Quote from JulianBianchi

can

I agree that the sole transition into UDH with the release of hundreds of eV is the way to go. However the comprehension of the energetics of UDH remains incomplete. For example, in Badiei et al (2010): Production of ultradense deuterium: A compact future fusion fuel, a continuous interconversion between RM and UDH is claimed, something that would be hardly possible for a difference of hundreds of eV between the two states.

Holmlid:

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The time variation of the collector signals was initially assumed to be due to time-of-flight of the ejected particles from the target to the collectors. Even the relatively low particle velocity of 10–20 MeV u-1 found with this assumption [21–23] is not explainable as originating in ordinary nuclear fusion. The highest energy particles from normal D+D fusion are neutrons with 14.1 MeV and protons with 14.7 MeV [57]. The high-energy protons are only formed by the D + 3He reaction step, which is relatively unlikely and for example not observed in our laser-induced D+D fusion study in D(0) [14]. Any high-energy neutrons would not be observed in the present experiments. Thus, ordinary fusion D+D cannot give the observed particle velocities. Further, similar particle velocities are obtained also from the laser-induced processes in p(0) as seen in Figs 4, 6 and 7 etc, where no ordinary fusion process can take place. Thus, it is apparent that the particle energy observed is derived from other nuclear processes than ordinary fusion. It is clear that such laser-induced nuclear processes exist in p(0) as well as in D(0). The low laser intensity used here, of the order of 3×1012 W cm-2 makes it impossible to directly accelerate the particles (especially the neutral ones) to high energies. For example, in Refs. [58,59] more than 1019 W cm-2 was used to accelerate heavy ions to > 1 MeV u-1 energies, thus close to 107higher intensity than used here.

If the energy of fusion (14 MeV) does not pack enough energy to accelerate the particles seen at 3/4 the speed of light. How can this 1 kev energy derived from UHD formation do it? It is time for you'll to go back and rethink things.

"I agree that the sole transition into UDH with the release of hundreds of eV is the way to go." This is wrong.

JulianBianchi

My understanding from what Holmlid writes (see an excerpt below, from the same paper I previously cited) is that H(0) absorbs efficiently any kind of energy input and can thus at least in part revert easily to the H(1) form, but since the H(0) is at a lower energy level it will tend to go back to this form again.

Another possible explanation based on more recent observations from his group however could be that while the hydrogen is in that form, something else also happens that causes at least part of the H(0) to temporarily revert back to the H(1) form. For example, In Charged particle energy spectra from laser-induced processes: Nuclear fusion in ultra-dense deuterium D(0) it's suggested that the state s=1 can be spontaneously formed at a low rate from it. There, the H-H distance is 0.56 pm, where it's assumed that fusion (and perhaps other processes interpreted for example as muon emission) is instantaneous, at least for deuterium. The transition of H(0) from the state s=2 (the normal state) to s=1 too liberates energy. The 2.3pm distance of the normal state should also be short enough to cause some amount of D-D fusion by tunneling.

If this is the case however, it would imply that whenever H(0) is formed some sort of nuclear emission would be inevitable. A possible way out would be if this was not the case with protium. So far Holmlid et al. have (to my knowledge) only documented spontaneous emission with D(0) and it's not clear if the same would still happen with p(0). Fusion shouldn't, at the very least.

* * *

Excerpt from Detection of MeV particles from ultra-dense protium p(-1): Laser-initiated self-compression from p(1) (submitted 2012-01)

Quote from Can

JulianBianchi

My understanding from what Holmlid writes (see an excerpt below, from the same paper I previously cited) is that H(0) absorbs efficiently any kind of energy input and can thus at least in part revert easily to the H(1) form, but since the H(0) is at a lower energy level it will tend to go back to that form again.

