Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

http://arxiv.org/pdf/1207.0079

These authors showed how to approach one of the fundamental problems of
hadronic physics, the calculation of the baryon masses from the Lagrangian
and the vacuum condensates of QCD.


Cheers: Axil


On Fri, Jan 25, 2013 at 11:50 PM, Eric Walker  wrote:

> I wrote:
>
>>
>> What is it that is causing the proton in this model to vary in mass, and
>> is the range of possible masses discrete or continuous?
>>
>
> I should anticipate one possible answer, which seems like a good
> explanation -- a proton is not a point particle, like a photon, and it does
> not travel at the speed of light.  It has mass and it has a speed that is
> less than c.  So the mass will vary with its speed; when it is stationary
> it will have a rest mass, and when it is travelling
> at relativistic velocities, it has a larger mass.
>
> Assuming the above is true, and assuming your model of a proton having an
> average mass is true, the question for me now becomes, is the (rest) mass a
> continuous value or discrete across a range?
>
> Eric
>
>

Re: [Vo]:Lattice Energy posting on recent Li-battery failures

[pagnucco]

Rob,

Those were my (too quickly chosen) words - not Lattice's.

Certainly a definitive test must use very-clean, totally
characterized materials.  If the reports of measurable amounts of
transmutations in several LENR experiments are true, then (hopefully)
some measurable residues could remain after subjecting Li-battery
materials to large currents (especially if fractal surface and
bulk feature were present?)

The investigators are still puzzled by the failures.  See -

"U.S. regulators: Boeing 787 probe far from complete"
http://www.reuters.com/article/2013/01/24/us-boeing-dreamliner-idUSBRE90M0ZO20130124

This sort of reminds me of the 1951 Jimmy Stewart movie,
"No Highway in the Sky".

I hope the current investigation comes up with a conclusive explanation.

-- Lou Pagnucco


Rob Dingemans wrote:
> Hi,
>
> On 24-1-2013 22:58, Jed Rothwell wrote:
>> I guess it was Lattice Energy who wrote:
>>
>> The LENR theory should be easily testable by autopsies on some
>> failed
>> batteries, looking for evidence of transmutations, i.e., unusual
>> isotopes or elements.
>>
>>
>> This would not be an "easy" test. There would be only microscopic
>> amounts of anomalous elements, and a burned battery is about as
>> contaminated and filthy as anything can be.
>
> I could be wrong, but is this not fairly easy to be determined by
> comparing the contents of several "clean" and "burned" batteries with a
> Mass Spectrometer?
>
> Kind regards,
>
> Rob
>

Re: [Vo]:Chemonuclear Transitions

[Eric Walker]

I wrote:

>
> What is it that is causing the proton in this model to vary in mass, and
> is the range of possible masses discrete or continuous?
>

I should anticipate one possible answer, which seems like a good
explanation -- a proton is not a point particle, like a photon, and it does
not travel at the speed of light.  It has mass and it has a speed that is
less than c.  So the mass will vary with its speed; when it is stationary
it will have a rest mass, and when it is travelling
at relativistic velocities, it has a larger mass.

Assuming the above is true, and assuming your model of a proton having an
average mass is true, the question for me now becomes, is the (rest) mass a
continuous value or discrete across a range?

Eric

Re: [Vo]:Chemonuclear Transitions

[Eric Walker]

On Fri, Jan 25, 2013 at 9:42 AM, Jones Beene  wrote:

In the end - it’s hard enough to convince observers that proton mass varies
> between atoms in any population - instead is an “average mass” which is not
> quantized.


One question I have about this approach has to do with a seeming move away
from quantization.  I take no position on whether this is possible or not,
and I may have misunderstood, so just trying to better understand.  In
order for the proton to have an average mass and not a fixed one, I think
there would need to be a degree of freedom that is not quantum, but
possibly discrete across a range of values or even continuous?  I think
you've mentioned the spin magnon in the past.  I believe this is a quasi
particle that is made up of the spins of the three quarks that make up the
proton?

What is it that is causing the proton in this model to vary in mass, and is
the range of possible masses discrete or continuous?

Eric

[Vo]:video: Peter Hagelstein Cold Fusion 101 Lectures

[Ruby]

They are difficult to hear, better with earphones, but what a special 
treat to hear the soft-spoken Dr. Peter Hagelstein describe his research 
in MIT's IAP short course Cold Fusion 101.  The slides alone show the 
line of thinking.


http://www.youtube.com/user/ColdFusionNow/videos?flow=grid&view=0

Thanks to Jeremy Rys who is attending the course for sharing this video.

Apparently, some MIT students wandered in, too.  Word's getting around 


--
Ruby Carat
r...@coldfusionnow.org 
Skype ruby-carat
www.coldfusionnow.org 

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

A thought occurred to me after the brief discussion that was conducted about 
the subject of D + D fusion.  The wikipedia article on fusion of this type 
suggests that there is always either a neutron or proton emitted from the 
reaction when hot fusion takes place.  This of course makes sense from the 
conservation of momentum and energy perspective as Dr. Storms has pointed out.


I commented that a measurement of the actual energy released to the alpha 
particles of cold fusion reactions would allow someone to calculate the energy 
and momentum that had to be left behind for the numbers to make sense.  My 
first thoughts on the matter were that this was going to require a large 
reactionary force if conservation of momentum was to be maintained.  I did not 
actually calculate the magnitude of the momentum or the energy associated with 
that mass conversion.


My choice of a central location from which to observe the reaction made it 
clear that the alpha particle would be frozen in place pending the release of 
this mass.  With this in mind I think that it would be wise for us to give very 
serious consideration to the prospect that direct fusion of D + D is unlikely.  
It would be a good idea to explore different paths that ultimately lead to the 
release of one or more alpha particles.  Of course the source for the reaction 
must be deuterium.  I am confident that this suggestion has been covered before 
and I am curious about the possible paths that are available.  Do any of these 
fit into place when a review of the active cold fusion metals is considered?  
Would the addition of a deuterium nuclei be encouraged by Pd for example?


Dave



-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Fri, Jan 25, 2013 9:18 pm
Subject: Re: [Vo]:Chemonuclear Transitions


Energy can be transferred from one molecule to another threw a quantum 
mechanical mechanism.
Yes
http://lightyears.blogs.cnn.com/2011/12/07/diamonds-entangled-in-physics-feat/
In the case of Walmsley's study, photons were showing up in two spots at the 
same time and causing vibrations within a pair of diamonds. The researchers 
made it happen by placing two diamonds about 15 centimeters (about 6 inches) 
apart on a table and then shooting a series of photons at a device called a 
beam splitter. Most of them went toward one diamond or the other, but a few of 
the photons went both ways at the same time. When those multitasking photons 
struck the pair of diamonds, they caused vibrations called phonons with each of 
the crystals.
The light from each of the beams recombines after exiting the crystals. And 
sometimes when the light is leaving the crystals, it has less energy than when 
it entered. That's how the researchers could tell that the photon had caused 
some vibrations.
"We know that one diamond is vibrating, but we don't know which one," Walmsley 
said. "In fact, the universe doesn't know which diamond is vibrating – the 
diamonds are entangled, with one vibration shared between them, even though 
they are separated in space."
 
Cheers:   Axil


On Fri, Jan 25, 2013 at 6:10 PM, Edmund Storms  wrote:


On Jan 25, 2013, at 3:49 PM,  
 wrote:


Excuse my grammar. English is not my native language.




I will try to answer your questions as simply as possible.



Can energy and momentum be transferred from the new He4 to another nucleus at 
some distains?




No


Energy can be transferred from one molecule to another threw a quantum 
mechanical mechanism.


Yes, at chemical levels of energy


This occurs in photo synthesis there excitations can jump between electrons in 
different molecules.




Yes


>From an older tread.

 http://www.mail-archive.com/vortex-l@eskimo.com/msg75294.html

Maybe a similar phenomenon can occur between nucleus?  This means the 
excitation from a He4 and momentum can be transferred




The amount energy generated by a nuclear reaction requires direct emission of a 
particle, which can include a photon. This is observed fact. The magnitude is 
too great to use mechanisms available in a chemical structure.  That is why 
most nuclear reactions are almost totally independent of the chemical 
environment.


to one or more receiver nucleus. These receiver nucleus must be a special 
nuclide suitable for  receive the energy and have a mechanism to

get rid of it. If several nucleus can get energy from one He4 it may radiate it 
as UV. If this not is possible I suggest that the receiver nucleus is a C12

how decay to 3 He4 as an reversed triple alpha.

