Re: [Vo]:understanding the relationship between internal conversion and the far field

[Eric Walker]

On Sun, Jul 6, 2014 at 3:41 PM,  wrote:

You can get an idea of this from the HUP where delta E x delta t >= h_bar/2.
>  If delta E is the energy of the reaction (about 4 MeV), then you get a
> time of
> at least 8E-23 sec.
> (I think this is the way it is normally calculated.)
> In any event it is obvious that a process that only takes order 1E-22
> seconds is
> far more likely to occur than one which takes 1E-9 seconds (I think this is
> actually more like 1E-17 BTW, but I'm not sure whether these times can be
> measured or this is just calculated.)
>

So that's how conjugate variables are used.  Note that the difference
between 1E-22 s and 1E-17 s is significant for what we're talking about.  A
photon can travel atomic distances in 1E-17 s, while it will not get very
far in 1E-22 s.  I think the shortest duration measured (according to
Wikipedia) is 12 attoseconds = 1.2E-17 s, so the 1E-22 must be a
calculation or a lower bound.

A photon is only virtual if the separation distance is less than the
> wavelength
> of the photon (actually I suspect that this should be wavelength/2*Pi).
>

According to this page [1], "in the case of photons, power and information
transfer by virtual particles is a relatively short-range phenomenon
(existing only within a few wavelengths of the field-disturbance, which
carries information or transferred power)".  Unless the statement is in
error, it seems, then, the virtual photon can exist within several
wavelengths rather than less than a wavelength.  Note that the wavelength
of a 24 MeV photon is 51 fermis.  Several wavelengths would be perhaps
100-200 fermis.  It seems that such a photon would not travel very far
before having to become a real photon.  If the 24 MeV could take the form
of thousands of virtual photons, by contrast, they would each have a much
smaller energy (24 MeV / 1000 = 24,000 eV, say), and a corresponding
wavelength of 51,666 fm.  Multiplying this wavelength by three to get
several times the wavelength, the virtual photons would extend out to
around 155 pm, which is on the order of the lattice spacing in a metal.

The reason I like virtual photons over real ones is that it's nicer to have
"messages being passed between interacting particles" (e.g., Feynman
diagrams) than to have to have "gammas being intercepted" (e.g.,
Widom-Larsen).

Eric


[1] http://en.wikipedia.org/wiki/Virtual_particle

Re: [Vo]:understanding the relationship between internal conversion and the far field

[mixent]

In reply to  Axil Axil's message of Sun, 6 Jul 2014 13:43:13 -0400:
Hi,
[snip]
>If the cause of LENR is the excitation of the vacuum through the injection of 
>very energetic EMF (magnetic), it might be possible that the energy intensive 
>magnetic fields supported by exothermic gainful nuclear reactions might be 
>injected into the vacuum enclosing the nucleus which might possibly supply the 
>energy needed to make endothermic nuclear reactions possible. 

You don't need to do this. Just fuse everything together first, then it fission
it to the detected end products. As long as the total mass out < total mass in,
the reaction as whole is exothermic.

Regards,

Robin van Spaandonk

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

Re: [Vo]:understanding the relationship between internal conversion and the far field

[mixent]

In reply to  Eric Walker's message of Sun, 6 Jul 2014 15:56:14 -0700:
Hi,
[snip]
>Are there any textbooks you can recommend that touch on some of these areas
>in some detail?
>
>Eric
Sorry, I don't know. Google is your friend (sometimes;)

Regards,

Robin van Spaandonk

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

Re: [Vo]:understanding the relationship between internal conversion and the far field

[Eric Walker]

Thank you, Bob.  If you have any textbooks you particularly like (e.g.,
touching on nuclear spin states and nuclear transitions), feel free to
recommend them.

