, but they are definitely self-serving. Have you read them? I have and the papers simply show the consequences of particle emission from any source. His arguments are correct and place a limit on the energy compared to what is observed. This is less self-serving than your arguments. And that no matter what is said about the Takahashi theory, it must be correct because it does not emit strong gamma and it must be better than my theory. Any theory of deuterium to 4He fusion is more likely to be correct to the degree that it does not “wish away” a strong gamma. Yes and I accept that you obsessed with this argument. I now give up trying to show you how this opinion is not correct and is not consistent with what is observed You apparently do not acknowledge any fact of nature independent of personal motivation. Amazing. You have presented no fact of nature to consider, and no indisputable fact of any kind - so my personal motivation does not enter into the discussion. Once again, you defect the issue and ignore what I have provided. Are you a lawyer, Jones? Ed Storms Jones
Re: [Vo]:Christopher H. Cooper
My usual take on this subject is that a bullet behaves much like a small explosive device. The energy delivered by the retardation of the fast moving bullet was deposited within his skull and must find a way out. Energy does not know direction since it is scalar which leads to damage in all directions. This type of demonstration is quite evident when a high speed bullet impacts a plastic container of water. In that case the water is rapidly expelled in all directions. Conservation of momentum is required in this special case along with conservation of energy. As you mention, plenty of material travels along with the spent bullet through the forward exit point. It needs to be proven that the momentum contained within this forward exiting mass is greater than the initial bullet momentum so that a negative momentum is generated that is large enough to send his head backwards. This may be possible, but it is not evident. I have fired plenty of rifles and have been subject to the kick due to the bullet being fired. In this case we are attempting to accept the notion that the brain matter and bullet leaving the front of his head actually has more kick(momentum) than if Oswald had held the rifle butt against his head(Oswald's) as the gun was fired. I must say that this seems highly unlikely after a bit of consideration. My conclusion at this time is that some other force must have been involved to make JFK's head react so strongly backwards. I believe some say that your muscles might tense due to damage of the brain which might be the explanation. Dave -Original Message- From: James Bowery jabow...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 5:31 pm Subject: Re: [Vo]:Christopher H. Cooper The incommensurability of momentum and energy plays tricks on people's intuition. A graphic example is the way the movie JFK used this in its climactic courtroom scene where the Zapruder film shows JFK's head going backwards giving the appearance of a second shooter coming from another direction than the Book Depository. If a bullet entered at high velocity from the back and dissipated its energy in JFK's brain in such a way as to pressurize it, then when it exited the forward side it would have exited at a lower velocity making a larger hole which would have been the preferred route of escape of the brain matter -- yielding a high mass flow in the forward direction. High mass flow at the same energy yields higher thrust. JFK's skull was a bit like a combustion chamber in a rocket and the larger hole at the front was the nozzle of the rocket engine. On Wed, Mar 5, 2014 at 4:18 PM, David Roberson dlrober...@aol.com wrote: Bob, Momentum in a linear product of mass and velocity. Energy is a non linear product with velocity being squared in the equations. The two are not compatible. There should be no problem taking two non spinning particles and ending up with opposite spins due to internal forces. These could independently interact with other particles to transmit the energy. Of course the initial spin energy of the two static particles must be derived from some other potential source of energy. It is important to keep the concept of angular energy and angular momentum separate just as with linear momentum and kinetic energy. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Mar 5, 2014 5:01 pm Subject: Re: [Vo]:Christopher H. Cooper Dave-- I think there is a large number of particles involved in the fractionation of energy resulting from LENR. Otherwise the structure would be damaged so as not to produce LENR anymore. I agree that angular momentum can not be generated, however, if two particles with equal but opposite spin--angular momentum--in the same system come together the net angular momentum is zero. How the spin energy for a system couples and excanges with potential energy is where better understanding is required. You noted the following: I have difficulty accepting the notion that potential energy can be converted into angular momentum. What is the basis for this lack of acceptance? Bob - Original Message - From: David Roberson To: vortex-l@eskimo.com Sent: Wednesday, March 05, 2014 1:27 PM Subject: Re: [Vo]:Christopher H. Cooper Bob, I agree with you that two particles are not required to conserve linear momentum. I have difficulty accepting the notion that potential energy can be converted into angular momentum. Angular momentum can not be generated in a closed system IIRC unless an equal amount of the opposite sign is co generated. The net system AM remains constant. If your assumed reaction includes a larger system of particles than the two initial particles then energy and momentum can be traded among the larger number. Dave
Re: [Vo]:Christopher H. Cooper
On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind that potential energy may be changed to the energy of angular momentum/spin energy in LENR. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 12:06 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 12:28 PM, Jones Beene wrote: From: Edmund Storms Jones, bremsstrahlung or slowing down radiation is not produced by photons. Who said it was? I'm not answering a claim. I'm simply giving information. You brought up photons by talking about gamma emissions, which are photons. You then added the production of bremsstrahlung, which I simply pointed out is not produced by gamma. You brought up photons. I asked for adequate documentation of intense photon emission - and am still waiting. I sent a list of references. If you want
Re: [Vo]:Christopher H. Cooper
Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 3:34 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support for the idea. Ed Storms On Mar 5, 2014, at 2:01 PM, Bob Cook wrote: Ed-- You said: Yes, that is what I'm saying. LENR can not result in a single alpha because two particles are required to conserve momentum when energy is released. I note that, if there is no linear momentum to start, two particles would not be required. I do not believe conservation of angular momentum requires two particles either. And keep in mind
Re: [Vo]:Christopher H. Cooper
It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. ***Isn't Reversible Proton Fusion (RPF) Gamma free? It's the most common fusion event in our solar system. I thought you were the one bringing it up every so often as a plausible theory... On Wed, Mar 5, 2014 at 8:21 AM, Jones Beene jone...@pacbell.net wrote: *From:* Eric Walker This working assumption (of a known fusion reaction) is not justifiable by facts, logic or common sense. Sure. That's you're opinion. You're entitled to an opinion. Sorry to have made this blanket statement in regard to your prior post specifically, Eric, since it is a generic criticism to many of the posts on Vortex and not personal - but... No, it's not opinion when 100% of the available proof is on your side. It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. QED. Since 1989, there have been assertions and claims, but they are only assertions, that LENR is proof of a gammaless nuclear reaction, but that is circular logic. LENR is proof of a thermal anomaly, and helium is seen in the ash, but that is all that can be said logically. Even if helium is seen in proportion to the excess heat, which is in dispute, that does not raise LENR to the level of a known fusion reaction which is gammaless, at least not so long as there are other valid explanations. To be raised to this level the claimant must also demonstrate in an experiment not involving LENR that 24 MeV gammas can be completely suppressed by any mechanism. Any mechanism will suffice. This has not been done, even with 1 MeV gammas since there is always leakage - even with lead shielding. By definition, cold fusion cannot be the same known reaction as deuterium fusion to helium, which was known prior to 1989 - if it is gammaless - unless and until it can be shown that there is a real physical mechanism for not only for suppressing gammas, but for suppressing 100% of them without exception. How is that opinion? Jones
Re: [Vo]:Christopher H. Cooper
https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 3:34 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy
Re: [Vo]:Christopher H. Cooper
Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 3:34 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm Subject: Re: [Vo]:Christopher H. Cooper Bob, we are discussing a basic and fundamental concept. The energy generated when mass-energy is released requires emission of at least two particles for the energy to be dissipated. I know of no example in nature where this requirement does not operate when energy is released. If energy is not released immediately, but is retained in the nucleus, this nucleus is found to be unstable and will eventually release energy over a period of time by emission of a particle, including a photon. This is how nature is found to behave. Imagining otherwise is not useful unless you have observed support
Re: [Vo]:Christopher H. Cooper