In absence of receiver nucleus there will be no reactions. But this did not 
explain the overcome of the coulomb barrier

and why its not works in absence of receiver nucleus.


I have heard that the conservation of momentum in LENR is commonly explained to 
"something"

how would be like the Mössbauer effect. But I understand this not so easily to 
explain more exactly.




The Mossbauer effect involves a very small energy change. It works only because 
the 

Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

*Energy can be transferred from one molecule to another threw a quantum
mechanical mechanism.*

Yes

http://lightyears.blogs.cnn.com/2011/12/07/diamonds-entangled-in-physics-feat/

In the case of Walmsley's study, photons were showing up in two spots at
the same time and causing vibrations within a pair of diamonds. The
researchers made it happen by placing two diamonds about 15 centimeters
(about 6 inches) apart on a table and then shooting a series of photons at
a device called a beam splitter. Most of them went toward one diamond or
the other, but a few of the photons went both ways at the same time. When
those multitasking photons struck the pair of diamonds, they caused
vibrations called phonons with each of the crystals.

The light from each of the beams recombines after exiting the crystals. And
sometimes when the light is leaving the crystals, it has less energy than
when it entered. That's how the researchers could tell that the photon had
caused some vibrations.

"We know that one diamond is vibrating, but we don't know which one,"
Walmsley said. "In fact, the universe doesn't know which diamond is
vibrating – the diamonds are entangled, with one vibration shared between
them, even though they are separated in space."



Cheers:   Axil

On Fri, Jan 25, 2013 at 6:10 PM, Edmund Storms wrote:

>
> On Jan 25, 2013, at 3:49 PM,  <
> torulf.gr...@bredband.net> wrote:
>
>  Excuse my grammar. English is not my native language.
>>
>>
> I will try to answer your questions as simply as possible.
>
>
>> Can energy and momentum be transferred from the new He4 to another
>> nucleus at some distains?
>>
>>
> No
>
>  Energy can be transferred from one molecule to another threw a quantum
>> mechanical mechanism.
>>
>>  Yes, at chemical levels of energy
>
>  This occurs in photo synthesis there excitations can jump between
>> electrons in different molecules.
>>
>>
> Yes
>
>  From an older tread.
>>
>>  
>> http://www.mail-archive.com/**vortex-l@eskimo.com/msg75294.**html
>>
>> Maybe a similar phenomenon can occur between nucleus?  This means the
>> excitation from a He4 and momentum can be transferred
>>
>>
> The amount energy generated by a nuclear reaction requires direct emission
> of a particle, which can include a photon. This is observed fact. The
> magnitude is too great to use mechanisms available in a chemical structure.
>  That is why most nuclear reactions are almost totally independent of the
> chemical environment.
>
>  to one or more receiver nucleus. These receiver nucleus must be a special
>> nuclide suitable for  receive the energy and have a mechanism to
>>
>> get rid of it. If several nucleus can get energy from one He4 it may
>> radiate it as UV. If this not is possible I suggest that the receiver
>> nucleus is a C12
>>
>> how decay to 3 He4 as an reversed triple alpha.
>>
>> In absence of receiver nucleus there will be no reactions. But this did
>> not explain the overcome of the coulomb barrier
>>
>> and why its not works in absence of receiver nucleus.
>>
>>
>> I have heard that the conservation of momentum in LENR is commonly
>> explained to "something"
>>
>> how would be like the Mössbauer effect. But I understand this not so
>> easily to explain more exactly.
>>
>>
> The Mossbauer effect involves a very small energy change. It works only
> because the target nucleus is very sensitive to the energy of the
> bombarding gamma. Therefore, the slight effect produced by the chemical
> lattice become visible. This effect is too small to influence energy being
> emitted by a fusion reaction in any meaningful way.
>
> Ed
>
>>
>> TG
>>
>>
>>
>

RE: [Vo]:Chemonuclear Transitions

[Jones Beene]

Oops - let me correct a  major typo. 

 

The proton-proton chain reaction on the sun is mostly “reversible fusion”.  P+P 
<-> H2

 

This, of course, should be: P+P < - > 2He (the helium-2 nucleus, which is 
unstable).

 

It has been posted here many times that the strong force is overwhelming at 
close range - and will bring too protons in a cavity together, despite Pauli 
(Pauli exclusion principle). 

 

But almost always Pauli prevails and the He2 nucleus which forms, immediately 
breaks up into the same two protons as if it was an elastic collision.  Thus, 
99.99+ % of all the fusion reactions, on all the stars in the Universe, can be 
said to be reversible, and do not produce much energy. 

 

The bigger question for NiH on earth is this: does reversible proton fusion 
produce any net energy at all? The currently favored model based on remote 
solar fusion from protium says NO, but there is really little way to be sure – 
except via P+P experiments on earth.

 

RE: [Vo]:Chemonuclear Transitions

[Jack Harbach-O'Sullivan]

Torulf:
 
*The first modern accomplishment of a relative high-power(but easily 
containable) Chemonuclear Transition
cascading reaction was accomplished by a kid named David Adair who designed and 
fabricated what
he called a 'Controlled Fusion Rocket-Reactor.'
 
No kidding; exactly what kid-Adair claimed happened and it eclipsed 
expectations when tested at
White Sand NM and handily pin-point landed at Groom Lake Nevada runway.  He had 
been mentored
by Werner von Braun.  Adair's been designing for NASA ever since.
 
He simply used quasi-conventional chemical fuel reactions but 'bottled' within 
a powerful Electro-Magnetic Sleeved
Reactor Chamber. . .  they build and use large versions of this a MIT often and 
it's all tantamount to 'cutting'
a 'chunk' from the Magneto-Track of Fermi or Hadron Collider in miniature.  
Using H2-O2 as fuel has let
us accomplish much of the 'not as conventional as it looked'  NASA rocketry 
success.  But they also got a yeild
of notable transitory He/Helium. . . we getting this yet?
 
Per my description:  any reaction~uh like 'cold fusion'~would reach easily the 
energy thresh-holds to accomplish
the cascading energy exchange at the atomic-molecular level that we're looking 
for with the 'high-yield' of the
transitional nuclear bonding energy that we're looking for.  The atomic 
'proton-micro-singularity' eyes are
DIALATED within this type of Adair-Bottle which ingress and then yield the 
desired 'higher quasi-unstable'
energy levels that even at the level of prozaic chemical reactions can achieve 
what Dave Adair called 'Controlled
Fusion' controlled reaction 'profound' energy output levels. . . these actually 
have been accomplish functional
overunity for years but nobody bothered to use scrutinized exactly what the 
Mag-Bottle Reactors were actually
accomplishing so did not subsequently exploit it. . .
 
But this is the NATURAL marriage of technologies that 'Cold Fusion' has been 
waiting for that has been staring
us all in the face. . . Cheers Dudes~:) Jack

 



Date: Fri, 25 Jan 2013 23:49:32 +0100
From: torulf.gr...@bredband.net
To: vortex-l@eskimo.com
CC: stor...@ix.netcom.com
Subject: Re: [Vo]:Chemonuclear Transitions


Excuse my grammar. English is not my native language.
 
Can energy and momentum be transferred from the new He4 to another nucleus at 
some distains?
Energy can be transferred from one molecule to another threw a quantum 
mechanical mechanism.
This occurs in photo synthesis there excitations can jump between electrons in 
different molecules.
>From an older tread.
 http://www.mail-archive.com/vortex-l@eskimo.com/msg75294.html
Maybe a similar phenomenon can occur between nucleus?  This means the 
excitation from a He4 and momentum can be transferred
to one or more receiver nucleus. These receiver nucleus must be a special 
nuclide suitable for  receive the energy and have a mechanism to
get rid of it. If several nucleus can get energy from one He4 it may radiate it 
as UV. If this not is possible I suggest that the receiver nucleus is a C12
how decay to 3 He4 as an reversed triple alpha.
In absence of receiver nucleus there will be no reactions. But this did not 
explain the overcome of the coulomb barrier
and why its not works in absence of receiver nucleus.
 
I have heard that the conservation of momentum in LENR is commonly explained to 
"something"
how would be like the Mössbauer effect. But I understand this not so easily to 
explain more exactly.
 
TG
  

Re: [Vo]:Lattice Energy posting on recent Li-battery failures

[Rob Dingemans]

Hi,

On 24-1-2013 22:58, Jed Rothwell wrote:

I guess it was Lattice Energy who wrote:

The LENR theory should be easily testable by autopsies on some failed
batteries, looking for evidence of transmutations, i.e., unusual
isotopes or elements.


