Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

The latest analog black hole experiment done in  Israel  has extracted
phonic energy as Hawking radiation from the vacuum.



https://www.sciencealert.com/scientists-have-stimulated-hawking-radiation-in-a-lab-analogue-of-a-black-hole



We Just Got Lab-Made Evidence of Stephen Hawking's Greatest Prediction
About Black Holes

So, cue trying to recreate it in a lab using black hole analogues. These
can be built from things that produce waves, such as fluid and sound waves
in a special tank , from
Bose-Einstein
condensates
, or
from light contained in optical fibre.

Does that help?

On Fri, Jul 26, 2019 at 6:06 PM  wrote:

> In reply to  bobcook39...@hotmail.com's message of Fri, 26 Jul 2019
> 02:28:29
> +:
> Hi Bob,
> [snip]
> >Robin-
> >
> >
> >
> >During NMR isomeric transitions, nuclear species are stimulated with a
> radio frequency EM field to gain kinetic spin energy in the form of
> increased angular momentum in small quanta of angular momentum—each quantum
> being equal to h/2pie.  An ambient magnetic field would change the allowed
> states for such nuclear spin energy states.
> >
> >Thus, change in a coherent system’s angular momentum occur in 0 or more
> quanta of angular momentum .   However the total angular momentum must be
> conserved just as energy is conserved in a coherent system phase change.
> (The reaction does not involve release of any particles with kinetic
> energy, including no photons or neutrinos)  Only an increase of phonic
> lattice energy and a decrease of nuclear orbital angular momentum happens
> associated with a different meta stable or stable nuclear
> configuration—even ones with a transmuted configuration, but withunchangedt
> sums of protons and neutrons.
>
> 1) What is this "coherent system", and specifically, in what respect is it
> coherent, i.e. which property of the system?
> 2) How do you propose that the nuclear energy is actually coupled to the
> phonic
> energy?
> 3) Changes in angular momentum of nuclei are usually paired with emission
> of a
> gamma ray or particle to conserve angular momentum. If you want to avoid
> this,
> then you need to provide an actual physical mechanism by which the angular
> momentum is transferred to the lattice, and specifically what it is in the
> lattice that it couples to. Furthermore, what is it that makes this method
> preferable above the usual methods (e.g. gamma emission)?
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  bobcook39...@hotmail.com's message of Fri, 26 Jul 2019 02:28:29
+:
Hi Bob,
[snip]
>Robin-
>
>
>
>During NMR isomeric transitions, nuclear species are stimulated with a radio 
>frequency EM field to gain kinetic spin energy in the form of increased 
>angular momentum in small quanta of angular momentum—each quantum being equal 
>to h/2pie.  An ambient magnetic field would change the allowed states for such 
>nuclear spin energy states.
>
>Thus, change in a coherent system’s angular momentum occur in 0 or more quanta 
>of angular momentum .   However the total angular momentum must be conserved 
>just as energy is conserved in a coherent system phase change.  (The reaction 
>does not involve release of any particles with kinetic energy, including no 
>photons or neutrinos)  Only an increase of phonic lattice energy and a 
>decrease of nuclear orbital angular momentum happens associated with a 
>different meta stable or stable nuclear configuration—even ones with a 
>transmuted configuration, but withunchangedt sums of protons and neutrons.

1) What is this "coherent system", and specifically, in what respect is it
coherent, i.e. which property of the system?
2) How do you propose that the nuclear energy is actually coupled to the phonic
energy?
3) Changes in angular momentum of nuclei are usually paired with emission of a
gamma ray or particle to conserve angular momentum. If you want to avoid this,
then you need to provide an actual physical mechanism by which the angular
momentum is transferred to the lattice, and specifically what it is in the
lattice that it couples to. Furthermore, what is it that makes this method
preferable above the usual methods (e.g. gamma emission)?
Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Robin-



During NMR isomeric transitions, nuclear species are stimulated with a radio 
frequency EM field to gain kinetic spin energy in the form of increased angular 
momentum in small quanta of angular momentum—each quantum being equal to 
h/2pie.  An ambient magnetic field would change the allowed states for such 
nuclear spin energy states.

Thus, change in a coherent system’s angular momentum occur in 0 or more quanta 
of angular momentum .   However the total angular momentum must be conserved 
just as energy is conserved in a coherent system phase change.  (The reaction 
does not involve release of any particles with kinetic energy, including no 
photons or neutrinos)  Only an increase of phonic lattice energy and a decrease 
of nuclear orbital angular momentum happens associated with a different meta 
stable or stable nuclear configuration—even ones with a transmuted 
configuration, but withunchangedt sums of protons and neutrons.



Bob Cook





Sent from Mail for Windows 10




From: mix...@bigpond.com 
Sent: Wednesday, July 24, 2019 1:29:51 PM
To: vortex-l@eskimo.com 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

In reply to  bobcook39...@hotmail.com's message of Wed, 24 Jul 2019 13:19:02
+:
Hi,
[snip]
>For example, spin energy transitions within a coupled “coherent” system may 
>not entail any radiation at all, if there is a perfect conservation of angular 
>momentum during the LENR event.  Of course radiant heat may be emitted in a 
>follow-up reaction involving the decay of the phonic energy of the coherent 
>system’s lattice.

Can you explain in detail what you have in mind?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

D* owns a deep orbit 2.15pm (Holmlid)In SO(4)this halve of that (1.07pm) 
like electron too. From the 7-Li-H* reaction we know that the deep shell 
electron binds to the nuclear flux too. Of course there are still many 
open questions but if you understand the mass structure e.g. of 
Deuterium I show in NPP 2.1.6 then you see which orbits can be used for 
deeper bindings. One revealing thing for fan's of classic physics would 
be to search for the shell electron of gold. Please tell me if you find 
a paper about e.g. Ag X or Ag XX or deeper states! Jürg Wyttenbach True, 
I did miss that point, but your statement raises another. In that case, 
you are only supplying a single extra electron from the neutron of the 
D, so the other K shell vacancy remains unfilled, and will cause a 
higher level electron to drop into the vacancy releasing an x-ray. Of 
course, for light elements this will only be a soft x-ray, but for 
mid-range or heavy atoms this can be quite energetic. Regards, Robin van 
Spaandonk



Am 24.07.19 um 23:16 schrieb mix...@bigpond.com:

In reply to  Jürg Wyttenbach's message of Wed, 24 Jul 2019 14:15:44 +0200:
Hi,
[snip]

K shells are not usually vacant, so such an electron would still upset
things. Regards,

You miss the point! If you increase the nuclear charge by +2 then
exactly 2 k-shell electrons are missing!

True, I did miss that point, but your statement raises another. In that case,
you are only supplying a single extra electron from the neutron of the D, so the
other K shell vacancy remains unfilled, and will cause a higher level electron
to drop into the vacancy releasing an x-ray.
Of course, for light elements this will only be a soft x-ray, but for mid-range
or heavy atoms this can be quite energetic.
Regards,


Robin van Spaandonk

local asymmetry = temporary success





--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  bobcook39...@hotmail.com's message of Wed, 24 Jul 2019 13:19:02
+:
Hi,
[snip]
>For example, spin energy transitions within a coupled “coherent” system may 
>not entail any radiation at all, if there is a perfect conservation of angular 
>momentum during the LENR event.  Of course radiant heat may be emitted in a 
>follow-up reaction involving the decay of the phonic energy of the coherent 
>system’s lattice.

Can you explain in detail what you have in mind?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Jürg Wyttenbach's message of Wed, 24 Jul 2019 14:15:44 +0200:
Hi,
[snip]
>K shells are not usually vacant, so such an electron would still upset 
>things. Regards,
>
>You miss the point! If you increase the nuclear charge by +2 then 
>exactly 2 k-shell electrons are missing!