Ed; The fact is that during cold fusion NO energetic gamma is emitted, This absolute statement is not true. Your theory of causation is weak in regards to not accounting for conditional gamma and high energy particle production. On Wed, Mar 5, 2014 at 8:14 PM, Axil Axil janap...@gmail.com wrote: https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.comwrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 3:34 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction is known to happen less than 1% of the time during hot fusion and it produces a 23 MeV gamma that is required to conserve momentum. This reaction is clearly not observed. We know this for a fact. Therefore, this idea is irrelevant. Ed Storms On Mar 5, 2014, at 2:34 PM, David Roberson wrote: Ed, the energy can be released in the form of a particle, such as an alpha, and a gamma ray. Energy and momentum can be conserved in that manner. The bulk of the energy will be given to the gamma ray due to the large difference in masses.Think of a rifle firing a bullet. Most of the energy ends up in the bullet while linear momentum is conserved. Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 4:09 pm
Re: [Vo]:Christopher H. Cooper
More http://newenergytimes.com/v2/library/2004/2004Focardi-EvidenceOfElectromagneticRadiation.pdf *Evidence of electromagnetic radiation from Ni-H Systems* We report evidence of photon emission in three experiments with hydrogen loading of Ni slabs, during the degassing phase, when hydrogen was introduced into the cell, and during thermal cycling. In the first experiment we obtained excess power of about 20 W, while in the second experiment photon emission was observed instead of power production. In the third experiment, a Ni sample in hydrogen underwent thermal excitation and showed an increasing photon emission for a few hours. This experiments shows that the energy production from LENR either comes out as gamma radiation or heat but not both. On Wed, Mar 5, 2014 at 8:20 PM, Axil Axil janap...@gmail.com wrote: Ed; The fact is that during cold fusion NO energetic gamma is emitted, This absolute statement is not true. Your theory of causation is weak in regards to not accounting for conditional gamma and high energy particle production. On Wed, Mar 5, 2014 at 8:14 PM, Axil Axil janap...@gmail.com wrote: https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.comwrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 3:34 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude that LENR is impossible, which of course is the skeptical conclusion. Nevertheless, the effect is real and therefore it must have an explanation. Until people actually search where the keys are located rather than under the lamppost, success will be impossible. Ed Storms Dave -Original Message- From: Edmund Storms stor...@ix.netcom.com To: vortex-l vortex-l@eskimo.com Cc: Edmund Storms stor...@ix.netcom.com Sent: Wed, Mar 5, 2014 5:29 pm Subject: Re: [Vo]:Christopher H. Cooper Yes Dave, that is true, but that is not what is observed. This reaction
Re: [Vo]:Christopher H. Cooper
Ed Storms deals with this piantelli finding in this paper as follows: http://lenr-canr.org/acrobat/StormsEnatureofen.pdf Ed says: This study raises many questions that are not answered and demonstrates some very unexpected behavior. Having this behavior made known without delay is more important than taking time to answer all questions before publication. Therefore, this paper should be viewed as a progress report about an important behavior. Ed, it is time to address the conditional nature of high energy radiation from LENR head on and avoid absolute statements about the subject until theory has addressed this unexpected behavior. On Wed, Mar 5, 2014 at 8:26 PM, Axil Axil janap...@gmail.com wrote: More http://newenergytimes.com/v2/library/2004/2004Focardi-EvidenceOfElectromagneticRadiation.pdf *Evidence of electromagnetic radiation from Ni-H Systems* We report evidence of photon emission in three experiments with hydrogen loading of Ni slabs, during the degassing phase, when hydrogen was introduced into the cell, and during thermal cycling. In the first experiment we obtained excess power of about 20 W, while in the second experiment photon emission was observed instead of power production. In the third experiment, a Ni sample in hydrogen underwent thermal excitation and showed an increasing photon emission for a few hours. This experiments shows that the energy production from LENR either comes out as gamma radiation or heat but not both. On Wed, Mar 5, 2014 at 8:20 PM, Axil Axil janap...@gmail.com wrote: Ed; The fact is that during cold fusion NO energetic gamma is emitted, This absolute statement is not true. Your theory of causation is weak in regards to not accounting for conditional gamma and high energy particle production. On Wed, Mar 5, 2014 at 8:14 PM, Axil Axil janap...@gmail.com wrote: https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.comwrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 3:34 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory
Re: [Vo]:Christopher H. Cooper
Axil-- Rossi and Focardi figured out how to make the reaction produce heat in stead of radiation. The source of the various radiation peaks observed would be nice to know. Bob - Original Message - From: Axil Axil To: vortex-l Sent: Wednesday, March 05, 2014 5:26 PM Subject: Re: [Vo]:Christopher H. Cooper More http://newenergytimes.com/v2/library/2004/2004Focardi-EvidenceOfElectromagneticRadiation.pdf Evidence of electromagnetic radiation from Ni-H Systems We report evidence of photon emission in three experiments with hydrogen loading of Ni slabs, during the degassing phase, when hydrogen was introduced into the cell, and during thermal cycling. In the first experiment we obtained excess power of about 20 W, while in the second experiment photon emission was observed instead of power production. In the third experiment, a Ni sample in hydrogen underwent thermal excitation and showed an increasing photon emission for a few hours. This experiments shows that the energy production from LENR either comes out as gamma radiation or heat but not both. On Wed, Mar 5, 2014 at 8:20 PM, Axil Axil janap...@gmail.com wrote: Ed; The fact is that during cold fusion NO energetic gamma is emitted, This absolute statement is not true. Your theory of causation is weak in regards to not accounting for conditional gamma and high energy particle production. On Wed, Mar 5, 2014 at 8:14 PM, Axil Axil janap...@gmail.com wrote: https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.com wrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Wednesday, March 05, 2014 3:34 PM Subject: Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too much energy for the secondary radiation to be missed. Therefore, this proposed reaction does not occur. Each theory suggested so far can be eliminated by identifying these conflicts with observation. If the observations were not so many and so strong, a person might conclude
Re: [Vo]:Christopher H. Cooper
Yes, Rossi added a secondary heater that brought the reactor up to operational temperature before startup. A cold reactor produces gamma and a hot one does not. This simple relationship is good input to LENR theory. This shows the central role that infrared photons play in the LENR reaction. Another possibility is the importance of getting above the Curie temperature of nickel. On Wed, Mar 5, 2014 at 9:08 PM, Bob Cook frobertc...@hotmail.com wrote: Axil-- Rossi and Focardi figured out how to make the reaction produce heat in stead of radiation. The source of the various radiation peaks observed would be nice to know. Bob - Original Message - *From:* Axil Axil janap...@gmail.com *To:* vortex-l vortex-l@eskimo.com *Sent:* Wednesday, March 05, 2014 5:26 PM *Subject:* Re: [Vo]:Christopher H. Cooper More http://newenergytimes.com/v2/library/2004/2004Focardi-EvidenceOfElectromagneticRadiation.pdf *Evidence of electromagnetic radiation from Ni-H Systems* We report evidence of photon emission in three experiments with hydrogen loading of Ni slabs, during the degassing phase, when hydrogen was introduced into the cell, and during thermal cycling. In the first experiment we obtained excess power of about 20 W, while in the second experiment photon emission was observed instead of power production. In the third experiment, a Ni sample in hydrogen underwent thermal excitation and showed an increasing photon emission for a few hours. This experiments shows that the energy production from LENR either comes out as gamma radiation or heat but not both. On Wed, Mar 5, 2014 at 8:20 PM, Axil Axil janap...@gmail.com wrote: Ed; The fact is that during cold fusion NO energetic gamma is emitted, This absolute statement is not true. Your theory of causation is weak in regards to not accounting for conditional gamma and high energy particle production. On Wed, Mar 5, 2014 at 8:14 PM, Axil Axil janap...@gmail.com wrote: https://docs.google.com/file/d/0B6id5Hf-xMWOYXVjekJCN1ZkQk0/edit?pli=1 Ed: I suggest that you integrate these experimental results from Paintilli into your thinking. I see high energy particle tracks an many counts in the gamma ray energy range. The last two slides show gammas. On Wed, Mar 5, 2014 at 8:09 PM, Edmund Storms stor...@ix.netcom.comwrote: Bob, 23.8 MeV of energy must be released for each He made. Each emitted He4 from Be8 needs to carry 23.8 MeV of energy. Please explain how even a small fraction of that energy can appear as spin. When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. Ed On Mar 5, 2014, at 5:07 PM, Bob Cook wrote: Ed-- You said: However, the resulting alpha would have too much energy for the secondary radiation to be missed. If the alphas are in high spin states upon the decomposition of Be-8, then small amounts of energy associated with transition from one state to the next lower state would never be seen. If many electrons are involved in the reaction it seems likely only small energy packets would be released. The secondary radiation may be missed. Why do you imply the secondary radiation should necessarily be a high energy photon(s)? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Wednesday, March 05, 2014 3:34 PM *Subject:* Re: [Vo]:Christopher H. Cooper On Mar 5, 2014, at 3:45 PM, David Roberson wrote: Ed, I was not suggesting that this reaction is the main one, I was merely pointing out that it is possible. Someone made a blanket statement that this path was not possible and I wanted to clear the air. Dave, none of us has the time to describe every aspect of the issue in each e-mail. We all have to assume the reader has done some homework and knows that the statement is not complete and that the writer also know this. In any case, emission of a photon makes the process two body, not one body as I was describing. The conservation of energy and momentum does not prevent this from happening as was stated. Had the original proposition been that it was not likely or observed I would have remained silent. The fact is that during cold fusion NO energetic gamma is emitted, which was known in 1989. Therefore, this issue is not relevant. People propose the He4 is emitted as an alpha, which means the helium has translational energy. This is not possible when one particle is involved, which is what I said. Takahashi proposes Be8 forms and decomposes into two alpha, which does conserve energy and momentum and is not inconsistent with the basic requirements. However, the resulting alpha would have too