This would not be an "easy" test. There would be only microscopic 
amounts of anomalous elements, and a burned battery is about as 
contaminated and filthy as anything can be.


I could be wrong, but is this not fairly easy to be determined by 
comparing the contents of several "clean" and "burned" batteries with a 
Mass Spectrometer?


Kind regards,

Rob

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

On Jan 25, 2013, at 3:49 PM,  > wrote:



Excuse my grammar. English is not my native language.



I will try to answer your questions as simply as possible.


Can energy and momentum be transferred from the new He4 to another  
nucleus at some distains?




No
Energy can be transferred from one molecule to another threw a  
quantum mechanical mechanism.



Yes, at chemical levels of energy
This occurs in photo synthesis there excitations can jump between  
electrons in different molecules.




Yes

From an older tread.

 http://www.mail-archive.com/vortex-l@eskimo.com/msg75294.html

Maybe a similar phenomenon can occur between nucleus?  This means  
the excitation from a He4 and momentum can be transferred




The amount energy generated by a nuclear reaction requires direct  
emission of a particle, which can include a photon. This is observed  
fact. The magnitude is too great to use mechanisms available in a  
chemical structure.  That is why most nuclear reactions are almost  
totally independent of the chemical environment.
to one or more receiver nucleus. These receiver nucleus must be a  
special nuclide suitable for  receive the energy and have a  
mechanism to


get rid of it. If several nucleus can get energy from one He4 it may  
radiate it as UV. If this not is possible I suggest that the  
receiver nucleus is a C12


how decay to 3 He4 as an reversed triple alpha.

In absence of receiver nucleus there will be no reactions. But this  
did not explain the overcome of the coulomb barrier


and why its not works in absence of receiver nucleus.


I have heard that the conservation of momentum in LENR is commonly  
explained to "something"


how would be like the Mössbauer effect. But I understand this not so  
easily to explain more exactly.




The Mossbauer effect involves a very small energy change. It works  
only because the target nucleus is very sensitive to the energy of the  
bombarding gamma. Therefore, the slight effect produced by the  
chemical lattice become visible. This effect is too small to influence  
energy being emitted by a fusion reaction in any meaningful way.


Ed


TG

Re: [Vo]:Chemonuclear Transitions

[torulf.greek]

Excuse my grammar. English is not my native language. 

Can energy
and momentum be transferred from the new He4 to another nucleus at some
distains? 

Energy can be transferred from one molecule to another threw
a quantum mechanical mechanism. 

This occurs in photo synthesis there
excitations can jump between electrons in different molecules. 

>From an
older tread. 


http://www.mail-archive.com/vortex-l@eskimo.com/msg75294.html 

Maybe a
similar phenomenon can occur between nucleus? This means the excitation
from a He4 and momentum can be transferred 

to one or more receiver
nucleus. These receiver nucleus must be a special nuclide suitable for
receive the energy and have a mechanism to 

get rid of it. If several
nucleus can get energy from one He4 it may radiate it as UV. If this not
is possible I suggest that the receiver nucleus is a C12 

how decay to
3 He4 as an reversed triple alpha. 

In absence of receiver nucleus
there will be no reactions. But this did not explain the overcome of the
coulomb barrier 

and why its not works in absence of receiver nucleus.


I have heard that the conservation of momentum in LENR is commonly
explained to "something" 

how would be like the Mössbauer effect. But I
understand this not so easily to explain more exactly. 

TG 

 

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

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, t

Re: [Vo]:new experiment (nitinol)

[Jack Cole]

That is what I suspect as a strong possibility.  I'm hoping a little
thicker wire will hold up better.  It is interesting to watch the nitinol
wire in electrolysis as the hydrogen production seems pretty vigorous down
the entire length of the wire.


On Fri, Jan 25, 2013 at 3:42 PM, Alan Fletcher  wrote:

> > From: "Jack Cole" 
> > Sent: Thursday, January 24, 2013 3:46:39 AM
>
> > Thanks Chuck,
> >
> >
> > It's encouraging to know we've had the same ideas! You may not have
> > had the polarity wrong. I've gone through two wires with it so far.
> > I've thought maybe I was putting too much power through it, but it
> > also may be that the hydrogen loading is very rough on the wire. After
> > running ~5 hrs the wire broke and you could touch it, and it would
> > disintegrate. I may need to try thicker wire (using .009" currently).
>
> Could you be seeing "hydrogen embrittlement" --- I'd just finished reading
> this article in one of my trade mags :
>
> Danger: When Hydrogen Embrittlement Strikes
> http://www.designnews.com/author.asp?section_id=1365&doc_id=248433
>
>

Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

Gold can come in many colors. Since ancient times, glass artists and
alchemists alike have known how to grind the metal into fine particles that
would take on hues such as red or mauve. Carbon nanotubes are the same,
different sizes shade water in different colors.

At scales even smaller, clusters of just a few dozen atoms display even
more outlandish behavior. Gold and other transition metals when combined
with certain other atoms often tend to aggregate in specific numbers and
highly symmetrical geometries, and sometimes these clusters can mimic the
chemistry of single atoms of a completely different element. They become,
as some researchers say, superatoms.


Recently researchers have reported successes in creating new superatoms and
deciphering their structures. In certain conditions, even familiar
molecules such as buckyballs--the soccer-ball-shaped cages made of 60
carbon atoms—can unexpectedly turn into superatoms.

Superatoms can be formed from large aggregation of atoms numbering in the
10s of thousands. The is no limit to the shells that can form.

Today at the cutting edge of science, researchers are already studying how
superatoms bind to each other and to organic molecules. Tracking superatoms
can help researchers learn how biological molecules move inside cells and
tissues, or determine the structure of those molecules precisely using
electron microscopes.

And by assembling superatoms of elements such as gold, carbon, aluminum,
titanium and tungsten researchers may soon be able to create entirely new
materials. Such materials could store hydrogen fuel in solid form at room
temperature, make more powerful rocket fuels or lead to computer chips with
molecule-sized features.

"Designer" materials made of superatoms could have combinations of physical
properties that don't exist in nature. As Kit Bowen, a chemical physicist
at Johns Hopkins University in Baltimore, puts it, it's as if you felt like
eating something hot and something cold at the same time, and could have it
both ways. "Like a hot-fudge sundae."

Small numbers of atoms often form structures as symmetrical, and almost as
intricate, as those of snowflakes. But while no two snowflakes, even if
they have the same number of water molecules, are identical, a small,
specific number of atoms of the same element typically will assemble into
the same, specific shape. The quintessential example is how 60 carbon atoms
form buckyballs.

The strange behavior of atoms in small groupings has been known for a long
time, though only recently have scientists begun to understand it in
detail.

The whole idea is that small is different, The physical properties of a
material, such as hardness and color, are the same for a l-pound lump of
the stuff as they are for a 100-ton chunk. But when you get to specks made
of a few million atoms or less, properties usually begin to change.


A job for superatoms

For larger clusters, it's not always clear when atoms will aggregate into
regular structures or into shapeless blobs with any number of atoms.

For example, in clusters of gold atoms each cluster member donates an
electron to the cluster, just as inside larger chunks of metal, where
mobile electrons can conduct electricity. Forty-four of those electrons get
immobilized in bonds between gold atoms, leaving 58 electrons free to roam.
These 58 electrons then orbit the cluster's core--made of positive gold
ions--just as they would orbit the nucleus of a stand-alone atom. And 58
happens to be a "magic number." It's the number of electrons needed to fill
a shell around the superatom, so that it won't feel a desire to add or shed
electrons, which would destabilize its structure.

This process is similar to what happens in noble gases, which are
chemically inert because they have just the right number of electrons to
fill a shell around the atom.


By tweaking the conditions in  lab vials, researchers can obtain clusters
of different numbers of gold atoms  although they haven't determined the
precise structure in those cases yet.

It is these cluster orbiting electrons that are important to the LENR
process.


Spreading jellium

The story of the superatom begins when two physicists walk into a barber
shop. Marvin Cohen of the University of California, Berkeley recalls how he
and a colleague, the late Walter Knight, ran into each other at their
favorite barber's one afternoon in 1984.

While waiting for his haircut, Knight talked about some surprising data
from an experiment in which he had baked a block of sodium and then
measured the masses, and thus the sizes, of vaporized particles that came
out.