True, I did miss that point, but your statement raises another. In that case,
you are only supplying a single extra electron from the neutron of the D, so the
other K shell vacancy remains unfilled, and will cause a higher level electron
to drop into the vacancy releasing an x-ray.
Of course, for light elements this will only be a soft x-ray, but for mid-range
or heavy atoms this can be quite energetic.
Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Robin—



If the electron is slow enough, I would guess it very well may react with other 
atomic electrons and be absorbed producing a charge on the system and a small 
increase in phonic energy.  For all practical purposes it would not be observed 
except for a small charge , accumulated with repeated like LENR reactions.



Bob Cook



Sent from Mail for Windows 10




From: mix...@bigpond.com 
Sent: Tuesday, July 23, 2019 6:46:04 PM
To: vortex-l@eskimo.com 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

In reply to  Jürg Wyttenbach's message of Sun, 21 Jul 2019 18:39:48 +0200:
Hi,
[snip]
>Bob
>
>One reason why the D* path is working like adding +2p/2e could be that
>the internal electron from the neutron only needs to do a little push to
>get to the k-shell. Thus no need to emit an electron!

K shells are not usually vacant, so such an electron would still upset things.
Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Axil—

Gamma radiation is by definition generated by a nuclear transition from one 
energy state to another.   I is not necessarily of Mev intensity.  In contrast 
x-rays are by definition  are produced by atomic electronic transitions from 
one energy state to another.

However a nuclear transition may happen within a coherent system of coupled 
particles with no gammas being produced. Not even low intensity irradiation as 
is produced in NMR events  may not happen in some LENR phenomena.

For example, spin energy transitions within a coupled “coherent” system may not 
entail any radiation at all, if there is a perfect conservation of angular 
momentum during the LENR event.  Of course radiant heat may be emitted in a 
follow-up reaction involving the decay of the phonic energy of the coherent 
system’s lattice.

Bob Cook

From: Axil Axil 
Sent: Tuesday, July 23, 2019 12:39:57 PM
To: vortex-l 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

If a nuclear reaction (fusion) was responsible for the transmutation, wouldn't 
gamma radiation be produced?

On Tue, Jul 23, 2019 at 4:35 PM mailto:mix...@bigpond.com>> 
wrote:
In reply to  Axil Axil's message of Mon, 22 Jul 2019 00:41:51 -0400:
Hi,
[snip]
>http://www.jmcchina.org/html/2019/1/20190101.htm
>
>Replication of biologic transmutation using a chemical reaction.
>
>The productivity of the transmutation was a function of the ambient
>temperature of the solution. 75C produced the most transmutation. Note that
>there was no report of a heating effect or other energy release that
>accompanied the transmutation.

The actual reported change was in the ppm range, so you should be able to
calculate whether or not any normal nuclear reaction energy release would have
been noticeable.
[snip]
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

K shells are not usually vacant, so such an electron would still upset 
things. Regards,


You miss the point! If you increase the nuclear charge by +2 then 
exactly 2 k-shell electrons are missing!

(If you understand the energy levels ...)

Jürg Wyttenbach

Am 24.07.19 um 04:46 schrieb mix...@bigpond.com:

In reply to  Jürg Wyttenbach's message of Sun, 21 Jul 2019 18:39:48 +0200:
Hi,
[snip]

Bob

One reason why the D* path is working like adding +2p/2e could be that
the internal electron from the neutron only needs to do a little push to
get to the k-shell. Thus no need to emit an electron!

K shells are not usually vacant, so such an electron would still upset things.
Regards,


Robin van Spaandonk

local asymmetry = temporary success





--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Jürg Wyttenbach's message of Sun, 21 Jul 2019 18:39:48 +0200:
Hi,
[snip]
>Bob
>
>One reason why the D* path is working like adding +2p/2e could be that 
>the internal electron from the neutron only needs to do a little push to 
>get to the k-shell. Thus no need to emit an electron!

K shells are not usually vacant, so such an electron would still upset things.
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Axil Axil's message of Tue, 23 Jul 2019 16:39:57 -0400:
Hi,
[snip]
>If a nuclear reaction (fusion) was responsible for the transmutation,
>wouldn't gamma radiation be produced?

Not necessarily as much as you might expect. It depends on the actual reaction.
If there are particles available to carry away the reaction energy, then very
few gammas are likely*. If those particles are electrons then some
bremsstrahlung is to be expected, but not so if they are heavy particles.
Positrons would of course result in annihilation gammas.

* This is because particle emission happens on the order of 1E-23 seconds,
whereas gamma emission is more on the order of 1E-17 seconds, so particles
usually carry the reaction energy away before a gamma has time to form.
This can result in a million times less gammas than one might otherwise expect.
If the number of reactions is small to start with, then the gammas may get lost
in the background noise, particularly if any such gammas are also shielded to
some extent by the apparatus itself.

Then of course, the experimenter also has to be looking for them.
[snip]
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

If a nuclear reaction (fusion) was responsible for the transmutation,
wouldn't gamma radiation be produced?

On Tue, Jul 23, 2019 at 4:35 PM  wrote:

> In reply to  Axil Axil's message of Mon, 22 Jul 2019 00:41:51 -0400:
> Hi,
> [snip]
> >http://www.jmcchina.org/html/2019/1/20190101.htm
> >
> >Replication of biologic transmutation using a chemical reaction.
> >
> >The productivity of the transmutation was a function of the ambient
> >temperature of the solution. 75C produced the most transmutation. Note
> that
> >there was no report of a heating effect or other energy release that
> >accompanied the transmutation.
>
> The actual reported change was in the ppm range, so you should be able to
> calculate whether or not any normal nuclear reaction energy release would
> have
> been noticeable.
> [snip]
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Axil Axil's message of Mon, 22 Jul 2019 00:41:51 -0400:
Hi,
[snip]
>http://www.jmcchina.org/html/2019/1/20190101.htm
>
>Replication of biologic transmutation using a chemical reaction.
>
>The productivity of the transmutation was a function of the ambient
>temperature of the solution. 75C produced the most transmutation. Note that
>there was no report of a heating effect or other energy release that
>accompanied the transmutation.

The actual reported change was in the ppm range, so you should be able to
calculate whether or not any normal nuclear reaction energy release would have
been noticeable.
[snip]
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

http://www.jmcchina.org/html/2019/1/20190101.htm

Replication of biologic transmutation using a chemical reaction.

The productivity of the transmutation was a function of the ambient
temperature of the solution. 75C produced the most transmutation. Note that
there was no report of a heating effect or other energy release that
accompanied the transmutation.

