subject:"Re\: \[Vo\]\:mainstream physics paper bout the Hot Cat, co\-author Andrea Rossi"

-Original Message-

 The argument can be made that there was NEVER enough lithium present in
the Lugano reactor to provide the reported net energy gain (1.5 MW-hrs) over
32 hours- even if 100% of the lithium was consumed and converted into
helium.

For the record - The total Lugano Fuel sample had a reported mass of 1 gram.

Element % by Weight
Nickel 55.0
Iron 39.0
Aluminum 4.3
Lithium 1.1
Hydrogen (no Deuterium) 0.6
Total 100.0

Therefore, there was .011 grams of lithium at the start. The average mass of
the lithium = 6.93 amu or 7 grams per mole = .0016 moles. If all of this
lithium, 100%, had fused with protons, giving 17 MeV per fused lithium atom,
then it would have been marginally sufficient to provide the energy reported
(10^28 eV). That assumes that every atom has been consumed - and assuming
that no energy was lost to x-ray radiation. BUT. 

.there was lots of lithium left over in the ash, so all of it could not have
reacted and possibly as much as 90% of the bremsstrahlung should have been
lost in an alumina reactor.

As for the argument that 8 MeV alpha particles produce bremsstrahlung which
is mostly thermalized, consider the case of Uranium decay. 

U is an alpha emitter, where the alpha has an average kinetic energy of only
5 MeV, yet this corresponds to a velocity which is 5% of the speed of light
producing substantial radiation, and despite the extremely high ability of U
to absorb such radiation - hundreds of times greater than alumina, most of
it escapes - which is why even small pitchblende samples make the Geiger
counter go wild. It is likely that only a few percent of 8 MeV alpha
bremsstrahlung will be completely thermalized by alumina absorption, since
alumina is fairly transparent to x-rays in this range. IOW most of that
putative 8 MeV should be lost as x-rays and not recorded as heat.

In short, I am having a hard time imagining how Cook and Rossi can believe
that lithium proton fusion is responsible for the energy gain - even if
there is a spin mechanism which bypasses the problem of x-rays from
bremsstrahlung.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-10 Thread Axil Axil

The amount of nickel Ni62 in the fuel load doubled from some unknown
combination of lighter elements. This fusion process should have released a
huge amount of nuclear binding energy. It is possible that the only thing
that lithium did was donate its neutron to the Nickel 58 to turn it into
Nickel 62.

On Fri, Apr 10, 2015 at 3:48 PM, Jones Beene jone...@pacbell.net wrote:

  -Original Message-

  The argument can be made that there was NEVER enough lithium present in
 the Lugano reactor to provide the reported net energy gain (1.5 MW-hrs) over
 32 hours- even if 100% of the lithium was consumed and converted into
 helium…

 For the record - The total Lugano Fuel sample had a reported mass of 1
 gram.

 *Element % by Weight*
 Nickel 55.0
 Iron 39.0
 Aluminum 4.3
 Lithium 1.1
 Hydrogen (no Deuterium) 0.6
 Total 100.0

 Therefore, there was .011 grams of lithium at the start. The average mass
 of the lithium = 6.93 amu or 7 grams per mole = .0016 moles. If all of
 this lithium, 100%, had fused with protons, giving 17 MeV per fused lithium
 atom, then it would have been marginally sufficient to provide the energy
 reported (10^28 eV). That assumes that every atom has been consumed - and 
 assuming
 that no energy was lost to x-ray radiation… BUT…

 …there was lots of lithium left over in the ash, so all of it could not
 have reacted and possibly as much as 90% of the bremsstrahlung should
 have been lost in an alumina reactor.

 As for the argument that 8 MeV alpha particles produce bremsstrahlung
 which is mostly thermalized, consider the case of Uranium decay.

 U is an alpha emitter, where the alpha has an average kinetic energy of
 only 5 MeV, yet this corresponds to a velocity which is 5% of the speed
 of light producing substantial radiation, and despite the extremely high
 ability of U to absorb such radiation – hundreds of times greater than
 alumina, most of it escapes - which is why even small pitchblende samples
 make the Geiger counter go wild. It is likely that only a few percent of 8
 MeV alpha bremsstrahlung will be completely thermalized by alumina
 absorption, since alumina is fairly transparent to x-rays in this range.
 IOW most of that putative 8 MeV should be lost as x-rays and not recorded
 as heat.

 In short, I am having a hard time imagining how Cook and Rossi can believe
 that lithium proton fusion is responsible for the energy gain – even if
 there is a spin mechanism which bypasses the problem of x-rays from
 bremsstrahlung.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

This is not what Cook and Rossi are now saying. They are claiming in this 
so-called “mainstream physics” paper, that lithium + proton fusion to helium 
accounts for the gain. 

 

If neutrons were involved there would be neutron activation, a widely known 
phenomenon - not seen at Lugano.

 

It looks like they have backed themselves in a corner with bad science …

 

From: Axil Axil 

 

Ø  The amount of nickel Ni62 in the fuel load doubled from some unknown 
combination of lighter elements. 

 

… more likely, pure isotope was added, since a pure specimen turned up in the 
ash.

 

This fusion process should have released a huge amount of nuclear binding 
energy. It is possible that the only thing that lithium did was donate its 
neutron to the Nickel 58 to turn it into Nickel 62.

 

 

Jones Beene wrote:

-Original Message-

 The argument can be made that there was NEVER enough lithium present in the 
 Lugano reactor to provide the reported net energy gain (1.5 MW-hrs) over 32 
 hours- even if 100% of the lithium was consumed and converted into helium…

For the record - The total Lugano Fuel sample had a reported mass of 1 gram. 

Element % by Weight
Nickel 55.0
Iron 39.0
Aluminum 4.3
Lithium 1.1
Hydrogen (no Deuterium) 0.6
Total 100.0

Therefore, there was .011 grams of lithium at the start. The average mass of 
the lithium = 6.93 amu or 7 grams per mole = .0016 moles. If all of this 
lithium, 100%, had fused with protons, giving 17 MeV per fused lithium atom, 
then it would have been marginally sufficient to provide the energy reported 
(10^28 eV). That assumes that every atom has been consumed - and assuming that 
no energy was lost to x-ray radiation… BUT… 

…there was lots of lithium left over in the ash, so all of it could not have 
reacted and possibly as much as 90% of the bremsstrahlung should have been lost 
in an alumina reactor.

As for the argument that 8 MeV alpha particles produce bremsstrahlung which is 
mostly thermalized, consider the case of Uranium decay. 

U is an alpha emitter, where the alpha has an average kinetic energy of only 5 
MeV, yet this corresponds to a velocity which is 5% of the speed of light 
producing substantial radiation, and despite the extremely high ability of U to 
absorb such radiation – hundreds of times greater than alumina, most of it 
escapes - which is why even small pitchblende samples make the Geiger counter 
go wild. It is likely that only a few percent of 8 MeV alpha bremsstrahlung 
will be completely thermalized by alumina absorption, since alumina is fairly 
transparent to x-rays in this range. IOW most of that putative 8 MeV should be 
lost as x-rays and not recorded as heat.

In short, I am having a hard time imagining how Cook and Rossi can believe that 
lithium proton fusion is responsible for the energy gain – even if there is a 
spin mechanism which bypasses the problem of x-rays from bremsstrahlung.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Bob Higgins's message of Fri, 10 Apr 2015 08:45:31 -0600:
Hi,
[snip]
Well, Piantelli may not be saying it is a hydrino because he doesn't look
at it that way.  He has black box evidence that the Ni and the H- anion
nuclei coalesce producing a specific set of branched outcomes, one of which
is ejection of a high energy proton.  However, he is not willing to say he
has evidence of compact forms of the H- anion.  I cannot envision how the
coalescence of the nuclei would occur without a DDL compact form of the
H- anion, but that is my limited vision.  If I understood Piantelli
correctly, he believes that this coalescence occurs so quickly as to make
the actual mechanism somewhat irrelevant.  This sort of reinforces Dennis
Cravens' accusation that I do too much ball and stick thinking.  It is my
thought process that tried to place a means on the coalescence rather
than Piantelli's.  Piantelli is too good a scientist to speculate on the
mechanism without having some evidence.

I asked Dr. Jerry Va'vra at Stanford if he knew of any analysis of the
possibility of a DDL state for the H- anion.  He replied that he had not
seen such an analysis.  Do you know if Dr. Mills done an analysis of the
shrunken states of the H- anion?

Bob

He calls them Hydrinohydride. The smallest is for p = 24. I.e. 24 times
smaller than normal H-. For greater values of p (i.e. further shrunken), the
second electron is unbound, according to his formula, so there is no
Hydrinohydride for larger p values.

(p = 1/n)



Regards,

Robin van Spaandonk

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

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

-Original Message-
From: mix...@bigpond.com 

 He calls them Hydrinohydride. The smallest is for p = 24. I.e. 24 times
smaller than normal H-. For greater values of p (i.e. further shrunken), the
second electron is unbound, according to his formula, so there is no
Hydrinohydride for larger p values.


BTW - it looks like Meulenberg is calling the lowest state version
femtohydrogen. If the number of names were an indicator of future fame -
this species is probably going to be quite well-known one of these days.
Lets see:

Dense hydrogen  dense-cluster hydrogen
Hydrino
IRH (inverted Rydberg hydrogen)
f/H (fractional hydrogen)
pychno-hydrogen
DDL (Deep Dirac Level)
femtohydrogen
virtual neutron
Dark Matter
Metallic hydrogen

... any others?

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Bob Higgins's message of Fri, 10 Apr 2015 09:04:29 -0600:
Hi,
[snip]
I cannot pretend to be a spokesman for Dr. Piantelli's theory. I have a
couple of observations from this theory that I still cannot internally
justify (from my own limited understanding of nuclear physics).  The first,
is that the H- anion is large - larger than a neutral H atom, and almost as
large as the Ni atom.  So, how could the H- anion still appear like a
negative ion as it approaches the Ni nucleus to 2-100 fm?  The H- nucleus
would no longer be screened at a distance of half of the H- diameter, so
the H- would never have been able to penetrate as a screened or attracted
particle - that is to say, by my thinking, unless the H- anion became a
compact atomic object, such as in a DDL state.  

...and this is what has led me to reject his theory out of hand.


However, once the H- begins
to enter the Ni atom, Piantelli says that the coalescence of the H- anion
and the Ni atom occur extremely quickly, all the way to the nucleus (I
think of galaxies colliding and coalescing).  All I can say is that I don't
understand exactly how it happens.

Horace's version makes more sense. I.e. an electron and a proton tunnel into the
Ni together as a neutral object. Hence no Coulomb barrier, which makes the
central charge irrelevant, and explains why transmutation reactions are seen
even for heavy atoms.

The same might be said for severely shrunken Hydrinos.


The second observation is that you cannot eject a 6 MeV proton by having an
H- anion fall into proximity with the Ni nucleus and then be
electrostatically ejected, unless some matter is converted to energy (the
fall should be conservative).  This suggests that something nuclear is
happening in the branch of the reaction that results in the ejection of the
6 MeV proton to supply the proton with its 6 MeV of energy.  Then,
shouldn't one expect that something nuclear happened to the Ni nucleus
during that branch as well?  It probably does.

...however if two protons approach the Ni at the same time, and one of them
fuses with the Ni releasing about 6 MeV of energy, and that energy is carried
away by the second proton, then it makes perfect sense.

What, two protons at the same time!?! Unlikely! - Not if they are already bound
together in a shrunken Hydrogen molecule.

However the problem with this entire scenario is that it creates radioactive Cu
nuclei, and no such radiation is detectedunless of course the electron is
captured by the proton inside the Cu nucleus converting it into a neutron in a
fast electron capture reaction. That would result in a Ni isotope increased in
mass by 1 neutron. The energy of the weak decay reaction could be carried away
by the neutrino, and go unnoticed.
Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Jones Beene's message of Fri, 10 Apr 2015 12:48:35 -0700:
Hi,
[snip]
which is why even small pitchblende samples make the Geiger
counter go wild.

Try putting a sheet of paper between the Geiger counter and the pitchblende. I
think you will find that it makes a huge difference. Most of the activity
detected is due to alphas emitted from the surface.

(Got your saw handy? ;)

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Jones Beene's message of Fri, 10 Apr 2015 13:53:13 -0700:
Hi,
[snip]
If neutrons were involved there would be neutron activation, a widely known 
phenomenon - not seen at Lugano.
[snip]
Indeed, the reaction:

Al27 + Li7 = Al28 + Li6 + 0.475 MeV would produce radioactive Al28 with a half
life of about 2 minutes and a 1.8 MeV gamma, besides a fast beta particle with
attendant bremsstrahlung.

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Jones Beene's message of Fri, 10 Apr 2015 15:54:53 -0700:
Hi,
[snip]
-Original Message-
From: mix...@bigpond.com 

 He calls them Hydrinohydride. The smallest is for p = 24. I.e. 24 times
smaller than normal H-. For greater values of p (i.e. further shrunken), the
second electron is unbound, according to his formula, so there is no
Hydrinohydride for larger p values.


BTW - it looks like Meulenberg is calling the lowest state version
femtohydrogen. If the number of names were an indicator of future fame -
this species is probably going to be quite well-known one of these days.
Lets see:

Dense hydrogen  dense-cluster hydrogen
Hydrino
IRH (inverted Rydberg hydrogen)
f/H (fractional hydrogen)
pychno-hydrogen
DDL (Deep Dirac Level)
femtohydrogen
virtual neutron
Dark Matter
Metallic hydrogen

... any others?

One of your own: sub-orbital Hydrogen. :)

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-10 Thread torulf.greek

Some pitchblende contains radium how emits gammas.


On Sat, 11 Apr 2015 09:31:35 +1000, mix...@bigpond.com wrote:
 In reply to  Jones Beene's message of Fri, 10 Apr 2015 12:48:35 -0700:
 Hi,
 [snip]
which is why even small pitchblende samples make the Geiger
counter go wild.
 
 Try putting a sheet of paper between the Geiger counter and the
 pitchblende. I
 think you will find that it makes a huge difference. Most of the activity
 detected is due to alphas emitted from the surface.
 
 (Got your saw handy? ;)
 
 Regards,
 
 Robin van Spaandonk
 
 http://rvanspaa.freehostia.com/project.html

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

A sheet of paper makes no difference whatsoever. A sheet of aluminum foil
doesn't make much difference but a nickel coin blocks most of this small
sample. The Gamma Scout is picking up x-rays  which are not stopped by paper
(even though alpha particles themselves would be). However, to be fair - a
fair fraction of the counts, at least in minerals, is probably due to
so-called daughters some of which emit gammas.


-Original Message-
From: mix...@bigpond.com 

In reply to  Jones Beene's message:
Hi,

which is why even small pitchblende samples make the Geiger counter go
wild.

Try putting a sheet of paper between the Geiger counter and the pitchblende.
I think you will find that it makes a huge difference. Most of the activity
detected is due to alphas emitted from the surface.

(Got your saw handy? ;)

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In follow-up hypothetical analysis of the Lugano measurements, consider
this.  Look at what it means for the ICP-MS assay of the fuel to have 94.1%
7Li and 5.9% 6Li.  With 100mg of LiAlH4 fuel source, the fuel source had
17.2mg of 7Li and 1.08mg of 6Li.  If one *presumes* that 6Li is not being
created and doesn't participate in the reaction; then in the ash there will
still be 1.08mg of 6Li.  The ICP-MS analysis of the ash shows that there is
42.5% of 7Li and 57.5% of 6Li.  Since (by presumption) there is still
1.08mg of 6Li left in the ash, there is only 0.79 mg of 7Li in the ash.
The amount of 7Li has decreased from 17.2mg to 0.79mg from fuel to ash.
Thus, only 0.79mg/17.2mg or only about 1/22 of the original 7Li remains in
the ash - based on the presumption that no 6Li was created.
Because the reaction showed no major output heat decline due to only 1/22
of the original 7Li being present by the end of the reaction, it suggests
to me that the 7Li may not be the primary source of heat in the reaction.
As an aside, if the heat produced over the course of the experiment was due
solely to the burning of 7Li, the consumption of (17.2 - 0.79) = 16.41 mg
of 7Li would require the reaction to produce ~8.4 MeV per atom of burned
7Li (based on the revised heat output of the Lugano experiment).
More likely the hypothesis that 6Li is not created and 7Li burning produces
the heat is not correct.  This hypothetical argument suggests that the 7Li
is participating in the reaction (perhaps producing some excess heat), some
6Li is probably being created in the reaction, and much of the heat is
coming from some other reaction - perhaps the transmutation/isotopic shift
in the Ni which was not depleted by the end of the reaction.

Piantelli's theory supports this.  He uses Li as a booster for his reactors
- using the Li to create excess heat from the 6 MeV protons being produced
(resulting in more than 6 MeV of heat per proton).  However, he does have
excess heat without the Li.

Bob Higgins

On Thu, Apr 9, 2015 at 1:30 AM, mix...@bigpond.com wrote:

 In reply to  Bob Higgins's message of Wed, 8 Apr 2015 10:12:11 -0600:
 Hi,
 [snip]
 The ICP-MS analysis is a full volume analysis and showed both Li isotopes
 near equal in percentage in the ash.

 Just a thought: If the Li was acting as a nuclear catalyst, shuttling back
 and
 forth between Li6-Li7, then a roughly equal distribution on the whole
 might be
 expected, since a preponderance of one over the other would lead to an
 increase
 in the number of reactions of the predominant isotope, resulting in more
 of them
 being converted to the other.

 i.e. an excess of Li7 would yield more reactions converting Li7 to Li6,
 and an
 excess of Li6 would result in more reactions converting Li6 to Li7.

 Regards,

 Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-09 Thread Daniel Rocha

Hi Bob, I have a different interpretation of what happens. Maybe this will
be useful for you:

The idea is simple. What is the easiest way to make small clusters of
atoms? Use the analogy with water vapor. Heat, say, Ni or Pd. Very small
clusters have its shape determined by the boundary, just like a liquid.

So, you heat the metal until it melts. Then, depending on the vapor
pressure of nickel it will condense a cloud. I don't have a book with
details, but it would be nice to use something that could help the vapor
condense. Lithium would be a nice choice, since it is a like Na, but it is
less electro positive and reactive, which would be too reducing at that
temperature.

Then, you apply some discharges, which will pinch some of the clusters. Or,
maybe the Li it self will cause the discharge, by acting through its charge
on the clusters. It will compress the hydrogen to extremely high pressures,
yielding solid metastable  hydrogen. If you have, say 4 atoms together, you
will obtain a TSC, from akito. Heat after death is when a lot of solid
metastable hydrogen. As it evaporates, parts with 4 atoms will be left,
which will cold fuse.

My theory about cold fusion is slightly different from Akito's, though.
It's the same until 1fm of distance between H. But then, HUP kicks in, and
won't let a closer approach. Then, a femto solution is found. It is very
different from others, though. The first level is 3.7KeV, and this is found
by simply solving the Hidrogen atom for another negative solutions, what
allows its existance, though (since what is actually obtained is masse
energy of the electron - 3.7KeV), its is that the boundary is no 2pi, but
2pi*137, that is, the electron gives a lot of turns to go back to the
starting point.
Note that this is only the enegy de broglie wavelength, actually, what you
get is a beating, that is a modulated frequency between the actual, very
spread location of the electron, which works as the envelope of the de
broglie wavelength.

Then, the end part is similar to Akito's, though there are stages before
that happen. Section 4 of the second paper of the file (The CN paper is not
representative of my idea, it's just the figure!)

http://www.roxit.ax/CN.pdf

The nuclei are squeezed closer and closer, as the system is more stable
with higher energy electrons in lower levels, due a new pseudo nuclei.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Bob Higgins's message of Wed, 8 Apr 2015 10:12:11 -0600:
Hi,
[snip]
The ICP-MS analysis is a full volume analysis and showed both Li isotopes near 
equal in percentage in the ash.

Just a thought: If the Li was acting as a nuclear catalyst, shuttling back and
forth between Li6-Li7, then a roughly equal distribution on the whole might be
expected, since a preponderance of one over the other would lead to an increase
in the number of reactions of the predominant isotope, resulting in more of them
being converted to the other.

i.e. an excess of Li7 would yield more reactions converting Li7 to Li6, and an
excess of Li6 would result in more reactions converting Li6 to Li7.

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Bob--

Does Piantelli say there are energetic EM photons seen in the reaction?  6 Mev 
protons would probably produce observable EM radiation, unless they were all 
consumed in the subsequent LENR reaction.  That seems unlikely to me.

Bob Cook
  - Original Message - 
  From: Bob Higgins 
  To: vortex-l@eskimo.com 
  Sent: Thursday, April 09, 2015 8:09 AM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

  In follow-up hypothetical analysis of the Lugano measurements, consider this. 
 Look at what it means for the ICP-MS assay of the fuel to have 94.1% 7Li and 
5.9% 6Li.  With 100mg of LiAlH4 fuel source, the fuel source had 17.2mg of 7Li 
and 1.08mg of 6Li.  If one presumes that 6Li is not being created and doesn't 
participate in the reaction; then in the ash there will still be 1.08mg of 6Li. 
 The ICP-MS analysis of the ash shows that there is 42.5% of 7Li and 57.5% of 
6Li.  Since (by presumption) there is still 1.08mg of 6Li left in the ash, 
there is only 0.79 mg of 7Li in the ash.  The amount of 7Li has decreased from 
17.2mg to 0.79mg from fuel to ash.  Thus, only 0.79mg/17.2mg or only about 1/22 
of the original 7Li remains in the ash - based on the presumption that no 6Li 
was created.
  Because the reaction showed no major output heat decline due to only 1/22 of 
the original 7Li being present by the end of the reaction, it suggests to me 
that the 7Li may not be the primary source of heat in the reaction.  As an 
aside, if the heat produced over the course of the experiment was due solely to 
the burning of 7Li, the consumption of (17.2 - 0.79) = 16.41 mg of 7Li would 
require the reaction to produce ~8.4 MeV per atom of burned 7Li (based on the 
revised heat output of the Lugano experiment).
  More likely the hypothesis that 6Li is not created and 7Li burning produces 
the heat is not correct.  This hypothetical argument suggests that the 7Li is 
participating in the reaction (perhaps producing some excess heat), some 6Li is 
probably being created in the reaction, and much of the heat is coming from 
some other reaction - perhaps the transmutation/isotopic shift in the Ni which 
was not depleted by the end of the reaction.

  Piantelli's theory supports this.  He uses Li as a booster for his reactors - 
using the Li to create excess heat from the 6 MeV protons being produced 
(resulting in more than 6 MeV of heat per proton).  However, he does have 
excess heat without the Li.

  Bob Higgins

  On Thu, Apr 9, 2015 at 1:30 AM, mix...@bigpond.com wrote:

In reply to  Bob Higgins's message of Wed, 8 Apr 2015 10:12:11 -0600:
Hi,
[snip]
The ICP-MS analysis is a full volume analysis and showed both Li isotopes 
near equal in percentage in the ash.