Knight's particles came in a range of sizes. But those made of eight, 20,
40, 58 (remember 58?) or 92 atoms were a lot more abundant. Cohen guessed
what might be happening, and he started scribbling some
back-of-the-envelope calculations. "Tony, the barber, thought we were
figuring out a way to beat the stock market," Cohen recalls.

Sodium is a metal, with a propen

Re: [Vo]:new experiment (nitinol)

[Alan Fletcher]

> From: "Jack Cole" 
> Sent: Thursday, January 24, 2013 3:46:39 AM

> Thanks Chuck,
> 
> 
> It's encouraging to know we've had the same ideas! You may not have
> had the polarity wrong. I've gone through two wires with it so far.
> I've thought maybe I was putting too much power through it, but it
> also may be that the hydrogen loading is very rough on the wire. After
> running ~5 hrs the wire broke and you could touch it, and it would
> disintegrate. I may need to try thicker wire (using .009" currently).

Could you be seeing "hydrogen embrittlement" --- I'd just finished reading this 
article in one of my trade mags :

Danger: When Hydrogen Embrittlement Strikes
http://www.designnews.com/author.asp?section_id=1365&doc_id=248433

Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

*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 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%

RE: EXTERNAL: Re: [Vo]:Chemonuclear Transitions

[Roarty, Francis X]

"Perhaps quantum mechanics is the process that arranges the loan." - Well said, 
and perhaps cavity QED allows these thousands of electrons to participate as a 
virtual body in the reaction.
Fran

From: David Roberson [mailto:dlrober...@aol.com]
Sent: Friday, January 25, 2013 3:40 PM
To: vortex-l@eskimo.com
Subject: EXTERNAL: Re: [Vo]:Chemonuclear Transitions

2000 electrons?  I expect that this many would do the trick.  If one can help a 
bit, then 2000 would help a lot more.  The end result I suspect is that the 
Coulomb energy must be absorbed from this group by some means if only for a 
brief period.  The fusion event would repay the loan with interest.  Perhaps 
quantum mechanics is the process that arranges the loan.

Dave

-Original Message-
From: Axil Axil mailto:janap...@gmail.com>>
To: vortex-l mailto:vortex-l@eskimo.com>>
Sent: Fri, Jan 25, 2013 3:31 pm
Subject: Re: [Vo]:Chemonuclear Transitions
Quantum mechanics lives in the realm of the wave. The electron will exert it 
influence on the positive charge nucleus in bits and pieces.

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.

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.



On Fri, Jan 25, 2013 at 3:01 PM, David Roberson 
mailto:dlrober...@aol.com>> 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 mailto:janap...@gmail.com>>
To: vortex-l mailto:vortex-l@eskimo.com>>
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 h

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

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   
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 
To: vortex-l 
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 s

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

2000 electrons?  I expect that this many would do the trick.  If one can help a 
bit, then 2000 would help a lot more.  The end result I suspect is that the 
Coulomb energy must be absorbed from this group by some means if only for a 
brief period.  The fusion event would repay the loan with interest.  Perhaps 
quantum mechanics is the process that arranges the loan.


Dave



-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Fri, Jan 25, 2013 3:31 pm
Subject: Re: [Vo]:Chemonuclear Transitions


Quantum mechanics lives in the realm of the wave. Theelectron will exert it 
influence on the positive charge nucleus in bits and pieces.
 
Take a look at this to give your imagination a brake:
 
http://en.wikipedia.org/wiki/Thomas%E2%80%93Fermi_screening
 
The Thomas-Fermiformula is a more general potential than the Coulomb'slaw.
 
For the nonlinearThomas-Fermi formula, solving these simultaneously can be 
difficult, and usuallythere is no analytical solution. However, the linearized 
formula has a simplesolution:
 
 R= (Q/r)((e)exp(-kr))
 
With k=0(no screening), this becomes the familiar Coulomb'slaw.
 
The infuence of about 2000electrons near the site of fusion will lower the 
coulomb barrier.
 
 



On Fri, Jan 25, 2013 at 3:01 PM, David Roberson  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 
To: vortex-l 
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 

Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

Quantum mechanics lives in the realm of the wave. The electron will exert
it influence on the positive charge nucleus in bits and pieces.



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
.



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.




On Fri, Jan 25, 2013 at 3:01 PM, David Roberson  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 
> To: vortex-l 
> 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 wrote:
>
>> Sometimes the emails do get crossed up with the number of responses.  In
>> this particular case I think that my inpu

Re: [Vo]:Stremmenos has answered

[Terry Blanton]

On Fri, Jan 25, 2013 at 2:59 PM, Peter Gluck  wrote:
> My dear Friends.
>
> Rules have to be respected, including those
> of the duels. I have published:
> http://egooutpeters.blogspot.ro/2013/01/answer-from-professor-stremmenos.html

Most countries choose ambassadors who are slow to emote.  He who
emotes first loses.

Re: [Vo]:Chemonuclear Transitions

[mixent]

In reply to  MarkI-ZeroPoint's message of Wed, 23 Jan 2013 13:07:46 -0800:
Hi,
[snip]
>systems, the rate enhancement of 2x10e44 is expected via coherent collapse

This is properly written 2E44. The "E" implies 10^.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

[Vo]:Sankaranarayanan tritium paper uploaded

[Jed Rothwell]

Sankaranarayanan, T.K., et al., *Investigation of low-level tritium
generation in Ni-H2O electrolytic cells.* Fusion Technol., 1996. *30*: p.
349.

http://lenr-canr.org/acrobat/Sankaranarinvestigatb.pdf


Some earlier papers:


Srinivasan, M., et al. *Tritium and Excess Heat Generation During
Electrolysis of Aqueous Solutions of Alkali Salts With Nickel
Cathode*. in *Third
International Conference on Cold Fusion*, "Frontiers of Cold Fusion". 1992.
Nagoya Japan: Universal Academy Press, Inc., Tokyo, Japan.

http://lenr-canr.org/acrobat/IkegamiHthirdinter.pdf (p. 123)

Srinivasan later retracted the excess heat results reported in ICCF3. He
spent 6 months at SRI trying to replicate the heat. He finally concluded
that it was caused by recombination (a shuttle reaction, to be exact). I do
not have a paper describing that, but he did retract.


Sankaranarayanan, T.K., et al. *Evidence for Tritium Generation in
Self-Heated Nickel Wires Subjected to Hydrogen Gas Absorption/Desorption
Cycles*. in *5th International Conference on Cold Fusion*. 1995.
Monte-Carlo, Monaco: IMRA Europe, Sophia Antipolis Cedex, France.

http://lenr-canr.org/acrobat/Sankaranarevidencefo.pdf

- Jed

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

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 
To: vortex-l 
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  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

Re: [Vo]:Chemonuclear Transitions

[Axil Axil]

*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  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 
> To: vortex-l 
> Cc: Edmund Storms 
> 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

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

Thanks Dave, I welcome the opportunity. Please forgive my brief style  
and frequent typos. This results from slow typing skill and an  
assumption that much of what I might say is already known by the  
reader, requiring only a hint to reach the answer.  Also, I do not  
encourage discussion about detail or arguments about basic ideas.  We  
all know that a lot is missing in our understanding of Nature, but I  
do not have the time to address any of these interesting issues except  
LENR.  My policy is fight only one war at a time.:-)


Ed


On Jan 25, 2013, at 11:38 AM, David Roberson wrote:

Thanks Ed, I think we are pretty much in agreement at this time.  I  
tend to view processes from the "other side" which sometimes can  
simplify understanding of complex events and that is why I  
commented.  Perhaps I got a bit too riled at the suggestion that my  
post was a total waste of time!


I greatly honor your contributions to and knowledge of this  
important field and I look forward to receiving additional guidance  
from your inputs to vortex.  We all appreciate the opportunity to  
converse with you when you join us.


Dave


-Original Message-
From: Edmund Storms 
To: vortex-l 
Cc: Edmund Storms 
Sent: Fri, Jan 25, 2013 1:27 pm
Subject: Re: [Vo]:Chemonuclear Transitions


On Jan 25, 2013, at 11:17 AM, David Roberson 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.


I agree

 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.


Yes, Dave that is the basic conclusion that results from the law of  
conservation of momentum. Thanks for making this clearer.


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.


Your point was not a waste. However, everyone should read every  
message before replying.