RE: [Vo]:Christopher H. Cooper
From: Kevin O'Malley It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. ***Isn't Reversible Proton Fusion (RPF) Gamma free? It's the most common fusion event in our solar system. I thought you were the one bringing it up every so often as a plausible theory... Cough... cough. Yes and Yes and Yes. But there is a timely caveat. ... is reversible fusion really fusion when the fusion bond lasts for only a few femtoseconds? Can we not agree that there is a fundamental difference between fusion which is permanent and fusion which is transitory? Therefore RPF is not really heavy-duty fusion-fusion, only FINO fusion (fusion in name only). That is my answer and I'm sticking to it... attachment: winmail.dat
Re: [Vo]:Christopher H. Cooper
On Wed, Mar 5, 2014 at 9:39 AM, Bob Cook frobertc...@hotmail.com wrote: I have had a similar notion relative to the Pd-D system. Specifically two D come together to form a virtual excited He particle with high spin energy that fractionates its high spin energy to electrons and other coupled particles to attain the desired low energy associated with the stable He particle. Only many low energy photons are involved. to balance the lower mass of the He compared to the starting material. Yes -- this is the system I'm rooting for right now as well in the context of PdD. This system is not too dissimilar from what I gather is Hagelstein's system, where the excited [dd]* resonance binds (indirectly) with the phonon modes, but instead of phonon modes, in this system the [dd]* is binding with sources of electrostatic charge (electrons and ion cores). One question I have is why Hagelstein has not explored this avenue. I will hazard a guess that it is because he wants an oscillator and coherent feedback, and this system does not oscillate and just dumps energy instead in one big transfer. Eric
Re: [Vo]:Christopher H. Cooper
On Wed, Mar 5, 2014 at 2:01 PM, Bob Cook frobertc...@hotmail.com wrote: I think there is a large number of particles involved in the fractionation of energy resulting from LENR. Otherwise the structure would be damaged so as not to produce LENR anymore. I like this line of approach. It reminds me of what Bob Higgins recently discussed [1]. Eric [1] https://www.mail-archive.com/vortex-l@eskimo.com/msg89992.html
Re: [Vo]:Christopher H. Cooper
On Wed, Mar 5, 2014 at 5:09 PM, Edmund Storms stor...@ix.netcom.com wrote: When alpha particles pass through material, a series of nuclear reactions can occur that emit radiation. In addition, bremsstrahlung radiation is emitted as the alpha slows down. Hagelstrin describes these processes in the papers I attached previously. I suggest you read them. If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. There would be the bath of photons from the fractionation, the nearly stationary 4He daughter, and no Bremsstrahlung from collisions by a fast particle. Eric
Re: [Vo]:Christopher H. Cooper
I wrote: If an alpha is born from a [dd]* resonance in which the mass energy is fractionated among a large number of sinks (e.g., nearby electrons and ion cores), the 4He daughter would have no or almost no energy. This was stated incorrectly. To the extent that there is binding between the [dd]* state and one or more nearby ion cores, I assume the daughter 4He would be imparted kinetic energy in corresponding measure. So if this system is anywhere near what is really going on, we have a parameter that we can play with and adjust to match the actual kinetic energies that are seen (not very much). The more there is interaction with the electronic structure, and the less there is interaction with the ion cores, the less kinetic energy imparted to the daughter 4He. Eric
Re: [Vo]:Christopher H. Cooper
Jones: I gather I don't really understand what you're getting at. My responses are designated by 4 embedded asterisks. On Wed, Mar 5, 2014 at 8:28 PM, Jones Beene jone...@pacbell.net wrote: From: Kevin O'Malley It is fact that LENR is not and cannot be a known fusion reaction, since it is fact that no known nuclear fusion reaction is gamma free. ***Isn't Reversible Proton Fusion (RPF) Gamma free? It's the most common fusion event in our solar system. I thought you were the one bringing it up every so often as a plausible theory... Cough... cough. Yes and Yes and Yes. But there is a timely caveat. ... is reversible fusion really fusion when the fusion bond lasts for only a few femtoseconds? My impression is that this is enough for the Sun to generate photons, Helium, and other stuff. Now, maybe that's only because it is so huge compared to the earth, but it is also gaseous, where we're dealing with condensed matter. Can we not agree that there is a fundamental difference between fusion which is permanent and fusion which is transitory? ***Perhaps that fundamental difference is between gaseous state and solid state... or even the proposed 5th state of matter: BECs. Basically, this is your main statement that I do not understand. Therefore RPF is not really heavy-duty fusion-fusion, only FINO fusion (fusion in name only). That is my answer and I'm sticking to it... ***Perhaps RPF is nature's way of desperately seeking equilibrium. Once fusion has taken place, it wrestles with the outcome until the atoms are in their most restful state, which could even be partial hydrogen...
RE: [Vo]:Christopher H. Cooper
From: Eric Walker Wikipedia has a discussion of Nickel hydride with several references to recent papers. I'm thinking more in relative terms -- I believe it takes quite a lot of energy to dissolve hydrogen into nickel in comparison to the relative ease with which hydrogen dissolves into palladium (which is sometimes called a hydrogen sponge). In his Arata replication, Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogen than Pd alone. But he also found that hydrogen concentration did NOT correlate to excess energy. However, this was with protium, not deuterium. The highest absorber was not the most active and a low absorber was actually superior. There is a known correlation of excess heat to deuterium concentration in Pd-D experiments, which is completely absent in Ni-H. This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. In many ways, protium and deuterium as so extremely different in physical properties (especially nuclear properties) that they should be considered to be different elements instead of isotopes of the same element.
Re: [Vo]:Christopher H. Cooper
Jones: You noted: In many ways, protium and deuterium as so extremely different in physical properties (especially nuclear properties) that they should be considered to be different elements instead of isotopes of the same element. Well said, especially regarding their different nuclear magnetic properties and spin. The two isotopes must couple with reactants, whatever the environment, in significantly different ways. I wonder how the magnetic susceptibility of the Pd-Ni alloy changes relative to Pd? More unpaired electrons may make for significantly different susceptibility and different electron interaction with D. I wonder if a Ni-low-Pd alloy still allows the Rossi reaction to happen with hydrogen? The Pd should strain the Ni lattice and could give some information on just where the Rossi reaction takes place, surface or in side the lattice. From the experiments on NiH it seems that it is pretty difficult to get protium inside the lattice--unlike Pd. This seems to point to surface reactions for Ni and bulk reaction for Pd. Bob - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Tuesday, March 04, 2014 6:27 AM Subject: RE: [Vo]:Christopher H. Cooper From: Eric Walker Wikipedia has a discussion of Nickel hydride with several references to recent papers. I'm thinking more in relative terms -- I believe it takes quite a lot of energy to dissolve hydrogen into nickel in comparison to the relative ease with which hydrogen dissolves into palladium (which is sometimes called a hydrogen sponge). In his Arata replication, Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogen than Pd alone. But he also found that hydrogen concentration did NOT correlate to excess energy. However, this was with protium, not deuterium. The highest absorber was not the most active and a low absorber was actually superior. There is a known correlation of excess heat to deuterium concentration in Pd-D experiments, which is completely absent in Ni-H. This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. In many ways, protium and deuterium as so extremely different in physical properties (especially nuclear properties) that they should be considered to be different elements instead of isotopes of the same element.