Just a thought: If the Li was acting as a nuclear catalyst, shuttling back 
and
forth between Li6-Li7, then a roughly equal distribution on the whole might 
be
expected, since a preponderance of one over the other would lead to an 
increase
in the number of reactions of the predominant isotope, resulting in more of 
them
being converted to the other.

i.e. an excess of Li7 would yield more reactions converting Li7 to Li6, and 
an
excess of Li6 would result in more reactions converting Li6 to Li7.

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Robin--

Another issue to consider is the effect of the over-lying magnetic field on 
the differential energy of the respective spin states and their angular 
momentum.  A magnetic field will separate the energies of the respective 
spin states of the He* IMO.  Any photon of energy would have to have a 
resonance  matching the original spin state and also match the energy of the 
transition from spin state to spin state.  It seems easier to distribute 
integral quanta of angular momentum via distribution in small quanta to many 
receptors.


Furthermore I am not sure that spin energy can change into photons with 
their linear momentum.  This assumes a transfer of linear momentum from 
angular momentum.  Potentially two back-to-back photons could be generated 
with 0 net linear momentum.  Such a transition may require two coherent He* 
in anti-parallel configuration to achieve conservation of angular momentum 
for the transition involving the photon emissions.


As you have noted the idea of a gamma is not consistent with the 
experimental evidence, in any case.


Bob Cook
- Original Message - 
From: mix...@bigpond.com

To: vortex-l@eskimo.com
Sent: Thursday, April 09, 2015 1:45 AM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi



In reply to  Bob Cook's message of Tue, 7 Apr 2015 22:57:00 -0700:
Hi,

Eric--

One additional idea.

What we have been considering is the formation of 8Be and its decay into 
two alpha particles with only spin energy involved.


As I have suggested before, two anti-parallel spin He* particles may form 
in adjacent fcc Pd lattice locations that are stuffed tight with  2 
deuterium nuclei.  The net spin of the two new He* particles is high--24 
mev--but amounts to 0 net angular momentum when considered as one item. 
However, each He* within the coherent system may be able to distribute its 
spin energy to the electrons in the vicinity, much as may happen with the 
decay of the 8Be nucleus.  The two LENR processes would be similar in this 
regard.


Bob

[snip]
The p+Li7 reaction yields 17.35 MeV, not 24 MeV.

Based on this, and an assumption that the radius of a Helium nucleus is 
about 2
fm, I calculated the angular momentum and found it to be about 2.5 times 
that of
a photon, so in theory, a couple of photons could be emitted before the 
nucleus

lost too much angular momentum. As to the energy of those photons that would
depend on the frequency, and that is where the ground gets a bit squishy. If 
you
base it on the rotational frequency of the nucleus, then the first photon 
has an
energy of about 6 MeV. This is a powerful gamma and would be easily 
detected.

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-09 Thread Alain Sepeda

maybe is it intepreted from the budget that Elforsk allocated for those
tests...
230k (eur) per year 2013 and 2014 as far as I remember.
more precisely
Budget ECAT 2012 200 kkr, 2013-2015 2000 kkr/year.

I imagine that even if fully given for the team, it goes to their
university/lab?

2015-04-09 7:15 GMT+02:00 Eric Walker eric.wal...@gmail.com:

 Hi,

 On Wed, Apr 8, 2015 at 1:18 PM, Jones Beene jone...@pacbell.net wrote:

 Levi and his team were reportedly paid half a million bucks ...


 Do you have a source for this that goes into more detail?

 Eric

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-09 Thread Axil Axil

The Lagano transmutation mysteries

https://www.facebook.com/MartinFleischmannMemorialProject/posts/952626724768027?comment_id=95298075143

The above referenced analysis describes the amount of increase or decrease
in the masses of the various components of the fuel change. Certain
components of the fuel increased and others components decreased. This
analysis also speculates about how these changes in masses of the various
fuel and ash components might be brought about by movement of material from
the hot parts of the reactor to colder regions, and also how lithium
migrates to the surface of the nickel particles for transmutation and then
when processed are somehow transported to other parts of the reactors
volume.
All this movement must be happening through some means of chemical
transport. But the mix of fuel and ash forms a solid mixture that seems
incomparable with the movement of elements in a transport medium.

I see one possibility to explain how the moment of these elements and
chemical compounds might be possible. These fuel and ash components could
move around in the hydrogen gas atmosphere as nanoparticles. What is hard
to explain is how the mass of nickel component of the fuel can increase in
the ash to double it weight as originally configured in the fuel load. The
nickel particles had a very intricate surface tubercle structure that makes
it impossible to structurally modify the particle without affecting the
tubercles.

But the isotopic composition of the nickel particles originally in the fuel
changed throughout their entire volume as if subatomic particles traveled
through the volume of these particles. A hydrogen negative ion as
postulated by Piantelli cannot penetrate into the center of the nickel
particle to produce isotopic transmutation. The active agent in this
isotopic transformation must be a subatomic particle.

The other type of nickel particle found was the smooth surface kind that
must have been formed by gradual accumulation of transmuted Ni62. What and
how this type of nickel particle is produced is an open question. The same
type of gradual accumulation must have had to produce the iron micro
particles.

On Thu, Apr 9, 2015 at 12:13 PM, Bob Higgins rj.bob.higg...@gmail.com
wrote:

Piantelli said that he has seen ~500 keV gammas (didn't say how many and
if always present) and he tested for beta+/beta- annihilation and did not
find the dual photon signature for that. I was asking if he had seen a DDL
signature for compaction of the H- anion.

The Bremsstrahlung from heavy particles like protons is not as prominent
as for light particles like electrons. The Bremsstrahlung has to do with
the deceleration and light particles stop much more quickly. Heavy
particles of a given MeV are going slower, and they stop more slowly with
various Compton scatterings. So I don't know if you would readily see
Bremsstrahlung from a 6 MeV proton.

On Thu, Apr 9, 2015 at 10:04 AM, Bob Cook frobertc...@hotmail.com wrote:

Bob--

Does Piantelli say there are energetic EM photons seen in the reaction?
6 Mev protons would probably produce observable EM radiation, unless they
were all consumed in the subsequent LENR reaction. That seems unlikely to
me.

Bob Cook

- Original Message -
*From:* Bob Higgins rj.bob.higg...@gmail.com
*To:* vortex-l@eskimo.com
*Sent:* Thursday, April 09, 2015 8:09 AM
*Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
Andrea Rossi

In follow-up hypothetical analysis of the Lugano measurements, consider
this. Look at what it means for the ICP-MS assay of the fuel to have
94.1% 7Li and 5.9% 6Li. With 100mg of LiAlH4 fuel source, the fuel source
had 17.2mg of 7Li and 1.08mg of 6Li. If one *presumes* that 6Li is not
being created and doesn't participate in the reaction; then in the ash
there will still be 1.08mg of 6Li. The ICP-MS analysis of the ash shows
that there is 42.5% of 7Li and 57.5% of 6Li. Since (by presumption) there
is still 1.08mg of 6Li left in the ash, there is only 0.79 mg of 7Li in the
ash. The amount of 7Li has decreased from 17.2mg to 0.79mg from fuel to
ash. Thus, only 0.79mg/17.2mg or only about 1/22 of the original 7Li
remains in the ash - based on the presumption that no 6Li was created.
Because the reaction showed no major output heat decline due to only 1/22
of the original 7Li being present by the end of the reaction, it suggests
to me that the 7Li may not be the primary source of heat in the reaction.
As an aside, if the heat produced over the course of the experiment was due
solely to the burning of 7Li, the consumption of (17.2 - 0.79) = 16.41 mg
of 7Li would require the reaction to produce ~8.4 MeV per atom of burned
7Li (based on the revised heat output of the Lugano experiment).
More likely the hypothesis that 6Li is not created and 7Li burning
produces the heat is not correct. This hypothetical argument suggests that
the 7Li is participating

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Robin--

I agree with your observation.  As far as I know there are no cross sections 
established for slow protons.


I think there may be some resonance cross sections for dipole and 
quadrupole electric and magnetic stimulation however.   They may even be 
published.


It takes a fine tuned photon beam from 2 directions in a magnetic field to 
achieve the quadrupole stimulation.  A broad spectrum radiation field from 
all directions may be able to achieve the stimulation on occasion depending 
upon the density of the electrons shielding the Li nuclei.  The electric or 
magnetic dipole stimulation may be easier.  N. Cook's energy charts may help 
determine the mode that is most likely.


If you recall, laser stimulation of Craven and Letts device seemed to cause 
excess heat consistently.


Bob Cook
- Original Message - 
From: mix...@bigpond.com

To: vortex-l@eskimo.com
Sent: Thursday, April 09, 2015 12:02 AM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi



In reply to  Jones Beene's message of Tue, 7 Apr 2015 12:24:18 -0700:
Hi,
[snip]
The Li nucleus becomes excited, but it cannot simply convert directly to 
beryllium without an energetic emission to compensate for the kinetic 
energy which caused the fusion.


All p+Li7 reactions that have been measured are caused by bombarding Li with
fast protons. If I'm not mistaken, it's the energy of the fast proton that
results in the gamma. However in the case of LENR there is no fast proton, 
since
the protons are room temperature or a couple of eV at most. The actual 
fusion
process being mediated purely by tunneling. So perhaps gamma-less p+Li7 
fusion

is indeed possible?

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Robin--

I doubt your assumption about the size of the excited He* entity is correct. 
It would have a distorted shape with the high angular momentum.  It's outer 
reach may extend far into the surrounding electron cloud and only return to 
the  lesser size of a photon upon loss of its spin energy.  I am not sure 
what the Pauli UP has to say about angular momentum/spin energy relative to 
location parameters.   I am assuming that the extent of the He* wave 
function is the effective size of that entity at any current quantum spin 
state.


Bob

- Original Message - 
From: mix...@bigpond.com

To: vortex-l@eskimo.com
Sent: Thursday, April 09, 2015 1:45 AM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi



In reply to  Bob Cook's message of Tue, 7 Apr 2015 22:57:00 -0700:
Hi,

Eric--

One additional idea.

What we have been considering is the formation of 8Be and its decay into 
two alpha particles with only spin energy involved.


As I have suggested before, two anti-parallel spin He* particles may form 
in adjacent fcc Pd lattice locations that are stuffed tight with  2 
deuterium nuclei.  The net spin of the two new He* particles is high--24 
mev--but amounts to 0 net angular momentum when considered as one item. 
However, each He* within the coherent system may be able to distribute its 
spin energy to the electrons in the vicinity, much as may happen with the 
decay of the 8Be nucleus.  The two LENR processes would be similar in this 
regard.


Bob

[snip]
The p+Li7 reaction yields 17.35 MeV, not 24 MeV.

Based on this, and an assumption that the radius of a Helium nucleus is 
about 2
fm, I calculated the angular momentum and found it to be about 2.5 times 
that of
a photon, so in theory, a couple of photons could be emitted before the 
nucleus

lost too much angular momentum. As to the energy of those photons that would
depend on the frequency, and that is where the ground gets a bit squishy. If 
you
base it on the rotational frequency of the nucleus, then the first photon 
has an
energy of about 6 MeV. This is a powerful gamma and would be easily 
detected.

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Jones Beene's message of Tue, 7 Apr 2015 12:24:18 -0700:
Hi,
[snip]
 The Li nucleus becomes excited, but it cannot simply convert directly to 
 beryllium without an energetic emission to compensate for the kinetic energy 
 which caused the fusion.

All p+Li7 reactions that have been measured are caused by bombarding Li with
fast protons. If I'm not mistaken, it's the energy of the fast proton that
results in the gamma. However in the case of LENR there is no fast proton, since
the protons are room temperature or a couple of eV at most. The actual fusion
process being mediated purely by tunneling. So perhaps gamma-less p+Li7 fusion
is indeed possible?

Regards,

Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Piantelli's theory says that H- anions are responsible for the Ni-H LENR
reaction.  According to his theory, the H- anion, as a composite fermion,
enters the Ni atom much as would a muon.  Somehow (and Piantelli doesn't
say how) the large H- anion must become a compact negatively charged object
(like a DDL state) and have a tiny orbital around the Ni nucleus.  The
resulting proximity of the H- anion to the Ni nucleus causes a nuclear
reaction with a number of branches.  Piantelli has measured 6 MeV protons
exiting the reaction as one of the branches, and various
transmutation/isotopic shifts of the large Ni nucleus.  Thus, high energy
protons come from one of the branches of this reaction.

Note that LiH is an ionic hydride, and the hydrogen in the liquid hydride
exists as hydrogen anions, H-.

On Thu, Apr 9, 2015 at 1:02 AM, mix...@bigpond.com wrote:

 In reply to  Jones Beene's message of Tue, 7 Apr 2015 12:24:18 -0700:
 Hi,
 [snip]
  The Li nucleus becomes excited, but it cannot simply convert directly to
 beryllium without an energetic emission to compensate for the kinetic
 energy which caused the fusion.

 All p+Li7 reactions that have been measured are caused by bombarding Li
 with
 fast protons. If I'm not mistaken, it's the energy of the fast proton that
 results in the gamma. However in the case of LENR there is no fast proton,
 since
 the protons are room temperature or a couple of eV at most. The actual
 fusion
 process being mediated purely by tunneling. So perhaps gamma-less p+Li7
 fusion
 is indeed possible?

 Regards,

 Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

It would not be fair to criticize the Lugano team for not measuring the gas
in their experiment.  Rossi lent them a reactor that was not designed for
sampling the product gas.  After the experiment, the seal had to be broken
open (there are various descriptions of how this happened).  To have
measured the gas left in the reactor at the end, the whole reactor would
have to be placed in a large ulta-high vacuum system with mechanical
feedthru attachments designed to break the seal on the reactor while inside
the UHV.  This would have been a large undertaking.  Having viewed their
setup, it was probably considerably out of their budget and scope.

On the other hand, the Parkhomov-like experiments, particularly with the
way MFMP has modified the seals, is well suited for gas sampling.  It is
highly desirable to sample the gas while the reactor is still hot and
before much of the hydrogen (isotope) gas can be re-absorbed into the metal
as LiH.  This is in my experimental plan.

Historically, credibly measuring He in the electrolytic PdD experiments was
hard - you have to prove it could not have come from atmospheric
contamination.  When the experiment is performed in high alumina reactor
tubes, as Alan Goldwater has shown, the gas pressure can be quite high
-200-600 PSI.  When sampled, the resulting pressure in the sample vessel
will be less than in the reaction tube due to the volume of the sample
container, but the sample pressure could easily be in the 30-100 PSI
range.  Finding a measurable percentage of the sample gas to be He could
confidently be determined to be a reaction product rather than
contamination due to the higher pressure of the sample container which
could not have been produced from atmospheric contamination.

Bob Higgins

On Wed, Apr 8, 2015 at 8:44 PM, Jones Beene jone...@pacbell.net wrote:

 Well on second look, at day 32, the internal helium pressure at 1200 C is
 about 2000 psi if indeed the Lugano excess heat calculation was correct (it
 wasn’t) which could arguably have been tolerated by the reactor. About 0.03
 moles of helium would have been produced at 8 MeV per atom to give the 1.5
 MW-hrs of dissipated excess heat, but as we know the Lugano excess heat
 calculation was grossly inflated by the incompetence of the Levi team.



 If the COP was closer to 1.5 as I suspect, then there would have been far
 less internal pressure from the accumulated helium – if lithium fusion was
 responsible. Thus, lithium fusion is not ruled out by pressure
 considerations. (but is ruled out by lack of gammas)



 *From:* Jones Beene



 Blaze- Disregard previous numbers. I’ll try to calculate the internal
 pressure at day 30 another way. The point remains that if lithium fusion is
 responsible for the gain, lots of helium needs to have been produced and
 the reactor probably could not have tolerated the pressure.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Bob Cook's message of Tue, 7 Apr 2015 22:57:00 -0700:
Hi,
Eric--

One additional idea.

What we have been considering is the formation of 8Be and its decay into two 
alpha particles with only spin energy involved.  

As I have suggested before, two anti-parallel spin He* particles may form in 
adjacent fcc Pd lattice locations that are stuffed tight with  2 deuterium 
nuclei.  The net spin of the two new He* particles is high--24 mev--but 
amounts to 0 net angular momentum when considered as one item.  However, each 
He* within the coherent system may be able to distribute its spin energy to 
the electrons in the vicinity, much as may happen with the decay of the 8Be 
nucleus.  The two LENR processes would be similar in this regard.

Bob   
[snip]
The p+Li7 reaction yields 17.35 MeV, not 24 MeV.

Based on this, and an assumption that the radius of a Helium nucleus is about 2
fm, I calculated the angular momentum and found it to be about 2.5 times that of
a photon, so in theory, a couple of photons could be emitted before the nucleus
lost too much angular momentum. As to the energy of those photons that would
depend on the frequency, and that is where the ground gets a bit squishy. If you
base it on the rotational frequency of the nucleus, then the first photon has an
energy of about 6 MeV. This is a powerful gamma and would be easily detected.
Regards,

Robin van Spaandonk

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

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-09 Thread Jones Beene

From: torulf.gr...@bredband.net 
May be of interest.
https://fys.kuleuven.be/iks/ns/files/thesis/raabephdthesis.pdf
-

This is a provocative paper on the He-6 halo nucleus. The spin energy of the 
halo is found in the 500 keV range. Maybe that is why Terry (of the spin 
cartel, Dirac wing) chose this value – as a reference to Hotson. 

It occurs to me - on trying to fit all of this new information into the dogbone 
mechanics, and combined with Meulenberg’s hypothesis on the availability of the 
DDL interaction - vis-à-vis the lithium nucleus in this case – that we could be 
seeing a glimmer of the answer. Many disparate details are falling into place. 
We also must add a revised version of “neutron hopping” (neutron transfer 
reactions) into the mix. The complete hypothesis is complex, but manageable. If 
it was simple, it would have been accepted years ago. Here is the “hopping” 
paper, but we need to add the DDL as a species which will substitute for a 
neutron in energy-neutral transfers.

http://www.newenergytimes.com/v2/library/1993/1993Hagelstein-Neutron-Transfer-Reactions-ICCF4.pdf

Part of the gain in the dogbone is spin-based, nuclear, non-fusion and 
gamma-free. Another part comes from hydrogen – in dropping to ground state 
redundancy. Actually, that gain is spin-based as well. The total gain involves 
dense hydrogen in the dark-matter (DDL) state which interacts lithium. 

Part of the answer, essentially, could be that DDL hydrogen atoms (highest 
state of redundancy, identifiable as dark matter) are substituting (hopping) as 
if they were neutrons, in Li-6 converting it into He-6 and then back again. 
Spin energy is released.

This hybrid mechanism is worth fleshing out in a dedicated thread. The best 
part of this hypothesis is that -- like other variants which depend on the 
reality of DDL/dark matter, this hypothesis is falsifiable. Easily falsifiable.

There is a soft x-ray signature for DDL/dark matter. The ways to detect it are 
available, but not with standard Geiger counters and monitors. This signature 
has actually been seen in LENR as far back as the mid-1990s. It can be detected 
with the proper device.

https://www.mail-archive.com/vortex-l%40eskimo.com/msg101783.html

Jones

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In reply to  Jones Beene's message of Wed, 8 Apr 2015 11:07:16 -0700:
Hi,

Li7 may give up a neutron to become Li6. If Li6 also gives up a neutron, it
would become Li5 which immediately decays into He4 plus a proton.


 

Surprise, surprise. 

 

Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

 

http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

 

I not believe this new revelation is credible, based on the appearance of the 
paper and the timing, since  he has never before said that helium was 
discovered. 

 

The guy is desperate for credibility.

 

 

 

From: Bob Higgins 

 

Jones,  What is your evidence for your statement:

 

The Lugano isotope data, even if it could be believed, completely negates the 
entire scenario since Li-7 is NOT depleted according to the Lugano report - 
but instead is converted to Li-6. 

 

First of all, there is a crude assay based on the size of the pure sphere - 
and no evidence of large imbalance of Li-7 elsewhere. More importantly, 85 
years of nuclear physics can present no thermal process where the bulk 
isotopic distribution varies more than a few percent per stage, yet the Lugano 
report, if it can be believed shows extremely pure Li-6 appearing in what is 
essentially one stage in one sample  many orders of magnitude purer than any 
know process can deliver. 

 

There are three possibilities  either the starting material was enriched in 
pure Li-6, which is most likely, or else the process of heat generation has 
converted the missing Li-7 into Li-6, which is endothermic, and unlikely to 
have happened in a process where excess heat is generated. The third 
possibility is that the ash was spiked with pure isotope.

 

Neither of these possibilities can in any way support a conclusion of 
lithium-7 plus proton fusion, especially with the lack of the expected gamma, 
and no indication of helium. 

 

To say that Levis crew did not test for helium is a complete cop-out and only 
indicative of further incompetence on the part of this team. With this claimed 
excess heat over 30 days there should have been a large amount of helium, 
actual overpressure: that is - if lithium fusion were taking place. A sample 
of gas should at least have been stored for later testing.

 

Most likely conclusion  Rossi understood from the start that lithium-6 is the 
active isotope, and he provided fuel which was highly enriched, and at the 
same time, provided a different fuel for the testing of the before sample. 
Only Rossi handled this fuel. He had complete control, and no one complained. 
BTW - The cost of that much lithium-6 (about 50 milligrams) available from 
several suppliers, is about $10.

 

Jones

 

 

What I drew from the report was the only thing that can be concluded was that 
the 7Li is more commensurate to the 6Li in the ash as compared to the fuel.  
There was no mass assay that determined how much total Li was present in the 
ash compared to the fuel.  We know that physically, a lot of the Li will be on 
the walls of the alumina tube, so we don't have any idea of the absolute 
depletion of Li mass in the reaction.

 

While it is possible that the 7Li is converted to 6Li, it is only one of the 
possibilities.  The ICP-MS analysis is a full volume analysis and showed both 
Li isotopes near equal in percentage in the ash.  How these isotopes became 
nearly equal is just blind speculation at the moment without further 
experimental data.  All of the possibilities for the ratio change from fuel to 
ash should be laid out and the plausibility of each examined.

 

Bob
Regards,

Robin van Spaandonk

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

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-09 Thread Jones Beene

From: Bob Higgins 

 

Ø  To have measured the gas left in the reactor at the end, the whole reactor 
would have to be placed in a large ulta-high vacuum system with mechanical 
feedthru attachments designed to break the seal on the reactor while inside the 
UHV. 

 

No way. Hot tapping machines are available everywhere. The only accommodation 
would be a diamond bit to go through the alumina. Any skilled plumber could do 
this.

 

Ø  This would have been a large undertaking.  Having viewed their setup, it was 
probably considerably out of their budget and scope.