Ed


 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 
To: vortex-l 
Cc: Edmund Storms 
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 consi

RE: EXTERNAL: RE: [Vo]:Chemonuclear Transitions

[Roarty, Francis X]

Just a small caveat regarding your statement “relative motion between the 
devices”… inside this environment you can have equivalent accelerations 
/gravitational changes to these devices that don’t obey the square law where 
tiny spatial displacements can result in huge changes in inertial frames due to 
suppression by geometry changes. This is why I see the quantum geometry as a 
contributing party to the multibodies under discussion.
Fran
_
From: Jones Beene [mailto:jone...@pacbell.net]
Sent: Friday, January 25, 2013 12:43 PM
To: vortex-l@eskimo.com
Subject: EXTERNAL: RE: [Vo]:Chemonuclear Transitions



  From: David Roberson

I find the P+P <-> H2 fusion reaction to be an interesting concept to speculate 
upon… Unless energy of an adequate quantity is released by some mechanism at 
the precise time of the collision, the kinetic energy of the relative motion 
between the devices is restored and they fly apart.


Correct. That is the problem in a nutshell. In fact, the kinetic energy is 
largely restored! It is spin energy of bosons in the proton which is slightly 
depleted. The effect from that, on kinetic energy, is negligible.

It is very difficult, at this point in a discussion, to introduce “QCD color 
change”, but it is the mechanism which must be involved in reversible strong 
force reactions - for there to be a small amount of gain (derived from the 
transitory 2He nucleus, as it flies apart without diminished kinetic energy). 
QCD is about as popular a topic, even among non-specialist scientists - as 
modern poetry, aka rap.

In the end - it’s hard enough to convince observers that proton mass varies 
between atoms in any population - instead is an “average mass” which is not 
quantized. But there are hundreds of precise measurement over time (and 
especially in other countries) where mass value does not correspond to the 
currently accepted value in the USA. Close but not the same. Efforts to 
quantize the proton like this one:

http://arxiv.org/abs/physics/0512108

are hopeless, and actually make a strong case for the opposite conclusion.

Jones

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

Thanks Ed, I think we are pretty much in agreement at this time.  I tend to 
view processes from the "other side" which sometimes can simplify understanding 
of complex events and that is why I commented.  Perhaps I got a bit too riled 
at the suggestion that my post was a total waste of time!


I greatly honor your contributions to and knowledge of this important field and 
I look forward to receiving additional guidance from your inputs to vortex.  We 
all appreciate the opportunity to converse with you when you join us.


Dave



-Original Message-
From: Edmund Storms 
To: vortex-l 
Cc: Edmund Storms 
Sent: Fri, Jan 25, 2013 1:27 pm
Subject: Re: [Vo]:Chemonuclear Transitions




On Jan 25, 2013, at 11:17 AM, David Roberson 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. 


I agree


 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.



Yes, Dave that is the basic conclusion that results from the law of 
conservation of momentum. Thanks for making this clearer. 

 

 
 
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. 



Your point was not a waste. However, everyone should read every message before 
replying.


Ed




 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 
 To: vortex-l 
 Cc: Edmund Storms 
 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 r

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

On Jan 25, 2013, at 11:17 AM, David Roberson 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.


I agree

 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.


Yes, Dave that is the basic conclusion that results from the law of  
conservation of momentum. Thanks for making this clearer.


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.


Your point was not a waste. However, everyone should read every  
message before replying.


Ed


 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 
To: vortex-l 
Cc: Edmund Storms 
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 
To: vortex-l 
Cc: Edmund Storms 
Sent: Fri, Jan 25, 2013 10:38 am
Subject: Re: [Vo]:Chemonuclear Transitions

The human mind is able to imagine endless 

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

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 
To: vortex-l 
Cc: Edmund Storms 
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 
 To: vortex-l 
 Cc: Edmund Storms 
 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 bo

RE: [Vo]:Chemonuclear Transitions

[Jones Beene]

From: David Roberson 

I find the P+P <-> H2 fusion reaction to be an interesting concept to
speculate upon… Unless energy of an adequate quantity is released by some
mechanism at the precise time of the collision, the kinetic energy of the
relative motion between the devices is restored and they fly apart. 


Correct. That is the problem in a nutshell. In fact, the kinetic energy is
largely restored! It is spin energy of bosons in the proton which is
slightly depleted. The effect from that, on kinetic energy, is negligible.

It is very difficult, at this point in a discussion, to introduce “QCD color
change”, but it is the mechanism which must be involved in reversible strong
force reactions - for there to be a small amount of gain (derived from the
transitory 2He nucleus, as it flies apart without diminished kinetic
energy). QCD is about as popular a topic, even among non-specialist
scientists - as modern poetry, aka rap.

In the end - it’s hard enough to convince observers that proton mass varies
between atoms in any population - instead is an “average mass” which is not
quantized. But there are hundreds of precise measurement over time (and
especially in other countries) where mass value does not correspond to the
currently accepted value in the USA. Close but not the same. Efforts to
quantize the proton like this one:

http://arxiv.org/abs/physics/0512108

are hopeless, and actually make a strong case for the opposite conclusion.

Jones


<>

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

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 
To: vortex-l 
Cc: Edmund Storms 
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 
Yes, people try to explain LENR using the behavior described in the  
paper.



--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

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.


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 
To: vortex-l 
Cc: Edmund Storms 
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 

 
Yes, people try to explain LENR using the behavior described in the paper.  





-- 
Daniel Rocha - RJ
 danieldi...@gmail.com
 



 

RE: [Vo]:Chemonuclear Transitions

[MarkI-ZeroPoint]

Ed:

"..the resulting single body splits into two bodies. These two bodies go off
in opposite directions."

 

Just how close to 'opposite'?  Exactly 180 degrees opposite?  180 degs +-
sigma? What is sigma in these reactions?

 

-Mark

 

From: Edmund Storms [mailto:stor...@ix.netcom.com] 
Sent: Friday, January 25, 2013 8:29 AM
To: vortex-l@eskimo.com
Cc: Edmund Storms
Subject: Re: [Vo]:Chemonuclear Transitions

 

Yes, they are "forever associated with lattice and geometry defects " but
that is not relevant. You need to understand what happens at the site of the
nuclear reaction. The site of a hot fusion reaction has two d coming
together with enough energy to overcome the Coulomb barrier and cause the
two d to fuse. Then the resulting single body splits into two bodies. These
two bodies go off in opposite directions while carrying the energy and
momentum.  This is conventional behavior.

 

For cold fusion to occur, the 2 d must come together without extra energy,
but nevertheless overcome the Coulomb barrier. How this process can occur is
being debated. Nevertheless, the result is a single He4 with 23.8 MeV of
energy. How does this energy get released and communicated to the world as
heat, which it does, while conserving momentum? That is the ONLY issue.

 

Ed

 

On Jan 25, 2013, at 9:11 AM, Roarty, Francis X wrote:





On Fri Jan 25th Ed Storms said [snip] Cold fusion is a 2-body to 1 body
reaction that violates this condition[/snip]. That might be correct from a
purely syntax perspective but is an unfair oversimplification, LENR and cold
fusion are forever associated with lattice and geometry defects in said
lattice - this is a quantum effect /extreme - multibody in the equivalent
sense where gas atoms react to changes in nano geometry. There is literature
regarding cavity QED that indicates these changes in cavity geometry violate
the square law and break the isotropy.

Regards

Fran

 

From: Edmund Storms [mailto:stor...@ix.netcom.com] 
Sent: Friday, January 25, 2013 10:38 AM
To: vortex-l@eskimo.com
Cc: Edmund Storms
Subject: EXTERNAL: 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 

Yes, people try to explain LENR using the behavior described in the paper.  




 

-- 
Daniel Rocha - RJ

danieldi...@gmail.com

 

 

[Vo]:off topic, More education, More over time, Less money

[fznidarsic]

I went to business school in the 1990's.  We studied ways to keep up with the 
Japanese?


We need more school.  Kids need to start young, have tutoring in the evening, 
never stop till they drop.  Study 
more and all of the time.  Give up your youth to the system.  It is a total 
quality system, every move you make is 
so that you can be employed within the system.  Were did this all lead?  Not 
were we expected.
The men just gave up.


I see the men hanging out in USA Starbucks today in the US.  Downsized, 
outsourced and in their mid to late 50's.  Highly educated men waiting to get 
onto the system at 62.  No, they do not want and cannot afford troubled women.




http://www.slate.com/articles/news_and_politics/foreigners/2009/06/the_herbivores_dilemma.html





Frank Znidarsic

[Vo]:Re: [Vo]:Mainstream scientific research is looking into LENR and doesn’t know it yet.

