Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Eric Walker]

I wrote:

I have not yet looked closely at Holmlid's results, but I don't write them
> off.  I'm keeping a distinction in my mind between his experimental
> observations and his theoretical speculations.
>

I have now had a chance to look more closely at Leif Holmlid's 2013 paper,
"Direct observation of particles with energy >10 MeV/u from laser-induced
fusion in ultra-dense deuterium", from which I quoted a few days ago about
there being ~ MeV particles being detected.  Bob Higgins corrected me and
pointed out that it was time-of-flight spectrometry and that the radiations
being detected were claimed to be consistent with ~ MeV u-1 particles, on
the basis of the nanosecond times of flight.

Here is what I can say after reading through the paper more carefully.
Holmlid is irradiating a target of copper or nickel with a laser.  Above
the target is a tube through which hydrogen or deuterium gas issues after
having passed through a plug of porous potassium catalyst.  Holmlid
assumes, without elaboration, that the catalyst has created "ultra-dense
deuterium" at that point.

It seems Holmlid has constructed his own custom time-of-flight spectrometer
using an oscilloscope together with a charged collector plate attached to a
chamber he has used in other experiments.  He does not report a calibration
step for his custom time-of-flight spectrometer, e.g., using americium.  It
is a little hard to imagine trusting one's own time-of-flight spectrometer
if one is not already an expert in this kind of measurement.  It is yet
harder to imagine trusting a custom instrument that has not been
calibrated.  Experts in the field using purchased spectrometers will surely
calibrate using natural radiation sources of known energy.

Holmlid shows great courage in reasoning from first principles in this
paper and in others of his that I've had a chance to read.  In this paper
he assumes, without much discussion, that he is seeing dd fusion.  He does
not think he is seeing neutrons, as weak interactions are expected to occur
with the aluminum degrader material (his meaning here was a little unclear).

It is possible and perhaps even likely that Holmlid is really seeing some
kind of radiation.  When he places an aluminum degrader before the
radiation, or uses a strong magnet, he still gets a signal, which probably
means the particles are not electrons, consistent with his conclusion.  But
I will not trust any of his conclusions until he brings in someone who is
skilled at measuring charged particle radiation and who is not a frequent
collaborator.

Eric

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

You might be right Axil, These days I certainly tend to think there is some 
kind of collective disruptive or resonant behaviour that is exciting the nuclei 
or causing them to act this way. I acknowledge your good arguments and evidence 
for the formation of SPP's in these devices. It also seems some kind of trigger 
for collective behaviour is required. 
I think its interesting to look at the possibilities of low energy virtual 
resonant meson exchange rather than nucleon exchange. If it can occur.
I'm certainly no expert on this but I suppose if the resonance increases slowly 
(but still fast in atomic scales) the first real single particles to be 
generated would be pions? unless particle pairs involving electrons, positrons, 
muons and associated neutrinos are be generated before. Is this correct 
thinking?
pion0 has slightly less mass than pion+ or pion- and has much shorter half life 
so it is curious if we do not see gamma from pion0 decay. Could it be that the 
longer half life of virtual Pion+ and Pion- means in theory are more likely to 
tunnel? Or electrons, muons and pion0 are suppressed somehow so that pion+ and 
pion- are the first to be generated.
I suppose generating a real meson would have higher energy consequences, but 
this could lead to the observed muons. 
I'm imagining if the energy is a "slowly" building resonant effect maybe as 
soon as a pion is manifested if it is a pion- perhaps its wave function 
occupies the S orbital to form Pionium until it interacts with the nucleus or 
decays via muon decay. If it is a pion+ perhaps it is ejected with sufficient 
energy to interact with another Deuterium nucleus to form a diproton that 
decays to 2 high energy protons, or it decays to a muon of characteristic 
energy that is later detected.
I suppose a real nuclear physicist will correct me on a lot of my assumptions.
Date: Tue, 20 Oct 2015 04:16:37 -0400
Subject: Re: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
From: janap...@gmail.com
To: vortex-l@eskimo.com

I believe that what you imagine is what is happening in one of the many cases 
involving SPP extreme magnetic projections and entanglement,

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Eric Walker]

On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke 
wrote:

If any was produced we would need to balance this against those the energy
> required for pion production.
>

The amount of energy needed to create a free pion is large; the rest mass
for a pion is ~ 135 MeV.  Consider that the largest amount of energy
typically discussed in the context of cold fusion up to now has been ~ 24
MeV.  Holmlid's observations are likely to go back to something other than
the generation of pions.