For anyone who is interested, I have found the following helpful in getting
a broad overview:

   - Turner, "Atoms, Radiation, and Radiation Protection,"
   http://www.amazon.com/Atoms-Radiation-Protection-James-Turner/dp/3527406069
   - Magill and Galy, "Radioactivity, Radionuclides, Radiation,"
   
http://www.amazon.com/Radioactivity-Radionuclides-Radiation-Joseph-Magill/dp/product-description/3540211160

The book by Turner is fantastic and goes into a range of topics, including
a detailed discussion about how various detectors work.  I found the book
in several MIT course syllabi.  Turner was at Oak Ridge National
Laboratory, and the book appears to be an important one.  The book by
Magill and Galy is a nice refresher, but they do not go into much detail
and make some assumptions about the reader's knowledge of the field, and I
wouldn't have been able to follow the discussion without having first read
Turner's book.  Next up is a textbook by Kenneth Krane called "Introductory
Nuclear Physics," which has also shown up in several MIT course syllabi.

Eric



On Sun, Jul 6, 2014 at 5:39 PM, Bob Cook  wrote:

>  Eric--
>
> I have several items that may pertain to your effort to understand
> internal conversion of nuclei.
>
>
>1. Various isotopes have dipole and quadrapole moments existing in the
>stable nucleus as well as the excited nucleus.   Isomers and radioactive
>nuclei may have these moments.  The moments may be electric dipole or
>quadrapole as well as magnetic dipole and quadrapole moments.
>
>
>  2. Modern electronics can produce both magnetic and electric
> dipole and quadrapole input oscillating   fields that can be in resonance
> with the respective resonance of any given isotope whether it is in its
> ground state, an isomeric condition or a radioactive state with a long half
> life.
>
> 3.  Spin states of the various nuclei and their excited states control the
> allowed transitions of the  nucleus from state to state.
>
> 4. Spin with orbital electrons in a lattice  and the electrons’ intrinsic
> spin as well as other particles within the lattice may also be involved in
> the reaction and provide the conditions that make the transitions in energy
> possible.   This idea is consistent with a paper recently published  in the
>   “Proceedings of the 14th Meeting of Japan CF Research Society, JCF14
> December 7 - 8, 2013 Tokyo Institute of Technology, Japan”( ISSN
> 2187-2260)
>
> 5.  The last paper of these proceedings   describes a theory that entails
> the extended quantum mechanical system connecting lattice nuclei as well as
> D or H within the lattice of transition metals.   Other papers in the
> proceedings discuss experimental data, mostly accomplished by various
> researchers presenting at the JCF14  meeting.  The idea involves bands of
> neutrons that extend between the metal nuclei of the lattice.  It suggests
> that the reaction is mediated by the H or D in the lattice.
>
> 6.  Back on June 22 Axil Axil wrote, “ It would be great if we could find
> one mechanism that provides the total set of transmutation produces seen in
> LENR. ... The Philips reaction is not a "one reaction fit all" solution.”
>  The paper cited above fits Axil’s wish.
>
> Jed identified these JCF proceedings as available at:
>
> *http://www.jcfrs.org/file/jcf14-proceedings.pdf* 
> 
>
> Bob
>
>
>
>
>
>
>
>
> Sent from Windows Mail
>
> *From:* Eric Walker 
> *Sent:* ‎Sunday‎, ‎July‎ ‎6‎, ‎2014 ‎2‎:‎56‎ ‎PM
> *To:* vortex-l@eskimo.com
>
> Are there any textbooks you can recommend that touch on some of these
> areas in some detail?
>
> Eric
>
>
> On Sun, Jul 6, 2014 at 3:41 PM,  wrote:
>
> In reply to  Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700:
>> Hi Eric,
>> [snip]
>> >Here's where my understanding starts to get fuzzy.  The above description
>> >talked about isomeric transitions, which involve the decay of a
>> metastable
>> >isomer to the ground state of the isotope.  Metastable isomers are
>> >long-lived excited states of nuclei, ones that have significant
>> half-lives.
>> > Similar, shorter-lived nuclei are not considered metastable and are
>> >instead referred to as compound nuclei.  For example, in dd fusion, the
>> >short-lived compound nucleus [dd]* is not an isomer of 4He because it
>> >decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?.
>>  (The
>> >4He+? branch is orders of magnitude less likely than the p+t and n+3He
>> >branches, whose likelihoods are roughly split 50-50.)  I understand from
>> >reading around that the emission of a gamma photon during the
>> deexcitation
>> >of a metastable isomer can be on the order of 10E-9 seconds, and that the
>> >time required for the emission is something that depends upon the spin of
>> >the ex

Re: [Vo]:understanding the relationship between internal conversion and the far field

[Bob Cook]

Eric--




I have several items that may pertain to your effort to understand internal 
conversion of nuclei.


