ALEX,
I can not address all you say because so much is based on imagination,
not fact. But let me make a few points.
The two process (1 and 2) must happen at the same time because the
evidence shows that the energy is released, not stored. No method is
known that can store 23.8 MeV in a nucleus. A nucleus having this much
stored energy would immediately get rid of it by some process. This is
well understood behavior.
Second, Kim speculates about the totally unsupported claim for
transmutation being the source of energy from the Rossi reactor. This
exercise shows that clever people can "explain" anything whether it is
real or not. Nothing Rossi claims about the process can be believed
until this process is in the hands of someone who knows how to make
proper measurements. The claimed Cu clearly did not result from a
nuclear reaction and even Rossi has backed off from this claim. Yet,
people can "explain" this process. Yes, transmutation can happen but
never enough to produce detectable energy.
You need to be careful of what you try to "explain".
Ed
On Jan 25, 2013, at 2:17 PM, Axil Axil wrote:
Yes, screening occurs. The question is, "Is this process alone
sufficient to create LENR at over 10^11 times/sec and how does it
allow the resulting energy be dissipated? Please answer this question.
There are two processes going on in cold fusion when it is working
properly: one(1) is charge accumulation that shields the coulomb
barrier of atoms, and two(2), the other is quantum mechanical
entanglement of protons from ionized hydrogen atoms that thermalize
the radiation produced by fusion.
It is possible for one(1) to be active when two(2) is not.
Early on, Rossi had trouble with his 100 gram reactor when it was
starting up and shutting down because it was too cold during those
times.
Dr. Kim explains the nuclear energy side of this entanglement
mechanism in this paper:
http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf
Kim shows how cold fusion of a cooper pair of protons (two protons
stuck together) will produce certain types of nuclear reactions.
In more detail in the old Rossi reactor design, at startup, a large
amount of gamma radiation appears before proton entanglement has
established itself since the temperature of the nickel has not
gotten to the relatively low Curie temperature (nickel has the Curie
temperature of 631 K (~358 C)). Formation of the proton condensate
is sensitive to the magnetic nature of nickel. When nickel is
ferromagnetic it won’t let the protons form and join the proton
assemblages. In such a collection of identical and entangled
protons, all the protons in the collection share in the nuclear
energy that any given member is exposed to. Nickel must first be
made paramagnetic by heat so that the protons can join the
superconductive proton assemblage. This entanglement process makes
the heat output conversion of the cold fusion reaction possible.
Rossi fixed this problem when he added a secondary heater to his old
design to preheat the reactor structure before the Ni-H reaction
begins.
The coherent and entangle wave forms of these many protons that
comprise the proton condensate will all work in concert through a
quantum mechanical wave based summation process to form a combined,
entangled and coherent single de-Broglie wave form. The whole proton
condensate then participates in nuclear fusion. But the proton
condensate can be spread out in the nickel lattice and also in the
hydrogen envelope and even inside the walls of the reaction vessel.
Because of its very large coherent de-Broglie wave form, the
effective quantum mechanical range at which this condensate operates
may be very large, being spread out anywhere up to hundreds of nano-
meters which always include the proton pair that has participated in
the fusion reaction.
It seems to me that when copper or tungsten is used as the lattice
material, the cold temperature problem in the lattice with regard to
gamma production is not as pronounced because of the paramagnetic
nature of these metals.
Superconductivity and ferromagnetism just do not go well together.
The research of Piantelli has shown that 6 MeV protons are coming
out of the nickel after these bars are immediately removed from the
Piantelli reactor.
This is a solid indicator to me that double proton fusion is
occurring in the nickel lattice. When these bars are removed from
the reactor they cool rapidly. This rapid cooling of these bars
takes their temperature quickly below the Curie temperature of
nickel. The energy of the cold fusion reaction is no longer being
thermalized by entanglement of the protons, so all 6 MeV of the
reaction is being produced by the nuclear relaxation process of the
excited nucleus.
As another example, when 100 or more protons work as a team in a
condensate, then I would estimate that as example a gamma ray with
an energy of 8 MeV would instead distribute the energy into an
average of 80 keV slices.
The binding energy made available by the fusion reaction is
transferred to the coherent and entangled ensemble of protons when
the fusion process completes. Whenever energy on any kind is
transferred within an entangled ensemble, this assemblage becomes
decoherent.
As Dr. Kim states, this thermalization process can be proven when
the nuclear reaction products from the Ni-H reaction are
characterized. These products of double proton fusion are unique and
are easily described.
Cheers: Axil
On Fri, Jan 25, 2013 at 3:41 PM, Edmund Storms
<[email protected]> wrote:
On Jan 25, 2013, at 1:31 PM, Axil Axil wrote:
Quantum mechanics lives in the realm of the wave. The electron will
exert it influence on the positive charge nucleus in bits and pieces.
