Please Axil, stay in contact with what is observed. Helium is the
observed nuclear product. This requires two d to fuse. These d are in
isolated and random sublattice sites. They have to all get into one
site to fuse. Therefore, they MUST assemble. The BEC is proposed to
result from this assembly. I do not know what you have in mind but it
is not related to what is known or what is being proposed.
Ed
On Jun 10, 2013, at 1:14 PM, Axil Axil wrote:
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
<[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.
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/[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
On 6/7/13, Axil Axil <[email protected]> wrote:
References:
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
*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.