Re: [Vo]:Christopher H. Cooper
On Mar 4, 2014, at 8:02 AM, Bob Cook wrote: From the experiments on NiH it seems that it is pretty difficult to get protium inside the lattice--unlike Pd. This seems to point to surface reactions for Ni and bulk reaction for Pd. Bob Bob, all the evidence shows that the nuclear reaction using Pd occurs on or near the surface. The fact that Pd absorbs hydrogen is not relevant. This ability to absorb has two effects. It allows hydrogen to leave the surface, which lowers the amount of D on the surface available for fusion, thereby limiting the reaction. And, the D in the lattice can supply the surface with D when D is not available from the gas or electrolytic action. In short, the process is more complex than you assume. Ed Storms
Re: [Vo]:Christopher H. Cooper
With palladium, deuterium serves two functions. It produces NAE by cracking it and it also provides a surface dielectric SPP cover the permeates the cracks. Any deuterium that penetrates deeply into the lattice is lost to the reaction. With NiH, the NAE is premade, or produced in an ongoing process by heat and/or spark. Hydrogen does not play a role in producing the NAE in NiH. There is no loss of hydrogen through too deep penetration of the bulk material so there is more pressure to enhance dielectric SPP performance. On Tue, Mar 4, 2014 at 10:14 AM, Edmund Storms stor...@ix.netcom.comwrote: On Mar 4, 2014, at 8:02 AM, Bob Cook wrote: From the experiments on NiH it seems that it is pretty difficult to get protium inside the lattice--unlike Pd. This seems to point to surface reactions for Ni and bulk reaction for Pd. Bob Bob, all the evidence shows that the nuclear reaction using Pd occurs on or near the surface. The fact that Pd absorbs hydrogen is not relevant. This ability to absorb has two effects. It allows hydrogen to leave the surface, which lowers the amount of D on the surface available for fusion, thereby limiting the reaction. And, the D in the lattice can supply the surface with D when D is not available from the gas or electrolytic action. In short, the process is more complex than you assume. Ed Storms
Re: [Vo]:Christopher H. Cooper
Jones, since you mention how different protium and deuterium are perhaps it is an excellent time to discuss the differences: Obviously the mass of deuterium is approximately double that of protium. At the same temperature protium would be moving between collisions at around the square root of two times faster. Deuterium is physically larger than protium but the size difference may not make a difference in the chemical behavior directly. Here I am referring to the nucleus and not a neutral atom. The magnetic properties of the two should be different but I leave that decision up to others with an opportunity to look into the issue. The spin differences must be important. The behavior of each of these isotopes to electromagnetic radiation would be quite different due to the large mass to charge ratio variation. Deuterium can combine with another of its same isotope to form a stable nucleus whereas protium generally does not. This may be the main physical difference affecting LENR behavior. Deuterium can supply a neutron to any nuclear reaction that protium can not. The proton and neutron can be separated and individually expelled under certain conditions. I suspect that protium diffuses more rapidly than deuterium through metals due to its lower mass and perhaps smaller physical size. I have barely breached the list of differences and I am confident that others can correct and improve this beginning. Take a moment to add factors that you may have knowledge of that clearly pertain to behavior separating these isotopes. Thanks, Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, Mar 4, 2014 9:28 am Subject: RE: [Vo]:Christopher H. Cooper From:Eric Walker Wikipedia has adiscussion of Nickel hydride with several references to recent papers. I'm thinking more in relative terms -- I believe ittakes quite a lot of energy to dissolve hydrogen into nickel in comparison tothe relative ease with which hydrogen dissolves into palladium (which issometimes called a hydrogen sponge). In his Arata replication,Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogenthan Pd alone. But he also found thathydrogen concentration did NOT correlate to excess energy. However, this waswith protium, not deuterium. The highest absorber was not the most active and alow absorber was actually superior. There is a known correlation of excess heatto deuterium concentration in Pd-D experiments, which is completely absent inNi-H. This is yet another reason,one of many - why consideration of all the evidence, giving no preference toPd-D, points to many different routes to gain in LENR. In many ways, protium anddeuterium as so extremely different in physical properties (especially nuclearproperties) that they should be considered to be different elements instead ofisotopes of the same element.
Re: [Vo]:Christopher H. Cooper
Ed- In the Navy's SPAWAR experiments, was it clear there was no He found in the bulk Pd? It seems I remember they noted He production. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Tuesday, March 04, 2014 7:14 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 4, 2014, at 8:02 AM, Bob Cook wrote: From the experiments on NiH it seems that it is pretty difficult to get protium inside the lattice--unlike Pd. This seems to point to surface reactions for Ni and bulk reaction for Pd. Bob Bob, all the evidence shows that the nuclear reaction using Pd occurs on or near the surface. The fact that Pd absorbs hydrogen is not relevant. This ability to absorb has two effects. It allows hydrogen to leave the surface, which lowers the amount of D on the surface available for fusion, thereby limiting the reaction. And, the D in the lattice can supply the surface with D when D is not available from the gas or electrolytic action. In short, the process is more complex than you assume. Ed Storms
Re: [Vo]:Christopher H. Cooper
Bob, you need to read more. At least 18 studies of He in Pd are available. In addition the issue has been discussed in detail in my book and in The status of cold fusion (2010), Naturwissenschaften, 97, 861 (2010). I sent a copy to your personal address. Ed Storms On Mar 4, 2014, at 8:24 AM, Bob Cook wrote: Ed- In the Navy's SPAWAR experiments, was it clear there was no He found in the bulk Pd? It seems I remember they noted He production. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Tuesday, March 04, 2014 7:14 AM Subject: Re: [Vo]:Christopher H. Cooper On Mar 4, 2014, at 8:02 AM, Bob Cook wrote: From the experiments on NiH it seems that it is pretty difficult to get protium inside the lattice--unlike Pd. This seems to point to surface reactions for Ni and bulk reaction for Pd. Bob Bob, all the evidence shows that the nuclear reaction using Pd occurs on or near the surface. The fact that Pd absorbs hydrogen is not relevant. This ability to absorb has two effects. It allows hydrogen to leave the surface, which lowers the amount of D on the surface available for fusion, thereby limiting the reaction. And, the D in the lattice can supply the surface with D when D is not available from the gas or electrolytic action. In short, the process is more complex than you assume. Ed Storms
RE: [Vo]:Christopher H. Cooper
Good start on a list. It is clear that the two isotopes are so very different in nuclear properties that they should be considered different elements- yet the chemical properties are identical or similar - so the profound nuclear differences are masked by chemical similarity. To add: one nucleus is Bosonic, the other is Fermionic. Boson statistics are probably the biggest difference of all. Near-field charge of the nuclei is another. NMR and magnetic susceptibility you mention. From: David Roberson Jones, since you mention how different protium and deuterium are perhaps it is an excellent time to discuss the differences: Obviously the mass of deuterium is approximately double that of protium. At the same temperature protium would be moving between collisions at around the square root of two times faster. Deuterium is physically larger than protium but the size difference may not make a difference in the chemical behavior directly. Here I am referring to the nucleus and not a neutral atom. The magnetic properties of the two should be different but I leave that decision up to others with an opportunity to look into the issue. The spin differences must be important. The behavior of each of these isotopes to electromagnetic radiation would be quite different due to the large mass to charge ratio variation. Deuterium can combine with another of its same isotope to form a stable nucleus whereas protium generally does not. This may be the main physical difference affecting LENR behavior. Deuterium can supply a neutron to any nuclear reaction that protium can not. The proton and neutron can be separated and individually expelled under certain conditions. I suspect that protium diffuses more rapidly than deuterium through metals due to its lower mass and perhaps smaller physical size. I have barely breached the list of differences and I am confident that others can correct and improve this beginning. Take a moment to add factors that you may have knowledge of that clearly pertain to behavior separating these isotopes. In his Arata replication, Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogen than Pd alone. But he also found that hydrogen concentration did NOT correlate to excess energy. However, this was with protium, not deuterium. The highest absorber was not the most active and a low absorber was actually superior. There is a known correlation of excess heat to deuterium concentration in Pd-D experiments, which is completely absent in Ni-H. This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. In many ways, protium and deuterium as so extremely different in physical properties (especially nuclear properties) that they should be considered to be different elements instead of isotopes of the same element.
