Re: [Vo]:understanding the relationship between internal conversion and the far field
On Sun, Jul 6, 2014 at 3:41 PM, mix...@bigpond.com wrote: You can get an idea of this from the HUP where delta E x delta t = h_bar/2. If delta E is the energy of the reaction (about 4 MeV), then you get a time of at least 8E-23 sec. (I think this is the way it is normally calculated.) In any event it is obvious that a process that only takes order 1E-22 seconds is far more likely to occur than one which takes 1E-9 seconds (I think this is actually more like 1E-17 BTW, but I'm not sure whether these times can be measured or this is just calculated.) So that's how conjugate variables are used. Note that the difference between 1E-22 s and 1E-17 s is significant for what we're talking about. A photon can travel atomic distances in 1E-17 s, while it will not get very far in 1E-22 s. I think the shortest duration measured (according to Wikipedia) is 12 attoseconds = 1.2E-17 s, so the 1E-22 must be a calculation or a lower bound. A photon is only virtual if the separation distance is less than the wavelength of the photon (actually I suspect that this should be wavelength/2*Pi). According to this page [1], in the case of photons, power and information transfer by virtual particles is a relatively short-range phenomenon (existing only within a few wavelengths of the field-disturbance, which carries information or transferred power). Unless the statement is in error, it seems, then, the virtual photon can exist within several wavelengths rather than less than a wavelength. Note that the wavelength of a 24 MeV photon is 51 fermis. Several wavelengths would be perhaps 100-200 fermis. It seems that such a photon would not travel very far before having to become a real photon. If the 24 MeV could take the form of thousands of virtual photons, by contrast, they would each have a much smaller energy (24 MeV / 1000 = 24,000 eV, say), and a corresponding wavelength of 51,666 fm. Multiplying this wavelength by three to get several times the wavelength, the virtual photons would extend out to around 155 pm, which is on the order of the lattice spacing in a metal. The reason I like virtual photons over real ones is that it's nicer to have messages being passed between interacting particles (e.g., Feynman diagrams) than to have to have gammas being intercepted (e.g., Widom-Larsen). Eric [1] http://en.wikipedia.org/wiki/Virtual_particle
RE: [Vo]:understanding the relationship between internal conversion and the far field
-Original Message- From: mix...@bigpond.com In reply to Axil Axil's message: If the cause of LENR is the excitation of the vacuum through the injection of very energetic EMF (magnetic), it might be possible that the energy intensive magnetic fields supported by exothermic gainful nuclear reactions might be injected into the vacuum enclosing the nucleus which might possibly supply the energy needed to make endothermic nuclear reactions possible. You don't need to do this. Just fuse everything together first, then it fission it to the detected end products. There is a simpler explanation than either of these - relating to magnetism, spin and symmetry - which can operate without fusion ... thanks to applying the findings of Dr. Steven Jones... not muon-Steve but thermite-Steve. BTW he was probably correct on both issues IMO. As was known long before 9/11, the most powerful non-nuclear explosive is a version of thermite, once made with nano-iron-oxide but now more exotic. Tons of this material were found in dust near the Twin Towers site. There was a document circulating on the Web, purporting to be Top-Secret, claiming the gain in nano-thermite is 50 times chemical. That document is probably typical Photoshop BS, and can be discounted, but... it could be intentional disinformation (in a convoluted scheme to discredit Dr Jones). Here is the sanitized Wiki version which does not mention superferromagnetism aka nanomagnetism. http://en.wikipedia.org/wiki/Nano-thermite The usual explanation is that the higher surface area of nano is responsible for the large increase in power. This could be part of the case, however, there is likely to be something else, far more energetic, besides a redox reaction going on in these explosions, related to superferromagnetism and superparamagnetism both of which are intrinsic to the main component of nanothermite. http://en.wikipedia.org/wiki/Superferromagnetism http://en.wikipedia.org/wiki/Superparamagnetism What Wiki glosses over is the interaction between spin/magnetic behavior of nanoparticles breaking known and hidden symmetry - which taps into Dirac negative energy. Thus we find that dimensional symmetry is overcome by intense magnetic pulses and conservation of energy falls by the way-side. Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Jones
RE: [Vo]:understanding the relationship between internal conversion and the far field
And to a lesser degree may relate to granary explosions, pyrophoricuty and maybe even Papp noble gas engine in that we may be dealing with Casimir geometries forming in ambient gases that break the isotropy creating the NAE, literally on the fly, as the reaction expands or contracts the gases caught between these grains acting as Casimir boundaries? Fran -Original Message- From: Jones Beene [mailto:jone...@pacbell.net] Sent: Monday, July 07, 2014 9:42 AM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:understanding the relationship between internal conversion and the far field -Original Message- From: mix...@bigpond.com In reply to Axil Axil's message: If the cause of LENR is the excitation of the vacuum through the injection of very energetic EMF (magnetic), it might be possible that the energy intensive magnetic fields supported by exothermic gainful nuclear reactions might be injected into the vacuum enclosing the nucleus which might possibly supply the energy needed to make endothermic nuclear reactions possible. You don't need to do this. Just fuse everything together first, then it fission it to the detected end products. There is a simpler explanation than either of these - relating to magnetism, spin and symmetry - which can operate without fusion ... thanks to applying the findings of Dr. Steven Jones... not muon-Steve but thermite-Steve. BTW he was probably correct on both issues IMO. As was known long before 9/11, the most powerful non-nuclear explosive is a version of thermite, once made with nano-iron-oxide but now more exotic. Tons of this material were found in dust near the Twin Towers site. There was a document circulating on the Web, purporting to be Top-Secret, claiming the gain in nano-thermite is 50 times chemical. That document is probably typical Photoshop BS, and