[Rich Murray]

http://phys.org/news/2013-01-physicists-surprisingly-small-proton-radius.html#nwlt

On Thu, Jan 24, 2013 at 10:47 PM, Finlay MacNab
 wrote:

> I was thinking the same thing while looking at this research.
>
> 
> Date: Fri, 25 Jan 2013 00:29:26 -0500
> From: janap...@gmail.com
> To: vortex-l@eskimo.com
> Subject: [Vo]:Mainstream scientific research is looking into LENR and
> doesn’t know it yet.
>
>
> http://www.sciencedaily.com/releases/2013/01/130124140704.htm
>
> Proton Size Puzzle: Surprisingly Small Proton Radius Confirmed With Laser
> Spectroscopy of Exotic Hydrogen
> Accepted scientific research is looking into LENR and doesn’t know it yet.
> Muons behave a lot like electrons, except for their mass: muons are 200
> times heavier than electrons. The atomic orbit of the muon is therefore much
> closer to the proton than the electron's orbit in a regular hydrogen atom.
> Because the Muon obits closer to the proton, the proton feels more negative
> charge.
> The proton “charge radius” is reduced as a result indicating a reduction in
> the intensity of the positive charge of the proton (hydrogen nucleus).
> This may show that the strength of the negative charge felt by the proton
> diminishes the intensity of its positive charge; a kind of electrical charge
> screening.
> This experiment is similar to an experiment that Mills might run.
> Electron screening produced in Mills chemical concoctions may also reduce
> the power of the positive charge of the nucleus resulting in lowered orbits
> for electrons (aka hydrinos).
>
> Cheers:   Axil

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

Instead, I suggest you consult any physics text about the law of  
conservation of momentum.


Ed
On Jan 25, 2013, at 9:16 AM, Daniel Rocha wrote:


No, we are certainly not. I let this Sisyphean task to Abd.

2013/1/25 Edmund Storms 
Daniel, we are not communicating.

--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[David Roberson]

I find the P+P <-> H2 fusion reaction to be an interesting concept to speculate 
upon.  A simple way that I use to have a possible understanding of why the 
fusion breaks up is to view the collision as basically an elastic collision 
between particles.  Unless energy of an adequate quantity is released by some 
mechanism at the precise time of the collision,  the kinetic energy of the 
relative motion between the devices is restored and they fly apart.


Consider yourself as an observer located at a position exactly between two 
equal energy P's heading for a direct collision.  When they are far apart you 
calculate the kinetic energy of each P to be the same and obtain three possible 
ranges of values.  One calculation reveals that the sum of the two kinetic 
energies is greater than that required to overcome the Coulomb barrier.  A 
second calculation shows that the kinetic energy of the pair before collision 
is exactly equal to the barrier energy, and the third calculation implies that 
there is not enough energy.


In the case where there is not sufficient energy, the two will approach, but 
immediately depart from each other with most the action dominated by the 
Coulomb forces.  When the energy is exactly that required to barely overcome 
the Coulomb barrier, the protons then begin to be be influenced mainly by the 
strong force.  This force is super powerful so the two P's accelerate toward 
each other until they collide.  Since I am assuming an elastic collision with 
no release of energy, the two rebound apart back to the point where the Coulomb 
barrier takes over.  The two P's will be in close contact for the most time 
possible under this set of conditions and have the best opportunity to fuse.  I 
consider them to fuse if a particle or quanta of energy is released that 
results in a reduction of stored energy so that they now do not have adequate 
energy to break free of each other.  The larger the quantity of energy 
released, the more likely the two P's remain close.  If a beta + decay can be 
arranged, that is sufficient to perform the function well.


If the original energy of  the two protons is greater than that required to 
exactly match the Coulomb barrier, then the two will have less time in close 
proximity and it becomes less likely for an adequate release of binding energy 
and for fusion to hold.


I generally assume that radiation is emitted on a continued basis from the 
protons as they decelerate towards each other since they carry a charge.  This 
represents energy being taken out of the pairs kinetic sum that might help 
improve the chance of fusing if emitted just after the Coulomb barrier is 
breached.  Unfortunately, the amount of radiation is small compared to the 
binding energy between two protons and would only have effect for an extremely 
tiny proportion of the collisions.  Furthermore, an excited pair of protons so 
loosely bound would easily fall prey to being disrupted by collisions with 
other protons due to the high temperature.  On the other hand, it might be 
advantageous in some collisions with the other particles.  Additional energy 
could possibly be transferred to these other impactors from the bound pair 
allowing them to become more bound.  Any process that allows the protons to 
remain near each other for a longer period of time would enhance the chance of 
a large energy release that completes the binding.


This hypothesis assumes that fusion would be optimized for an extremely tiny 
range of relative kinetic energies.  If also would suggest that there is a 
minimum temperature below which the likelihood of collisions between protons of 
the correct energies becomes rare and fusion is non productive.  It would 
predict that relatively large energy releases such as beta + decays would be 
the dominate indicator of successful fusion.  I would expect to detect a 
continuous flux of radiation from the acceleration and deceleration of the 
protons as they collide.  Also, energy would be expected to be transferred into 
the proton plasma in the form of heat from loosely bound protons as they bind 
tighter heading toward eventual fusion.   And, when a beta + decay occurs, the 
fusion process is completed between a proton pair and that event is locked into 
place.


This represents my current views toward fusion and do not imply that I consider 
the above hypothesis original as it seems to be obvious behavior.  Perhaps 
someone with more knowledge about the actual ash of proton to proton fusion 
would help me to understand what is proven to occur in real life.


Dave


 



-Original Message-
From: Jones Beene 
To: vortex-l 
Sent: Fri, Jan 25, 2013 10:17 am
Subject: RE: [Vo]:Chemonuclear Transitions



The proton-proton chain reaction on the sun is mostly “reversiblefusion”.  P+P 
<-> H2
 
It has been posted here many times that the strong force is overwhelmingat 
close range - and will bring too protons together , despite Pauli. But 
almostalways th

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

Yes, they are "forever associated with lattice and geometry defects "  
but that is not relevant. You need to understand what happens at the  
site of the nuclear reaction. The site of a hot fusion reaction has  
two d coming together with enough energy to overcome the Coulomb  
barrier and cause the two d to fuse. Then the resulting single body  
splits into two bodies. These two bodies go off in opposite directions  
while carrying the energy and momentum.  This is conventional behavior.


For cold fusion to occur, the 2 d must come together without extra  
energy, but nevertheless overcome the Coulomb barrier. How this  
process can occur is being debated. Nevertheless, the result is a  
single He4 with 23.8 MeV of energy. How does this energy get released  
and communicated to the world as heat, which it does, while conserving  
momentum? That is the ONLY issue.


Ed

On Jan 25, 2013, at 9:11 AM, Roarty, Francis X wrote:

On Fri Jan 25th Ed Storms said [snip] Cold fusion is a 2-body to 1  
body reaction that violates this condition[/snip]. That might be  
correct from a purely syntax perspective but is an unfair  
oversimplification, LENR and cold fusion are forever associated with  
lattice and geometry defects in said lattice – this is a quantum  
effect /extreme - multibody in the equivalent sense where gas atoms  
react to changes in nano geometry. There is literature regarding  
cavity QED that indicates these changes in cavity geometry violate  
the square law and break the isotropy.

Regards
Fran

From: Edmund Storms [mailto:stor...@ix.netcom.com]
Sent: Friday, January 25, 2013 10:38 AM
To: vortex-l@eskimo.com
Cc: Edmund Storms
Subject: EXTERNAL: 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 
Yes, people try to explain LENR using the behavior described in the  
paper.



--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Daniel Rocha]

No, we are certainly not. I let this Sisyphean task to Abd.

2013/1/25 Edmund Storms 

> Daniel, we are not communicating.
>

-- 
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

Daniel, we are not communicating. Do you understand the law of  
conservation of momentum that applies to all nuclear reactions? That  
is the only thing I'm discussing. When a nuclear reaction occurs, the  
energy must be communicated to the rest of the world and momentum must  
be conserved in the process.  Quarks and gluons have no role in this  
requirement. These are particles within the nucleus and are not  
emitted as separate entities.


Ed


On Jan 25, 2013, at 8:49 AM, Daniel Rocha wrote:



The number of elements is not an issue. You can just have increase  
the precision by considering an arbitrarily high quantity of  
particles, like quarks and gluons and whatever particle of the SM  
you want. So, there is no rule restricting the number of bodies  
taking part in the problem.