Various isotopes have dipole and quadrapole moments existing in the stable 
nucleus as well as the excited nucleus.   Isomers and radioactive nuclei may 
have these moments.  The moments may be electric dipole or quadrapole as well 
as magnetic dipole and quadrapole moments.






2. Modern electronics can produce both magnetic and electric dipole 
and quadrapole input oscillating   fields that can be in resonance with the 
respective resonance of any given isotope whether it is in its ground state, an 
isomeric condition or a radioactive state with a long half life.  




3.  Spin states of the various nuclei and their excited states control the 
allowed transitions of the  nucleus from state to state.  




4. Spin with orbital electrons in a lattice  and the electrons’ intrinsic spin 
as well as other particles within the lattice may also be involved in the 
reaction and provide the conditions that make the transitions in energy 
possible.   This idea is consistent with a paper recently published  in the   
“Proceedings of the 14th Meeting of Japan CF Research Society, JCF14   
December 7 - 8, 2013 Tokyo Institute of Technology, Japan”( ISSN 2187-2260) 
  




5.  The last paper of these proceedings   describes a theory that entails the 
extended quantum mechanical system connecting lattice nuclei as well as D or H 
within the lattice of transition metals.   Other papers in the  proceedings 
discuss experimental data, mostly accomplished by various researchers 
presenting at the JCF14  meeting.  The idea involves bands of neutrons that 
extend between the metal nuclei of the lattice.  It suggests that the reaction 
is mediated by the H or D in the lattice.  


6.  Back on June 22 Axil Axil wrote, “ It would be great if we could find one 
mechanism that provides the total set of transmutation produces seen in LENR. 
... The Philips reaction is not a "one reaction fit all" solution.”

The paper cited above fits Axil’s wish.


Jed identified these JCF proceedings as available 
at:http://www.jcfrs.org/file/jcf14-proceedings.pdf
Bob


















Sent from Windows Mail





From: Eric Walker
Sent: ‎Sunday‎, ‎July‎ ‎6‎, ‎2014 ‎2‎:‎56‎ ‎PM
To: vortex-l@eskimo.com







Are there any textbooks you can recommend that touch on some of these areas in 
some detail?




Eric







On Sun, Jul 6, 2014 at 3:41 PM,  wrote:



In reply to  Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700:
Hi Eric,
[snip]

>Here's where my understanding starts to get fuzzy.  The above description
>talked about isomeric transitions, which involve the decay of a metastable
>isomer to the ground state of the isotope.  Metastable isomers are
>long-lived excited states of nuclei, ones that have significant half-lives.
> Similar, shorter-lived nuclei are not considered metastable and are
>instead referred to as compound nuclei.  For example, in dd fusion, the
>short-lived compound nucleus [dd]* is not an isomer of 4He because it

>decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?.  (The
>4He+? branch is orders of magnitude less likely than the p+t and n+3He

>branches, whose likelihoods are roughly split 50-50.)  I understand from
>reading around that the emission of a gamma photon during the deexcitation
>of a metastable isomer can be on the order of 10E-9 seconds, and that the
>time required for the emission is something that depends upon the spin of
>the excited nucleus.  Excited nuclei with certain spins will take
>significantly longer to emit a gamma photon than nuclei with other spins.


By first order approximation, metastable nuclei can't decay. In order to do so,
I think they need some form of external interaction. That's why they have
comparatively long lifetimes.


> Am I correct in thinking that the same principles apply to a compound
>nucleus such as [dd]*?  I.e., in the gamma photon branch, the [dd]*
>de-excitation is on the order of 10E-9 seconds, or perhaps longer?  Also, I
>haven't found a reference that gives the approximate times needed for the
>other branches (p+t and n+3He), in which the compound nucleus splits up
>into fragments.