Eric

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Axil Axil]

I believe that what you imagine is what is happening in one of the many
cases involving SPP extreme magnetic projections and entanglement,

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

In case it is interesting I have found a couple of interesting papers on pionic 
deuterium.

http://www.slac.stanford.edu/econf/C070910/PDF/290.pdf

http://www.researchgate.net/profile/Joao_Veloso2/publication/226690552_Pionic_Deuterium/links/0fcfd509900d147bc600.pdf?inViewer=true=true=true=publication_detail


The papers are quite technical and apparently Pionic Deuterium has been studied 
quite extensively. There are some interesting points about pion production and 
absorption with nuclei and nucleon pairs, ground state s orbital interaction 
with the nucleus, x-Ray production and experimental set up. The second paper is 
much more detailed.

It seems to me if we see effects in the experiments described in these papers, 
if pions are indeed generated then we might even be more likely to see 
something in ultra dense material.

I wonder what kinds of energies would be yielded in these kind of reactions. If 
any was produced we would need to balance this against those the energy 
required for pion production. But perhaps the excess energy for pion production 
comes from muon catalysed fusion and / or external forces and resonance effects.

Sent from my iPhone

> On 19 Oct 2015, at 22:18, Stephen Cooke  wrote:
> 
> I guess there is a chance that they saw X-ray spectra from Pionic Deuterium 
> as well as inferring from muons of specific energy. It will be interesting to 
> see what he says on Thursday, I hope someone can ask these kind of questions. 
> 
> Sent from my iPad
> 
>> On 19 okt. 2015, at 18:18, Jones Beene  wrote:
>> 
>> The lifetime of pions is a factor of 100 times shorter than the muon, which 
>> is itself short.
>>  
>> How pions can be detected at all, with so short a lifetime is a question 
>> worth asking.
>>  
>> There is a good possibility that they are inferred – from finding muons, 
>> which can be detected. Hopefully, Ólafsson will address this issue of 
>> detection on Thursday.
>>  
>>  
>> From: Stephen Cooke
>>  
>> … Is It possible that the nuclei are sufficiently close that those pions or 
>> virtual ones get generated in association with one nucleus and absorbed by 
>> another either directly or by tunneling at lower energy?
>> 
>>  

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Axil Axil]

These SPP are inherently coherent. This means that they all live in a state
of bose einstein condensation(BEC). When a nuclear reaction occurs within
this BEC, the nature of the nuclear reaction changes. Everything is
entangled. Super absorption and super-radiation comes into play. The rules
that work for neutron based reactions no longer apply.

The decay energies from meson decay go right back into storage within the
SPP.

Using neutron based thinking to understand how LENR reactions work is a
mistake. All these LENR reactions need to be understood in their own rite.
This study will be one of the major LENR research efforts during this
century. There will be a lot of work required to get through all this LENR
research involing the ins and outs of the LENR nuclear reaction..

On Tue, Oct 20, 2015 at 11:42 AM, Stephen Cooke <stephen_coo...@hotmail.com>
wrote:

> One negative point to this idea I suppose is that if a pion is absorbed by
> a nucleon pair in a nucleus the whole mass energy in the pion will be
> released so if even a stationary pion was absorbed the final 2 nucleons in
> the pair will be ejected with kinetic energy about 63 MeV each. I suppose
> it is difficult to see how these nucleons would not generate gamma or
> neutrons on interaction with other nuclei.
>
> Also on the negative side: If a pion is implicated and needs to be
> generated within a nucleus, i suppose if this is not from a high energy
> collision it would need to be created in a nucleus with higher total
> binding energy than the pion mass energy. This would be a medium weight
> nucleus so if we assume 8 MeV binding energy per nucleon it would probably
> require a nucleus heavier than Oxygen at least.
>
> --
> From: stephen_coo...@hotmail.com
> To: vortex-l@eskimo.com
> Subject: RE: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
> Date: Tue, 20 Oct 2015 16:16:09 +0200
>
>
> 'The amount of energy needed to create a free pion is large; the rest mass
> for a pion is ~ 135 MeV'
>
> Very true this is also true for the muon which has a rest mass for a pion
> is ~ 106 MeV.
>
> I'm not sure if muons can be generated without pions? muon pair production
> would require even more energy.
>
> The energy is also quite high compared to the binding energy of light
> Nuclei.
>
> If I am right I think the laser produces much less thermal energy too.
>
> It is difficult to imagine how either muons (or pions) can form with out
> some kind of collective resonance effect or an additional high MeV energy
> source such as sufficient energy from a high energy fusion event or more
> even strangely a nucleon decay to mesons.
>
> It will be interesting what Holmlids observations and explanations say,
> I'm quite curious as you say he could well have another explanation.
>
>
> ----------
> From: eric.wal...@gmail.com
> Date: Tue, 20 Oct 2015 08:47:08 -0500
> Subject: Re: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
> To: vortex-l@eskimo.com
>
> On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke <stephen_coo...@hotmail.com
> > wrote:
>
> If any was produced we would need to balance this against those the energy
> required for pion production.
>
>
> The amount of energy needed to create a free pion is large; the rest mass
> for a pion is ~ 135 MeV.  Consider that the largest amount of energy
> typically discussed in the context of cold fusion up to now has been ~ 24
> MeV.  Holmlid's observations are likely to go back to something other than
> the generation of pions.
>
> Eric
>
>

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Eric Walker]

On Tue, Oct 20, 2015 at 9:16 AM, Stephen Cooke 
wrote:

Very true this is also true for the muon which has a rest mass for a pion
> is ~ 106 MeV.


