subject:"Re\: \[Vo\]\:Mizuno, Rossi \& copper transmutation"

 states (such as s = 0, l 
= 1). Therefore the quantum state of the deuterium is a superposition (a linear 
combination) of the s = 1, l = 0 state and the s = 1, l = 2 state, even though 
the first component is much bigger. Since the total angular momentum j is also 
a good quantum number (it is a constant in time), both components must have the 
same j, and therefore j = 1. This is the total spin of the deuterium nucleus. 

To summarize, the deuterium nucleus is antisymmetric in terms of isospin, and 
has spin 1 and even (+1) parity. The relative angular momentum of its nucleons 
l is not well defined, and the deuteron is a superposition of mostly l = 0 with 
some l = 2. 

The measured value of the deuterium magnetic dipole moment, is 0.857 μN. This 
suggests that the state of the deuterium is indeed only approximately s = 1, l 
= 0 state, and is actually a linear combination of (mostly) this state with s = 
1, l = 2 state. 

The electric dipole is zero as usual. 

The measured electric quadropole of the deuterium is 0.2859 e·fm2. While the 
order of magnitude is reasonable, since the deuterium radius is of order of 1 
femtometer (see below) and its electric charge is e, the above model does not 
suffice for its computation. More specifically, the electric quadrupole does 
not get a contribution from the l =0 state (which is the dominant one) and does 
get a contribution from a term mixing the l =0 and the l =2 states, because the 
electric quadrupole operator does not commute with angular momentum. The latter 
contribution is dominant in the absence of a pure l = 0 contribution, but 
cannot be calculated without knowing the exact spatial form of the nucleons 
wavefunction inside the deuterium. 
Source:
1. Elements of Nuclear Physics, W. E Burcham, Longman,1979. 
2. http://en.wikipedia.org/wiki/Hydrogen-2 

Note the last paragraph and the conclusion about electric quadrupoles not 
commuting with angular momentum. 

 It suggests that there may be a way to stimulate the D via an electric 
quadrupole input signal.   Also with a magnetic moment the D must respond to a 
magnetic field and fine tuning of an oscillating magnetic field may very well 
excite the D to flip up and down in the field.  The composite particles of the 
D should have sligltly different magnetic moments that can respond and create 
an excited state IMHO on a transient short lived time frame.   However in a 
coherent system such a transient may be enough to cause other transitions of 
similar energy states to occur with mass energy being changed to angular 
momentum energy.

Bob Cook

  
  - Original Message - 
  From: Eric Walker 
  To: vortex-l@eskimo.com 
  Sent: Friday, October 17, 2014 9:54 PM
  Subject: Re: [Vo]:Mizuno, Rossi  copper transmutation


  On Fri, Oct 17, 2014 at 10:41 AM, Bob Cook frobertc...@hotmail.com wrote: 


Do you know if the experiments looked at excited spin energy states that 
may be  possible at higher spin quanta?


  Unfortunately I don't have any other details and don't know of a particular 
experiment to refer to.  Here is the quote from a textbook I recently finished 
reading:


For nuclear physicists, the deuteron should be what the hydrogen atom is 
for atomic physicists.  Just as the measured Balmer series of electromagnetic 
transitions between the excited states of hydrogen led to an understanding of 
the structure of hydrogen, so should the electromagnetic transitions between 
the excited states of the deuteron lead to an understanding of its structure.  
Unfortunately, there are no excited states of the deuteron—it is such a weakly 
bound system that the only excited states are unbound systems consisting of a 
free proton and neutron. [1]


  Eric


  [1] Kenneth S. Krane, Introductory Nuclear Physics, pp. 80-81; author's 
emphasis.

Re: [Vo]:Mizuno, Rossi copper transmutation

Eric--

One additional comment on the D excited state question.

Suppose in stead of one D you realize a pair of D's with their spin vectors 
pointing in opposite directions in a coherent system like one may have in the 
middle of a body centered cubic (BCC) cell of a Pd metal lattice or a Ni metal 
lattice with a strong B magnetic field.  Such a pair may act like a Cooper pair 
with a 0 spin and hence a 4He nucleus.  The transition from 2 D's to 4He  could 
occur  via a distribution of mass energy to excited spin states of the lattice 
electrons and/or metal nuclei.  Proper alignment initially of the the D's may 
be important to obtain antiparallel conditions and could be encouraged 
statistically with varying magnetic fields and /or temperature of the lattice.  
 A quadruple oscillating electric field may also help to excite the D's to shed 
their excess mass relative to the developing 4He particle.  

The magnetic field should actually reduce the spatial options available in the 
BCC cell for the D's that happen to be there and improve the statistics for 
their arriving at the same location to form a Cooper pair.  The spin coupling 
may be a strong tendency in such a situation.  It is with electrons as Pauli 
pointed out.  

Bob 
  - Original Message - 
  From: Eric Walker 
  To: vortex-l@eskimo.com 
  Sent: Friday, October 17, 2014 9:54 PM
  Subject: Re: [Vo]:Mizuno, Rossi  copper transmutation


  On Fri, Oct 17, 2014 at 10:41 AM, Bob Cook frobertc...@hotmail.com wrote: 


Do you know if the experiments looked at excited spin energy states that 
may be  possible at higher spin quanta?


  Unfortunately I don't have any other details and don't know of a particular 
experiment to refer to.  Here is the quote from a textbook I recently finished 
reading:


For nuclear physicists, the deuteron should be what the hydrogen atom is 
for atomic physicists.  Just as the measured Balmer series of electromagnetic 
transitions between the excited states of hydrogen led to an understanding of 
the structure of hydrogen, so should the electromagnetic transitions between 
the excited states of the deuteron lead to an understanding of its structure.  
Unfortunately, there are no excited states of the deuteron—it is such a weakly 
bound system that the only excited states are unbound systems consisting of a 
free proton and neutron. [1]


  Eric


  [1] Kenneth S. Krane, Introductory Nuclear Physics, pp. 80-81; author's 
emphasis.

RE: [Vo]:Mizuno, Rossi copper transmutation

Bob,
I have cherry-picked three major “spin facts” from this compendium which
indicate that if one wants to apply a nano-magnetism or spin-coupling
modality to LENR, it is highly preferable to use deuterium, as opposed to
hydrogen. That may be why Mizuno chose the deuterium-nickel combination. All
eyes will be shifting to Mizuno in less than three weeks.

From: Bob Cook 
[snip] The deuteron, being an isospin singlet, is antisymmetric under
nucleon exchange due to isospin, and therefore must be symmetric under the
double exchange of their spin and location. Therefore it can be in either of
the following two different states: Symmetric spin and symmetric under
parity. In this case, the exchange of the two nucleons will multiply the
deuterium wavefunction by (-1) from isospin exchange, (+1) from spin
exchange and (+1) from parity (location exchange), for a total of (-1) as
needed for antisymmetry…. In this case, the exchange of the two nucleons
will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1)
from spin exchange and (-1) from parity (location exchange), again for a
total of (-1) as needed for antisymmetry. [snip]

…suggesting that there may be a way to stimulate the D via an electric
quadrupole input signal.   Also with a magnetic moment the D must respond to
a magnetic field and fine tuning of an oscillating magnetic field may very
well excite the D to flip up and down in the field.  The composite particles
of the D should have slightly different magnetic moments that can respond
and create an excited state IMHO on a transient short lived time frame.
However in a coherent system such a transient may be enough to cause other
transitions of similar energy states to occur with mass energy being changed
to angular momentum energy.

The quadrupole input is a strong clue.
 

attachment: winmail.dat

Re: [Vo]:Mizuno, Rossi copper transmutation


Jones--

Was there any indication that the Mizuno experiments used quadrupole 
electric or magnetic inputs?


I was not aware of this, if it happened.

Also keep in mind that D is a Bose particle (as is 4HE) and can form a BEC 
or a duplex BEC with two different Bose particles.  This may be a reality in 
a strong magnetic field, temperature be damned.


The question I have is how a BEC can shed energy and change the mass of its 
constituents without disrupting the condensate.  Maybe it is a series of 
condensation and disruption that controls the reaction.  The dynamics of 
this process would be key to controlling the rate of the process.


Bob Cook


- Original Message - 
From: Jones Beene jone...@pacbell.net

To: vortex-l@eskimo.com
Sent: Saturday, October 18, 2014 6:34 AM
Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation


Bob,
I have cherry-picked three major “spin facts” from this compendium which
indicate that if one wants to apply a nano-magnetism or spin-coupling
modality to LENR, it is highly preferable to use deuterium, as opposed to
hydrogen. That may be why Mizuno chose the deuterium-nickel combination. All
eyes will be shifting to Mizuno in less than three weeks.

From: Bob Cook
[snip] The deuteron, being an isospin singlet, is antisymmetric under
nucleon exchange due to isospin, and therefore must be symmetric under the
double exchange of their spin and location. Therefore it can be in either of
the following two different states: Symmetric spin and symmetric under
parity. In this case, the exchange of the two nucleons will multiply the
deuterium wavefunction by (-1) from isospin exchange, (+1) from spin
exchange and (+1) from parity (location exchange), for a total of (-1) as
needed for antisymmetry…. In this case, the exchange of the two nucleons
will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1)
from spin exchange and (-1) from parity (location exchange), again for a
total of (-1) as needed for antisymmetry. [snip]

…suggesting that there may be a way to stimulate the D via an electric
quadrupole input signal.   Also with a magnetic moment the D must respond to
a magnetic field and fine tuning of an oscillating magnetic field may very
well excite the D to flip up and down in the field.  The composite particles
of the D should have slightly different magnetic moments that can respond
and create an excited state IMHO on a transient short lived time frame.
However in a coherent system such a transient may be enough to cause other
transitions of similar energy states to occur with mass energy being changed
to angular momentum energy.

The quadrupole input is a strong clue.

RE: [Vo]:Mizuno, Rossi copper transmutation

-Original Message-
From: Bob Cook 

 Was there any indication that the Mizuno experiments used quadrupole 
electric or magnetic inputs? I was not aware of this, if it happened.

This is an interesting point, and the Mizuno experiment may not have been 
optimized. Hopefully the next iteration will tell us more. One interesting 
thing about superparamagnetism is that a quadrupole is a natural outgrowth of 
any changing field, and if the deuterium becomes superparamagnetic that is all 
we need.

 Also keep in mind that D is a Bose particle (as is 4HE) and can form a BEC 
or a duplex BEC with two different Bose particles. This may be a reality in 
a strong magnetic field, temperature be damned.

Deuterium in normally diamagnetic as a molecule and we really do not need for 
it to a BEC to see excess energy. As a bare nucleus, it may be 
superparamagnetic, and that could be enough for spin coupling.

Superparamagnetism in somewhat new to consideration in the LENR field, since 
it is an outgrowth of nanotechnology. We have much to learn about spin-coupling 
but in my mind, superparamagnetism will be a major piece of the puzzle, but 
certainly bosons which are superparamagnetic are more likely to participate in 
energy transfers than fermions.

 The question I have is how a BEC can shed energy and change the mass of its 
constituents without disrupting the condensate. 

Well - Deuterium cannot even become a condensate in normal LENR, since it 
requires cryogenic temperatures. However, there are boson condensates in SPP 
which form at elevated temperature but they are not massive. Axil and myself 
have presented scholarly articles to that effect. An answer for original 
question is that an SPP condensate allows energy to be coupled from A to B 
(where A is a boson like the deuteron) without itself necessarily participating 
in the reaction.

 Maybe it is a series of condensation and disruption that controls the 
 reaction. The dynamics of this process would be key to controlling the rate 
 of the process.

Still not sure why you think that deuterium can become a BEC above a few 
degrees of absolute zero? Is there any evidence for this?

Jones

- Original Message - 
Bob,

I have cherry-picked three major “spin facts” from this compendium which
indicate that if one wants to apply a nano-magnetism or spin-coupling
modality to LENR, it is highly preferable to use deuterium, as opposed to
hydrogen. That may be why Mizuno chose the deuterium-nickel combination. All
eyes will be shifting to Mizuno in less than three weeks.

From: Bob Cook
[snip] The deuteron, being an isospin singlet, is antisymmetric under
nucleon exchange due to isospin, and therefore must be symmetric under the
double exchange of their spin and location. Therefore it can be in either of
the following two different states: Symmetric spin and symmetric under
parity. In this case, the exchange of the two nucleons will multiply the
deuterium wavefunction by (-1) from isospin exchange, (+1) from spin
exchange and (+1) from parity (location exchange), for a total of (-1) as
needed for antisymmetry…. In this case, the exchange of the two nucleons
will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1)
from spin exchange and (-1) from parity (location exchange), again for a
total of (-1) as needed for antisymmetry. [snip]

…suggesting that there may be a way to stimulate the D via an electric
quadrupole input signal.   Also with a magnetic moment the D must respond to
a magnetic field and fine tuning of an oscillating magnetic field may very
well excite the D to flip up and down in the field.  The composite particles
of the D should have slightly different magnetic moments that can respond
and create an excited state IMHO on a transient short lived time frame.
However in a coherent system such a transient may be enough to cause other
transitions of similar energy states to occur with mass energy being changed
to angular momentum energy.

The quadrupole input is a strong clue.

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-18 Thread Eric Walker

I wrote:

Unfortunately I don't have any other details and don't know of a particular
 experiment to refer to.  Here is the quote from a textbook I recently
 finished reading: ...


It was late last night, and the paragraph I found and quoted pertained to
deuterium, not 4He, which you were asking about, Bob.  I recall reading
that 4He does not have an excited state either, but I will have to see if I
can find where I saw that (it might also have been a mistaken impression).

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation


Jones-

I do not know of data on high temperature BEC's of particles of high mass.

Y. Kims theory for the reaction of Bose particles, including duplex 
compinations of different Bose particles, may be of some importance.  I have 
not studied his theory in depth, but the fact that he seems to think BEC 
may be involved in the LENR process suggests that room temperature BEC may 
happen.


The following is a list of Kim's theories of recent times.

a.. Y.E. Kim, Theoretical Analysis and Reaction Mechanisms for Experimental 
Results of Hydrogen-Oxygen-Metal Systems, Purdue Nuclear and Many-Body 
Theory Group (PNMBTG) preprint — PNMBTG-05-2014 (May 2014). (PDF)
a.. Y.E. Kim, Hartree-Fock Theory with Correlation Effects Applied to 
Nuclear Reaction Rates for Charged Bose Nuclei Confined in a Harmonic Trap, 
Purdue Nuclear and Many-Body Theory Group (PNMBTG) preprint PNMBTG-8-2013 
(August, 2013). (PDF)
a.. Y.E. Kim, Conventional Nuclear Theory of Low-Energy Nuclear Reactions 
in Metals: Alternative Approach to Clean Fusion Energy Generation, Purdue 
Nuclear and Many Body Theory Group (PNMBTG) Preprint PNMBTG-12—7 (July 
2012). Invited paper presented at the 17th International Conference on Cold 
Fusion (ICCF-17), Daejeon, Korea, August 12-17, 2012. (PDF), to be published 
in the ICCF-17 Proceedings.
a.. Y.E. Kim, Cryogenic Ignition of Deuteron Fusion in Micro/Nano-Scale 
Metal Particles, Purdue Nuclear and Many Body Theory Group (PNMBTG) 
Preprint PNMBTG-11-2011 (November 2011). Invited paper presented at Topical 
Meeting of the 2012 Nuclear and Emerging Technologies for Space (NETS), the 
43rd Lunar and Planetary Science Conference, March 19-23, 2012, the 
Woodlands, Texas. (PDF)
a.. Y.E. Kim, Nuclear Reactions in Micro/Nano-Scale Metal Particles, 
Few-Body Systems 54, 25-30 (2013), invited paper presented at the 5th 
Asia-Pacific Conference on Few-Body Problems in Physics (APFB2011), Seoul, 
Korea, August 22-26, 2011. (PDF)
a.. Y.E. Kim, Generalized Theory of Bose-Einstein Condensation Nuclear 
Fusion for Hydrogen-Metal System, Purdue Nuclear and Many Body Theory Group 
(PNMBTG) Preprint PNMBTG-6-2011 (June 2011). (PDF)


They can be copied as PDF documents from the following web page:

http://www.physics.purdue.edu/people/faculty/yekim.html

Another paper by Kim has the following abstract:

ABSTRACT

Generalized theory of Bose-Einstein condensation nuclear fusion (BECNF) is 
used to carry out theoretical analyses of recent experimental results of 
Rossi et al. for hydrogen-nickel system. Based on incomplete experimental 
information currently available, preliminary theoretical explanations of the 
experimental results are presented in terms of the generalized BECNF theory. 
Additional accurate experimental data are needed for obtaining more complete 
theoretical descriptions and predictions, which can be tested by further 
experiments.