 

Again – no way - $1000 max plus time. All it takes is competence. We are 
talking about massive amounts of helium which would have been produced which is 
completely different from electrolysis.

 

Ø  Historically, credibly measuring He in the electrolytic PdD experiments was 
hard - you have to prove it could not have come from atmospheric contamination. 
 

 

Geeze-Louise, those were subwatt systems. This is a 1.5 megawatt-hour system – 
millions of time more output - which would have produced .03 moles of helium 
over the run. There would be no contamination from helium in air. If most of 
the gas were leaking out during the run - any good helium leak detector would 
have seen it.

 

Finding a measurable percentage of the sample gas to be He could confidently be 
determined to be a reaction product rather than contamination due to the higher 
pressure of the sample container which could not have been produced from 
atmospheric contamination.

Bob – you are living in the dark ages of subwatt electrolysis systems - and 
trying to defend a bunch of incompetent experimenters. 

There would be no problem finding helium – when as ash, it was responsible for 
1.5 MW-hrs of energy over 32 days – and this is essentially what Cook and Rossi 
are now saying. Levi’s work was beyond bad. It cannot be defended.

Jones

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Piantelli said that he has seen ~500 keV gammas (didn't say how many and if
always present) and he tested for beta+/beta- annihilation and did not find
the dual photon signature for that.  I was asking if he had seen a DDL
signature for compaction of the H- anion.

The Bremsstrahlung from heavy particles like protons is not as prominent as
for light particles like electrons.  The Bremsstrahlung has to do with the
deceleration and light particles stop much more quickly.  Heavy particles
of a given MeV are going slower, and they stop more slowly with various
Compton scatterings.  So I don't know if you would readily see
Bremsstrahlung from a 6 MeV proton.

On Thu, Apr 9, 2015 at 10:04 AM, Bob Cook frobertc...@hotmail.com wrote:

  Bob--

 Does Piantelli say there are energetic EM photons seen in the reaction?  6
 Mev protons would probably produce observable EM radiation, unless they
 were all consumed in the subsequent LENR reaction.  That seems unlikely to
 me.

 Bob Cook

 - Original Message -
 *From:* Bob Higgins rj.bob.higg...@gmail.com
 *To:* vortex-l@eskimo.com
 *Sent:* Thursday, April 09, 2015 8:09 AM
 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

 In follow-up hypothetical analysis of the Lugano measurements, consider
 this.  Look at what it means for the ICP-MS assay of the fuel to have
 94.1% 7Li and 5.9% 6Li.  With 100mg of LiAlH4 fuel source, the fuel source
 had 17.2mg of 7Li and 1.08mg of 6Li.  If one *presumes* that 6Li is not
 being created and doesn't participate in the reaction; then in the ash
 there will still be 1.08mg of 6Li.  The ICP-MS analysis of the ash shows
 that there is 42.5% of 7Li and 57.5% of 6Li.  Since (by presumption) there
 is still 1.08mg of 6Li left in the ash, there is only 0.79 mg of 7Li in the
 ash.  The amount of 7Li has decreased from 17.2mg to 0.79mg from fuel to
 ash.  Thus, only 0.79mg/17.2mg or only about 1/22 of the original 7Li
 remains in the ash - based on the presumption that no 6Li was created.
 Because the reaction showed no major output heat decline due to only 1/22
 of the original 7Li being present by the end of the reaction, it suggests
 to me that the 7Li may not be the primary source of heat in the reaction.
 As an aside, if the heat produced over the course of the experiment was due
 solely to the burning of 7Li, the consumption of (17.2 - 0.79) = 16.41 mg
 of 7Li would require the reaction to produce ~8.4 MeV per atom of burned
 7Li (based on the revised heat output of the Lugano experiment).
  More likely the hypothesis that 6Li is not created and 7Li burning
 produces the heat is not correct.  This hypothetical argument suggests that
 the 7Li is participating in the reaction (perhaps producing some excess
 heat), some 6Li is probably being created in the reaction, and much of the
 heat is coming from some other reaction - perhaps the
 transmutation/isotopic shift in the Ni which was not depleted by the end of
 the reaction.

 Piantelli's theory supports this.  He uses Li as a booster for his
 reactors - using the Li to create excess heat from the 6 MeV protons being
 produced (resulting in more than 6 MeV of heat per proton).  However, he
 does have excess heat without the Li.

 Bob Higgins

 On Thu, Apr 9, 2015 at 1:30 AM, mix...@bigpond.com wrote:

 In reply to  Bob Higgins's message of Wed, 8 Apr 2015 10:12:11 -0600:
 Hi,
 [snip]
 The ICP-MS analysis is a full volume analysis and showed both Li
 isotopes near equal in percentage in the ash.

 Just a thought: If the Li was acting as a nuclear catalyst, shuttling
 back and
 forth between Li6-Li7, then a roughly equal distribution on the whole
 might be
 expected, since a preponderance of one over the other would lead to an
 increase
 in the number of reactions of the predominant isotope, resulting in more
 of them
 being converted to the other.

 i.e. an excess of Li7 would yield more reactions converting Li7 to Li6,
 and an
 excess of Li6 would result in more reactions converting Li6 to Li7.

 Regards,

 Robin van Spaandonk

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

It is my understanding that even quark spin can be aligned in a strong magnetic 
field, since a magnetic field can penetrate right through a nucleus.  It does 
not stop at the boundary by some magnetic stop sign.  Thus I would say even 
your isospin particles can be polarized in a strong magnetic field in that  
their constituent quarks would be polarized.  Somebody who has a better 
knowledge of quark spin may be able to correct this understanding.  

Bob

Bob 
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:47 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  isospin is a product of the strong force and of the quarks inside the protons 
and neutrons. It is fixed no matter how the atoms spins. An atom might be 
induced to spin using EMF but usually that spin cannot effect the isospin of 
the nucleus. with zero isospin. Only non zero isospins are effected by RF. 


  On Wed, Apr 8, 2015 at 1:37 AM, Bob Cook frobertc...@hotmail.com wrote:

Two particles spinning anti-parallel equal 0 spin if they each have an 
equal spin energy.  Angular momentum is a vector quantity, not a scalar one.  

Bob
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:16 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi


  I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin. 
How can a nuclear reaction involving them have  huge annular momentum?


  On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com 
wrote:

Hi Bob, 


The possibility you've been drawing attention to, that the result of 
the decay of the [8Be]* compound nucleus into two 4He nuclei with little linear 
momentum and a great deal of angular momentum makes for an interesting thought 
experiment.  Out of curiosity, I calculated the energy that would be needed to 
break up an alpha particle into either tritium and a proton or 3He and a 
neutron, which would be the reverse of these two reactions:


3He + n → 4He + Q (19.3 MeV)
t + p → 4He + Q (20.5 MeV)


As I understand it, this implies that angular momentum sufficient to 
produce ~ 19 MeV of centripetal force would be needed to break apart a 4He into 
either 3He and a neutron or tritium and a proton.  This suggests that a 4He can 
carry a large amount of angular momentum before it is likely to break apart. (I 
assume the process is probabilistic and that the force needed lies along a 
distribution.)


Further comments inline.


Eric




On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com 
wrote:


  However, I know of know reason why the light nuclei cannot have any 
spin quantum number--high or low.  Any spin quantum is available.


Further to the thought experiment, I think we should make a clear 
distinction between two types of spin -- there's the actual spinning motion 
of a nucleus (e.g., 4He), and there is the spin state of the nucleus.  At 
higher rates of rotation, a heavy nucleus such as an isotope of nickel will 
reconfigure into a higher spin state, presumably through deformation.  In such 
a state a photon may be emitted, with the nucleus relaxing into a lower spin 
state.  Here my mental model is of neodymium magnets spinning around in a 
clump.  When they snap together into a lower-energy configuration, a photon is 
emitted through the movement of the magnets as they snap together.  The photon 
is emitted in a direction and carries away energy in such a way as to slow the 
angular movement of the spinning nucleus a little (by the amount of energy 
carried away by the photon).  The participants involved in such a transition 
are the nucleons, and the energy of the photon that is emitted will 
correspondingly be in the keV or MeV range, which is in the nuclear range.


A light nucleus, such as 4He, does not have a bound excited state.  My 
understanding is that it cannot deform under high angular momentum into a 
higher energy state which will emit a photon when it relaxes.  The 4He will 
either break apart into lighter constituents under centrifugal forces or it 
will not.  But I'm guessing that the actual moment-to-moment velocity of the 
4He about its axis of motion is in principle a continuous quantity.  If this is 
true, perhaps the energy could be released to the environment in small amounts.


Where the thought experiment gets interesting is in the supposition 
that you and others have already offered in this thread, that charged body such 
as a 4He nucleus that is spinning at an incredible rate will set up a magnetic 
field.  This magnetic field could disturb nearby electrons, causing them to 
emit lower energy photons in the process.


Although I do not see anything special in the 7Li+p to 8Be transition 
that has been

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Axil--

Note the following conclusion of the Wiki item on neutron magnetic moment:

  While the results of this calculation are encouraging, the masses of 
the up or down quarks were assumed to be 1/3 the mass of a nucleon,[31] whereas 
the masses of these quarks are only about 1% that of a nucleon.[32] The 
discrepancy stems from the complexity of the Standard Model for nucleons, where 
most of their mass originates in the gluon fields and virtual particles that 
are essential aspects of the strong force.[32] Further, the complex system of 
quarks and gluons that constitute a neutron requires a relativistic treatment. 
A calculation of nucleon magnetic moments from first principles is not yet 
available.  


How do you know about the magnetic properties of isospin particles and their 
reaction to  a magnetic field?

Bob
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:47 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  isospin is a product of the strong force and of the quarks inside the protons 
and neutrons. It is fixed no matter how the atoms spins. An atom might be 
induced to spin using EMF but usually that spin cannot effect the isospin of 
the nucleus. with zero isospin. Only non zero isospins are effected by RF. 


  On Wed, Apr 8, 2015 at 1:37 AM, Bob Cook frobertc...@hotmail.com wrote:

Two particles spinning anti-parallel equal 0 spin if they each have an 
equal spin energy.  Angular momentum is a vector quantity, not a scalar one.  

Bob
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:16 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi


  I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin. 
How can a nuclear reaction involving them have  huge annular momentum?


  On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com 
wrote:

Hi Bob, 


The possibility you've been drawing attention to, that the result of 
the decay of the [8Be]* compound nucleus into two 4He nuclei with little linear 
momentum and a great deal of angular momentum makes for an interesting thought 
experiment.  Out of curiosity, I calculated the energy that would be needed to 
break up an alpha particle into either tritium and a proton or 3He and a 
neutron, which would be the reverse of these two reactions:


3He + n → 4He + Q (19.3 MeV)
t + p → 4He + Q (20.5 MeV)


As I understand it, this implies that angular momentum sufficient to 
produce ~ 19 MeV of centripetal force would be needed to break apart a 4He into 
either 3He and a neutron or tritium and a proton.  This suggests that a 4He can 
carry a large amount of angular momentum before it is likely to break apart. (I 
assume the process is probabilistic and that the force needed lies along a 
distribution.)


Further comments inline.


Eric




On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com 
wrote:


  However, I know of know reason why the light nuclei cannot have any 
spin quantum number--high or low.  Any spin quantum is available.


Further to the thought experiment, I think we should make a clear 
distinction between two types of spin -- there's the actual spinning motion 
of a nucleus (e.g., 4He), and there is the spin state of the nucleus.  At 
higher rates of rotation, a heavy nucleus such as an isotope of nickel will 
reconfigure into a higher spin state, presumably through deformation.  In such 
a state a photon may be emitted, with the nucleus relaxing into a lower spin 
state.  Here my mental model is of neodymium magnets spinning around in a 
clump.  When they snap together into a lower-energy configuration, a photon is 
emitted through the movement of the magnets as they snap together.  The photon 
is emitted in a direction and carries away energy in such a way as to slow the 
angular movement of the spinning nucleus a little (by the amount of energy 
carried away by the photon).  The participants involved in such a transition 
are the nucleons, and the energy of the photon that is emitted will 
correspondingly be in the keV or MeV range, which is in the nuclear range.


A light nucleus, such as 4He, does not have a bound excited state.  My 
understanding is that it cannot deform under high angular momentum into a 
higher energy state which will emit a photon when it relaxes.  The 4He will 
either break apart into lighter constituents under centrifugal forces or it 
will not.  But I'm guessing that the actual moment-to-moment velocity of the 
4He about its axis of motion is in principle a continuous quantity.  If this is 
true, perhaps the energy could be released to the environment in small amounts.


Where the thought experiment gets interesting is in the supposition

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Bob Higgins

Jones,  What is your evidence for your statement:

The Lugano isotope data, even if it could be believed, completely negates
the entire scenario since Li-7 is NOT depleted according to the Lugano
report - but instead is converted to Li-6. 


What I drew from the report was the only thing that can be concluded was
that the 7Li is more commensurate to the 6Li in the ash as compared to the
fuel.  There was no mass assay that determined how much total Li was
present in the ash compared to the fuel.  We know that physically, a lot of
the Li will be on the walls of the alumina tube, so we don't have any idea
of the absolute depletion of Li mass in the reaction.

While it is possible that the 7Li is converted to 6Li, it is only one of
the possibilities.  The ICP-MS analysis is a full volume analysis and
showed both Li isotopes near equal in percentage in the ash.  How these
isotopes became nearly equal is just blind speculation at the moment
without further experimental data.  All of the possibilities for the ratio
change from fuel to ash should be laid out and the plausibility of each
examined.

Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Bob,

 

Axil is correct in that the Be-8 and He-4 cannot project large spin energy 
transfer. It does not help that spin can be anti-parallel when it is based on 0 
spin particles to begin with (they would not decay if that was the case) … 
however…

 

There is an interesting isotope of helium, generally neglected- which is indeed 
a high-spin halo nucleus – He-6. The isotope has been overlooked in LENR - but 
is so important in physics that a book has been written about it.

 

“A Cluster Model of Helium-6 and Lithium-6” by Jeremy Robert Armstrong - 
Michigan State University, Department of Physics and Astronomy, 2007 - 372 
pages. Obviously the focus is cosmology, but the fact that this high spin 
isotope is known physics is important.

 

Helium-6 is short-lived with a sub-second lifetime, but it is a borromean 
system, and thus is remarkably stable for such a high spin nucleus, and could 
convert back to lithium-6 after it has transferred MeV of spin energy to 
magnons without the timing problems faced by Hagelstein’s magic phonons.

 

http://books.google.com/books/about/A_Cluster_Model_of_Helium_6_and_Lithium.html?id=8MBxqPpWAF8C

 

I think this idea of spin energy transfer from Lithium-6 - Helium-6 in a 
reversible reaction powered by the strong force – and possibly even completely 
divorced from the nuclear fusion reaction, could actually be salvaged -- by the 
basic idea that lithium converts to high spin helium, and then back. Lithium-6 
is a singularity in many ways. 

 

BUT, correspondingly, the helium at the start is not Li-7, as Cook and Rossi 
claim – it is Li-6. In fact the Rossi premise is looking so bad that it is 
completely unsalvageable. The Lugano isotope data, even if it could be 
believed, completely negates the entire scenario since Li-7 is NOT depleted 
according to the Lugano report - but instead is converted to Li-6. 

 

However, a novel QCD reaction where Li-6 - He-6 oscillate in a reversible 
reaction, powered by the strong force, is worth pursuing – as it can provide 
spin energy to magnons at the expense of gluon mass. 

 

It’s too bad that Armstrong’s book on the above topic is not available as an 
ebook. It could be important to the idea that spin energy without high energy 
radiation, is the key to understanding this type of reaction.

 

From: Bob Cook 

 

Two particles spinning anti-parallel equal 0 spin if they each have an equal 
spin energy.  Angular momentum is a vector quantity, not a scalar one.  

 

From: Axil Axil mailto:janap...@gmail.com  

 

I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin. How 
can a nuclear reaction involving them have  huge annular momentum?

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Axil Axil

In the one example in which we have a full accounting of the element
percentage in  the fuel and also as transmuted in the ash is the DGT
transmutation assay provided in the ICCF-17 paper.  In that assay, there
was a large percentage increase in light elements including lithium,
beryllium, and boron.

If past is prolong, I would expect that the Lithium 6 seen in the Lagano
ash assay was produced through some undefined LENR nuclear fusion reaction
involving hydrogen.

On Wed, Apr 8, 2015 at 12:12 PM, Bob Higgins rj.bob.higg...@gmail.com
wrote:

 Jones,  What is your evidence for your statement:

 The Lugano isotope data, even if it could be believed, completely negates
 the entire scenario since Li-7 is NOT depleted according to the Lugano
 report - but instead is converted to Li-6. 


 What I drew from the report was the only thing that can be concluded was
 that the 7Li is more commensurate to the 6Li in the ash as compared to the
 fuel.  There was no mass assay that determined how much total Li was
 present in the ash compared to the fuel.  We know that physically, a lot of
 the Li will be on the walls of the alumina tube, so we don't have any idea
 of the absolute depletion of Li mass in the reaction.

 While it is possible that the 7Li is converted to 6Li, it is only one of
 the possibilities.  The ICP-MS analysis is a full volume analysis and
 showed both Li isotopes near equal in percentage in the ash.  How these
 isotopes became nearly equal is just blind speculation at the moment
 without further experimental data.  All of the possibilities for the ratio
 change from fuel to ash should be laid out and the plausibility of each
 examined.

 Bob

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

I am not surprised that He has not been reported from the Lugano E-Cat test 
heretofore.

Helium is hard to collect, being an inert gas, and at temperatures it diffuses 
rapidly in porous materials.  I would have said much of the He in the Lugano 
test would have escaped the reactor, either during operation or upon opening 
for inspection.  Rossi may have gone to some extent to collect the He that he 
is now reporting to confirm the Rossi--Cook theory of its generation.  

I am surprised at the suggested incredible occurrence of He in the Hot Cat 
test.  It has been reported by SPAWARS and several others in early LENR 
experiments and has been associated with excess heat.  Some of these 
experiments included Li in the reaction, making it a not-uncommon possible 
reactant in cases where He was actually identified as a product.  

This conversation leads me to guess at another mechanism to get to the high 
Li-6 ratio Jones indicates is difficult to reach by any known means, including 
expensive isotope separation processes.  That mechanism would be the generation 
of Li-6 from deuterium and or protium directly in the Ni and Pd lattices.  It 
may be that Li-6 was not necessary to produce  He in the Pd lattice, but is 
necessary in the Li--Ni lattice system.  In other words in the Ni system to 
arrive at the stable He nuclei it is necessary to go through the Be-8 
configuration, using every bit of Li-7 available, and producing new Li-7 via 
He-6 and Li-6 or some other route making use of the available protium as the 
feed stock.

If the Lugano analysis of the ash for Li-6 ratio is accurate, I do not believe 
that Li-6 could have been added after the completion of the test, given the 
difficulty noted above to make such a highly concentrated Li-6 batch of metal 
by any known means--Jones's observation, with which I agree.  

It is clear that Li-7 is a lot better liked by nature than Li-6 given their 
natural ratios.  The wonder is that there is any Li-6 around if the natural 
generation was via He-4.  As Jones apply points out He-6 may be the smoking gun 
to get to Li-6 and hence back to He the stable entity which nature likes 
because of its high binding energy.  The Second Law has strange ways of 
expressing itself, particularly when it comes to nuclear reactions and coherent 
systems. 

Jones has suggested the coupling of Spin energy of a composite particle with 
the strong force/energy field provided by gluons and the effective mass they 
add to composite particles.  It is suggested that the two sources of 
composite particle energy may be exchangeable in terms of mass.  I have not 
heard of this, however, it may be the case.  

Assuming a wave function exists for composite particles, then this coupling I 
assume would be evident in the mathematics of the wave function.  Does anyone 
have knowledge of papers relative to this issue of the mechanism for the 
exchange of spin energy to mass.  (It should involve the conservation of 
angular momentum as well as energy.)

Bob Cook

  


  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Wednesday, April 08, 2015 10:51 AM
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  From: Bob Higgins 

   

  Jones,  What is your evidence for your statement:

   

The Lugano isotope data, even if it could be believed, completely negates 
the entire scenario since Li-7 is NOT depleted according to the Lugano report - 
but instead is converted to Li-6. 

   

  First of all, there is a crude assay based on the size of the pure sphere - 
and no evidence of large imbalance of Li-7 elsewhere. More importantly, 85 
years of nuclear physics can present no thermal process where the bulk isotopic 
distribution varies more than a few percent per stage, yet the Lugano report, 
if it can be believed shows extremely pure Li-6 appearing in what is 
essentially one stage in one sample – many orders of magnitude purer than any 
know process can deliver. 

   

  There are three possibilities – either the starting material was enriched in 
pure Li-6, which is most likely, or else the process of heat generation has 
converted the missing Li-7 into Li-6, which is endothermic, and unlikely to 
have happened in a process where excess heat is generated. The third 
possibility is that the ash was spiked with pure isotope.

   

  Neither of these possibilities can in any way support a conclusion of 
lithium-7 plus proton fusion, especially with the lack of the expected gamma, 
and no indication of helium. 

   

  To say that Levi’s crew did not test for helium is a complete cop-out and 
only indicative of further incompetence on the part of this team. With this 
claimed excess heat over 30 days there should have been a large amount of 
helium, actual overpressure: that is - if lithium fusion were taking place. A 
sample of gas should at least have been stored for later testing.

   

  Most likely conclusion – Rossi

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

I would have said Rossi is the most credible person in the field of LENR and 
not desperate for credibility given his apparent RD knowhow.  

Bob Cook  
  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Wednesday, April 08, 2015 11:07 AM
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

  Surprise, surprise. 

  Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

  http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

  I not believe this new revelation is credible, based on the appearance of the 
paper and the timing, since  he has never before said that helium was 
discovered. 

  The guy is desperate for credibility.

  From: Bob Higgins 

  Jones,  What is your evidence for your statement:

The Lugano isotope data, even if it could be believed, completely negates 
the entire scenario since Li-7 is NOT depleted according to the Lugano report - 
but instead is converted to Li-6. 

  First of all, there is a crude assay based on the size of the pure sphere - 
and no evidence of large imbalance of Li-7 elsewhere. More importantly, 85 
years of nuclear physics can present no thermal process where the bulk isotopic 
distribution varies more than a few percent per stage, yet the Lugano report, 
if it can be believed shows extremely pure Li-6 appearing in what is 
essentially one stage in one sample – many orders of magnitude purer than any 
know process can deliver. 

  There are three possibilities – either the starting material was enriched in 
pure Li-6, which is most likely, or else the process of heat generation has 
converted the missing Li-7 into Li-6, which is endothermic, and unlikely to 
have happened in a process where excess heat is generated. The third 
possibility is that the ash was spiked with pure isotope.