This is one of the reasons I don't think muons are involved either.
Another reason -- if muons were being generated, you'd get muon catalyzed
fusion, and muon catalyzed fusion has all of the normal branching ratios
(e.g., for dd fusion, you'd get equal parts neutrons and tritium).

It will be interesting what Holmlids observations and explanations say, I'm
> quite curious as you say he could well have another explanation.


I have not yet looked closely at Holmlid's results, but I don't write them
off.  I'm keeping a distinction in my mind between his experimental
observations and his theoretical speculations.

Eric

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

One negative point to this idea I suppose is that if a pion is absorbed by a 
nucleon pair in a nucleus the whole mass energy in the pion will be released so 
if even a stationary pion was absorbed the final 2 nucleons in the pair will be 
ejected with kinetic energy about 63 MeV each. I suppose it is difficult to see 
how these nucleons would not generate gamma or neutrons on interaction with 
other nuclei.
Also on the negative side: If a pion is implicated and needs to be generated 
within a nucleus, i suppose if this is not from a high energy collision it 
would need to be created in a nucleus with higher total binding energy than the 
pion mass energy. This would be a medium weight nucleus so if we assume 8 MeV 
binding energy per nucleon it would probably require a nucleus heavier than 
Oxygen at least. 

From: stephen_coo...@hotmail.com
To: vortex-l@eskimo.com
Subject: RE: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
Date: Tue, 20 Oct 2015 16:16:09 +0200




'The amount of energy needed to create a free pion is large; the rest mass for 
a pion is ~ 135 MeV'
Very true this is also true for the muon which has a rest mass for a pion is ~ 
106 MeV. 
I'm not sure if muons can be generated without pions? muon pair production 
would require even more energy. 
The energy is also quite high compared to the binding energy of light Nuclei.
If I am right I think the laser produces much less thermal energy too.
It is difficult to imagine how either muons (or pions) can form with out some 
kind of collective resonance effect or an additional high MeV energy source 
such as sufficient energy from a high energy fusion event or more even 
strangely a nucleon decay to mesons.
It will be interesting what Holmlids observations and explanations say, I'm 
quite curious as you say he could well have another explanation.

From: eric.wal...@gmail.com
Date: Tue, 20 Oct 2015 08:47:08 -0500
Subject: Re: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
To: vortex-l@eskimo.com

On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke <stephen_coo...@hotmail.com> 
wrote:
If any was produced we would need to balance this against those the energy 
required for pion production.
The amount of energy needed to create a free pion is large; the rest mass for a 
pion is ~ 135 MeV.  Consider that the largest amount of energy typically 
discussed in the context of cold fusion up to now has been ~ 24 MeV.  Holmlid's 
observations are likely to go back to something other than the generation of 
pions.
Eric

  

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Axil Axil]

The SPP is also a analog black hole. This quasiparticle accumulates energy
in huge amounts and may become  as massive as 1 million gigavolts. This
particle is also a tachyon that forms a condensate called a tachyon
condensate  according to theory produces particles such as mesons.

On Tue, Oct 20, 2015 at 6:31 AM, Stephen Cooke <stephen_coo...@hotmail.com>
wrote:

> You might be right Axil, These days I certainly tend to think there is
> some kind of collective disruptive or resonant behaviour that is exciting
> the nuclei or causing them to act this way. I acknowledge your good
> arguments and evidence for the formation of SPP's in these devices. It also
> seems some kind of trigger for collective behaviour is required.
>
> I think its interesting to look at the possibilities of low energy virtual
> resonant meson exchange rather than nucleon exchange. If it can occur.
>
> I'm certainly no expert on this but I suppose if the resonance increases
> slowly (but still fast in atomic scales) the first real *single*
> particles to be generated would be pions? unless particle pairs involving
> electrons, positrons, muons and associated neutrinos are be generated
> before. Is this correct thinking?
>
> pion0 has slightly less mass than pion+ or pion- and has much shorter half
> life so it is curious if we do not see gamma from pion0 decay. Could it be
> that the longer half life of virtual Pion+ and Pion- means in theory are
> more likely to tunnel? Or electrons, muons and pion0 are suppressed somehow
> so that pion+ and pion- are the first to be generated.
>
> I suppose generating a real meson would have higher energy consequences,
> but this could lead to the observed muons.
>
> I'm imagining if the energy is a "slowly" building resonant effect maybe
> as soon as a pion is manifested if it is a pion- perhaps its wave function
> occupies the S orbital to form Pionium until it interacts with the nucleus
> or decays via muon decay. If it is a pion+ perhaps it is ejected with
> sufficient energy to interact with another Deuterium nucleus to form a
> diproton that decays to 2 high energy protons, or it decays to a muon of
> characteristic energy that is later detected.
>
> I suppose a real nuclear physicist will correct me on a lot of my
> assumptions.
>
> --
> Date: Tue, 20 Oct 2015 04:16:37 -0400
> Subject: Re: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
> From: janap...@gmail.com
> To: vortex-l@eskimo.com
>
> I believe that what you imagine is what is happening in one of the many
> cases involving SPP extreme magnetic projections and entanglement,
>