On Fri, Jul 19, 2019 at 11:14 AM JonesBeene  wrote:

> *“The *energy release per atom would be useful, to narrow down the
> possibilities.”
>
>
>
> Yes. No doubt this detail would be very useful to know, but is it even
> possible to know?
>
> Probably NOT as of now – since it makes a fundamental assumption which is
> not proved.
>
> That fundamental assumption is that energy release happens only once per
> atom – as in fusion. At first this seems to be a logical assumption, but
> fusion is not yet proved. If atoms produce lesser energy sequentially
> (still giving up mass)  then the energy per atom would not be relevant
> since any atom could radiate excess energy several times or several million
> times during the run.
>
> At this point we do no need to be specific about the details of the
> alternative mechanism to show the logical error, but there are several
> recognized possibilities that actually make as much sense as fusion
> including a version of the Hotson theory.
>
> One particular  operative mechanism  which could change perceptions is
> related to  the experimental findings which have been provided by Hora,
> Miley, Winterberg and Holmlid, et al. going back many years, which involve
> Bose-Einstein clustering. There is no apparent limitation on how many times
> an individual atom can give up mass-energy in the Coulomb explosion if and
> when they occur sequentially.
>
> To complicated matters – these experts suggest that the BEC cluster can
> act as an extremely efficient fusion target to be imploded with a laser. In
> that case the energy release per atom in the cluster would be less than the
> fusion of two deuterons – on average but the helium is thereafter
> unreactive so energy per atom would be useful to know.
>
> There are other alternative mechanisms for gain not involving fusion.
> These researchers  also suggest or imply that clustering “alone” can
> produce significant excess energy with no fusion  and/or a delayed nucleon
> annihilation event. Here, we find  the sequential Coulomb explosion where
> atoms can participate many times.
>
> Moreover, the Coulomb explosion is presently a proved mechanism with a
> signature emission which has been documented via experiment. In contrast
> there is no documented fusion evidence from the Mizuno breakthrough - as of
> now. It is a mistake to assume that this proof is just around the corner.
> It may not happen. I predict it will not.
>
> If one is firmly convinced that deuterium fusion must be happening in the
> new Mizuno breakthrough due to the robustness of the output or their own
> per theory or patent -  be prepared to jump- ship since there is NO report
> of  helium which is an absolute requirement to prove that particular
> mechanism .
>
> Until that time that substantial helium-4 is detected – the only gainful
> outcomes we know of  now from the published record are  non-fusion and one
> of them relates to the ~630 eV emission from Coulomb explosions. This gain
> is probably nuclear related but also probably not related to nuclear
> fusion, unless fusion is time-shifted in the QM sense so as to replace a
> deficit.
>
> Jones
>
>
>
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

Bob

One reason why the D* path is working like adding +2p/2e could be that 
the internal electron from the neutron only needs to do a little push to 
get to the k-shell. Thus no need to emit an electron!


But LENR definitely is fascinating for people doing real experiments as 
it might be the last real chance to add something important to our 
knowledge.


On on the other side it's a pity to see that 1-ends of physicists 
(mathematicians)  waste their live with reshuffling SO(3)xSU(2)xU(1) 
equations and try to find new answers that never will shine up. LENR 
would need this brain power!


Jürg

Am 21.07.19 um 15:58 schrieb bobcook39...@hotmail.com:


Jurg—

Electrons are also emitted to conserve charge,

Bob Cook

Sent from Mail  for 
Windows 10



*From:* Jürg Wyttenbach 
*Sent:* Saturday, July 20, 2019 8:09:51 AM
*To:* vortex-l@eskimo.com 
*Subject:* Re: [Vo]:If Mizuno is correct, this design is likely 
tobetheprecursor to all future devices

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

To learn about this use.
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

Most of the time the A+D* reaction is followed by a beta+ decay. But
there are some exceptions

Jürg

Am 20.07.19 um 00:39 schrieb mix...@bigpond.com:
> In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 
+0200:

> Hi,
> [snip]
>> In nuclear transformation (LENR) D* adds like a double proton and H*
>> adds like a neutron. That's what we see (exactly measure) from the 
gamma

>> radiation signature of complex reactions.
> Could you give a couple of example reactions? (I'm a little unsure 
of what you

> mean when you say "adds like".)
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>
>

--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06



--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Jurg—



Electrons are also emitted to conserve charge,



Bob Cook



Sent from Mail for Windows 10




From: Jürg Wyttenbach 
Sent: Saturday, July 20, 2019 8:09:51 AM
To: vortex-l@eskimo.com 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

To learn about this use.
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

Most of the time the A+D* reaction is followed by a beta+ decay. But
there are some exceptions

Jürg

Am 20.07.19 um 00:39 schrieb mix...@bigpond.com:
> In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 +0200:
> Hi,
> [snip]
>> In nuclear transformation (LENR) D* adds like a double proton and H*
>> adds like a neutron. That's what we see (exactly measure) from the gamma
>> radiation signature of complex reactions.
> Could you give a couple of example reactions? (I'm a little unsure of what you
> mean when you say "adds like".)
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>
>

--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Jurg—



Those are good LENR examples.



Thanks for the clarification.



Bob Cook



Sent from Mail for Windows 10




From: Jürg Wyttenbach 
Sent: Saturday, July 20, 2019 8:09:51 AM
To: vortex-l@eskimo.com 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

To learn about this use.
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

Most of the time the A+D* reaction is followed by a beta+ decay. But
there are some exceptions

Jürg

Am 20.07.19 um 00:39 schrieb mix...@bigpond.com:
> In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 +0200:
> Hi,
> [snip]
>> In nuclear transformation (LENR) D* adds like a double proton and H*
>> adds like a neutron. That's what we see (exactly measure) from the gamma
>> radiation signature of complex reactions.
> Could you give a couple of example reactions? (I'm a little unsure of what you
> mean when you say "adds like".)
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>
>

--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

This is my favorite experiment: Radioactive decay half-life acceleration
cause by time dilation

Reference: https://arxiv.org/abs/1112


Accelerated alpha-decay of 232U isotope achieved by exposure of its aqueous
solution with gold nanoparticles to laser radiation

A.V. Simakin, G.A. Shafeev

This reference sheds light on the posit that an electromagnetic black hole
accelerates radioactive decay through relativistic time dilation. With this
posit in mind, experimental data elucidates what the optimum duration of
the pumping EMF time-frame should be.

"Experimental results are presented on laser-induced accelerated
alpha-decay of Uranium-232 nuclei under laser exposure of Au nanoparticles
in aqueous solutions of its salt. It is demonstrated that the decrease of
alpha-activity strongly depends on the peak intensity of the laser
radiation in the liquid and is highest at several terawatt per square
centimeter. The decrease of alpha-activity of the exposed solutions is
accompanied by the deviation of gamma-activities of daughter nuclides of
Uranium-232 from their equilibrium values. Possible mechanisms of the laser
influence on the alpha-activity are discussed on the basis of the
amplification of the electric field of laser wave on metallic
nanoparticles."

The mechanism of polariton Bose condensation causation involves the
generation of plasma and associated intense EMF fields around a
nanoparticle that generates a critical density of surface plasmon
polaritons (SPP) which acts as a micro analog black hole in which
relativistic time dilation occurs in and around its vortex flux tubes. The
SPP and Its vortex tube are pictured below along with associated theory.

In the LION experiment, we saw how spin vortex flux tubes produced
transmutation of elements. This article explains how those tubes form.

https://advances.sciencemag...


Vortex and half-vortex dynamics in a nonlinear spinor quantum fluid

The A.V. Simakin article shows that the polariton Bose condensate reduces
the 69 year half-life of U232 to 5 microseconds. It addition, the
acceleration of radioactive decay of the 4.5 billion year half-life of
uranium is sufficiently time accelerated to produce significant detectable
uranium decay products.

Both the power of the laser pulse and its duration are important factors in
producing optimum half-life decay acceleration. The nanosecond laser pulse
length is too short to optimize radioactive decay acceleration as well of
the 350 ps laser source. The 150 ps laser source seems to be a better fit
to match the 50 ps life span of the polariton which is the resonance value
for optimization of decay.

Another insight gleaned for this experiment is that the length of the laser
pulse does not produce continuing radioactive decay acceleration. The LENR
activity is constrained by the lifetime of the polariton and is active only
as long as the polariton is active.