  Neither of these possibilities can in any way support a conclusion of 
lithium-7 plus proton fusion, especially with the lack of the expected gamma, 
and no indication of helium. 

  To say that Levi’s crew did not test for helium is a complete cop-out and 
only indicative of further incompetence on the part of this team. With this 
claimed excess heat over 30 days there should have been a large amount of 
helium, actual overpressure: that is - if lithium fusion were taking place. A 
sample of gas should at least have been stored for later testing.

  Most likely conclusion – Rossi understood from the start that lithium-6 is 
the active isotope, and he provided fuel which was highly enriched, and at the 
same time, provided a different fuel for the testing of the “before” sample. 
Only Rossi handled this fuel. He had complete control, and no one complained. 
BTW - The cost of that much lithium-6 (about 50 milligrams) available from 
several suppliers, is about $10.

  Jones

  What I drew from the report was the only thing that can be concluded was that 
the 7Li is more commensurate to the 6Li in the ash as compared to the fuel.  
There was no mass assay that determined how much total Li was present in the 
ash compared to the fuel.  We know that physically, a lot of the Li will be on 
the walls of the alumina tube, so we don't have any idea of the absolute 
depletion of Li mass in the reaction.

  While it is possible that the 7Li is converted to 6Li, it is only one of the 
possibilities.  The ICP-MS analysis is a full volume analysis and showed both 
Li isotopes near equal in percentage in the ash.  How these isotopes became 
nearly equal is just blind speculation at the moment without further 
experimental data.  All of the possibilities for the ratio change from fuel to 
ash should be laid out and the plausibility of each examined.

  Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

 

Surprise, surprise. 

 

Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

 

http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

 

I not believe this new revelation is credible, based on the appearance of the 
paper and the timing, since  he has never before said that helium was 
discovered. 

 

The guy is desperate for credibility.

 

 

 

From: Bob Higgins 

 

Jones,  What is your evidence for your statement:

 

The Lugano isotope data, even if it could be believed, completely negates the 
entire scenario since Li-7 is NOT depleted according to the Lugano report - but 
instead is converted to Li-6. 

 

First of all, there is a crude assay based on the size of the pure sphere - and 
no evidence of large imbalance of Li-7 elsewhere. More importantly, 85 years of 
nuclear physics can present no thermal process where the bulk isotopic 
distribution varies more than a few percent per stage, yet the Lugano report, 
if it can be believed shows extremely pure Li-6 appearing in what is 
essentially one stage in one sample – many orders of magnitude purer than any 
know process can deliver. 

 

There are three possibilities – either the starting material was enriched in 
pure Li-6, which is most likely, or else the process of heat generation has 
converted the missing Li-7 into Li-6, which is endothermic, and unlikely to 
have happened in a process where excess heat is generated. The third 
possibility is that the ash was spiked with pure isotope.

 

Neither of these possibilities can in any way support a conclusion of lithium-7 
plus proton fusion, especially with the lack of the expected gamma, and no 
indication of helium. 

 

To say that Levi’s crew did not test for helium is a complete cop-out and only 
indicative of further incompetence on the part of this team. With this claimed 
excess heat over 30 days there should have been a large amount of helium, 
actual overpressure: that is - if lithium fusion were taking place. A sample of 
gas should at least have been stored for later testing.

 

Most likely conclusion – Rossi understood from the start that lithium-6 is the 
active isotope, and he provided fuel which was highly enriched, and at the same 
time, provided a different fuel for the testing of the “before” sample. Only 
Rossi handled this fuel. He had complete control, and no one complained. BTW - 
The cost of that much lithium-6 (about 50 milligrams) available from several 
suppliers, is about $10.

 

Jones

 

 

What I drew from the report was the only thing that can be concluded was that 
the 7Li is more commensurate to the 6Li in the ash as compared to the fuel.  
There was no mass assay that determined how much total Li was present in the 
ash compared to the fuel.  We know that physically, a lot of the Li will be on 
the walls of the alumina tube, so we don't have any idea of the absolute 
depletion of Li mass in the reaction.

 

While it is possible that the 7Li is converted to 6Li, it is only one of the 
possibilities.  The ICP-MS analysis is a full volume analysis and showed both 
Li isotopes near equal in percentage in the ash.  How these isotopes became 
nearly equal is just blind speculation at the moment without further 
experimental data.  All of the possibilities for the ratio change from fuel to 
ash should be laid out and the plausibility of each examined.

 

Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

From: Bob Higgins 

 

Jones,  What is your evidence for your statement:

 

The Lugano isotope data, even if it could be believed, completely negates the 
entire scenario since Li-7 is NOT depleted according to the Lugano report - but 
instead is converted to Li-6. 

 

First of all, there is a crude assay based on the size of the pure sphere - and 
no evidence of large imbalance of Li-7 elsewhere. More importantly, 85 years of 
nuclear physics can present no thermal process where the bulk isotopic 
distribution varies more than a few percent per stage, yet the Lugano report, 
if it can be believed shows extremely pure Li-6 appearing in what is 
essentially one stage in one sample – many orders of magnitude purer than any 
know process can deliver. 

 

There are three possibilities – either the starting material was enriched in 
pure Li-6, which is most likely, or else the process of heat generation has 
converted the missing Li-7 into Li-6, which is endothermic, and unlikely to 
have happened in a process where excess heat is generated. The third 
possibility is that the ash was spiked with pure isotope.

 

Neither of these possibilities can in any way support a conclusion of lithium-7 
plus proton fusion, especially with the lack of the expected gamma, and no 
indication of helium. 

 

To say that Levi’s crew did not test for helium is a complete cop-out and only 
indicative of further incompetence on the part of this team. With this claimed 
excess heat over 30 days there should have been a large amount of helium, 
actual overpressure: that is - if lithium fusion were taking place. A sample of 
gas should at least have been stored for later testing.

 

Most likely conclusion – Rossi understood from the start that lithium-6 is the 
active isotope, and he provided fuel which was highly enriched, and at the same 
time, provided a different fuel for the testing of the “before” sample. Only 
Rossi handled this fuel. He had complete control, and no one complained. BTW - 
The cost of that much lithium-6 (about 50 milligrams) available from several 
suppliers, is about $10.

 

Jones

 

 

What I drew from the report was the only thing that can be concluded was that 
the 7Li is more commensurate to the 6Li in the ash as compared to the fuel.  
There was no mass assay that determined how much total Li was present in the 
ash compared to the fuel.  We know that physically, a lot of the Li will be on 
the walls of the alumina tube, so we don't have any idea of the absolute 
depletion of Li mass in the reaction.

 

While it is possible that the 7Li is converted to 6Li, it is only one of the 
possibilities.  The ICP-MS analysis is a full volume analysis and showed both 
Li isotopes near equal in percentage in the ash.  How these isotopes became 
nearly equal is just blind speculation at the moment without further 
experimental data.  All of the possibilities for the ratio change from fuel to 
ash should be laid out and the plausibility of each examined.

 

Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread torulf.greek



May be of interest.


https://fys.kuleuven.be/iks/ns/files/thesis/raabephdthesis.pdf 

On
Wed, 8 Apr 2015 10:51:24 -0700, Jones Beene  wrote:   

FROM: Bob
Higgins  

Jones, What is your evidence for your statement: 

The
Lugano isotope data, even if it could be believed, completely negates
the entire scenario since Li-7 is NOT depleted according to the Lugano
report - but instead is converted to Li-6.

First of all, there is
a crude assay based on the size of the pure sphere - and no evidence of
large imbalance of Li-7 elsewhere. More importantly, 85 years of nuclear
physics can present no thermal process where the bulk isotopic
distribution varies more than a few percent per stage, yet the Lugano
report, if it can be believed shows extremely pure Li-6 appearing in
what is essentially one stage in one sample - many orders of magnitude
purer than any know process can deliver.  

There are three
possibilities - either the starting material was enriched in pure Li-6,
which is most likely, or else the process of heat generation has
converted the missing Li-7 into Li-6, which is endothermic, and unlikely
to have happened in a process where excess heat is generated. The third
possibility is that the ash was spiked with pure isotope. 

Neither of
these possibilities can in any way support a conclusion of lithium-7
plus proton fusion, especially with the lack of the expected gamma, and
no indication of helium.  

To say that Levi's crew did not test for
helium is a complete cop-out and only indicative of further incompetence
on the part of this team. With this claimed excess heat over 30 days
there should have been a large amount of helium, actual overpressure:
that is - if lithium fusion were taking place. A sample of gas should at
least have been stored for later testing. 

Most likely conclusion -
Rossi understood from the start that lithium-6 is the active isotope,
and he provided fuel which was highly enriched, and at the same time,
provided a different fuel for the testing of the before sample. Only
Rossi handled this fuel. He had complete control, and no one complained.
BTW - The cost of that much lithium-6 (about 50 milligrams) available
from several suppliers, is about $10. 

Jones 

What I drew from the
report was the only thing that can be concluded was that the 7Li is more
commensurate to the 6Li in the ash as compared to the fuel. There was no
mass assay that determined how much total Li was present in the ash
compared to the fuel. We know that physically, a lot of the Li will be
on the walls of the alumina tube, so we don't have any idea of the
absolute depletion of Li mass in the reaction.  

While it is possible
that the 7Li is converted to 6Li, it is only one of the possibilities.
The ICP-MS analysis is a full volume analysis and showed both Li
isotopes near equal in percentage in the ash. How these isotopes became
nearly equal is just blind speculation at the moment without further
experimental data. All of the possibilities for the ratio change from
fuel to ash should be laid out and the plausibility of each examined. 


Bob

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread torulf.greek



There have been p/Ni lenr with K and no Li. 

On Wed, 8 Apr 2015
12:18:36 -0700, Bob Cook  wrote:   
I am not surprised that He has not
been reported from the Lugano E-Cat test heretofore. 

Helium is hard to
collect, being an inert gas, and at temperatures it diffuses rapidly in
porous materials. I would have said much of the He in the Lugano test
would have escaped the reactor, either during operation or upon opening
for inspection. Rossi may have gone to some extent to collect the He
that he is now reporting to confirm the Rossi--Cook theory of its
generation.  

I am surprised at the suggested incredible occurrence of
He in the Hot Cat test. It has been reported by SPAWARS and several
others in early LENR experiments and has been associated with excess
heat. Some of these experiments included Li in the reaction, making it a
not-uncommon possible reactant in cases where He was actually identified
as a product.  

This conversation leads me to guess at another
mechanism to get to the high Li-6 ratio Jones indicates is difficult to
reach by any known means, including expensive isotope separation
processes. That mechanism would be the generation of Li-6 from deuterium
and or protium directly in the Ni and Pd lattices. It may be that Li-6
was not necessary to produce He in the Pd lattice, but is necessary in
the Li--Ni lattice system. In other words in the Ni system to arrive at
the stable He nuclei it is necessary to go through the Be-8
configuration, using every bit of Li-7 available, and producing new Li-7
via He-6 and Li-6 or some other route making use of the available
protium as the feed stock. 

If the Lugano analysis of the ash for Li-6
ratio is accurate, I do not believe that Li-6 could have been added
after the completion of the test, given the difficulty noted above to
make such a highly concentrated Li-6 batch of metal by any known
means--Jones's observation, with which I agree.  

It is clear that Li-7
is a lot better liked by nature than Li-6 given their natural ratios.
The wonder is that there is any Li-6 around if the natural generation
was via He-4. As Jones apply points out He-6 may be the smoking gun to
get to Li-6 and hence back to He the stable entity which nature likes
because of its high binding energy. The Second Law has strange ways of
expressing itself, particularly when it comes to nuclear reactions and
coherent systems.  

Jones has suggested the coupling of Spin energy of
a composite particle with the strong force/energy field provided by
gluons and the effective mass they add to composite particles. It is
suggested that the two sources of composite particle energy may be
exchangeable in terms of mass. I have not heard of this, however, it may
be the case.  

Assuming a wave function exists for composite particles,
then this coupling I assume would be evident in the mathematics of the
wave function. Does anyone have knowledge of papers relative to this
issue of the mechanism for the exchange of spin energy to mass. (It
should involve the conservation of angular momentum as well as energy.)


Bob Cook 

- Original Message - 
FROM: Jones Beene [1] 
TO:
vortex-l@eskimo.com [2] 
SENT: Wednesday, April 08, 2015 10:51 AM

SUBJECT: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author
Andrea Rossi 

FROM: Bob Higgins  

Jones, What is your evidence for
your statement: 

The Lugano isotope data, even if it could be
believed, completely negates the entire scenario since Li-7 is NOT
depleted according to the Lugano report - but instead is converted to
Li-6.

First of all, there is a crude assay based on the size of
the pure sphere - and no evidence of large imbalance of Li-7 elsewhere.
More importantly, 85 years of nuclear physics can present no thermal
process where the bulk isotopic distribution varies more than a few
percent per stage, yet the Lugano report, if it can be believed shows
extremely pure Li-6 appearing in what is essentially one stage in one
sample - many orders of magnitude purer than any know process can
deliver.  

There are three possibilities - either the starting material
was enriched in pure Li-6, which is most likely, or else the process of
heat generation has converted the missing Li-7 into Li-6, which is
endothermic, and unlikely to have happened in a process where excess
heat is generated. The third possibility is that the ash was spiked with
pure isotope. 

Neither of these possibilities can in any way support a
conclusion of lithium-7 plus proton fusion, especially with the lack of
the expected gamma, and no indication of helium.  

To say that Levi's
crew did not test for helium is a complete cop-out and only indicative
of further incompetence on the part of this team. With this claimed
excess heat over 30 days there should have been a large amount of
helium, actual overpressure: that is - if lithium fusion were taking
place. A sample of gas should at least have been stored for later
testing. 

Most likely conclusion - Rossi understood

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Bob Higgins

Jones, we DO know that there is a large imbalance in the distribution of
7Li in the ash.  Look at the difference between the SIMS results which
provide isotopic analysis of the material near the surface, and the results
of ICP-MS which is a bulk analysis of the particle.  The surface shows
change of 7Li from 92% in fuel to 8% in ash, with the 6Li going from 8% in
fuel to 92% in the ash.  The ICP-MS bulk analysis shows that the overall
fuel particle had 94% 7Li and 6% 6Li in the fuel going to 58% 7Li and 42%
6Li in the ash.  Clearly there is much more change of 7Li / 6Li on the
surface than in the overall bulk.  Whatever is happening to the Li isotopes
is happening at an apparent greater rate at the surface of the particle.

Since the reaction is likely to occurring in a thin liquid LiH-Al alloy
film whetted to the surface of a Ni particle, it is logical to assume the
reaction may be occurring at the liquid interface to the Ni particle.  Yet
it is the solidified composite surface that shows the greatest depletion of
the 7Li.  Could the 6Li stratify to the surface during the high temperature
liquid phase of the film, owing to some physical difference between 6Li and
7Li (for example liquid density of 6LiH vs 7LiH)?

I am not questioning that the isotopic ratio is changing, only that the
change appears bigger at the surface than in the particle as a whole.

Bob

On Wed, Apr 8, 2015 at 11:51 AM, Jones Beene jone...@pacbell.net wrote:

 *From:* Bob Higgins



 Jones,  What is your evidence for your statement:



 The Lugano isotope data, even if it could be believed, completely negates
 the entire scenario since Li-7 is NOT depleted according to the Lugano
 report - but instead is converted to Li-6. 



 First of all, there is a crude assay based on the size of the pure sphere
 - and no evidence of large imbalance of Li-7 elsewhere. More importantly,
 85 years of nuclear physics can present no thermal process where the bulk
 isotopic distribution varies more than a few percent per stage, yet the
 Lugano report, if it can be believed shows extremely pure Li-6 appearing in
 what is essentially one stage in one sample – many orders of magnitude
 purer than any know process can deliver.



 There are three possibilities – either the starting material was enriched
 in pure Li-6, which is most likely, or else the process of heat generation
 has converted the missing Li-7 into Li-6, which is endothermic, and
 unlikely to have happened in a process where excess heat is generated. The
 third possibility is that the ash was spiked with pure isotope.



 Neither of these possibilities can in any way support a conclusion of
 lithium-7 plus proton fusion, especially with the lack of the expected
 gamma, and no indication of helium.



 To say that Levi’s crew did not test for helium is a complete cop-out and
 only indicative of further incompetence on the part of this team. With this
 claimed excess heat over 30 days there should have been a large amount of
 helium, actual overpressure: that is - if lithium fusion were taking place.
 A sample of gas should at least have been stored for later testing.



 Most likely conclusion – Rossi understood from the start that lithium-6 is
 the active isotope, and he provided fuel which was highly enriched, and at
 the same time, provided a different fuel for the testing of the “before”
 sample. Only Rossi handled this fuel. He had complete control, and no one
 complained. BTW - The cost of that much lithium-6 (about 50 milligrams)
 available from several suppliers, is about $10.



 Jones





 What I drew from the report was the only thing that can be concluded was
 that the 7Li is more commensurate to the 6Li in the ash as compared to the
 fuel.  There was no mass assay that determined how much total Li was
 present in the ash compared to the fuel.  We know that physically, a lot of
 the Li will be on the walls of the alumina tube, so we don't have any idea
 of the absolute depletion of Li mass in the reaction.



 While it is possible that the 7Li is converted to 6Li, it is only one of
 the possibilities.  The ICP-MS analysis is a full volume analysis and
 showed both Li isotopes near equal in percentage in the ash.  How these
 isotopes became nearly equal is just blind speculation at the moment
 without further experimental data.  All of the possibilities for the ratio
 change from fuel to ash should be laid out and the plausibility of each
 examined.



 Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

From: Bob Cook 

 

I am not surprised that He has not been reported from the Lugano E-Cat test 
heretofore.

 

 

Bob – As someone who is dedicated to seeking answers to the most important 
problem facing society in the coming decades – how to get off of addiction to 
fossil fuel - you should be both surprised and disappointed. Look at Levi’s CV. 
Numerous papers related to helium detection. Was he instructed not to test? Or 
did he test and withhold that information until now?

 

Levi and his team were reportedly paid half a million bucks for what amounts to 
pretending to do science. They did not calibrate. They royally screwed-up 
thermal measurement. They failed to collect thermocouple data. They apparently 
did not test the gas evolution either during the run (with a leak detector) or 
after. They let Rossi handle all the samples. The list of failures and gross 
incompetence goes on and on. 

 

In the end, there is little assurance that there was thermal gain at Lugano - 
and had it not been for Parkhomov, things would be much worse.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

 

 

Blaze- Disregard previous numbers. I’ll try to calculate the internal pressure 
at day 30 another way. The point remains that if lithium fusion is responsible 
for the gain, lots of helium needs to have been produced and the reactor 
probably could not have tolerated the pressure.

 

From: Blaze Spinnaker 

 

Ø  Jones, it is possible that helium was observed and was originally discounted 
as error.  That happens.

 

Not when this much claimed energy has been seen. 

 

Think about the implications. The Lugano experiment supposedly generated 2 kW 
excess for 30+ days. This is about 10^28 eV equivalent. If all this energy was 
coming from helium, as a result of lithium fusion, at 16 MeV a pop, then it 
amounts to several moles of gas. A mole of helium fills about 25 liters at room 
temp - so this would have been about 50 liters of helium. Even if they 
overestimated the gain by a factor of 10, and the excess was 200 watts, the 
reactor could not have survived the internal pressure.

 

Anyway – the Lugano report was supposed to be a scientific paper. You do not 
discount or hide anything – you report and let the chips fall where they may.

 

It is highly improbably that lithium fusion to helium is the power source 
behind this reactor, but it looks like pure Li-6 was intentionally added to 
natural LAH. That narrows the possibilities.

 

--Surprise,
 surprise.  

Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

I do not believe this new revelation is credible, based on the appearance of 
the paper and the timing, since he has never before said that helium was 
discovered.  

The guy is desperate for credibility.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

From: Bob Higgins 

 

Jones, we DO know that there is a large imbalance in the distribution of 7Li in 
the ash.  

 

 

I agree with that – as far as it goes. 

 

The problem is that the imbalance is entirely consistent with having mixed pure 
Li-6 isotope with LAH containing the natural ratio – and then running the 
reactor for days.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Well on second look, at day 32, the internal helium pressure at 1200 C is about 
2000 psi if indeed the Lugano excess heat calculation was correct (it wasn’t) 
which could arguably have been tolerated by the reactor. About 0.03 moles of 
helium would have been produced at 8 MeV per atom to give the 1.5 MW-hrs of 
dissipated excess heat, but as we know the Lugano excess heat calculation was 
grossly inflated by the incompetence of the Levi team.

 

If the COP was closer to 1.5 as I suspect, then there would have been far less 
internal pressure from the accumulated helium – if lithium fusion was 
responsible. Thus, lithium fusion is not ruled out by pressure considerations. 
(but is ruled out by lack of gammas)

 

From: Jones Beene 

 

Blaze- Disregard previous numbers. I’ll try to calculate the internal pressure 
at day 30 another way. The point remains that if lithium fusion is responsible 
for the gain, lots of helium needs to have been produced and the reactor 
probably could not have tolerated the pressure.

 

From: Blaze Spinnaker 

 

Ø  Jones, it is possible that helium was observed and was originally discounted 
as error.  That happens.

 

Not when this much claimed energy has been seen. 

 

Think about the implications. The Lugano experiment supposedly generated 2 kW 
excess for 30+ days. This is about 10^28 eV equivalent. If all this energy was 
coming from helium, as a result of lithium fusion, at 16 MeV a pop, then it 
amounts to several moles of gas. A mole of helium fills about 25 liters at room 
temp - so this would have been about 50 liters of helium. Even if they 
overestimated the gain by a factor of 10, and the excess was 200 watts, the 
reactor could not have survived the internal pressure.

 

Anyway – the Lugano report was supposed to be a scientific paper. You do not 
discount or hide anything – you report and let the chips fall where they may.

 

It is highly improbably that lithium fusion to helium is the power source 
behind this reactor, but it looks like pure Li-6 was intentionally added to 
natural LAH. That narrows the possibilities.

 

--Surprise,
 surprise.  

Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

I do not believe this new revelation is credible, based on the appearance of 
the paper and the timing, since he has never before said that helium was 
discovered.  

The guy is desperate for credibility.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Eric Walker

Hi,

On Wed, Apr 8, 2015 at 1:18 PM, Jones Beene jone...@pacbell.net wrote:

Levi and his team were reportedly paid half a million bucks ...


Do you have a source for this that goes into more detail?

Eric

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Blaze Spinnaker

Jones, it is possible that helium was observed and was originally
discounted as error.  That happens.

On Wed, Apr 8, 2015 at 11:07 AM, Jones Beene jone...@pacbell.net wrote:



 Surprise, surprise.



 Fresh on the heels of a paper which suggests that lots of helium should
 have been found, Rossi suddenly reveals that yes, we found it but are just
 now taking the opportunity to reveal that we found it.



 http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/



 I not believe this new revelation is credible, based on the appearance of
 the paper and the timing, since  he has never before said that helium was
 discovered.



