Axil--
That is an interesting observation regarding photo-fusion. EMF photons would carry spin to the reaction I would say. How is it there is no spin associated with the incoming photon? unless there are two oppositely polarized photons arriving at the same time. Bob Sent from Windows Mail From: Axil Axil Sent: Saturday, July 12, 2014 9:26 AM To: vortex-l@eskimo.com One characteristic of photo-fusion is that no change in angular momentum occurs. The EMF carries no angular momentum into the reaction. If the nucleus goes into the reaction with zero spin, it will come out of the reaction with zero spin. On Sat, Jul 12, 2014 at 1:22 PM, David Roberson <dlrober...@aol.com> wrote: Bob, A careful choice of your reference frame can help resolve many of the linear momentum issues. I like to choose one that is located at a point where the net linear momentum of the particles is zero before the reaction. Under that condition it is relatively easy to follow the reactions since the final momentum must also remain zero and avoids nasty math errors. Dave -----Original Message----- From: Bob Cook <frobertc...@hotmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Sat, Jul 12, 2014 3:50 am Subject: Re: [Vo]:Dynamic nuclear polarization Dave-- I would assume in the hot fusion regime that significant linear momentum must be conserved in addition to the conservation of energy associated with kinetic energy of colliding particles. In cold fusion LENR there is know momentum other than angular momentum to conserve. Gammas and other linear momentum carrying particles are not needed and in fact not possible because of their of their necessary of carrying linear momentum. it is for this basic reason that I do not anticipate the existence of ganmas or any energetic particle to be associated with LENR. If any has a good physical explanation of the mechanism for the distribution of the linear momentum between decay products I would love to see it. Bob Sent from Windows Mail From: David Roberson Sent: Friday, July 11, 2014 11:31 AM To: vortex-l@eskimo.com When I take a step back I realize that it appears like a miracle for the energy to always come out in small fractions of the total available. I have to ask whether or not this unusual situation may be related to the conditions upon which the reaction occurs. Is anyone aware of an experiment that actually involves fusion of D x D at low temperatures while the radiation is monitored? We do have data describing what is released at very high kinetic energies, but is there a threshold below which our preferred path may be exclusive? I suppose the closest analogy would be muon fusion. If I recall, that pretty much matches what is emitted under hot fusion conditions. Perhaps your point is valid and there is zero chance that D x D fusion is taking place directly. If true, some sneak path is being followed and it is common for alpha radiation to be generated in nuclear reactions. Plenty of energy can be deposited by alpha radiation into the structure. Keeping that under control without generating gammas is quite a trick. And, what other nuclear ash should we be seeing? I hope that Rossi and the future report from the long term experiment will help to answer many of our questions. Dave -----Original Message----- From: Jones Beene <jone...@pacbell.net> To: vortex-l <vortex-l@eskimo.com> Sent: Fri, Jul 11, 2014 2:19 pm Subject: RE: [Vo]:Dynamic nuclear polarization From: David Roberson I think Bob is hoping that energy can be taken away in smaller chunks and that is what I would want to see as well….Has anyone identified exactly where the large MeV energy from a D x D fusion is stored? It remains in place for a short duration until released. Perhaps it can be taken in many portions instead of one dangerous gamma. Dave, Once again, the relevant question is not whether energy can be released piecemeal, in many small undetectable portions. We can assume that it can. The relevant question is this: can a new and previously unknown mechanism accomplish this incredible feat 100% of the time, to the complete exclusion of the known mechanism? Clearly – that is most unlikely. The 23 MeV would need to come out in packets of no more than about 6 keV each. Anything above this level would show up on the kind of meters which have been used for many years, and which have already proved that strong radiation above background level is seldom seen. Think about it. That lack of any radiation signature in most experiments of this kind means the large amount of energy (from the formative alpha particle) comes out in at least 4,000 individual packets, none of which can ever be larger than what is detectable. And furthermore, never ever do we see the “known release mechanism” of standard physics. If true, this proposition is moving towards an “intelligent” release of radiation, in which packets must be monitored and rejected if they are too energetic. That kind of control is absurd, of course, but it highlights the larger absurdity of suggesting that this reaction must involve the fusion of deuterons to helium with no gamma signature. There are better alternatives. Jones