-----Original Message----- From: Bob Cook > Was there any indication that the Mizuno experiments used quadrupole electric or magnetic inputs? I was not aware of this, if it happened.
This is an interesting point, and the Mizuno experiment may not have been optimized. Hopefully the next iteration will tell us more. One interesting thing about superparamagnetism is that a quadrupole is a natural outgrowth of any changing field, and if the deuterium becomes superparamagnetic that is all we need. > Also keep in mind that D is a Bose particle (as is 4HE) and can form a BEC or a duplex BEC with two different Bose particles. This may be a reality in a strong magnetic field, temperature be damned. Deuterium in normally diamagnetic as a molecule and we really do not need for it to a BEC to see excess energy. As a bare nucleus, it may be superparamagnetic, and that could be enough for spin coupling. Superparamagnetism in somewhat "new" to consideration in the LENR field, since it is an outgrowth of nanotechnology. We have much to learn about spin-coupling but in my mind, superparamagnetism will be a major piece of the puzzle, but certainly bosons which are superparamagnetic are more likely to participate in energy transfers than fermions. > The question I have is how a BEC can shed energy and change the mass of its constituents without disrupting the condensate. Well - Deuterium cannot even become a condensate in normal LENR, since it requires cryogenic temperatures. However, there are boson condensates in SPP which form at elevated temperature but they are not massive. Axil and myself have presented scholarly articles to that effect. An answer for original question is that an SPP condensate allows energy to be coupled from A to B (where A is a boson like the deuteron) without itself necessarily participating in the reaction. > Maybe it is a series of condensation and disruption that controls the > reaction. The dynamics of this process would be key to controlling the rate > of the process. Still not sure why you think that deuterium can become a BEC above a few degrees of absolute zero? Is there any evidence for this? Jones ----- Original Message ----- Bob, I have cherry-picked three major “spin facts” from this compendium which indicate that if one wants to apply a nano-magnetism or spin-coupling modality to LENR, it is highly preferable to use deuterium, as opposed to hydrogen. That may be why Mizuno chose the deuterium-nickel combination. All eyes will be shifting to Mizuno in less than three weeks. From: Bob Cook [snip] The deuteron, being an isospin singlet, is antisymmetric under nucleon exchange due to isospin, and therefore must be symmetric under the double exchange of their spin and location. Therefore it can be in either of the following two different states: Symmetric spin and symmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (+1) from spin exchange and (+1) from parity (location exchange), for a total of (-1) as needed for antisymmetry…. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1) from spin exchange and (-1) from parity (location exchange), again for a total of (-1) as needed for antisymmetry. [snip] …suggesting that there may be a way to stimulate the D via an electric quadrupole input signal. Also with a magnetic moment the D must respond to a magnetic field and fine tuning of an oscillating magnetic field may very well excite the D to flip up and down in the field. The composite particles of the D should have slightly different magnetic moments that can respond and create an "excited" state IMHO on a transient short lived time frame. However in a coherent system such a transient may be enough to cause other transitions of similar energy states to occur with mass energy being changed to angular momentum energy. The quadrupole input is a strong clue.