 The guy is desperate for credibility.







 *From:* Bob Higgins



 Jones,  What is your evidence for your statement:



 The Lugano isotope data, even if it could be believed, completely negates
 the entire scenario since Li-7 is NOT depleted according to the Lugano
 report - but instead is converted to Li-6. 



 First of all, there is a crude assay based on the size of the pure sphere
 - and no evidence of large imbalance of Li-7 elsewhere. More importantly,
 85 years of nuclear physics can present no thermal process where the bulk
 isotopic distribution varies more than a few percent per stage, yet the
 Lugano report, if it can be believed shows extremely pure Li-6 appearing in
 what is essentially one stage in one sample – many orders of magnitude
 purer than any know process can deliver.



 There are three possibilities – either the starting material was enriched
 in pure Li-6, which is most likely, or else the process of heat generation
 has converted the missing Li-7 into Li-6, which is endothermic, and
 unlikely to have happened in a process where excess heat is generated. The
 third possibility is that the ash was spiked with pure isotope.



 Neither of these possibilities can in any way support a conclusion of
 lithium-7 plus proton fusion, especially with the lack of the expected
 gamma, and no indication of helium.



 To say that Levi’s crew did not test for helium is a complete cop-out and
 only indicative of further incompetence on the part of this team. With this
 claimed excess heat over 30 days there should have been a large amount of
 helium, actual overpressure: that is - if lithium fusion were taking place.
 A sample of gas should at least have been stored for later testing.



 Most likely conclusion – Rossi understood from the start that lithium-6 is
 the active isotope, and he provided fuel which was highly enriched, and at
 the same time, provided a different fuel for the testing of the “before”
 sample. Only Rossi handled this fuel. He had complete control, and no one
 complained. BTW - The cost of that much lithium-6 (about 50 milligrams)
 available from several suppliers, is about $10.



 Jones





 What I drew from the report was the only thing that can be concluded was
 that the 7Li is more commensurate to the 6Li in the ash as compared to the
 fuel.  There was no mass assay that determined how much total Li was
 present in the ash compared to the fuel.  We know that physically, a lot of
 the Li will be on the walls of the alumina tube, so we don't have any idea
 of the absolute depletion of Li mass in the reaction.



 While it is possible that the 7Li is converted to 6Li, it is only one of
 the possibilities.  The ICP-MS analysis is a full volume analysis and
 showed both Li isotopes near equal in percentage in the ash.  How these
 isotopes became nearly equal is just blind speculation at the moment
 without further experimental data.  All of the possibilities for the ratio
 change from fuel to ash should be laid out and the plausibility of each
 examined.



 Bob

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

From: Blaze Spinnaker 

 

Ø  Jones, it is possible that helium was observed and was originally discounted 
as error.  That happens.

 

Not when this much claimed energy has been seen. 

 

Think about the implications. The Lugano experiment supposedly generated 2 kW 
excess for 30+ days. This is about 10^28 eV equivalent. If all this energy was 
coming from helium, as a result of lithium fusion, at 16 MeV a pop, then it 
amounts to several moles of gas. A mole of helium fills about 25 liters at room 
temp - so this would have been about 50 liters of helium. Even if they 
overestimated the gain by a factor of 10, and the excess was 200 watts, the 
reactor could not have survived the internal pressure.

 

Anyway – the Lugano report was supposed to be a scientific paper. You do not 
discount or hide anything – you report and let the chips fall where they may.

 

It is highly improbably that lithium fusion to helium is the power source 
behind this reactor, but it looks like pure Li-6 was intentionally added to 
natural LAH. That narrows the possibilities.

 

--Surprise,
 surprise.  

Fresh on the heels of a paper which suggests that lots of helium should have 
been found, Rossi suddenly reveals that yes, we found it but are just now 
taking the opportunity to reveal that we found it.

http://www.e-catworld.com/2015/04/08/rossi-helium-found-in-e-cat-reaction/

I do not believe this new revelation is credible, based on the appearance of 
the paper and the timing, since he has never before said that helium was 
discovered.  

The guy is desperate for credibility.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-08 Thread Terry Blanton

Amazing that some folks have no trouble accepting 16 MeV is the source of
LENR energy but quickly discount that 500 keV of spin energy can impart
linear momentum due to centripetal disruption of weakly bound particles.

-Spin Cartel

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Roarty, Francis X

I would be surprised if the paths  outlined in their paper were not already 
suggested here on vortex since… according to Rossi we only sit back and make 
countless guesses without ever studying or investing in actual experiment … I 
think he is feeling the heat on his toes now and casting aspersions while he 
cherry picks ideas from vortex authors and tries to rewrap them in fresh terms 
so we won’t recognize them:_) but he also admits he doesn’t know the source of 
the energy either – his paper pushes the quantum features established by the 
lattice on the IPM and effects on spin coupling..sound familiar? I do like that 
he pushes the field down to the angstrom level but still believe this is 
accomplished via relativistic effect [Naudt’s 05 paper] and the angstrom level 
he is promoting is only a relative measure not really occurring from the 
perspective of the local tiny observer.

[snipbut theory alone provides no clue
on how sufficient energy could lead to these
nuclear reactions. While this theoretical
stalemate remains unresolved, however, we
demonstrate below how specific isotopic
structures in the lattice IPM could in principle
lead to the strong depletion of 7Li4,[/snip]

From: Peter Gluck [mailto:peter.gl...@gmail.com]
Sent: Tuesday, April 07, 2015 5:18 AM
To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker; 
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff 
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon; Roberto 
Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin; Steve Katinski; 
David Daggett; Valerio Ciampoli; Peter Bjorkbom
Subject: EXTERNAL: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi

Dear Friends,
Again a  midday edition of Ego Out- - bis dat qui cito dat-
I know you all are curious to see it

 http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html

Let's read and discuss the paper

I already know what will it change.

I hope the theorists will speak.

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Daniel Rocha

Cook's theory is good, indeed, It would be wonderful if Cook's theory was
mainstream. But it isn't, even though he is fighting for it for 4
decades...

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Oops meant to say Li7 and not Li6…

 

 

From: Jones Beene 

 

Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging his 
bets and perpetuating a myth – mentions the nickel-to-copper nonsense - and 
makes no real progress here.

 

In short - this paper is cannot be taken seriously. A waste of time.

 

Norman Cook is using Rossi to promote his own fringe theory, and that is why we 
have such a confused mélange of disparate ideas.

 

Sadly, since the paper is electronic, it cannot even be used as fishwrap. 

 

 

From: Frank Znidarsic 

 

Interesting paper 

 

Jones ideas about lithium are reaffirmed.

 

It mentions the liquid drop model.  It does not mention that sound propagates 
in all liquids.

 

It clearly shows the tetrahedral structure of the nucleus.  It makes not 
mention of the 1.36 fm spacing

within the tetrahedral.  It makes no mention of an elastic constant and does 
not attempt to compute the velocity of sound within the nuclear structure.

 

Frank Znidarsic

 

 

 

-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

I would be surprised if the paths  outlined in their paper were not already 
suggested here on vortex since… according to Rossi we only sit back and make 
countless guesses without ever studying or investing in actual experiment … I 
think he is feeling the heat on his toes now and casting aspersions while he 
cherry picks ideas from vortex authors and tries to rewrap them in fresh terms 
so we won’t recognize them:_) but he also admits he doesn’t know the source of 
the energy either – his paper pushes the quantum features established by the 
lattice on the IPM and effects on spin coupling..sound familiar? I do like that 
he pushes the field down to the angstrom level but still believe this is 
accomplished via relativistic effect [Naudt’s 05 paper] and the angstrom level 
he is promoting is only a relative measure not really occurring from the 
perspective of the local tiny observer.

 

[snipbut theory alone provides no clue

on how sufficient energy could lead to these

nuclear reactions. While this theoretical

stalemate remains unresolved, however, we

demonstrate below how specific isotopic

structures in the lattice IPM could in principle

lead to the strong depletion of 7Li4,[/snip]

 

From: Peter Gluck [mailto:peter.gl...@gmail.com mailto:peter.gl...@gmail.com? 
] 
Sent: Tuesday, April 07, 2015 5:18 AM
To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker; 
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff 
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon; Roberto 
Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin; Steve Katinski; 
David Daggett; Valerio Ciampoli; Peter Bjorkbom
Subject: EXTERNAL: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi

 

Dear Friends,

Again a  midday edition of Ego Out- - bis dat qui cito dat-

I know you all are curious to see it

 

 http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html

 

Let's read and discuss the paper


 

I already know what will it change.

 

I hope the theorists will speak.

 

Peter

-- 

Dr. Peter Gluck

Cluj, Romania

http://egooutpeters.blogspot.com

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Peter--

Thanks for the heads up on the Cook-Rossi paper.

The most important observation IMHO is the caption for Figure 5, specifically, 
...two alpha particles (D), which are released with 17 MeV of angular 
momentum, but without gamma radiation.  

The smoking gun is the huge energy of the angular momentum.  If angular 
momentum is conserved, it must be distributed in the system as angular 
momentum--spin energy of Ni isotopes, the electrons or maybe the Al. 

The paper ends with the suggestion that the electronic structure of the system 
is involved as in the interaction of electrons and nuclei in the Mossbauer 
Effect.  This interaction may very well be spin coupling IMHO.  

Bob Cook

- Original Message - 
  From: Peter Gluck 
  To: Arik El Boher ; Bo Hoistadt ; Brian Ahern ; CMNS ; Dagmar Kuhn ; doug 
marker ; Dr. Braun Tibor ; eCatNews ; Gabriel Moagar-Poladian ; Gary ; Haiko 
Lietz ; jeff aries ; Lewan Mats ; Nicolaie N. Vlad ; Peter Mobberley ; Pierre 
Clauzon ; Roberto Germano ; Roy Virgilio ; Sunwon Park ; vlad ; VORTEX ; Mark 
Tsirlin ; Steve Katinski ; David Daggett ; Valerio Ciampoli ; Peter Bjorkbom 
  Sent: Tuesday, April 07, 2015 2:17 AM
  Subject: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Dear Friends,
  Again a  midday edition of Ego Out- - bis dat qui cito dat-
  I know you all are curious to see it


   http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html


  Let's read and discuss the paper



  I already know what will it change.


  I hope the theorists will speak.


  Peter
  -- 

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

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Frank Znidarsic

Interesting paper


Jones ideas about lithium are reaffirmed.


It mentions the liquid drop model.  It does not mention that sound propagates 
in all liquids.


It clearly shows the tetrahedral structure of the nucleus.  It makes not 
mention of the 1.36 fm spacing
within the tetrahedral.  It makes no mention of an elastic constant and does 
not attempt to compute the velocity of sound within the nuclear structure.


Frank Znidarsic










-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  
   
I would be surprised if the paths  outlined in their paper were not already 
suggested here on vortex since… according to Rossi we only sit back and make 
countless guesses without ever studying or investing in actual experiment … I 
think he is feeling the heat on his toes now and casting aspersions while he 
cherry picks ideas from vortex authors and tries to rewrap them in fresh terms 
so we won’t recognize them:_) but he also admits he doesn’t know the source of 
the energy either – his paper pushes the quantum features established by the 
lattice on the IPM and effects on spin coupling..sound familiar? I do like that 
he pushes the field down to the angstrom level but still believe this is 
accomplished via relativistic effect [Naudt’s 05 paper] and the angstrom level 
he is promoting is only a relative measure not really occurring from the 
perspective of the local tiny observer.
   
 
   
[snipbut theory alone provides no clue
   
on how sufficient energy could lead to these
   
nuclear reactions. While this theoretical
   
stalemate remains unresolved, however, we
   
demonstrate below how specific isotopic
   
structures in the lattice IPM could in principle
   
lead to the strong depletion of 7Li4,[/snip]
   
 
   
From: Peter Gluck [mailto:peter.gl...@gmail.com] 
 Sent: Tuesday, April 07, 2015 5:18 AM
 To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker; 
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff 
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon; Roberto 
Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin; Steve Katinski; 
David Daggett; Valerio Ciampoli; Peter Bjorkbom
 Subject: EXTERNAL: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi
   
 
   

Dear Friends,

 
Again a  midday edition of Ego Out- - bis dat qui cito dat-


 
I know you all are curious to see it


 
 


 
 http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html


 
 


 
Let's read and discuss the paper
 
 
  
 
 
 
  
I already know what will it change.
 
 
  
 
 
 
  
I hope the theorists will speak.
 
 
  
 
 
 
  
Peter
 
 
-- 
 
  
Dr. Peter Gluck
  
   
Cluj, Romania
  
  
   
http://egooutpeters.blogspot.com

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging his 
bets and perpetuating a myth – mentions the nickel-to-copper nonsense - and 
makes no real progress here.

 

In short - this paper is cannot be taken seriously. A waste of time.

 

Norman Cook is using Rossi to promote his own fringe theory, and that is why we 
have such a confused mélange of disparate ideas.

 

Sadly, since the paper is electronic, it cannot even be used as fishwrap. 

 

 

From: Frank Znidarsic 

 

Interesting paper 

 

Jones ideas about lithium are reaffirmed.

 

It mentions the liquid drop model.  It does not mention that sound propagates 
in all liquids.

 

It clearly shows the tetrahedral structure of the nucleus.  It makes not 
mention of the 1.36 fm spacing

within the tetrahedral.  It makes no mention of an elastic constant and does 
not attempt to compute the velocity of sound within the nuclear structure.

 

Frank Znidarsic

 

 

 

-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

I would be surprised if the paths  outlined in their paper were not already 
suggested here on vortex since… according to Rossi we only sit back and make 
countless guesses without ever studying or investing in actual experiment … I 
think he is feeling the heat on his toes now and casting aspersions while he 
cherry picks ideas from vortex authors and tries to rewrap them in fresh terms 
so we won’t recognize them:_) but he also admits he doesn’t know the source of 
the energy either – his paper pushes the quantum features established by the 
lattice on the IPM and effects on spin coupling..sound familiar? I do like that 
he pushes the field down to the angstrom level but still believe this is 
accomplished via relativistic effect [Naudt’s 05 paper] and the angstrom level 
he is promoting is only a relative measure not really occurring from the 
perspective of the local tiny observer.

 

[snipbut theory alone provides no clue

on how sufficient energy could lead to these

nuclear reactions. While this theoretical

stalemate remains unresolved, however, we

demonstrate below how specific isotopic

structures in the lattice IPM could in principle

lead to the strong depletion of 7Li4,[/snip]

 

From: Peter Gluck [mailto:peter.gl...@gmail.com mailto:peter.gl...@gmail.com? 
] 
Sent: Tuesday, April 07, 2015 5:18 AM
To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker; 
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff 
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon; Roberto 
Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin; Steve Katinski; 
David Daggett; Valerio Ciampoli; Peter Bjorkbom
Subject: EXTERNAL: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi

 

Dear Friends,

Again a  midday edition of Ego Out- - bis dat qui cito dat-

I know you all are curious to see it

 

 http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html

 

Let's read and discuss the paper


 

I already know what will it change.

 

I hope the theorists will speak.

 

Peter

-- 

Dr. Peter Gluck

Cluj, Romania

http://egooutpeters.blogspot.com

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Jones--

The paper was understandable for me.  The lack of data on the Ni isotopic 
ratios is significant and in keeping with a respectful, unknowing, position.  
The only issue I had with the paper was editorial--the use of Ni instead of N 
to stand for nitrogen.  I was impressed that there was no hand waving to 
suggest understanding, only straight forward statements about lack of 
understanding where appropriate. 

Bob
  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Tuesday, April 07, 2015 7:57 AM
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Oops meant to say Li7 and not Li6…

   

   

  From: Jones Beene 

   

  Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging his 
bets and perpetuating a myth – mentions the nickel-to-copper nonsense - and 
makes no real progress here.

   

  In short - this paper is cannot be taken seriously. A waste of time.

   

  Norman Cook is using Rossi to promote his own fringe theory, and that is why 
we have such a confused mélange of disparate ideas.

   

  Sadly, since the paper is electronic, it cannot even be used as fishwrap. 

   

   

  From: Frank Znidarsic 

   

  Interesting paper 

   

  Jones ideas about lithium are reaffirmed.

   

  It mentions the liquid drop model.  It does not mention that sound propagates 
in all liquids.

   

  It clearly shows the tetrahedral structure of the nucleus.  It makes not 
mention of the 1.36 fm spacing

  within the tetrahedral.  It makes no mention of an elastic constant and does 
not attempt to compute the velocity of sound within the nuclear structure.

   

  Frank Znidarsic

   

   

   

  -Original Message-
  From: Roarty, Francis X francis.x.roa...@lmco.com
  To: vortex-l vortex-l@eskimo.com
  Sent: Tue, Apr 7, 2015 6:38 am
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

  I would be surprised if the paths  outlined in their paper were not already 
suggested here on vortex since… according to Rossi we only sit back and make 
countless guesses without ever studying or investing in actual experiment … I 
think he is feeling the heat on his toes now and casting aspersions while he 
cherry picks ideas from vortex authors and tries to rewrap them in fresh terms 
so we won’t recognize them:_) but he also admits he doesn’t know the source of 
the energy either – his paper pushes the quantum features established by the 
lattice on the IPM and effects on spin coupling..sound familiar? I do like that 
he pushes the field down to the angstrom level but still believe this is 
accomplished via relativistic effect [Naudt’s 05 paper] and the angstrom level 
he is promoting is only a relative measure not really occurring from the 
perspective of the local tiny observer.

   

  [snipbut theory alone provides no clue

  on how sufficient energy could lead to these

  nuclear reactions. While this theoretical

  stalemate remains unresolved, however, we

  demonstrate below how specific isotopic

  structures in the lattice IPM could in principle

  lead to the strong depletion of 7Li4,[/snip]

   

  From: Peter Gluck [mailto:peter.gl...@gmail.com] 
  Sent: Tuesday, April 07, 2015 5:18 AM
  To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker; 
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff 
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon; Roberto 
Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin; Steve Katinski; 
David Daggett; Valerio Ciampoli; Peter Bjorkbom
  Subject: EXTERNAL: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi

   

  Dear Friends,

  Again a  midday edition of Ego Out- - bis dat qui cito dat-

  I know you all are curious to see it

   

   http://egooutpeters.blogspot.ro/2015/04/rossi-is-co-author-of-mainstream.html

   

  Let's read and discuss the paper


   

  I already know what will it change.

   

  I hope the theorists will speak.

   

  Peter

  -- 

  Dr. Peter Gluck

  Cluj, Romania

  http://egooutpeters.blogspot.com

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Axil--

You said Cook said this: Cook says that high energy alpha particles exit the 
NAE at high energy and deliver their energy to the far field at an some 
indeterminate distance from the NAE that produced the energy.  

I did not see this statement.  Where was Cook's statement made?

What I saw in the new paper was that the energy of the alphas from the Be-8 
decay was in the form of 17 Mev of angular momentum (spin energy)--not kinetic 
energy.  (The slowing-down of 17 MeV alphas would cause noticeable x-rays and 
other high energy EM radiation.)  The alphas apparently stays put and transfers 
its excess energy via spin coupling, one spin quanta or so at a time.

Bob Cook


  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 8:57 AM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Like so many LENR theories, the Cook theory of the LENR reaction is not 
fundamental. Like almost all other LENR  theories, it deals with the emergent 
results of the fundamental LENR reaction without explaining the cause of the 
observed experimental results.


  If a theory cannot explain EVERY aspect of the experimental results in every 
dimension, it is not valid.
  In particular, the way energy of these high powered alpha particles are 
converted to heat is not addressed, even though that part of the LENR theory is 
central to how the energy of the nuclear reaction is converted to soft x-rays 
and extreme ultraviolet light.


  I have concluded from the experimental results derived from many LENR systems 
that the gamma suppression and the basic LENR nuclear reaction is tightly 
coupled together so that if a LENR based nuclear event occurs, no gamma is ever 
seen in a environment that has gotten hot enough (500C).


  Gamma suppression is an essential part of the LENR reaction.  So Gamma 
suppression is an essential part of what is going on inside the Nuclear Active 
Environment. If energy is carried away from the NAE, it cannot be converted to 
its final moderated form (soft x-rays and extreme ultraviolet light.) by the 
LENR reaction.  


  Cook says that high energy alpha particles exit the NAE at high energy and 
deliver their energy to the far field at an some indeterminate distance from 
the NAE that produced the energy. If this were true, there is always a slight 
chance that the alpha particle could exit the gas envelop and deposit its 
kinetic energy in the Alumina shell where a gamma ray would result. This gamma 
ray is never seen. So if an alpha particle is produced it must have little or 
no kinetic energy that is transferred to the far field.


  All the energy of the nuclear reaction is carried away from the NAE by the 
LENR reaction itself. The gamma emission is an intrinsic part of the LENR 
reaction energy transfer mechanism.




  On Tue, Apr 7, 2015 at 11:21 AM, a.ashfield a.ashfi...@verizon.net wrote:

Jones Beene writes.  this paper is cannot be taken seriously. A waste of 
time.

I wish you wouldn't just damn the paper out of hand but give some reasons 
of just why it is wrong.  I don't have the knowledge of nuclear reactions that 
some others do here, but most of the theories seem far from solid to me and 
this one is no worse.  It should at least be considered.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Bob Cook,
Could that possibly explain the high magnetic field reported by the 
discredited DGT experiments, or was that discredited too?

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Like so many LENR theories, the Cook theory of the LENR reaction is not
fundamental. Like almost all other LENR  theories, it deals with the
emergent results of the fundamental LENR reaction without explaining the
cause of the observed experimental results.

If a theory cannot explain EVERY aspect of the experimental results in
every dimension, it is not valid.
In particular, the way energy of these high powered alpha particles are
converted to heat is not addressed, even though that part of the LENR
theory is central to how the energy of the nuclear reaction is converted to
soft x-rays and extreme ultraviolet light.

I have concluded from the experimental results derived from many LENR
systems that the gamma suppression and the basic LENR nuclear reaction is
tightly coupled together so that if a LENR based nuclear event occurs, *no
gamma is ever seen* in a environment that has gotten hot enough (500C).

Gamma suppression is an essential part of the LENR reaction.  So Gamma
suppression is an essential part of what is going on inside the Nuclear
Active Environment. If energy is carried away from the NAE, it cannot be
converted to its final moderated form (soft x-rays and extreme ultraviolet
light.) by the LENR reaction.

Cook says that high energy alpha particles exit the NAE at high energy and
deliver their energy to the far field at an some indeterminate distance
from the NAE that produced the energy. If this were true, there is always a
slight chance that the alpha particle could exit the gas envelop and
deposit its kinetic energy in the Alumina shell where a gamma ray would
result. This gamma ray is never seen. So if an alpha particle is produced
it must have little or no kinetic energy that is transferred to the far
field.