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

'The amount of energy needed to create a free pion is large; the rest mass for 
a pion is ~ 135 MeV'
Very true this is also true for the muon which has a rest mass for a pion is ~ 
106 MeV. 
I'm not sure if muons can be generated without pions? muon pair production 
would require even more energy. 
The energy is also quite high compared to the binding energy of light Nuclei.
If I am right I think the laser produces much less thermal energy too.
It is difficult to imagine how either muons (or pions) can form with out some 
kind of collective resonance effect or an additional high MeV energy source 
such as sufficient energy from a high energy fusion event or more even 
strangely a nucleon decay to mesons.
It will be interesting what Holmlids observations and explanations say, I'm 
quite curious as you say he could well have another explanation.

From: eric.wal...@gmail.com
Date: Tue, 20 Oct 2015 08:47:08 -0500
Subject: Re: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
To: vortex-l@eskimo.com

On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke <stephen_coo...@hotmail.com> 
wrote:
If any was produced we would need to balance this against those the energy 
required for pion production.
The amount of energy needed to create a free pion is large; the rest mass for a 
pion is ~ 135 MeV.  Consider that the largest amount of energy typically 
discussed in the context of cold fusion up to now has been ~ 24 MeV.  Holmlid's 
observations are likely to go back to something other than the generation of 
pions.
Eric
  

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

This is definitely a crazy train of thought on my part but it's got me 
wondering: Bearing in mind Holmlids detection of muons and possibility they 
come from decay of positive or negative pions along with the fact that they are 
generated in ultra dense  deuterium. Is It possible that the nuclei are 
sufficiently close that those pions or virtual ones get generated in 
association with one nucleus and absorbed by another either directly or by 
tunneling at lower energy? If this occurs would this then change neutrons to 
protons and visa versa in the different nuclei? Could energy be exchanged 
between nuclei with transfer of neutral pions. I appreciate this would be very 
strange if this could occur. So I suppose the very short half life and extent 
of the pion wave function is not sufficient even in ultra dense material to 
allow this before the pion converts to a muon? 


> On 12 Oct 2015, at 17:25, Jones Beene  wrote:
> 
> MMPD  Muon Mediated Deuteron Disintegration
> 
> 
> The work of Leif Holmlid and others has opened up the possibility of 
> understanding what appears to be a new kind of nuclear reaction – a limited 
> type of chain reaction which is not fusion nor fission. The result of this 
> reaction is the complete disintegration of deuteron into quarks -- and then 
> into muons. The continuing reaction is propagated and catalyzed by muons 
> before they decay. Most of the net energy of the reaction is lost in the form 
> of neutrinos, but the fraction which is thermalized is still significant.
> 
> This nuclear reaction is dependent on the prior formation of a population of 
> “ultra-dense deuterium” which is an isomer of hydrogen which forms as a 2D 
> (two dimensional) layer on selected surfaces. The densification process has 
> been recognized for many years and rigorously described in the important 
> paper from 2009 of Nabil Lawandy entitled “Interactions of Charged Particles 
> on Surfaces.”
> 
> www.lenr-canr.org/acrobat/LawandyNMinteractio.pdf
> 
> 
> Individual deuterons are bosons which can occupy the same quantum state, so 
> long as their electrons are delocalized. This delocalization of electrons is 
> the key feature of ultra-dense deuterium, which becomes a condensate at 
> elevated temperature, thus allowing this novel reaction.
> 
> Upon application of a laser pulse which irradiates the condensate, a few 
> muons will be emitted which then proceed as a limited chain-reaction to 
> catalyze further reactions. In this reaction of relatively cold deuterons, 
> gamma emission cannot proceed, and fusion to deuterium is suppressed in favor 
> of complete disintegration of protons and neutrons into quarks.
> 
> The excess energy which would normally be expressed as very energetic gammas 
> is internalized to dislocate quarks from the bound state. Almost immediately, 
> quarks decay into muons – which have a greatly increased lifetime (but still 
> short) and muons are capable of catalyzing and  propagating the further 
> continuity of the reaction in a way reminiscent of nuclear fission of uranium 
> (in which neutrons are the mediator). Most of the net energy of this reaction 
> is lost to neutrino formation - but thermal gain is still possible.
> 
> More details to follow…
> 
> Jones

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Jones Beene]

The lifetime of pions is a factor of 100 times shorter than the muon, which is 
itself short. 