On Sat, Jul 20, 2019 at 6:01 PM Jürg Wyttenbach  wrote:

> So which gammas are emitted in the first case, and what would you expect if 
> the
> reaction were 7-Li + H* --> 8-Be --> 2 4-He?
>
> H* is p+e that get both added. This was my first key finding about 3 years 
> ago.
>
> You can look this explanation (gammas) up in my very old writeup about
> LENR or in the Lipinski patent.
>
> This is a very good teaching example and I would prefer that everybody
> interested in real LENR does this as an exercise.  It is absolute key to
> know/understand the details about H*, D*.
>
> Juerg
>
> Am 20.07.19 um 22:49 schrieb mix...@bigpond.com:
>
> In reply to  Jürg Wyttenbach's message of Sat, 20 Jul 2019 18:09:51 +0200:
> Hi,
> [snip]
>
> Two simple samples:
>
> 7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)
>
> 105Pd +D* -->107Cd --> 107Ag classic P
>
> You can identify the decay paths by the typical gammas emitted.
>
> So which gammas are emitted in the first case, and what would you expect if 
> the
> reaction were 7-Li + H* --> 8-Be --> 2 4-He?
>
> ...and how do you know that the first reaction isn't really:-
>
> 7-Li + D* --> 8Li (neutron transfer) + p (wanders off) --> 8-Be --> 2 4-He,
> making use of trace D in the H?
> Does the reaction rate change if you incrementally add D to the H?
>
>
> To learn about this use. 
> https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html
>
> Most of the time the A+D* reaction is followed by a beta+ decay. But
> there are some exceptions
>
> ...so you detect pairs of positron annihilation gammas?
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>
>
>
>
> --
> Jürg Wyttenbach
> Bifangstr.22
> 8910 Affoltern a.A.
> 044 760 14 18
> 079 246 36 06
>
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

So which gammas are emitted in the first case, and what would you expect 
if the reaction were 7-Li + H* --> 8-Be --> 2 4-He?


H* is p+e that get both added. This was my first key finding about 3 years ago.

You can look this explanation (gammas) up in my very old writeup about 
LENR or in the Lipinski patent.


This is a very good teaching example and I would prefer that everybody 
interested in real LENR does this as an exercise.  It is absolute key to 
know/understand the details about H*, D*.


Juerg

Am 20.07.19 um 22:49 schrieb mix...@bigpond.com:

In reply to  Jürg Wyttenbach's message of Sat, 20 Jul 2019 18:09:51 +0200:
Hi,
[snip]

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

So which gammas are emitted in the first case, and what would you expect if the
reaction were 7-Li + H* --> 8-Be --> 2 4-He?

...and how do you know that the first reaction isn't really:-

7-Li + D* --> 8Li (neutron transfer) + p (wanders off) --> 8-Be --> 2 4-He,
making use of trace D in the H?
Does the reaction rate change if you incrementally add D to the H?


To learn about this use.
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

Most of the time the A+D* reaction is followed by a beta+ decay. But
there are some exceptions

...so you detect pairs of positron annihilation gammas?
Regards,


Robin van Spaandonk

local asymmetry = temporary success





--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Jürg Wyttenbach's message of Sat, 20 Jul 2019 18:09:51 +0200:
Hi,
[snip]
>Two simple samples:
>
>7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)
>
>105Pd +D* -->107Cd --> 107Ag classic P
>
>You can identify the decay paths by the typical gammas emitted.

So which gammas are emitted in the first case, and what would you expect if the
reaction were 7-Li + H* --> 8-Be --> 2 4-He?

...and how do you know that the first reaction isn't really:-

7-Li + D* --> 8Li (neutron transfer) + p (wanders off) --> 8-Be --> 2 4-He,
making use of trace D in the H?
Does the reaction rate change if you incrementally add D to the H?

>
>To learn about this use. 
>https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html
>
>Most of the time the A+D* reaction is followed by a beta+ decay. But 
>there are some exceptions

...so you detect pairs of positron annihilation gammas?
Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Arnaud Kodeck]

Hello Jürg,

How do you witness the reaction 107Cd->107Ag ?

Arnaud

-Original Message-
From: Jürg Wyttenbach  
Sent: Saturday, 20 July 2019 18:10
To: vortex-l@eskimo.com
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

To learn about this use. 
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

Most of the time the A+D* reaction is followed by a beta+ decay. But 
there are some exceptions

Jürg

Am 20.07.19 um 00:39 schrieb mix...@bigpond.com:
> In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 +0200:
> Hi,
> [snip]
>> In nuclear transformation (LENR) D* adds like a double proton and H*
>> adds like a neutron. That's what we see (exactly measure) from the gamma
>> radiation signature of complex reactions.
> Could you give a couple of example reactions? (I'm a little unsure of what you
> mean when you say "adds like".)
> Regards,
>
>
> Robin van Spaandonk
>
> local asymmetry = temporary success
>
>
>

-- 
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

Two simple samples:

7-Li + H* --> 8Li --> 8-Be --> 2 4-He. (Lipinski reaction)

105Pd +D* -->107Cd --> 107Ag classic P

You can identify the decay paths by the typical gammas emitted.

To learn about this use. 
https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html


Most of the time the A+D* reaction is followed by a beta+ decay. But 
there are some exceptions


Jürg

Am 20.07.19 um 00:39 schrieb mix...@bigpond.com:

In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 +0200:
Hi,
[snip]

In nuclear transformation (LENR) D* adds like a double proton and H*
adds like a neutron. That's what we see (exactly measure) from the gamma
radiation signature of complex reactions.

Could you give a couple of example reactions? (I'm a little unsure of what you
mean when you say "adds like".)
Regards,


Robin van Spaandonk

local asymmetry = temporary success





--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Jürg Wyttenbach's message of Fri, 19 Jul 2019 23:05:52 +0200:
Hi,
[snip]
>In nuclear transformation (LENR) D* adds like a double proton and H* 
>adds like a neutron. That's what we see (exactly measure) from the gamma 
>radiation signature of complex reactions.

Could you give a couple of example reactions? (I'm a little unsure of what you
mean when you say "adds like".)
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  JonesBeene's message of Fri, 19 Jul 2019 08:14:19 -0700:
Hi,
[snip]
>“The energy release per atom would be useful, to narrow down the 
>possibilities.”
>
>Yes. No doubt this detail would be very useful to know, but is it even 
>possible to know?

I think that with careful work, it is possible. 

1) It should be possible to measure the total energy release over a long period.
2) The amount of D used should be able to be calculated by subtracting what is
left over at the end from what was made available during the course of the run.
The difference (if any) is what was used.

I deliberately specified a "long" run, because if the difference is small, then
the error can potentially be large, because there is always some measurement
error, and a very small signal may be lost in the measurement error.

If at the end of such a long run the difference is still small or near to
non-existent, then we are obviously dealing with one of the very energetic
options.
[snip]
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

Am 19.07.19 um 19:32 schrieb Jones Beene:
Which is to say: until we get mass spec readings for significant 
amounts of helium after a long run, fusion remains just a fall-back 
assumption based on old electrolysis results - and possibly 
unjustified for anything else.


Surprisingly, even with a reactor operating at 3 kw for extended 
periods - there is no report of tritium or transmutation of any kind.



In nuclear transformation (LENR) D* adds like a double proton and H* 
adds like a neutron. That's what we see (exactly measure) from the gamma 
radiation signature of complex reactions.


Tritium is not possible with most D-D LENR. Only 3-He will be seen. (If 
you would use neutron rich isotopes then the range of results would be 
much larger. But this is for large state labs.)


Transmutations always happen as side reaction with about 10E-8 of the 
reaction power equivalent. We at Ecalox measure LENR gamma radiation 
since about 2 years now. (> 1 spectra stored now)



However, Your answer does not answer Robins question about your model’s 
accurate energy calculations.