Re: [Vo]:Christopher H. Cooper
In reply to Axil Axil's message of Sun, 2 Mar 2014 13:23:09 -0500: Hi, [snip] I was under the impression that DGT started with a 1 Tesla field that they created themselves, and that the experiment itself increased this to 1.6 T. IOW a 60% increase. It is common for ferromagnetic materials to increase the field strength of a magnet when inserted into the core of a coil, due to the materials electron spins aligning themselves with the field, so I would not be surprised to find that the imposed field increased by 60% when Ni was introduced to the imposed field. Have I misunderstood the DGT report? Like you, any one of us can only do so much of what is required. To come up with an all inclusive theory, we must trust the word and the work done by others. I must admit that I trust DGT. So far, their experimental observation about magnetic field strength has no impact on the theory (HEMI) that they put forward. They have no theroritical based interest in misleading us to advance their theory base on Dr. Kims work. Like us, DGT is simply amazed at the magnetic nature of their experimental find but have not connected it to HEMI in any way. This is hard to understand. On the part of DGT, there is no self interest in tossing an almost unbelievable finding into their finding and in fact this finding undercuts HEMI. In fact such a finding is a major distraction. They really need to do a major rethink of their experimental position on HEMI and BEC as per Dr. Kim. On Sun, Mar 2, 2014 at 1:03 PM, Edmund Storms stor...@ix.netcom.com wrote: On Mar 2, 2014, at 10:47 AM, Axil Axil wrote: These Nanoplasmonic experiments with uranium can be done inexpensively, why can't Ed replicate these experiments? Because I have only two hands and no financial support. If you want this replicated, I suggest you hire someone to do this. Ed Storms Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
DGT: After each triggering duty cycle (the triggering sequences producing excess heat), the magnetic fields at ~18 cm from the reactor at all three locations rose from ~0.6 Tesla to ~1.6 Tesla (DC peak) during each reaction period. Such anomalous peak signals were maintained for approximately 3-4 sec after the HV currents were cut off. Axil: The duty cycle is triggered by a high voltage current. After the HV spark, the magnetic field measures .6 tesla. The magnetic field grows along with the reaction for 3-4 seconds. At the peak of the reaction cycle, the magnetic field is 1.6 tesla. Read it yourself here: http://www.physics.purdue.edu/people/faculty/yekim/ICCF-18-JCMNS-KH-Pre-2.pdf On Tue, Mar 4, 2014 at 4:16 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Sun, 2 Mar 2014 13:23:09 -0500: Hi, [snip] I was under the impression that DGT started with a 1 Tesla field that they created themselves, and that the experiment itself increased this to 1.6 T. IOW a 60% increase. It is common for ferromagnetic materials to increase the field strength of a magnet when inserted into the core of a coil, due to the materials electron spins aligning themselves with the field, so I would not be surprised to find that the imposed field increased by 60% when Ni was introduced to the imposed field. Have I misunderstood the DGT report? Like you, any one of us can only do so much of what is required. To come up with an all inclusive theory, we must trust the word and the work done by others. I must admit that I trust DGT. So far, their experimental observation about magnetic field strength has no impact on the theory (HEMI) that they put forward. They have no theroritical based interest in misleading us to advance their theory base on Dr. Kims work. Like us, DGT is simply amazed at the magnetic nature of their experimental find but have not connected it to HEMI in any way. This is hard to understand. On the part of DGT, there is no self interest in tossing an almost unbelievable finding into their finding and in fact this finding undercuts HEMI. In fact such a finding is a major distraction. They really need to do a major rethink of their experimental position on HEMI and BEC as per Dr. Kim. On Sun, Mar 2, 2014 at 1:03 PM, Edmund Storms stor...@ix.netcom.com wrote: On Mar 2, 2014, at 10:47 AM, Axil Axil wrote: These Nanoplasmonic experiments with uranium can be done inexpensively, why can't Ed replicate these experiments? Because I have only two hands and no financial support. If you want this replicated, I suggest you hire someone to do this. Ed Storms Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
In reply to Axil Axil's message of Tue, 4 Mar 2014 16:54:13 -0500: Hi, It seems I got the magnitudes reversed. However consider the following:- Below the Curie Temperature Ni behaves as a Ferromagnetic material, and increases the field strength when a current is applied, as it's magnetic domains align with the field, and one another. However as energy is generated in the reactor, and the temperature rises above the Curie temperature, the magnet is destroyed leaving the magnetic domains in a disordered state, where they remain as the metal cools toward the end of the cycle. When the next cycle begins, they become ordered once again. DGT: After each triggering duty cycle (the triggering sequences producing excess heat), the magnetic fields at ~18 cm from the reactor at all three locations rose from ~0.6 Tesla to ~1.6 Tesla (DC peak) during each reaction period. Such anomalous peak signals were maintained for approximately 3-4 sec after the HV currents were cut off. Axil: The duty cycle is triggered by a high voltage current. After the HV spark, the magnetic field measures .6 tesla. The magnetic field grows along with the reaction for 3-4 seconds. At the peak of the reaction cycle, the magnetic field is 1.6 tesla. Read it yourself here: http://www.physics.purdue.edu/people/faculty/yekim/ICCF-18-JCMNS-KH-Pre-2.pdf [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
Here is my take on nickel and the Curie temperature. First, the Ni/H reactor will not work well if its operating temperature is below the Curie temperature. A cold reactor will radiate gamma rays. At low temperatures, the nuclear reaction is not part of the magnetic based positive feedback loop and gamma radiation that is produced escapes to the far field and is not thermalized. When the operating temperature rises above the Curie temperature, organized global ferrimagnetism is destroyed leaving the magnetic domains in a localized though organize state of magnetic vortex formation, where they remain until the metal cools at shutdown. These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. These soliton waveforms produce anapole magnetic field on the atomic scale of about 10^16 tesla. These fields' condense mesons from the vacuum which produces nuclear cluster fusion reactions in the surrounding matter. On Tue, Mar 4, 2014 at 8:58 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 4 Mar 2014 16:54:13 -0500: Hi, It seems I got the magnitudes reversed. However consider the following:- Below the Curie Temperature Ni behaves as a Ferromagnetic material, and increases the field strength when a current is applied, as it's magnetic domains align with the field, and one another. However as energy is generated in the reactor, and the temperature rises above the Curie temperature, the magnet is destroyed leaving the magnetic domains in a disordered state, where they remain as the metal cools toward the end of the cycle. When the next cycle begins, they become ordered once again. DGT: After each triggering duty cycle (the triggering sequences producing excess heat), the magnetic fields at ~18 cm from the reactor at all three locations rose from ~0.6 Tesla to ~1.6 Tesla (DC peak) during each reaction period. Such anomalous peak signals were maintained for approximately 3-4 sec after the HV currents were cut off. Axil: The duty cycle is triggered by a high voltage current. After the HV spark, the magnetic field measures .6 tesla. The magnetic field grows along with the reaction for 3-4 seconds. At the peak of the reaction cycle, the magnetic field is 1.6 tesla. Read it yourself here: http://www.physics.purdue.edu/people/faculty/yekim/ICCF-18-JCMNS-KH-Pre-2.pdf [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Christopher H. Cooper
On Tue, Mar 4, 2014 at 6:27 AM, Jones Beene jone...@pacbell.net wrote: In his Arata replication, Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogen than Pd alone. This is interesting. But now we're talking about an Ni-Pd alloy, and neither Ni nor Pd. Perhaps there is a mismatch of some kind that causes the lattice spacing to increase. This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. Sure. It would seem that there are different reactants and byproducts in NiH and PdD; for example, in the case of PdD we know about 4He and occasionally tritium, and we have no evidence that I know of for either of these in the case of NiH. I still see similarities between the two systems, though. My working assumption is that both NiH and PdD (as well as W, Ti, etc.) involve fusion in some way. Both are without gammas. Both are systems involving hydrogen and transition metals. There's reason to think that reactions in both systems occur at the surface or near it. None of this is to say that there's not a complex series of steps involved or a large parameter space. But I have not seen any compelling reason to conclude that the systems are different at a basic level, and much to suggest that what is at work in both of them is similar or analogous, with different inputs and different parameters which influence the outcomes. Eric