You can get an idea of this from the HUP where delta E x delta t >= h_bar/2.
If delta E is the energy of the reaction (about 4 MeV), then you get a time of
at least 8E-23 sec.
(I think this is the way it is normally calculated.)
In any event it is obvious that a process that only takes order 1E-22 seconds is
far more likely to occur than one which takes 1E-9 seconds (I think this is
actually more like 1E-17 BTW, but I'm not sure whether these times can be
measured or this is just calculated.)


>
>Returning to internal conversion, one explanation for it focuses on the
>fact that inner shell electrons have a high probability of passing through
>the nucleus.


Yes but they are only present for a very short time.


>The idea is that dur

Re: [Vo]:understanding the relationship between internal conversion and the far field

[Eric Walker]

Are there any textbooks you can recommend that touch on some of these areas
in some detail?

Eric


On Sun, Jul 6, 2014 at 3:41 PM,  wrote:

In reply to  Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700:
> Hi Eric,
> [snip]
> >Here's where my understanding starts to get fuzzy.  The above description
> >talked about isomeric transitions, which involve the decay of a metastable
> >isomer to the ground state of the isotope.  Metastable isomers are
> >long-lived excited states of nuclei, ones that have significant
> half-lives.
> > Similar, shorter-lived nuclei are not considered metastable and are
> >instead referred to as compound nuclei.  For example, in dd fusion, the
> >short-lived compound nucleus [dd]* is not an isomer of 4He because it
> >decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?.
>  (The
> >4He+? branch is orders of magnitude less likely than the p+t and n+3He
> >branches, whose likelihoods are roughly split 50-50.)  I understand from
> >reading around that the emission of a gamma photon during the deexcitation
> >of a metastable isomer can be on the order of 10E-9 seconds, and that the
> >time required for the emission is something that depends upon the spin of
> >the excited nucleus.  Excited nuclei with certain spins will take
> >significantly longer to emit a gamma photon than nuclei with other spins.
>
> By first order approximation, metastable nuclei can't decay. In order to
> do so,
> I think they need some form of external interaction. That's why they have
> comparatively long lifetimes.
>
> > Am I correct in thinking that the same principles apply to a compound
> >nucleus such as [dd]*?  I.e., in the gamma photon branch, the [dd]*
> >de-excitation is on the order of 10E-9 seconds, or perhaps longer?  Also,
> I
> >haven't found a reference that gives the approximate times needed for the
> >other branches (p+t and n+3He), in which the compound nucleus splits up
> >into fragments.
>
> You can get an idea of this from the HUP where delta E x delta t >=
> h_bar/2.
> If delta E is the energy of the reaction (about 4 MeV), then you get a
> time of
> at least 8E-23 sec.
> (I think this is the way it is normally calculated.)
> In any event it is obvious that a process that only takes order 1E-22
> seconds is
> far more likely to occur than one which takes 1E-9 seconds (I think this is
> actually more like 1E-17 BTW, but I'm not sure whether these times can be
> measured or this is just calculated.)
>
> >
> >Returning to internal conversion, one explanation for it focuses on the
> >fact that inner shell electrons have a high probability of passing through
> >the nucleus.
>
> Yes but they are only present for a very short time.
>
> >The idea is that during the time that the electron is within
> >the nucleus there is a nontrivial probability that it will interact with
> >the excited state, which, if this happens, will result in the energy of
> the
> >excited state being passed on to the electron.  The implication is that
> the
> >less likely an electron is to be found within the nucleus, the less likely
> >that the electron will be ejected as a result of internal conversion.  So
> >the probability of IC is highest with K-shell electrons and decreases the
> >further you go out.  One question I have about this explanation is that IC
> >is mediated via the electromagnetic interaction; my understanding of this
> >is that there is a virtual photon that passes from the nucleus to the
> >electron.  I do not see why the electron would particularly need to be
> >passing through the nucleus for such a virtual photon to reach it, for the
> >electromagnetic interaction is long-range.  Anywhere a virtual photon can
> >reach, it seems, there would be a nontrivial probability for an internal
> >conversion decay to occur.
>
> A photon is only virtual if the separation distance is less than the
> wavelength
> of the photon (actually I suspect that this should be wavelength/2*Pi).
>
>
> >Another challenge I have with this explanation
> >is that I think the de Broglie wavelength of an orbital electron is going
> >to be far larger than the nucleus or any than particular nucleon; my
> >understanding is that this is a problem because the de Broglie wavelengths
> >have to be roughly comparable for an interaction of some kind to be
> >probable.
>
> ...but it isn't very probable, that's why it can barely compete with "slow"
> gamma emission. However I don't know what part of the probability is due to
> duration of the interaction, and what part due to Be Broglie wavelength
> mismatch. Perhaps you also need to take into consideration the be Broglie
> wavelength of the nucleons. Also consider that by the time an orbital
> electron
> passes through the nucleus it has gained considerable kinetic energy from
> the
> electric field, so it's De Broglie wavelength is much shorter. (Still too
> long,
> but not by many orders of magnitude. :)
>
> >
> >One question I have has to do with the energy of the v