Another possible explanation based on more recent observations from his group however could be that while the hydrogen is in that form, something else also happens that causes at least part of the H(0) to temporarily revert back to the H(1) form. For example, In Charged particle energy spectra from laser-induced processes: Nuclear fusion in ultra-dense deuterium D(0) it's suggested that the state s=1 can be spontaneously formed at a low rate from it. There, the H-H distance is 0.56 pm, where it's assumed that fusion (and perhaps other processes interpreted for example as muon emission) is instantaneous, at least for deuterium. The transition of H(0) from the state s=2 (the normal state) to s=1 too liberates energy. The 2.3pm distance of the normal state should also be short enough to cause some amount of D-D fusion by tunneling.

If this is the case however, it would imply that whenever H(0) is formed some sort of nuclear emission would be inevitable. A possible way out would be if this was not the case with protium. So far Holmlid et al. have (to my knowledge) only documented spontaneous emission with D(0) and it's not clear if the same would still happen with p(0). Fusion shouldn't, at the very least.

* * *

Excerpt from Detection of MeV particles from ultra-dense protium p(-1): Laser-initiated self-compression from p(1) (submitted 2012-01)

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Relation between ultra-dense hydrogen H(0) and other forms of hydrogen.

The blue arrow indicates the real-time switching between the two forms H(1) and H(0). The axes are not to scale.

https://doi.org/10.1371/journal.pone.0169895.g001

axil

H(0) is an umbrella term implying all hydrogen isotopes, i.e. p(0), D(0) and even T(0) (although I don't know if he's ever tested that).

From Phase transition temperatures of 405-725 K in superfluid ultra-dense hydrogen clusters on metal surfaces (submitted 2016-01)

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[…] H may indicate all isotopes of hydrogen, or the material is indicated more precisely by using p, D or T.

Emission of subatomic particles has indeed been observed inferred with p(0) - for example in the very same paper you linked above - but only using relatively powerful laser impulses. if you check out Holmlid's papers published over the past 2-3 years where spontaneous emission (= without a laser) is mentioned, deuterium is always used, unless I missed something (I can't exclude that it could be the case, admittedly). It would be nice to also have confirmation of spontaneous emission with ultra-dense protium.

Quote from Can

axil

H(0) is an umbrella term implying all hydrogen isotopes, i.e. p(0), D(0) and even T(0) (although I don't know if he's ever tested that).

From Phase transition temperatures of 405-725 K in superfluid ultra-dense hydrogen clusters on metal surfaces (submitted 2016-01)

Emission of subatomic particles has indeed been observed inferred with p(0) - for example in the very same paper you linked above - but only using relatively powerful laser impulses. if you check out Holmlid's papers published over the past 2-3 years where spontaneous emission (= without a laser) is mentioned, deuterium is always used, unless I missed something (I can't exclude that it could be the case, admittedly). It would be nice to also have confirmation of spontaneous emission with ultra-dense protium.

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Recently, much more precise inter-nuclear distance measurements have been done by rotational spectroscopy, giving distances in D(0) with a precision of ±0.003 pm [10]. MeV particles are ejected by laser-induced processes in both D(0) and p(0) [11–13]

Emission of subatomic particles have indeed been observed inferred with p(0). This means that the nuclear processes produced by H(0) is NOT fusion.

As in LENR, once H(0) has be activated by a weak laser shot, it extracts and stores energy from the surrounding nuclear matter and generates mesons with that energy through subatomic particle production.

See

http://english.pku.edu.cn/news_events/news/research/5335.htm

Physicists observe spontaneous symmetry breaking in an optical microcavity

This explains how the H(0) (and LENR) is activated through a change in state of polaritons. These polaritons

form in the electron spin wave that surrounds the H(0).

Quote from axil

This means that the nuclear processes produced by H(0) is NOT fusion.

Not ONLY fusion (in the case of deuterium).

Quote from axil

As in LENR, once H(0) has be activated by a weak laser shot ...

But in the preceding messages I was clearly referring to the spontaneous signal occurring without a laser (or other energetic impulses or shocks) that Holmlid has documented in some publications using deuterium.

I was making a completely different point than what you're trying to write above: that a spontaneous nuclear signal with protium (mesons and muon emission without the application of energetic impulses) has not been documented yet, from what I've been able to read so far.

Quote from Can

Not ONLY fusion (in the case of deuterium).

