On Jun 11, 2013, at 12:26 PM, Kevin O'Malley wrote:
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.
So, you propose that PdD contains BEC in the lattice as it forms.
Nevertheless, all PdD is not nuclear active and PdD has been studied
very carefully for the last 100 years with no indication of any kind
of cluster being present. These facts conflict with your assumption.
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.
Why does this collapse happen? How many d must be present in the BEC
before fusion takes place?
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.
Yes, and the number is huge. How can so many d assemble in any
reasonable time and not be noticed when materials are examined?
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.
Nice assumption, but what process determines when this happens?
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.
A BEC is a unique structure that has to exist as a unique unit or
group. The Ni has no role in its composition. There are no "boxes".
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.
So, you assume BEC form in the crack. Why not assume that a Hydroton
forms there? Why invoke a structure that is known not to form at high
temperature while ignoring a structure that can form at high
temperature?
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
This model totally ignores the consequence of isolated fusion. This
kind of fusion creates a conservation of momentum problem and easily
detected radiation, which is not produced.
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.
All ideas have to be tested against logic and observation. Otherwise,
the sky's the limit. The theories that pass the test are then worth
considering further. Otherwise, a lot of time is wasted talking about
ideas that have no reality. The BEC, so far, has no reality.
Ed
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.