Re: [Vo]:understanding the relationship between internal conversion and the far field

[mixent]

In reply to  Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700:
Hi Eric,
[snip]
>Here's where my understanding starts to get fuzzy.  The above description
>talked about isomeric transitions, which involve the decay of a metastable
>isomer to the ground state of the isotope.  Metastable isomers are
>long-lived excited states of nuclei, ones that have significant half-lives.
> Similar, shorter-lived nuclei are not considered metastable and are
>instead referred to as compound nuclei.  For example, in dd fusion, the
>short-lived compound nucleus [dd]* is not an isomer of 4He because it
>decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+?.  (The
>4He+? branch is orders of magnitude less likely than the p+t and n+3He
>branches, whose likelihoods are roughly split 50-50.)  I understand from
>reading around that the emission of a gamma photon during the deexcitation
>of a metastable isomer can be on the order of 10E-9 seconds, and that the
>time required for the emission is something that depends upon the spin of
>the excited nucleus.  Excited nuclei with certain spins will take
>significantly longer to emit a gamma photon than nuclei with other spins.

By first order approximation, metastable nuclei can't decay. In order to do so,
I think they need some form of external interaction. That's why they have
comparatively long lifetimes.

> Am I correct in thinking that the same principles apply to a compound
>nucleus such as [dd]*?  I.e., in the gamma photon branch, the [dd]*
>de-excitation is on the order of 10E-9 seconds, or perhaps longer?  Also, I
>haven't found a reference that gives the approximate times needed for the
>other branches (p+t and n+3He), in which the compound nucleus splits up
>into fragments.

You can get an idea of this from the HUP where delta E x delta t >= h_bar/2.
If delta E is the energy of the reaction (about 4 MeV), then you get a time of
at least 8E-23 sec.
(I think this is the way it is normally calculated.)
In any event it is obvious that a process that only takes order 1E-22 seconds is
far more likely to occur than one which takes 1E-9 seconds (I think this is
actually more like 1E-17 BTW, but I'm not sure whether these times can be
measured or this is just calculated.)

>
>Returning to internal conversion, one explanation for it focuses on the
>fact that inner shell electrons have a high probability of passing through
>the nucleus.  

Yes but they are only present for a very short time.

>The idea is that during the time that the electron is within
>the nucleus there is a nontrivial probability that it will interact with
>the excited state, which, if this happens, will result in the energy of the
>excited state being passed on to the electron.  The implication is that the
>less likely an electron is to be found within the nucleus, the less likely
>that the electron will be ejected as a result of internal conversion.  So
>the probability of IC is highest with K-shell electrons and decreases the
>further you go out.  One question I have about this explanation is that IC
>is mediated via the electromagnetic interaction; my understanding of this
>is that there is a virtual photon that passes from the nucleus to the
>electron.  I do not see why the electron would particularly need to be
>passing through the nucleus for such a virtual photon to reach it, for the
>electromagnetic interaction is long-range.  Anywhere a virtual photon can
>reach, it seems, there would be a nontrivial probability for an internal
>conversion decay to occur.  

A photon is only virtual if the separation distance is less than the wavelength
of the photon (actually I suspect that this should be wavelength/2*Pi).


>Another challenge I have with this explanation
>is that I think the de Broglie wavelength of an orbital electron is going
>to be far larger than the nucleus or any than particular nucleon; my
>understanding is that this is a problem because the de Broglie wavelengths
>have to be roughly comparable for an interaction of some kind to be
>probable.

