On Mon, Jun 10, 2013 at 6:51 AM, Edmund Storms <[email protected]>wrote:
> Kevin, when you suggest involvement of a BEC, you need to consider the > sequence of the process. First deuterons have to assemble into a BEC of a > increasingly larger size. Two d must first come together and remain > together because they are able to form a BEC rather than a D2 molecule. > However, these two d do not fuse. Then another d arrives and joins the > group. This assembly continues to be a BEC, but does not fuse. Only when > hundreds of d have assembled does two of the members of the BEC fuse and > communicate all of the fusion energy equally to all other members of the > group. > > You have to answer a series of questions if you go down this path. ***Well, I'm no nuclear physicist, but I'll take an inductive 40,000 foot SWAG at it. > For example, if two d can form a BEC and be stable, why do these BEC not > grown in number until they can be detected? ***I think the BEC is formed with multiple d atoms within the Palladium matrix, involving the Palladium AND Deuterium, or in the other case, Hydrogen and Nickel. > How big can a BEC get before it is unstable in a lattice or living cell? ***I don't think BECs have a problem with instability. They are low energy physical phenomenon. Atoms that participate in a BEC tend to lose energy AFAIK. It is this loss of energy that is what causes 2 atoms to fuse, because they get close enough that the Coulomb barrier starts to collapse. > How big must this limit be before the energy from a single d-d fusion > acquired by each emitted d is too small to detect? ***According to the article cited, the energy absorbed by the BEC is directly related to the number of atoms within the BEC, so the "too small to detect" energy would be thermal. From the thermal energy detected, we can proceed backwards to how many atoms were involved with the BEC. > Why does the BEC wait to fuse? ***It would be 2 atoms within a BEC of, say, a thousand atoms, which includes atoms from the Palladium or Nickel Matrix. The BEC basically slows down the energy of the atoms participating until the point where 2 of those involved get close enough and "tired enough" to slip together past the Coulomb Barrier, and those 2 atoms fuse, emit the gamma ray and destroy that particular BEC. > Why do only 2 d fuse in a large group? ***It could be for multiple reasons. Here is one way I imagine it might happen. As the BEC forms the atoms start moving in a very coherent fashion and a hydrogen atom goes from one boxed in Nickel position to another, along with all the others. But if one of them "gets stuck" in its box, as the other hydrogen atom arrives in the box get stuck together. It would be those 2 atoms which eventually fuse. Or it could happen that 2 BECs form next to eachother, start vibrating and doing their thing and collide with each other, with 2 of the atoms on the edge of each participating BEC fusing because the BEC forces are stronger than the Coulomb Barrier. Or, to use your cracks as the observed trigger, the BEC forms with hundreds of hydrogen atoms and Nickel atoms but at the surface, the atoms are in disarray due to a crack. The BEC runs into those atoms which are on the surface stuck in a crack and this collision causes 2 to fuse. http://www.youtube.com/watch?v=SoiteXBb1mA&feature=player_embedded About 3:40 into the animation. I found it at Superwaves's site Jed's comment: I believe all of the claims up to 3:40 are based on conventional electrochemistry. At that point the narrator claims that that D ions at high concentration in the lattice begin moving together "like a school of fish," and then they fuse > How is the Coulomb barrier reduced? ***AFAIK, when a BEC forms, it reduces energy to participating atoms. All of the atoms within that BEC start to function as one entity, as one atom per se. It is this loss of energy that makes the hydrogen atoms docile enough to let down their Coulomb Barrier guard. > How is energy communicated to all members of the BEC? ***I would suspect through phonon interaction. But I really have no idea. > I suspect these answers are not easy to justify. If the answers are not > provided, this mechanism can not be a solution to the CF problem. ***Interesting posture. It omits the possibility that BECs are the solution to the CF problem, but the people suggesting it as the solution are unable to come up with those answers. > > Ed > > On Jun 9, 2013, at 9:46 PM, Kevin O'Malley wrote: > > On another thread, Edmund Storms posted how many nuclear fusion atoms >> must take place to generate 1 Watt of power. We can work backwords >> from that number, knowing that a certain number of Watts are >> generated. Then we know how many atoms/second are fusing. From that >> calculation you can figure out how many OTHER atoms need to be >> involved with the BEC in order for it to have the frequency being >> observed. I doubt the entire device needs to be involve. I think it >> would be hundreds of BECs forming with thousands of atoms. >> >> >> http://www.mail-archive.com/**[email protected]/msg81244.**html<http://www.mail-archive.com/[email protected]/msg81244.html> >> >> This paper verifies that a photon eradiated Bose-Einstein condensate will >> cut the frequency of incoming photons by dividing that frequency between N >> numbers of atoms. >> >> >> http://arxiv.org/pdf/1203.**1261v1.pdf<http://arxiv.org/pdf/1203.1261v1.pdf> >> >> >> >> On 6/7/13, Axil Axil <[email protected]> wrote: >> >>> References: >>> >>> >>> http://phys.org/news/2013-05-**einstein-spooky-action-common-** >>> large.html<http://phys.org/news/2013-05-einstein-spooky-action-common-large.html> >>> >>> >>> *Einstein's 'spooky action' common in large quantum systems, >>> mathematicians >>> find* >>> >>> >>> If you like mathematics that can choke an elephant try this as follows: >>> >>> >>> http://arxiv.org/pdf/1106.**2264v3.pdf<http://arxiv.org/pdf/1106.2264v3.pdf> >>> >>> >>> *ENTANGLEMENT THRESHOLDS FOR RANDOM INDUCED STATES* >>> >>> Why does a Ni/H reactor form a Bose-Einstein condensate throughout its >>> entire volume? STANIS LAW J. SZAREK provides the answer; the dipoles >>> throughout the reactor are forced to become totally entangled when the >>> percentage of dipole entanglement exceeds 20%. >>> >>> >>> >>> The Ni/H reactor will formulate a very large entangled system when it is >>> in >>> operation. As a large system, it has no choice but to become totally >>> entangled. >>> >>> >>> Infrared Photon tunneling between the individual Nano-cavities is the >>> method by which quantum entanglement is spread Josephson like from one >>> nano-cavity to its immediate neighbors. >>> >>> >>> When the Ni/H reactor is not totally entangled, it renders the nuclear >>> energy it produces from the decoherent nano-cavities as gamma radiation. >>> However, if the 20% entanglement threshold is reached, the energy >>> produced >>> by the LENR reaction is thermalized through the process of frequency >>> sharing as in a large super atom. >>> >>> When a Ni/H reactor is not yet totally entangled, it will produce gamma >>> radiation. This can happen when the reactor is heating up upon startup or >>> cooling down at shutdown. >>> >>> In the LeClair reactor, the 20% entanglement threshold is never reached >>> and >>> a significant proportion of its energy output is rendered as gamma >>> radiation. >>> >>> A Ni/H reactor must exceed this 20% dipole entanglement threshold before >>> its energy production phase is initiated to avoid the inconvenience of >>> gamma production. >>> >>> >> >