2013/1/25 Edmund Storms 
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.



--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Roarty, Francis X]

On Fri Jan 25th Ed Storms said [snip] Cold fusion is a 2-body to 1 body 
reaction that violates this condition[/snip]. That might be correct from a 
purely syntax perspective but is an unfair oversimplification, LENR and cold 
fusion are forever associated with lattice and geometry defects in said lattice 
- this is a quantum effect /extreme - multibody in the equivalent sense where 
gas atoms react to changes in nano geometry. There is literature regarding 
cavity QED that indicates these changes in cavity geometry violate the square 
law and break the isotropy.
Regards
Fran

From: Edmund Storms [mailto:stor...@ix.netcom.com]
Sent: Friday, January 25, 2013 10:38 AM
To: vortex-l@eskimo.com
Cc: Edmund Storms
Subject: EXTERNAL: 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 mailto:stor...@ix.netcom.com>>
Yes, people try to explain LENR using the behavior described in the paper.


--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Daniel Rocha]

The number of elements is not an issue. You can just have increase the
precision by considering an arbitrarily high quantity of particles, like
quarks and gluons and whatever particle of the SM you want. So, there is no
rule restricting the number of bodies taking part in the problem.

2013/1/25 Edmund Storms 

> 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.
>


-- 
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

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 
Yes, people try to explain LENR using the behavior described in the  
paper.



--
Daniel Rocha - RJ
danieldi...@gmail.com

Re: [Vo]:Chemonuclear Transitions

[Daniel Rocha]

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 

> Yes, people try to explain LENR using the behavior described in the paper.
>
>


-- 
Daniel Rocha - RJ
danieldi...@gmail.com

RE: [Vo]:Chemonuclear Transitions

[Jones Beene]

The proton-proton chain reaction on the sun is mostly “reversible fusion”.  P+P 
<-> H2

 

It has been posted here many times that the strong force is overwhelming at 
close range - and will bring too protons together , despite Pauli. But almost 
always the He2 nucleus which forms then immediately breaks up.  Thus, 99.99+ % 
of all fusion reactions, on all stars in the Universe, can be said to be 
reversible, and do not produce much energy. The bigger question for NiH is 
this: does reversible proton fusion produce any net energy? The currently 
favored model for solar fusion says NO.

 

He2 does form from the interaction however, and it disappear rapidly - but ever 
so often there is a beta decay. Only one reversible reaction in 10^20 proceeds 
to beta decay. Thus the solar model is not compatible with Ni-H. 

 

Ed Storms clearly states that he is suggesting a novel form of this reaction - 
mediated by another particle such as an electron, deflated electron or so on. 
He is aware of the rarity of the beta decay.

 

There is another hypothesis, or model, which I’ve been airing for about 6 
months. It can operate along side of other models or alone. It suggests that 
proton reversible fusion does produce a small amount of heat due to QCD “color 
change”. The mass of the proton is slightly reduced in the process. That solves 
many theoretical problems, but admittedly there is no proof (unless NiH is the 
proof).

 

The proton - in this model is not quantized. Its “known mass” is an average 
mass, and can vary slightly up or down from average. In addition to shedding 
small amounts of energy via QCD, depleted protons can also capture small 
amounts of mass-energy via free electrons on the sun, under gravity 
compression. This energy transfer in either case comes from QM - spin transfer 
via magnons. 

 

The mediating quasi-particle for this process is the magnon. That is important 
for NiH. If nickel were not ferromagnetic, there would probably be no energy 
transfer from reversible fusion.

 

Before you ask – yes palladium is ferromagnetic in alloy form, and as a hydride:

 

http://cpb.iphy.ac.cn/EN/abstract/abstract25888.shtml 

 

 

From: Eric Walker 

 

Chuck Sites wrote:

 

The proton-proton chain reaction is initiated with a strong interaction between 
two protons,  that binds to form a diproton, the diproton then decays via weak 
interaction (a W boson) into a deuteron + electron + electron neutrino  and 
0.42 MeV of energy.  

Wikipedia has a very good description of this processes:

 

The proton-proton chain does seem promising at first, especially when one takes 
into account some of the difficulties with the kind of activation that would 
occur if there were a lot of neutron-moderated reactions.  But the 
proton-proton chain has its own difficulties.  See [1], below, for an earlier 
discussion.

 

Briefly, the diproton lasts for a vanishingly small amount of time before it 
breaks up.  Only a very small fraction of diprotons go on to form deuterium; in 
the sun, this process is a limiting one that prevents it from rapidly burning 
through its fuel.  In known cases, the rate of deuterium formation is small 
because the weak force requires that a very high energy barrier be surpassed 
before a proton will convert to a neutron. Widom and Larsen have other ideas on 
this particular point, and it is part of what makes their writings difficult 
for physicist types (of which I am not one) to get a handle on.  See also the 
comments to this physics.SE question for more details [2].  I believe Ed Storms 
proposes an alternate form of weak-force moderated nuclear reaction, along the 
lines of a slow p-e-p reaction, and I would assume that similar difficulties 
must be addressed in this instance as well.

 

Assuming the weak interaction really does provide a limiting barrier, any 
fusion-like reaction is presumably going to have to occur either through the 
action of deuterium or higher, on one hand, or through proton capture within a 
larger nucleus, on the other, unless a non-fusion reaction along the lines of 
what Jones or Mills describes is going on.  Obviously there is also the matter 
of the Coulomb barrier, but I think we've gotten used to ignoring it for the 
sake of convenience. ;)

 

Eric

 

 

[1] http://www.mail-archive.com/vortex-l@eskimo.com/msg67691.html

[2] 
http://physics.stackexchange.com/questions/23640/what-interactions-would-take-place-between-a-free-proton-and-a-dipolariton

 

 

Re: [Vo]:Chemonuclear Transitions

[Edmund Storms]

Yes, people try to explain LENR using the behavior described in the  
paper.  However, the behavior being caused is not cold fusion but hot  
fusion.  Hot fusion is an entirely different reaction with different  
nuclear products and a different mechanism.  Hot fusion REQUIRES high  
energy, cold fusion does not. Hot fusion produces neutrons, cold  
fusion does not. Hot fusion generates energetic particles, cold fusion  
does not. Hot fusion produces energy by two reactions d+d=n+He3 and d 
+d=t+p, cold fusion makes energy by d+d=He4 + ?.  It is a waste of  
time trying to explain cold fusion using hot fusion as an example.


This approach is as pointless as trying to explain cold fusion using  
the conditions in the Sun. Nuclear reactions, like all kinds of  
reactions, are influenced by the conditions. Change the condition and  
the reaction mechanism and products change. An explanation that  
applies to one condition will not apply to another, so why do this?


Ed Storms
On Jan 24, 2013, at 10:05 PM, Daniel Rocha wrote:

I don't know what you mean by "This study has no relationship to  
cold fusion because the same nuclear products are not formed." See  
page p.14, section 13. He tries to explain Rossi's reactor. See p.  
18, table II. This context shows he's trying to explain CF and  
Rossi's reactor.



2013/1/23 Edmund Storms 
This paper and many others like it describe how HOT fusion is  
enhanced when it occurs in a chemical lattice. This study has no  
relationship to cold fusion because the same nuclear products are  
not formed.  While the lattice enhances the hot fusion rate, it does  
so only at very low energy where the rate is already very small.   
Here are some other studies.


Ed


1.Dignan, T.G., et al., A search for neutrons from  
fusion in a highly deuterated cooled palladium thin film. J. Fusion  
Energy, 1990. 9(4): p. 469.


2.Durocher, J.J.G., et al., A search for evidence of  
cold fusion in the direct implantation of palladium and indium with  
deuterium. Can. J. Phys., 1989. 67: p. 624.


3. Gu, A.G., et al., Experimental study on cold fusion using  
deuterium gas and deuterium ion beam with palladium. J. Fusion  
Energy, 1990. 9(3): p. 329.


4. Gu, A.G., et al., Preliminary experimental study on cold  
fusion using deuterium gas and deuterium plasma in the presence of  
palladium. Fusion Technol., 1989. 16: p. 248.


5.Kosyakhkov, A.A., et al., Neutron yield in the  
deuterium ion implantation into titanium. Fiz. Tverd. Tela, 1990.  
32: p. 3672 (in Russian).


6.Kosyakhkov, A.A., et al., Mass-spectrometric study of  
the products of nuclear reactions occurring by ion-plasma saturation  
of titanium with deuterium. Dokl. Akad. Nauk. [Tekh. Fiz.), 1990.  
312(1): p. 96 (in Russian).


