Re: [Vo]:Curious irony
On Wed, Apr 24, 2013 at 10:07 PM, Jones Beene jone...@pacbell.net wrote: Harry In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. This is not accurate. The diproton fissions back into two protons the vast majority of the time. The Wiki article is not well-worded on this point but later on it corrects the misunderstanding. It is only the rare occasion where the positron is emitted - otherwise the Sun would burn up its fuel too quickly. yes In RPF, Reversible Proton Fusion - the two protons which are immediately split from nascent He-2 are technically not the original two protons which fused, since there has been color charge alteration in the quarks during the brief instant when they were fused. I am interested in rookie protons which haven't formed anything beyond a diproton or a deuteron. Harry
RE: [Vo]:Curious irony
-Original Message- From: mix...@bigpond.com Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. The mass of two protons is 2.014552933 amu. The mass of a deuteron is 2.01355362 amu. RVS: Note that the deuteron is actually lighter than the two protons. IOW this reaction is exothermic. Not exactly true, Robin. Once again - the deuteron does NOT form directly from two protons! Never. There is a required step which you are leaving out, where outside energy is brought in. The deuteron forms only from a diproton, which itself has formed from two protons PLUS added mass from outside the reactants. If that mass has not been added, which is the vast majority of the time, there is no reaction. IOW without added mass-energy which is brought in above the rest mass of the reactants, the reaction is endothermic. With the added mass the reaction appears exothermic, but that is due to added mass-energy from outside the reactants. Semantics allows either depiction - exothermic or endothermic, depending on whether one is looking at protons or diprotons. Since you were looking at protons, the reaction is endothermic based on their rest mass. This is similar to the situation with an accelerator - where a beam of atoms forms a new element which has an unfavorable energy balance - but the beam momentum adds the necessary mass-energy to make it happen. Protons in the sun can be accelerated locally to high energies. Jones attachment: winmail.dat
Re: [Vo]:Curious irony
In reply to Jones Beene's message of Thu, 25 Apr 2013 06:38:34 -0700: Hi Jones,Terry, I'll respond to both posts at the same time. [snip] -Original Message- From: mix...@bigpond.com Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. The mass of two protons is 2.014552933 amu. The mass of a deuteron is 2.01355362 amu. RVS: Note that the deuteron is actually lighter than the two protons. IOW this reaction is exothermic. Not exactly true, Robin. Once again - the deuteron does NOT form directly from two protons! Never. There is a required step which you are leaving out, where outside energy is brought in. Both you and Terry are correct in that doing work by bringing the protons together should add mass to the system. However even without that, the reaction would be exothermic. (BTW, you get that energy back again when the reaction happens). There are two reasons the reaction is so slow. 1) It's a weak force reaction. 2) Tunneling. I have recently come to wonder whether or not tunneling can even happen in this case. That's because particles don't tunnel unless there is a net energy advantage in doing so, and there is none for two protons. So I am forced to consider the possibility that electron capture may be the sole means by which this reaction occurs in the Sun. I don't see how physicists could possibly measure the ratio of beta+ to electron capture reactions occurring in the core of the Sun anyway (see e.g. http://www.sns.ias.edu/~jnb/SNviewgraphs/NuclearFusion/nucfusion.html). (Any electron annihilation gammas created in the core would be thermalized long before they got to the surface.) Perhaps the ratio is determined from fusion experiments here on Earth, where annihilation gammas are more readily detected? Or perhaps it's just a calculated value? Or maybe this really is a WL reaction, where a proton is converted into a neutron momentarily while emitting a positron, and then the neutron fuses with a free proton in time to repay the loan before the thugs come knocking? The deuteron forms only from a diproton, which itself has formed from two protons PLUS added mass from outside the reactants. If that mass has not been added, which is the vast majority of the time, there is no reaction. IOW without added mass-energy which is brought in above the rest mass of the reactants, the reaction is endothermic. With the added mass the reaction appears exothermic, but that is due to added mass-energy from outside the reactants. Semantics allows either depiction - exothermic or endothermic, depending on whether one is looking at protons or diprotons. Since you were looking at protons, the reaction is endothermic based on their rest mass. So according to you, 2.01355362 2.01455293 ? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
In reply to Jones Beene's message of Thu, 25 Apr 2013 06:38:34 -0700: Hi, I wrote: I don't see how physicists could possibly measure the ratio of beta+ to electron capture reactions occurring in the core of the Sun anyway (see e.g. http://www.sns.ias.edu/~jnb/SNviewgraphs/NuclearFusion/nucfusion.html). (Any electron annihilation gammas created in the core would be thermalized long before they got to the surface.) Perhaps the ratio is determined from fusion experiments here on Earth, where annihilation gammas are more readily detected? Or perhaps it's just a calculated value? Or maybe this really is a WL reaction, where a proton is converted into a neutron momentarily while emitting a positron, and then the neutron fuses with a free proton in time to repay the loan before the thugs come knocking? Closer examination of John's website shows that the two reactions (beta+ electron capture) create neutrinos of different energy, so presumably neutrino measurements were used to determine the ratio. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Curious irony