The full paper can be found at the following web page:

http://www.freerepublic.com/focus/chat/2746057/posts

The latter link suggested the BEC that I have talked about at above 
cryogenic temperatures.


Bob Cook


Bob Cook



- Original Message - 
From: Jones Beene jone...@pacbell.net

To: vortex-l@eskimo.com
Sent: Saturday, October 18, 2014 7:33 AM
Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation


-Original Message-
From: Bob Cook


Was there any indication that the Mizuno experiments used quadrupole

electric or magnetic inputs? I was not aware of this, if it happened.

This is an interesting point, and the Mizuno experiment may not have been 
optimized. Hopefully the next iteration will tell us more. One interesting 
thing about superparamagnetism is that a quadrupole is a natural outgrowth 
of any changing field, and if the deuterium becomes superparamagnetic that 
is all we need.



Also keep in mind that D is a Bose particle (as is 4HE) and can form a BEC

or a duplex BEC with two different Bose particles. This may be a reality in
a strong magnetic field, temperature be damned.

Deuterium in normally diamagnetic as a molecule and we really do not need 
for it to a BEC to see excess energy. As a bare nucleus, it may be 
superparamagnetic, and that could be enough for spin coupling.


Superparamagnetism in somewhat new to consideration in the LENR field, 
since it is an outgrowth of nanotechnology. We have much to learn about 
spin-coupling but in my mind, superparamagnetism will be a major piece of 
the puzzle, but certainly bosons which are superparamagnetic are more likely 
to participate in energy transfers than fermions.


The question I have is how a BEC can shed energy and change the mass of 
its

constituents without disrupting the condensate.

Well - Deuterium cannot even become a condensate in normal LENR, since it 
requires cryogenic temperatures. However, there are boson condensates in SPP 
which form at elevated temperature but they are not massive. Axil and myself 
have presented scholarly articles to that effect. An answer

RE: [Vo]:Mizuno, Rossi copper transmutation

-Original Message-
From: Bob Cook 

 I do not know of data on high temperature BEC's of particles of high mass.

 Y. Kims theory for the reaction of Bose particles, including duplex 
compinations of different Bose particles, may be of some importance.  I have 
not studied his theory in depth, but the fact that he seems to think BEC 
may be involved in the LENR process suggests that room temperature BEC may 
happen.


Yes - to a degree, the BEC is likely to be involved, but not in a high mass 
particle like the deuteron. There are dozens of papers like this 

http://www.nature.com/nmat/journal/v13/n3/full/nmat3825.html

It would appear that the limit for BEC at high temperature is about a few 
thousand times less than the mass of the deuteron.

Re: [Vo]:Mizuno, Rossi copper transmutation

it is highly preferable to use deuterium, as opposed to hydrogen.

I disagree.

Deuterium has a non zero spin whereas hydrogen has a zero spin which is
required in low powered LENR reactions.

On Sat, Oct 18, 2014 at 9:34 AM, Jones Beene jone...@pacbell.net wrote:

 Bob,
 I have cherry-picked three major “spin facts” from this compendium which
 indicate that if one wants to apply a nano-magnetism or spin-coupling
 modality to LENR, it is highly preferable to use deuterium, as opposed to
 hydrogen. That may be why Mizuno chose the deuterium-nickel combination.
 All
 eyes will be shifting to Mizuno in less than three weeks.

 From: Bob Cook
 [snip] The deuteron, being an isospin singlet, is antisymmetric under
 nucleon exchange due to isospin, and therefore must be symmetric under the
 double exchange of their spin and location. Therefore it can be in either
 of
 the following two different states: Symmetric spin and symmetric under
 parity. In this case, the exchange of the two nucleons will multiply the
 deuterium wavefunction by (-1) from isospin exchange, (+1) from spin
 exchange and (+1) from parity (location exchange), for a total of (-1) as
 needed for antisymmetry…. In this case, the exchange of the two nucleons
 will multiply the deuterium wavefunction by (-1) from isospin exchange,
 (-1)
 from spin exchange and (-1) from parity (location exchange), again for a
 total of (-1) as needed for antisymmetry. [snip]

 …suggesting that there may be a way to stimulate the D via an electric
 quadrupole input signal.   Also with a magnetic moment the D must respond
 to
 a magnetic field and fine tuning of an oscillating magnetic field may very
 well excite the D to flip up and down in the field.  The composite
 particles
 of the D should have slightly different magnetic moments that can respond
 and create an excited state IMHO on a transient short lived time frame.
 However in a coherent system such a transient may be enough to cause other
 transitions of similar energy states to occur with mass energy being
 changed
 to angular momentum energy.

 The quadrupole input is a strong clue.

RE: [Vo]:Mizuno, Rossi copper transmutation

From: Axil 

 

it is highly preferable to use deuterium, as opposed to hydrogen.

 

I disagree.

 

Deuterium has a non zero spin whereas hydrogen has a zero spin which is 
required in low powered LENR reactions. 

 

 

Says who? What is your evidence?

Re: [Vo]:Mizuno, Rossi copper transmutation

Ni61 is non reactive as stated by DGT and confirmed by Mizuno as presented
in Cook's !CCF-18 presentation

On Sat, Oct 18, 2014 at 3:17 PM, Jones Beene jone...@pacbell.net wrote:

  *From:* Axil



 it is highly preferable to use deuterium, as opposed to hydrogen.



 I disagree.



 Deuterium has a non zero spin whereas hydrogen has a zero spin which is
 required in low powered LENR reactions.





 Says who? What is your evidence?

Re: [Vo]:Mizuno, Rossi copper transmutation

See

https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/36817/SimulationNuclearTransmutationPresentation.pdf?sequence=2

On Sat, Oct 18, 2014 at 3:24 PM, Axil Axil janap...@gmail.com wrote:

 Ni61 is non reactive as stated by DGT and confirmed by Mizuno as presented
 in Cook's !CCF-18 presentation

 On Sat, Oct 18, 2014 at 3:17 PM, Jones Beene jone...@pacbell.net wrote:

  *From:* Axil



 it is highly preferable to use deuterium, as opposed to hydrogen.



 I disagree.



 Deuterium has a non zero spin whereas hydrogen has a zero spin which is
 required in low powered LENR reactions.





 Says who? What is your evidence?

RE: [Vo]:Mizuno, Rossi copper transmutation

So what, that’s not general evidence? 

Even if true, it relates to nickel, and not to hydrogen/deuterium.
Everything since 1989 in LENR points to deuterium being as active if not
more than hydrogen.

From: Axil Axil 
*   Ni61 is non reactive as stated by DGT and confirmed by Mizuno as
presented in Cook's ICCF-18 presentation … Deuterium has a non zero spin
whereas hydrogen has a zero spin which is required in low powered LENR
reactions. 
Says who? What is your evidence?
 

attachment: winmail.dat

Re: [Vo]:Mizuno, Rossi copper transmutation

I would disagree with the spins reported by Axil for D and a Proton.  D is +1 
and the Proton is +1/2 in non excited states or ground states.  The neutron 
also has a +1/2 spin. The proton and neutron spins seem to add to make up the 
+1 spin of the D.

Bob
  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Saturday, October 18, 2014 12:17 PM
  Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation

  From: Axil 

  it is highly preferable to use deuterium, as opposed to hydrogen.

  I disagree.

  Deuterium has a non zero spin whereas hydrogen has a zero spin which is 
required in low powered LENR reactions. 

  Says who? What is your evidence?

Re: [Vo]:Mizuno, Rossi copper transmutation

 have been much more stable
(although still beta decaying to deuterium) had the strong force been 2%
greater. Its instability is due to spin-spin interactions in the nuclear
force, and the Pauli exclusion principle, which forces the two protons to
have anti-aligned spins and gives the diproton a negative binding energy.*



*By the way, the ash produced by the LENR reaction will have a non-zero
nuclear spin such as lithium, boron, and beryllium. This is due to the fact
that the ash is at the end of the LENR reaction chain that terminates with
an isotope featuring a non-zero nuclear spin.*



*Furthermore, all the stable isotopes of copper have a non-zero nuclear
spin. This may be way these isotopes are found in the ash assay of Rossi’s
reactor.*



*One last correlation remains.*



*It seems that the popular wet LENR catalyst acts like a superconductor for
protons where protons pair up into a cooper pair.*



*See*



*http://arxiv.org/pdf/0807.1386.pdf http://arxiv.org/pdf/0807.1386.pdf*



*This work emphasizes that atoms in the crystal-field of KHCO3 are not
individual particles possessing properties in their own right. They merge
into macroscopic states and exhibit all features of quantum mechanics:
non-locality, entanglement, spin-symmetry, superposition and interferences.
There is every reason to suppose that similar quantum effects should occur
in many hydrogen bonded crystals undergoing structural phase transitions.*



*I understand spin-symmetry to mean a zero spin.*



*This catalyst provides a proton dimer of zero spin to the wet LENR
reaction. This is the reason why this catalyst enhances electrolytic LENR
in water. *

On Sat, Oct 18, 2014 at 6:38 PM, Bob Cook frobertc...@hotmail.com wrote:

  I would disagree with the spins reported by Axil for D and a Proton.  D
 is +1 and the Proton is +1/2 in non excited states or ground states.  The
 neutron also has a +1/2 spin. The proton and neutron spins seem to add to
 make up the +1 spin of the D.

 Bob

 - Original Message -
 *From:* Jones Beene jone...@pacbell.net
 *To:* vortex-l@eskimo.com
 *Sent:* Saturday, October 18, 2014 12:17 PM
 *Subject:* RE: [Vo]:Mizuno, Rossi  copper transmutation

  *From:* Axil



 it is highly preferable to use deuterium, as opposed to hydrogen.



 I disagree.



 Deuterium has a non zero spin whereas hydrogen has a zero spin which is
 required in low powered LENR reactions.





 Says who? What is your evidence?

Re: [Vo]:Mizuno, Rossi copper transmutation

Axil--

Take a look at the Norman Cook discussion that you just posted as to the 
depletion of the various isotopes of Ni in LENR testing that Cook evaluates.

IT INCLUDES DEPLETION OF ODD NUCLEON  ISOTOPES WITH NET SPIN  (NI-59 WITH -3/2) 
AS WELL AS EVEN NUCLEON  ISOTOPES WITH 0 SPIN.   ONLY NI-61 (ODD NUMBER OF 
NUCLEONS) DOES NOT SEEM TO REACT MUCH.  

Bob Cook 
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Saturday, October 18, 2014 4:11 PM
  Subject: Re: [Vo]:Mizuno, Rossi  copper transmutation


  There is good reason to believe that magnetism is the prime mover in LENR. 
Under this speculative paradigm, it is interesting to consider the options and 
consequences of this conjecture. In such a paradigm, any technology that is 
friendly to magnetism would be good for LENR, and conversely, a technology that 
undercuts the strength of magnetism is bad.



  The Pd/D wet technology is more unfriendly to magnetism than nickel because 
it makes magnetism more difficult to maintain. Firstly as a general 
technological principle, an isotope must have a nuclear spin of zero to enable 
the LENR reaction. There is much experimental evidence to support this 
conjecture. For an explanation see below.   In this respect, palladium has a 
nuclear spin profile that is about 78% effective. 105Pd has a non-zero spin and 
is 22% of the isotopic contents of run of the mill palladium. 



  On the other hand, Nickel is much more efficient in terms of supporting 
magnetism. 61Ni has a non-zero nuclear spin, but that isotope is only 1.14% of 
the isotopic content of Nickel.



  Palladium is paramagnetic and Nickel is ferromagnetic. So nickel is more 
desirable than palladium as a magnetic reaction catalyst.



  In more detail, this thinking is underpinned by a speculative LENR reaction 
rule that is interesting to explore. That rule is that the LENR reaction must 
occur among atomic ions that have zero nuclear spin.


  In explanation, Nuclear magnetic resonance (NMR) is a physical phenomenon in 
which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. 
This energy is at a specific resonance frequency which depends on the strength 
of the magnetic field and the magnetic properties of the isotope of the atoms; 
in practical applications, the frequency is similar to old style VHF and UHF 
television broadcasts (60–1000 MHz). NMR allows the observation of specific 
quantum mechanical magnetic properties of the atomic nucleus. 



  All isotopes that contain an odd number of protons and/or of neutrons have an 
intrinsic magnetic moment and angular momentum, in other words a nonzero spin, 
while all nuclides with even numbers of both have a total spin of zero. The 
most commonly studied NMR active nuclei are 1H and 13C, although nuclei from 
isotopes of many other elements (e.g. 2H, 6Li, 10B, 11B, 14N, 15N, 17O, 19F, 
23Na, 29Si, 31P, 35Cl, 113Cd, 129Xe, 195Pt) have been studied by high-field NMR 
spectroscopy as well.



  It is now known that Ni61 does not participate in the LENR reaction. Ni61 is 
a NMR active isotope. When a magnetic field is applied to an NMR active 
isotope, the magnetic energy imparted to the nucleus is dissipated by induced 
nuclear vibrational energy which is radiated away as rf energy. The non-zero 
spin of the the nucleus shields the nucleus from the external magnetic field 
not allowing that field to penetrate into it. External magnetic fields catalyze 
changes in the protons and neutrons in the nucleus as well as enabling 
accelerated quantum mechanical tunneling. If this external magnetic field is 
shielded by NMR activity, LENR transmutation of the protons and neutrons in the 
nucleus is made more difficult.



  Therefore, during the course of an extended LENR reaction cycle, isotope 
depletion will tend to favor the enrichment and buildup of NMR active elements.



  Hydrogen with non-zero spin will not participate in the LENR reaction whereas 
cooper pairs of protons will. Expect LENR reactions centered on pairs of 
protons with zero spin.



  Also, as the LERN reaction matures and more NMR active isotopes accumulate, 
the LENR reactor will put out increasing levels or rf radiation derived from 
the nuclear vibrations of the NMR isotope.



   

  This NMR thinking also applies to the nature of the various isotopes of 
hydrogen.



  Molecular hydrogen occurs in two isomeric forms, one with its two proton 
spins aligned parallel (orthohydrogen), the other with its two proton spins 
aligned antiparallel (parahydrogen). At room temperature and thermal 
equilibrium, hydrogen consists of approximately 75% orthohydrogen and 25%  
parahydrogen.





  Orthohydrogen hydrogen has non zero spin, this is bad for Ni/H LENR because 
the non zero spin wastes magnetic energy by producing RF radiation. 
Parahydrogen hydrogen has zero spin. This is good for Ni/H LENR because this 
type of hydrogen is magnetically

Re: [Vo]:Mizuno, Rossi copper transmutation

Ni59 is not stable. I don't see it mentioned in the Cook report. You have a
case for Pd105 but this non zero spin conjecture is not a hard cast
rule. Non zero spin is just unfavorable in the LENR reaction based on the
strength of the magnetic field producing the reaction. Non zero spin just
discourages the LENR reaction, not prevents it. A strong enough magnetic
field will transmute anything no matter what the spin of the nucleus is.

On Sat, Oct 18, 2014 at 7:29 PM, Bob Cook frobertc...@hotmail.com wrote:

  Axil--

 Take a look at the Norman Cook discussion that you just posted as to the
 depletion of the various isotopes of Ni in LENR testing that Cook
  evaluates.

 IT INCLUDES DEPLETION OF ODD NUCLEON  ISOTOPES WITH NET SPIN  (NI-59 WITH
 -3/2) AS WELL AS EVEN NUCLEON  ISOTOPES WITH 0 SPIN.   ONLY NI-61 (ODD
 NUMBER OF NUCLEONS) DOES NOT SEEM TO REACT MUCH.