can be discounted, but... it could be intentional disinformation (in a convoluted scheme to discredit Dr Jones). Here is the sanitized Wiki version which does not mention superferromagnetism aka nanomagnetism. http://en.wikipedia.org/wiki/Nano-thermite The usual explanation is that the higher surface area of nano is responsible for the large increase in power. This could be part of the case, however, there is likely to be something else, far more energetic, besides a redox reaction going on in these explosions, related to superferromagnetism and superparamagnetism both of which are intrinsic to the main component of nanothermite. http://en.wikipedia.org/wiki/Superferromagnetism http://en.wikipedia.org/wiki/Superparamagnetism What Wiki glosses over is the interaction between spin/magnetic behavior of nanoparticles breaking known and hidden symmetry - which taps into Dirac negative energy. Thus we find that dimensional symmetry is overcome by intense magnetic pulses and conservation of energy falls by the way-side. Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Jones
RE: [Vo]:understanding the relationship between internal conversion and the far field
One further point relative to... it would surprise no one to learn that a nickel version is also used (nominally NiOOH). As mentioned before wrt Rydberg holes in nickel and iron, the combination of the two (found in the Rossi reactor as well) is especially energetic for ground state redundancy, since all of these deeper levels are covered: 54.4 eV 108.5 eV 190.4 eV 299.2 eV And the interesting deduction from those values is that average gain for the proton of NiOOH would be in the previously mentioned range - where the bulk material would appear to actually have about 50 time more energy than chemical, which was similar to a claim by some proponents of ballotechnics... Of interest in this regard is this report: http://www.proton21.com.ua/publ/Proton21_Energy_EN.pdf
Re: [Vo]:understanding the relationship between internal conversion and the far field
Bob, The way you are describing the electromagnetic interactions below suggest that they are a normal two way process. By this I mean that energy can be exchanged in both directions so that it is possible for a nuclei in the low energy state to be brought up to an excited state by external means. Is the possible for normal metastates like Eric is discussing? I realize this is true for MRI systems, but is it possible for most similar cases? Should one assume that those reactions which emit a particle(neutrino) destroy the future interactions and put an end to the couplings? Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Jul 6, 2014 9:36 pm Subject: Re: [Vo]:understanding the relationship between internal conversion and the far field Eric-- I have several items that may pertain to your effort to understand internal conversion of nuclei. Various isotopes have dipole and quadrapole moments existing in the stable nucleus as well as the excited nucleus. Isomers and radioactive nuclei may have these moments. The moments may be electric dipole or quadrapole as well as magnetic dipole and quadrapole moments. 2. Modern electronics can produce both magnetic and electric dipole and quadrapole input oscillating fields that can be in resonance with the respective resonance of any given isotope whether it is in its ground state, an isomeric condition or a radioactive state with a long half life. 3. Spin states of the various nuclei and their excited states control the allowed transitions of the nucleus from state to state. 4. Spin with orbital electrons in a lattice and the electrons’ intrinsic spin as well as other particles within the lattice may also be involved in the reaction and provide the conditions that make the transitions in energy possible. This idea is consistent with a paper recently published in the “Proceedings of the 14th Meeting of Japan CF Research Society, JCF14 December 7 - 8, 2013 Tokyo Institute of Technology, Japan”( ISSN 2187-2260) 5. The last paper of these proceedings describes a theory that entails the extended quantum mechanical system connecting lattice nuclei as well as D or H within the lattice of transition metals. Other papers in the proceedings discuss experimental data, mostly accomplished by various researchers presenting at the JCF14 meeting. The idea involves bands of neutrons that extend between the metal nuclei of the lattice. It suggests that the reaction is mediated by the H or D in the lattice. 6. Back on June 22 Axil Axil wrote, “ It would be great if we could find one mechanism that provides the total set of transmutation produces seen in LENR. ... The Philips reaction is not a one reaction fit all solution.” The paper cited above fits Axil’s wish. Jed identified these JCF proceedings as available at: http://www.jcfrs.org/file/jcf14-proceedings.pdf Bob Sent from Windows Mail From: Eric Walker Sent: Sunday, July 6, 2014 2:56 PM To: vortex-l@eskimo.com Are there any textbooks you can recommend that touch on some of these areas in some detail? Eric On Sun, Jul 6, 2014 at 3:41 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?. (The 4He+? branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take significantly longer to emit a gamma photon than nuclei with other spins. By first order approximation, metastable nuclei can't decay. In order to do so, I think they need some form of external interaction. That's why they have comparatively long lifetimes. Am I correct in thinking that the same principles apply to a compound nucleus such as [dd]*? I.e., in the gamma photon branch, the [dd]* de-excitation is on the order of 10E-9 seconds, or perhaps longer? Also, I haven't found a reference that gives the approximate times needed for the other branches (p+t
Re: [Vo]:understanding the relationship between internal conversion and the far field
