I think you are being very dismissive of the way quantum mechanics works with in the nuclear realm. It all boils down to PSI and if the nuclear force is point charge with a probability of interacting defined by PSI, or that PSI is blurred motion where the nuclear force is spread over space describing PSI. Is it a wave or is a particle probability? It's a very fundamental question with respect to BECs. The BEC comes about by the overlapping wave functions of integral spin. By it's nature bose particles when chilled they like to fall towards ground states and as they do, their PSI's will completely overlap making one big PSI(n) where PSI(n) describes all of the properties of that mix. The PSI is the matter wave, and with the matter wave all of the other attributes of a particle are carried along, so the electric force and the nuclear force(s) are just aspects of that PSI(n). The overlap of the PSI is where there is a probability of interaction. That's why I mentioned the Gamow factor is that it describes perfectly what the collision of two PSI's with nuclear interactions looks like. At very high energies, it looks like CERN, but at very low energies it looks like solid state.
Eventually you have to have PSI(x) describing the model. If that wave function overlap doesn't occur, there is no probability for interaction and nothing will occur. I think maybe a hybrid of Chubbs' and Kim's theory could be very compelling. Specifically with the Nano scale BECs or 100 atom Bose-band states. Best Regards, Chuck On Wed, Feb 13, 2013 at 10:15 AM, Edmund Storms <stor...@ix.netcom.com>wrote: > Eric, the details do not matter. The basic idea is wrong. The details are > just a series of arbitrary assumptions to avoid dropping the initial > premise. We are simply playing whack-a-mole. He strings a collection of > words together that have no logical relationship, but because the > vocabulary of QM mathematics is used, no one questions the statements. If > Ron wants to make a contribution, he needs to apply his ideas to what > actually exists in the real world based on what material science has agreed > is real based on much study. Simply making up concepts to which math is > applied is not useful except as a game. Also, we are describing a > mechanism. Describing one part in isolation is not useful. This is like > saying an automobile works by turning the key in the ignition and then go > on to describe the key in great detail. > > On Feb 12, 2013, at 10:42 PM, Eric Walker wrote: > > On Tue, Feb 12, 2013 at 7:13 AM, Edmund Storms <stor...@ix.netcom.com>wrote: > > There is no alpha. The helium CAN NOT MOVE spontaneously. The helium >> contains extra energy as mass. This mass must be converted to energy before >> it can appear as reaction energy. The He is fixed in space. Normally the He >> nucleus explodes into fragments producing hot fusion. Or it emits a gamma >> which releases the mass-energy. This conversion CAN NOT OCCUR outside of >> the nucleus simply by being near a Pd. >> > > I suspect that you are very busy and haven't had time to read Ron's > writeup closely. Here is what he says about the production of the alpha: > > The fusion of deuterons always happens through unstable intermediate > states, and the cross section to alpha particle is only small because of > the same non-relativistic issue. To get an alpha, you need to emit a > gamma-ray photon, and emissions of photons are suppressed by 1/c factors. > > > Yes, this is why the hot fusion products occur rather than helium. Even > this statement is ambiguous - what does 1/c factors mean? In fact, the > explanation is much easier to understand simply by noting that energy can > be lost by the nucleus exploding into its parts faster than it can be > released by gamma emission. The issue is based on relative rates. Why is > gamma emission slow? It is slow for the same reason it is slow when photons > are emitted from any energetic nucleus. Many explanations have been > suggested including the need to assemble the required energy and spin in > the nucleus before the photon can be emitted. The statement of 1/c factors > has no relationship to this process. > > When there is a nucleus nearby, it can be kicked electrostatically, and > this process is easier than kicking out a photon, because it is > nonrelativistic (the same holds for an electron, but with much smaller > cross section due to the smaller charge, and there is no reason to suspect > concentration of wavefunction around electron density, as there is for a > nucleus). > > > Here Ron makes an assumption that has no justification. The nearest nuclei > is many Å away and surrounded by electrons. Any nuclear-nuclear interaction > is impossible. That is why spontaneous nuclear reactions are so rare. > > > The time-scale for kicking a nucleus is the lifetime of the two-deuteron > resonance, which is not very long, in terms of distance, it is about 100 > fermis, this is about the same size as the inner shell. If the deuterons > are kicking about at random, this coincidence is not significant, but if > the deuteron-hole excitations are banded, it is plausible that nearly all > the energetic deuteron-deuteron collisions take place very close to a > nucleus, as explained above. > > > This is word salad without meaning in the real world. He makes up a number > and then assumes it applies it to an imagined process. Yes, the d must be > bonded (or as he says banded), but how? > > > There are conservation laws broken when a nucleus is nearby. The nucleus > breaks parity, so it might open up a fusion channel, by allowing deuteron > pairs to decay to an alpha from a parity odd state. Such a transition would > never be observed in a dilute beam fusion, because these fusions happen far > away from anything else. This hypothesis is not excluded by alpha particle > spectroscopy (there are a lot of relevant levels of different parities), > but it is not predicted either. > > > This is word salad. His statement about beams reveals an ignorance about > how beams are used. They are used to bombard a solid in which many > interactions take place resulting in hot fission. > > > Here there is a concept of a "two-deuteron resonance," i.e., the > metastable 4He you're talking about following upon the d+d fusion, which > will not last long and must shed some energy. Ron states or alludes to the > following in the above paragraph: > > 1. There is a metastable "two-deuteron resonance" that will decay. > This is the energetic 4He you're referring to, which will then go and do > something else. > 2. There are three channels for the decay of the > two-deuteron resonance: (a) d+d → [2d]* → 3He+n, (b) d+d → [2d]* → t+p, (c) > d+d → [2d]* → 4He+ɣ. Normally (a) and (b) predominate and (c) is rare. > But the reason that (c) is rare is that it takes a while for the photon to > be produced (my reading, anyway, of "emission of photons are suppressed by > 1/c factors"). > > This is a restatement of the earlier comment, which is correct. > > > 1. When there is a palladium nucleus (not atom) nearby, however, the > energy that would have been dumped as a photon will instead be kicked to a > proton in the palladium nucleus, a process that occurs quickly rather than > slowly. Because this occurs quickly, branch (c) is enhanced and branches > (a) and (b) are suppressed in direct proportion. > > > This is an impossible assumption. > > > 1. When the mass deficit of the two-deuteron resonance is > electrostatically dumped into the proton in the nearby palladium nucleus on > the order of 24 MeV, you will get a palladium nucleus with additional > kinetic energy an energetically stable recoil alpha, moving quite quickly. > > In his original description Ron has touched on points that address nearly > every objection you have raised so far. His description may well be > incorrect, but I suspect it is not incorrect for the reasons you have > mentioned so far. > > > Eric, this discussion is a waste of time simply because the concept has no > relationship to reality. Clever people can create all kinds of personal > realities that are useful as games or as a guide for their lives. But in > science, the reality has to be shared based on centuries of hard work by > millions of people. New ideas have to fit into what is known and must be > described using words that have common meaning. People seem free to > imagine anything about CF that would be laughable if applied to any other > field of study. > > > I don't mean to press this issue. I just think Ron's theory should be > read closely before objections are raised; some very good objections have > already been raised in earlier threads. I understand if you're too busy or > if this lead does not seem to merit your time. There may be interest among > others here. It is also entirely possible that while Ron knows something > about the math involved, he knows nothing about what happens with these > things in real-life. I am wary of drawing this conclusion myself without > further evidence. > > > You are on the right tract. Just have more courage to call a spade a > spade, or more exactly call nonsense what it is. > > Ed > > > Eric > > >