 Bob Cook

 - Original Message -
 *From:* Axil Axil janap...@gmail.com
 *To:* vortex-l vortex-l@eskimo.com
 *Sent:* Saturday, October 18, 2014 4:11 PM
 *Subject:* Re: [Vo]:Mizuno, Rossi  copper transmutation

  *There is good reason to believe that magnetism is the prime mover in
 LENR. Under this speculative paradigm, it is interesting to consider the
 options and consequences of this conjecture. In such a paradigm, any
 technology that is friendly to magnetism would be good for LENR, and
 conversely, a technology that undercuts the strength of magnetism is bad.*



 *The Pd/D wet technology is more unfriendly to magnetism than nickel
 because it makes magnetism more difficult to maintain. Firstly as a general
 technological principle, an isotope must have a nuclear spin of zero to
 enable the LENR reaction. There is much experimental evidence to support
 this conjecture. For an explanation see below.   In this respect, palladium
 has a nuclear spin profile that is about 78% effective. 105Pd has a
 non-zero spin and is 22% of the isotopic contents of run of the mill
 palladium. *



 *On the other hand, Nickel is much more efficient in terms of supporting
 magnetism. 61Ni has a non-zero nuclear spin, but that isotope is only 1.14%
 of the isotopic content of Nickel.*



 *Palladium is paramagnetic and Nickel is ferromagnetic. So nickel is more
 desirable than palladium as a magnetic reaction catalyst.*


 *In more detail, this thinking is underpinned by a speculative LENR
 reaction rule that is interesting to explore. That rule is that the LENR
 reaction must occur among atomic ions that have zero nuclear spin.*

  *In explanation, Nuclear magnetic resonance (NMR) is a physical
 phenomenon in which nuclei in a magnetic field absorb and re-emit
 electromagnetic radiation. This energy is at a specific resonance frequency
 which depends on the strength of the magnetic field and the magnetic
 properties of the isotope of the atoms; in practical applications, the
 frequency is similar to old style VHF and UHF television broadcasts
 (60–1000 MHz). NMR allows the observation of specific quantum mechanical
 magnetic properties of the atomic nucleus. *



 *All isotopes that contain an odd number of protons and/or of neutrons
 have an intrinsic magnetic moment and angular momentum, in other words a
 nonzero spin, while all nuclides with even numbers of both have a total
 spin of zero. The most commonly studied NMR active nuclei are 1H and 13C,
 although nuclei from isotopes of many other elements (e.g. 2H, 6Li, 10B,
 11B, 14N, 15N, 17O, 19F, 23Na, 29Si, 31P, 35Cl, 113Cd, 129Xe, 195Pt) have
 been studied by high-field NMR spectroscopy as well.*



 *It is now known that Ni61 does not participate in the LENR reaction. Ni61
 is a NMR active isotope. When a magnetic field is applied to an NMR active
 isotope, the magnetic energy imparted to the nucleus is dissipated by
 induced nuclear vibrational energy which is radiated away as rf energy. The
 non-zero spin of the the nucleus shields the nucleus from the external
 magnetic field not allowing that field to penetrate into it. External
 magnetic fields catalyze changes in the protons and neutrons in the nucleus
 as well as enabling accelerated quantum mechanical tunneling. If this
 external magnetic field is shielded by NMR activity, LENR transmutation of
 the protons and neutrons in the nucleus is made more difficult.*



 *Therefore, during the course of an extended LENR reaction cycle, isotope
 depletion will tend to favor the enrichment and buildup of NMR active
 elements.*



 *Hydrogen with non-zero spin will not participate in the LENR reaction
 whereas cooper pairs of protons will. Expect LENR reactions centered on
 pairs of protons with zero spin.*



 *Also, as the LERN reaction matures and more NMR active isotopes
 accumulate, the LENR reactor will put out increasing levels or rf radiation
 derived from the nuclear vibrations of the NMR isotope.*





 *This NMR thinking also applies to the nature of the various isotopes of
 hydrogen.*



 *Molecular hydrogen occurs in two isomeric forms

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-18 Thread Eric Walker

On Sat, Oct 18, 2014 at 6:20 AM, Bob Cook frobertc...@hotmail.com wrote:

 A quadruple oscillating electric field may also help to excite the D's to
 shed their excess mass relative to the developing 4He particle.


This sounds a little bit wishful to me.  :)

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-17 Thread Bob Cook

Eric--

You wrote the following some time ago:
  - Original Message - 
  From: Eric Walker 
  To: vortex-l@eskimo.com 
  Sent: Thursday, October 02, 2014 7:58 PM
  Subject: Re: [Vo]:Mizuno, Rossi  copper transmutation

  On Thu, Oct 2, 2014 at 12:17 AM, frobertcook frobertc...@hotmail.com wrote:

I like the idea of excited He as an intermediate.

  I recall reading that 4He does not have a bound excited nuclear state, 
although it may have a resonance for a very brief period of time.

  Eric

  Do you know if the experiments looked at excited spin energy states that may 
be  possible at higher spin quanta?  It probably would take the form of a 
deformed nucleus with some moment of inertia or dipole arrangement created by 
high magnetic or electric fields at resonant condition, like that associated 
with a tuned high energy laser beam shinning on double ionized 4He nuclei.  

  Bob

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-17 Thread Eric Walker

On Fri, Oct 17, 2014 at 10:41 AM, Bob Cook frobertc...@hotmail.com wrote:

Do you know if the experiments looked at excited spin energy states that
 may be  possible at higher spin quanta?


Unfortunately I don't have any other details and don't know of a particular
experiment to refer to.  Here is the quote from a textbook I recently
finished reading:

For nuclear physicists, the deuteron should be what the hydrogen atom is
for atomic physicists.  Just as the measured Balmer series of
electromagnetic transitions between the excited states of hydrogen led to
an understanding of the structure of hydrogen, so should the
electromagnetic transitions between the excited states of the deuteron lead
to an understanding of its structure.  Unfortunately, there are *no excited
states* of the deuteron—it is such a weakly bound system that the only
excited states are unbound systems consisting of a free proton and
neutron. [1]


Eric

[1] Kenneth S. Krane, *Introductory Nuclear Physics*, pp. 80-81; author's
emphasis.

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-02 Thread frobertcook

Harry
I like the idea of excited He as an intermediate. However, I would indicate
low energy radiation associted will He* spin energy deflation with angular
momentum an the spin energy distributed throughout the quantum dot mattrix and
its electrons.

The He* may occur in pairs aligned antiparallel in the local
magnetic field to allow conservation of AM.

THERE ARE NO GAMMAS.

Bob

Sent from my Verizon Wireless 4G LTE SmartphoneH Veeder hveeder...@gmail.com
wrote:
The link has two drawings on the same page. The top drawing, which is the
one I found, doesn't challenge FP research.

The bottom drawing is my modified version and it is intended to show that
the fusion process can be considered reversible as long as it does not
reach the final stage.
Are you asking yourself why is he proposing a hypothetical fusion process
that does not result in the production energy?

Harry

On Sun, Sep 28, 2014 at 5:10 AM, Alain Sepeda alain.sep...@gmail.com
wrote:

I don't see how it challenge FP,
it is theory?

2014-09-28 2:34 GMT+02:00 H Veeder hveeder...@gmail.com:

On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

In reply to H Veeder's message of Wed, 24 Sep 2014 23:04:12 -0400:
Hi Harry,
[snip]
Since we are dealing in impossibilities from the outset, it seems like
false logic to argue that the probability of endothermic reactions
is improbable.
[snip]
I have told you what I think and why. Whether or not you choose to
accept it is
up to you.

Regards,

Robin van Spaandonk

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

I found this drawing on a site which happened to be extremely critical of
PF's research.

https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

What I like about the drawing is that it shows the three d-d fusion
pathways all passing through the same intermediate stage of high energy
helium 4. I modified the drawing to show the reaction going in both
directions before the excited intermediate stage has a chance to decay. I
think that the lattice facilitates the initiation of fusion but it also
tends to inhibits the completion the fusion process. The question of course
is of what relevancy is this scenario if it does not produce energy? If it
can form an epicatalytic
process then it is very relevant.

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-02 Thread Eric Walker

On Thu, Oct 2, 2014 at 12:17 AM, frobertcook frobertc...@hotmail.com
wrote:

 I like the idea of excited He as an intermediate.


I recall reading that 4He does not have a bound excited nuclear state,
although it may have a resonance for a very brief period of time.

Eric

Re: Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-01 Thread mixent

In reply to  H Veeder's message of Wed, 1 Oct 2014 00:25:05 -0400:
Hi,
[snip]
Since the second nickel nucleus has an extra neutron it is
now in an excited state. While it is excited the hydrogen nucleus on the
left retreats and the hydrogen nucleus on the right  is
approaches. 

Timing problem again. Gamma emission in approx. 1E-17 sec. Oscillation rate of
the H atoms in the THz range. That means that the cycle time of the H atoms is
about 1E-12 sec. Gamma decay is about 10 times faster, so most of the time
the energy will be emitted as a gamma.
Furthermore, I don't think the Nickel is going to be all that willing to part
with it's new toy anyway. ;)

Regards,

Robin van Spaandonk

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

Re: Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-10-01 Thread H Veeder

On Wed, Oct 1, 2014 at 5:52 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Wed, 1 Oct 2014 00:25:05 -0400:
 Hi,
 [snip]
 Since the second nickel nucleus has an extra neutron it is
 now in an excited state. While it is excited the hydrogen nucleus on the
 left retreats and the hydrogen nucleus on the right  is
 approaches.

 Timing problem again. Gamma emission in approx. 1E-17 sec. Oscillation
 rate of
 the H atoms in the THz range. That means that the cycle time of the H
 atoms is
 about 1E-12 sec. Gamma decay is about 10 times faster, so most of the
 time
 the energy will be emitted as a gamma.
 Furthermore, I don't think the Nickel is going to be all that willing to
 part
 with it's new toy anyway. ;)

 Regards,

 Robin van Spaandonk

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

 
Every theorist begins by choosing to accept some impossibilities and to
reject other impossibilities.
It seems to me that the choice is based as much on logic and evidence as it
is based on the theorist's particular training, personal experiences and
intuition.
Since I can't draw on a wealth of knowledge about chemistry, nuclear
physics or condensed matter to lend credibility to my choices I will hence
forth not
theorize about this phenomena.

Harry

Harry









Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread H Veeder

On Tue, Sep 30, 2014 at 1:07 AM, Eric Walker eric.wal...@gmail.com wrote:

 On Mon, Sep 29, 2014 at 2:54 PM, mix...@bigpond.com wrote:

 If it happened nobody would notice.


 Yes.  I think it would be indistinguishable from an elastic collision (if
 the two situations are different).

 Eric


That analogy assumes the excited nucleus immediately reverts or fissions
back into the original parts.
However, if there is a significant time delay before fission occurs and the
excited nucleus is able to migrate to different site during that delay,
then when fission does occur it will cause a local temperature increase at
the different site.

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread mixent

In reply to  H Veeder's message of Tue, 30 Sep 2014 17:39:12 -0400:
Hi,
[snip]
On Tue, Sep 30, 2014 at 1:07 AM, Eric Walker eric.wal...@gmail.com wrote:

 On Mon, Sep 29, 2014 at 2:54 PM, mix...@bigpond.com wrote:

 If it happened nobody would notice.


 Yes.  I think it would be indistinguishable from an elastic collision (if
 the two situations are different).

 Eric


That analogy assumes the excited nucleus immediately reverts or fissions
back into the original parts.
However, if there is a significant time delay before fission occurs and the
excited nucleus is able to migrate to different site during that delay,
then when fission does occur it will cause a local temperature increase at
the different site.

There isn't time for it migrate. The fission to either He3 + n or T + p happens
in about 1E-22 sec. For this not to happen, it would have to fission back to D+D
in less time than that.

Regards,

Robin van Spaandonk

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

Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread H Veeder

On Tue, Sep 30, 2014 at 5:51 PM, mix...@bigpond.com wrote:

In reply to H Veeder's message of Tue, 30 Sep 2014 17:39:12 -0400:
Hi,
[snip]
On Tue, Sep 30, 2014 at 1:07 AM, Eric Walker eric.wal...@gmail.com
wrote:

On Mon, Sep 29, 2014 at 2:54 PM, mix...@bigpond.com wrote:

If it happened nobody would notice.

Yes. I think it would be indistinguishable from an elastic collision
(if
the two situations are different).

Eric

That analogy assumes the excited nucleus immediately reverts or fissions
back into the original parts.
However, if there is a significant time delay before fission occurs and
the
excited nucleus is able to migrate to different site during that delay,
then when fission does occur it will cause a local temperature increase at
the different site.

There isn't time for it migrate. The fission to either He3 + n or T + p
happens
in about 1E-22 sec. For this not to happen, it would have to fission back
to D+D
in less time than that.

Regards,

Robin van Spaandonk

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

I wonder if the decay time of 1E-22 secs is theoretically derived or if
it is empirically derived. If it is empirically derived then it might not
always be true, but I will accept your point for the time being and switch
to a process involving neutron stripping, which is where our exchange began.

Imagine a line of nickel nuclei with one deuterium nucleus in the gap
between the first two nickel nuclei. The remaining gaps are each occupied
with a hydrogen nucleus.
Imagine just the deuterium and hydrogen nuclei oscillating back and forth
in unison in the gaps. When the deuterium nucleus gets close enough to
connect with the second Nickel nucleus it gives up its neutron to that
nickel nucleus. Since the second nickel nucleus has an extra neutron it is
now in an excited state. While it is excited the hydrogen nucleus on the
left retreats and the hydrogen nucleus on the right is
approaches. Eventually the hydrogen nucleus on the right connects with the
excited nickel nucleus and the extra neutron in the excited nickel nucleus
is transferred to it. (Technically speaking this is not a reverse reaction
since it involves a new association, but this is a work in progress which
you and others are helping to complete so forgive me if I do not use always
use the best terms). The neutron transfers continue so that energy is
moved from the beginning of the line to the end of the line.

I illustrated the process here:
https://docs.google.com/document/d/1dzUFl91yhYGk5CTnAX_eXCYPgIemqlTF3XuQkRQf_hA/edit?usp=sharing

The process is like a bucket brigade but instead of water being transferred
it is fire. Incidentally while looking at some youtube videos of bucket
brigades I stumbled on a video where fire is moved instead of water.
https://www.youtube.com/watch?v=tsfJZfHARLk

Anyway, if the general conept is not inane, I am sure there are other
possible bucket brigades involving different nuclei.

Harry

Re: Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread Eric Walker

On Tue, Sep 30, 2014 at 9:25 PM, H Veeder hveeder...@gmail.com wrote:

When the deuterium nucleus gets close enough to connect with the second
 Nickel nucleus it gives up its neutron to that nickel nucleus.


I think you're going to need a powerful force to make this part happen.
Think of the proton that is part of the deuteron and the nickel nucleus as
extremely powerful, oppositely magnetized metal spheres.  They're going to
do whatever they can to avoid each other, including sending the deuteron
along a curved path out of the line of collision with the nickel nucleus if
such a path is allowed by the velocity of the deuteron.

Eric

Re: Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread Eric Walker

I wrote:

Think of the proton that is part of the deuteron and the nickel nucleus as
 extremely powerful, oppositely magnetized metal spheres.


I didn't say that very well.  They're like two magnets with the same poles
facing each other (these magnets are monopoles, so there's no other pole to
allow them to flip around).  Also, magnetism isn't the force involved,
technically speaking, but the general physical interaction is how I think
about it.

Eric

Re: Nuclear bucket brigade - was Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-30 Thread H Veeder

Now I'll give *you* something to believe. I'm just one hundred and one,
five months and a day.'

'I can't believe *that!*' said Alice.

'Can't you?' the Queen said in a pitying tone. 'Try again: draw a long
breath, and shut your eyes.'

Alice laughed. 'There's no use trying,' she said 'one *can't* believe
impossible things.'

'I daresay you haven't had much practice,' said the Queen. 'When I was your
age, I always did it for half-an-hour a day. Why, sometimes I've believed
as many as six impossible things before breakfast.

On Wed, Oct 1, 2014 at 12:39 AM, Eric Walker eric.wal...@gmail.com wrote:

 I wrote:

 Think of the proton that is part of the deuteron and the nickel nucleus as
 extremely powerful, oppositely magnetized metal spheres.


 I didn't say that very well.  They're like two magnets with the same poles
 facing each other (these magnets are monopoles, so there's no other pole to
 allow them to flip around).  Also, magnetism isn't the force involved,
 technically speaking, but the general physical interaction is how I think
 about it.

 Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-29 Thread mixent

In reply to  H Veeder's message of Sun, 28 Sep 2014 19:18:08 -0400:
Hi Harry,
The link has two drawings on the same page. The top drawing, which is the
one I found, doesn't challenge FP research.

The bottom drawing is my modified version and it is intended to show that
the fusion process can be considered reversible as long as it does not
reach the final stage.
Are you asking yourself why is he proposing a hypothetical fusion process
that does not result in the production energy?

Harry
[snip]
If it happened nobody would notice.
Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-29 Thread Eric Walker

On Mon, Sep 29, 2014 at 2:54 PM, mix...@bigpond.com wrote:

If it happened nobody would notice.


Yes.  I think it would be indistinguishable from an elastic collision (if
the two situations are different).

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-28 Thread Alain Sepeda

I don't see how it challenge FP,
it is theory?