I believe that this type of mechanism is the source of the amazing megawatt level heat production seen in the Rossi meltdown process where the nickel powder has melted down and the high temperature insulation is breaking down into nanoparticles at 2000C. On Mon, Jul 7, 2014 at 9:54 AM, Roarty, Francis X francis.x.roa...@lmco.com wrote: And to a lesser degree may relate to granary explosions, pyrophoricuty and maybe even Papp noble gas engine in that we may be dealing with Casimir geometries forming in ambient gases that break the isotropy creating the NAE, literally on the fly, as the reaction expands or contracts the gases caught between these grains acting as Casimir boundaries? Fran -Original Message- From: Jones Beene [mailto:jone...@pacbell.net] Sent: Monday, July 07, 2014 9:42 AM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:understanding the relationship between internal conversion and the far field -Original Message- From: mix...@bigpond.com In reply to Axil Axil's message: If the cause of LENR is the excitation of the vacuum through the injection of very energetic EMF (magnetic), it might be possible that the energy intensive magnetic fields supported by exothermic gainful nuclear reactions might be injected into the vacuum enclosing the nucleus which might possibly supply the energy needed to make endothermic nuclear reactions possible. You don't need to do this. Just fuse everything together first, then it fission it to the detected end products. There is a simpler explanation than either of these - relating to magnetism, spin and symmetry - which can operate without fusion ... thanks to applying the findings of Dr. Steven Jones... not muon-Steve but thermite-Steve. BTW he was probably correct on both issues IMO. As was known long before 9/11, the most powerful non-nuclear explosive is a version of thermite, once made with nano-iron-oxide but now more exotic. Tons of this material were found in dust near the Twin Towers site. There was a document circulating on the Web, purporting to be Top-Secret, claiming the gain in nano-thermite is 50 times chemical. That document is probably typical Photoshop BS, and can be discounted, but... it could be intentional disinformation (in a convoluted scheme to discredit Dr Jones). Here is the sanitized Wiki version which does not mention superferromagnetism aka nanomagnetism. http://en.wikipedia.org/wiki/Nano-thermite The usual explanation is that the higher surface area of nano is responsible for the large increase in power. This could be part of the case, however, there is likely to be something else, far more energetic, besides a redox reaction going on in these explosions, related to superferromagnetism and superparamagnetism both of which are intrinsic to the main component of nanothermite. http://en.wikipedia.org/wiki/Superferromagnetism http://en.wikipedia.org/wiki/Superparamagnetism What Wiki glosses over is the interaction between spin/magnetic behavior of nanoparticles breaking known and hidden symmetry - which taps into Dirac negative energy. Thus we find that dimensional symmetry is overcome by intense magnetic pulses and conservation of energy falls by the way-side. Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Jones
Re: [Vo]:understanding the relationship between internal conversion and the far field
In reply to Jones Beene's message of Mon, 7 Jul 2014 06:41:53 -0700: Hi, [snip] Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Not only that but Mills is using chemicals with crystalline water in his latest version because, under heating, the water is released as molecular water rather than liquid water, and according to him, molecular water is a Mills catalyst (m=3). (Note that 2 FeOOH = Fe2O3 + H2O when heated). Also 2Al + 3H2O = Al2O3 + 3H2 as the source of the Hydrogen). Furthermore a 50 fold increase in energy over normal chemical energy falls right in the Hydrino ballpark. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:understanding the relationship between internal conversion and the far field
This is an interesting discussion about a powerful explosive material(50x normal). Could it be that the military will go to lengths to keep this out of the public use? I suspect that they would stagnate progress completely if they have their chance. It is good that cars, air planes, and all of the other technological advancements that could give them an edge in future engagements found their way into common usage. In today's environment It would not surprise me to see a black curtain fall over any new discovery which could have military implications, and unfortunately LENR is in that category. The lack of public funding implies unusual treatment for a scientific subject, but perhaps there is a massive undertaking hidden from our view. I am not aware of any rumors to that effect however. How is this for a conspiracy theory? :-) Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jul 7, 2014 6:21 pm Subject: Re: [Vo]:understanding the relationship between internal conversion and the far field In reply to Jones Beene's message of Mon, 7 Jul 2014 06:41:53 -0700: Hi, [snip] Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Not only that but Mills is using chemicals with crystalline water in his latest version because, under heating, the water is released as molecular water rather than liquid water, and according to him, molecular water is a Mills catalyst (m=3). (Note that 2 FeOOH = Fe2O3 + H2O when heated). Also 2Al + 3H2O = Al2O3 + 3H2 as the source of the Hydrogen). Furthermore a 50 fold increase in energy over normal chemical energy falls right in the Hydrino ballpark. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:understanding the relationship between internal conversion and the far field
Papp had the patent on this exploding water technology. It is in the public domain now. On Mon, Jul 7, 2014 at 6:41 PM, David Roberson dlrober...@aol.com wrote: This is an interesting discussion about a powerful explosive material(50x normal). Could it be that the military will go to lengths to keep this out of the public use? I suspect that they would stagnate progress completely if they have their chance. It is good that cars, air planes, and all of the other technological advancements that could give them an edge in future engagements found their way into common usage. In today's environment It would not surprise me to see a black curtain fall over any new discovery which could have military implications, and unfortunately LENR is in that category. The lack of public funding implies unusual treatment for a scientific subject, but perhaps there is a massive undertaking hidden from our view. I am not aware of any rumors to that effect however. How is this for a conspiracy theory? :-) Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jul 7, 2014 6:21 pm Subject: Re: [Vo]:understanding the relationship between internal conversion and the far field In reply to Jones Beene's message of Mon, 7 Jul 2014 06:41:53 -0700: Hi, [snip] Not quite the same as LENR but close, since apparently with the military version of nano-thermite, instead of finding the usual nanoparticles of magnetite, we find that a ferrihydrite (nominally FeOOH) is used, which apparently has hydrogen content in order to provide the vehicle for gain. It would surprise no one to learn that a nickel version is also used (nominally NiOOH). As Robin will note, either can fit into a Mills explanation as well, since both metals have Rydberg IP holes. Not only that but Mills is using chemicals with crystalline water in his latest version because, under heating, the water is released as molecular water rather than liquid water, and according to him, molecular water is a Mills catalyst (m=3). (Note that 2 FeOOH = Fe2O3 + H2O when heated). Also 2Al + 3H2O = Al2O3 + 3H2 as the source of the Hydrogen). Furthermore a 50 fold increase in energy over normal chemical energy falls right in the Hydrino ballpark. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:understanding the relationship between internal conversion and the far field