 

How pions can be detected at all, with so short a lifetime is a question worth 
asking. 

 

There is a good possibility that they are inferred – from finding muons, which 
can be detected. Hopefully, Ólafsson will address this issue of detection on 
Thursday.

 

 

From: Stephen Cooke 

 

… Is It possible that the nuclei are sufficiently close that those pions or 
virtual ones get generated in association with one nucleus and absorbed by 
another either directly or by tunneling at lower energy? 

 

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

I guess there is a chance that they saw X-ray spectra from Pionic Deuterium as 
well as inferring from muons of specific energy. It will be interesting to see 
what he says on Thursday, I hope someone can ask these kind of questions. 

Sent from my iPad

> On 19 okt. 2015, at 18:18, Jones Beene  wrote:
> 
> The lifetime of pions is a factor of 100 times shorter than the muon, which 
> is itself short.
>  
> How pions can be detected at all, with so short a lifetime is a question 
> worth asking.
>  
> There is a good possibility that they are inferred – from finding muons, 
> which can be detected. Hopefully, Ólafsson will address this issue of 
> detection on Thursday.
>  
>  
> From: Stephen Cooke
>  
> … Is It possible that the nuclei are sufficiently close that those pions or 
> virtual ones get generated in association with one nucleus and absorbed by 
> another either directly or by tunneling at lower energy?
> 
>  

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Stephen Cooke]

 environment such as a 
collection of nuclei in a Rydberg matter or UDD or an an atom with ionised 
inner orbitals such a transition of a pion and muon decay can be more likely. 
Could it be in Rydberg matter the nuclei are too closely packed for beta decay 
to occur due to the size of the electron wave function in the first electron 
orbital but pion-muon decay would still be possible? In normal matter with 
electrons in occupied inner orbitals could this prevent muon decay occurring 
and instead favour nucleon integrity from a conservation of energy point of 
view and beta decay? Could such a behaviour be evaluated and measured in terms 
of half lives and size of wave functions and quantum tunnelling effects? A 
crazy question... Could a bound nucleon such as a neutron theoretically decay 
into to or temporarily exist as 3 pions? EDIT: Interestingly 3 pions would have 
less than half the mass of a nucleon but I suppose other conservation rules 
would need to be respected, i'm not sure if this is possible. But if it was 
could this be an alternative source of energy?
I wonder if the lasers in Holmlids experiments are required to produce the 
rydberg matter, cause it to form UDD or initiate its "muon fusion" type 
behaviour?I remember a while back Axil explained to me about how Rydberg 
Hydrogen matter forms in 2d crystals and in fact they can stack into threads. I 
wonder if Deuterium is used if this the same as UDD? Could threads of Rydberg 
Matter like this resonate with particular frequencies and have a "thermal" 
phonon effect as has been discussed elsewhere? And would this have a 
characteristic frequency? Could the laser used by Holmlid excite this resonance 
at higher frequency compared to thermal resonance in this ultra dense material 
for example?On a sperate point would Muonic deuterium be special in some way? 
The orbital muon would spend relatively more time in the nucleus when it does 
would there be a net charge impact in the nucleus. Could this also disturb 
significantly the coulomb barrier, and perhaps even perturb the nucleus.I'm 
also speculating a lot as an amateur enthusiast… and probably sprouting rubbish 
in my enthusiasm. So I hope someone with more knowledge can clarify and knock 
some holes in what i just said.

Thanks
Stephen
From: jone...@pacbell.net
To: vortex-l@eskimo.com
Subject: RE: [Vo]:MMDD  Muon Mediated Deuteron Disintegration
Date: Mon, 12 Oct 2015 10:15:30 -0700






RE: [Vo]:MMDD  Muon Mediated Deuteron Disintegration




Correction

Ø   

Ø   In this reaction of relatively cold deuterons, gamma emission cannot 
proceed, and fusion to deuterium is suppressed in favor of complete 
disintegration of protons and neutrons into quarks. 

… should read: “gamma emission cannot proceed, and fusion of deuterium to 
helium is suppressed in favor of complete disintegration of protons and 
neutrons into quarks.”

BTW – there is some support for this view - showing up in a paper by Granados 
on the photodisintegration of deuterium the within the QCD hard rescattering 
model (HRM). According to the HRM, the process develops in three steps: a 
photon (in this case, the 24 MeV internalized photon) knocks a quark from the 
nucleon; the struck quark rescatters off a quark from another nucleon; then the 
energetic quarks recombine into two outgoing baryons which have large 
transverse momenta. 