...but it isn't very probable, that's why it can barely compete with "slow"
gamma emission. However I don't know what part of the probability is due to
duration of the interaction, and what part due to Be Broglie wavelength
mismatch. Perhaps you also need to take into consideration the be Broglie
wavelength of the nucleons. Also consider that by the time an orbital electron
passes through the nucleus it has gained considerable kinetic energy from the
electric field, so it's De Broglie wavelength is much shorter. (Still too long,
but not by many orders of magnitude. :)

>
>One question I have has to do with the energy of the virtual photon.
> Internal conversion is less likely, other things being equal, if the
>energy of the transition is large (e.g., on the order of MeVs).  

Perhaps because high energy reactions often decay via other faster paths?
The HUP logic applied here above would indicate that gamma decay times decrease
with incre

[Vo]:Gluons Chip in for Proton Spin

[Axil Axil]

http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.113.012001

Gluons Chip in for Proton Spin

It looks like polarized gluons produce most of the spin of the proton. That
means that the gluons are magnetic entities.

A magnetic field applied to the proton could disrupt the polarization of
the gluons and therefore the strong force that keeps protons and neutrons
together in the nucleus.

There is an intimate relationship between the strong force, magnetic force,
and the gluon that might underpin LENR reactions at the most basic level.

.

Re: [Vo]:understanding the relationship between internal conversion and the far field

[Axil Axil]

One of the insights that we might possibly draw from the Mizuno experiment
is that there might be a mix of exothermic and endothermic nuclear
reactions going on simultaneously in LERN transmutation reactions.



If the cause of LENR is the excitation of the vacuum through the injection
of very energetic EMF (magnetic), it might be possible that the energy
intensive magnetic fields supported by exothermic gainful nuclear reactions
might be injected into the vacuum enclosing the nucleus which might
possibly supply the energy needed to make endothermic nuclear reactions
possible.



Such a situation would make analyzing how energy is produced in a LENR
reaction very complicated and confusing.



For example, the iron and helium seen in some LENR experiments could be
derived from an endothermic alpha decay of nickel. In a similar way,
chromium and titanium could be produced through the same endothermic alpha
decay mechanism from higher Z elements in a series of alpha emission based
reactions to lower Z elements. This alpha decay reaction could be the
source of helium seen in many LENR experiments.



The energy needed to cause such alpha decays could come from the exothermic
fusion of hydrogen and or helium into boron, lithium and beryllium.



As long as the LENR reaction is exothermic on the average, any sort of
nuclear reaction can occur because the energy storehouse of the highly
energetic magnetic solitons which catalyze transmutation could
supply stored magnetic energy that could be shared later in a subsequent
endothermic reaction.






On Sun, Jul 6, 2014 at 12:33 PM, Eric Walker  wrote:

> Interesting proposal, Jones.  I'll have to think about it.  Thanks for the
> pointer to the paper.
>
> On Sun, Jul 6, 2014 at 6:34 AM, Jones Beene  wrote:
>
> If Mizuno is correct we could add a fourth pathway for the dd reaction in
>> LENR. That would be p+p+p+p. It is not clear if this reaction should be
>> called fusion or fission or a new form of IC :-) but as of July 2014, it
>> looms as the most important unexplained experiment on the horizon for
>> deuterium-based LENR.
>>
>
> For this one, my working hypothesis is that Mizuno is seeing these two
> reactions in tandem:
>
> (X)Ni + d → (X+1)Ni + p (Oppenheimer-Phillips process)
> p + d → p + p + n → p + p + p (about 10 min. later)
>
> This tack obviously has several objections to worry about, such as the
> lack of detected neutrons and the lack of activation.  My initial strategy
> for dealing with these objections is to wonder whether the channel is
> actually pretty small, in comparison to any energy that is developed.
>
> Eric
>
>

Re: [Vo]:understanding the relationship between internal conversion and the far field

[Eric Walker]

Interesting proposal, Jones.  I'll have to think about it.  Thanks for the
pointer to the paper.