Re: [Vo]:Christopher H. Cooper
It might be correct to say that there is one basic cause with many possible effects. Take the acceleration in the decay of radioactive isotopes. Such an effect is a hard one to explain. On Tue, Mar 4, 2014 at 11:48 PM, Eric Walker eric.wal...@gmail.com wrote: On Tue, Mar 4, 2014 at 6:27 AM, Jones Beene jone...@pacbell.net wrote: In his Arata replication, Ahern found that an alloy of mostly nickel with less than 10% Pd takes up more hydrogen than Pd alone. This is interesting. But now we're talking about an Ni-Pd alloy, and neither Ni nor Pd. Perhaps there is a mismatch of some kind that causes the lattice spacing to increase. This is yet another reason, one of many - why consideration of all the evidence, giving no preference to Pd-D, points to many different routes to gain in LENR. Sure. It would seem that there are different reactants and byproducts in NiH and PdD; for example, in the case of PdD we know about 4He and occasionally tritium, and we have no evidence that I know of for either of these in the case of NiH. I still see similarities between the two systems, though. My working assumption is that both NiH and PdD (as well as W, Ti, etc.) involve fusion in some way. Both are without gammas. Both are systems involving hydrogen and transition metals. There's reason to think that reactions in both systems occur at the surface or near it. None of this is to say that there's not a complex series of steps involved or a large parameter space. But I have not seen any compelling reason to conclude that the systems are different at a basic level, and much to suggest that what is at work in both of them is similar or analogous, with different inputs and different parameters which influence the outcomes. Eric
Re: [Vo]:Christopher H. Cooper
That just the CMFV theory of fusion. - Original Message - From: Axil Axil To: vortex-l Sent: Tuesday, March 04, 2014 6:37 PM Subject: Re: [Vo]:Christopher H. Cooper Here is my take on nickel and the Curie temperature. First, the Ni/H reactor will not work well if its operating temperature is below the Curie temperature. A cold reactor will radiate gamma rays. At low temperatures, the nuclear reaction is not part of the magnetic based positive feedback loop and gamma radiation that is produced escapes to the far field and is not thermalized. When the operating temperature rises above the Curie temperature, organized global ferrimagnetism is destroyed leaving the magnetic domains in a localized though organize state of magnetic vortex formation, where they remain until the metal cools at shutdown. These local vortex formations provide templates upon which the solitons will condense. These quantum cavities absorbed both gamma radiation from nuclear reactions and infrared radiation from the reactor structure and amalgamate these waves into a XUV soliton waveform resonant with the diameter of the quantum cavity: about 1 to 2 nanometers. These soliton waveforms produce anapole magnetic field on the atomic scale of about 10^16 tesla. These fields' condense mesons from the vacuum which produces nuclear cluster fusion reactions in the surrounding matter. On Tue, Mar 4, 2014 at 8:58 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 4 Mar 2014 16:54:13 -0500: Hi, It seems I got the magnitudes reversed. However consider the following:- Below the Curie Temperature Ni behaves as a Ferromagnetic material, and increases the field strength when a current is applied, as it's magnetic domains align with the field, and one another. However as energy is generated in the reactor, and the temperature rises above the Curie temperature, the magnet is destroyed leaving the magnetic domains in a disordered state, where they remain as the metal cools toward the end of the cycle. When the next cycle begins, they become ordered once again. DGT: After each triggering duty cycle (the triggering sequences producing excess heat), the magnetic fields at ~18 cm from the reactor at all three locations rose from ~0.6 Tesla to ~1.6 Tesla (DC peak) during each reaction period. Such anomalous peak signals were maintained for approximately 3-4 sec after the HV currents were cut off. Axil: The duty cycle is triggered by a high voltage current. After the HV spark, the magnetic field measures .6 tesla. The magnetic field grows along with the reaction for 3-4 seconds. At the peak of the reaction cycle, the magnetic field is 1.6 tesla. Read it yourself here: http://www.physics.purdue.edu/people/faculty/yekim/ICCF-18-JCMNS-KH-Pre-2.pdf [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: EXTERNAL: RE: [Vo]:Christopher H. Cooper
Jones, I was going to add something to that effect.. I have a good feeling regarding SPP as the bootstrap energy source. It is one less miracle compared to my theory of runaway discounts of the disassociation threshold allowing fractional hydrogen to oscillate between bond states powered by random motion.. with SPP the geometry only needs to form the fractional hydrogen states.. I am very ok with SPP being the missing piece of the puzzle - I was working to hard to explain the initial source of energy and this solution is not only elegant but certainly fits the wider range of geometries Axil notes in the Rossi tubules.. should we be looking for similar geometries in Mills skeletal cats? Fran _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: Sunday, March 02, 2014 12:32 PM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:Christopher H. Cooper From: Frank roarty Again..the nanotube is only going to be active at the openings and defects.. It is a macro example of the difference between Casimir and dynamic Casimir effect and we clearly need a robust dynamic effect along with robust thermal linkage to prevent it from self destructing. Fran, This may be partly true (that there is a Casimir connection, and anytime there is a Casimir geometry this is likely), nevertheless, at least in Cooper's patent/experiment CNT alone is not enough - with or without a Casimir contribution. Not even close. CNT and electrical current will NOT come close to a nuclear effect either. Thus, CNT is not a substitute for a palladium lattice in any way shape or form. We are dealing with a completely different form of LENR with plasmons, and not the same type which is found in Pd-D. The must be an significant power input to trigger the LENR reaction - and if the only apparent input is low power, such as visible light photons - then clearly there must be an amplification mechanism for that input. The amplification must be in the range of 100,000:1 or more. SPP can do that and perhaps the Casimir force is contributory - since the geometry is in the correct range. This is why the patent application is appealing even if Cooper himself did not realize what he had stumbled upon with SPP. Which is to say that even the inventor may have missed the key point of the light source, and thus the experiment begs to be replicated with a focus on SPP and a coherent light source. Note that I am not saying that the Casimir force cannot be contributory, but only that CNT and Casimir alone are not enough, even if you add electrical input (there will be no LERN). BTW - CNT were added to an electrolysis cell 5 years ago in an experiment with light water - and there was no gain whatsoever. There was a video of that failed effort on YouTube and this was known for many years - so the bottom line is: what we must have to achieve LENR is an extreme amplification mechanism for the power input. Unfortunately, it appears that Ed may have attempted to replicate only part of the experiment, the CNT part only - and that is because the inventor did not recognize SPP, not did Ed - since he is convinced, despite NASA's support - that SPP do not represent an effective amplification mechanism. If I had to guess, since Ed cannot talk about his attempt, my conjecture is that he tried to use CNT in heavy water with electrical current and an electrolyte, but with no coherent light source. That approach is almost guaranteed to fail, and it was shown to have failed as far back as 2008. All the RD out there seems to support the idea that surface plasmons do indeed constitute an extraordinary amplification mechanism - so why not take advantage of the expertise of the scientific literature on this particular point, including the support of NASA and others (W-L jumped on the SPP band-wagon). In the end, I think the issue of failure to replicate Cooper's patent application may be one of intransigence, based on an incorrect mindset from the start- one that failed to understand the advantages of SPP. That is forgivable since the inventor himself did not recognize it either - but what is not forgivable is continuing intransigence now that this issue has been highlighted. From: Edmund Storms Nice thought Kevin. Chris and I collaborated to see if CNT were nuclear active. They were not, at least when using our methods. I suspect the conditions in the tube are not correct to form the Hydroton. Well, it is good to know that you and Chris collaborated, but not so good to learn that his technique may not work, as claimed. Can you describe what methods were used? Did you use a coherent or nearly coherent light source? Without a source of coherent light, SPP are unlikely to form. Jones