But in the preceding messages I was clearly referring to the spontaneous signal occurring without a laser (or other energetic impulses or shocks) that Holmlid has documented in some publications using deuterium.

I was making a completely different point than what you're trying to write above: that a spontaneous nuclear signal with protium (mesons and muon emission without the application of energetic impulses) has not been documented yet, from what I've been able to read so far.

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It sounds like you are correct in that laser irradiation is not required to begin muon production, but the intensity of muon production is low.

It is likely that the initial production of the H(0) generates an amount of latent stored energy, an energy of excitation from which muons form and that energy is dissipated over time.

It could be that the formation of the Bose Condensate in H(0) generates energy in the electrons that are expelled from the hydrogen orbits when the meissner effect sets in. Those electrons travel outward from the center of the H(0) at the a speed close to the speed of light.

Fusion through muon catalyzed reactions are produced.

The muon production reaction is based on an optical mechanism that is amplified by the application of light energy.

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Two different sources for producing H(0) have been used

for this study. They are similar to a source described in a

previous publication.28 Potassium-doped iron oxide catalyst

samples (cylindric pellets)32,33 in the sources produce the ultradense

H(0) from hydrogen or deuterium gas flow at pressures

of 10−5–100 mbars. The sources give a slowly decaying muon

signal for several hours and days after being used for producing

H(0). They can be triggered to increase the muon production

by laser irradiation inside the chambers or sometimes even by

turning on the fluorescent lamps in the laboratory for a short

time

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Quote from axil

It is likely that the initial production of the H(0) generates an amount of latent stored energy, an energy of excitation from which muons form and that energy is dissipated over time.

As H(0) forms? I think that is simply a matter that the closer the electrons are drawn in to the core, the tighter the atom is bound together. The change of energy level causes heat to be released.

Muon (and meson) production is a different process that can occur later on when more energy is added to the system. I personally suspect that when deuterium is used some of this energy can also come from the D-D fusion events occurring spontaneously due to the short interatomic distance of this ultra-dense hydrogen matter. Hence, the spontaneous muon signal. There is a brief description of the laser-induced process in the latest paper that Holmlid has submitted (open access; pdf available, excerpt below), but also on another recent one (also open access):

It's certainly interesting from the excerpt from the paper you've cited that just the diffuse fluorescent light of Holmlid's laboratory when turned on (the cell has a transparent window for the laser beam, when used) can reinvigorate the muon signal. However it's not clear if it only enhances a signal that has already started or if it's on its own able to start an inactive cell that has not been triggered yet for muon emission. I think the wording appears to imply the former; if it was actually the latter it would mean that H(0) (generally speaking) is not as stable as sometimes suggested.

On a loosely related note, in https://arxiv.org/abs/1302.2781 it's suggested that D(0) - at the time called D(-1) - and by extension H(0) will react with oxygen and form water when exposed to air. Perhaps this could be a good way to "deal with it" if one doesn't want to have anything to do with possible nuclear reactions/emissions and only take advantage of the heat of formation of UDH which could almost be considered a supra-chemical effect (although I guess that water formation from H(0) would be an equally strong endothermic reaction and therefore would have to be done in a different location).

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[...] The properties of D(-1) have been amply described in the literature cited above and elsewhere. This material is stable on metal surfaces and inside the catalyst used for its formation during several days (Badiei et al. 2010c) in a vacuum (high or medium). However, D(-1) will react with oxygen and form water if exposed to air. It is not created by the interaction of the laser with deuterium in any form which is clear from experiments using just one pulse (Andersson & Holmlid 2011), but is formed by a catalytic process.

Regarding "I think that is simply a matter that the closer the electrons are drawn in to the core, the tighter the atom is bound together."

You are not considering the formation of superconductivity in H(0) as follows:

They may all be characterized as spin-based Rydberg Matter (RM) [2]. This model is based on a theoretical description by J.E. Hirsch [7].