I'm in the process of trying to better understand internal conversion and it's cross section vis-a-vis inner shell electrons and sources of charge in the far field. I'm hoping someone (Robin?) can help me to get the terminology right and point me to further reading. Here is my understanding so far. Internal conversion is a process in which an inner shell electron is expelled from an atom as the result of a nuclear transition. It is mediated by the electromagnetic force (in contrast to the weak or strong interactions) and results from electromagnetic coupling between the electron and an excited nucleus. The kind of nuclear transition that leads to internal conversion is generally an isomeric transition; i.e., the deexcitation of a metastable isomer to a lower energy level. Internal conversion competes with gamma emission, and there is an internal conversion coefficient, which for a competing pair of branches, one IC and the other gamma photon emitting, gives you the ratio of internal conversion electrons to photons. In some cases the internal conversion coefficient can be quite high, meaning that IC is greatly favored over gamma photon emission. There are a number of factors that are thought to go into a high IC coefficient -- when the energy of the transition is small, when the nucleus is large, and when the daughter nucleus has zero spin, for example. Unlike in the case of beta emission, the energies for internal conversion electrons are not broadband and show up in line spectra as sharp peaks. This is because unlike in the case of beta decay there is no neutrino to take part of the energy of the decay away from the emitted electron. Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+ɣ. (The 4He+ɣ branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take significantly longer to emit a gamma photon than nuclei with other spins. Am I correct in thinking that the same principles apply to a compound nucleus such as [dd]*? I.e., in the gamma photon branch, the [dd]* de-excitation is on the order of 10E-9 seconds, or perhaps longer? Also, I haven't found a reference that gives the approximate times needed for the other branches (p+t and n+3He), in which the compound nucleus splits up into fragments. Returning to internal conversion, one explanation for it focuses on the fact that inner shell electrons have a high probability of passing through the nucleus. The idea is that during the time that the electron is within the nucleus there is a nontrivial probability that it will interact with the excited state, which, if this happens, will result in the energy of the excited state being passed on to the electron. The implication is that the less likely an electron is to be found within the nucleus, the less likely that the electron will be ejected as a result of internal conversion. So the probability of IC is highest with K-shell electrons and decreases the further you go out. One question I have about this explanation is that IC is mediated via the electromagnetic interaction; my understanding of this is that there is a virtual photon that passes from the nucleus to the electron. I do not see why the electron would particularly need to be passing through the nucleus for such a virtual photon to reach it, for the electromagnetic interaction is long-range. Anywhere a virtual photon can reach, it seems, there would be a nontrivial probability for an internal conversion decay to occur. Another challenge I have with this explanation is that I think the de Broglie wavelength of an orbital electron is going to be far larger than the nucleus or any than particular nucleon; my understanding is that this is a problem because the de Broglie wavelengths have to be roughly comparable for an interaction of some kind to be probable. One question I have has to do with the energy of the virtual photon. Internal conversion is less likely, other things being equal, if the energy of the transition is large (e.g., on the order of MeVs). Not having read this detail, I would have thought that the energy of the decay would factor into the
RE: [Vo]:understanding the relationship between internal conversion and the far field
Eric - By now you fully appreciate the impossibility of reconciling mainstream fusion details with LENR. There is little way to rationalize all of the contradictions, based on the data now available… but that does not keep us from trying. Let me add this, which would clear up some of the contradiction, if it were true. You state: For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+ɣ. If Mizuno is correct we could add a fourth pathway for the dd reaction in LENR. That would be p+p+p+p. It is not clear if this reaction should be called fusion or fission or a new form of IC :-) but as of July 2014, it looms as the most important unexplained experiment on the horizon for deuterium-based LENR. But then again we must ask, can these results be applied to retrospectively explain experiments going back 25 years? …those who have followed the field for so long would complain: “what about all the reports of 4He correlating to excess heat ?” The answer to that would be something like this. Deuterium, in condensed matter will often form into a stable Cooper pair of deuterons (or dense deuterium), which is stable enough to pass through a spectrometer and fool the experimenter into the belief that the species is 4He. Either way, some form of dense paired deuterium happens often enough to explain the tiny amount of 4He which seems to have been documented in the past. From there on, we must explain why this so-called “4He” or mass-4 signal on a mass spectrometer is some other species instead of helium. That is far easier than to rationalize the lack of 24 MeV gammas. Here is a paper of interest from Winterberg, but it does not go far enough: http://arxiv.org/ftp/arxiv/papers/0912/0912.5414.pdf The basic concept is that a paired species of deuterons can form and linger in condensed matter for weeks, but actual decay can occur after a delay. When decay occurs, only protons remain plus excess energy which is below gamma intensity but much more than chemical. Thus we explain all the main details of “helium LENR” but without helium. 