2014-09-28 2:34 GMT+02:00 H Veeder hveeder...@gmail.com:



 On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Wed, 24 Sep 2014 23:04:12 -0400:
 Hi Harry,
 [snip]
 Since we are dealing in impossibilities from the outset, it seems like
 false logic to argue that the probability of endothermic reactions
 is improbable.
 [snip]
 I have told you what I think and why. Whether or not you choose to accept
 it is
 up to you.

 Regards,

 Robin van Spaandonk

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



 I found this drawing on a site which happened to be extremely critical of
 PF's research.


 https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

 What I like about the drawing is that it shows the three d-d fusion
 pathways all passing through the same intermediate stage of high energy
 helium 4. I modified the drawing to show the reaction going in both
 directions before the excited intermediate stage has a chance to decay. I
 think that the lattice facilitates the initiation of fusion but it also
 tends to inhibits the completion the fusion process. The question of course
 is of what relevancy is this scenario if it does not produce energy? If it
 can form an epicatalytic
 process then it is very relevant.


 Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-28 Thread H Veeder

The link has two drawings on the same page. The top drawing, which is the
one I found, doesn't challenge FP research.

Harry

On Sun, Sep 28, 2014 at 5:10 AM, Alain Sepeda alain.sep...@gmail.com
wrote:

I don't see how it challenge FP,
it is theory?

2014-09-28 2:34 GMT+02:00 H Veeder hveeder...@gmail.com:

On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

Regards,

Robin van Spaandonk

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

I found this drawing on a site which happened to be extremely critical of
PF's research.

https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-27 Thread H Veeder

On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Wed, 24 Sep 2014 23:04:12 -0400:
 Hi Harry,
 [snip]
 Since we are dealing in impossibilities from the outset, it seems like
 false logic to argue that the probability of endothermic reactions
 is improbable.
 [snip]
 I have told you what I think and why. Whether or not you choose to accept
 it is
 up to you.

 Regards,

 Robin van Spaandonk

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



I found this drawing on a site which happened to be extremely critical of
PF's research.

https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

What I like about the drawing is that it shows the three d-d fusion
pathways all passing through the same intermediate stage of high energy
helium 4. I modified the drawing to show the reaction going in both
directions before the excited intermediate stage has a chance to decay. I
think that the lattice facilitates the initiation of fusion but it also
tends to inhibits the completion the fusion process. The question of course
is of what relevancy is this scenario if it does not produce energy? If it
can form an epicatalytic
process then it is very relevant.


Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-27 Thread Axil Axil

I suggest that a very fast monolithic reaction process will allow Helium-2
(diproton) to form. Then immediately, before the positrons decay can take
place producing neutrons, two diproton atoms will fuse to the latent helium
4 intermediate product will take place comprised of 4 protons.

After positron emission produces 2 neutrons to form helium 4,, the
characteristic LENR emission of 1,02 MeV gammas will manifest,

Upon the collision of a particle and an anti-particle, e.g. electron and
positron, these are annihilated as particles and the mass of these
particles converted into energy. Electron and positron have a rest mass
which is together equal to an energy of 1.02 MeV. Upon the annihilation
of both particles, two gamma,

The usual radiation product of a LENR reaction is positrons and lots of
them.

On Sat, Sep 27, 2014 at 8:34 PM, H Veeder hveeder...@gmail.com wrote:

On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

Regards,

Robin van Spaandonk

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

I found this drawing on a site which happened to be extremely critical of
PF's research.

https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-27 Thread Axil Axil

Upon the annihilation of both particles, two gamma,

should read

Upon the annihilation of both particles, two 512 KeV gamma are produced
that travel in an antiparallel direction away from the point of
annihilation .

On Sat, Sep 27, 2014 at 9:04 PM, Axil Axil janap...@gmail.com wrote:

After positron emission produces 2 neutrons to form helium 4,, the
characteristic LENR emission of 1,02 MeV gammas will manifest,

The usual radiation product of a LENR reaction is positrons and lots of
them.

On Sat, Sep 27, 2014 at 8:34 PM, H Veeder hveeder...@gmail.com wrote:

On Thu, Sep 25, 2014 at 5:42 PM, mix...@bigpond.com wrote:

Regards,

Robin van Spaandonk

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

I found this drawing on a site which happened to be extremely critical of
PF's research.

https://docs.google.com/document/d/1OpDKkgdQKrgP29Nxa0N_biIsLz0qeY8UGDGpFJCFSy0/edit?usp=sharing

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-25 Thread mixent

In reply to  H Veeder's message of Wed, 24 Sep 2014 23:04:12 -0400:
Hi Harry,
[snip]
Since we are dealing in impossibilities from the outset, it seems like
false logic to argue that the probability of endothermic reactions
is improbable.
[snip]
I have told you what I think and why. Whether or not you choose to accept it is
up to you.

Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-24 Thread H Veeder

I suspect both endothermic and exothermic reactions occur even inside the
tokamak, but on balance more exothermic reactions occur.

Harry


On Sun, Sep 21, 2014 at 6:30 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Sun, 21 Sep 2014 17:35:34 -0400:
 Hi,

 Nuclear energies are 6 orders of magnitude larger than chemical energies,
 which
 I would expect to reduce the chances to the point where it's not even worth
 considering.
 However, that said, it should be noted that the same is not always true for
 reactions where D is converted into T.

 e.g. the following reaction is exothermic:-

 9Be+2H = 4He + 4He + 3H + 4.684 MeV


 On Sat, Sep 20, 2014 at 10:59 PM, mix...@bigpond.com wrote:
 
  In reply to  H Veeder's message of Sat, 20 Sep 2014 20:53:37 -0400:
  Hi,
  [snip]
  If hydrinos and deuterinos are both present, perhaps it is possible for
  the
  neutron stripping to work in two directions such that a deuterino can
 give
  up a neutron to a heavy nucleus and a heavy nucleus can give up a
 neutron
  to hydrino. ( I am thinking of a nuclear version of epicatalysis.)
  
  Harry
  A heavy nucleus won't give up a neutron to a Hydrino, because in doing
 so
  it
  would lose about 5-10 MeV, but only gain 2.2 MeV from the formation of
 the
  deuteron.
 
 
 
 ?That means it is an endothermic reaction, but that doesn't mean it is
 impossible?.
 I am not implying that neutron stripping should be discarded ?if the
 reverse reaction is possible.
 ?
 I
 ? ?
 mentioned epicatalysis because
 ?theoretical research on?
 
 ?the subject
 
 ?was recently ?
 published in Physical Review E. Along with some empirical evidence the
 research suggests that deviations
 ?of practical significance ?
 from the 2nd law of law thermodynamics are possible
 ?with epicatalysis
 ?:?
 
 https://www.facebook.com/ParadigmEnergy/posts/249600938581128
 
 Now the theory of epicatalysis is based on chemical activity, but I don't
 see why the theory could not be broadened to include nuclear activity or
 other unconventional high energy activity if a given heat anomaly is too
 large to explain by just chemical activity.
 
 A tacit assumption of CF/LENR research is that an anomalous thermal signal
 will have practical significance if it results from the conversion of
 potential energy into kinetic energy in a one way process. The assumption
 holds whether the source of energy is nuclear or chemical or some other.
 Consequently, measured temperature anomalies are suspect until they are
 supported by additional calorimetry which yields a global temperature
 rise.
 If this global temperature rise (excess heat signal) is not found, and
 measurement error is ruled out, then the temperature anomaly will be
 classified as a local fluctuation with no practical significance. This
 interpretation of temperature signals is motivated by the demands of the
 2nd law of thermodynamics.
 
 However, if a process like epicatalysis is creating the temperature
 anomalies then the methods used to measure an excess heat signal need to
 be
 reconsidered. Detecting an excess heat signal ordinarily means looking for
 a global temperature rise which requires that the source of an anomaly be
 placed in a thermally closed environment since it is assumed the
 temperature rise is based on the creation of kinetic energy from inside
 the
 system. In contradistinction epicatalysis transfers energy from a lower
 temperature region to a higher temperature region. If the purpose of the
 enclosure is to detect a global temperature rise none will be found.
 
 Harry
 Regards,

 Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-24 Thread mixent

In reply to  H Veeder's message of Wed, 24 Sep 2014 21:46:11 -0400:
Hi,
[snip]
I suspect both endothermic and exothermic reactions occur even inside the
tokamak, but on balance more exothermic reactions occur.

Harry

Endothermic reactions only happen when ingoing particles have enough kinetic
energy to make the reaction happen. IOW they are not really endothermic when all
energy sources are taken into account.

However most nuclear reactions where a neutron transfers to an external proton
to create deuterium would be genuinely endothermic, and thus would not occur,
unless of course the proton had high kinetic energy to start with.

Of course it's possible that this happens, but the reactions going the other way
are going to outnumber them by many thousands to 1, because very few of the
energetic protons created are immediately going to encounter another heavy
nucleus before losing some energy to ionization, and of those that do
immediately encounter another heavy nucleus, only a small percentage are going
to produce D.
Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-24 Thread H Veeder

On Wed, Sep 24, 2014 at 10:13 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Wed, 24 Sep 2014 21:46:11 -0400:
 Hi,
 [snip]
 I suspect both endothermic and exothermic reactions occur even inside the
 tokamak, but on balance more exothermic reactions occur.
 
 Harry

 Endothermic reactions only happen when ingoing particles have enough
 kinetic
 energy to make the reaction happen. IOW they are not really endothermic
 when all
 energy sources are taken into account.

 However most nuclear reactions where a neutron transfers to an external
 proton
 to create deuterium would be genuinely endothermic, and thus would not
 occur,
 unless of course the proton had high kinetic energy to start with.

 Of course it's possible that this happens, but the reactions going the
other way

 are going to outnumber them by many thousands to 1, because very few of the
 energetic protons created are immediately going to encounter another heavy
 nucleus before losing some energy to ionization, and of those that do
 immediately encounter another heavy nucleus, only a small percentage are
 going
 to produce D.


Ok now lets return to condensed matter systems.


All the nuclear
reactions
 offer
ed
for reports of excess heat
in such 
systems 
are suppose to be
 theoretically
impossible.
Since we are dealing in impossibilities from the outset, it seems like
false logic to argue that the probability of endothermic reactions
is improbable.

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-24 Thread H Veeder

On Wed, Sep 24, 2014 at 10:59 PM, H Veeder hveeder...@gmail.com wrote:



 On Wed, Sep 24, 2014 at 10:13 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Wed, 24 Sep 2014 21:46:11 -0400:
 Hi,
 [snip]
 I suspect both endothermic and exothermic reactions occur even inside the
 tokamak, but on balance more exothermic reactions occur.
 
 Harry

 Endothermic reactions only happen when ingoing particles have enough
 kinetic
 energy to make the reaction happen. IOW they are not really endothermic
 when all
 energy sources are taken into account.

 However most nuclear reactions where a neutron transfers to an external
 proton
 to create deuterium would be genuinely endothermic, and thus would not
 occur,
 unless of course the proton had high kinetic energy to start with.

 Of course it's possible that this happens, but the reactions going the
 other way

 are going to outnumber them by many thousands to 1, because very few of
 the
 energetic protons created are immediately going to encounter another heavy
 nucleus before losing some energy to ionization, and of those that do
 immediately encounter another heavy nucleus, only a small percentage are
 going
 to produce D.


 
 Ok now lets return to condensed matter systems.

 
 All the nuclear
 reactions
  offer
 ed
 for reports of excess heat
 in such 
 systems 
 are suppose to be
  theoretically
 impossible.
 
 Since we are dealing in impossibilities from the outset, it seems like
 false logic to argue that the probability of endothermic reactions
 is improbable.

 Harry



That should read:

Ok now lets return to condensed matter systems.

All the nuclear _explanations_ offered for reports of excess heat in such
systems are supose to theoretically impossible.

Since we are dealing in impossibilities from the outset, it seems like
false logic to argue that the probability of endothermic reactions
is improbable.

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-23 Thread Axil Axil

I found it.

https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/36817/SimulationNuclearTransmutationPresentation.pdf?sequence=2



On Mon, Sep 22, 2014 at 11:48 PM, Eric Walker eric.wal...@gmail.com wrote:

 On Mon, Sep 22, 2014 at 9:26 AM, Axil Axil janap...@gmail.com wrote:

 If you look at the ICCF-18 transmutation study of nickel and palladium
 study by Cook, you will see that Mizuno shows the same isotopic shifts in
 nickel that DGT shows. Ni61 does not participate in the reaction but all
 other isotopes of nickel do.


 I'm having trouble finding the transmutation study by Cook.  I have found
 this:

 http://iccf18.research.missouri.edu/files/Poster/Cook.pdf

 Is Cook's study on transmutations in nickel and palladium available
 online?  I take it that it is a summary of experiments and not a set of ab
 initio calculations?

 Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-23 Thread Roarty, Francis X

In reply to  Jones Beene's message of Mon, 22 Sep 2014 16:32:02 -0700:  [snip] 
The usual lame rationalizations we have used is that the energy was
borrowed in advance to overcome the Coulomb barrier or shed in advance to
achieve the redundancy ...[/snip] 

IMHO the  lock step motion of gas atoms in a loaded lattice provide the bank 
that loans energy to the reaction. I see this as ZPE underpinning where the 
geometry organizes random motion even extending down into the smaller NAE 
region where the very local defects in the lattice oppose the quantum forces of 
cavity  geometry all occurring unbeknownst to the gas atom who always sees his 
local environment full of virtual particles unchanged and simply responds to 
HUP to derive his normal random motion much like local and global geometry 
establishes local and regional weather patterns.

My point is that I agree it may not be fusion but disagree wrt there not being 
a source of energy to borrow.. the lock step motion of that many atoms 
represent a bank or a hammer and the closed cavity of the NAE is the loan 
seeking anvil.

Fran

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-23 Thread Axil Axil

Fran:

As is have posted many times, it is hard to tell what emerges from what.
What comes first the chicken of the egg.

Did you know that the Casmir force and zero point energy can be completely
controlled by polariton condensation.

*New regime in the Casimir force observed*

http://phys.org/news/2013-12-regime-casimir.html

and

*Dymamical Casimir emission from polariton condensates*

http://arxiv.org/abs/1303.1027

The Casmir effect can be eliminated or greatly amplified by polariton
condensation. This new understanding in among the new hot topics in cavity
physics.

On Tue, Sep 23, 2014 at 6:56 AM, Roarty, Francis X 
francis.x.roa...@lmco.com wrote:

 In reply to  Jones Beene's message of Mon, 22 Sep 2014 16:32:02 -0700:
 [snip] The usual lame rationalizations we have used is that the energy was
 borrowed in advance to overcome the Coulomb barrier or shed in advance to
 achieve the redundancy ...[/snip]

 IMHO the  lock step motion of gas atoms in a loaded lattice provide the
 bank that loans energy to the reaction. I see this as ZPE underpinning
 where the geometry organizes random motion even extending down into the
 smaller NAE region where the very local defects in the lattice oppose the
 quantum forces of cavity  geometry all occurring unbeknownst to the gas
 atom who always sees his local environment full of virtual particles
 unchanged and simply responds to HUP to derive his normal random motion
 much like local and global geometry establishes local and regional weather
 patterns.

 My point is that I agree it may not be fusion but disagree wrt there not
 being a source of energy to borrow.. the lock step motion of that many
 atoms represent a bank or a hammer and the closed cavity of the NAE is the
 loan seeking anvil.

 Fran

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-23 Thread Roarty, Francis X

Thanks Axil, good citation and not surprising – I have been following this near 
field region since day one,  convinced that there is relativistic magic 
occurring as you approach the focal limits described by Liptschitz et all where 
the quantum forces from the solid geometry can initially focus. The author is 
calling for further research in light of data that doesn’t agree with our 
present models, instead the data suggests the force falls much faster with 
separation for this configuration[corrugated plane / sphere]. The geometry is 
not the standard  perpendicular plate arrangement. I assume he applied the 
appropriate Casimir formulas for this geometry but the data – by disagreeing 
with calculated value  indicates the better documented case for parallel plates 
is just coincidental [perhaps some missing parameter that cancels for the 
parallel case]. We knew that pv1=pv2 for a long time before the parameters were 
contained and we realized pv1/t1=pv2/t2 and the era of refrigeration was born. 
My gut still screams relativistic effects even though there is no dv/dt 
approaching C there is the intriguing possibility that the longer 
wavelength/larger virtual particles supposedly restricted from occurring 
between boundaries of smaller separation are actually still able to occupy the 
region by shrinking from our perspective such that any gas atoms present also 
seem to shrink [ie hydrino /fh]. My bet is that one day some math guy will 
prove a relationship between the Casimir formula and the dilation formula made 
famous in the Twin Paradox.  I firmly believe these f/h can actually slip 
between separations we perceive as  less than atomic orbital diameters because 
they are temporally displaced and still riding that focal distance between 
solid boundaries because they perceive a spatial separation due to 
dilation/contraction [ the fractional hydrogen is temporally displaced without 
the need for near C velocity]
Fran


From: Axil Axil [mailto:janap...@gmail.com]
Sent: Tuesday, September 23, 2014 11:59 AM
To: vortex-l
Subject: EXTERNAL: Re: [Vo]:Mizuno, Rossi  copper transmutation

Fran:

As is have posted many times, it is hard to tell what emerges from what. What 
comes first the chicken of the egg.