Alex, you are using the wave model and I'm using the particle model.
Both are accepted by science and are useful. However, it is best to
stick to one or the other in a discussion. Otherwise, the discussion
gets too confusing to be useful.
Take a look at this to give your imagination a brake:
http://en.wikipedia.org/wiki/Thomas%E2%80%93Fermi_screening
The Thomas-Fermi formula is a more general potential than the
Coulomb's law.
Yes, screening occurs. The question is, "Is this process alone
sufficient to create LENR at over 10^11 times/sec and how does it
allow the resulting energy be dissipated? Please answer this question.
For the nonlinear Thomas-Fermi formula, solving these
simultaneously can be difficult, and usually there is no analytical
solution. However, the linearized formula has a simple solution:
R= (Q/r)((e)exp(-kr))
With k=0 (no screening), this becomes the familiar Coulomb's law.
The infuence of about 2000 electrons near the site of fusion will
lower the coulomb barrier.
No material has 2000 electrons at any nucleus where they must be
located to lower the barrier.
Ed
On Fri, Jan 25, 2013 at 3:01 PM, David Roberson
<[email protected]> wrote:
That is an interesting complication Axil. There is no doubt that
the electrons can act as a screen of the electric field to an
extent. Once, I tried to get a handle upon the magnitude of this
effect from a simple mental model point of view and a few things
seemed to show up. The COE and COM like to make it difficult to
visualize. I placed an electron between two protons and realized
that as long as the electron was in the middle, there was no
Coulomb barrier to counter since the negative charge exerted a
slightly larger pull than the opposite positive charge repelled as
the combination gets smaller.
This model leads to an interesting idea. If the electron could be
judiciously placed precisely between the protons, there would be no
net force acting upon it. If we then allow the protons to slowly
come together, there would be no net energy imparted upon the
electron as the system shrinks. Each proton would actually be
drawn towards the other one and a small amount of energy would be
imparted upon each. This is due to the fact that the electron
charge is closer to the proton charge than is the other positive
repelling charge.
This process could be continued until something gives. A net
amount of energy is given to the protons as they head towards each
other. The electron is merely kept in the center without expending
any energy.
Now, if the electron squirts out of the line at right angles to the
axis between the protons, then it must be given energy equal to the
amount of Coulomb energy that it helped overcome as the protons
came towards each other. This would be expected if the electron
were to escape the vicinity. The protons would then possess the
same amount of energy that they would have obtained had they not
had the electron to help.
If an electron could be coaxed into this behavior and remain
between the proton pair until the group merges, then fusion would
be common. Since this is not true, one must assume that the
electron diverts at some point. Perhaps a gamma ray comes along
to set it free, but more likely, quantum mechanics intervenes and
the electron begins some form of orbital motion around one or both
protons. Unless the orbit that it settles within allows for the
release of extremely high energy, then the protons are not close
enough to fuse. I suspect that a process of this general nature
might lower the net Coulomb barrier to a degree, but I have no idea
how much.
I began to think of a multiple electron case, but grew weary as my
mind wasted away.
Dave
-----Original Message-----
From: Axil Axil <[email protected]>
To: vortex-l <[email protected]>
Sent: Fri, Jan 25, 2013 2:21 pm
Subject: Re: [Vo]:Chemonuclear Transitions
For one, it is not possible for an alpha with that total energy to
be released.
I would like to introduce a complicating factor: electron screening..
Both the cross section of alpha decay and nuclear fusion can be
significantly reduced by electron screening.
In fact I believe that the helium 4 seen in cold fusion experiments
are many times derived from enhanced alpha emissions from high Z
elements rather than fusion of hydrogen.
In the presence of an electron cloud, the consideration of the
coulomb barrier potential must be replaced by the Tomas Fermi
potential to account for electron screening.
Furthermore In astrophysics, cross sections of low energy fusion
events can increase by a factor of one million based on the extent
of electron screening around the fusion site. In fact, it is
impossible to experimentally produce correct stellar fusion
reaction cross sections because both theory and experiment is not
able to explain astrophysical fusion based observations due to the
electron screening problem.
Astrophysics uses the Trojan horse approximation to get around this
electron screening conundrum.
Cheers: Axil
On Fri, Jan 25, 2013 at 1:17 PM, David Roberson
<[email protected]> wrote:
Sometimes the emails do get crossed up with the number of
responses. In this particular case I think that my input helped to
clarify the problem to many others who may be following this
discussion. My choice of observation locations proves that there
are two bodies or body equivalents that must exit the reaction.