All the energy of the nuclear reaction is carried away from the NAE by the
LENR reaction itself. The gamma emission is an intrinsic part of the LENR
reaction energy transfer mechanism.


On Tue, Apr 7, 2015 at 11:21 AM, a.ashfield a.ashfi...@verizon.net wrote:

  Jones Beene writes.  this paper is cannot be taken seriously. A waste
 of time.

 I wish you wouldn't just damn the paper out of hand but give some reasons
 of just why it is wrong.  I don't have the knowledge of nuclear reactions
 that some others do here, but most of the theories seem far from solid to
 me and this one is no worse.  It should at least be considered.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Bob,

There are other silly errors besides the confusion over nitrogen. Any paper,
even a preprint, can be understandable and yet meaningless – when the major
premise is ignored. You may be too willing to find spin - or angular
momentum, in a place that it cannot exist.

First and foremost, and UNFORGIVEABLE - the paper fails to address the lack
of relevant evidence or data for its major premise:  Quote: “Although the
main source of energy appears to be the 7Li4 (p,α) α reaction”…

Yet, they show no helium ash! They show no bremsstrahlung! They show no
x-rays! In short there is NO physical evidence whatsoever for the existence
of the major premise- which, if it were true, should supply an unbelievable
17 MeV in energy – so why even waste time with this geometric model that
relates best to nickel to copper? BTW this excess energy simply cannot be
“only” angular momentum. 17 MeV of angular momentum ! in a nucleus – think
about it.

The Norman Cook model could be relevant to nickel fusion, or not, but they
show no good evidence of nickel fusion, so it is superfluous for lithium to
helium. Yet that is why Cook is involved – to promote a theory which he has
been promoting in Infinite Energy for years (maybe a decade) – and few have
found it intuitive for LENR.

Lastly, I’d be surprised if Rossi has ever met Cook - this looks like some
kind of marriage of convenience to promote his fringe theory, and possibly
to give Rossi some semblance of credibility – but only for those who do not
look very deeply at what is missing here. This paper is a complete waste of
time.

From: Bob Cook 

Jones--
 
The paper was understandable for me.  The lack of data on the Ni isotopic
ratios is significant and in keeping with a respectful, unknowing, position.
The only issue I had with the paper was editorial--the use of Ni instead of
N to stand for nitrogen.  I was impressed that there was no hand waving to
suggest understanding, only straight forward statements about lack of
understanding where appropriate. 
 
Bob
- Original Message - 
From: Jones Beene mailto:jone...@pacbell.net  

Oops meant to say Li7 and not Li6…


From: Jones Beene 

Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging
his bets and perpetuating a myth – mentions the nickel-to-copper nonsense -
and makes no real progress here.

In short - this paper is cannot be taken seriously. A waste of time.

Norman Cook is using Rossi to promote his own fringe theory, and that is why
we have such a confused mélange of disparate ideas.

Sadly, since the paper is electronic, it cannot even be used as fishwrap. 


From: Frank Znidarsic 

Interesting paper 

Jones ideas about lithium are reaffirmed.

It mentions the liquid drop model.  It does not mention that sound
propagates in all liquids.

It clearly shows the tetrahedral structure of the nucleus.  It makes not
mention of the 1.36 fm spacing
within the tetrahedral.  It makes no mention of an elastic constant and does
not attempt to compute the velocity of sound within the nuclear structure.

Frank Znidarsic



-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author
Andrea Rossi
I would be surprised if the paths  outlined in their paper were not already
suggested here on vortex since… according to Rossi we only sit back and make
countless guesses without ever studying or investing in actual experiment …
I think he is feeling the heat on his toes now and casting aspersions while
he cherry picks ideas from vortex authors and tries to rewrap them in fresh
terms so we won’t recognize them:_) but he also admits he doesn’t know the
source of the energy either – his paper pushes the quantum features
established by the lattice on the IPM and effects on spin coupling..sound
familiar? I do like that he pushes the field down to the angstrom level but
still believe this is accomplished via relativistic effect [Naudt’s 05
paper] and the angstrom level he is promoting is only a relative measure not
really occurring from the perspective of the local tiny observer.
 
[snipbut theory alone provides no clue
on how sufficient energy could lead to these
nuclear reactions. While this theoretical
stalemate remains unresolved, however, we
demonstrate below how specific isotopic
structures in the lattice IPM could in principle
lead to the strong depletion of 7Li4,[/snip]
 
From: Peter Gluck [mailto:peter.gl...@gmail.com
mailto:peter.gl...@gmail.com? ] 
Sent: Tuesday, April 07, 2015 5:18 AM
To: Arik El Boher; Bo Hoistadt; Brian Ahern; CMNS; Dagmar Kuhn; doug marker;
Dr. Braun Tibor; eCatNews; Gabriel Moagar-Poladian; Gary; Haiko Lietz; jeff
aries; Lewan Mats; Nicolaie N. Vlad; Peter Mobberley; Pierre Clauzon;
Roberto Germano; Roy Virgilio; Sunwon Park; vlad; VORTEX; Mark Tsirlin;
Steve Katinski; David Daggett; Valerio Ciampoli; Peter Bjorkbom
Subject: EXTERNAL: [Vo

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Jones--

Do you know of any papers that measure the decay energy of Be-8 decays with 
no gammas.  (This no-gamma decay is nearly one of a kind.)  It would be 
interesting to see how it is explained--the  no gammas that is.


If I find a good reference, I will let you know.

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

To: vortex-l@eskimo.com
Sent: Tuesday, April 07, 2015 9:41 AM
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi



Bob,

There are other silly errors besides the confusion over nitrogen. Any paper,
even a preprint, can be understandable and yet meaningless – when the major
premise is ignored. You may be too willing to find spin - or angular
momentum, in a place that it cannot exist.

First and foremost, and UNFORGIVEABLE - the paper fails to address the lack
of relevant evidence or data for its major premise:  Quote: “Although the
main source of energy appears to be the 7Li4 (p,α) α reaction”…

Yet, they show no helium ash! They show no bremsstrahlung! They show no
x-rays! In short there is NO physical evidence whatsoever for the existence
of the major premise- which, if it were true, should supply an unbelievable
17 MeV in energy – so why even waste time with this geometric model that
relates best to nickel to copper? BTW this excess energy simply cannot be
“only” angular momentum. 17 MeV of angular momentum ! in a nucleus – think
about it.

The Norman Cook model could be relevant to nickel fusion, or not, but they
show no good evidence of nickel fusion, so it is superfluous for lithium to
helium. Yet that is why Cook is involved – to promote a theory which he has
been promoting in Infinite Energy for years (maybe a decade) – and few have
found it intuitive for LENR.

Lastly, I’d be surprised if Rossi has ever met Cook - this looks like some
kind of marriage of convenience to promote his fringe theory, and possibly
to give Rossi some semblance of credibility – but only for those who do not
look very deeply at what is missing here. This paper is a complete waste of
time.

From: Bob Cook

Jones--

The paper was understandable for me.  The lack of data on the Ni isotopic
ratios is significant and in keeping with a respectful, unknowing, position.
The only issue I had with the paper was editorial--the use of Ni instead of
N to stand for nitrogen.  I was impressed that there was no hand waving to
suggest understanding, only straight forward statements about lack of
understanding where appropriate.

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


Oops meant to say Li7 and not Li6…


From: Jones Beene

Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging
his bets and perpetuating a myth – mentions the nickel-to-copper nonsense -
and makes no real progress here.

In short - this paper is cannot be taken seriously. A waste of time.

Norman Cook is using Rossi to promote his own fringe theory, and that is why
we have such a confused mélange of disparate ideas.

Sadly, since the paper is electronic, it cannot even be used as fishwrap.


From: Frank Znidarsic

Interesting paper

Jones ideas about lithium are reaffirmed.

It mentions the liquid drop model.  It does not mention that sound
propagates in all liquids.

It clearly shows the tetrahedral structure of the nucleus.  It makes not
mention of the 1.36 fm spacing
within the tetrahedral.  It makes no mention of an elastic constant and does
not attempt to compute the velocity of sound within the nuclear structure.

Frank Znidarsic



-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author
Andrea Rossi
I would be surprised if the paths  outlined in their paper were not already
suggested here on vortex since… according to Rossi we only sit back and make
countless guesses without ever studying or investing in actual experiment …
I think he is feeling the heat on his toes now and casting aspersions while
he cherry picks ideas from vortex authors and tries to rewrap them in fresh
terms so we won’t recognize them:_) but he also admits he doesn’t know the
source of the energy either – his paper pushes the quantum features
established by the lattice on the IPM and effects on spin coupling..sound
familiar? I do like that he pushes the field down to the angstrom level but
still believe this is accomplished via relativistic effect [Naudt’s 05
paper] and the angstrom level he is promoting is only a relative measure not
really occurring from the perspective of the local tiny observer.

[snipbut theory alone provides no clue
on how sufficient energy could lead to these
nuclear reactions. While this theoretical
stalemate remains unresolved, however, we
demonstrate below how specific isotopic
structures in the lattice IPM could in principle
lead to the strong depletion of 7Li4,[/snip]

From: Peter

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Jones Beene writes.  this paper is cannot be taken seriously. A waste 
of time.


I wish you wouldn't just damn the paper out of hand but give some 
reasons of just why it is wrong.  I don't have the knowledge of nuclear 
reactions that some others do here, but most of the theories seem far 
from solid to me and this one is no worse.  It should at least be 
considered.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Yes - that is another error on the part of Cook. I would call it major instead 
of minor.

Rather it is a major misrepresentation instead of minor error - since - 
although the Be8 decays with no gamma (that much is true) the proton plus Li7 
reaction, which he neglects to mention - definitely does produce substantial 
gamma radiation - which must show up.

In short, the net reaction of going from lithium to helium via proton fusion 
will ALWAYS produce some gamma radiation, as well as bremsstrahlung.

-Original Message-
From: Bob Cook 

Jones--

Do you know of any papers that measure the decay energy of Be-8 decays with no 
gammas.  (This no-gamma decay is nearly one of a kind.)  It would be 
interesting to see how it is explained--the  no gammas that is.

If I find a good reference, I will let you know.

Bob
- Original Message -
From: Jones Beene jone...@pacbell.net
To: vortex-l@eskimo.com
Sent: Tuesday, April 07, 2015 9:41 AM
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


Bob,

There are other silly errors besides the confusion over nitrogen. Any paper, 
even a preprint, can be understandable and yet meaningless – when the major 
premise is ignored. You may be too willing to find spin - or angular momentum, 
in a place that it cannot exist.

First and foremost, and UNFORGIVEABLE - the paper fails to address the lack of 
relevant evidence or data for its major premise:  Quote: “Although the main 
source of energy appears to be the 7Li4 (p,α) α reaction”…

Yet, they show no helium ash! They show no bremsstrahlung! They show no x-rays! 
In short there is NO physical evidence whatsoever for the existence of the 
major premise- which, if it were true, should supply an unbelievable
17 MeV in energy – so why even waste time with this geometric model that 
relates best to nickel to copper? BTW this excess energy simply cannot be 
“only” angular momentum. 17 MeV of angular momentum ! in a nucleus – think 
about it.

The Norman Cook model could be relevant to nickel fusion, or not, but they show 
no good evidence of nickel fusion, so it is superfluous for lithium to helium. 
Yet that is why Cook is involved – to promote a theory which he has been 
promoting in Infinite Energy for years (maybe a decade) – and few have found it 
intuitive for LENR.

Lastly, I’d be surprised if Rossi has ever met Cook - this looks like some kind 
of marriage of convenience to promote his fringe theory, and possibly to give 
Rossi some semblance of credibility – but only for those who do not look very 
deeply at what is missing here. This paper is a complete waste of time.

From: Bob Cook

Jones--

The paper was understandable for me.  The lack of data on the Ni isotopic 
ratios is significant and in keeping with a respectful, unknowing, position.
The only issue I had with the paper was editorial--the use of Ni instead of N 
to stand for nitrogen.  I was impressed that there was no hand waving to 
suggest understanding, only straight forward statements about lack of 
understanding where appropriate.

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

Oops meant to say Li7 and not Li6…


From: Jones Beene

Well, according to Rossi – Ni7 and not Ni6 is active -  but he is hedging his 
bets and perpetuating a myth – mentions the nickel-to-copper nonsense - and 
makes no real progress here.

In short - this paper is cannot be taken seriously. A waste of time.

Norman Cook is using Rossi to promote his own fringe theory, and that is why we 
have such a confused mélange of disparate ideas.

Sadly, since the paper is electronic, it cannot even be used as fishwrap.


From: Frank Znidarsic

Interesting paper

Jones ideas about lithium are reaffirmed.

It mentions the liquid drop model.  It does not mention that sound propagates 
in all liquids.

It clearly shows the tetrahedral structure of the nucleus.  It makes not 
mention of the 1.36 fm spacing within the tetrahedral.  It makes no mention of 
an elastic constant and does not attempt to compute the velocity of sound 
within the nuclear structure.

Frank Znidarsic



-Original Message-
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Apr 7, 2015 6:38 am
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi I would be surprised if the paths  outlined in their paper were not 
already suggested here on vortex since… according to Rossi we only sit back and 
make countless guesses without ever studying or investing in actual experiment 
… I think he is feeling the heat on his toes now and casting aspersions while 
he cherry picks ideas from vortex authors and tries to rewrap them in fresh 
terms so we won’t recognize them:_) but he also admits he doesn’t know the 
source of the energy either – his paper pushes the quantum features established 
by the lattice on the IPM and effects on spin coupling..sound familiar? I do 
like

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In my understanding about the LENR process, I have borrowed Dr. Kims idea
about the Bosenova as the way that the energy contents of the soliton is
recycled back into the thermal invirnment of the reactor. After the
bosenova explodes, the electrons and the soft x-rays and XUV are decoupled
whereupon these photons are thermalized further by standard electron shell
processes.

On Tue, Apr 7, 2015 at 1:10 PM, a.ashfield a.ashfi...@verizon.net wrote:

  Bob Cook,
 Could that possibly explain the high magnetic field reported by the
 discredited DGT experiments, or was that discredited too?

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Figure 5 depicts the alpha creation process. See how the alpha particles
are moving away in opposite directions?

Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice
structure to which an additional proton will produce a two-tetrahedron
structure, giving 8 Be4 (B). The double alpha lattice structure (C) can
then break into independent two alpha particles (D), which are released
with 17 MeV of angular momentum, but without gamma radiation.

On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:

  Axil--

 You said Cook said this: Cook says that high energy alpha particles exit
 the NAE at high energy and deliver their energy to the far field at an some
 indeterminate distance from the NAE that produced the energy.

 I did not see this statement.  Where was Cook's statement made?

 What I saw in the new paper was that the energy of the alphas from the
 Be-8 decay was in the form of 17 Mev of angular momentum (spin energy)--not
 kinetic energy.  (The slowing-down of 17 MeV alphas would cause noticeable
 x-rays and other high energy EM radiation.)  The alphas apparently stays
 put and transfers its excess energy via spin coupling, one spin quanta or
 so at a time.

 Bob Cook



 - Original Message -
 *From:* Axil Axil janap...@gmail.com
 *To:* vortex-l vortex-l@eskimo.com
 *Sent:* Tuesday, April 07, 2015 8:57 AM
 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

  Like so many LENR theories, the Cook theory of the LENR reaction is not
 fundamental. Like almost all other LENR  theories, it deals with the
 emergent results of the fundamental LENR reaction without explaining the
 cause of the observed experimental results.

 If a theory cannot explain EVERY aspect of the experimental results in
 every dimension, it is not valid.
  In particular, the way energy of these high powered alpha particles are
 converted to heat is not addressed, even though that part of the LENR
 theory is central to how the energy of the nuclear reaction is converted to
 soft x-rays and extreme ultraviolet light.

 I have concluded from the experimental results derived from many LENR
 systems that the gamma suppression and the basic LENR nuclear reaction is
 tightly coupled together so that if a LENR based nuclear event occurs, *no
 gamma is ever seen* in a environment that has gotten hot enough (500C).

 Gamma suppression is an essential part of the LENR reaction.  So Gamma
 suppression is an essential part of what is going on inside the Nuclear
 Active Environment. If energy is carried away from the NAE, it cannot be
 converted to its final moderated form (soft x-rays and extreme ultraviolet
 light.) by the LENR reaction.

 Cook says that high energy alpha particles exit the NAE at high energy and
 deliver their energy to the far field at an some indeterminate distance
 from the NAE that produced the energy. If this were true, there is always a
 slight chance that the alpha particle could exit the gas envelop and
 deposit its kinetic energy in the Alumina shell where a gamma ray would
 result. This gamma ray is never seen. So if an alpha particle is produced
 it must have little or no kinetic energy that is transferred to the far
 field.

 All the energy of the nuclear reaction is carried away from the NAE by the
 LENR reaction itself. The gamma emission is an intrinsic part of the LENR
 reaction energy transfer mechanism.


 On Tue, Apr 7, 2015 at 11:21 AM, a.ashfield a.ashfi...@verizon.net
 wrote:

 Jones Beene writes.  this paper is cannot be taken seriously. A waste of
 time.

 I wish you wouldn't just damn the paper out of hand but give some reasons
 of just why it is wrong.  I don't have the knowledge of nuclear reactions
 that some others do here, but most of the theories seem far from solid to
 me and this one is no worse.  It should at least be considered.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Axil--

Note that the paper says the energy is angular momentum not kinetic energy of 
the alphas.  Angular momentum energy is spin energy.  The alphas move away with 
essentially no kinetic energy normally associated with non-solid state or 
non-coherent systems.  It is my conclusion from what Cook claims, that the 
electronic cloud must shield the alphas as they are formed from being repulsed 
from each other, or their charge does not materialize until the spin energy is 
fractionated and their distance is sufficient so as to impart only a small 
kinetic energy to each particle, if they have not already become neutral He 
atoms. The Pauli Uncertainty Principle may come into play to spread the wave 
function of the spin energy of the excited He* to a large radius compared to 
the radius associated with a ground state He nucleus.  The enlarged wave 
function may also act to couple to the rest of the particles in the locale 
(coherent system), including the electrons.  

Bob
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:52 AM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Figure 5 depicts the alpha creation process. See how the alpha particles are 
moving away in opposite directions?


  Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice structure 
to which an additional proton will produce a two-tetrahedron structure, giving 
8 Be4 (B). The double alpha lattice structure (C) can then break into 
independent two alpha particles (D), which are released with 17 MeV of angular 
momentum, but without gamma radiation. 


  On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:

Axil--

You said Cook said this: Cook says that high energy alpha particles exit 
the NAE at high energy and deliver their energy to the far field at an some 
indeterminate distance from the NAE that produced the energy.  

I did not see this statement.  Where was Cook's statement made?

What I saw in the new paper was that the energy of the alphas from the Be-8 
decay was in the form of 17 Mev of angular momentum (spin energy)--not kinetic 
energy.  (The slowing-down of 17 MeV alphas would cause noticeable x-rays and 
other high energy EM radiation.)  The alphas apparently stays put and transfers 
its excess energy via spin coupling, one spin quanta or so at a time.

Bob Cook


  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 8:57 AM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi


  Like so many LENR theories, the Cook theory of the LENR reaction is not 
fundamental. Like almost all other LENR  theories, it deals with the emergent 
results of the fundamental LENR reaction without explaining the cause of the 
observed experimental results.


  If a theory cannot explain EVERY aspect of the experimental results in 
every dimension, it is not valid.
  In particular, the way energy of these high powered alpha particles are 
converted to heat is not addressed, even though that part of the LENR theory is 
central to how the energy of the nuclear reaction is converted to soft x-rays 
and extreme ultraviolet light.


  I have concluded from the experimental results derived from many LENR 
systems that the gamma suppression and the basic LENR nuclear reaction is 
tightly coupled together so that if a LENR based nuclear event occurs, no gamma 
is ever seen in a environment that has gotten hot enough (500C).


  Gamma suppression is an essential part of the LENR reaction.  So Gamma 
suppression is an essential part of what is going on inside the Nuclear Active 
Environment. If energy is carried away from the NAE, it cannot be converted to 
its final moderated form (soft x-rays and extreme ultraviolet light.) by the 
LENR reaction.  


  Cook says that high energy alpha particles exit the NAE at high energy 
and deliver their energy to the far field at an some indeterminate distance 
from the NAE that produced the energy. If this were true, there is always a 
slight chance that the alpha particle could exit the gas envelop and deposit 
its kinetic energy in the Alumina shell where a gamma ray would result. This 
gamma ray is never seen. So if an alpha particle is produced it must have 
little or no kinetic energy that is transferred to the far field.


  All the energy of the nuclear reaction is carried away from the NAE by 
the LENR reaction itself. The gamma emission is an intrinsic part of the LENR 
reaction energy transfer mechanism.




  On Tue, Apr 7, 2015 at 11:21 AM, a.ashfield a.ashfi...@verizon.net 
wrote:

Jones Beene writes.  this paper is cannot be taken seriously. A waste 
of time.

I wish you wouldn't just damn the paper out of hand but give some 
reasons of just why it is wrong.  I don't have the knowledge of nuclear

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi


Jones Beene,
You are deliberately misleading about Rossi getting his degree 
on-line.  He got his PhD from Milan University.   He did take an 
on-line course in chemical engineering later, in order to learn about 
that.  Seems to be a reasonable thing to do.


You make much of the lack of theory for how the Li7 gathers a proton.  
The authors admit they don't know.  BUT the idea of the Be splitting 
into two alphas and not emitting gamma radiation suggests to me that it 
is worth looking for a mechanism for the Li to gain the necessary 
proton.  I assume no one knows at present.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Jones--

I tend to agree with the issues you identify.  However, I know of know reason 
why the light nuclei cannot have any spin quantum number--high or low.  Any 
spin quantum is available.  They may have never been observed in light nuclei 
because the proper coherent system and magnetic field was not available to 
establish conditions to get to the high spin states.  Options were not 
available to do the fractionation and natures pathway to lower total energy of 
the coherent system. (I consider the second law applies to coherent 
systems--i.e., they strive to reach the lowest energy possible given the QM 
system and the available resonant coupling coeff's.)

I agree with you that N. Cook does not take on the spin orbit issue and 
coupling to the coherent system.  I think one of the references by Mullenberg 
etal. may in part address this coupling. 

If small packets of spin energy can be distributed, I see no reason why a large 
numbers of small packets cannot be distributed at the same time.  It only means 
you must have a coherent resonant system.  

This seems to be a miracle, however, it is my experience that most miracles 
actually turn out to be due to a law of nature which was not understood at the 
time the miracle was noted. 