Did you know that the Casmir force and zero point energy can be completely 
controlled by polariton condensation.

New regime in the Casimir force observed

http://phys.org/news/2013-12-regime-casimir.html

and

Dymamical Casimir emission from polariton condensates

http://arxiv.org/abs/1303.1027

The Casmir effect can be eliminated or greatly amplified by polariton 
condensation. This new understanding in among the new hot topics in cavity 
physics.

On Tue, Sep 23, 2014 at 6:56 AM, Roarty, Francis X 
francis.x.roa...@lmco.commailto:francis.x.roa...@lmco.com wrote:
In reply to  Jones Beene's message of Mon, 22 Sep 2014 16:32:02 -0700:  [snip] 
The usual lame rationalizations we have used is that the energy was
borrowed in advance to overcome the Coulomb barrier or shed in advance to
achieve the redundancy ...[/snip]

IMHO the  lock step motion of gas atoms in a loaded lattice provide the bank 
that loans energy to the reaction. I see this as ZPE underpinning where the 
geometry organizes random motion even extending down into the smaller NAE 
region where the very local defects in the lattice oppose the quantum forces of 
cavity  geometry all occurring unbeknownst to the gas atom who always sees his 
local environment full of virtual particles unchanged and simply responds to 
HUP to derive his normal random motion much like local and global geometry 
establishes local and regional weather patterns.

My point is that I agree it may not be fusion but disagree wrt there not being 
a source of energy to borrow.. the lock step motion of that many atoms 
represent a bank or a hammer and the closed cavity of the NAE is the loan 
seeking anvil.

Fran

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-23 Thread mixent

In reply to  Eric Walker's message of Mon, 22 Sep 2014 22:42:04 -0700:
Hi Eric,

On the face of it this sounds reasonable, but real life is seldom so simple.
Some deuterons will bounce off lattice nuclei in elastic collisions and head off
in completely different directions, so I would expect at least some reactions to
produce protons that end up injected into the lattice. Note that at the atomic
level material surfaces are often rough.

On Mon, Sep 22, 2014 at 10:33 PM, mix...@bigpond.com wrote:

...but wouldn't you expect 1/2 to fly away from the surface, and half to fly
 into it?


I would expect there to be an anisotropy.  As I envision it, there's an
electric arc pulling a mass of protons into a recess.  For a fraction of a
moment, the pressure is astronomical.  During this brief moment a deuteron
(the smaller species are all ionized within the arc) is forced up against a
lattice site, coming from the direction of the open area and the current
towards the wall of the substrate.  Unless there's some kind of rotation
during the moment of contact, if the lattice site is on the left and the
deuteron is coming from the right to the left, I would expect the daughter
proton to push off of the daughter nickel and be expelled back out to the
right, which is the open area.  I assume this would all happen too quickly
for any kind of rotation of the nickel/deuteron system.

Eric
Regards,

Robin van Spaandonk

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

RE: [Vo]:Mizuno, Rossi copper transmutation

I've looked through the isotope charts again - searching for reactions that
rapidly decay back to the starting element or to any stable isotope which
has already been reported to be there, and have not found any other
possibility...

...other than Ni58 (d,Cu59) - Ni60  which happens by EC or positron
emission, with a half-life of 20 minutes or so, and which fits the facts as
reported in the most robust experiments (Rossi, DGT, Thermacore, Mills).

1)  No or few gamma
2)  No or little radioactive ash
3)  No tritium, helium or positron annihilation
4)  No or little bremsstrahlung 
5)  Excess energy which is at least 1000 times more than chemical

Since nickel absorbs a deuteron and decays back to nickel in minutes, with
low energy release, this reaction fits the bill. You may be thinking - what
about the positron (beta positive) decay? No problem there, since nuclei
which decay by positron emission also decay by electron capture in a known
branching ratio which is dependant on the net energy of reaction.

According to wiki-the-wonderful, in low-energy decays, electron capture is
energetically favored by reactions below 1.022 MeV. The final state will
have an electron added or a positron removed - and so the energy released is
determinative of what can happen in the branching. As the energy of the
decay goes up, so does the branching ratio towards positron emission.
However, if the energy difference is low, then positron emission cannot
occur, and electron capture is the sole decay mode. This would seem to be
ready-made for the DDDL or deuteron-deep-Dirac-level species, which uses
its tight electron for more than one purpose and probably reduces the net
energy of the reaction as well.

This still leaves spin conservation as the major problem. The end products
of this reaction would be Ni60, and the starting nickel would be Ni58, so
that is no problem. Both are spin 0.

But the intermediary isotope, with short half-life would be Cu60 which is
spin 2+ and the deuterium can only add is 1+ spin, and the EC electron
another ½ spin. This over-simplification of spin issues - probably means
that the reaction can only happen if a neutrino is captured, or else the
inherent spin deficit decreases the half-life even more than its short
nature. Probably the neutrino.

Best of all - as a general working hypothesis which would make this relevant
to LENR but is not expected to be seen anywhere else (which explains why it
is not documented in the physics literature, as of now) there is NO other
isotope in the periodic table (other than Ni58) - which is both a proton
conductor and demonstrably neutron-deficient ! (the proof of that being that
Ni-58 is lower amu than the preceding lower Z element (cobalt-59). That's
right it is a perfect storm scenario. If this evolving explanation is
correct, it will be seen nowhere else in the periodic table, since it
demands conditions which do not exist anywhere else.

This means, anthropomorphically speaking - that Ni58 desperately wants
two more neutrons, and to get them, it essentially steals from its
surroundings, whenever a deuteron comes too close... especially a DDDL.

Falsifiability? Yes, this is falsifiable in three different way, which is a
big advantage. Give me a working Rossi reactor :-) and a few months: if the
[Ni-Ni] explanation is true, if will be proved beyond all reasonable
doubt. 

P.S. do I get to keep the reactor?
_

One more thing to add ... wrt the overdue suggestion (Doh,
slaps forehead) that Rossi's secret sauce is looking like it is deuterium.
Thank you, Clean Planet.

The reaction would probably work best if it is started with
regular hydrogen, and then deuterium is added later. This is because the
exchange reaction between hydrogen and deuterium itself is so robust. In
fact, many of the early critics of LENR thought that the entire phenomenon
could be related to deuterium exchange. It is that energetic.

As we know, Rossi has this mysterious system - which he
calls cat-and-mouse. He has been intentionally vague on how it functions.
Yet in reappraisal, this system is fully consistent with having two
chambers, the main one containing hydrogen and the nickel reactant - and the
smaller one deuterium (or a mix of H and D). The metering response can be
simply by voltage to a window, since deuterium will diffuse through many
proton conductors in direct proportion to negative charge. Positive charge
stops the diffusion, which is easily controllable by a sensor.

The purpose of the small chamber (mouse) is to meter D into
the main chamber at a controlled rate, to avoid a runaway. If Rossi can be
believed, he suffered several runaways with the HotCat which we can imagine
did not have this kind of metering device.

This seems to fit into everything we know, so long as one
ignores Rossi's own

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Arnaud Kodeck

Jones,

Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a spin
0.

_
From: Jones Beene [mailto:jone...@pacbell.net] 
Sent: lundi 22 septembre 2014 17:34
To: vortex-l@eskimo.com
Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation

Typo- the suggested reaction is Ni58 + D - Cu60 - Ni60

_

I've looked through the isotope charts again -
searching for reactions that rapidly decay back to the starting element or
to any stable isotope which has already been reported to be there, and have
not found any other possibility...

...other than Ni58 (d,Cu59) - Ni60  which
happens by EC or positron emission, with a half-life of 20 minutes or so,
and which fits the facts as reported in the most robust experiments (Rossi,
DGT, Thermacore, Mills).

1)  No or few gamma
2)  No or little radioactive ash
3)  No tritium, helium or positron annihilation
4)  No or little bremsstrahlung 
5)  Excess energy which is at least 1000 times more than chemical

Since nickel absorbs a deuteron and decays back to
nickel in minutes, with low energy release, this reaction fits the bill. You
may be thinking - what about the positron (beta positive) decay? No problem
there, since nuclei which decay by positron emission also decay by electron
capture in a known branching ratio which is dependant on the net energy of
reaction.

According to wiki-the-wonderful, in low-energy
decays, electron capture is energetically favored by reactions below 1.022
MeV. The final state will have an electron added or a positron removed - and
so the energy released is determinative of what can happen in the branching.
As the energy of the decay goes up, so does the branching ratio towards
positron emission. However, if the energy difference is low, then positron
emission cannot occur, and electron capture is the sole decay mode. This
would seem to be ready-made for the DDDL or deuteron-deep-Dirac-level
species, which uses its tight electron for more than one purpose and
probably reduces the net energy of the reaction as well.

This still leaves spin conservation as the major
problem. The end products of this reaction would be Ni60, and the starting
nickel would be Ni58, so that is no problem. Both are spin 0.

But the intermediary isotope, with short half-life
would be Cu60 which is spin 2+ and the deuterium can only add is 1+ spin,
and the EC electron another ½ spin. This over-simplification of spin issues
- probably means that the reaction can only happen if a neutrino is
captured, or else the inherent spin deficit decreases the half-life even
more than its short nature. Probably the neutrino.

Best of all - as a general working hypothesis which
would make this relevant to LENR but is not expected to be seen anywhere
else (which explains why it is not documented in the physics literature, as
of now) there is NO other isotope in the periodic table (other than Ni58) -
which is both a proton conductor and demonstrably neutron-deficient ! (the
proof of that being that Ni-58 is lower amu than the preceding lower Z
element (cobalt-59). That's right it is a perfect storm scenario. If this
evolving explanation is correct, it will be seen nowhere else in the
periodic table, since it demands conditions which do not exist anywhere
else.

This means, anthropomorphically speaking - that
Ni58 desperately wants two more neutrons, and to get them, it essentially
steals from its surroundings, whenever a deuteron comes too close...
especially a DDDL.

Falsifiability? Yes, this is falsifiable in three
different way, which is a big advantage. Give me a working Rossi reactor :-)
and a few months: if the [Ni-Ni] explanation is true, if will be proved
beyond all reasonable doubt. 

P.S. do I get to keep the reactor?

_

One more thing to add ... wrt the overdue
suggestion (Doh, slaps forehead) that Rossi's secret sauce is looking like
it is deuterium. Thank you, Clean Planet.

The reaction would probably work best if it
is started with regular hydrogen, and then deuterium is added later. This is
because the exchange reaction between hydrogen and deuterium itself is so
robust. In fact, many of the early critics of LENR thought that the entire
phenomenon could be related to deuterium exchange. It is that energetic.

As we know, Rossi has this mysterious system
- which he calls cat-and-mouse. He has been intentionally vague

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Axil Axil

If you look at the ICCF-18 transmutation study of nickel and palladium
study by Cook, you will see that Mizuno shows the same isotopic shifts in
nickel that DGT shows. Ni61 does not participate in the reaction but all
other isotopes of nickel do.

Sorry, that link to this reference is broken.

On Mon, Sep 22, 2014 at 12:01 PM, Arnaud Kodeck arnaud.kod...@lakoco.be
wrote:

 Jones,

 Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60 has a
 spin
 0.

 _
 From: Jones Beene [mailto:jone...@pacbell.net]
 Sent: lundi 22 septembre 2014 17:34
 To: vortex-l@eskimo.com
 Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation

 Typo- the suggested reaction is Ni58 + D - Cu60 - Ni60

 _

 I've looked through the isotope charts again -
 searching for reactions that rapidly decay back to the starting element or
 to any stable isotope which has already been reported to be there, and have
 not found any other possibility...

 ...other than Ni58 (d,Cu59) - Ni60  which
 happens by EC or positron emission, with a half-life of 20 minutes or so,
 and which fits the facts as reported in the most robust experiments (Rossi,
 DGT, Thermacore, Mills).

 1)  No or few gamma
 2)  No or little radioactive ash
 3)  No tritium, helium or positron annihilation
 4)  No or little bremsstrahlung
 5)  Excess energy which is at least 1000 times more than chemical

 Since nickel absorbs a deuteron and decays back to
 nickel in minutes, with low energy release, this reaction fits the bill.
 You
 may be thinking - what about the positron (beta positive) decay? No problem
 there, since nuclei which decay by positron emission also decay by electron
 capture in a known branching ratio which is dependant on the net energy of
 reaction.

 According to wiki-the-wonderful, in low-energy
 decays, electron capture is energetically favored by reactions below 1.022
 MeV. The final state will have an electron added or a positron removed -
 and
 so the energy released is determinative of what can happen in the
 branching.
 As the energy of the decay goes up, so does the branching ratio towards
 positron emission. However, if the energy difference is low, then positron
 emission cannot occur, and electron capture is the sole decay mode. This
 would seem to be ready-made for the DDDL or deuteron-deep-Dirac-level
 species, which uses its tight electron for more than one purpose and
 probably reduces the net energy of the reaction as well.

 This still leaves spin conservation as the major
 problem. The end products of this reaction would be Ni60, and the starting
 nickel would be Ni58, so that is no problem. Both are spin 0.

 But the intermediary isotope, with short half-life
 would be Cu60 which is spin 2+ and the deuterium can only add is 1+ spin,
 and the EC electron another ½ spin. This over-simplification of spin issues
 - probably means that the reaction can only happen if a neutrino is
 captured, or else the inherent spin deficit decreases the half-life even
 more than its short nature. Probably the neutrino.

 Best of all - as a general working hypothesis which
 would make this relevant to LENR but is not expected to be seen anywhere
 else (which explains why it is not documented in the physics literature, as
 of now) there is NO other isotope in the periodic table (other than Ni58) -
 which is both a proton conductor and demonstrably neutron-deficient ! (the
 proof of that being that Ni-58 is lower amu than the preceding lower Z
 element (cobalt-59). That's right it is a perfect storm scenario. If this
 evolving explanation is correct, it will be seen nowhere else in the
 periodic table, since it demands conditions which do not exist anywhere
 else.

 This means, anthropomorphically speaking - that
 Ni58 desperately wants two more neutrons, and to get them, it essentially
 steals from its surroundings, whenever a deuteron comes too close...
 especially a DDDL.

 Falsifiability? Yes, this is falsifiable in three
 different way, which is a big advantage. Give me a working Rossi reactor
 :-)
 and a few months: if the [Ni-Ni] explanation is true, if will be proved
 beyond all reasonable doubt.

 P.S. do I get to keep the reactor?

 _

 One more thing to add ... wrt the overdue
 suggestion (Doh, slaps forehead) that Rossi's secret sauce is looking
 like
 it is deuterium. Thank you, Clean Planet.

 The reaction would probably work best if it
 is started with regular hydrogen, and then deuterium is added later

RE: [Vo]:Mizuno, Rossi copper transmutation


From: Arnaud Kodeck 

Jones,

Why not consider also the Ni58 + 2p - Zn60 - Cu60 - Ni60? Zn60
has a spin 0.

_
the suggested reaction is Ni58 + D - Cu60 - Ni60

Arnaud, This would be a three body reaction, no? 

You may be suggesting this reaction - in the event that Rossi does not use
deuterium. That is wise to consider - since he professes not to, despite a
tank of it being seen in his lab, early on. 

There is an even better possibility when two protons densified as a DDL
molecule, and would act like the two needed neutrons, to make this reaction
work. 

If my understanding is correct, nickel-58 is active ONLY because it is
neutron-deficient, and the two protons do not help the immediate situation,
at least not on the surface - even if both protons decay to neutrons,
eventually. However, all bets are off with the DDL, since it allows the
protons to look like virtual neutrons.

There is nothing out there, which fits all of the parameters seamlessly, so
in the end - we need reliable data. But it looks like we are framing a
workable situation with enough variable to accommodate either D, H or H+D as
the active gases. 

In short, your suggestion may work well - an especially if Rossi uses
hydrogen only, and even more so - if the signature of the DDL formation
(soft x-ray) is documented.