Separately there are tables of numerous measured isomeric energy states, 
which could be compared directly with the detailed calculations of your 
physical 6-D modeling.   Are you or anyone else doing such calculations 
in way of validating you models of nuclei?


Bob Cook

Currently I have other priorities than extending my model. But, as a 
strongly handicapped person I would welcome any help. Such a person 
would of course learn much more than has been written so far!


There is good reason to believe that a nice piece of software that 
models the EM collapse of two ring currents in SO(4) could give some 
deep insight into the nature of physical constants.


Currently I follow a refined approach to model the D-D fusion process 
that seems to give the same resonance 1000eV as the magnetic moment 
model and is in good agreement with Mills measurements.



But for this you will have to wait as there are to many open tasks now.

And to remind everybody: A model is only as exact as experiments tell 
you. Thus I give you "the most likely value" you can see in a simple 
experiment. An exact value makes no sense as nobody has an exact 
measurement...


Jürg


--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

The elements transmuted by the LENR reaction are usually calcium,
magnesium, sulfur, carbon, iron, aluminum. Helium is not usually found. In
nature,  supernova nucleosynthesis: the nucleosynthesis of chemical
elements in supernova explosions are the usual nature generators of these
heavier elements. Shock-wave based supernova nucleosynthesis and
hydrostatic-burning processes create most of the isotopes of the elements
carbon (Z = 6), oxygen (Z = 8), and elements with Z = 10–28 (from neon to
nickel).

How can LENR reproduce the conditions inside an exploding supernova using a
compression of elements in the fusion process?

Any detection of elements heavier that helium precludes the fusion process
as the cause of LENR transmutation.

By the way, living thing have be found to produce  supernova
nucleosynthesis type elements as a usual byproduct of their life cycle.

On Fri, Jul 19, 2019 at 1:34 PM Jones Beene  wrote:

> Jürg Wyttenbach wrote:
>
> > In the Mizuno case we can exclude this behavior, as clustered D-D,
> inside larger clusters, always will undergo fusion...
>
>
> Always? ... doubt it. There is no evidence from Mizuno of helium and it
> makes no sense to be dogmatic on the issue until evidence arrives.
>
> Which is to say: until we get mass spec readings for significant amounts
> of helium after a long run, fusion remains just a fall-back assumption
> based on old electrolysis results - and possibly unjustified for anything
> else.
>
> Surprisingly, even with a reactor operating at 3 kw for extended periods -
> there is no report of tritium or transmutation of any kind. Mizuno is
> reportedly an expert at radiation detection so the lack of any mention by
> him is curious, to say the least. Even if the branching ratio of LENR in
> general favors mostly helium - at this high level of output there is little
> logical way to claim that absolute quenching of the normal branching ratio
> all the way back to zero tritium; and tritium could not be missed by him in
> small amounts, if it was present.
>
> Home usage of the reactor almost guarantees he assumes no tritium even at
> very high thermal output. He would not jeopardize his family's health.
>
> I agree that most of the LENR experts think deuterium fusion is
> responsible for the excess heat, but as of now that seems like little more
> than speculation to me - especially since in the earlier runs at 4000 Pa,
> with only protium as fuel (but with rubbed Pd on nickel), more excess
> energy was seen with protium-- than with deuterium under the same
> conditions. This favors a non-fusion modus operandi.
>
> Jones
>
>
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jones Beene]

 Jürg Wyttenbach wrote:  
> In the Mizuno case we can exclude this behavior, as clustered D-D, 
inside larger clusters, always will undergo fusion...

Always? ... doubt it. There is no evidence from Mizuno of helium and it makes 
no sense to be dogmatic on the issue until evidence arrives. 

Which is to say: until we get mass spec readings for significant amounts of 
helium after a long run, fusion remains just a fall-back assumption based on 
old electrolysis results - and possibly unjustified for anything else.
Surprisingly, even with a reactor operating at 3 kw for extended periods - 
there is no report of tritium or transmutation of any kind. Mizuno is 
reportedly an expert at radiation detection so the lack of any mention by him 
is curious, to say the least. Even if the branching ratio of LENR in general 
favors mostly helium - at this high level of output there is little logical way 
to claim that absolute quenching of the normal branching ratio all the way back 
to zero tritium; and tritium could not be missed by him in small amounts, if it 
was present. 

Home usage of the reactor almost guarantees he assumes no tritium even at very 
high thermal output. He would not jeopardize his family's health.

I agree that most of the LENR experts think deuterium fusion is responsible for 
the excess heat, but as of now that seems like little more than speculation to 
me - especially since in the earlier runs at 4000 Pa, with only protium as fuel 
(but with rubbed Pd on nickel), more excess energy was seen with protium-- than 
with deuterium under the same conditions. This favors a non-fusion modus 
operandi.
Jones


  

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Jones—

I agree with your observation  regarding multiple simultaneous events and  the 
apparent lack of the involvement of single nuclei in a LENR event.

Magnetic coupling between multiple particles (nucleons and atomic and or 
plasmonic electrons or other magnetic dipoles) can allow sharing  their angular 
momentum (spin energy) simultaneously without the production of energetic 
charged particles.  This is a desirable characteristic of LENR, since it 
significantly reduces or eliminates hazardous radiation and nuclear activation 
associated with releases of neutrons.

  Mundane heat energy results in the small kinetic energy increases of many 
electrons and multiple nucleons in a crystal lattice.

Bob Cook




From: JonesBeene 
Sent: Friday, July 19, 2019 7:14:19 AM
To: vortex-l@eskimo.com
Subject: RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

“The energy release per atom would be useful, to narrow down the possibilities.”

Yes. No doubt this detail would be very useful to know, but is it even possible 
to know?
Probably NOT as of now – since it makes a fundamental assumption which is not 
proved.
That fundamental assumption is that energy release happens only once per atom – 
as in fusion. At first this seems to be a logical assumption, but fusion is not 
yet proved. If atoms produce lesser energy sequentially (still giving up mass)  
then the energy per atom would not be relevant since any atom could radiate 
excess energy several times or several million times during the run.
At this point we do no need to be specific about the details of the alternative 
mechanism to show the logical error, but there are several recognized 
possibilities that actually make as much sense as fusion including a version of 
the Hotson theory.
One particular  operative mechanism  which could change perceptions is related 
to  the experimental findings which have been provided by Hora, Miley, 
Winterberg and Holmlid, et al. going back many years, which involve 
Bose-Einstein clustering. There is no apparent limitation on how many times an 
individual atom can give up mass-energy in the Coulomb explosion if and when 
they occur sequentially.
To complicated matters – these experts suggest that the BEC cluster can act as 
an extremely efficient fusion target to be imploded with a laser. In that case 
the energy release per atom in the cluster would be less than the fusion of two 
deuterons – on average but the helium is thereafter unreactive so energy per 
atom would be useful to know.
There are other alternative mechanisms for gain not involving fusion. These 
researchers  also suggest or imply that clustering “alone” can produce 
significant excess energy with no fusion  and/or a delayed nucleon annihilation 
event. Here, we find  the sequential Coulomb explosion where atoms can 
participate many times.
Moreover, the Coulomb explosion is presently a proved mechanism with a 
signature emission which has been documented via experiment. In contrast there 
is no documented fusion evidence from the Mizuno breakthrough - as of now. It 
is a mistake to assume that this proof is just around the corner. It may not 
happen. I predict it will not.
If one is firmly convinced that deuterium fusion must be happening in the new 
Mizuno breakthrough due to the robustness of the output or their own per theory 
or patent -  be prepared to jump- ship since there is NO report of  helium 
which is an absolute requirement to prove that particular mechanism .
Until that time that substantial helium-4 is detected – the only gainful 
outcomes we know of  now from the published record are  non-fusion and one of 
them relates to the ~630 eV emission from Coulomb explosions. This gain is 
probably nuclear related but also probably not related to nuclear fusion, 
unless fusion is time-shifted in the QM sense so as to replace a deficit.
Jones

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

Am 19.07.19 um 17:14 schrieb JonesBeene:
There are other alternative mechanisms for gain not involving fusion. 
These researchers  also suggest or imply that clustering “alone” can 
produce significant excess energy with no fusion  and/or a delayed 
nucleon annihilation event. Here, we 


Clustering energy in the range of 500-1000eV is seen in Takahashi's Ni-H 
process where he produces about 50 watt's of excess energy. Also R. 
Mills process leads to toroidial H* clusters with the same "low" energy 
gain.