Bob

 

 
  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Tuesday, April 07, 2015 1:14 PM
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  From: Bob Cook 

   

  Note that the paper says the energy is angular momentum not kinetic energy of 
the alphas.  Angular momentum energy is spin energy.  The alphas move away with 
essentially no kinetic energy normally associated with non-solid state or 
non-coherent systems

   

   

  Bob,

   

  But Norman Cook has not published on high spin nuclei nor does his basic 
theory go that far - so how can you trust his pronouncement on this important 
detail?

   

  High spin nuclei are well-known in the literature – they are generally heavy 
nuclei - and helium-4 is NOT one of them, and even if it was, the spin energy 
is never much more than 1 MeV. Can you find any reference in the literature to 
angular momentum of any nuclei in excess of 2 MeV? Almost always, the spin 
converts rapidly into gamma rays – not seen in the dogbone.

   

  To me it is absurd for him to suggest, without any reference to the 
literature or experiment, that so much energy can be carried by an alpha 
particle as spin energy and then taper off gradually. As you know, I am 
completely on board with your hypothesis that the gain in this type of reactor 
could come from spin – just not this much spin coming from nuclear fusion. 

   

  Furthermore, since x-rays are not seen – the putative high-spin alpha would 
have to deposit the energy without exception in thousands of perfectly small 
sequential distributions (ala Hagelstein’s magic phonons) which adds another 
miracle. 

   

  That much energy, carried away as angular momentum, is much harder to 
justify, compared to smaller packets of spin being cohered at Terahertz rates.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

From: Bob Cook 

 

Note that the paper says the energy is angular momentum not kinetic energy of 
the alphas.  Angular momentum energy is spin energy.  The alphas move away with 
essentially no kinetic energy normally associated with non-solid state or 
non-coherent systems

 

 

Bob,

 

But Norman Cook has not published on high spin nuclei nor does his basic theory 
go that far - so how can you trust his pronouncement on this important detail?

 

High spin nuclei are well-known in the literature – they are generally heavy 
nuclei - and helium-4 is NOT one of them, and even if it was, the spin energy 
is never much more than 1 MeV. Can you find any reference in the literature to 
angular momentum of any nuclei in excess of 2 MeV? Almost always, the spin 
converts rapidly into gamma rays – not seen in the dogbone.

 

To me it is absurd for him to suggest, without any reference to the literature 
or experiment, that so much energy can be carried by an alpha particle as spin 
energy and then taper off gradually. As you know, I am completely on board with 
your hypothesis that the gain in this type of reactor could come from spin – 
just not this much spin coming from nuclear fusion. 

 

Furthermore, since x-rays are not seen – the putative high-spin alpha would 
have to deposit the energy without exception in thousands of perfectly small 
sequential distributions (ala Hagelstein’s magic phonons) which adds another 
miracle. 

 

That much energy, carried away as angular momentum, is much harder to justify, 
compared to smaller packets of spin being cohered at Terahertz rates.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Here is where Norman Cook and Rossi demonstrate a basic lack of understanding 
of nuclear physics.

To be precise – as mentioned before, two alpha particles released from Be decay 
are indeed gamma free. That detail is not in question – as far as it goes, but 
it does not cover the complete fusion reaction, only part of it. Unfortunately 
many write-ups do not detail the complete reaction, and Wiki is an offender in 
this case. 

Ironically, however - the proton + lithium-7 reaction has historically been 
used as a source of gamma radiation, going back half a century ! But if you 
obtained your PhD degree by mail order, as did AR - then you may not have known 
that.

What is being missed here, and in Norman Cook’s explanation - is the prompt 
gamma which occurs (statistically) at the time the proton interacts with the 
lithium nucleus to form beryllium and then later two alphas. The Li nucleus 
becomes excited, but it cannot simply convert directly to beryllium without an 
energetic emission to compensate for the kinetic energy which caused the 
fusion. There are known signatures for these gammas and statistically they 
occur when protons fuse with lithium at low energy.

Proton + Li-7 → Be-8 + γ (gamma) → alpha + alpha (no gamma)

Note: This gamma does NOT derive from the beryllium decay itself - but from the 
fusion of the proton with the lithium nucleus. This is not always depicted in 
the reaction graphics, and if you depend on Wiki as your sole authority on 
physics, then you may miss it. 

From: Axil Axil 

Figure 5 depicts the alpha creation process. See how the alpha particles are 
moving away in opposite directions?

Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice structure 
to which an additional proton will produce a two-tetrahedron structure, giving 
8 Be4 (B). The double alpha lattice structure (C) can then break into 
independent two alpha particles (D), which are released with 17 MeV of angular 
momentum, but without gamma radiation. 

On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:
Axil--
 
You said Cook said this: Cook says that high energy alpha particles exit the 
NAE at high energy and deliver their energy to the far field at an some 
indeterminate distance from the NAE that produced the energy.  
 
I did not see this statement.  Where was Cook's statement made?
 
What I saw in the new paper was that the energy of the alphas from the Be-8 
decay was in the form of 17 Mev of angular momentum (spin energy)--not kinetic 
energy.  (The slowing-down of 17 MeV alphas would cause noticeable x-rays and 
other high energy EM radiation.)  The alphas apparently stays put and transfers 
its excess energy via spin coupling, one spin quanta or so at a time.
 
Bob Cook
 
 
- Original Message - 
From: Axil Axil mailto:janap...@gmail.com  
To: vortex-l mailto:vortex-l@eskimo.com  
Sent: Tuesday, April 07, 2015 8:57 AM
Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

Like so many LENR theories, the Cook theory of the LENR reaction is not 
fundamental. Like almost all other LENR  theories, it deals with the emergent 
results of the fundamental LENR reaction without explaining the cause of the 
observed experimental results.

If a theory cannot explain EVERY aspect of the experimental results in every 
dimension, it is not valid.
In particular, the way energy of these high powered alpha particles are 
converted to heat is not addressed, even though that part of the LENR theory is 
central to how the energy of the nuclear reaction is converted to soft x-rays 
and extreme ultraviolet light.

I have concluded from the experimental results derived from many LENR systems 
that the gamma suppression and the basic LENR nuclear reaction is tightly 
coupled together so that if a LENR based nuclear event occurs, no gamma is ever 
seen in a environment that has gotten hot enough (500C).

Gamma suppression is an essential part of the LENR reaction.  So Gamma 
suppression is an essential part of what is going on inside the Nuclear Active 
Environment. If energy is carried away from the NAE, it cannot be converted to 
its final moderated form (soft x-rays and extreme ultraviolet light.) by the 
LENR reaction.  

Cook says that high energy alpha particles exit the NAE at high energy and 
deliver their energy to the far field at an some indeterminate distance from 
the NAE that produced the energy. If this were true, there is always a slight 
chance that the alpha particle could exit the gas envelop and deposit its 
kinetic energy in the Alumina shell where a gamma ray would result. This gamma 
ray is never seen. So if an alpha particle is produced it must have little or 
no kinetic energy that is transferred to the far field.

All the energy of the nuclear reaction is carried away from the NAE by the LENR 
reaction itself. The gamma emission is an intrinsic part of the LENR reaction 
energy transfer mechanism.


On Tue, Apr 7, 2015

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
RossiJones--

I agree with what you say regarding the Li, P reaction and the source of low 
energy gammas (about .5 mev).  The gammas were only called gammas back in the 
60's because they were associated with nuclear reactions.  They are soft gammas 
IMHO and may have been what Rossi and Focardi thought were from the Beta+, 
Beta- reaction.  

I concluded that Cook surely knows about that reaction and the Li,P reaction 
you note.  I conceded he did not consider the old style Li,P reaction was the 
same one the current paper identifies.  I would bet that all the old tests of 
the Li,P reaction were done with energetic protons with the necessity of 
conserving energy and momentum.  How else would they have been 
accomplished--with slow protons?  The coulomb barrier was there and would not 
allow a proton to get near a Li nucleus--this would have been the main stream 
thought.  

Solid state Li,P reaction papers without energetic protons is what I would look 
for. I have found none.

Bob
  - Original Message - 
  From: Jones Beene 
  To: vortex-l@eskimo.com 
  Sent: Tuesday, April 07, 2015 12:24 PM
  Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Here is where Norman Cook and Rossi demonstrate a basic lack of understanding 
of nuclear physics.

  To be precise – as mentioned before, two alpha particles released from Be 
decay are indeed gamma free. That detail is not in question – as far as it 
goes, but it does not cover the complete fusion reaction, only part of it. 
Unfortunately many write-ups do not detail the complete reaction, and Wiki is 
an offender in this case. 

  Ironically, however - the proton + lithium-7 reaction has historically been 
used as a source of gamma radiation, going back half a century ! But if you 
obtained your PhD degree by mail order, as did AR - then you may not have known 
that.

  What is being missed here, and in Norman Cook’s explanation - is the prompt 
gamma which occurs (statistically) at the time the proton interacts with the 
lithium nucleus to form beryllium and then later two alphas. The Li nucleus 
becomes excited, but it cannot simply convert directly to beryllium without an 
energetic emission to compensate for the kinetic energy which caused the 
fusion. There are known signatures for these gammas and statistically they 
occur when protons fuse with lithium at low energy.


  Proton + Li-7 → Be-8 + γ (gamma) → alpha + alpha (no gamma)

  Note: This gamma does NOT derive from the beryllium decay itself - but from 
the fusion of the proton with the lithium nucleus. This is not always depicted 
in the reaction graphics, and if you depend on Wiki as your sole authority on 
physics, then you may miss it. 

  From: Axil Axil 


  Figure 5 depicts the alpha creation process. See how the alpha particles are 
moving away in opposite directions?


  Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice structure 
to which an additional proton will produce a two-tetrahedron structure, giving 
8 Be4 (B). The double alpha lattice structure (C) can then break into 
independent two alpha particles (D), which are released with 17 MeV of angular 
momentum, but without gamma radiation. 


  On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:

  Axil--



  You said Cook said this: Cook says that high energy alpha particles exit the 
NAE at high energy and deliver their energy to the far field at an some 
indeterminate distance from the NAE that produced the energy.  



  I did not see this statement.  Where was Cook's statement made?



  What I saw in the new paper was that the energy of the alphas from the Be-8 
decay was in the form of 17 Mev of angular momentum (spin energy)--not kinetic 
energy.  (The slowing-down of 17 MeV alphas would cause noticeable x-rays and 
other high energy EM radiation.)  The alphas apparently stays put and transfers 
its excess energy via spin coupling, one spin quanta or so at a time.



  Bob Cook





  - Original Message - 

  From: Axil Axil 

  To: vortex-l 

  Sent: Tuesday, April 07, 2015 8:57 AM

  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Like so many LENR theories, the Cook theory of the LENR reaction is not 
fundamental. Like almost all other LENR  theories, it deals with the emergent 
results of the fundamental LENR reaction without explaining the cause of the 
observed experimental results.


  If a theory cannot explain EVERY aspect of the experimental results in every 
dimension, it is not valid.

  In particular, the way energy of these high powered alpha particles are 
converted to heat is not addressed, even though that part of the LENR theory is 
central to how the energy of the nuclear reaction is converted to soft x-rays 
and extreme ultraviolet light.


  I have concluded from the experimental results derived from

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Daniel Rocha

The gamma from Be8 is weak, only 46 KeV.

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Charles Francis

In Italy the title ‘Dr’ does not imply a PhD or medical degree, but only a 
basic (undergraduate) degree. 

From: a.ashfield [mailto:a.ashfi...@verizon.net] 
Sent: 07 April 2015 22:20
To: vortex-l@eskimo.com
Subject: RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

Jones Beene,
You are deliberately misleading about Rossi getting his degree on-line.  He 
got his PhD from Milan University.   He did take an on-line course in chemical 
engineering later, in order to learn about that.  Seems to be a reasonable 
thing to do.

You make much of the lack of theory for how the Li7 gathers a proton.  The 
authors admit they don't know.  BUT the idea of the Be splitting into two 
alphas and not emitting gamma radiation suggests to me that it is worth looking 
for a mechanism for the Li to gain the necessary proton.  I assume no one knows 
at present.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

I wanted to show that the assertions and assumptions made in the Cook paper
were inconsistent with generally accepted posits of LENR not to criticize
those assertions and assumptions against know nuclear reactions. To little
is known about how nuclear reaction in LENR occur to find fault with their
description in a comparison with those accepted by nuclear physics.

On Tue, Apr 7, 2015 at 3:24 PM, Jones Beene jone...@pacbell.net wrote:

  Here is where Norman Cook and Rossi demonstrate a basic lack of understanding
 of nuclear physics.

 To be precise – as mentioned before, two alpha particles released from Be
 decay are indeed gamma free. That detail is not in question – as far as
 it goes, but it does not cover the complete fusion reaction, only part of
 it. Unfortunately many write-ups do not detail the complete reaction, and
 Wiki is an offender in this case.

 Ironically, however - the proton + lithium-7 reaction has historically
 been used as a source of gamma radiation, going back half a century ! But
 if you obtained your PhD degree by mail order, as did AR - then you may
 not have known that.

 What is being missed here, and in Norman Cook’s explanation - is the
 prompt gamma which occurs (statistically) at the time the proton
 interacts with the lithium nucleus to form beryllium and then later two
 alphas. The Li nucleus becomes excited, but it cannot simply convert
 directly to beryllium without an energetic emission to compensate for the
 kinetic energy which caused the fusion. There are known signatures for
 these gammas and statistically they occur when protons fuse with lithium
 at low energy.

 Proton + Li-7 → Be-8 + γ (gamma) → alpha + alpha (no gamma)

 Note: This gamma does NOT derive from the beryllium decay itself - but
 from the fusion of the proton with the lithium nucleus. This is not
 always depicted in the reaction graphics, and if you depend on Wiki as you
 r sole authority on physics, then you may miss it.

 *From:* Axil Axil

 Figure 5 depicts the alpha creation process. See how the alpha particles
 are moving away in opposite directions?

 Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice
 structure to which an additional proton will produce a two-tetrahedron
 structure, giving 8 Be4 (B). The double alpha lattice structure (C) can
 then break into independent two alpha particles (D), which are released
 with 17 MeV of angular momentum, but without gamma radiation.

 On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook *frobertc...@hotmail.com*
 frobertc...@hotmail.com wrote:

 Axil--



 You said Cook said this: Cook says that high energy alpha particles exit
 the NAE at high energy and deliver their energy to the far field at an some
 indeterminate distance from the NAE that produced the energy.



 I did not see this statement.  Where was Cook's statement made?



 What I saw in the new paper was that the energy of the alphas from the
 Be-8 decay was in the form of 17 Mev of angular momentum (spin energy)--not
 kinetic energy.  (The slowing-down of 17 MeV alphas would cause noticeable
 x-rays and other high energy EM radiation.)  The alphas apparently stays
 put and transfers its excess energy via spin coupling, one spin quanta or
 so at a time.



 Bob Cook





 - Original Message -

 *From:* *Axil Axil* janap...@gmail.com

 *To:* *vortex-l* vortex-l@eskimo.com

 *Sent:* Tuesday, April 07, 2015 8:57 AM

 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

 Like so many LENR theories, the Cook theory of the LENR reaction is not
 fundamental. Like almost all other LENR  theories, it deals with the
 emergent results of the fundamental LENR reaction without explaining the
 cause of the observed experimental results.

 If a theory cannot explain EVERY aspect of the experimental results in
 every dimension, it is not valid.

 In particular, the way energy of these high powered alpha particles are
 converted to heat is not addressed, even though that part of the LENR
 theory is central to how the energy of the nuclear reaction is converted to
 soft x-rays and extreme ultraviolet light.

 I have concluded from the experimental results derived from many LENR
 systems that the gamma suppression and the basic LENR nuclear reaction is
 tightly coupled together so that if a LENR based nuclear event occurs,*
 no gamma is ever seen* in a environment that has gotten hot enough (500C).

 Gamma suppression is an essential part of the LENR reaction.  So Gamma
 suppression is an essential part of what is going on inside the Nuclear
 Active Environment. If energy is carried away from the NAE, it cannot be
 converted to its final moderated form (soft x-rays and extreme ultraviolet
 light.) by the LENR reaction.

 Cook says that high energy alpha particles exit the NAE at high energy and
 deliver their energy to the far field at an some indeterminate distance
 from the NAE that produced the energy. If this were true, there is always a
 slight

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

The following link is to a paper that Cook and Rossi may be familiar with--

 http://www.roxit.ax/CN.pdf

It deals with the interaction of electrons with the nuclei and their influence 
on nuclear reactions.  Very interesting.  Note the connection to the Lugano 
professors.  It parallels some of the ideas Cook and Rossi have presented. 

Bob Cook (no relation to N. Cook)
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 2:25 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  I wanted to show that the assertions and assumptions made in the Cook paper 
were inconsistent with generally accepted posits of LENR not to criticize those 
assertions and assumptions against know nuclear reactions. To little is known 
about how nuclear reaction in LENR occur to find fault with their description 
in a comparison with those accepted by nuclear physics.


  On Tue, Apr 7, 2015 at 3:24 PM, Jones Beene jone...@pacbell.net wrote:

Here is where Norman Cook and Rossi demonstrate a basic lack of 
understanding of nuclear physics.

To be precise – as mentioned before, two alpha particles released from Be 
decay are indeed gamma free. That detail is not in question – as far as it 
goes, but it does not cover the complete fusion reaction, only part of it. 
Unfortunately many write-ups do not detail the complete reaction, and Wiki is 
an offender in this case. 

Ironically, however - the proton + lithium-7 reaction has historically been 
used as a source of gamma radiation, going back half a century ! But if you 
obtained your PhD degree by mail order, as did AR - then you may not have known 
that.

What is being missed here, and in Norman Cook’s explanation - is the prompt 
gamma which occurs (statistically) at the time the proton interacts with the 
lithium nucleus to form beryllium and then later two alphas. The Li nucleus 
becomes excited, but it cannot simply convert directly to beryllium without an 
energetic emission to compensate for the kinetic energy which caused the 
fusion. There are known signatures for these gammas and statistically they 
occur when protons fuse with lithium at low energy.


Proton + Li-7 → Be-8 + γ (gamma) → alpha + alpha (no gamma)

Note: This gamma does NOT derive from the beryllium decay itself - but from 
the fusion of the proton with the lithium nucleus. This is not always depicted 
in the reaction graphics, and if you depend on Wiki as your sole authority on 
physics, then you may miss it. 

From: Axil Axil 


Figure 5 depicts the alpha creation process. See how the alpha particles 
are moving away in opposite directions?


Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice 
structure to which an additional proton will produce a two-tetrahedron 
structure, giving 8 Be4 (B). The double alpha lattice structure (C) can then 
break into independent two alpha particles (D), which are released with 17 MeV 
of angular momentum, but without gamma radiation. 


On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:

Axil--



You said Cook said this: Cook says that high energy alpha particles exit 
the NAE at high energy and deliver their energy to the far field at an some 
indeterminate distance from the NAE that produced the energy.  



I did not see this statement.  Where was Cook's statement made?



What I saw in the new paper was that the energy of the alphas from the Be-8 
decay was in the form of 17 Mev of angular momentum (spin energy)--not kinetic 
energy.  (The slowing-down of 17 MeV alphas would cause noticeable x-rays and 
other high energy EM radiation.)  The alphas apparently stays put and transfers 
its excess energy via spin coupling, one spin quanta or so at a time.



Bob Cook





- Original Message - 

From: Axil Axil 

To: vortex-l 

Sent: Tuesday, April 07, 2015 8:57 AM

Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author 
Andrea Rossi


Like so many LENR theories, the Cook theory of the LENR reaction is not 
fundamental. Like almost all other LENR  theories, it deals with the emergent 
results of the fundamental LENR reaction without explaining the cause of the 
observed experimental results.


If a theory cannot explain EVERY aspect of the experimental results in 
every dimension, it is not valid.

In particular, the way energy of these high powered alpha particles are 
converted to heat is not addressed, even though that part of the LENR theory is 
central to how the energy of the nuclear reaction is converted to soft x-rays 
and extreme ultraviolet light.


I have concluded from the experimental results derived from many LENR 
systems that the gamma suppression and the basic LENR nuclear reaction is 
tightly coupled together so that if a LENR based nuclear event occurs, no gamma 
is ever seen in a environment that has gotten

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

In LENR angular momentum can be converted back and forth to linear momentum
many times  before that energy is projected as magnetic force through EMF
photons.

Metal refection converts spin to linear force to drive dipole motion of the
electron hole pair.

http://arxiv.org/ftp/arxiv/papers/1308/1308.0547.pdf

*Extraordinary momentum and spin in evanescent waves*

Then the energy of the dipole is converted back to angular momentum by a
surface boundary to form a magnetic soliton. Then the soliton converts
angular momentum of the amplified spin to an anapole magnetic field that
delivers energy at a distance to the nucleus.

This is how the energy content of the spin of infrared photons are
converted to energy that increases the mass of the proton to convert that
proton into a neutron.

On Tue, Apr 7, 2015 at 3:29 PM, Bob Cook frobertc...@hotmail.com wrote:

  Axil--

 Note that the paper says the energy is angular momentum not kinetic energy
 of the alphas.  Angular momentum energy is spin energy.  The alphas move
 away with essentially no kinetic energy normally associated with non-solid
 state or non-coherent systems.  It is my conclusion from what Cook
 claims, that the electronic cloud must shield the alphas as they are formed
 from being repulsed from each other, or their charge does not materialize
 until the spin energy is fractionated and their distance is sufficient so
 as to impart only a small kinetic energy to each particle, if they have not
 already become neutral He atoms. The Pauli Uncertainty Principle may come
 into play to spread the wave function of the spin energy of the excited He*
 to a large radius compared to the radius associated with a ground state He
 nucleus.  The enlarged wave function may also act to couple to the rest of
 the particles in the locale (coherent system), including the electrons.

 Bob

 - Original Message -
 *From:* Axil Axil janap...@gmail.com
 *To:* vortex-l vortex-l@eskimo.com
 *Sent:* Tuesday, April 07, 2015 10:52 AM
 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

 Figure 5 depicts the alpha creation process. See how the alpha particles
 are moving away in opposite directions?

 Figure 5: The lowest-lying excited-state of 7 Li4 (A) has a lattice
 structure to which an additional proton will produce a two-tetrahedron
 structure, giving 8 Be4 (B). The double alpha lattice structure (C) can
 then break into independent two alpha particles (D), which are released
 with 17 MeV of angular momentum, but without gamma radiation.

 On Tue, Apr 7, 2015 at 12:39 PM, Bob Cook frobertc...@hotmail.com wrote:

  Axil--

 You said Cook said this: Cook says that high energy alpha particles
 exit the NAE at high energy and deliver their energy to the far field at an
 some indeterminate distance from the NAE that produced the energy.

 I did not see this statement.  Where was Cook's statement made?

 What I saw in the new paper was that the energy of the alphas from the
 Be-8 decay was in the form of 17 Mev of angular momentum (spin energy)--not
 kinetic energy.  (The slowing-down of 17 MeV alphas would cause noticeable
 x-rays and other high energy EM radiation.)  The alphas apparently stays
 put and transfers its excess energy via spin coupling, one spin quanta or
 so at a time.

