I will try this explanation once again and see it is understood this time.
*“First deuterons have to assemble into a BEC”* This is the first assumption that is misleading you in to an incorrect conclusion. Deuterons are not necessarily involved in the BEC formation process. It involves dipole MOTION. The motion of charge separation becomes coherent. See this video of periodic motion becoming synchronized* * * * * * *32 out of sync metronomes end up synchronizing* * * * * http://www.youtube.com/watch?v=kqFc4wriBvE When there is strong coupling between dipoles they will become coherent. If you cannot believe your own eyes, it has been demonstrated in science many times in many ways. In the context of electron motion in a metal lattice, dipole movement is defined as follows: The product of magnitude of charge and the distance of separation between the charges; *Dipole moment* may refer to the measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system's overall polarity. *“Two d must first come together and remain together because they are able to form a BEC rather than a D2 molecule.”* This is a concept that is wholly formed in your own mind and has not been experimentally verified by many experiments in the science of Nanoplasmonics. Kevin is referring to an experiment that shows how the coherent MOTION of Rydberg atoms share large photons of laser energy. This experiment is directly applicable example of how large photons of energy derived from gainful subatomic transitions in atoms are shared in a metal lattice through the action of coherent motion of charge separation. *“You have to answer a series of questions if you go down this path. 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?”* * * * * They have been detected in experiments by both George Miley and Francesco Celani when they detected a drop of electrical resistance in areas of dipole condensation formation. *“Why does the BEC wait to fuse?”* A BEC is just one part of an EMF amplification cascade where EMF is concentrated to levels so high that these force fields disrupt the normal configurations within the nucleus of the atom and allow the nucleus to reconfigure into a system of lower energy potential. Usually, heavy nuclei become lighter in this process; however the possibility that a proton will be incorporated in the new nuclear configuration is possible but less likely. *“I suspect these answers are not easy to justify. If the answers are not provided, this mechanism cannot be a solution to the CF problem.”* This mechanism of EMF amplification has been seen is Nanoplasmonic experiments where the EMF produced in: “hot spots” become so intense that the chemicals used to determine field EMF strength were destroyed. LENR is just an intensification of the Nanoplasmonic “Hot Spot” mechanism where EMF disruption becomes so strong that nuclear processes are affected. >From the standpoint of physical principles LENR vs. Nanoplasmonics, it is more a matter of quantity rather than quality of the physical processes. On Mon, Jun 10, 2013 at 9:51 AM, Edmund Storms <stor...@ix.netcom.com>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. 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? How big can a BEC get before it is > unstable in a lattice or living cell? 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? Why does the BEC wait to fuse? Why do only 2 d fuse in a large > group? How is the Coulomb barrier reduced? How is energy communicated to > all members of the BEC? 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. > > 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/**vortex-l@eskimo.com/msg81244.**html<http://www.mail-archive.com/vortex-l@eskimo.com/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 <janap...@gmail.com> 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. >>> >>> >> >