Hi Robin, Well I am including the mass-energy of the positron and the neutrino, which are emitted - added to the mass of the deuteron to suggest that all of these weigh considerably more than two protons. Therefore outside energy (momentum) would have to be employed, even though real energy is emitted and that looks like exotherm. But even then, I admit that there is a math problem in that this particular electron neutrino from the diproton reaction apparently may not have the mass-energy which is seen and documented by experiments on earth -like SNO - where the mass-energy is in the several MeV range. Almost all solar neutrinos come from the diproton reaction but the Wiki entry suggests are much lower in energy than what is actually measured at SNO. Even with oscillation, I do not see how the energy could be lower on the sun where they are formed but higher when measured at SNO. In fact, it is only a guess as to what they are on the sun, despite what Wiki states as fact - since we obviously cannot measure them there. Bottom line: if they are 1 MeV neutrinos on the sun from two proton and beta decay, then the net diproton reaction is endothermic when we look at only the rest mass of the protons, and if they are 400 keV mass-energy, the reaction is balanced, and if they are 100 keV the reaction is exothermic. Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. The mass of two protons is 2.014552933 amu. The mass of a deuteron is 2.01355362 amu. RVS: Note that the deuteron is actually lighter than the two protons. IOW this reaction is exothermic. Not exactly true, Robin. Once again - the deuteron does NOT form directly from two protons! Never. There is a required step which you are leaving out, where outside energy is brought in. Both you and Terry are correct in that doing work by bringing the protons together should add mass to the system. However even without that, the reaction would be exothermic. (BTW, you get that energy back again when the reaction happens). There are two reasons the reaction is so slow. 1) It's a weak force reaction. 2) Tunneling. I have recently come to wonder whether or not tunneling can even happen in this case. That's because particles don't tunnel unless there is a net energy advantage in doing so, and there is none for two protons. So I am forced to consider the possibility that electron capture may be the sole means by which this reaction occurs in the Sun. I don't see how physicists could possibly measure the ratio of beta+ to electron capture reactions occurring in the core of the Sun anyway (see e.g. http://www.sns.ias.edu/~jnb/SNviewgraphs/NuclearFusion/nucfusion.html). (Any electron annihilation gammas created in the core would be thermalized long before they got to the surface.) Perhaps the ratio is determined from fusion experiments here on Earth, where annihilation gammas are more readily detected? Or perhaps it's just a calculated value? Or maybe this really is a WL reaction, where a proton is converted into a neutron momentarily while emitting a positron, and then the neutron fuses with a free proton in time to repay the loan before the thugs come knocking? The deuteron forms only from a diproton, which itself has formed from two protons PLUS added mass from outside the reactants. If that mass has not been added, which is the vast majority of the time, there is no reaction. IOW without added mass-energy which is brought in above the rest mass of the reactants, the reaction is endothermic. With the added mass the reaction appears exothermic, but that is due to added mass-energy from outside the reactants. Semantics allows either depiction - exothermic or endothermic, depending on whether one is looking at protons or diprotons. Since you were looking at protons, the reaction is endothermic based on their rest mass. So according to you, 2.01355362 2.01455293 ? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html attachment: winmail.dat
Re: [Vo]:Curious irony
In reply to Jones Beene's message of Thu, 25 Apr 2013 17:36:50 -0700: Hi, [snip] Hi Robin, Well I am including the mass-energy of the positron and the neutrino, which are emitted - added to the mass of the deuteron to suggest that all of these weigh considerably more than two protons. Therefore outside energy (momentum) would have to be employed, even though real energy is emitted and that looks like exotherm. The difference in rest mass energy between two protons and a deuteron is 930.854 keV. The rest mass of an electron/positron is 511 keV. That leaves 930.854 - 511 = 419.8 keV left over to be shared as kinetic energy between the neutrino and the positron. Since the actual distribution will vary from one fusion reaction to the next, the neutrino has a maximum energy of 419.8 keV. John's web page quotes a maximum of 423 keV (first line in the p-p nuclear fusion reactions diagram), the difference due to slight inaccuracies in masses quoted. When the positron annihilates an electron (later on), another 1.02 MeV is liberated. In the electron capture version of the reaction, the electron takes part in the initial reaction directly, rather than in a delayed annihilation reaction, so the net energy from an electron capture reaction is 1.02 MeV + 423 keV = 1.44 MeV, all of which is carried by the neutrino. (See the second line in the p-p nuclear fusion reactions diagram on John's web page.) The neutrino gets the lot, because there is no excited state of the deuteron with energy less than 1.44 MeV, so no gamma emission is possible. The only particle leaving the reaction is the neutrino. (And the deuteron itself of course, but the mass of the neutrino must be so slight that it gets nearly 100% of the energy.) But even then, I admit that there is a math problem in that this particular electron neutrino from the diproton reaction apparently may not have the mass-energy which is seen and documented by experiments on earth -like SNO - where the mass-energy is in the several MeV range. Almost all solar neutrinos come from the diproton reaction but the Wiki entry suggests are much lower in energy than what is actually measured at SNO. I think the SNO neutrinos are from other fusion reactions. Take a look at John's web page. http://www.sns.ias.edu/~jnb/SNviewgraphs/NuclearFusion/nucfusion.html Even with oscillation, I do not see how the energy could be lower on the sun where they are formed but higher when measured at SNO. In fact, it is only a guess as to what they are on the sun, despite what Wiki states as fact - since we obviously cannot measure them there. Bottom line: if they are 1 MeV neutrinos on the sun from two proton and beta decay, then the net diproton reaction is endothermic when we look at only the rest mass of the protons, and if they are 400 keV mass-energy, the reaction is balanced, and if they are 100 keV the reaction is exothermic. Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. As mentioned previously, nucleons bound in nuclei have less mass than free nucleons. It works like this: Take two blocks. Weigh them separately. Add the weights together. Glue the blocks together. Weigh the glued combination. It weighs less than the blocks did initially (well it does when you use nucleons ;). The missing mass has been converted to energy - nuclear fusion energy. However the two blocks are still present in the glued mass. Conclusion:- The blocks in the glued mass weigh less than separate autonomous blocks. Therefore, you can't just use the mass of a free neutron when talking about the mass of a neutron bound in a deuterium nucleus. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 5:45 PM, mix...@bigpond.com wrote: In reply to Harry Veeder's message of Tue, 23 Apr 2013 14:28:00 -0400: Hi, [snip] If a neutron can be made decay while in a deuteron then it seems to me the warming of the lattice is best explained by the motion arising from the mutual repulsion of the protons. Thermalization of gammas is not necessary. When a deuteron is formed from a neutron and a proton, mass is converted into energy (2.2 MeV). It is this loss of energy that prevents a neutron already in a deuteron from decaying. Decay only happens when it results in the release of energy, and the neutron bound in D has already lost too much for it to decay. Regards, In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. However, occasionally one of the protons transforms into a neutron by emitting a beta and a neutrino before fission occurs. This results in a stable deuteron. If this is correct, then a deuteron is stable because it is in a lower energy state than the diproton. (Remember a diproton has been pushed together under high pressures and temperatures so it contains more potential energy than two isolated protons or an isolated an proton and a isolated neutron.) Therefore a deuteron will return to the same level of instability as a diproton if it absorbs enough energy again. The energy profile of the deuteron and proton can be characterised by using the visual aid of mountain with depression at the top. The potential energy of a diproton corresponds with two protons resting on opposite sides of the depression. The potential energy of the deuteron corresponds to a proton and neutron resting in the depression. Harry
Re: [Vo]:Curious irony
In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. Darn, I forgot to correct that. It should say deuteron formation begins with the fusion of two protons and the emission of positron and neutrino. Since the diproton is very unstable it usually fissions soon after. Harry
RE: [Vo]:Curious irony
Harry In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. This is not accurate. The diproton fissions back into two protons the vast majority of the time. The Wiki article is not well-worded on this point but later on it corrects the misunderstanding. It is only the rare occasion where the positron is emitted - otherwise the Sun would burn up its fuel too quickly. In RPF, Reversible Proton Fusion - the two protons which are immediately split from nascent He-2 are technically not the original two protons which fused, since there has been color charge alteration in the quarks during the brief instant when they were fused. However, occasionally one of the protons transforms into a neutron by emitting a beta and a neutrino before fission occurs. This results in a stable deuteron. If this is correct, then a deuteron is stable because it is in a lower energy state than the diproton. Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. attachment: winmail.dat
Re: [Vo]:Curious irony
On Wed, Apr 24, 2013 at 10:07 PM, Jones Beene jone...@pacbell.net wrote: Harry In stars deuterons formation begins with the fusion of two protons into a diproton. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Since the diproton is very unstable it usually fissions soon after by emitting a positron and a neutrino. This is not accurate. The diproton fissions back into two protons the vast majority of the time. The Wiki article is not well-worded on this point but later on it corrects the misunderstanding. It is only the rare occasion where the positron is emitted - otherwise the Sun would burn up its fuel too quickly. Yes. I caught my mistake and made another post correcting it. In RPF, Reversible Proton Fusion - the two protons which are immediately split from nascent He-2 are technically not the original two protons which fused, since there has been color charge alteration in the quarks during the brief instant when they were fused. However, occasionally one of the protons transforms into a neutron by emitting a beta and a neutrino before fission occurs. This results in a stable deuteron. If this is correct, then a deuteron is stable because it is in a lower energy state than the diproton. Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. There something weird here. Usually a system is considered stable when it is in lower energy state. Mass can't be equivalent energy when used as measure of stability. Harry
Re: [Vo]:Curious irony
On Wed, Apr 24, 2013 at 10:19 PM, Harry Veeder hveeder...@gmail.com wrote: On Wed, Apr 24, 2013 at 10:07 PM, Jones Beene jone...@pacbell.net wrote: In RPF, Reversible Proton Fusion - the two protons which are immediately split from nascent He-2 are technically not the original two protons which fused, since there has been color charge alteration in the quarks during the brief instant when they were fused. However, occasionally one of the protons transforms into a neutron by emitting a beta and a neutrino before fission occurs. This results in a stable deuteron. If this is correct, then a deuteron is stable because it is in a lower energy state than the diproton. Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. Jones, You are consider the combined mass of two isolated protons. However, the mass of a diproton is greater than this and it is greater than the mass of one deuteron. Harry
RE: [Vo]:Curious irony
Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. Jones You are considering the combined mass of two isolated protons. However, the mass of a diproton is greater than this and it is greater than the mass of one deuteron. Harry The energy necessary to make up the mass difference between two protons and the deuteron - which occasionally derives from their endothermic fusion (in the solar proton chain reaction sequence) is supplied by either the momentum of the protons or by absorbed gamma radiation - which is intense in the solar core. What you are saying essentially is that proton momentum can add mass to the diproton - and that is the same thing. A diproton has too short a lifetime to accurately measure its mass, but we can assume that it gains mass from the collision energy - and that is the missing mass-energy which is necessary to balance the equation.