This is a stretch… of course … if it were not, someone would certainly have 
suggested it before now.

Jones 

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Jones Beene]

From: Stephen Cooke 

*   Could there be characteristic photon emission from transitions in
muon shell levels similar to those from electrons and at what frequencies
these occur. Could these be observed experimentally? If characteristic
radiation can be seen from muon energy level transitions then it could be
interesting to see if radiation of these frequencies occur astronomically. 

A fraction of typical Cherenkov radiation in fission reactor spent fuel
pools comes from muons. Someone out there probably knows the exact
signature frequency of light originating from muons in that situation, both
the initiating frequency and the downshifted, but I do not. This signal is
detected astronomically as well. There are Cherenkov detectors made
specifically for atmospheric detection.

And yes - this signature could probably be used as further evidence of muons
- by replicators of Holmlid - most of whom do not have access to muon
detectors. Although observed to be blue, most Cherenkov radiation is
actually in the ultraviolet spectrum, downshifted to visible blue by the
water interaction.

It would be interesting to position a "glow tube" experiment in a water bath
and try to isolate the characteristic signal of muons using filters and
spectrometers, but the signal could be overwhelmed by the intensity of IR.

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Bob Higgins]

Jones,

While this is interesting speculation, I have come to assess probabilities
of of heretofore unknown reactions based inversely on binding energy.  If
we look at molecular binding energy, it is less than atomic binding.
Nuclear binding energy is greater than atomic binding.  Sub-nucleon binding
would have to be even higher energy than nuclear.  Between each of these,
there seems to be a factor of somewhere between 10^3 and 10^6 in binding
energy.  With nuclear binding in the MeV range, sub-nucleon binding would
be in the GeV-TeV range.  These binding characteristics are part of the
nature of the stable universe.

So, to me, the probability of LENR being related to shenanigans in
sub-nucleonic physics is something like 10^3-10^6 less likely than
something happening in nuclear physics.

With sub-nucleonic binding in the GeV-TeV range, how can something like a
laser with photons in the eV range have an effect?

On Mon, Oct 12, 2015 at 9:25 AM, Jones Beene  wrote:

> MMPD  Muon Mediated Deuteron Disintegration
>
> The work of Leif Holmlid and others has opened up the possibility of 
> understanding
> what appears to be a new kind of nuclear reaction – a limited type of chain
> reaction which is not fusion nor fission. The result of this reaction is
> the complete disintegration of deuteron into quarks -- and then into muons.
> The continuing reaction is propagated and catalyzed by muons before they
> decay. Most of the net energy of the reaction is lost in the form of
> neutrinos, but the fraction which is thermalized is still significant.
>
> This nuclear reaction is dependent on the prior formation of a population
> of “ultra-dense deuterium” which is an isomer of hydrogen which forms as
> a 2D (two dimensional) layer on selected surfaces. The densification
> process has been recognized for many years and rigorously described in
> the important paper from 2009 of Nabil Lawandy entitled “Interactions of C
> harged Particles on Surfaces.”
>
> *www.**lenr**-**canr**.org/acrobat/LawandyNMinteractio.pdf*
> 
>
> Individual deuterons are bosons which can occupy the same quantum state,
> so long as their electrons are delocalized. This delocalization of
> electrons is the key feature of ultra-dense deuterium, which becomes a
> condensate at elevated temperature, thus allowing this novel reaction.
>
> Upon application of a laser pulse which irradiates the condensate, a few
> muons will be emitted which then proceed as a limited chain-reaction to
> catalyze further reactions. In this reaction of relatively cold deuterons,
> gamma emission cannot proceed, and fusion to deuterium is suppressed in
> favor of complete disintegration of protons and neutrons into quarks.
>
> The excess energy which would normally be expressed as very energetic gamma
> s is internalized to dislocate quarks from the bound state. Almost
> immediately, quarks decay into muons – which have a greatly increased
> lifetime (but still short) and muons are capable of catalyzing and
> propagating the further continuity of the reaction in a way reminiscent
> of nuclear fission of uranium (in which neutrons are the mediator). Most
> of the net energy of this reaction is lost to neutrino formation - but thermal
> gain is still possible.
>
> More details to follow…
>
> Jones
>

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Jones Beene]

Correction
*   
*   In this reaction of relatively cold deuterons, gamma emission cannot
proceed, and fusion to deuterium is suppressed in favor of complete
disintegration of protons and neutrons into quarks. 
. should read: "gamma emission cannot proceed, and fusion of deuterium to
helium is suppressed in favor of complete disintegration of protons and
neutrons into quarks."
BTW - there is some support for this view - showing up in a paper by
Granados on the photodisintegration of deuterium the within the QCD hard
rescattering model (HRM). According to the HRM, the process develops in
three steps: a photon (in this case, the 24 MeV internalized photon) knocks
a quark from the nucleon; the struck quark rescatters off a quark from
another nucleon; then the energetic quarks recombine into two outgoing
baryons which have large transverse momenta. 
This is a stretch. of course . if it were not, someone would certainly have
suggested it before now.
Jones

RE: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Jones Beene]

Bob,

 

Good analysis. Subnuclear binding energy is significantly higher than nuclear 
binding energy, in general. As we know, it takes many terawatts to show 
evidence of the Higgs boson. After that, we must bootstrap power into energy to 
make this happen.