Jones


attachment: winmail.dat

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Arnaud Kodeck

Yes, in my view, the DDL state diatomic hydrogen (shrunken hydrogen) reacts
with Ni58. Should both atoms be in shrunken state? Is the DDL atoms small
enough to go in the lattice?

We can consider as well with pD or DD DDL state if natural hydrogen is used.

_
From: Jones Beene [mailto:jone...@pacbell.net] 
Sent: lundi 22 septembre 2014 18:28
To: vortex-l@eskimo.com
Subject: RE: [Vo]:Mizuno, Rossi  copper transmutation

From: Arnaud Kodeck 

Jones,

Why not consider also the Ni58 + 2p - Zn60 - Cu60
- Ni60? Zn60 has a spin 0.

_
the suggested reaction is Ni58 + D - Cu60
- Ni60

Arnaud, This would be a three body reaction, no? 

You may be suggesting this reaction - in the event that Rossi does
not use deuterium. That is wise to consider - since he professes not to,
despite a tank of it being seen in his lab, early on. 

There is an even better possibility when two protons densified as a
DDL molecule, and would act like the two needed neutrons, to make this
reaction work. 

If my understanding is correct, nickel-58 is active ONLY because it
is neutron-deficient, and the two protons do not help the immediate
situation, at least not on the surface - even if both protons decay to
neutrons, eventually. However, all bets are off with the DDL, since it
allows the protons to look like virtual neutrons.

There is nothing out there, which fits all of the parameters
seamlessly, so in the end - we need reliable data. But it looks like we are
framing a workable situation with enough variable to accommodate either D, H
or H+D as the active gases. 

In short, your suggestion may work well - an especially if Rossi
uses hydrogen only, and even more so - if the signature of the DDL formation
(soft x-ray) is documented.

Jones

attachment: winmail.dat

RE: [Vo]:Mizuno, Rossi copper transmutation

_
From: Arnaud Kodeck 

Yes, in my view, the DDL state diatomic hydrogen (shrunken hydrogen)
reacts with Ni58. Should both atoms be in shrunken state? 

Yes, that would seem to be highly beneficial. The reaction looks less like a
three-body reaction if it happens with a tight DDL molecule, which is
possibly in the 10s of Fermi size range.

Is the DDL small enough to go in the lattice?

Easily. The beauty of nickel as the host - from the Rydberg orbital
viewpoint, is that it has two orbitals which are located at very good match
in energy level for the correct hole, both of which are in its valence
band at IP5 and IP6 ! This would essentially permit the DDL molecule, which
has two electrons of a set Rydberg value, to find stability inside the shell
- by replacing two normal electrons of nickel at a moderately deep level.

Very few proton conductors have two deep orbitals which are adjoining in
Rydberg values. Curiously, cobalt and iron are the others. This means any
ferromagnetic material could be substituted for some of the nickel and host
the DDL. From there, the excursion of the DDL to the nucleus on an
occasional basis would seem to be highly favored. Only nickel has the
neutron deficient isotope, however.

As you can see, this is a mix of Mills CQM, the DDL version of other
theorists and a few new additions. It would not require that excess energy
is given up in shrinkage, as does Mills theory, since the progression goes
to fusion eventually, which never happens according to CQM. This version
does borrow the idea that the orbital electrons must have Rydberg values
if they are to give up a proper hole for substitution (with the two
electrons of the DDL). 

But AFAIK - Mills has not recognized the novelty of this suggestion, which
is that adjoining holes in the valence shell of a ferromagnetic element
like nickel, is the special parameter for LENR. Why would he? ...since he
denies LENR is real, it has not occurred to him. In fact he uses other
metals besides nickel these days instead in his own experiments. Maybe he
intentionally avoids nickel :-) 

Jones
attachment: winmail.dat

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread mixent

In reply to  Jones Beene's message of Mon, 22 Sep 2014 08:33:43 -0700:
Hi,
Typo- the suggested reaction is  - Ni60
[snip]
Ni58 + D - Cu60 + 11.252 MeV

Normally one would expect prompt gammas from this reaction totaling 11.25 MeV.
If no gammas are detected, what do you propose happens to the energy?

Regards,

Robin van Spaandonk

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

RE: [Vo]:Mizuno, Rossi copper transmutation

The usual lame rationalizations we have used is that the energy was
borrowed in advance to overcome the Coulomb barrier or shed in advance to
achieve the redundancy ...

But you're right - fusion numbers simply don't work well for the reality of
a Rossi type reaction, as there is too much excess energy to hide... and
this rationalization is no better than the factionalized gamma.

So there you have it... back to no-fusion in LENR it is...

-Original Message-
From: mix...@bigpond.com 
Hi,
Typo- the suggested reaction is  - Ni60
[snip]
Ni58 + D - Cu60 + 11.252 MeV

Normally one would expect prompt gammas from this reaction totaling 11.25
MeV.
If no gammas are detected, what do you propose happens to the energy?

Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread mixent

In reply to  Jones Beene's message of Mon, 22 Sep 2014 16:32:02 -0700:
Hi,
[snip]
The usual lame rationalizations we have used is that the energy was
borrowed in advance to overcome the Coulomb barrier or shed in advance to
achieve the redundancy ...

But you're right - fusion numbers simply don't work well for the reality of
a Rossi type reaction, as there is too much excess energy to hide... and
this rationalization is no better than the factionalized gamma.

So there you have it... back to no-fusion in LENR it is...

I would still be inclined to consider reactions that produce heavy charged
particles. The heavier and slower, the better. E.g. fusion/fission reactions.

I think the secondary gammas from heavily charged slow moving daughter nuclei
might have been shielded.

One possibility is the p-B11 reaction which produces quite low energy alphas
because there are three of them. Furthermore the double charge on the alpha
particles means both rapid energy loss to electrons, and strong repulsion from
other nuclei, thus strongly reducing the chances of creating secondary gammas.
Also, if such a reaction were to occur within the mass of the Boron, then the
short range of the alphas would mostly keep them in the Boron itself, and the
lowest excited state of B11 is 2.1247 MeV. This is not much less than the energy
of the alphas, so in order to produce any secondary gammas at all, such an alpha
would need to collide directly with another B11 nucleus almost immediately after
creation, i.e. before it lost too much energy through ionization. That thus
works to reduce the intensity of any secondary radiation.

(B10 OTOH has excited states at 718  740 keV, but then B10 is only 20% of
natural Boron).
Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Eric Walker

On Mon, Sep 22, 2014 at 9:26 AM, Axil Axil janap...@gmail.com wrote:

If you look at the ICCF-18 transmutation study of nickel and palladium
 study by Cook, you will see that Mizuno shows the same isotopic shifts in
 nickel that DGT shows. Ni61 does not participate in the reaction but all
 other isotopes of nickel do.


I'm having trouble finding the transmutation study by Cook.  I have found
this:

http://iccf18.research.missouri.edu/files/Poster/Cook.pdf

Is Cook's study on transmutations in nickel and palladium available
online?  I take it that it is a summary of experiments and not a set of ab
initio calculations?

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Eric Walker

On Mon, Sep 22, 2014 at 8:01 PM, mix...@bigpond.com wrote:

I would still be inclined to consider reactions that produce heavy charged
 particles. The heavier and slower, the better. E.g. fusion/fission
 reactions.


The reactions I've been looking at recently have charged particles as
daughters as well.  But the daughters are generally protons in the 5-10 MeV
range.  The way I propose that gammas from excited nuclei are avoided is to
suggest that the reactions occur at the surface and that the daughters fly
out from the surface:

+ d

+++ p p p

+++ p p p d p p
++ p --- p d p
+++ p p d p p p

+++ p p p

+ p


Here the (+)'s are nickel lattice sites.  The p results from an Ni(d,p)Ni
reaction.  The arrow represents the momentum.  Although the p is born with
~ 5-10 MeV of energy, it burrows into the other p's at the surface, quickly
thermalizing to a much lower energy.  Occasionally there is a d that is
broken apart through spallation.  This wouldn't happen very often with a
normal hydrogen mix, because there are only ~ 1/6000 parts deuterium, and
only a fraction of these would be encountered (and only a fraction of the
neutrons resulting from such spallations would exit the system).

I think the secondary gammas from heavily charged slow moving daughter
 nuclei
 might have been shielded.


By this I take it you mean gammas from lattice sites excited through
inelastic collisions?

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread mixent

In reply to  Eric Walker's message of Mon, 22 Sep 2014 21:08:59 -0700:
Hi Eric,
[snip]
On Mon, Sep 22, 2014 at 8:01 PM, mix...@bigpond.com wrote:

I would still be inclined to consider reactions that produce heavy charged
 particles. The heavier and slower, the better. E.g. fusion/fission
 reactions.


The reactions I've been looking at recently have charged particles as
daughters as well.  But the daughters are generally protons in the 5-10 MeV
range.  The way I propose that gammas from excited nuclei are avoided is to
suggest that the reactions occur at the surface and that the daughters fly
out from the surface:

...but wouldn't you expect 1/2 to fly away from the surface, and half to fly
into it?


+ d

+++ p p p

+++ p p p d p p
++ p --- p d p
+++ p p d p p p

+++ p p p

+ p


Here the (+)'s are nickel lattice sites.  The p results from an Ni(d,p)Ni
reaction.  The arrow represents the momentum.  Although the p is born with
~ 5-10 MeV of energy, it burrows into the other p's at the surface, quickly
thermalizing to a much lower energy.  Occasionally there is a d that is
broken apart through spallation.  This wouldn't happen very often with a
normal hydrogen mix, because there are only ~ 1/6000 parts deuterium, and
only a fraction of these would be encountered (and only a fraction of the
neutrons resulting from such spallations would exit the system).

I think the secondary gammas from heavily charged slow moving daughter
 nuclei
 might have been shielded.


By this I take it you mean gammas from lattice sites excited through
inelastic collisions?

Yes, that's what I meant.

Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-22 Thread Eric Walker

On Mon, Sep 22, 2014 at 10:33 PM, mix...@bigpond.com wrote:

...but wouldn't you expect 1/2 to fly away from the surface, and half to fly
 into it?


I would expect there to be an anisotropy.  As I envision it, there's an
electric arc pulling a mass of protons into a recess.  For a fraction of a
moment, the pressure is astronomical.  During this brief moment a deuteron
(the smaller species are all ionized within the arc) is forced up against a
lattice site, coming from the direction of the open area and the current
towards the wall of the substrate.  Unless there's some kind of rotation
during the moment of contact, if the lattice site is on the left and the
deuteron is coming from the right to the left, I would expect the daughter
proton to push off of the daughter nickel and be expelled back out to the
right, which is the open area.  I assume this would all happen too quickly
for any kind of rotation of the nickel/deuteron system.

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Eric Walker

On Sat, Sep 20, 2014 at 10:42 AM, Jones Beene jone...@pacbell.net wrote:

Secondly, and most importantly - the neutron of the deuteron offers Coulomb
 shielding.


Can you elaborate on this?  I would have expected the neutron to be more or
less invisible, as far as the Coulomb field is concerned.

Eric

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Jones Beene

From: Eric Walker 

 

Secondly, and most importantly - the neutron of the deuteron offers Coulomb 
shielding.

 

Can you elaborate on this?  I would have expected the neutron to be more or 
less invisible, as far as the Coulomb field is concerned.

 

Eric

 

Of course, the neutron looks to be neutral – from a distance, so it's not 
generally expected to shield anything massively in the same way that magnetism 
is shielded by a high mu metal or a dielectric shields against an electrostatic 
field. But both of those mechanisms may offer a clue.

 

Unless, of course, the neutron has a significant dipole moment or a significant 
near-field charge, so that is aligns geometrically between the proton and the 
approaching nickel nucleus – to partially shield in the same way a dielectric 
would.

 

The strong force is so much stronger than electrical repulsion, that any small 
effect can make a difference at close range.

 

This may an open matter, as to whether current theory suggests the dipole 
moment of the neutron may be nonzero (does anyone know for sure?) but the 
neutron is reported to have a negative near-field, which means that it could 
provide some degree of shielding effect – certainly the electrostatic field 
lines are much different than with a bare proton. I do not have a handy 
reference for the proof of a neutron near-field, so this should be double 
checked.

 

Jones

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Eric Walker

On Sun, Sep 21, 2014 at 8:35 AM, Jones Beene jone...@pacbell.net wrote:

The strong force is so much stronger than electrical repulsion, that any
 small effect can make a difference at close range.


If the possible Coulomb shielding effect from the neutron works at the same
range as the strong force (i.e., is quite short-range), then I think the
deuterons needed for the proposed bosonic deuteron capture will require a
kinetic energy that is on the same order as that for neutron stripping via
the Oppenheimer-Phillips process.

Eric

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Jones Beene

Eric,

 

It really gets down to whether the gainful reaction is thermonuclear or quantum 
mechanical. If you look in the archives, “stripping” was favored by me for many 
years, and I first introduced it here - but opinions change. 

 

Your opinion may not change, but here is what convinced me that there is a 
better model than stripping.

 

If the gain is QM based – which is to say a type of nuclear tunneling which is 
different than electron tunneling in semiconductors, then bosons are highly 
favored to begin with. Horace Heffner use to talk about a QM “slingshot” effect 
where the positive end (of a cold deuteron) enters the electron cloud of the 
nickel. This could work with a proton as well but the dynamics change at some 
point since the proton can go only as deep as the inner electron orbitals, to 
which it is strongly attracted.

 

Because the deuteron is shaped like a barbell, and has a positive end with long 
separation to the neutron end, which has a negative near-field, the deuteron 
can be whipped around in a spinning motion like a slingshot by the inner 
orbital.

 

Most of the time the positive end of the deuteron is eventually repelled by the 
heavy nucleus, as the proton is all of the time - but on occasion the 
neutron-end of the deuteron is aligned perfectly to lead the way into the 
femtometer geometry of the strong force. The probability of this precise 
geometry is low, but the transaction rate is high.

 

From: Eric Walker 

 

The strong force is so much stronger than electrical repulsion, that any small 
effect can make a difference at close range.

 

If the possible Coulomb shielding effect from the neutron works at the same 
range as the strong force (i.e., is quite short-range), then I think the 
deuterons needed for the proposed bosonic deuteron capture will require a 
kinetic energy that is on the same order as that for neutron stripping via the 
Oppenheimer-Phillips process.

 

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Eric Walker

On Sun, Sep 21, 2014 at 9:42 AM, Jones Beene jone...@pacbell.net wrote:

If you look in the archives, “stripping” was favored by me for many years,
 and I first introduced it here - but opinions change.


The first reference I saw to the OP process was from a thread between you
and Abd Lomax, in 2010, in which you appeared to have introduced the
possibility.  In these posts I give credit to you:

https://www.mail-archive.com/vortex-l@eskimo.com/msg92455.html
https://www.mail-archive.com/vortex-l@eskimo.com/msg92381.html


 If the gain is QM based – which is to say a type of nuclear tunneling
 which is different than electron tunneling in semiconductors, then bosons
 are highly favored to begin with.


Right now I like neutron stripping and the OP process.  In the past, in
approximate chronological order, I've argued for a kind of nano-Polywell;
an ill-conceived dipolariton-based bosonic fusion; Widom-Larsen; p+d fusion
in nickel without thought given to the gammas; hidden d+d fusion and
Pd-attenuated gammas; deuteron and/or proton capture in nickel;
non-equilibrium disruption of the electronic structure of the metal and
attending Coulomb screening; d+d fusion through z-pinch in electric arcs
together with a new kind of electromagnetic channel that short-circuits the
formation of gammas; and now OP and neutron stripping.  As I learn more
about the relevant physics and see insurmountable problems, I'm willing to
switch to a new hypothesis.  (I continue to take seriously some of the more
recent thought experiments even as I give attention to OP + neutron
stripping in the context of nickel.)

In this particular case it's not so much about arguing against something
that is QM based, in which spin is central, in favor of neutron
stripping.  I'm addressing an objection you raised earlier on in this
thread:

Note that stripping is closer to brute force thermodynamics, and unlikely
 to happen in condensed matter.


I'm saying that the same objection applies to the bosonic deuteron capture
reaction that you've proposed, because the neutron, as you have clarified,
will only screen at short distances.

Eric

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread Jones Beene

Eric - In the end, there’s nothing new under the sun and the best we can do
is try to get it right at least once along the pathway. 

One good thing about a long-running forum, with a heated give-and-take of
ideas - is that if you can grasp everyone’s position, even for a few hours,
and evolve your own thinking often enough with improvement over time, then
eventually … it should be possible to pick and choose among old posts and
find one that makes the writer look like a genius. :-)

The problem is in making that brilliant post the most recent one!