Re: [Vo]:Curious irony
In reply to Jones Beene's message of Wed, 24 Apr 2013 19:07:24 -0700: Hi, [snip] Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. The mass of two protons is 2.014552933 amu. The mass of a deuteron is 2.01355362 amu. Note that the deuteron is actually lighter than the two protons. IOW this reaction is exothermic. It is this mass difference that is responsible for deuterium being bound. Bound nucleons weigh less than free particles, and the amount by which they weigh less varies with the nucleus they are in. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
On Thu, Apr 25, 2013 at 12:07 AM, mix...@bigpond.com wrote: In reply to Jones Beene's message of Wed, 24 Apr 2013 19:07:24 -0700: Hi, [snip] Actually The neutron has mass slightly larger than that of a proton: 939.565378 MeV compared to 938.272046 MeV. Consequently, a deuteron has slightly more mass than a diproton. That is one of the many reasons why the reaction on the Sun, the one that results in a deuteron is extraordinarily rare. It is basically endothermic. The mass of two protons is 2.014552933 amu. The mass of a deuteron is 2.01355362 amu. Note that the deuteron is actually lighter than the two protons. IOW this reaction is exothermic. It is this mass difference that is responsible for deuterium being bound. Bound nucleons weigh less than free particles, and the amount by which they weigh less varies with the nucleus they are in. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html My argument that a diproton is more massive than a deuteron was based on adding mass from doing work to overcome electrostatic repulsion. I forgot that the strong force reduces the mass of the particles, but on balance the mass of a diproton should still be greater than a deuteron. Do you agree? Harry
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 12:30 AM, Jones Beene jone...@pacbell.net wrote: -Original Message- From: Harry Veeder * Here is an idea related to the natural propensity for a diproton to fission which you have previously mentionedSuppose a neutron within a deuteron is converted into proton. The subsequent motion of the protons due to mutual repulsion would heat the lattice. HARRY, when a neutron decays there is a lot more energy than you may suspect. Free neutrons are unstable with a half life of about 650-1000 seconds but the neutron in deuterium is assumed to be stable. If the neutron could be made to decay, the energy yield should be averaging close to 1.3 MeV so you do not have to worry about the mutual repulsion of protons. Their energy would be insignificant by comparison. according to Wikipedia free neutrons can decay in two ways 1) n → p + e− + neutrino 2) n → p + e− + neutrino + gamma In first case it is not clear how or if the energy of the decay products can thermalized. In the second case a Hagelstein process is required to thermalize the gammas. However, it is hard to see a neutron decay happening with any regularity, and the cross-section for gamma capture would be extraordinarily low. However, we have talked before about NMR techniques - which could conceivably decouple the neutron and thereby allow it to decay naturally as a free neutron. This is arguably possible because the electron of the deuteron supplies a local field of about 12 T. if memory serves, and the NRM frequencies of the proton and deuteron are very different in that field - so it might be possible to do some kind of resonant splitting - to free the neutron. The frequencies needed are not extreme, below microwave actually. Jones If a neutron can be made decay while in a deuteron then it seems to me the warming of the lattice is best explained by the motion arising from the mutual repulsion of the protons. Thermalization of gammas is not necessary. Harry
Re: [Vo]:Curious irony
To use a chemical analogy Ni62 is inert. It is not prone to change through fusion or fission. Perhaps this is the ideal context for getting other nuclei to change. harry On Mon, Apr 22, 2013 at 11:23 AM, Jones Beene jone...@pacbell.net wrote: Naïve metaphorical approach to Rossi's claim. Imagine a number of strong springs subject to compressive loads. The strongest spring gives the fastest return to normal geometry following compression, but it is always less than a full 100% return. What is the limiting factor on how close to 100% return of energy is available? Whatever that factor consists of, arguably makes the spring more subject to catastrophic failure. This kind of 5th year logic explains why it is true that in Nature - the nucleus with the highest binding strength is found in such low enrichment. By all rights Ni-62 should represent more than 3.6 percent of all nickel atoms since it has what appears to be the highest bonding strength. But there are other factors involved. Anyway - most ductile metals, like nickel, are tough because the atoms are forced together by a sea of electrons, not to be confused with the sea of Dirac. OTOH maybe the two should be confused. The negative charge agglomeration (glue) is subject to self-limiting Coulomb forces. At the limit of electron cohesive strength, we may also find a coupling to nuclear stability - and we may also find the beginning of the next plateau of friability (to continue the metaphor). Thus Ni-62 having reached the pinnacle of nuclear strength among all elements, could be in a slot where it can fail catastrophically in a way that is triggered by electron collapse, which forces an adjacent proton to merge with into new nucleus. Oops, we must first make that proton become bosonic - which is the DDL atom (deep Dirac layer), so as to appear bosonic. Roger and out, wave function collapse, the new magic - no problemo g. http://www.mail-archive.com/vortex-l@eskimo.com/msg72566.html _ If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation.
