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 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 distance at which the electron would need to be in order for an interaction to be likely. At low energies (in the keV), the electron would need to be nearby, and at higher energies (MeV), the electron would need to be further away. What details of the underlying mechanics am I missing in thinking this? Eric
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