Ed, these are very good questions. At the risk of reiterating points made in older threads, I'll attempt to address each question as I am able.
On Sat, Jun 22, 2013 at 6:11 AM, Edmund Storms <stor...@ix.netcom.com>wrote: In your theory, how is the energy released as kinetic energy without > particles being emitted? > It's not my theory -- it's Ron Maimon's. He's saying that in a Pd/D system, specifically, there is a set of conditions in which two deuterons will approach a palladium nucleus simultaneously. The close proximity of the deuterons to the palladium nucleus will have two effects. The first effect is to "focus" their de Broglie waves in a way that will make it more likely for them to overlap. The second effect is to cause the d+d→4He+Q branch to become much preferred over the d(d,p)t and d(d,n)3He branches seen in plasma fusion. The reason it becomes preferred is that the "Q" is dumped as an electrostatic impulse that is shared between the daughter 4He and spectator palladium nucleus once the metastable [4He]* transitions to ground, rather than being emitted as a gamma ray. Electrostatic dumping happens quickly, and hence is more probable, while the emission of a photon takes a long time. This electrostatic dumping of the energy translates into kinetic energy, as might happen with an Auger electron in other contexts. The reason the other two d+d branches are competitively disfavored, as far as I can tell, is that the modified d(d,Q)4He branch becomes all the more likely. The reason it becomes very likely is that there are 46 protons in the palladium nucleus with which to interact via the electrostatic force. Ron talks about the Pd/D system, and I have graciously borrowed his ideas and attempted to apply them to other systems such as Ni/H. This account does imply the emission of particles. So an important question is under what conditions they are produced and whether we have done adequate work in ruling out prompt radiation when excess heat is underway. There are plenty of experiments showing only marginal levels of prompt radiation emerging from the substrate. There are paltry few experiments, as far as I can tell, showing that when there is excess heat there is no prompt radiation taking place at some depth within the substrate. How is momentum conserved? > The momentum of a fusion reaction is shared between the daughter or daughters and the metal spectator nucleus. So in branches where a photon would normally be emitted, there is instead recoil of the metastable daughter during the transition to ground and no need for photon emission. Kinetic energy is defined as something moving with a velocity. How is this > velocity created from initially still objects while momentum is conserved. > In the case of Pd/D, this is understood to happen electrostatically with the decay of the metastable [4He]* daughter. The 4He is pushed off of the nearby palladium nucleus, like a bullet from the back of the chamber of a rifle. > Also, why does the system choose to release energy this way? What rule > makes this the easiest way? > I'm not sure. This is one of the many questions I have. I have been trying to understand the system sufficiently to gain insight into these questions, but it's been a slow learning process. Just to anticipate an objection that this account implies energetic particles, and energetic particles and their side effects are not seen, I should mention that I'm reading the older papers and am trying to collect more data on this topic so that I can better understand this objection. Eric