From: Eric Walker 

If you look in the archives, “stripping” was favored by me
for many years, and I first introduced it here - but opinions change.

The first reference I saw to the OP process was from a
thread between you and Abd Lomax, in 2010, in which you appeared to have
introduced the possibility.  In these posts I give credit to you:


https://www.mail-archive.com/vortex-l@eskimo.com/msg92455.html

https://www.mail-archive.com/vortex-l@eskimo.com/msg92381.html
 
If the gain is QM based – which is to say a type of nuclear
tunneling which is different than electron tunneling in semiconductors, then
bosons are highly favored to begin with.

Right now I like neutron stripping and the OP process.  In
the past, in approximate chronological order, I've argued for a kind of
nano-Polywell; an ill-conceived dipolariton-based bosonic fusion;
Widom-Larsen; p+d fusion in nickel without thought given to the gammas;
hidden d+d fusion and Pd-attenuated gammas; deuteron and/or proton capture
in nickel; non-equilibrium disruption of the electronic structure of the
metal and attending Coulomb screening; d+d fusion through z-pinch in
electric arcs together with a new kind of electromagnetic channel that
short-circuits the formation of gammas; and now OP and neutron stripping.
As I learn more about the relevant physics and see insurmountable problems,
I'm willing to switch to a new hypothesis.  (I continue to take seriously
some of the more recent thought experiments even as I give attention to OP +
neutron stripping in the context of nickel.)

In this particular case it's not so much about arguing
against something that is QM based, in which spin is central, in favor of
neutron stripping.  I'm addressing an objection you raised earlier on in
this thread:

Note that stripping is closer to brute force thermodynamics,
and unlikely to happen in condensed matter.

I'm saying that the same objection applies to the bosonic
deuteron capture reaction that you've proposed, because the neutron, as you
have clarified, will only screen at short distances.

Eric

attachment: winmail.dat

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread H Veeder

On Sat, Sep 20, 2014 at 10:59 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Sat, 20 Sep 2014 20:53:37 -0400:
 Hi,
 [snip]
 If hydrinos and deuterinos are both present, perhaps it is possible for
 the
 neutron stripping to work in two directions such that a deuterino can give
 up a neutron to a heavy nucleus and a heavy nucleus can give up a neutron
 to hydrino. ( I am thinking of a nuclear version of epicatalysis.)
 
 Harry
 A heavy nucleus won't give up a neutron to a Hydrino, because in doing so
 it
 would lose about 5-10 MeV, but only gain 2.2 MeV from the formation of the
 deuteron.



That means it is an endothermic reaction, but that doesn't mean it is
impossible.
I am not implying that neutron stripping should be discarded if the
reverse reaction is possible.

I
 
mentioned epicatalysis because
theoretical research on

the subject

was recently 
published in Physical Review E. Along with some empirical evidence the
research suggests that deviations
of practical significance 
from the 2nd law of law thermodynamics are possible
with epicatalysis
:

https://www.facebook.com/ParadigmEnergy/posts/249600938581128

Now the theory of epicatalysis is based on chemical activity, but I don't
see why the theory could not be broadened to include nuclear activity or
other unconventional high energy activity if a given heat anomaly is too
large to explain by just chemical activity.

A tacit assumption of CF/LENR research is that an anomalous thermal signal
will have practical significance if it results from the conversion of
potential energy into kinetic energy in a one way process. The assumption
holds whether the source of energy is nuclear or chemical or some other.
Consequently, measured temperature anomalies are suspect until they are
supported by additional calorimetry which yields a global temperature rise.
If this global temperature rise (excess heat signal) is not found, and
measurement error is ruled out, then the temperature anomaly will be
classified as a local fluctuation with no practical significance. This
interpretation of temperature signals is motivated by the demands of the
2nd law of thermodynamics.

However, if a process like epicatalysis is creating the temperature
anomalies then the methods used to measure an excess heat signal need to be
reconsidered. Detecting an excess heat signal ordinarily means looking for
a global temperature rise which requires that the source of an anomaly be
placed in a thermally closed environment since it is assumed the
temperature rise is based on the creation of kinetic energy from inside the
system. In contradistinction epicatalysis transfers energy from a lower
temperature region to a higher temperature region. If the purpose of the
enclosure is to detect a global temperature rise none will be found.

Harry

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread mixent

In reply to  H Veeder's message of Sun, 21 Sep 2014 17:35:34 -0400:
Hi,

Nuclear energies are 6 orders of magnitude larger than chemical energies, which
I would expect to reduce the chances to the point where it's not even worth
considering. 
However, that said, it should be noted that the same is not always true for
reactions where D is converted into T. 

e.g. the following reaction is exothermic:-

9Be+2H = 4He + 4He + 3H + 4.684 MeV


On Sat, Sep 20, 2014 at 10:59 PM, mix...@bigpond.com wrote:

 In reply to  H Veeder's message of Sat, 20 Sep 2014 20:53:37 -0400:
 Hi,
 [snip]
 If hydrinos and deuterinos are both present, perhaps it is possible for
 the
 neutron stripping to work in two directions such that a deuterino can give
 up a neutron to a heavy nucleus and a heavy nucleus can give up a neutron
 to hydrino. ( I am thinking of a nuclear version of epicatalysis.)
 
 Harry
 A heavy nucleus won't give up a neutron to a Hydrino, because in doing so
 it
 would lose about 5-10 MeV, but only gain 2.2 MeV from the formation of the
 deuteron.



?That means it is an endothermic reaction, but that doesn't mean it is
impossible?.
I am not implying that neutron stripping should be discarded ?if the
reverse reaction is possible.
?
I
? ?
mentioned epicatalysis because
?theoretical research on?

?the subject

?was recently ?
published in Physical Review E. Along with some empirical evidence the
research suggests that deviations
?of practical significance ?
from the 2nd law of law thermodynamics are possible
?with epicatalysis
?:?

https://www.facebook.com/ParadigmEnergy/posts/249600938581128

Now the theory of epicatalysis is based on chemical activity, but I don't
see why the theory could not be broadened to include nuclear activity or
other unconventional high energy activity if a given heat anomaly is too
large to explain by just chemical activity.

A tacit assumption of CF/LENR research is that an anomalous thermal signal
will have practical significance if it results from the conversion of
potential energy into kinetic energy in a one way process. The assumption
holds whether the source of energy is nuclear or chemical or some other.
Consequently, measured temperature anomalies are suspect until they are
supported by additional calorimetry which yields a global temperature rise.
If this global temperature rise (excess heat signal) is not found, and
measurement error is ruled out, then the temperature anomaly will be
classified as a local fluctuation with no practical significance. This
interpretation of temperature signals is motivated by the demands of the
2nd law of thermodynamics.

However, if a process like epicatalysis is creating the temperature
anomalies then the methods used to measure an excess heat signal need to be
reconsidered. Detecting an excess heat signal ordinarily means looking for
a global temperature rise which requires that the source of an anomaly be
placed in a thermally closed environment since it is assumed the
temperature rise is based on the creation of kinetic energy from inside the
system. In contradistinction epicatalysis transfers energy from a lower
temperature region to a higher temperature region. If the purpose of the
enclosure is to detect a global temperature rise none will be found.

Harry
Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-21 Thread mixent

In reply to  mix...@bigpond.com's message of Mon, 22 Sep 2014 08:30:04 +1000:
Hi,
[snip]
9Be+2H = 4He + 4He + 3H + 4.684 MeV

BTW, this reaction also works for H (but only just).

1H+9Be = 4He + 4He + 2H + 0.651 MeV

..and I suspect that 9Be is the only naturally occurring isotope for which it
will work.

Regards,

Robin van Spaandonk

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

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Jones Beene

One more thing to add ... wrt the overdue suggestion (Doh, slaps forehead)
that Rossi's secret sauce is looking like it is deuterium. Thank you,
Clean Planet.

The reaction would probably work best if it is started with regular
hydrogen, and then deuterium is added later. This is because the exchange
reaction between hydrogen and deuterium itself is so robust. In fact, many
of the early critics of LENR thought that the entire phenomenon could be
related to deuterium exchange. It is that energetic.

As we know, Rossi has this mysterious system - which he calls cat-and-mouse.
He has been intentionally vague on how it functions. Yet in reappraisal,
this system is fully consistent with having two chambers, the main one
containing hydrogen and the nickel reactant - and the smaller one deuterium
(or a mix of H and D). The metering response can be simply by voltage to a
window, since deuterium will diffuse through many proton conductors in
direct proportion to negative charge. Positive charge stops the diffusion,
which is easily controllable by a sensor.

The purpose of the small chamber (mouse) is to meter D into the main chamber
at a controlled rate, to avoid a runaway. If Rossi can be believed, he
suffered several runaways with the HotCat which we can imagine did not have
this kind of metering device.

This seems to fit into everything we know, so long as one ignores Rossi's
own denial of using deuterium. But deuterium is the one thing which, if true
- he would never admit to. That is, if Ni-D is indeed the essence of E-Cat,
in the same way that the change from palladium to nickel could be the
essence of the Mizuno reactor.

Things just keep getting curiouser and curiouser...
_

One interesting detail, in retrospect, about
Yoshino/Mizuno's MIT presentation and the switch to nickel (from palladium)
while keeping deuterium as the active gas may have been overlooked to date.
Apologies- if this slant on the underlying reaction has appeared before.

It is the copper connection. As we know, Focardi and Rossi
believed that the E-Cat is/was transmuting nickel into copper by fusing with
a proton. When one mentions a copper connection, seldom does Mizuno's
amazing new work come to mind. However, all reactions of nickel with a
proton result in a radioactive isotope with a half-life which is long enough
for it to have been seen. This kind of hot isotope is not reported in any
study of the Rossi reactor - but his proponents are hoping that the TIP2
report will find evidence of copper transmutation.

The same kind of signature radioactivity is not true with
deuterium as the active gas. In fact, the solution is so stunning - that we
have to wonder if Rossi may be using deuterium as his secret ingredient.
Terry will remember that in the very first image to come from Rossi, there
was a color-coded tank of deuterium in the Lab. Apparently it was not
intended to be noticed. When questioned about this later, Rossi glibly said
the purpose of D was to stop the reaction if it got out of hand ! 

With this new information... well... you can be the judge of
whether Rossi's excuse was ever true. Notably deuterium in never seen
again... 

Nickel 58 is the most abundant isotope of element 28, and as
recently mentioned is out-of-place in the periodic table, being lighter
than any stable cobalt isotope, the element to the left. By itself, that
factoid would be unique in that it only happens in one other place in the
entire periodic table, where elements routinely increase in average amu, in
step with z But wait there's more than relative lightness (putative
receptivity to nucleon addition).

Look at Copper-60 , the expected product of a deuteron
fusing to Ni-58. Cu60 has a short half-life and decays back to Ni60 in
minutes. It could escape detection in any reactor - so long as a reactor was
not opened for a few hours, since all one would see is a nickel isotope
which is expected to be there. The beta decay is fairly strong however.

The biggest problem with this scenario could be conservation
of spin. Ni58 is 0 spin, Cu60 is +2, and D is +1. A beta decay ostensibly
does not solve that problem. But the chance of this being the gainful
reaction in conjunction with nuclear spin-coupling as a predecessor is
otherwise worth looking at ways to get around conservation of spin.

This elegant possibility of a gainful reaction in which
stable nickel converts to stable nickel, giving up energy, is why my
prediction for the Mizuno presentation in November is to suggest that they
will see a relative decrease in Ni58 and a relative increase in Ni60.

The more intriguing idea is that Rossi has been using
deuterium all along in his E-Cat, but the only time the secret almost got
out was in the original demo !

Jones

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Eric Walker

On Sat, Sep 20, 2014 at 7:40 AM, Jones Beene jone...@pacbell.net wrote:

However, all reactions of nickel with a proton result in a
 radioactive isotope with a half-life which is long enough for it to have
 been seen. This kind of hot isotope is not reported in any study of the
 Rossi reactor - but his proponents are hoping that the TIP2 report will
 find
 evidence of copper transmutation.


I don't think anyone here has been advocating for proton capture for a
while.  Robin might still like the idea in connection with shrunken
hydrogen, for in that case the ejected electron can fill in for the gamma
and carry the momentum.  I've personally run with the idea of proton
capture in the past, but have stepped away from it.  Perhaps you're
referring to proponents in other forums?

This elegant possibility of a gainful reaction in which stable nickel
 converts to stable nickel, giving up energy, is why my prediction for the
 Mizuno presentation in November is to suggest that they will see a relative
 decrease in Ni58 and a relative increase in Ni60.


The nickel to nickel idea seems very promising.  I doubt there is deuteron
capture, because if there is deuteron capture, there is probably proton
capture as well, along with all of the nasty gammas.  This is what is
leading me to deuterium stripping -- e.g., 60Ni(d,p)61Ni.  Here the neutron
is stripped off of the deuteron and added to the nickel, and the proton
flies in the other direction, rather than there being a full capture.

Eric

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Eric Walker

On Sat, Sep 20, 2014 at 7:40 AM, Jones Beene jone...@pacbell.net wrote:

This elegant possibility of a gainful reaction in which stable nickel
 converts to stable nickel, giving up energy, is why my prediction for the
 Mizuno presentation in November is to suggest that they will see a relative
 decrease in Ni58 and a relative increase in Ni60.


As an alternative prediction, if there is deuteron stripping rather than
deuteron capture and then decay, one will see this:

58Ni + d → 59Ni + p
59Ni → 59Co + β+ ( 1 percent of the time)
β+ + β- → 2ɣ Q (511 keV each)


So there would be an excess of 59Co together with annihilation photons.
The annihilation photons would be difficult to fully shield, although their
rate is due to the half-life of the 59Ni decay and only indirectly tracks
the rate of the reaction itself.  Because the β+ decay rate is much smaller
than any inferred reaction rate, the annihilation photons will only
intermittently escape through the metal casing and make it into the
detector, which has a relatively small aperture (compared to a full solid
angle) and is less than 100 percent efficient (e.g., 26 percent
efficient).  With these things in mind, you wouldn't necessarily see
annihilation photons above background.  But the 59Co should increase
significantly above its normal amount.

The β+ decay occurs in only a very small number of cases.  Most of the time
(99 percent) the decay is via electron capture, a point I have missed up to
now.  So that will attenuate the expected number of annihilation photons in
my models by two orders of magnitude.  The decay, of course, is still to
59Co.

Eric

RE: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Jones Beene

 

 

From: Eric Walker 

 

*  This elegant possibility of a gainful reaction in which stable nickel
converts to stable nickel, giving up energy, is why my prediction for the
Mizuno presentation in November is to suggest that they will see a relative
decrease in Ni58 and a relative increase in Ni60.

 

*  The nickel to nickel idea seems very promising.  I doubt there is deuteron 
capture, because if there is deuteron capture, there is probably proton capture 
as well, along with all of the nasty gammas.  This is what is leading me to 
deuterium stripping -- e.g., 60Ni(d,p)61Ni.  Here the neutron is stripped off 
of the deuteron and added to the nickel, and the proton flies in the other 
direction, rather than there being a full capture.

 

Eric

 

Deuteron capture seems to be far and away the more likely scenario – at least 
more than proton capture for three reasons.

 

First the deuteron is a boson, as is the nickel nucleus. This is not invoking a 
condensate state or even a pseudo BEC, it relates to simple QM probability/ 
nuclear tunneling probability. 

 

Note that stripping is closer to brute force thermodynamics, and unlikely to 
happen in condensed matter.

 

Secondly, and most importantly - the neutron of the deuteron offers Coulomb 
shielding. 

 

This is related to isospin… I will allow Axil to elaborate on isospin since he 
first introduced it into the mix.

 

Thirdly –  the theory must reflect actual results. The main point of the 
previous post was to show that in the nuclear physics of Ni - Cu, there is 
apparently only one possibility which fits into the observation of 

1)no radioactive debris and 

2)no obvious transmutation product.

3)no gamma

 

Jones

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Terry Blanton

You've certainly been consistent Jones. Quoting you from 2011:

[Vo]:Deuterium kills the reaction?

Jones Beene jone...@pacbell.net via eskimo.com

1/19/11
to vortex-l

One detail worth exploring further was the statement from Rossi that
only hydrogen works, and that deuterium kills the reaction !

That is counter-intuitive to say the least. Everyone in hot fusion
knows for an absolute fact that deuterium is the more activenucleus,
right? And everyone in LENR knows that deuterium and palladium work,
whereas H2 is often used as the ‘control’ to show what doesn’t work.
Go figure.

Well, pondering this for a moment, the only possible property that
comes to mind to explain it was posted a few days ago –the “composite
boson” in the context of negative temperature. It is sounding better
and better as a rationale.

