On Mon, Feb 25, 2013 at 6:59 PM, <[email protected]> wrote: If the energy is released in the form of a fast particle, then it does not > have > to depend on diffusion. A fast particle will rip through a lattice at high > speed, leaving a trail of ionized atoms in it's wake. >
That provides a nice segue to a recent set of responses from Ron Maimon to questions that have been raised here about his Auger deuteron proposal. The questions he is responding to have been posted here [1], as is his reply in the comments section. Copied below is the text of the reply with brief notes referring to the questions. Eric "1+2 [concerning the probability of deuterons meeting and ROI in general]. The dynamics of charged particles in materials is not as "frictional" as you imagine. Charged particles don't slow down through bulk dynamics like a baseball in water, they have to transfer their energy through ionization of atoms, and in the case of 20KeV deuterons, they need to ionize other inner shells. They will eventually slow down, but it isn't clear how quickly, because it depends the precise density of states. If you have banded excitations, there are sometimes band-gaps where you have no states, so you can sometimes have extremely long-lived high energy states just because there aren't any available states to allow a simple de-excitation process. So if there are no available states at 17 KeV to transition to when you kick out 3KeV in exciting the n=1 p-wave electrons, these things could live a long time. I am not saying that this is necessarily so, I have no idea how long lived a 20 KeV deuteron is, but it can have a lifetime from a fraction of a second to minutes. But in any case, there is certainly not a barrier to having a 20KeV particle go microns or a millimeter through a metal before slowing down to 5KeV. The penetration depth of 20 KeV charged particles is not 10 atoms. "In order to have the cold-fusion proceed, you do need a region with high density of deuterons, high enough for 2 deuterons to fuse. When this is so, this first happens right near a nucleus, because this is where the wavefunction of 20KeV deuterons are most concentrated. The wavefunction of such a deuteron has turning points near the nucleus, and you can describe it semi-classically, and the thing is peaked at the K-shell radius, about 100 fermis from the nucleus. "The concentration of deuterons can plausibly happen in many ways, just from local electric fields concentrating the positively charged fluid into some region, it doesn't require conspiracy. The deuterons are charged and macroscopically flowing through the metal, like electrons, except at much higher energy. The result is that they can concentrate in regions where the electric field is big. Whether they flow in the direction of E or paradoxically against the direction of E depends on details of the energy as a function of wavenumber in the high-energy band, but my intuition is that they will flow like positively charged objects, in the direction of E, and opposite the direction of electron flow. This is a bit confusing, because this means they flow away from the surface of the cathode toward the interior. But they could also flow the other way, if their effective mass is negative, meaning if you produce them near a maximum of the band dispersion relation. "3 [concerning a novel, radiation-free d+d→4He+Q branch]. The whole point is that the process is d+d fusion right next to a Pd nucleus, so that the process completes with electrostatic energy transfer to the nucleus, not with the standard transfer of energy to an ejected proton or neutron. Then the result is an ejected He4 and a fragmented nucleus. We have no data on three-body fusion collisions, they never occur in dilute plasmas, so we don't even have an estimate for the cross section (except order of magnitude it should be not too many orders of magnitude larger than usual fusion, and this is ok for the theory to work). The point here is that due to the concentration of the wavefunction of the deuterons near the Pd nuclei, from the turning-point, you get a larger density of deuterium near the Pd, and you can have fusion when the local density of deuterium exceeds a certain threshhold. "4 [concerning the generation of neutrons]. The random bumping around fusion will occur in this theory, you will get occasional sporadic neutrons from the system. I don't consider this a problem, as it explains the neutron signal people see. Neutrons are not easy to detect anyway." [1] http://rolling-balance.blogspot.com/2013/02/ron-maimons-theory-2.html