Re: [Vo]:Curious irony
The penetration of the coulomb barrier by a neutron from the outside as per WL does not guaranty a LENR reaction. Many neutrons may need to be added to get some nuclear event to occur. For example, Nickel-58 is the most abundant isotope of nickel, making up 68.077% of the natural abundance. Ni58 may require many neutrons to be added to that nucleus before it becomes unstable. DGT says that NI58 works for LENR . Nickel isotopes can go up to Ni66 so we can estimate adding 8 neutrons before some energy poor beta decay happens through beta decay. This neutron process is not energy efficient. The reaction could stop if a stable nickel isotope is reached like Ni64. This loses more energy. Adding neutrons to Nickel from outside the coulomb barrier does not look like an efficient LENR mechanism. On Tue, Apr 23, 2013 at 3:05 PM, Harry Veeder hveeder...@gmail.com wrote: To use a chemical analogy Ni62 is inert. It is not prone to change through fusion or fission. Perhaps this is the ideal context for getting other nuclei to change. harry On Mon, Apr 22, 2013 at 11:23 AM, Jones Beene jone...@pacbell.net wrote: Naïve metaphorical approach to Rossi's claim. Imagine a number of strong springs subject to compressive loads. The strongest spring gives the fastest return to normal geometry following compression, but it is always less than a full 100% return. What is the limiting factor on how close to 100% return of energy is available? Whatever that factor consists of, arguably makes the spring more subject to catastrophic failure. This kind of 5th year logic explains why it is true that in Nature - the nucleus with the highest binding strength is found in such low enrichment. By all rights Ni-62 should represent more than 3.6 percent of all nickel atoms since it has what appears to be the highest bonding strength. But there are other factors involved. Anyway - most ductile metals, like nickel, are tough because the atoms are forced together by a sea of electrons, not to be confused with the sea of Dirac. OTOH maybe the two should be confused. The negative charge agglomeration (glue) is subject to self-limiting Coulomb forces. At the limit of electron cohesive strength, we may also find a coupling to nuclear stability - and we may also find the beginning of the next plateau of friability (to continue the metaphor). Thus Ni-62 having reached the pinnacle of nuclear strength among all elements, could be in a slot where it can fail catastrophically in a way that is triggered by electron collapse, which forces an adjacent proton to merge with into new nucleus. Oops, we must first make that proton become bosonic - which is the DDL atom (deep Dirac layer), so as to appear bosonic. Roger and out, wave function collapse, the new magic - no problemo g. http://www.mail-archive.com/vortex-l@eskimo.com/msg72566.html _ If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the
RE: [Vo]:Curious irony
Harry, If Rossi were the least bit credible, Ni-62 as an active ingredient would be worth digging into deeper. Most likely, this application itself is an elaborate tactic to keep competitors at bay while another real patent application remains unpublished. It seems very unlikely to me that Ni-62 is special for gain, and that is why I was trying to add some humor into the mix. The patent application is so poorly drafted that it must be an embarrassment to whoever produced it - and it cannot have been intended to protect anything of value... so the bottom line is this: if Rossi is as clever as some think him to be, then we are pretty much falling into his snare by giving this document credibility- when surely it deserves none. Having said that - there does seem to be an inexpensive way to enrich nickel in the heavier isotopes up to perhaps 5 times natural ratios, but it's not worth mentioning. -Original Message- From: Harry Veeder To use a chemical analogy Ni62 is inert. It is not prone to change through fusion or fission. Perhaps this is the ideal context for getting other nuclei to change. Naïve metaphorical approach to Rossi's claim. Imagine a number of strong springs subject to compressive loads. The strongest spring gives the fastest return to normal geometry following compression, but it is always less than a full 100% return. What is the limiting factor on how close to 100% return of energy is available? Whatever that factor consists of, arguably makes the spring more subject to catastrophic failure. This kind of 5th year logic explains why it is true that in Nature - the nucleus with the highest binding strength is found in such low enrichment. By all rights Ni-62 should represent more than 3.6 percent of all nickel atoms since it has what appears to be the highest bonding strength. But there are other factors involved. Anyway - most ductile metals, like nickel, are tough because the atoms are forced together by a sea of electrons, not to be confused with the sea of Dirac. OTOH maybe the two should be confused. The negative charge agglomeration (glue) is subject to self-limiting Coulomb forces. At the limit of electron cohesive strength, we may also find a coupling to nuclear stability - and we may also find the beginning of the next plateau of friability (to continue the metaphor). Thus Ni-62 having reached the pinnacle of nuclear strength among all elements, could be in a slot where it can fail catastrophically in a way that is triggered by electron collapse, which forces an adjacent proton to merge with into new nucleus. Oops, we must first make that proton become bosonic - which is the DDL atom (deep Dirac layer), so as to appear bosonic. Roger and out, wave function collapse, the new magic - no problemo g. http://www.mail-archive.com/vortex-l@eskimo.com/msg72566.html _ If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation.
Re: [Vo]:Curious irony
Makes perfect sense. excess heat is being generated by the motion of the particles involved, and becoming more tightly bound and higher forces to create the new atoms would move everything more, yes no? On Sun, Apr 21, 2013 at 9:21 PM, Jones Beene jone...@pacbell.net wrote: On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation.