On Sun, Jul 6, 2014 at 6:34 AM, Jones Beene  wrote:

If Mizuno is correct we could add a fourth pathway for the dd reaction in
> LENR. That would be p+p+p+p. It is not clear if this reaction should be
> called fusion or fission or a new form of IC :-) but as of July 2014, it
> looms as the most important unexplained experiment on the horizon for
> deuterium-based LENR.
>

For this one, my working hypothesis is that Mizuno is seeing these two
reactions in tandem:

(X)Ni + d → (X+1)Ni + p (Oppenheimer-Phillips process)
p + d → p + p + n → p + p + p (about 10 min. later)

This tack obviously has several objections to worry about, such as the lack
of detected neutrons and the lack of activation.  My initial strategy for
dealing with these objections is to wonder whether the channel is actually
pretty small, in comparison to any energy that is developed.

Eric

[Vo]:Clockwise, counterclockwise or translation?

[David Jonsson]

Look at this very special kind of motion
https://www.shadertoy.com/view/lds3Wl

First I thought the clover turned clockwise, then by looking at the dots I
thought it turned counterclockwise. Then I realized all dots are moving on
straight lines.

If the dots were electrons I wonder what the magnetic field could be? In
the direction through the screen the B-field could be positive, negative or
zero. I suppose it is zero since there is no net rotation of charges?

David

RE: [Vo]:understanding the relationship between internal conversion and the far field

[Jones Beene]

Species which look like helium on a mass spec - especially when subject to
variable mass interactions immediately prior to measurement. 
Helium - 4.002602 amu
D2 -  4.028203 amu
4 protons as Rydberg matter – 4.03176 amu

Next we must consider variable mass. To understand variable mass, this
article may not help most of us, but it is all there is for now – except for
Noether’s theorem and other imponderables like Fock-Stueckelberg.
http://en.wikipedia.org/wiki/On_shell_and_off_shell

What does it take to get a Cooper pair of D to look like 4He on a mass-spec
and yet decay energetically into protons and electrons, which should be
endothermic ?

http://www.scribd.com/doc/139182265/Theories-of-variable-mass-particles-and-
low-energy-nuclear-phenomena

"Theories of variable mass particles and low energy nuclear phenomena" by
Mark Davidson

Conclusion of Paper:  "We want to emphasize that there is no direct
experimental evidence yet that masses of electrons, nucleons, or nuclei can
change significantly in a condensed matter setting Nevertheless, it is
this author's opinion that Fock-Stueckelberg or other type of are a possible
explanation for such variations and that all of the
experiments in LENR can potentially be explained if they are occurring."

Another tid-bit: Scattering experiments indicate that the 4He nuclear
structure is C3v, which means that it has a tetrahedral arrangement of
neutrons and protons. An alpha particle can be thought of as a pair of
deuterons. However, the neutrons of the alpha particle form a pair and so do
the protons. The neutron-neutron pair involves a spin pairing and a strong
force bonding. The strong force between two neutrons is a repulsion which
pushes the separation distance between the centers of the neutrons to the
limit of spin pairing. The same thing occurs for the protons except there is
an electrostatic repulsion in addition to the strong force repulsion. It is
amazing that the alpha is so stable.

The strong force between a neutron and a proton is an attraction and there
has to be a rotation to maintain a separation between a neutron and a proton
in a neutron-proton pair. This can be a complex mix of nuclear dynamics in
condensed matter which is far different than in a plasma – thus leading to
the suspicion that variable mass plays a role, and that real fusion to
helium cannot happen except in a plasma.
_

The basic concept is that a paired species of deuterons can
form and linger in condensed matter for weeks, but actual decay can occur
after a delay. When decay occurs, only protons remain plus excess energy
which is below gamma intensity but much more than chemical. Thus we explain
all the main details of “helium LENR” but without helium.
1)  Excess heat
2)  Lack of gamma radiation
3)  What “appears to be” 4He on a mass-spec, but is not
4)  The so-called “heat after death” phenomenon… 

Of course, this explanation raises as many
questions as it answers, and leaves open the door that a small percentage of
reactions still must proceed to tritium, since tritium is documented; but
anything that removes the 24MeV millstone from the shoulders of LENR should
be given full consideration.


