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.