Re: [Vo]:Curious irony
Jones, Rossi may be playing games with his patents and the particular isotope may be inconsequential but your post has made me rethink the role of the binding energy of the lattice nuclei. Theoreticians have tended to fall into two camps. Camp 1) Excess energy comes from the loaded nuclei so the binding energy of the lattice nuclei is incidental. Camp 2) Excess energy comes the lattice nuclei so the binding energy of the loaded nuclei is incidental. In fact camp 2 is only interested in hydrogen which has no binding energy. ;-) I am proposing that the binding energy of the loaded nuclei and the lattice nuclei are both relevant but the role of binding energy of the lattice nuclei is different from that imagined by camp 2. Harry On Tue, Apr 23, 2013 at 3:49 PM, Jones Beene jone...@pacbell.net wrote: Harry, If Rossi were the least bit credible, Ni-62 as an active ingredient would be worth digging into deeper. Most likely, this application itself is an elaborate tactic to keep competitors at bay while another real patent application remains unpublished. It seems very unlikely to me that Ni-62 is special for gain, and that is why I was trying to add some humor into the mix. The patent application is so poorly drafted that it must be an embarrassment to whoever produced it - and it cannot have been intended to protect anything of value... so the bottom line is this: if Rossi is as clever as some think him to be, then we are pretty much falling into his snare by giving this document credibility- when surely it deserves none. Having said that - there does seem to be an inexpensive way to enrich nickel in the heavier isotopes up to perhaps 5 times natural ratios, but it's not worth mentioning. -Original Message- From: Harry Veeder To use a chemical analogy Ni62 is inert. It is not prone to change through fusion or fission. Perhaps this is the ideal context for getting other nuclei to change. Naïve metaphorical approach to Rossi's claim. Imagine a number of strong springs subject to compressive loads. The strongest spring gives the fastest return to normal geometry following compression, but it is always less than a full 100% return. What is the limiting factor on how close to 100% return of energy is available? Whatever that factor consists of, arguably makes the spring more subject to catastrophic failure. This kind of 5th year logic explains why it is true that in Nature - the nucleus with the highest binding strength is found in such low enrichment. By all rights Ni-62 should represent more than 3.6 percent of all nickel atoms since it has what appears to be the highest bonding strength. But there are other factors involved. Anyway - most ductile metals, like nickel, are tough because the atoms are forced together by a sea of electrons, not to be confused with the sea of Dirac. OTOH maybe the two should be confused. The negative charge agglomeration (glue) is subject to self-limiting Coulomb forces. At the limit of electron cohesive strength, we may also find a coupling to nuclear stability - and we may also find the beginning of the next plateau of friability (to continue the metaphor). Thus Ni-62 having reached the pinnacle of nuclear strength among all elements, could be in a slot where it can fail catastrophically in a way that is triggered by electron collapse, which forces an adjacent proton to merge with into new nucleus. Oops, we must first make that proton become bosonic - which is the DDL atom (deep Dirac layer), so as to appear bosonic. Roger and out, wave function collapse, the new magic - no problemo g. http://www.mail-archive.com/vortex-l@eskimo.com/msg72566.html _ If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very
Re: [Vo]:Curious irony
In reply to Harry Veeder's message of Tue, 23 Apr 2013 14:28:00 -0400: Hi, [snip] If a neutron can be made decay while in a deuteron then it seems to me the warming of the lattice is best explained by the motion arising from the mutual repulsion of the protons. Thermalization of gammas is not necessary. When a deuteron is formed from a neutron and a proton, mass is converted into energy (2.2 MeV). It is this loss of energy that prevents a neutron already in a deuteron from decaying. Decay only happens when it results in the release of energy, and the neutron bound in D has already lost too much for it to decay. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
In reply to Harry Veeder's message of Tue, 23 Apr 2013 15:05:59 -0400: Hi, [snip] To use a chemical analogy Ni62 is inert. It is not prone to change through fusion or fission. Perhaps this is the ideal context for getting other nuclei to change. harry IMO Rossi only concentrates on 62Ni because there is more of it in natural Ni that 64Ni, and because fusion with a proton results in 63Cu which is stable. Since it produces a stable isotope he can then claim that his reactor doesn't produce radioactive isotopes. In short, you need to consider his motivations when studying his claims. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 3:49 PM, Jones Beene jone...@pacbell.net wrote: Having said that - there does seem to be an inexpensive way to enrich nickel in the heavier isotopes up to perhaps 5 times natural ratios, but it's not worth mentioning. And why is it not worth mentioning?
Re: [Vo]:Curious irony
In reply to Jones Beene's message of Tue, 23 Apr 2013 12:49:27 -0700: Hi, [snip] Having said that - there does seem to be an inexpensive way to enrich nickel in the heavier isotopes up to perhaps 5 times natural ratios, but it's not worth mentioning. It might be worth considering if the natural selection that any putative fusion reaction employs isn't very strong. IOW if some radioactive isotopes are created when using natural Ni. In order to at least minimize the creation of radioisotopes, it might prove beneficial to enrich the Ni in heavier isotopes before use. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
In reply to Axil Axil's message of Tue, 23 Apr 2013 15:15:24 -0400: Hi, [snip] The penetration of the coulomb barrier by a neutron from the outside as per WL does not guaranty a LENR reaction. Many neutrons may need to be added to get some nuclear event to occur. For example, Nickel-58 is the most abundant isotope of nickel, making up 68.077% of the natural abundance. Ni58 may require many neutrons to be added to that nucleus before it becomes unstable. By unstable you appear to be referring to beta decay/alpha decay. However neither of these are necessary for the nucleus to release energy. e.g. 58Ni + n = 59Ni + 9 MeV. This results in a 59Ni nucleus in an excited state, and it soon loses the 9 MeV of energy in the form of gamma radiation as it decays to the ground state. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