1) Excess heat 2) Lack of gamma radiation 3) What “appears to be” 4He on a mass-spec, but is not 4) The so-called “heat after death” phenomenon… Of course, this explanation raises as many questions as it answers, and leaves open the door that a small percentage of reactions still must proceed to tritium, since tritium is documented; but anything that removes the 24MeV millstone from the shoulders of LENR should be given full consideration. From: Eric Walker I'm in the process of trying to better understand internal conversion and it's cross section vis-a-vis inner shell electrons and sources of charge in the far field. I'm hoping someone (Robin?) can help me to get the terminology right and point me to further reading. Here is my understanding so far. Internal conversion is a process in which an inner shell electron is expelled from an atom as the result of a nuclear transition. It is mediated by the electromagnetic force (in contrast to the weak or strong interactions) and results from electromagnetic coupling between the electron and an excited nucleus. The kind of nuclear transition that leads to internal conversion is generally an isomeric transition; i.e., the deexcitation of a metastable isomer to a lower energy level. Internal conversion competes with gamma emission, and there is an internal conversion coefficient, which for a competing pair of branches, one IC and the other gamma photon emitting, gives you the ratio of internal conversion electrons to photons. In some cases the internal conversion coefficient can be quite high, meaning that IC is greatly favored over gamma photon emission. There are a number of factors that are thought to go into a high IC coefficient -- when the energy of the transition is small, when the nucleus is large, and when the daughter nucleus has zero spin, for example. Unlike in the case of beta emission, the energies for internal conversion electrons are not broadband and show up in line spectra as sharp peaks. This is because unlike in the case of beta decay there is no neutrino to take part of the energy of the decay away from the emitted electron. Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of
RE: [Vo]:understanding the relationship between internal conversion and the far field
Species which look like helium on a mass spec - especially when subject to variable mass interactions immediately prior to measurement. Helium - 4.002602 amu D2 - 4.028203 amu 4 protons as Rydberg matter – 4.03176 amu Next we must consider variable mass. To understand variable mass, this article may not help most of us, but it is all there is for now – except for Noether’s theorem and other imponderables like Fock-Stueckelberg. http://en.wikipedia.org/wiki/On_shell_and_off_shell What does it take to get a Cooper pair of D to look like 4He on a mass-spec and yet decay energetically into protons and electrons, which should be endothermic ? http://www.scribd.com/doc/139182265/Theories-of-variable-mass-particles-and- low-energy-nuclear-phenomena Theories of variable mass particles and low energy nuclear phenomena by Mark Davidson Conclusion of Paper: We want to emphasize that there is no direct experimental evidence yet that masses of electrons, nucleons, or nuclei can change significantly in a condensed matter setting Nevertheless, it is this author's opinion that Fock-Stueckelberg or other type of are a possible explanation for such variations and that all of the experiments in LENR can potentially be explained if they are occurring. Another tid-bit: Scattering experiments indicate that the 4He nuclear structure is C3v, which means that it has a tetrahedral arrangement of neutrons and protons. An alpha particle can be thought of as a pair of deuterons. However, the neutrons of the alpha particle form a pair and so do the protons. The neutron-neutron pair involves a spin pairing and a strong force bonding. The strong force between two neutrons is a repulsion which pushes the separation distance between the centers of the neutrons to the limit of spin pairing. The same thing occurs for the protons except there is an electrostatic repulsion in addition to the strong force repulsion. It is amazing that the alpha is so stable. The strong force between a neutron and a proton is an attraction and there has to be a rotation to maintain a separation between a neutron and a proton in a neutron-proton pair. This can be a complex mix of nuclear dynamics in condensed matter which is far different than in a plasma – thus leading to the suspicion that variable mass plays a role, and that real fusion to helium cannot happen except in a plasma. _ The basic concept is that a paired species of deuterons can form and linger in condensed matter for weeks, but actual decay can occur after a delay. When decay occurs, only protons remain plus excess energy which is below gamma intensity but much more than chemical. Thus we explain all the main details of “helium LENR” but without helium. 1) Excess heat 2) Lack of gamma radiation 3) What “appears to be” 4He on a mass-spec, but is not 4) The so-called “heat after death” phenomenon… Of course, this explanation raises as many questions as it answers, and leaves open the door that a small percentage of reactions still must proceed to tritium, since tritium is documented; but anything that removes the 24MeV millstone from the shoulders of LENR should be given full consideration. attachment: winmail.dat
Re: [Vo]:understanding the relationship between internal conversion and the far field
Interesting proposal, Jones. I'll have to think about it. Thanks for the pointer to the paper. On Sun, Jul 6, 2014 at 6:34 AM, Jones Beene jone...@pacbell.net wrote: If Mizuno is correct we could add a fourth pathway for the dd reaction in LENR. That would be p+p+p+p. It is not clear if this reaction should be called fusion or fission or a new form of IC :-) but as of July 2014, it looms as the most important unexplained experiment on the horizon for deuterium-based LENR. For this one, my working hypothesis is that Mizuno is seeing these two reactions in tandem: (X)Ni + d → (X+1)Ni + p (Oppenheimer-Phillips process) p + d → p + p + n → p + p + p (about 10 min. later) This tack obviously has several objections to worry about, such as the lack of detected neutrons and the lack of activation. My initial strategy for dealing with these objections is to wonder whether the channel is actually pretty small, in comparison to any energy that is developed. Eric
Re: [Vo]:understanding the relationship between internal conversion and the far field