7. Liu, R., et al., Measurement of neutron energy spectra  
from the gas discharge facility. Yuanzi Yu Fenzi Wuli Xuebao, 1994.  
11(2): p. 115 (in Chinese).


8. Myers, S.M., et al., Superstoichiometry, accelerated  
diffusion, and nuclear reactions in deuterium-implanted palladium.  
Phys. Rev. B, 1991. 43: p. 9503.


9. Prelas, M., et al., Cold fusion experiments using  
Maxwellian plasmas and sub-atmospheric deuterium gas. J. Fusion  
Energy, 1990. 9(3): p. 309.


10.Takahashi, A. Results of experimental studies of  
excess heat vs nuclear products correlation and conceivable reaction  
model. in The Seventh International Conference on Cold Fusion. 1998.  
Vancouver, Canada: ENECO, Inc., Salt Lake City, UT. p. 378-382.


11.   Wang, T., et al. Anomalous phenomena in E<18 KeV hydrogen  
ion beam implantation experiments on Pd and Ti. in Sixth  
International Conference on Cold Fusion, Progress in New Hydrogen  
Energy. 1996. Lake Toya, Hokkaido, Japan: New Energy and Industrial  
Technology Development Organization, Tokyo Institute of Technology,  
Tokyo, Japan. p. 401.


12.McKee, J.S.C., et al. Neutron emission from low- 
energy deuteron injection of deuteron-implanted metal foils (Pd, Ti,  
and In). in Anomalous Nuclear Effects in Deuterium/Solid Systems,  
"AIP Conference Proceedings 228". 1990. Brigham Young Univ., Provo,  
UT: American Institute of Physics, New York. p. 275.


13.   Isobe, Y., et al. Search for coherent deuteron fusion by  
beam and electrolysis experiments. in 8th International Conference  
on Cold Fusion. 2000. Lerici (La Spezia), Italy: Italian Physical  
Society, Bologna, Italy. p. 17-22.


14.   Isobe, Y., et al., Search for multibody nuclear reactions  
in metal deuteride induced with ion beam and electrolysis methods.  
Jpn. J. Appl. Phys., 2002. 41(3): p. 1546-1556.


15.Zelenskii, V.F., et al., Experiments on cold nuclear  
fusion in Pd and Ti saturated with deuterium by ion implantation.  
Vopr. At. Nauki Tekh. Ser.: Fiz. Radiats. Povr. Radiats.  
Materialoved., 1990. 52(1): p. 65 (in Russian).


16.Martynov, M.I., 

Re: [Vo]:patent saga, Rossi enters the battle

[Peter Gluck]

an illness cured s...fast!

On Fri, Jan 25, 2013 at 4:23 PM, ChemE Stewart  wrote:

> You r a young dude!
>
>
> On Friday, January 25, 2013, Peter Gluck wrote:
>
>> let's try, how old are you Terry? I'm 75.25 
>> Peter
>>
>> On Fri, Jan 25, 2013 at 3:39 PM, Terry Blanton wrote:
>>
>>> On Fri, Jan 25, 2013 at 8:17 AM, Peter Gluck 
>>> wrote:
>>> > dry Ni/H systems high temperature, (4-800 C) Ni as rod with
>>> nanosurface or
>>> > powder, hydrogen from bottle- sounds different I think.
>>>
>>> May we both live to see the ensuing litigious joust results.
>>>
>>>
>>
>>
>> --
>> Dr. Peter Gluck
>> Cluj, Romania
>> http://egooutpeters.blogspot.com
>>
>


-- 
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com

Re: [Vo]:patent saga, Rossi enters the battle

[ChemE Stewart]

You r a young dude!

On Friday, January 25, 2013, Peter Gluck wrote:

> let's try, how old are you Terry? I'm 75.25 
> Peter
>
> On Fri, Jan 25, 2013 at 3:39 PM, Terry Blanton 
> 
> > wrote:
>
>> On Fri, Jan 25, 2013 at 8:17 AM, Peter Gluck 
>> >
>> wrote:
>> > dry Ni/H systems high temperature, (4-800 C) Ni as rod with nanosurface
>> or
>> > powder, hydrogen from bottle- sounds different I think.
>>
>> May we both live to see the ensuing litigious joust results.
>>
>>
>
>
> --
> Dr. Peter Gluck
> Cluj, Romania
> http://egooutpeters.blogspot.com
>

Re: [Vo]:patent saga, Rossi enters the battle

[Terry Blanton]

On Fri, Jan 25, 2013 at 8:47 AM, Peter Gluck  wrote:
> let's try, how old are you Terry? I'm 75.25 

Mentally, about 19; chronologically, 59.  But no one is promised tomorrow.

Re: [Vo]:patent saga, Rossi enters the battle

[Peter Gluck]

let's try, how old are you Terry? I'm 75.25 
Peter

On Fri, Jan 25, 2013 at 3:39 PM, Terry Blanton  wrote:

> On Fri, Jan 25, 2013 at 8:17 AM, Peter Gluck 
> wrote:
> > dry Ni/H systems high temperature, (4-800 C) Ni as rod with nanosurface
> or
> > powder, hydrogen from bottle- sounds different I think.
>
> May we both live to see the ensuing litigious joust results.
>
>


-- 
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com

Re: [Vo]:patent saga, Rossi enters the battle

[Terry Blanton]

On Fri, Jan 25, 2013 at 8:17 AM, Peter Gluck  wrote:
> dry Ni/H systems high temperature, (4-800 C) Ni as rod with nanosurface or
> powder, hydrogen from bottle- sounds different I think.

May we both live to see the ensuing litigious joust results.

Re: [Vo]:patent saga, Rossi enters the battle

[Peter Gluck]

dry Ni/H systems high temperature, (4-800 C) Ni as rod with nanosurface or
powder, hydrogen from bottle- sounds different I think.
Peter

On Fri, Jan 25, 2013 at 3:11 PM, Terry Blanton  wrote:

> On Fri, Jan 25, 2013 at 3:41 AM, Peter Gluck 
> wrote:
> > and this, dear Terry could be an obstacle for new Ni-H patents?
>
> Nickel wire (tube)
> Potassium carbonate
> Pressurized, heated, nascent hydrogen
> Excess heat
>
> Sounds familiar to me.
>
>


-- 
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com

Re: [Vo]:patent saga, Rossi enters the battle

[Terry Blanton]

On Fri, Jan 25, 2013 at 3:41 AM, Peter Gluck  wrote:
> and this, dear Terry could be an obstacle for new Ni-H patents?

Nickel wire (tube)
Potassium carbonate
Pressurized, heated, nascent hydrogen
Excess heat

Sounds familiar to me.

Re: [Vo]:patent saga, Rossi enters the battle

[Peter Gluck]

and this, dear Terry could be an obstacle for new Ni-H patents?
Peter

On Fri, Jan 25, 2013 at 12:32 AM, Terry Blanton  wrote:

>
>
> On Thu, Jan 24, 2013 at 3:46 PM, Peter Gluck 
> wrote:
> > Randy Mills says hsi process has nothing to do
> > with Rossi's or Piantelli's. And he is not interested
> > in ny communiction with these individuals.
> > The problem is that his CIHT is progressing very
> > slowly.
>
> I'm talking about the March 24, 1994 Thermacore report:
>
> *Anomalous heat was measured from a reaction of atomic hydrogen in
> contact with
> potassium carbonate on a nickel surface. The nickel surface consisted of
> 500 feet of
> 0.0625 inch diameter tubing wrapped in a coil. The coil was inserted into
> a pressure
> vessel containing a light water solution of potassium carbonate. The
> tubing and solution
> were heated to a steady state temperature of 249°C using an FR heater.
> Hydrogen at
> 1100 psig was applied to the inside of the tubing. After the application
> of hydrogen, a
> 32°C increase in temperature of the cell was measured which corresponds to
> 25 watts
> of heat. Heat production under these conditions is predicted by the theory
> of Mills where
> a new species of hydrogen is produced that has a lower energy state then
> normal
> hydrogen. ESCA analyses, done independently by Lehigh University, have
> found the
> predicted 55 eV signature of this new species of hydrogen. Work is
> continuing at
> Thermacore with internal funding to bring this technology to the
> marketplace.*
> *
> *
> http://lenr-canr.org/acrobat/GernertNnascenthyd.pdf
>



-- 
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com