In the Mizuno case we can exclude this behavior, as clustered D-D, 
inside larger clusters, always will undergo fusion The energy of the 
Mizuno process is in the region > 100'000keV /Pd or surface Ni, what 
also excludes H*/D* condensation as a final source of energy.



Pd-D or Ni-D fusion so long always produced 4-He with minor amounts of 
3-He (< 10E-6). There is more than enough experimental proof and to ask 
doing it again is interesting only for detecting how much Pd/Ni gets 
transmuted. Usually such side reactions that transmute Pd/Ni are below 
10E-8 compared to D-D fusion reactions.



I guess after the first successful replication of Mizuno will see a 
sudden death of ITER and similar approaches.



Jürg


--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

https://www.youtube.com/watch?v=wAsceiIMY2I

The question is "how can microorganisms produce enough power to transmute
elements".

The answer to this question could involve a global Bose condensate (BEC)
that forms throughout the entire extent of the bug colony. This BEC
connects each bug through quantum mechanical entanglement into a network
where any single bug can tap into the total power potential of the entire
colony through a process called superradience.

As the transmutation of elements takes place, the LENR reaction does not
expose the fradual bodies of the individual bugs to the destructive
radiation, energy, and particles produced by the transmutation process. But
the entangled connectivity provided by the BEC can distribute the
transmuted elements to each bug as required.

We have seen this sharing of transmuted elements between LENR active agents
in the LION reactor experiments.

The bugs aquire the elements they need to sustain their life cycle but LENR
hides the destructive potential that this life sustaining transmutation of
elements engenders.

On Fri, Jul 19, 2019 at 11:14 AM JonesBeene  wrote:

> *“The *energy release per atom would be useful, to narrow down the
> possibilities.”
>
>
>
> Yes. No doubt this detail would be very useful to know, but is it even
> possible to know?
>
> Probably NOT as of now – since it makes a fundamental assumption which is
> not proved.
>
> That fundamental assumption is that energy release happens only once per
> atom – as in fusion. At first this seems to be a logical assumption, but
> fusion is not yet proved. If atoms produce lesser energy sequentially
> (still giving up mass)  then the energy per atom would not be relevant
> since any atom could radiate excess energy several times or several million
> times during the run.
>
> At this point we do no need to be specific about the details of the
> alternative mechanism to show the logical error, but there are several
> recognized possibilities that actually make as much sense as fusion
> including a version of the Hotson theory.
>
> One particular  operative mechanism  which could change perceptions is
> related to  the experimental findings which have been provided by Hora,
> Miley, Winterberg and Holmlid, et al. going back many years, which involve
> Bose-Einstein clustering. There is no apparent limitation on how many times
> an individual atom can give up mass-energy in the Coulomb explosion if and
> when they occur sequentially.
>
> To complicated matters – these experts suggest that the BEC cluster can
> act as an extremely efficient fusion target to be imploded with a laser. In
> that case the energy release per atom in the cluster would be less than the
> fusion of two deuterons – on average but the helium is thereafter
> unreactive so energy per atom would be useful to know.
>
> There are other alternative mechanisms for gain not involving fusion.
> These researchers  also suggest or imply that clustering “alone” can
> produce significant excess energy with no fusion  and/or a delayed nucleon
> annihilation event. Here, we find  the sequential Coulomb explosion where
> atoms can participate many times.
>
> Moreover, the Coulomb explosion is presently a proved mechanism with a
> signature emission which has been documented via experiment. In contrast
> there is no documented fusion evidence from the Mizuno breakthrough - as of
> now. It is a mistake to assume that this proof is just around the corner.
> It may not happen. I predict it will not.
>
> If one is firmly convinced that deuterium fusion must be happening in the
> new Mizuno breakthrough due to the robustness of the output or their own
> per theory or patent -  be prepared to jump- ship since there is NO report
> of  helium which is an absolute requirement to prove that particular
> mechanism .
>
> Until that time that substantial helium-4 is detected – the only gainful
> outcomes we know of  now from the published record are  non-fusion and one
> of them relates to the ~630 eV emission from Coulomb explosions. This gain
> is probably nuclear related but also probably not related to nuclear
> fusion, unless fusion is time-shifted in the QM sense so as to replace a
> deficit.
>
> Jones
>
>
>
>

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[JonesBeene]

“The energy release per atom would be useful, to narrow down the possibilities.”

Yes. No doubt this detail would be very useful to know, but is it even possible 
to know?
Probably NOT as of now – since it makes a fundamental assumption which is not 
proved.
That fundamental assumption is that energy release happens only once per atom – 
as in fusion. At first this seems to be a logical assumption, but fusion is not 
yet proved. If atoms produce lesser energy sequentially (still giving up mass)  
then the energy per atom would not be relevant since any atom could radiate 
excess energy several times or several million times during the run. 
At this point we do no need to be specific about the details of the alternative 
mechanism to show the logical error, but there are several recognized 
possibilities that actually make as much sense as fusion including a version of 
the Hotson theory.
One particular  operative mechanism  which could change perceptions is related 
to  the experimental findings which have been provided by Hora, Miley, 
Winterberg and Holmlid, et al. going back many years, which involve 
Bose-Einstein clustering. There is no apparent limitation on how many times an 
individual atom can give up mass-energy in the Coulomb explosion if and when 
they occur sequentially.
To complicated matters – these experts suggest that the BEC cluster can act as 
an extremely efficient fusion target to be imploded with a laser. In that case 
the energy release per atom in the cluster would be less than the fusion of two 
deuterons – on average but the helium is thereafter unreactive so energy per 
atom would be useful to know.
There are other alternative mechanisms for gain not involving fusion. These 
researchers  also suggest or imply that clustering “alone” can produce 
significant excess energy with no fusion  and/or a delayed nucleon annihilation 
event. Here, we find  the sequential Coulomb explosion where atoms can 
participate many times.
Moreover, the Coulomb explosion is presently a proved mechanism with a 
signature emission which has been documented via experiment. In contrast there 
is no documented fusion evidence from the Mizuno breakthrough - as of now. It 
is a mistake to assume that this proof is just around the corner. It may not 
happen. I predict it will not.
If one is firmly convinced that deuterium fusion must be happening in the new 
Mizuno breakthrough due to the robustness of the output or their own per theory 
or patent -  be prepared to jump- ship since there is NO report of  helium 
which is an absolute requirement to prove that particular mechanism .
Until that time that substantial helium-4 is detected – the only gainful 
outcomes we know of  now from the published record are  non-fusion and one of 
them relates to the ~630 eV emission from Coulomb explosions. This gain is 
probably nuclear related but also probably not related to nuclear fusion, 
unless fusion is time-shifted in the QM sense so as to replace a deficit.
Jones

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

There is a natural tendency in the formulation of LENR theory to ignore or
flat out deny the existence of outlying or contraindicated  LENR processes
that are generally observed and proven by observation but conflict with
preconceived  notions of LENR reality.