One of the insights that we might possibly draw from the Mizuno experiment is that there might be a mix of exothermic and endothermic nuclear reactions going on simultaneously in LERN transmutation reactions. If the cause of LENR is the excitation of the vacuum through the injection of very energetic EMF (magnetic), it might be possible that the energy intensive magnetic fields supported by exothermic gainful nuclear reactions might be injected into the vacuum enclosing the nucleus which might possibly supply the energy needed to make endothermic nuclear reactions possible. Such a situation would make analyzing how energy is produced in a LENR reaction very complicated and confusing. For example, the iron and helium seen in some LENR experiments could be derived from an endothermic alpha decay of nickel. In a similar way, chromium and titanium could be produced through the same endothermic alpha decay mechanism from higher Z elements in a series of alpha emission based reactions to lower Z elements. This alpha decay reaction could be the source of helium seen in many LENR experiments. The energy needed to cause such alpha decays could come from the exothermic fusion of hydrogen and or helium into boron, lithium and beryllium. As long as the LENR reaction is exothermic on the average, any sort of nuclear reaction can occur because the energy storehouse of the highly energetic magnetic solitons which catalyze transmutation could supply stored magnetic energy that could be shared later in a subsequent endothermic reaction. On Sun, Jul 6, 2014 at 12:33 PM, Eric Walker eric.wal...@gmail.com wrote: Interesting proposal, Jones. I'll have to think about it. Thanks for the pointer to the paper. On Sun, Jul 6, 2014 at 6:34 AM, Jones Beene jone...@pacbell.net wrote: If Mizuno is correct we could add a fourth pathway for the dd reaction in LENR. That would be p+p+p+p. It is not clear if this reaction should be called fusion or fission or a new form of IC :-) but as of July 2014, it looms as the most important unexplained experiment on the horizon for deuterium-based LENR. For this one, my working hypothesis is that Mizuno is seeing these two reactions in tandem: (X)Ni + d → (X+1)Ni + p (Oppenheimer-Phillips process) p + d → p + p + n → p + p + p (about 10 min. later) This tack obviously has several objections to worry about, such as the lack of detected neutrons and the lack of activation. My initial strategy for dealing with these objections is to wonder whether the channel is actually pretty small, in comparison to any energy that is developed. Eric
Re: [Vo]:understanding the relationship between internal conversion and the far field
In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?. (The 4He+? branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take significantly longer to emit a gamma photon than nuclei with other spins. By first order approximation, metastable nuclei can't decay. In order to do so, I think they need some form of external interaction. That's why they have comparatively long lifetimes. Am I correct in thinking that the same principles apply to a compound nucleus such as [dd]*? I.e., in the gamma photon branch, the [dd]* de-excitation is on the order of 10E-9 seconds, or perhaps longer? Also, I haven't found a reference that gives the approximate times needed for the other branches (p+t and n+3He), in which the compound nucleus splits up into fragments. You can get an idea of this from the HUP where delta E x delta t = h_bar/2. If delta E is the energy of the reaction (about 4 MeV), then you get a time of at least 8E-23 sec. (I think this is the way it is normally calculated.) In any event it is obvious that a process that only takes order 1E-22 seconds is far more likely to occur than one which takes 1E-9 seconds (I think this is actually more like 1E-17 BTW, but I'm not sure whether these times can be measured or this is just calculated.) Returning to internal conversion, one explanation for it focuses on the fact that inner shell electrons have a high probability of passing through the nucleus. Yes but they are only present for a very short time. The idea is that during the time that the electron is within the nucleus there is a nontrivial probability that it will interact with the excited state, which, if this happens, will result in the energy of the excited state being passed on to the electron. The implication is that the less likely an electron is to be found within the nucleus, the less likely that the electron will be ejected as a result of internal conversion. So the probability of IC is highest with K-shell electrons and decreases the further you go out. One question I have about this explanation is that IC is mediated via the electromagnetic interaction; my understanding of this is that there is a virtual photon that passes from the nucleus to the electron. I do not see why the electron would particularly need to be passing through the nucleus for such a virtual photon to reach it, for the electromagnetic interaction is long-range. Anywhere a virtual photon can reach, it seems, there would be a nontrivial probability for an internal conversion decay to occur. A photon is only virtual if the separation distance is less than the wavelength of the photon (actually I suspect that this should be wavelength/2*Pi). Another challenge I have with this explanation is that I think the de Broglie wavelength of an orbital electron is going to be far larger than the nucleus or any than particular nucleon; my understanding is that this is a problem because the de Broglie wavelengths have to be roughly comparable for an interaction of some kind to be probable. ...but it isn't very probable, that's why it can barely compete with slow gamma emission. However I don't know what part of the probability is due to duration of the interaction, and what part due to Be Broglie wavelength mismatch. Perhaps you also need to take into consideration the be Broglie wavelength of the nucleons. Also consider that by the time an orbital electron passes through the nucleus it has gained considerable kinetic energy from the electric field, so it's De Broglie wavelength is much shorter. (Still too long, but not by many orders of magnitude. :) One question I have has to do with the energy of the virtual photon. Internal conversion is less likely, other things being equal, if the energy of the transition is large (e.g., on the order of MeVs). Perhaps because high energy reactions often decay via other faster paths? The HUP logic applied here above would indicate that gamma decay times decrease with increasing energy, while IC times would be unaffected,
Re: [Vo]:understanding the relationship between internal conversion and the far field