Nickel-59 is a long-lived cosmogenic radionuclide with a half-life of 76,000 years. Seems like a long time to wait. On Tue, Apr 23, 2013 at 6:22 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 23 Apr 2013 15:15:24 -0400: Hi, [snip] The penetration of the coulomb barrier by a neutron from the outside as per WL does not guaranty a LENR reaction. Many neutrons may need to be added to get some nuclear event to occur. For example, Nickel-58 is the most abundant isotope of nickel, making up 68.077% of the natural abundance. Ni58 may require many neutrons to be added to that nucleus before it becomes unstable. By unstable you appear to be referring to beta decay/alpha decay. However neither of these are necessary for the nucleus to release energy. e.g. 58Ni + n = 59Ni + 9 MeV. This results in a 59Ni nucleus in an excited state, and it soon loses the 9 MeV of energy in the form of gamma radiation as it decays to the ground state. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
In reply to Axil Axil's message of Tue, 23 Apr 2013 21:46:53 -0400: Hi, [snip] Nickel-59 is a long-lived cosmogenic radionuclide with a half-life of 76,000 years. Seems like a long time to wait. No, that's a beta decay once it's in the ground state. The transition from the excited state to the ground state, with emission of 9 MeV worth of gamma(s) usually happens within a tiny fraction of a second. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
thanks On Tue, Apr 23, 2013 at 10:03 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 23 Apr 2013 21:46:53 -0400: Hi, [snip] Nickel-59 is a long-lived cosmogenic radionuclide with a half-life of 76,000 years. Seems like a long time to wait. No, that's a beta decay once it's in the ground state. The transition from the excited state to the ground state, with emission of 9 MeV worth of gamma(s) usually happens within a tiny fraction of a second. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 2:48 PM, mix...@bigpond.com wrote: IMO Rossi only concentrates on 62Ni because there is more of it in natural Ni that 64Ni, and because fusion with a proton results in 63Cu which is stable. Since it produces a stable isotope he can then claim that his reactor doesn't produce radioactive isotopes. In short, you need to consider his motivations when studying his claims. He's obviously got an interest in emphasizing a stable decay product. But there is a table in Ed Storms's book that summarizes transmutations that have been observed in LENR experiments, and they are nearly all to stable isotopes. As something I've followed less than systematically in other contexts as well, this does seem to be the case most of the time. Eric
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 3:22 PM, mix...@bigpond.com wrote: 58Ni + n = 59Ni + 9 MeV. This results in a 59Ni nucleus in an excited state, and it soon loses the 9 MeV of energy in the form of gamma radiation as it decays to the ground state. I think you're addressing a specific point. It's interesting to note that unless there's a gamma thermalization mechanism, this kind of reaction will both be deadly and will not result in much heat. I suspect it is the kinetic energy of the decay products that causes the majority of heat. Eric
Re: [Vo]:Curious irony
In reply to Eric Walker's message of Tue, 23 Apr 2013 19:48:20 -0700: Hi, [snip] I think you're addressing a specific point. It's interesting to note that unless there's a gamma thermalization mechanism, this kind of reaction will both be deadly and will not result in much heat. I suspect it is the kinetic energy of the decay products that causes the majority of heat. 1) The excited state is likely to decay in a number of steps, the sum of which is 9 MeV. That means that for each reaction there would be multiple gammas produced with varying energies. 2) The lower the energy of a gamma, the greater the absorption coefficient. IOW a larger percentage of low energy gammas are absorbed than of high energy gammas for any given material thickness. 3) Rossi uses (or used) 2 cm of lead shielding, which would stop enough of the gammas to convert a useable percentage of the energy to heat. 4) The majority or the gammas would nevertheless escape, and have not been measured, and Rossi would be dead, which he isn't. 5) Consequently, it is highly unlikely that neutron capture is the energy generating mechanism (if indeed there is one) in Rossi's device. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
On Tue, Apr 23, 2013 at 8:19 PM, mix...@bigpond.com wrote: 5) Consequently, it is highly unlikely that neutron capture is the energy generating mechanism (if indeed there is one) in Rossi's device. Can a similar argument be made for proton capture? I got the impression somewhere that proton capture is fundamentally more benign than neutron capture, both in the immediate effects and in any unstable daughters, but this could be a misunderstanding. Eric
Re: [Vo]:Curious irony
In reply to Eric Walker's message of Tue, 23 Apr 2013 20:25:05 -0700: Hi, [snip] On Tue, Apr 23, 2013 at 8:19 PM, mix...@bigpond.com wrote: 5) Consequently, it is highly unlikely that neutron capture is the energy generating mechanism (if indeed there is one) in Rossi's device. Can a similar argument be made for proton capture? I got the impression somewhere that proton capture is fundamentally more benign than neutron capture, both in the immediate effects and in any unstable daughters, but this could be a misunderstanding. Eric Normally one would also expect gamma emission from proton capture, but there is more leeway here for alternative ways of getting rid of the energy. e.g. a Hydrino molecule brings 4 particles to the party. Two protons and two electrons, so it becomes possible for the energy of the reaction to be carried away by fast particles rather than as gamma emission. (The latter is relatively slow compared to particle emission; about 5-6 orders of magnitude difference in response time.) Even a lone Hydrino brings at least an extra electron. (Better than turning up empty handed, as neutrons always do. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Curious irony