One of these observations is that transmutation of elements in LENR does
not for the most part yield energy: gamma, heat, light, particles, and
loses mass.





These counter indicators include biological transmutation, the associated
very safe and benign environment that LENR occurs in, and the unexplained
loss of mass and/or energy where that loss is absolutely required by any
proposed theory.





I have in mind the production of ferrosilicon in India where an electric
arc processing method transmutes 4.5 tons of iron and silicon each day from
carbon, oxygen and other ambient elements inside the environment of the
electric arc blast furnace.





The outsized amount of transmutation of so much mass every day should
produce enough energy to meet the power needs of Europe for a year, and yet
that huge amount of energy production is not observed.





Transmutation in the LENR reaction for the most part occurs under a state
of quantum mechanical superposition where that energy produced and the mass
exposed to the LENR reaction is permanently lost to our reality.





This argument about hydrogen fusion is pointless because the energy
produced by this reaction is invisible to our observation. The issue is
that the weirdness of quantum mechanics is made manifest to our observation
and we cannot understand that it is happening.


On Tue, Jul 16, 2019 at 9:54 AM JonesBeene  wrote:

>
>
> *From: *H LV 
>
>
>
>- How much of the energy in a nuclear reaction is actually due to mass
>change?
>
>
>
> Is there any reason to think that it would not be all?
>
>
>
> Even if sequential hydrogen cluster formation is responsible for the gain,
> and there is no fusion at all - the ultimate source of that heat would
> still be nuclear mass.
>

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[H LV]

On Tue., Jul. 16, 2019, 9:54 a.m. JonesBeene,  wrote:

>
>
> *From: *H LV 
>
>
>
>- How much of the energy in a nuclear reaction is actually due to mass
>change?
>
>
>
> Is there any reason to think that it would not be all?
>
>
>
> Even if sequential hydrogen cluster formation is responsible for the gain,
> and there is no fusion at all - the ultimate source of that heat would
> still be nuclear mass.
>


I think there is large chunk of nuclear physics which is waiting to be
formulated in which mass-energy equivalence can be ignored to a first
approximation. The reason I think this is that chemistry for instance, as
far I know, is fully explicable without it. Remove mass from the ledger and
all there are molecular, atomic and nuclear forces which vary in magnitude.
Harry

>

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[JonesBeene]

From: H LV

➢ How much of the energy in a nuclear reaction is actually due to mass change?  

Is there any reason to think that it would not be all?

Even if sequential hydrogen cluster formation is responsible for the gain, and 
there is no fusion at all - the ultimate source of that heat would still be 
nuclear mass.

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jed Rothwell]

Axil Axil  wrote:


> If the reactor stops producing heat at some later time, the fuel
> consumption rate can be calculated and this data might indicate what energy
> production mechanism is producing energy in the mesh reactor.
>

I have a feeling that might be caused by contamination rather than fuel
exhaustion. It might be difficult to determine which it is. If air leaks
in, that will definitely contaminate, and cause the reaction to stop.
Mizuno has deliberately admitted air, and observed that is what happens. I
do not know how much air it takes.

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Axil Axil]

One possible way to shed some new light on the fusion issue is to run an
experiment were the pressure of deuterium is reduced slowly down toward the
1 pa limit to see how the rate of heat output behaves. Once a minimum
sustainable pressure is determined where the mesh reactor still produces
steady heat output, run the reactor at this  minimum  productive pressure
regime and monitor how the heat output varies over time.

If the reactor stops producing heat at some later time, the fuel
consumption rate can be calculated and this data might indicate what energy
production mechanism is producing energy in the mesh reactor.

On Mon, Jul 15, 2019 at 9:56 AM JonesBeene  wrote:

>
>
> Reality Check. Surprisingly, nuclear fusion of deuterium into helium seems 
> NOT sufficiently energetic to account for the Mizuno claim of heating his 
> home.
>
>
>
> Mass is apparently being converted into energy, but how? And what are the 
> ramifications of such a low reactor inventory of deuterium gas?
>
>
>
> The main contenders for excess energy production would be:
>
>
>
> 1) D+D -> He
>
> 2) Deflation of electrons – i.e. the Millsean approach
>
> 3) Disintegration of deuterons into muons – Holmlid’s theory - which is far 
> more energetic than fusion in terms of entropy per unit of mass
>
> 4) Sequential Coulomb explosions from cluster formation –hypothesis from 
> Hora, Miley etc.
>
> 5) Any combination or permutation of the above
>
>
>
> If fusion of D into He is your choice - then one gram of fused deuterium 
> yields 10^12J (one terajoule)of energy, but when based on the low operating 
> pressure of 100-300 Pa (100 Pa = .001 bar) and the need for low metal 
> loading, as stated in his paper - that set of factors represents a tiny fuel 
> inventory, such that when completely fused into helium would generate about 
> 278 kilowatt hours of equivalent heat.
>
>
>
> If Mizuno was producing close to 3 kW continuous to heat his house in a 
> Sapporo winter, he could run it for only about 100 hours without a refill if 
> the gain was from fusion and the inventory was at the low end of his specs. 
> At any rate, if the gain was from nuclear fusion only - then almost all of 
> the deuterium would be converted, and the helium ash should be easily 
> measurable.
>
>
>
> There should be no need for a cold trap to increase the helium ratio – the 
> residual gas after less than a week should be almost all helium, no? Even if 
> these calculations are off by a large factor, the helium content should be 
> obvious.
>
>
>
> IOW – in the naïve assessment of the breakthrough claim of Mizuno – 
> specifically the heating of his home – after 100 hours or so of operation, 
> there should be a whopping milligram of helium and little deuterium in the 
> reactor to measure.
>
>
>
> In contrast – Holmlid’s theory proposes deuteron disintegration (with
> inadvertent fusion). His theory suggests that about 4 GeV of mass-energy
> per every two atoms of deuterium lost could be converted into energy. This
> is about 150 times MORE potential energy per unit of mass (converted into
> energy) than can be derived from fusion into helium.
>
>
>
> On the surface, then – fusion of deuterium into helium appears to be too
> weak a reaction to account for the Mizuno claims of heating his home, and
> only the Holmlid effect would have an adequate output.
>
>
>
> Why isn’t the Holmlid effect the favored hypothesis?
>
>
>
> Jones
>

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[russ.george]

If one is working with a quadrapole mass spec, and especially a small one like 
an RGA it will be impossible to devolve the peaks of 4He and D2. Only by being 
certain that little D2 is present by trapping it in a cold or getter trap on 
the way to the mass spec can one ever be certain that the sample is 4He instead 
of D2. The practice is clearly informative as one learns the operation of the 
RGA with and without the cold trap.

 

Don’t be fooled by imaginary methods with such close masses. No one will ever 
believe 4He in such an instrument without proof that the D2 signal is 
suppressed.

 

 

From: Jürg Wyttenbach  
Sent: Monday, July 15, 2019 3:34 PM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

 

For 100kw/h about 1.2mg of deuterium are needed.

 

If pressure is lower then the relative density of D (D2 gas) increases, 
somewhere between 0.15 & 0.45g/l.The inventory is given by Ni/pd surface bound 
D, the volume (15l) of the reactor and the pressure factor (=0.003 for   300Pa) 
.

 

But Mizuno recommends to always let the bottle attached and of course he did 
feed additional deuterium if needed.

 

Holmid does not yet produce any energy. First he must avoid to mainly produce 
positive muons...

 

 

Jürg

 

 

 

Am 15.07.19 um 15:56 schrieb JonesBeene:

 
Reality Check. Surprisingly, nuclear fusion of deuterium into helium seems NOT 
sufficiently energetic to account for the Mizuno claim of heating his home.
 