Are there any textbooks you can recommend that touch on some of these areas in some detail? Eric On Sun, Jul 6, 2014 at 3:41 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?. (The 4He+? branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take significantly longer to emit a gamma photon than nuclei with other spins. By first order approximation, metastable nuclei can't decay. In order to do so, I think they need some form of external interaction. That's why they have comparatively long lifetimes. Am I correct in thinking that the same principles apply to a compound nucleus such as [dd]*? I.e., in the gamma photon branch, the [dd]* de-excitation is on the order of 10E-9 seconds, or perhaps longer? Also, I haven't found a reference that gives the approximate times needed for the other branches (p+t and n+3He), in which the compound nucleus splits up into fragments. You can get an idea of this from the HUP where delta E x delta t = h_bar/2. If delta E is the energy of the reaction (about 4 MeV), then you get a time of at least 8E-23 sec. (I think this is the way it is normally calculated.) In any event it is obvious that a process that only takes order 1E-22 seconds is far more likely to occur than one which takes 1E-9 seconds (I think this is actually more like 1E-17 BTW, but I'm not sure whether these times can be measured or this is just calculated.) Returning to internal conversion, one explanation for it focuses on the fact that inner shell electrons have a high probability of passing through the nucleus. Yes but they are only present for a very short time. The idea is that during the time that the electron is within the nucleus there is a nontrivial probability that it will interact with the excited state, which, if this happens, will result in the energy of the excited state being passed on to the electron. The implication is that the less likely an electron is to be found within the nucleus, the less likely that the electron will be ejected as a result of internal conversion. So the probability of IC is highest with K-shell electrons and decreases the further you go out. One question I have about this explanation is that IC is mediated via the electromagnetic interaction; my understanding of this is that there is a virtual photon that passes from the nucleus to the electron. I do not see why the electron would particularly need to be passing through the nucleus for such a virtual photon to reach it, for the electromagnetic interaction is long-range. Anywhere a virtual photon can reach, it seems, there would be a nontrivial probability for an internal conversion decay to occur. A photon is only virtual if the separation distance is less than the wavelength of the photon (actually I suspect that this should be wavelength/2*Pi). Another challenge I have with this explanation is that I think the de Broglie wavelength of an orbital electron is going to be far larger than the nucleus or any than particular nucleon; my understanding is that this is a problem because the de Broglie wavelengths have to be roughly comparable for an interaction of some kind to be probable. ...but it isn't very probable, that's why it can barely compete with slow gamma emission. However I don't know what part of the probability is due to duration of the interaction, and what part due to Be Broglie wavelength mismatch. Perhaps you also need to take into consideration the be Broglie wavelength of the nucleons. Also consider that by the time an orbital electron passes through the nucleus it has gained considerable kinetic energy from the electric field, so it's De Broglie wavelength is much shorter. (Still too long, but not by many orders of magnitude. :) One question I have has to do with the energy of the virtual photon. Internal conversion is less likely, other things being equal, if the energy of the transition is large (e.g.,
Re: [Vo]:understanding the relationship between internal conversion and the far field
Eric-- I have several items that may pertain to your effort to understand internal conversion of nuclei. Various isotopes have dipole and quadrapole moments existing in the stable nucleus as well as the excited nucleus. Isomers and radioactive nuclei may have these moments. The moments may be electric dipole or quadrapole as well as magnetic dipole and quadrapole moments. 2. Modern electronics can produce both magnetic and electric dipole and quadrapole input oscillating fields that can be in resonance with the respective resonance of any given isotope whether it is in its ground state, an isomeric condition or a radioactive state with a long half life. 3. Spin states of the various nuclei and their excited states control the allowed transitions of the nucleus from state to state. 4. Spin with orbital electrons in a lattice and the electrons’ intrinsic spin as well as other particles within the lattice may also be involved in the reaction and provide the conditions that make the transitions in energy possible. This idea is consistent with a paper recently published in the “Proceedings of the 14th Meeting of Japan CF Research Society, JCF14 December 7 - 8, 2013 Tokyo Institute of Technology, Japan”( ISSN 2187-2260) 5. The last paper of these proceedings describes a theory that entails the extended quantum mechanical system connecting lattice nuclei as well as D or H within the lattice of transition metals. Other papers in the proceedings discuss experimental data, mostly accomplished by various researchers presenting at the JCF14 meeting. The idea involves bands of neutrons that extend between the metal nuclei of the lattice. It suggests that the reaction is mediated by the H or D in the lattice. 6. Back on June 22 Axil Axil wrote, “ It would be great if we could find one mechanism that provides the total set of transmutation produces seen in LENR. ... The Philips reaction is not a one reaction fit all solution.” The paper cited above fits Axil’s wish. Jed identified these JCF proceedings as available at:http://www.jcfrs.org/file/jcf14-proceedings.pdf Bob Sent from Windows Mail From: Eric Walker Sent: Sunday, July 6, 2014 2:56 PM To: vortex-l@eskimo.com Are there any textbooks you can recommend that touch on some of these areas in some detail? Eric On Sun, Jul 6, 2014 at 3:41 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?. (The 4He+? branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take significantly longer to emit a gamma photon than nuclei with other spins. By first order approximation, metastable nuclei can't decay. In order to do so, I think they need some form of external interaction. That's why they have comparatively long lifetimes. Am I correct in thinking that the same principles apply to a compound nucleus such as [dd]*? I.e., in the gamma photon branch, the [dd]* de-excitation is on the order of 10E-9 seconds, or perhaps longer? Also, I haven't found a reference that gives the approximate times needed for the other branches (p+t and n+3He), in which the compound nucleus splits up into fragments. You can get an idea of this from the HUP where delta E x delta t = h_bar/2. If delta E is the energy of the reaction (about 4 MeV), then you get a time of at least 8E-23 sec. (I think this is the way it is normally calculated.) In any event it is obvious that a process that only takes order 1E-22 seconds is far more likely to occur than one which takes 1E-9 seconds (I think this is actually more like 1E-17 BTW, but I'm not sure whether these times can be measured or this is just calculated.) Returning to internal conversion, one explanation for it focuses on the fact that inner shell electrons have a high probability of passing through the nucleus. Yes but they are only present for a very short time. The idea is that during the