Now it is plain for all to see that it is not possible for an alpha
particle to be the only result since I have demonstrated that the
conservation of momentum would be violated it this were to happen.
Before my mental example, it was just a statement that was
difficult to defend. Now we can more readily understand the type
of reaction that must take place in this form of fusion. For one,
it is not possible for an alpha with that total energy to be
released. If we could get a measure of the energy of the alphas
that actually are emitted, then that information can be directly
used to calculate the transferred momentum and energy which is
received by the matrix. Now, I have shown that some reactionary
force is required through which the energy and momentum is
transferred to the system. This is an important observation in my
opinion.
It is good that the members of vortex-l can discuss issues of this
nature since much is not known about the reactions that take
place. Sometimes a small spark of incite at the correct moment
will lead to added knowledge. Perhaps others now will realize that
what I have written here is educational. The next time, they might
use my ideal observation location or something of a similar nature
to understand other physics problems. Had I written a paper, it is
likely that I would have overlooked this particular tidbit of
knowledge and left out a major issue that should have been
considered.
So, I suggest that we continue to engage in similar discussions
within vortex and enlarge our knowledge base since no one person is
required to be the holder of all that is important. Knowledge is
always advancing as more minds are engaged.
I vote for open discussion within vortex. And, my post was not a
waste of anybodies time. Proof of this assertion will be from this
point forth since most of those engaged in the current discussion
will now understand the issue of energy and momentum requirements.
Dave
-----Original Message-----
From: Edmund Storms <[email protected]>
To: vortex-l <[email protected]>
Cc: Edmund Storms <[email protected]>
Sent: Fri, Jan 25, 2013 12:12 pm
Subject: Re: [Vo]:Chemonuclear Transitions
The problem with such exchanges is that the messages to different
people cross so that I have to explain the same thing several
times, which is a waste of time. That is why I write papers so that
everyone can study the same explanation.
On Jan 25, 2013, at 9:51 AM, David Roberson wrote:
Ed, I am confused by your statement that cold fusion is a 2-body
to 1 body reaction. I see two reaction components unless I am
missing something. One is the alpha particle and the other
appears in the form of mass released as energy into the
surrounding structure.
The energy release must result from emission of something. Normally
in hot fusion, the release results from emission of a strong gamma
when He4 forms. This gamma is not present when He4 forms during
cold fusion. Why not? The mechanism of energy transfer is obviously
not conventional, yet it must be consistent with the law of
conservation of momentum. I try to solve this problem in my
theory. Most people ignore the issue.
Ed
Every observer must see that the laws of physics apply to what he
sees. My favorite point is to be located precisely between the
two protons as they head toward each other with exactly the same
energy. In this location an observer sees that a finite amount of
kinetic energy is measured for the two particles and that there is
exactly zero momentum for the equal velocity pair. When they
collide together, there is no motion required for the resulting
alpha particle until it releases the excess energy. When that
energy is finally emitted in some form, then a reaction force
would result in relative motion of the alpha particle. In this
manner, both conservation of energy as well as conservation of
momentum is shown.
In my experience, when these laws are seen by any one observer,
then they are true for all of the others. Do you see a hole in
this argument? How are the laws true for others but not for the
one ideally located?
Dave
-----Original Message-----
From: Edmund Storms <[email protected]>
To: vortex-l <[email protected]>
Cc: Edmund Storms <[email protected]>
Sent: Fri, Jan 25, 2013 10:38 am
Subject: Re: [Vo]:Chemonuclear Transitions
The human mind is able to imagine endless possibilities. In order
to make any progress, a triage must be done by eliminating the
ideas that are so improbable or so illogical that they have very
little chance of being correct. That is what I'm attempting to do.
In any case, several basic rules MUST be considered. Hot fusion is
a conventional 2 body-2 body reaction as is required to carry away
the energy and momentum. Cold fusion is a 2-body to 1 body
reaction that violates this condition. That violation MUST be
acknowledged and explained.
People are not free to imaginary any thing. Certain rules are
known to apply. These rules are so basic that they MUST not be
ignored.
Ed Storms
On Jan 25, 2013, at 8:22 AM, Daniel Rocha wrote:
d+d=n+He3 and d+d=t+p
What about d+d+...+d=? We don't know. This is what many many
particle models ends up being. Theyare hot fusion. The only
difference it is that there are many, more than 2>, incoming
nuclei to fuse. You cannot do that in experiments using
colliders, it is too unlikely. So, you cannot say that cold
fusion is any different than hot fusion that easily.
2013/1/25 Edmund Storms <[email protected]>
Yes, people try to explain LENR using the behavior described in
the paper.
--
Daniel Rocha - RJ
[email protected]