In reply to Eric Walker's message of Tue, 23 Apr 2013 20:25:05 -0700: Hi, [snip] I got the impression somewhere that proton capture is fundamentally more benign than neutron capture, both in the immediate effects and in any unstable daughters, but this could be a misunderstanding. Eric You probably have me to thank for that impression. I have been saying it long enough. :) The reason is that when fission ensues, the ratio of neutrons:protons in stable daughter nuclei is less than that in the parent nucleus[1], hence fission results in excess neutrons. That means that either free neutrons are produced (during the fission of very heavy nuclei), or at least one of the daughter nuclei ends up with too many neutrons, and is consequently radioactive. This is exacerbated when the fission reaction is brought about by adding a neutron. However if you could bring about the fission reaction by adding one or more protons instead of a neutron, then you already have a head start on balancing the neutrons and protons in the daughter nuclei, and thus improving their stability, so that they need not be radioactive. (Nature actually prefers to produce stable nuclei when it gets a chance). 1) If you look at the n:p ratio for the elements, you will see that it increases as you go up in the periodic table (once you get past about S). At the low end it tends to be 1:1, while at the high end it's about 1.6:1. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Curious irony
Naïve metaphorical approach to Rossi's claim. Imagine a number of strong springs subject to compressive loads. The strongest spring gives the fastest return to normal geometry following compression, but it is always less than a full 100% return. What is the limiting factor on how close to 100% return of energy is available? Whatever that factor consists of, arguably makes the spring more subject to catastrophic failure. This kind of 5th year logic explains why it is true that in Nature - the nucleus with the highest binding strength is found in such low enrichment. By all rights Ni-62 should represent more than 3.6 percent of all nickel atoms since it has what appears to be the highest bonding strength. But there are other factors involved. Anyway - most ductile metals, like nickel, are tough because the atoms are forced together by a sea of electrons, not to be confused with the sea of Dirac. OTOH maybe the two should be confused. The negative charge agglomeration (glue) is subject to self-limiting Coulomb forces. At the limit of electron cohesive strength, we may also find a coupling to nuclear stability - and we may also find the beginning of the next plateau of friability (to continue the metaphor). Thus Ni-62 having reached the pinnacle of nuclear strength among all elements, could be in a slot where it can fail catastrophically in a way that is triggered by electron collapse, which forces an adjacent proton to merge with into new nucleus. Oops, we must first make that proton become bosonic - which is the DDL atom (deep Dirac layer), so as to appear bosonic. Roger and out, wave function collapse, the new magic - no problemo g. http://www.mail-archive.com/vortex-l@eskimo.com/msg72566.html _ If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation. attachment: winmail.dat
Re: [Vo]:Curious irony
On Mon, Apr 22, 2013 at 12:21 AM, Jones Beene jone...@pacbell.net wrote: On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation. Jones, Here is an idea related to the natural propensity for a diproton to fission which you have previously mentioned. Suppose a neutron within a deuteron is converted into proton. The subsequent motion of the protons due to mutual repulsion would heat the lattice. The lattice does this by somehow collectively focus a gamma level of energy into the deuteron. This part would be endothermic. This represents an inversion of Haglestein's problem, where the release of gamma level energy is distributed over the lattice. Deuterium is present in Ni and Pd systems but the nature of a Ni-D system is such that a much lower ratio of D to H is optimal. Too much deuterium in a Ni system quenches the reaction as has been observed. Harry Harry
RE: [Vo]:Curious irony
-Original Message- From: Harry Veeder * Here is an idea related to the natural propensity for a diproton to fission which you have previously mentionedSuppose a neutron within a deuteron is converted into proton. The subsequent motion of the protons due to mutual repulsion would heat the lattice. HARRY, when a neutron decays there is a lot more energy than you may suspect. Free neutrons are unstable with a half life of about 650-1000 seconds but the neutron in deuterium is assumed to be stable. If the neutron could be made to decay, the energy yield should be averaging close to 1.3 MeV so you do not have to worry about the mutual repulsion of protons. Their energy would be insignificant by comparison. However, it is hard to see a neutron decay happening with any regularity, and the cross-section for gamma capture would be extraordinarily low. However, we have talked before about NMR techniques - which could conceivably decouple the neutron and thereby allow it to decay naturally as a free neutron. This is arguably possible because the electron of the deuteron supplies a local field of about 12 T. if memory serves, and the NRM frequencies of the proton and deuteron are very different in that field - so it might be possible to do some kind of resonant splitting - to free the neutron. The frequencies needed are not extreme, below microwave actually. Jones attachment: winmail.dat
RE: [Vo]:Curious irony
If Rossi's invention works - and for the reason supplied in the application, and if one wanted to apply standard logic to why the isotope with the highest binding energy per nucleon of all known nuclides is responsible, then perhaps one could pose the argument that: the one with the most - has the most to spare... To continue with a little more punagement, one could opine this kind of logic makes it Marx... ... but is it Karl or Groucho? Reason has always existed, but not always in a reasonable form Karl Marx A child of five would understand this. Send someone to fetch a child of five Groucho Marx _ On April 15th, an update has been made to the Rossi patent application at the European Patent Office - which was mentioned previously here. https://register.epo.org/espacenet/application?documentId=EUIP5C400118284nu mber=EP08873805lng=ennpl=false As you can see, Nickel-62 is featured in Claim One as the active species for the reaction, essentially making this patent very specific. The curious factoid ... or irony is that Ni-62 (NOT an iron isotope) - is a singularity in a way, being the isotope with the highest binding energy per nucleon of all known nuclides (~8.8 MeV per) and yet here it is being identified as active for the anomalous energy Rossi claims to have found with hydrogen. Jones On the one hand, if there is true gain in this device primarily due to properties of this isotope - being a singularity could be an important clue. OTOH it is most surprising that the physical property for which it derives its uniqueness - is the opposite of what one logically expects in the situation. attachment: winmail.dat