Re: [Vo]:understanding the relationship between internal conversion and the far field
Thank you, Bob. If you have any textbooks you particularly like (e.g., touching on nuclear spin states and nuclear transitions), feel free to recommend them. For anyone who is interested, I have found the following helpful in getting a broad overview: - Turner, Atoms, Radiation, and Radiation Protection, http://www.amazon.com/Atoms-Radiation-Protection-James-Turner/dp/3527406069 - Magill and Galy, Radioactivity, Radionuclides, Radiation, http://www.amazon.com/Radioactivity-Radionuclides-Radiation-Joseph-Magill/dp/product-description/3540211160 The book by Turner is fantastic and goes into a range of topics, including a detailed discussion about how various detectors work. I found the book in several MIT course syllabi. Turner was at Oak Ridge National Laboratory, and the book appears to be an important one. The book by Magill and Galy is a nice refresher, but they do not go into much detail and make some assumptions about the reader's knowledge of the field, and I wouldn't have been able to follow the discussion without having first read Turner's book. Next up is a textbook by Kenneth Krane called Introductory Nuclear Physics, which has also shown up in several MIT course syllabi. Eric On Sun, Jul 6, 2014 at 5:39 PM, Bob Cook frobertc...@hotmail.com wrote: Eric-- I have several items that may pertain to your effort to understand internal conversion of nuclei. 1. Various isotopes have dipole and quadrapole moments existing in the stable nucleus as well as the excited nucleus. Isomers and radioactive nuclei may have these moments. The moments may be electric dipole or quadrapole as well as magnetic dipole and quadrapole moments. 2. Modern electronics can produce both magnetic and electric dipole and quadrapole input oscillating fields that can be in resonance with the respective resonance of any given isotope whether it is in its ground state, an isomeric condition or a radioactive state with a long half life. 3. Spin states of the various nuclei and their excited states control the allowed transitions of the nucleus from state to state. 4. Spin with orbital electrons in a lattice and the electrons’ intrinsic spin as well as other particles within the lattice may also be involved in the reaction and provide the conditions that make the transitions in energy possible. This idea is consistent with a paper recently published in the “Proceedings of the 14th Meeting of Japan CF Research Society, JCF14 December 7 - 8, 2013 Tokyo Institute of Technology, Japan”( ISSN 2187-2260) 5. The last paper of these proceedings describes a theory that entails the extended quantum mechanical system connecting lattice nuclei as well as D or H within the lattice of transition metals. Other papers in the proceedings discuss experimental data, mostly accomplished by various researchers presenting at the JCF14 meeting. The idea involves bands of neutrons that extend between the metal nuclei of the lattice. It suggests that the reaction is mediated by the H or D in the lattice. 6. Back on June 22 Axil Axil wrote, “ It would be great if we could find one mechanism that provides the total set of transmutation produces seen in LENR. ... The Philips reaction is not a one reaction fit all solution.” The paper cited above fits Axil’s wish. Jed identified these JCF proceedings as available at: *http://www.jcfrs.org/file/jcf14-proceedings.pdf* http://www.jcfrs.org/file/jcf14-proceedings.pdf Bob Sent from Windows Mail *From:* Eric Walker eric.wal...@gmail.com *Sent:* Sunday, July 6, 2014 2:56 PM *To:* vortex-l@eskimo.com Are there any textbooks you can recommend that touch on some of these areas in some detail? Eric On Sun, Jul 6, 2014 at 3:41 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] Here's where my understanding starts to get fuzzy. The above description talked about isomeric transitions, which involve the decay of a metastable isomer to the ground state of the isotope. Metastable isomers are long-lived excited states of nuclei, ones that have significant half-lives. Similar, shorter-lived nuclei are not considered metastable and are instead referred to as compound nuclei. For example, in dd fusion, the short-lived compound nucleus [dd]* is not an isomer of 4He because it decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?. (The 4He+? branch is orders of magnitude less likely than the p+t and n+3He branches, whose likelihoods are roughly split 50-50.) I understand from reading around that the emission of a gamma photon during the deexcitation of a metastable isomer can be on the order of 10E-9 seconds, and that the time required for the emission is something that depends upon the spin of the excited nucleus. Excited nuclei with certain spins will take
Re: [Vo]:understanding the relationship between internal conversion and the far field
In reply to Eric Walker's message of Sun, 6 Jul 2014 15:56:14 -0700: Hi, [snip] Are there any textbooks you can recommend that touch on some of these areas in some detail? Eric Sorry, I don't know. Google is your friend (sometimes;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:understanding the relationship between internal conversion and the far field
In reply to Axil Axil's message of Sun, 6 Jul 2014 13:43:13 -0400: Hi, [snip] If the cause of LENR is the excitation of the vacuum through the injection of very energetic EMF (magnetic), it might be possible that the energy intensive magnetic fields supported by exothermic gainful nuclear reactions might be injected into the vacuum enclosing the nucleus which might possibly supply the energy needed to make endothermic nuclear reactions possible. You don't need to do this. Just fuse everything together first, then it fission it to the detected end products. As long as the total mass out total mass in, the reaction as whole is exothermic. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