 

Fortunately – terawatt pulses are (or will be) available with moderately costly 
lasers. Here is a story on a 10 terawatt laser for the well-equipped garage lab 
…

 

http://www.slashgear.com/10-terawatt-laser-fits-on-a-desktop-04300268/

 

You need high power to start things. After a high power laser pulse starts a 
reaction, the energy in play for the next round (for MMDD to work in this 
circumstance) – is limited to 24 MeV per incidence. That level is significant, 
but it may not be enough - without some kind of follow-on process, like a 
limited chain reaction, to add continuity. 

 

I’m trying to find further support for this alternative, but basically – if one 
can show that a 24 MeV photon can dislodge a quark in a situation where the 
strong force can be harnessed to do the rest, then maybe the concept will go 
somewhere. 

 

Otherwise – how do we explain the muons seen by Holmlid? My fear is that they 
are measurement error, but until that is determined, this suggestion of muon 
chain reaction –MMDD - is a an alternative to consider.

 

From: Bob Higgins 

 

While this is interesting speculation, I have come to assess probabilities of 
of heretofore unknown reactions based inversely on binding energy.  If we look 
at molecular binding energy, it is less than atomic binding.  Nuclear binding 
energy is greater than atomic binding.  Sub-nucleon binding would have to be 
even higher energy than nuclear.  Between each of these, there seems to be a 
factor of somewhere between 10^3 and 10^6 in binding energy.  With nuclear 
binding in the MeV range, sub-nucleon binding would be in the GeV-TeV range.  
These binding characteristics are part of the nature of the stable universe.  

 

So, to me, the probability of LENR being related to shenanigans in 
sub-nucleonic physics is something like 10^3-10^6 less likely than something 
happening in nuclear physics.  

 

With sub-nucleonic binding in the GeV-TeV range, how can something like a laser 
with photons in the eV range have an effect?

 

On Mon, Oct 12, 2015 at 9:25 AM, Jones Beene  wrote:

MMPD  Muon Mediated Deuteron Disintegration

The work of Leif Holmlid and others has opened up the possibility of 
understanding what appears to be a new kind of nuclear reaction – a limited 
type of chain reaction which is not fusion nor fission. The result of this 
reaction is the complete disintegration of deuteron into quarks -- and then 
into muons. The continuing reaction is propagated and catalyzed by muons before 
they decay. Most of the net energy of the reaction is lost in the form of 
neutrinos, but the fraction which is thermalized is still significant.

This nuclear reaction is dependent on the prior formation of a population of 
“ultra-dense deuterium” which is an isomer of hydrogen which forms as a 2D (two 
dimensional) layer on selected surfaces. The densification process has been 
recognized for many years and rigorously described in the important paper from 
2009 of Nabil Lawandy entitled “Interactions of Charged Particles on Surfaces.”

  
www.lenr-canr.org/acrobat/LawandyNMinteractio.pdf

Individual deuterons are bosons which can occupy the same quantum state, so 
long as their electrons are delocalized. This delocalization of electrons is 
the key feature of ultra-dense deuterium, which becomes a condensate at 
elevated temperature, thus allowing this novel reaction. 

Upon application of a laser pulse which irradiates the condensate, a few muons 
will be emitted which then proceed as a limited chain-reaction to catalyze 
further reactions. In this reaction of relatively cold deuterons, gamma 
emission cannot proceed, and fusion to deuterium is suppressed in favor of 
complete disintegration of protons and neutrons into quarks. 

The excess energy which would normally be expressed as very energetic gammas is 
internalized to dislocate quarks from the bound state. Almost immediately, 
quarks decay into muons – which have a greatly increased lifetime (but still 
short) and muons are capable of catalyzing and  propagating the further 
continuity of the reaction in a way reminiscent of nuclear fission of uranium 
(in which neutrons are the mediator). Most of the net energy of this reaction 
is lost to neutrino formation - but thermal gain is still possible.

More details to follow…

Jones

 

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Eric Walker]

On Mon, Oct 12, 2015 at 12:15 PM, Jones Beene  wrote:

> According to the HRM, the process develops in
> three steps: a photon (in this case, the 24 MeV internalized photon) knocks
> a quark from the nucleon; the struck quark rescatters off a quark from
> another nucleon; then the energetic quarks recombine into two outgoing
> baryons which have large transverse momenta.