RE: EXTERNAL: Re: [Vo]:Christopher H. Cooper
Dave, I think this is where geometry comes in, these anomalies are confined to fewer dimensions creating an imbalance to this normal cancelation you correctly identified. It is bordering on 2d when suppression is at its most robust as an inverse cube of the spacing between boundaries. Fran From: David Roberson [mailto:dlrober...@aol.com] Sent: Sunday, March 02, 2014 3:01 PM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Christopher H. Cooper Interesting. But how does the net field become large unless some mechanism coordinates the destruction of the balls? Many random direction vectors yields near zero sums. Dave -Original Message- From: Axil Axil janap...@gmail.commailto:janap...@gmail.com To: vortex-l vortex-l@eskimo.commailto:vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:55 pm Subject: Re: [Vo]:Christopher H. Cooper Yes, there is a load of fun in this sort of speculation. One possibility is that micro sized magnetic balls as described by DGT that start small and grow to huge power until they explode could produce a varying magnetic field that would induce a current through changing magnetic flux.. On Sun, Mar 2, 2014 at 2:46 PM, David Roberson dlrober...@aol.commailto:dlrober...@aol.com wrote: That brings back fond memories. He does say e.m.f. which makes me wonder how he performed that measurement. I would anticipate that he must use at least two probes to come to that conclusion and his active material hopefully does not short out the voltage. Another possibility is that he measured a large magnetic field which he assumes must be as a result of DC current flowing. Since DC current or AC for that matter requires a loop voltage in order to flow, it makes sense to believe that an e.m.f. is present. Actually, an e.m.f. should be present in that case and what Rossi states below about an expert observing it falls into line. I find myself wondering if there are other good ways to achieve very high strength magnetic fields without currents flowing. Permanent magnets offer a clue. I am guessing here and attempting to decode Rossi speak at the same time. That has its hazards! :-) Dave -Original Message- From: Axil Axil janap...@gmail.commailto:janap...@gmail.com To: vortex-l vortex-l@eskimo.commailto:vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:25 pm Subject: Re: [Vo]:Christopher H. Cooper Andrea Rossi December 30th, 2012 at 3:01 PM http://www.journal-of-nuclear-physics.com/?p=771cpage=4#comment-514345 Dear Bernie Koppenhofer: You are touching a very important point: during these very days, and also during the more recent tests, we are working on this issue. I think we will be able to produce directly e.m.f. , but much work has to be done. Actually, we already produced direct e.m.f. with the reactors at high temperature, and we measured it with the very precise measurement instrumentation introduced by the third party expert, but we are not ready for an industrial production, while we are at a high level of industrialization for the production of heat and, at this point , also of high temperature steam, which is the gate to the Carnot Cycle. Thank you for your good comment. Warm Regards, A.R. On Sun, Mar 2, 2014 at 2:04 PM, Axil Axil janap...@gmail.commailto:janap...@gmail.com wrote: I believe that heat is not the only product of the LENR reaction. It may not even the most important sink for LENR power generation. I believe that electron production is a major magnification of over unity power generation. Rossi indicated that there was an unknown source of current production in his reactor and he was looking into how this could happen. I know that the PAPP engine produced current out of whole cloth. The design of the engine depended on it. Here is my take on where these electrons are coming from. When the magnetic field strength gets strong enough, mesons are condensed out of the vacuum. The final decay products of mesons are electrons. On Sun, Mar 2, 2014 at 1:34 PM, David Roberson dlrober...@aol.commailto:dlrober...@aol.com wrote: I also find it amazing that DGT seems to overlook the implications of their discovery. It reminds me of not seeing the forest through the trees. Since Rossi made an earlier claim that he might be able to generate electricity directly by some obscure discovery, I suspect that he realized the importance of the large magnetic fields residing within his device. So far he has kept this type of information private, carefully leaking out the news of some non specific discovery. Rossi knows when to release findings that might assist competitors. Dave -Original Message- From: Axil Axil janap...@gmail.commailto:janap...@gmail.com To: vortex-l vortex-l@eskimo.commailto:vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 1:23 pm Subject: Re: [Vo]:Christopher H. Cooper Like you, any one of us can only do so much of what is required. To come up with an all inclusive
Re: [Vo]:Christopher H. Cooper
Agreed..and this field seems to require a careful balance upon the head of a pin to keep the active region heat sunk enough to draw off energy while not allowing the reaction to drop off or run away. This is why I posit that eventually there will be birth to grave precautions taken to safeguard the geometry and why I think so many previous tests have failed like MAHG and Patterson beads that could have been both more robust and lasted longer [repeatability] had the materials been created and maintained with an inert blanket.. Mills does this to a limited level by keeping Rayney Nickel wet but even there he has the potential for water vapor to react with the most active regions. Fran From: Axil Axil [mailto:janap...@gmail.com] Sent: Sunday, March 02, 2014 3:04 PM To: vortex-l Subject: EXTERNAL: Re: [Vo]:Christopher H. Cooper What is the course of an open ender positive feedback loop without limit. An eventual explosion. Nothing lasts forever in a positive feedback loop. There is always a limit to everything. On Sun, Mar 2, 2014 at 3:00 PM, David Roberson dlrober...@aol.commailto:dlrober...@aol.com wrote: Interesting. But how does the net field become large unless some mechanism coordinates the destruction of the balls? Many random direction vectors yields near zero sums. Dave -Original Message- From: Axil Axil janap...@gmail.commailto:janap...@gmail.com To: vortex-l vortex-l@eskimo.commailto:vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:55 pm Subject: Re: [Vo]:Christopher H. Cooper Yes, there is a load of fun in this sort of speculation. One possibility is that micro sized magnetic balls as described by DGT that start small and grow to huge power until they explode could produce a varying magnetic field that would induce a current through changing magnetic flux.. On Sun, Mar 2, 2014 at 2:46 PM, David Roberson dlrober...@aol.commailto:dlrober...@aol.com wrote: That brings back fond memories. He does say e.m.f. which makes me wonder how he performed that measurement. I would anticipate that he must use at least two probes to come to that conclusion and his active material hopefully does not short out the voltage. Another possibility is that he measured a large magnetic field which he assumes must be as a result of DC current flowing. Since DC current or AC for that matter requires a loop voltage in order to flow, it makes sense to believe that an e.m.f. is present. Actually, an e.m.f. should be present in that case and what Rossi states below about an expert observing it falls into line. I find myself wondering if there are other good ways to achieve very high strength magnetic fields without currents flowing. Permanent magnets offer a clue. I am guessing here and attempting to decode Rossi speak at the same time. That has its hazards! :-) Dave -Original Message- From: Axil Axil janap...@gmail.commailto:janap...@gmail.com To: vortex-l vortex-l@eskimo.commailto:vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:25 pm Subject: Re: [Vo]:Christopher H. Cooper Andrea Rossi December 30th, 2012 at 3:01 PM http://www.journal-of-nuclear-physics.com/?p=771cpage=4#comment-514345 Dear Bernie Koppenhofer: You are touching a very important point: during these very days, and also during the more recent tests, we are working on this issue. I think we will be able to produce directly e.m.f. , but much work has to be done. Actually, we already produced direct e.m.f. with the reactors at high temperature, and we measured it with the very precise measurement instrumentation introduced by the third party expert, but we are not ready for an industrial production, while we are at a high level of industrialization for the production of heat and, at this point , also of high temperature steam, which is the gate to the Carnot Cycle. Thank you for your good comment. Warm Regards, A.R. On Sun, Mar 2, 2014 at 2:04 PM, Axil Axil janap...@gmail.commailto:janap...@gmail.com wrote: I believe that heat is not the only product of the LENR reaction. It may not even the most important sink for LENR power generation. I believe that electron production is a major magnification of over unity power generation. Rossi indicated that there was an unknown source of current production in his reactor and he was looking into how this could happen. I know that the PAPP engine produced current out of whole cloth. The design of the engine depended on it. Here is my take on where these electrons are coming from. When the magnetic field strength gets strong enough, mesons are condensed out of the vacuum. The final decay products of mesons are electrons. On Sun, Mar 2, 2014 at 1:34 PM, David Roberson dlrober...@aol.commailto:dlrober...@aol.com wrote: I also find it amazing that DGT seems to overlook the implications of their discovery. It reminds me of not seeing the forest through the trees. Since Rossi made an earlier claim that he might be able to generate