Mass is apparently being converted into energy, but how? And what are the 
ramifications of such a low reactor inventory of deuterium gas?
 
The main contenders for excess energy production would be:
 
1) D+D -> He
2) Deflation of electrons – i.e. the Millsean approach
3) Disintegration of deuterons into muons – Holmlid’s theory - which is far 
more energetic than fusion in terms of entropy per unit of mass 
4) Sequential Coulomb explosions from cluster formation –hypothesis from Hora, 
Miley etc.
5) Any combination or permutation of the above
 
If fusion of D into He is your choice - then one gram of fused deuterium yields 
10^12J (one terajoule)of energy, but when based on the low operating pressure 
of 100-300 Pa (100 Pa = .001 bar) and the need for low metal loading, as stated 
in his paper - that set of factors represents a tiny fuel inventory, such that 
when completely fused into helium would generate about 278 kilowatt hours of 
equivalent heat. 
 
If Mizuno was producing close to 3 kW continuous to heat his house in a Sapporo 
winter, he could run it for only about 100 hours without a refill if the gain 
was from fusion and the inventory was at the low end of his specs. At any rate, 
if the gain was from nuclear fusion only - then almost all of the deuterium 
would be converted, and the helium ash should be easily measurable.
 
There should be no need for a cold trap to increase the helium ratio – the 
residual gas after less than a week should be almost all helium, no? Even if 
these calculations are off by a large factor, the helium content should be 
obvious.
 
IOW – in the naïve assessment of the breakthrough claim of Mizuno – 
specifically the heating of his home – after 100 hours or so of operation, 
there should be a whopping milligram of helium and little deuterium in the 
reactor to measure.
 

In contrast – Holmlid’s theory proposes deuteron disintegration (with 
inadvertent fusion). His theory suggests that about 4 GeV of mass-energy per 
every two atoms of deuterium lost could be converted into energy. This is about 
150 times MORE potential energy per unit of mass (converted into energy) than 
can be derived from fusion into helium.

 

On the surface, then – fusion of deuterium into helium appears to be too weak a 
reaction to account for the Mizuno claims of heating his home, and only the 
Holmlid effect would have an adequate output.

 

Why isn’t the Holmlid effect the favored hypothesis?

 

Jones

 

-- 
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

For 100kw/h about 1.2mg of deuterium are needed.

If pressure is lower then the relative density of D (D2 gas) increases, 
somewhere between 0.15 & 0.45g/l.The inventory is given by Ni/pd surface 
bound D, the volume (15l) of the reactor and the pressure factor (=0.003 
for 300Pa) .


But Mizuno recommends to always let the bottle attached and of course he 
did feed additional deuterium if needed.


Holmid does not yet produce any energy. First he must avoid to mainly 
produce positive muons...



Jürg



Am 15.07.19 um 15:56 schrieb JonesBeene:

Reality Check. Surprisingly, nuclear fusion of deuterium into helium seems NOT 
sufficiently energetic to account for the Mizuno claim of heating his home.
Mass is apparently being converted into energy, but how? And what are the 
ramifications of such a low reactor inventory of deuterium gas?
The main contenders for excess energy production would be:
1) D+D -> He
2) Deflation of electrons – i.e. the Millsean approach
3) Disintegration of deuterons into muons – Holmlid’s theory - which is far 
more energetic than fusion in terms of entropy per unit of mass
4) Sequential Coulomb explosions from cluster formation –hypothesis from Hora, 
Miley etc.
5) Any combination or permutation of the above
If fusion of D into He is your choice - then one gram of fused deuterium yields 
10^12J (one terajoule)of energy, but when based on the low operating pressure 
of 100-300 Pa (100 Pa = .001 bar) and the need for low metal loading, as stated 
in his paper - that set of factors represents a tiny fuel inventory, such that 
when completely fused into helium would generate about 278 kilowatt hours of 
equivalent heat.
If Mizuno was producing close to 3 kW continuous to heat his house in a Sapporo 
winter, he could run it for only about 100 hours without a refill if the gain 
was from fusion and the inventory was at the low end of his specs. At any rate, 
if the gain was from nuclear fusion only - then almost all of the deuterium 
would be converted, and the helium ash should be easily measurable.
There should be no need for a cold trap to increase the helium ratio – the 
residual gas after less than a week should be almost all helium, no? Even if 
these calculations are off by a large factor, the helium content should be 
obvious.
IOW – in the naïve assessment of the breakthrough claim of Mizuno – 
specifically the heating of his home – after 100 hours or so of operation, 
there should be a whopping milligram of helium and little deuterium in the 
reactor to measure.

In contrast – Holmlid’s theory proposes deuteron disintegration (with 
inadvertent fusion). His theory suggests that about 4 GeV of 
mass-energy per every two atoms of deuterium lost could be converted 
into energy. This is about 150 times MORE potential energy per unit of 
mass (converted into energy) than can be derived from fusion into helium.


On the surface, then – fusion of deuterium into helium appears to be 
too weak a reaction to account for the Mizuno claims of heating his 
home, and only the Holmlid effect would have an adequate output.


Why isn’t the Holmlid effect the favored hypothesis?

Jones



--
Jürg Wyttenbach
Bifangstr.22
8910 Affoltern a.A.
044 760 14 18
079 246 36 06

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[JonesBeene]

Reality Check. Surprisingly, nuclear fusion of deuterium into helium seems NOT 
sufficiently energetic to account for the Mizuno claim of heating his home.

Mass is apparently being converted into energy, but how? And what are the 
ramifications of such a low reactor inventory of deuterium gas?

The main contenders for excess energy production would be:

1) D+D -> He
2) Deflation of electrons – i.e. the Millsean approach
3) Disintegration of deuterons into muons – Holmlid’s theory - which is far 
more energetic than fusion in terms of entropy per unit of mass 
4) Sequential Coulomb explosions from cluster formation –hypothesis from Hora, 
Miley etc.
5) Any combination or permutation of the above

If fusion of D into He is your choice - then one gram of fused deuterium yields 
10^12J (one terajoule)of energy, but when based on the low operating pressure 
of 100-300 Pa (100 Pa = .001 bar) and the need for low metal loading, as stated 
in his paper - that set of factors represents a tiny fuel inventory, such that 
when completely fused into helium would generate about 278 kilowatt hours of 
equivalent heat. 

If Mizuno was producing close to 3 kW continuous to heat his house in a Sapporo 
winter, he could run it for only about 100 hours without a refill if the gain 
was from fusion and the inventory was at the low end of his specs. At any rate, 
if the gain was from nuclear fusion only - then almost all of the deuterium 
would be converted, and the helium ash should be easily measurable.

There should be no need for a cold trap to increase the helium ratio – the 
residual gas after less than a week should be almost all helium, no? Even if 
these calculations are off by a large factor, the helium content should be 
obvious.

IOW – in the naïve assessment of the breakthrough claim of Mizuno – 
specifically the heating of his home – after 100 hours or so of operation, 
there should be a whopping milligram of helium and little deuterium in the 
reactor to measure.

In contrast – Holmlid’s theory proposes deuteron disintegration (with 
inadvertent fusion). His theory suggests that about 4 GeV of mass-energy per 
every two atoms of deuterium lost could be converted into energy. This is about 
150 times MORE potential energy per unit of mass (converted into energy) than 
can be derived from fusion into helium.

On the surface, then – fusion of deuterium into helium appears to be too weak a 
reaction to account for the Mizuno claims of heating his home, and only the 
Holmlid effect would have an adequate output.

Why isn’t the Holmlid effect the favored hypothesis?

Jones