If we go with this explanation, I think we have to set aside color
confinement, unless there is no period of time during which the quark
is by itself.

Eric

Re: [Vo]:MMDD .... Muon Mediated Deuteron Disintegration

[Bob Higgins]

It takes many TeV to show the Higgs - not TeraWatts.  A 1.0 TeV particle
only has 1.6e-7 joules of kinetic energy, but that energy is in a single
particle and can create a TeV event in a collision.

A TW laser will not really help.  The individual laser photons are only eV
level, and a TW laser would simply apply a lot of photons in approx. the
same place at approx. the same time.  If the probability of a photon
interaction is X, where X is less than 1, then the probability of a
simultaneous interaction with 2 photons is on the order of X^2.  The
probability of simultaneous interaction with 10^9 photons is on the order
of X^10^9 - a really small number.  If  you had a condensate and
simultaneously absorbed something like 10^9 photons you may get somewhere,
but the problem is that before that happens you have probably already
absorbed 10^6 photons and have completely destroyed your condensate.
Actually the condensate may be completely destroyed by having absorbed a
much smaller number of photons.

I am more likely to explain muons seen by Holmlid as having been a mistaken
ID.  But I haven't concluded that yet - I am still digesting his papers.
It is a long chain of experimental evidence.

On Mon, Oct 12, 2015 at 11:50 AM, Jones Beene  wrote:

> Bob,
>
>
>
> Good analysis. Subnuclear binding energy is significantly higher than
> nuclear binding energy, in general. As we know, it takes many terawatts to
> show evidence of the Higgs boson. After that, we must bootstrap power into
> energy to make this happen.
>
>
>
> Fortunately – terawatt pulses are (or will be) available with moderately
> costly lasers. Here is a story on a 10 terawatt laser for the well-equipped
> garage lab …
>
>
>
> http://www.slashgear.com/10-terawatt-laser-fits-on-a-desktop-04300268/
>
>
>
> You need high power to start things. After a high power laser pulse starts
> a reaction, the energy in play for the next round (for MMDD to work in this
> circumstance) – is limited to 24 MeV per incidence. That level is
> significant, but it may not be enough - without some kind of follow-on
> process, like a limited chain reaction, to add continuity.
>
>
>
> I’m trying to find further support for this alternative, but basically –
> if one can show that a 24 MeV photon can dislodge a quark in a situation
> where the strong force can be harnessed to do the rest, then maybe the
> concept will go somewhere.
>
>
>
> Otherwise – how do we explain the muons seen by Holmlid? My fear is that
> they are measurement error, but until that is determined, this suggestion
> of muon chain reaction –MMDD - is a an alternative to consider.
>
>
>
> *From:* Bob Higgins
>
>
>
> While this is interesting speculation, I have come to assess probabilities
> of of heretofore unknown reactions based inversely on binding energy.  If
> we look at molecular binding energy, it is less than atomic binding.
> Nuclear binding energy is greater than atomic binding.  Sub-nucleon binding
> would have to be even higher energy than nuclear.  Between each of these,
> there seems to be a factor of somewhere between 10^3 and 10^6 in binding
> energy.  With nuclear binding in the MeV range, sub-nucleon binding would
> be in the GeV-TeV range.  These binding characteristics are part of the
> nature of the stable universe.
>
>
>
> So, to me, the probability of LENR being related to shenanigans in
> sub-nucleonic physics is something like 10^3-10^6 less likely than
> something happening in nuclear physics.
>
>
>
> With sub-nucleonic binding in the GeV-TeV range, how can something like a
> laser with photons in the eV range have an effect?
>
>
>
> On Mon, Oct 12, 2015 at 9:25 AM, Jones Beene  wrote:
>
> MMPD  Muon Mediated Deuteron Disintegration
>
> The work of Leif Holmlid and others has opened up the possibility of 
> understanding
> what appears to be a new kind of nuclear reaction – a limited type of chain
> reaction which is not fusion nor fission. The result of this reaction is
> the complete disintegration of deuteron into quarks -- and then into muons.
> The continuing reaction is propagated and catalyzed by muons before they
> decay. Most of the net energy of the reaction is lost in the form of
> neutrinos, but the fraction which is thermalized is still significant.
>
> This nuclear reaction is dependent on the prior formation of a population
> of “ultra-dense deuterium” which is an isomer of hydrogen which forms as
> a 2D (two dimensional) layer on selected surfaces. The densification
> process has been recognized for many years and rigorously described in the
> important paper from 2009 of Nabil Lawandy entitled “Interactions of
> Charged Particles on Surfaces.”
>
> www.*lenr*-*canr*.org/acrobat/LawandyNMinteractio.pdf
> 
>
> Individual deuterons are bosons which can occupy the same quantum state,
> so long as their electrons are delocalized. This delocalization of
> electrons