Re: [Vo]:Christopher H. Cooper
Axil-- Fission reactors with water cooling generally have a negative temperature feedback and are much safer than metal coolant reactors with positive temperature feedback. However, metal cooled reactors have been designed and worked ok. With good design even a positive temperature feedback may work. In a QM system things happen so fast it would be harder to control than in a fission reactor. The key for control may be to limit the size of the QM system that reacts at any time, or increase the response time of the initiator--may the on-off pulse of the magnetic field in the case of the Pd and Ni systems. Bob - Original Message - From: Axil Axil To: vortex-l Sent: Sunday, March 02, 2014 12:04 PM Subject: Re: [Vo]:Christopher H. Cooper What is the course of an open ender positive feedback loop without limit. An eventual explosion. Nothing lasts forever in a positive feedback loop. There is always a limit to everything. On Sun, Mar 2, 2014 at 3:00 PM, David Roberson dlrober...@aol.com wrote: Interesting. But how does the net field become large unless some mechanism coordinates the destruction of the balls? Many random direction vectors yields near zero sums. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:55 pm Subject: Re: [Vo]:Christopher H. Cooper Yes, there is a load of fun in this sort of speculation. One possibility is that micro sized magnetic balls as described by DGT that start small and grow to huge power until they explode could produce a varying magnetic field that would induce a current through changing magnetic flux.. On Sun, Mar 2, 2014 at 2:46 PM, David Roberson dlrober...@aol.com wrote: That brings back fond memories. He does say e.m.f. which makes me wonder how he performed that measurement. I would anticipate that he must use at least two probes to come to that conclusion and his active material hopefully does not short out the voltage. Another possibility is that he measured a large magnetic field which he assumes must be as a result of DC current flowing. Since DC current or AC for that matter requires a loop voltage in order to flow, it makes sense to believe that an e.m.f. is present. Actually, an e.m.f. should be present in that case and what Rossi states below about an expert observing it falls into line. I find myself wondering if there are other good ways to achieve very high strength magnetic fields without currents flowing. Permanent magnets offer a clue. I am guessing here and attempting to decode Rossi speak at the same time. That has its hazards! :-) Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:25 pm Subject: Re: [Vo]:Christopher H. Cooper Andrea Rossi December 30th, 2012 at 3:01 PM http://www.journal-of-nuclear-physics.com/?p=771cpage=4#comment-514345 Dear Bernie Koppenhofer: You are touching a very important point: during these very days, and also during the more recent tests, we are working on this issue. I think we will be able to produce directly e.m.f. , but much work has to be done. Actually, we already produced direct e.m.f. with the reactors at high temperature, and we measured it with the very precise measurement instrumentation introduced by the third party expert, but we are not ready for an industrial production, while we are at a high level of industrialization for the production of heat and, at this point , also of high temperature steam, which is the gate to the Carnot Cycle. Thank you for your good comment. Warm Regards, A.R. On Sun, Mar 2, 2014 at 2:04 PM, Axil Axil janap...@gmail.com wrote: I believe that heat is not the only product of the LENR reaction. It may not even the most important sink for LENR power generation. I believe that electron production is a major magnification of over unity power generation. Rossi indicated that there was an unknown source of current production in his reactor and he was looking into how this could happen. I know that the PAPP engine produced current out of whole cloth. The design of the engine depended on it. Here is my take on where these electrons are coming from. When the magnetic field strength gets strong enough, mesons are condensed out of the vacuum. The final decay products of mesons are electrons. On Sun, Mar 2, 2014 at 1:34 PM, David Roberson dlrober...@aol.com wrote: I also find it amazing that DGT seems to overlook the implications of their discovery. It reminds me of not seeing the forest through the trees. Since Rossi made
Re: [Vo]:Christopher H. Cooper
On Mon, Mar 3, 2014 at 10:06 AM, Bob Cook frobertc...@hotmail.com wrote: Axil-- Fission reactors with water cooling generally have a negative temperature feedback and are much safer than metal coolant reactors with positive temperature feedback. However, metal cooled reactors have been designed and worked ok. With good design even a positive temperature feedback may work. In a uranium reactor, U238 provides the negative temperature control through Doppler broadening. http://www.safetyinengineering.com/FileUploads/Nuclear%20reactor%20stability%20and%20controllability_1314016641_2.pdf Light water absorbs more neutrons then heavy water and sodium hardly absorbs any neutrons (fast ones) at all. Designing a fission reactor requires a lot of experience and education. Positive coefficient of reactivity can never be positive. That is inviting a possibility of super criticality. A reactor that can go super critical cannot be licensed. In a QM system things happen so fast it would be harder to control than in a fission reactor. The key for control may be to limit the size of the QM system that reacts at any time, or increase the response time of the initiator--may the on-off pulse of the magnetic field in the case of the Pd and Ni systems. The DGT LENR reactor is only supercritical when the spark is arcing. But when the spark is off, that reactor returns to sub criticality. DGT tossed Rossi out of their deal because his reactor can go super critical. DGT designed their home grown reactor to be inherently safe through sub criticality just like all fission reactors. Bob - Original Message - *From:* Axil Axil janap...@gmail.com *To:* vortex-l vortex-l@eskimo.com *Sent:* Sunday, March 02, 2014 12:04 PM *Subject:* Re: [Vo]:Christopher H. Cooper What is the course of an open ender positive feedback loop without limit. An eventual explosion. Nothing lasts forever in a positive feedback loop. There is always a limit to everything. On Sun, Mar 2, 2014 at 3:00 PM, David Roberson dlrober...@aol.com wrote: Interesting. But how does the net field become large unless some mechanism coordinates the destruction of the balls? Many random direction vectors yields near zero sums. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:55 pm Subject: Re: [Vo]:Christopher H. Cooper Yes, there is a load of fun in this sort of speculation. One possibility is that micro sized magnetic balls as described by DGT that start small and grow to huge power until they explode could produce a varying magnetic field that would induce a current through changing magnetic flux.. On Sun, Mar 2, 2014 at 2:46 PM, David Roberson dlrober...@aol.comwrote: That brings back fond memories. He does say e.m.f. which makes me wonder how he performed that measurement. I would anticipate that he must use at least two probes to come to that conclusion and his active material hopefully does not short out the voltage. Another possibility is that he measured a large magnetic field which he assumes must be as a result of DC current flowing. Since DC current or AC for that matter requires a loop voltage in order to flow, it makes sense to believe that an e.m.f. is present. Actually, an e.m.f. should be present in that case and what Rossi states below about an expert observing it falls into line. I find myself wondering if there are other good ways to achieve very high strength magnetic fields without currents flowing. Permanent magnets offer a clue. I am guessing here and attempting to decode Rossi speak at the same time. That has its hazards! :-) Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Mar 2, 2014 2:25 pm Subject: Re: [Vo]:Christopher H. Cooper Andrea Rossi December 30th, 2012 at 3:01 PM http://www.journal-of-nuclear-physics.com/?p=771cpage=4#comment-514345 Dear Bernie Koppenhofer: You are touching a very important point: during these very days, and also during the more recent tests, we are working on this issue. I think we will be able to produce directly e.m.f. , but much work has to be done. Actually, we already produced direct e.m.f. with the reactors at high temperature, and we measured it with the very precise measurement instrumentation introduced by the third party expert, but we are not ready for an industrial production, while we are at a high level of industrialization for the production of heat and, at this point , also of high temperature steam, which is the gate to the Carnot Cycle. Thank you for your good comment. Warm Regards, A.R. On Sun, Mar 2, 2014 at 2:04 PM, Axil Axil janap...@gmail.com wrote: I believe that heat is not the only product of the LENR reaction. It may not even the most important sink for LENR power generation. I