<>

RE: [Vo]:understanding the relationship between internal conversion and the far field

[Jones Beene]

Eric - By now you fully appreciate the impossibility of reconciling
mainstream fusion details with LENR. There is little way to rationalize all
of the contradictions, based on the data now available… but that does not
keep us from trying.

Let me add this, which would clear up some of the contradiction, if it were
true. You state: 

For example, in dd fusion, the short-lived compound nucleus [dd]* is not an
isomer of 4He because it decays quite rapidly in one of three ways -- to
p+t, n+3He and 4He+ɣ.  

If Mizuno is correct we could add a fourth pathway for the dd reaction in
LENR. That would be p+p+p+p. It is not clear if this reaction should be
called fusion or fission or a new form of IC :-) but as of July 2014, it
looms as the most important unexplained experiment on the horizon for
deuterium-based LENR. 

But then again we must ask, can these results be applied to retrospectively
explain experiments going back 25 years? …those who have followed the field
for so long would complain: “what about all the reports of 4He correlating
to excess heat ?”

The answer to that would be something like this. Deuterium, in condensed
matter will often form into a stable Cooper pair of deuterons (or dense
deuterium), which is stable enough to pass through a spectrometer and fool
the experimenter into the belief that the species is 4He. Either way, some
form of dense paired deuterium happens often enough to explain the tiny
amount of 4He which seems to have been documented in the past. 

>From there on, we must explain why this so-called “4He” or mass-4 signal on
a mass spectrometer is some other species instead of helium. That is far
easier than to rationalize the lack of 24 MeV gammas.

Here is a paper of interest from Winterberg, but it does not go far enough:
http://arxiv.org/ftp/arxiv/papers/0912/0912.5414.pdf

The basic concept is that a paired species of deuterons can form and linger
in condensed matter for weeks, but actual decay can occur after a delay.
When decay occurs, only protons remain plus excess energy which is below
gamma intensity but much more than chemical. Thus we explain all the main
details of “helium LENR” but without helium.
1)  Excess heat
2)  Lack of gamma radiation
3)  What “appears to be” 4He on a mass-spec, but is not
4)  The so-called “heat after death” phenomenon… 

Of course, this explanation raises as many questions as it
answers, and leaves open the door that a small percentage of reactions still
must proceed to tritium, since tritium is documented; but anything that
removes the 24MeV millstone from the shoulders of LENR should be given full
consideration.

From: Eric Walker 

I'm in the process of trying to better understand internal
conversion and it's cross section vis-a-vis inner shell electrons and
sources of charge in the far field.  I'm hoping someone (Robin?) can help me
to get the terminology right and point me to further reading.

Here is my understanding so far.  Internal conversion is a
process in which an inner shell electron is expelled from an atom as the
result of a nuclear transition.  It is mediated by the electromagnetic force
(in contrast to the weak or strong interactions) and results from
electromagnetic coupling between the electron and an excited nucleus.  The
kind of nuclear transition that leads to internal conversion is generally an
isomeric transition; i.e., the deexcitation of a metastable isomer to a
lower energy level.  Internal conversion competes with gamma emission, and
there is an internal conversion coefficient, which for a competing pair of
branches, one IC and the other gamma photon emitting, gives you the ratio of
internal conversion electrons to photons.  In some cases the internal
conversion coefficient can be quite high, meaning that IC is greatly favored
over gamma photon emission.  There are a number of factors that are thought
to go into a high IC coefficient -- when the energy of the transition is
small, when the nucleus is large, and when the daughter nucleus has zero
spin, for example.  Unlike in the case of beta emission, the energies for
internal conversion electrons are not broadband and show up in line spectra
as sharp peaks.  This is because unlike in the case of beta decay there is
no neutrino to take part of the energy of the decay away from the emitted
electron.

Here's where my understanding starts to get fuzzy.  The
above description talked about isomeric transitions, which involve the decay
of a metastable isomer to the ground state of the isotope.  Metastable
isomers are long-lived excited states of nuclei, ones that have significant
half-lives.  Similar, shorter-lived nuclei are not considered metastable and
are instead referred to as compound nuclei.  For example, in dd fusion, the
short-lived compound nucleus [dd]* is not an isomer of 4He because it decays
quite rapidly in one of