To rephrase, the complex argument goes like this. The heat anomaly,
whether it is fusion or not depends on “pycno” or dense hydrogen
clusters. Based on Lawandy’s paper and others, we see that spillover
catalysts operate by splitting molecular hydrogen into atomic hydrogen
without ionization. Dense hydrogen forms from atomic hydrogen if there
are adjoiningdielectric surfaces or cavities. Atomic hydrogen is a
composite boson. If there are internal defects (cavities) for atoms to
accumulate, they somehow seem to densify there without ever going
molecular.

We know that H is a composite boson which is a singularity in nature –
as it is composed of the minimum number of fermions (2) that permit
both states to oscillate back and forth… and furthermore having this
minimum number of quantum states to“align” (statistically) means that
it is exponentially easier to condense than deuterium at so-called
negative temperature (which are not “cold”) especially since spin can
be aligned magnetically...

Thanks to google books, we have access to an old issue of New
Scientist from 1981. On p. 205-6 there is clear indication that we
have known for nearly 30 years that hydrogen condensation can happen
at cryogenic temperatures – i.e. that monatomic hydrogen is a
composite boson independent of the molecular state - which has very
unusual properties as a condensate.

http://books.google.com/books?id=IbbMj56ht8sCpg=PA205lpg=PA205dq=composite-boson+monatomic-hydrogensource=blots=XlZyp6rE-9sig=AwMnZv-hCQzTfcbnkN2mQZ65VG0hl=enei=JFwaTab7Oon0tgPSpKjJCgsa=Xoi=book_resultct=resultresnum=1sqi=2ved=0CBwQ6AEwAA#v=onepageqf=false

This paper seems to have been largely forgotten, and offers no
indication that “negative temperature” could provide an alternative to
cryogenic temperature. And certainly no indication that the Casimir
cavity can provide a locus for negative temperature.

No one should be blamed at this juncture for being completely
skeptical that negative temperature in a cavity can do this, even on a
temporary time frame; and the only evidence of it today is the
implication from half a dozen papers which indicate that so-called
pycno-hydrogen exists (under many different names, even IRH or Inverse
Rydberg Hydrogen). Rossi’s results are consistent with this modality,
and Holmlid and Miley claim to have evidence of tiny bits of hydrogen
a million times denser than liquid H2.

Are they nuts too? Or is it all fitting together like a jigsaw puzzle?

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Terry Blanton

On Sat, Sep 20, 2014 at 10:40 AM, Jones Beene jone...@pacbell.net wrote:

 Terry will remember
 that in the very first image to come from Rossi, there was a color-coded
 tank of deuterium in the Lab.

It might be in this vid:

http://www.rainews.it/it/video.php?id=23074

The D2 gas might have been the one with the strips at the base of the valve.

BTW, didn't someone once claim that the Cu probably came from migration.  ;-)

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread Terry Blanton

strips = stripes

On Sat, Sep 20, 2014 at 2:38 PM, Terry Blanton hohlr...@gmail.com wrote:
 On Sat, Sep 20, 2014 at 10:40 AM, Jones Beene jone...@pacbell.net wrote:

 Terry will remember
 that in the very first image to come from Rossi, there was a color-coded
 tank of deuterium in the Lab.

 It might be in this vid:

 http://www.rainews.it/it/video.php?id=23074

 The D2 gas might have been the one with the strips at the base of the valve.

 BTW, didn't someone once claim that the Cu probably came from migration.  
 ;-)

Re: [Vo]:Mizuno, Rossi copper transmutation

Single proton capture will not work because the spin of a single proton is
non zero. Double proton capture will work because the spin of 2He is zero.

Piantelli shows a 6 MeV proton coming out of a nickel bar. This implies
that a proton pair entered the nickel nucleus: one to produce the 6 MeV via
fusion of nickel into copper and one proton to exit the nucleus to remove
that energy from the nucleus.

Also, the large amount of iron reported in Rossi’s ash assay, requires a
reaction involving two protons. The abundance of light elements in the DGT
ash assay requires fusion of multiple proton pairs with nickel.

It is a safe assumption that pairing of protons is occurring.


I

1H+1H+62Ni = 63Zn + n + 1.974 MeV
1H+1H+62Ni = 64Zn + 13.835 MeV
1H+1H+62Ni = 63Cu + 1H + 6.122 MeV
1H+1H+62Ni = 60Ni + 4He + 9.879 MeV
1H+1H+62Ni = 4He + 4He + 56Fe + 3.495 MeV   this one produces iron.
1H+1H+62Ni = 52Cr + 12C + 3.249 MeV
1H+1H+62Ni = 48Ti + 16O + 1.057 MeV
1H+1H+62Ni = 34S + 30Si + 2.197 MeV
1H+1H+1H+1H+62Ni = 65Ge + n + 10.750 MeV
1H+1H+1H+1H+62Ni = 66Ge + 24.037 MeV
1H+1H+1H+1H+62Ni = 63Ga + 3H + 4.007 MeV
1H+1H+1H+1H+62Ni = 64Ga + 2H + 8.108 MeV
1H+1H+1H+1H+62Ni = 65Ga + 1H + 17.778 MeV
1H+1H+1H+1H+62Ni = 61Zn + 5He + 7.372 MeV
1H+1H+1H+1H+62Ni = 62Zn + 4He + 21.156 MeV
1H+1H+1H+1H+62Ni = 63Zn + 3He + 9.692 MeV
1H+1H+1H+1H+62Ni = 59Cu + 7Li + 3.859 MeV
1H+1H+1H+1H+62Ni = 60Cu + 6Li + 6.667 MeV
1H+1H+1H+1H+62Ni = 61Cu + 5Li + 12.713 MeV
1H+1H+1H+1H+62Ni = 56Ni + 10Be + 3.707 MeV
1H+1H+1H+1H+62Ni = 57Ni + 9Be + 7.144 MeV
1H+1H+1H+1H+62Ni = 4He + 4He + 58Ni + 17.696 MeV
1H+1H+1H+1H+62Ni = 59Ni + 7Be + 7.795 MeV
1H+1H+1H+1H+62Ni = 60Ni + 6Be + 8.507 MeV
1H+1H+1H+1H+62Ni = 55Co + 11B + 7.769 MeV
1H+1H+1H+1H+62Ni = 56Co + 10B + 6.398 MeV
1H+1H+1H+1H+62Ni = 57Co + 9B + 9.338 MeV
1H+1H+1H+1H+62Ni = 52Fe + 14C + 7.721 MeV
1H+1H+1H+1H+62Ni = 53Fe + 13C + 10.230 MeV
1H+1H+1H+1H+62Ni = 54Fe + 12C + 18.662 MeV
1H+1H+1H+1H+62Ni = 55Fe + 11C + 9.239 MeV
1H+1H+1H+1H+62Ni = 56Fe + 10C + 7.316 MeV
1H+1H+1H+1H+62Ni = 51Mn + 15N + 10.550 MeV
1H+1H+1H+1H+62Ni = 52Mn + 14N + 10.252 MeV
1H+1H+1H+1H+62Ni = 53Mn + 13N + 11.752 MeV
1H+1H+1H+1H+62Ni = 54Mn + 12N + 0.627 MeV
1H+1H+1H+1H+62Ni = 48Cr + 18O + 6.010 MeV
1H+1H+1H+1H+62Ni = 49Cr + 17O + 8.549 MeV
1H+1H+1H+1H+62Ni = 50Cr + 16O + 17.406 MeV
1H+1H+1H+1H+62Ni = 51Cr + 15O + 11.003 MeV
1H+1H+1H+1H+62Ni = 52Cr + 14O + 9.819 MeV
1H+1H+1H+1H+62Ni = 47V + 19F + 5.899 MeV
1H+1H+1H+1H+62Ni = 48V + 18F + 6.011 MeV
1H+1H+1H+1H+62Ni = 49V + 17F + 8.415 MeV
1H+1H+1H+1H+62Ni = 50V + 16F + 0.951 MeV
1H+1H+1H+1H+62Ni = 44Ti + 22Ne + 7.983 MeV
1H+1H+1H+1H+62Ni = 45Ti + 21Ne + 7.147 MeV
1H+1H+1H+1H+62Ni = 46Ti + 20Ne + 13.575 MeV
1H+1H+1H+1H+62Ni = 47Ti + 19Ne + 5.591 MeV
1H+1H+1H+1H+62Ni = 48Ti + 18Ne + 5.580 MeV
1H+1H+1H+1H+62Ni = 41Sc + 25Na + 0.410 MeV
1H+1H+1H+1H+62Ni = 42Sc + 24Na + 2.949 MeV
1H+1H+1H+1H+62Ni = 43Sc + 23Na + 8.128 MeV
1H+1H+1H+1H+62Ni = 44Sc + 22Na + 5.408 MeV
1H+1H+1H+1H+62Ni = 45Sc + 21Na + 5.662 MeV
1H+1H+1H+1H+62Ni = 39Ca + 27Mg + 4.271 MeV
1H+1H+1H+1H+62Ni = 40Ca + 26Mg + 13.471 MeV
1H+1H+1H+1H+62Ni = 41Ca + 25Mg + 10.740 MeV
1H+1H+1H+1H+62Ni = 42Ca + 24Mg + 14.890 MeV
1H+1H+1H+1H+62Ni = 43Ca + 23Mg + 6.292 MeV
1H+1H+1H+1H+62Ni = 44Ca + 22Mg + 4.275 MeV
1H+1H+1H+1H+62Ni = 37K + 29Al + 5.425 MeV
1H+1H+1H+1H+62Ni = 38K + 28Al + 8.061 MeV
1H+1H+1H+1H+62Ni = 39K + 27Al + 13.413 MeV
1H+1H+1H+1H+62Ni = 40K + 26Al + 8.155 MeV
1H+1H+1H+1H+62Ni = 41K + 25Al + 6.885 MeV
1H+1H+1H+1H+62Ni = 34Ar + 32Si + 4.868 MeV
1H+1H+1H+1H+62Ni = 35Ar + 31Si + 8.406 MeV
1H+1H+1H+1H+62Ni = 36Ar + 30Si + 17.074 MeV
1H+1H+1H+1H+62Ni = 37Ar + 29Si + 15.252 MeV
1H+1H+1H+1H+62Ni = 38Ar + 28Si + 18.617 MeV
1H+1H+1H+1H+62Ni = 39Ar + 27Si + 8.036 MeV
1H+1H+1H+1H+62Ni = 40Ar + 26Si + 4.594 MeV
1H+1H+1H+1H+62Ni = 32Cl + 34P + 0.297 MeV
1H+1H+1H+1H+62Ni = 33Cl + 33P + 9.751 MeV
1H+1H+1H+1H+62Ni = 34Cl + 32P + 11.155 MeV
1H+1H+1H+1H+62Ni = 35Cl + 31P + 15.864 MeV
1H+1H+1H+1H+62Ni = 36Cl + 30P + 12.132 MeV
1H+1H+1H+1H+62Ni = 37Cl + 29P + 11.124 MeV
1H+1H+1H+1H+62Ni = 33S + 33S + 15.582 MeV
1H+1H+1H+1H+62Ni = 34S + 32S + 18.357 MeV
1H+1H+1H+1H+62Ni = 35S + 31S + 10.301 MeV
1H+1H+1H+1H+62Ni = 36S + 30S + 7.137 MeV

The last 4 produce lighter elements.

There are also similar reactions for the other Ni isotopes, and also for the
daughter products of the initial reactions, e.g. :-

1H+1H+64Zn = 66Ge + 10.202 MeV
1H+1H+64Zn = 65Ga + 1H + 3.942 MeV
1H+1H+64Zn = 62Zn + 4He + 7.321 MeV
1H+1H+64Zn = 4He + 4He + 58Ni + 3.860 MeV
1H+1H+64Zn = 54Fe + 12C + 4.827 MeV
1H+1H+64Zn = 50Cr + 16O + 3.571 MeV
1H+1H+64Zn = 42Ca + 24Mg + 1.055 MeV
1H+1H+64Zn = 36Ar + 30Si + 3.239 MeV
1H+1H+64Zn = 37Ar + 29Si + 1.417 MeV
1H+1H+64Zn = 38Ar + 28Si + 4.782 MeV
1H+1H+64Zn = 35Cl + 31P + 2.029 MeV
1H+1H+64Zn = 33S + 33S + 1.746 MeV
1H+1H+64Zn = 34S + 32S + 4.522 MeV



A polariton is a photon and an electron locked together in a pair. This
pair orbits around a cavity on its edge. The spin of all polaritons are
pointed such that the

Re: [Vo]:Mizuno, Rossi copper transmutation

Deuterium kills the reaction because its spin is non zero.

On Sat, Sep 20, 2014 at 2:35 PM, Terry Blanton hohlr...@gmail.com wrote:

You've certainly been consistent Jones. Quoting you from 2011:

[Vo]:Deuterium kills the reaction?

Jones Beene jone...@pacbell.net via eskimo.com

1/19/11
to vortex-l

One detail worth exploring further was the statement from Rossi that
only hydrogen works, and that deuterium kills the reaction !

Are they nuts too? Or is it all fitting together like a jigsaw puzzle?

Re: [Vo]:Mizuno, Rossi copper transmutation

This spin alignment of deuterium is why a plasma of hydrogen must be formed
to produce hydrogen crystals where deuterium must be reconfigured to a zero
spin alignment as the plasma cools.

Adding deuterium gas from a tank as Rossi has done will provide non zero
spin deuterium. It order for the deuterium to stay non zero spin, Rossi
must disable the plasma forming electrical circuit before the Deuterium gas
from the tank is added.

On Sat, Sep 20, 2014 at 2:35 PM, Terry Blanton hohlr...@gmail.com wrote:

You've certainly been consistent Jones. Quoting you from 2011:

[Vo]:Deuterium kills the reaction?

Jones Beene jone...@pacbell.net via eskimo.com

1/19/11
to vortex-l

One detail worth exploring further was the statement from Rossi that
only hydrogen works, and that deuterium kills the reaction !

Are they nuts too? Or is it all fitting together like a jigsaw puzzle?

Re: [Vo]:Mizuno, Rossi copper transmutation

One reason the Rossi requires high heat from external electric power input
is to produce molecules with zero spin. He now uses the Mouse to make
these special molecules.

The down side of high heat that can form a plasma is that such application
of heat can cause reactor-away. The Mouse was configured to produce zero
spin hydrogen but because it has a pronounced sub-critical nature, it will
not melt down no matter how much external heat is applied. The purpose of
the Mouse is to produce Rydberg hydrogen matter with zero spin through a
cooling plasma process. These solid crystals of hydrogen will then be feed
from the Mouse to the Cat where the main near critical high COP reaction
takes place using zero spin hydrogen.

On Sat, Sep 20, 2014 at 2:57 PM, Axil Axil janap...@gmail.com wrote:

This spin alignment of deuterium is why a plasma of hydrogen must be
formed to produce hydrogen crystals where deuterium must be reconfigured to
a zero spin alignment as the plasma cools.

On Sat, Sep 20, 2014 at 2:35 PM, Terry Blanton hohlr...@gmail.com wrote:

You've certainly been consistent Jones. Quoting you from 2011:

[Vo]:Deuterium kills the reaction?

Jones Beene jone...@pacbell.net via eskimo.com

1/19/11
to vortex-l

One detail worth exploring further was the statement from Rossi that
only hydrogen works, and that deuterium kills the reaction !

Are they nuts too? Or is it all fitting together like a jigsaw puzzle?

Re: [Vo]:Mizuno, Rossi copper transmutation

2014-09-20 Thread mixent

In reply to  Eric Walker's message of Sat, 20 Sep 2014 10:18:40 -0700:
Hi,
[snip]
The nickel to nickel idea seems very promising.  I doubt there is deuteron
capture, because if there is deuteron capture, there is probably proton
capture as well, along with all of the nasty gammas.  This is what is
leading me to deuterium stripping -- e.g., 60Ni(d,p)61Ni.  Here the neutron
is stripped off of the deuteron and added to the nickel, and the proton
flies in the other direction, rather than there being a full capture.

Actually, I rather like this idea. It's much easier for a neutron to tunnel than
for a proton, because the neutron has no Coulomb barrier opposing it. (Both the
neutron and the proton however need to find 2.2 MeV to escape the deuterium
nucleus, so parting is equally difficult for each.)

Note also that a severely shrunken Deuterino will have a much higher chance of
participating in such a reaction because it can get close to a target nucleus.
Regards,

Robin van Spaandonk

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

Re: [Vo]:Mizuno, Rossi copper transmutation