 Bob Cook



 - Original Message -
 *From:* Axil Axil janap...@gmail.com
 *To:* vortex-l vortex-l@eskimo.com
 *Sent:* Tuesday, April 07, 2015 8:57 AM
 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

  Like so many LENR theories, the Cook theory of the LENR reaction is not
 fundamental. Like almost all other LENR  theories, it deals with the
 emergent results of the fundamental LENR reaction without explaining the
 cause of the observed experimental results.

 If a theory cannot explain EVERY aspect of the experimental results in
 every dimension, it is not valid.
  In particular, the way energy of these high powered alpha particles are
 converted to heat is not addressed, even though that part of the LENR
 theory is central to how the energy of the nuclear reaction is converted to
 soft x-rays and extreme ultraviolet light.

 I have concluded from the experimental results derived from many LENR
 systems that the gamma suppression and the basic LENR nuclear reaction is
 tightly coupled together so that if a LENR based nuclear event occurs, *no
 gamma is ever seen* in a environment that has gotten hot enough (500C).

 Gamma suppression is an essential part of the LENR reaction.  So Gamma
 suppression is an essential part of what is going on inside the Nuclear
 Active Environment. If energy is carried away from the NAE, it cannot be
 converted to its final moderated form (soft x-rays and extreme ultraviolet
 light.) by the LENR reaction.

 Cook says that high energy alpha particles exit the NAE at high energy
 and deliver their energy to the far field at an some indeterminate distance
 from the NAE

RE: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Charles Francis 
http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=from:%22Charles+Francis%22 
Tue, 07 Apr 2015 14:00:04 -0700 
http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=date:20150407


In Italy the title ‘Dr’ does not imply a PhD or medical degree, but only a
basic (undergraduate) degree

I am well aware of that it is different.  In several ways: Rossi had to submit 
a thesis (on relativity).
Jones Beene used the ad hom that Rossi just had a degree from on line.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

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

Simulation of the Nuclear Transmutation Effects in LENR ... the
presentation form ICCF-18

Cook states as follows:  Interestingly, Ni61 was not strongly depleted–
apparently not participating in the LENR

What Cook has not yet recognized is that each LENR system produces a
different pattern of nuclear transmutations that is based of the chemistry
associated with that system.

There is an endless number of transmutation patterns that will be produced
by any given LENR system. It is useless to assume that these patterns are
meaningful as a cause of the LENR reaction. They are a RESULT of a
particular type of reaction happening at any given time in any given system.

On Tue, Apr 7, 2015 at 6:57 PM, a.ashfield a.ashfi...@verizon.net wrote:

  Charles Francis
 http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=from:%22Charles+Francis%22
  Tue,
 07 Apr 2015 14:00:04 -0700
 http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=date:20150407

 In Italy the title ‘Dr’ does not imply a PhD or medical degree, but only a
 basic (undergraduate) degree

 I am well aware of that it is different.  In several ways: Rossi had to 
 submit a thesis (on relativity).
 Jones Beene used the ad hom that Rossi just had a degree from on line.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

2015-04-07 Thread Eric Walker

Hi Bob,

The possibility you've been drawing attention to, that the result of the
decay of the [8Be]* compound nucleus into two 4He nuclei with little linear
momentum and a great deal of angular momentum makes for an interesting
thought experiment.  Out of curiosity, I calculated the energy that would
be needed to break up an alpha particle into either tritium and a proton or
3He and a neutron, which would be the reverse of these two reactions:

3He + n → 4He + Q (19.3 MeV)
t + p → 4He + Q (20.5 MeV)

As I understand it, this implies that angular momentum sufficient to
produce ~ 19 MeV of centripetal force would be needed to break apart a 4He
into either 3He and a neutron or tritium and a proton.  This suggests that
a 4He can carry a large amount of angular momentum before it is likely to
break apart. (I assume the process is probabilistic and that the force
needed lies along a distribution.)

Further comments inline.

Eric


On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:

However, I know of know reason why the light nuclei cannot have any spin
 quantum number--high or low.  Any spin quantum is available.


Further to the thought experiment, I think we should make a clear
distinction between two types of spin -- there's the actual spinning
motion of a nucleus (e.g., 4He), and there is the spin state of the
nucleus.  At higher rates of rotation, a heavy nucleus such as an isotope
of nickel will reconfigure into a higher spin state, presumably through
deformation.  In such a state a photon may be emitted, with the nucleus
relaxing into a lower spin state.  Here my mental model is of neodymium
magnets spinning around in a clump.  When they snap together into a
lower-energy configuration, a photon is emitted through the movement of the
magnets as they snap together.  The photon is emitted in a direction and
carries away energy in such a way as to slow the angular movement of the
spinning nucleus a little (by the amount of energy carried away by the
photon).  The participants involved in such a transition are the nucleons,
and the energy of the photon that is emitted will correspondingly be in the
keV or MeV range, which is in the nuclear range.

A light nucleus, such as 4He, does not have a bound excited state.  My
understanding is that it cannot deform under high angular momentum into a
higher energy state which will emit a photon when it relaxes.  The 4He will
either break apart into lighter constituents under centrifugal forces or it
will not.  But I'm guessing that the actual moment-to-moment velocity of
the 4He about its axis of motion is in principle a continuous quantity.  If
this is true, perhaps the energy could be released to the environment in
small amounts.

Where the thought experiment gets interesting is in the supposition that
you and others have already offered in this thread, that charged body such
as a 4He nucleus that is spinning at an incredible rate will set up a
magnetic field.  This magnetic field could disturb nearby electrons,
causing them to emit lower energy photons in the process.

Although I do not see anything special in the 7Li+p to 8Be transition that
has been proposed (and note Jones's point about the gamma that would be
omitted in the process), I think the more general notion of the energy of a
nuclear transition somehow being deposited in angular momentum and then
released in small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin.
How can a nuclear reaction involving them have  huge annular momentum?

On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com wrote:

 Hi Bob,

 The possibility you've been drawing attention to, that the result of the
 decay of the [8Be]* compound nucleus into two 4He nuclei with little linear
 momentum and a great deal of angular momentum makes for an interesting
 thought experiment.  Out of curiosity, I calculated the energy that would
 be needed to break up an alpha particle into either tritium and a proton or
 3He and a neutron, which would be the reverse of these two reactions:

 3He + n → 4He + Q (19.3 MeV)
 t + p → 4He + Q (20.5 MeV)

 As I understand it, this implies that angular momentum sufficient to
 produce ~ 19 MeV of centripetal force would be needed to break apart a 4He
 into either 3He and a neutron or tritium and a proton.  This suggests that
 a 4He can carry a large amount of angular momentum before it is likely to
 break apart. (I assume the process is probabilistic and that the force
 needed lies along a distribution.)

 Further comments inline.

 Eric


 On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:

 However, I know of know reason why the light nuclei cannot have any spin
 quantum number--high or low.  Any spin quantum is available.


 Further to the thought experiment, I think we should make a clear
 distinction between two types of spin -- there's the actual spinning
 motion of a nucleus (e.g., 4He), and there is the spin state of the
 nucleus.  At higher rates of rotation, a heavy nucleus such as an isotope
 of nickel will reconfigure into a higher spin state, presumably through
 deformation.  In such a state a photon may be emitted, with the nucleus
 relaxing into a lower spin state.  Here my mental model is of neodymium
 magnets spinning around in a clump.  When they snap together into a
 lower-energy configuration, a photon is emitted through the movement of the
 magnets as they snap together.  The photon is emitted in a direction and
 carries away energy in such a way as to slow the angular movement of the
 spinning nucleus a little (by the amount of energy carried away by the
 photon).  The participants involved in such a transition are the nucleons,
 and the energy of the photon that is emitted will correspondingly be in the
 keV or MeV range, which is in the nuclear range.

 A light nucleus, such as 4He, does not have a bound excited state.  My
 understanding is that it cannot deform under high angular momentum into a
 higher energy state which will emit a photon when it relaxes.  The 4He will
 either break apart into lighter constituents under centrifugal forces or it
 will not.  But I'm guessing that the actual moment-to-moment velocity of
 the 4He about its axis of motion is in principle a continuous quantity.  If
 this is true, perhaps the energy could be released to the environment in
 small amounts.

 Where the thought experiment gets interesting is in the supposition that
 you and others have already offered in this thread, that charged body such
 as a 4He nucleus that is spinning at an incredible rate will set up a
 magnetic field.  This magnetic field could disturb nearby electrons,
 causing them to emit lower energy photons in the process.

 Although I do not see anything special in the 7Li+p to 8Be transition that
 has been proposed (and note Jones's point about the gamma that would be
 omitted in the process), I think the more general notion of the energy of a
 nuclear transition somehow being deposited in angular momentum and then
 released in small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

The isospin in a nuclear reaction is conserved. The quantum law is called
the conservation of isospin.

On Wed, Apr 8, 2015 at 1:16 AM, Axil Axil janap...@gmail.com wrote:

 I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin.
 How can a nuclear reaction involving them have  huge annular momentum?

 On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com
 wrote:

 Hi Bob,

 The possibility you've been drawing attention to, that the result of the
 decay of the [8Be]* compound nucleus into two 4He nuclei with little linear
 momentum and a great deal of angular momentum makes for an interesting
 thought experiment.  Out of curiosity, I calculated the energy that would
 be needed to break up an alpha particle into either tritium and a proton or
 3He and a neutron, which would be the reverse of these two reactions:

 3He + n → 4He + Q (19.3 MeV)
 t + p → 4He + Q (20.5 MeV)

 As I understand it, this implies that angular momentum sufficient to
 produce ~ 19 MeV of centripetal force would be needed to break apart a 4He
 into either 3He and a neutron or tritium and a proton.  This suggests that
 a 4He can carry a large amount of angular momentum before it is likely to
 break apart. (I assume the process is probabilistic and that the force
 needed lies along a distribution.)

 Further comments inline.

 Eric


 On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:

 However, I know of know reason why the light nuclei cannot have any spin
 quantum number--high or low.  Any spin quantum is available.


 Further to the thought experiment, I think we should make a clear
 distinction between two types of spin -- there's the actual spinning
 motion of a nucleus (e.g., 4He), and there is the spin state of the
 nucleus.  At higher rates of rotation, a heavy nucleus such as an isotope
 of nickel will reconfigure into a higher spin state, presumably through
 deformation.  In such a state a photon may be emitted, with the nucleus
 relaxing into a lower spin state.  Here my mental model is of neodymium
 magnets spinning around in a clump.  When they snap together into a
 lower-energy configuration, a photon is emitted through the movement of the
 magnets as they snap together.  The photon is emitted in a direction and
 carries away energy in such a way as to slow the angular movement of the
 spinning nucleus a little (by the amount of energy carried away by the
 photon).  The participants involved in such a transition are the nucleons,
 and the energy of the photon that is emitted will correspondingly be in the
 keV or MeV range, which is in the nuclear range.

 A light nucleus, such as 4He, does not have a bound excited state.  My
 understanding is that it cannot deform under high angular momentum into a
 higher energy state which will emit a photon when it relaxes.  The 4He will
 either break apart into lighter constituents under centrifugal forces or it
 will not.  But I'm guessing that the actual moment-to-moment velocity of
 the 4He about its axis of motion is in principle a continuous quantity.  If
 this is true, perhaps the energy could be released to the environment in
 small amounts.

 Where the thought experiment gets interesting is in the supposition that
 you and others have already offered in this thread, that charged body such
 as a 4He nucleus that is spinning at an incredible rate will set up a
 magnetic field.  This magnetic field could disturb nearby electrons,
 causing them to emit lower energy photons in the process.

 Although I do not see anything special in the 7Li+p to 8Be transition
 that has been proposed (and note Jones's point about the gamma that would
 be omitted in the process), I think the more general notion of the energy
 of a nuclear transition somehow being deposited in angular momentum and
 then released in small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Eric--

One understanding that I have had is that in a quantum system angular momentum 
is quantized and must be multiples of h/2 that's Plank's constant (h).  Also 
angular momentum must be conserved in what ever reaction happens to my 
knowledge.  Also I do not know of any reason that He* could not happen with 
each new He* spinning in opposite direction with respect to a magnetic field 
and slow down incrementally with angular momentum distributed to the coherent 
system of electrons. The slowing down process may actually happen on time scale 
associated with nuclear transitions. 

I know of know other ways energy could be distributed with no apparent kinetic 
energy associated with the new He nuclei.  The total energy for both He* nuclei 
totals about 13 Mev, well below the amount you suggest it would take to cause a 
He-4 to fly apart.  

I am not familiar with the wave function that describes an alpha particle, but 
would guess the family of solutions  include spin quanta, energy quanta and 
other parameters and that there are some solutions with high quanta possible.  
Whatever it is, it is a complex mathematical function.  

Lets keep thinking.

Bob  
  - Original Message - 
  From: Eric Walker 
  To: vortex-l@eskimo.com 
  Sent: Tuesday, April 07, 2015 9:28 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi


  Hi Bob,


  The possibility you've been drawing attention to, that the result of the 
decay of the [8Be]* compound nucleus into two 4He nuclei with little linear 
momentum and a great deal of angular momentum makes for an interesting thought 
experiment.  Out of curiosity, I calculated the energy that would be needed to 
break up an alpha particle into either tritium and a proton or 3He and a 
neutron, which would be the reverse of these two reactions:


  3He + n → 4He + Q (19.3 MeV)
  t + p → 4He + Q (20.5 MeV)


  As I understand it, this implies that angular momentum sufficient to produce 
~ 19 MeV of centripetal force would be needed to break apart a 4He into either 
3He and a neutron or tritium and a proton.  This suggests that a 4He can carry 
a large amount of angular momentum before it is likely to break apart. (I 
assume the process is probabilistic and that the force needed lies along a 
distribution.)


  Further comments inline.


  Eric




  On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:


However, I know of know reason why the light nuclei cannot have any spin 
quantum number--high or low.  Any spin quantum is available.


  Further to the thought experiment, I think we should make a clear distinction 
between two types of spin -- there's the actual spinning motion of a nucleus 
(e.g., 4He), and there is the spin state of the nucleus.  At higher rates of 
rotation, a heavy nucleus such as an isotope of nickel will reconfigure into a 
higher spin state, presumably through deformation.  In such a state a photon 
may be emitted, with the nucleus relaxing into a lower spin state.  Here my 
mental model is of neodymium magnets spinning around in a clump.  When they 
snap together into a lower-energy configuration, a photon is emitted through 
the movement of the magnets as they snap together.  The photon is emitted in a 
direction and carries away energy in such a way as to slow the angular movement 
of the spinning nucleus a little (by the amount of energy carried away by the 
photon).  The participants involved in such a transition are the nucleons, and 
the energy of the photon that is emitted will correspondingly be in the keV or 
MeV range, which is in the nuclear range.


  A light nucleus, such as 4He, does not have a bound excited state.  My 
understanding is that it cannot deform under high angular momentum into a 
higher energy state which will emit a photon when it relaxes.  The 4He will 
either break apart into lighter constituents under centrifugal forces or it 
will not.  But I'm guessing that the actual moment-to-moment velocity of the 
4He about its axis of motion is in principle a continuous quantity.  If this is 
true, perhaps the energy could be released to the environment in small amounts.


  Where the thought experiment gets interesting is in the supposition that you 
and others have already offered in this thread, that charged body such as a 4He 
nucleus that is spinning at an incredible rate will set up a magnetic field.  
This magnetic field could disturb nearby electrons, causing them to emit lower 
energy photons in the process.


  Although I do not see anything special in the 7Li+p to 8Be transition that 
has been proposed (and note Jones's point about the gamma that would be omitted 
in the process), I think the more general notion of the energy of a nuclear 
transition somehow being deposited in angular momentum and then released in 
small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

Two particles spinning anti-parallel equal 0 spin if they each have an equal 
spin energy.  Angular momentum is a vector quantity, not a scalar one.  

Bob
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Tuesday, April 07, 2015 10:16 PM
  Subject: Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea 
Rossi

  I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin. How 
can a nuclear reaction involving them have  huge annular momentum?

  On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com wrote:

Hi Bob,

The possibility you've been drawing attention to, that the result of the 
decay of the [8Be]* compound nucleus into two 4He nuclei with little linear 
momentum and a great deal of angular momentum makes for an interesting thought 
experiment.  Out of curiosity, I calculated the energy that would be needed to 
break up an alpha particle into either tritium and a proton or 3He and a 
neutron, which would be the reverse of these two reactions:

3He + n → 4He + Q (19.3 MeV)
t + p → 4He + Q (20.5 MeV)

As I understand it, this implies that angular momentum sufficient to 
produce ~ 19 MeV of centripetal force would be needed to break apart a 4He into 
either 3He and a neutron or tritium and a proton.  This suggests that a 4He can 
carry a large amount of angular momentum before it is likely to break apart. (I 
assume the process is probabilistic and that the force needed lies along a 
distribution.)

Further comments inline.

Eric

On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:

  However, I know of know reason why the light nuclei cannot have any spin 
quantum number--high or low.  Any spin quantum is available.

Further to the thought experiment, I think we should make a clear 
distinction between two types of spin -- there's the actual spinning motion 
of a nucleus (e.g., 4He), and there is the spin state of the nucleus.  At 
higher rates of rotation, a heavy nucleus such as an isotope of nickel will 
reconfigure into a higher spin state, presumably through deformation.  In such 
a state a photon may be emitted, with the nucleus relaxing into a lower spin 
state.  Here my mental model is of neodymium magnets spinning around in a 
clump.  When they snap together into a lower-energy configuration, a photon is 
emitted through the movement of the magnets as they snap together.  The photon 
is emitted in a direction and carries away energy in such a way as to slow the 
angular movement of the spinning nucleus a little (by the amount of energy 
carried away by the photon).  The participants involved in such a transition 
are the nucleons, and the energy of the photon that is emitted will 
correspondingly be in the keV or MeV range, which is in the nuclear range.

A light nucleus, such as 4He, does not have a bound excited state.  My 
understanding is that it cannot deform under high angular momentum into a 
higher energy state which will emit a photon when it relaxes.  The 4He will 
either break apart into lighter constituents under centrifugal forces or it 
will not.  But I'm guessing that the actual moment-to-moment velocity of the 
4He about its axis of motion is in principle a continuous quantity.  If this is 
true, perhaps the energy could be released to the environment in small amounts.

Where the thought experiment gets interesting is in the supposition that 
you and others have already offered in this thread, that charged body such as a 
4He nucleus that is spinning at an incredible rate will set up a magnetic 
field.  This magnetic field could disturb nearby electrons, causing them to 
emit lower energy photons in the process.

Although I do not see anything special in the 7Li+p to 8Be transition that 
has been proposed (and note Jones's point about the gamma that would be omitted 
in the process), I think the more general notion of the energy of a nuclear 
transition somehow being deposited in angular momentum and then released in 
small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi

isospin is a product of the strong force and of the quarks inside the
protons and neutrons. It is fixed no matter how the atoms spins. An atom
might be induced to spin using EMF but usually that spin cannot effect the
isospin of the nucleus. with zero isospin. Only non zero isospins are
effected by RF.

On Wed, Apr 8, 2015 at 1:37 AM, Bob Cook frobertc...@hotmail.com wrote:

  Two particles spinning anti-parallel equal 0 spin if they each have an
 equal spin energy.  Angular momentum is a vector quantity, not a scalar
 one.

 Bob

 - Original Message -
 *From:* Axil Axil janap...@gmail.com
 *To:* vortex-l vortex-l@eskimo.com
 *Sent:* Tuesday, April 07, 2015 10:16 PM
 *Subject:* Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author
 Andrea Rossi

 I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin.
 How can a nuclear reaction involving them have  huge annular momentum?

 On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker eric.wal...@gmail.com
 wrote:

 Hi Bob,

 The possibility you've been drawing attention to, that the result of the
 decay of the [8Be]* compound nucleus into two 4He nuclei with little linear
 momentum and a great deal of angular momentum makes for an interesting
 thought experiment.  Out of curiosity, I calculated the energy that would
 be needed to break up an alpha particle into either tritium and a proton or
 3He and a neutron, which would be the reverse of these two reactions:

 3He + n → 4He + Q (19.3 MeV)
 t + p → 4He + Q (20.5 MeV)

 As I understand it, this implies that angular momentum sufficient to
 produce ~ 19 MeV of centripetal force would be needed to break apart a 4He
 into either 3He and a neutron or tritium and a proton.  This suggests that
 a 4He can carry a large amount of angular momentum before it is likely to
 break apart. (I assume the process is probabilistic and that the force
 needed lies along a distribution.)

 Further comments inline.

 Eric


 On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook frobertc...@hotmail.com wrote:

 However, I know of know reason why the light nuclei cannot have any spin
 quantum number--high or low.  Any spin quantum is available.


 Further to the thought experiment, I think we should make a clear
 distinction between two types of spin -- there's the actual spinning
 motion of a nucleus (e.g., 4He), and there is the spin state of the
 nucleus.  At higher rates of rotation, a heavy nucleus such as an isotope
 of nickel will reconfigure into a higher spin state, presumably through
 deformation.  In such a state a photon may be emitted, with the nucleus
 relaxing into a lower spin state.  Here my mental model is of neodymium
 magnets spinning around in a clump.  When they snap together into a
 lower-energy configuration, a photon is emitted through the movement of the
 magnets as they snap together.  The photon is emitted in a direction and
 carries away energy in such a way as to slow the angular movement of the
 spinning nucleus a little (by the amount of energy carried away by the
 photon).  The participants involved in such a transition are the nucleons,
 and the energy of the photon that is emitted will correspondingly be in the
 keV or MeV range, which is in the nuclear range.

 A light nucleus, such as 4He, does not have a bound excited state.  My
 understanding is that it cannot deform under high angular momentum into a
 higher energy state which will emit a photon when it relaxes.  The 4He will
 either break apart into lighter constituents under centrifugal forces or it
 will not.  But I'm guessing that the actual moment-to-moment velocity of
 the 4He about its axis of motion is in principle a continuous quantity.  If
 this is true, perhaps the energy could be released to the environment in
 small amounts.

 Where the thought experiment gets interesting is in the supposition that
 you and others have already offered in this thread, that charged body such
 as a 4He nucleus that is spinning at an incredible rate will set up a
 magnetic field.  This magnetic field could disturb nearby electrons,
 causing them to emit lower energy photons in the process.

 Although I do not see anything special in the 7Li+p to 8Be transition
 that has been proposed (and note Jones's point about the gamma that would
 be omitted in the process), I think the more general notion of the energy
 of a nuclear transition somehow being deposited in angular momentum and
 then released in small amounts is a very interesting one.

Re: [Vo]:mainstream physics paper bout the Hot Cat, co-author Andrea Rossi