Read J.E. Hirsch [7]

The origin of the Meissner effect in new and old superconductors

Electrons are not drawn into the core, they are expelled out of the core by the meissner effect. The positively charged core attracts electrons until the meissner effect counterbalances them from moving any further inward. The electrons then form a spin wave at this point of force equilibrium between coulomb attraction and the meissner effect.

Regarding: "The change of energy level causes heat to be released."

When the meissner effect pushes the electrons out of the core, gamma radiation is produced by bremsstrahlung. No heat is involved.

axil

I don't think I can competently reply to what you're implying in your latest comment other than saying that I feel it conflicts with Holmlid's explanation/model from his own papers, but since the binding energy of ultra-dense hydrogen pairs is in the order of < 1 keV, wouldn't the radiation produced by Bremsstrahlung, if any, during these transitions be in the form of extreme UV (EUV) / very soft X-rays?

(EDIT: which incidentally is also, if I recall correctly - although I've never studied his theories in detail so I could be wrong - what Randall Mills suggests to be occurring during Hydrino formation. EDIT2: for example see http://www.brilliantlightpower…/EUV-Mechanism-051817.pdf )

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Electrons are not drawn into the core, they are expelled out of the core by the meissner effect.

It's not true - the magnetic field is expelled to surface of superconductors, not electrons.

Quote from Can

axil

I don't think I can competently reply to what you're implying in your latest comment other than saying that I feel it conflicts with Holmlid's explanation/model from his own papers, but since the binding energy of ultra-dense hydrogen pairs is in the order of < 1 keV, wouldn't the radiation produced by Bremsstrahlung, if any, during these transitions be in the form of extreme UV (EUV) / very soft X-rays?

(EDIT: which incidentally is also, if I recall correctly - although I've never studied his theories in detail so I could be wrong - what Randall Mills suggests to be occurring during Hydrino formation. EDIT2: for example see http://www.brilliantlightpower…/EUV-Mechanism-051817.pdf )

MFMP has seen this Bremsstrahlung like radiation which ranges from 20 KeV to 1.2MeV that immediately preceded the onset of anomalous heat production.

See

This plot did not show any k line spikes which marks the identity of the radioactive isotope that is usually seen in the Bremsstrahlung radiation curve. This means that the MFMP Bremsstrahlung was not produced by a radioactive isotope or any target material interacting with electrons.

For example the spectrum of the X-rays emitted by an X-ray tube with a rhodium target, operated at 60 kV. The continuous curve is due to bremsstrahlung, and the spikes are characteristic K lines for rhodium. The curve goes to zero at 21 pm in agreement with the Duane–Hunt law.

I have contended that the 10 nanometer XUV radiation that leads R. Mills to the hydrino theory is produced by the optical cavities that typified polaritons with a 2 nanometer size. Mills might be seeing XUV produced by polaritons generated inside an optical cavity and not by electrons in a degenerate orbit around a nucleus.

This is how polaritons form inside pits, cracks and bumps on the surface of metal.

My vision of what is happening on the surface of the UDH nanowire is that polaritons are forming in the electron layer that surround the positive core of the UDH like they form on the surface of any metallic nanoparticle. These polariton solitons naturally form into many equally sized entangled pockets. The more energy that the polaritons absorb, the smaller that these pockets get and the higher the frequency of the EMF that the carry.

Finally, it seem to me that Holmlid ideas are evolving over time and his ideas are changing as his research proceeds. I wonder if Holmlid has fully integrated the Hole theory of superconductivity into his own ideas.

http://en.wikipedia.org/wiki/List_of_quasiparticles

We can understand from this list of quasiparticles that a quasiparticle can be simple in that it is comprised by only fundamental particles, or complex in that it can be made up of a mixture of fundamental and/or any number of less complicated quasiparticles.

In my opinion, there is a unrecognized complex quasiparticle that produces the LENR reaction. This quasiparticle has been identified in the tracks that it leaves in photo emulsions from LENR ash as a monopole by some LENR researchers or a tachyon monopole by other LENR researchers.

The generic term that is in common use by these LENR researchers for this type of LENR active quasiparticle is the "Exotic Neutral Particle".

One very important instance of this quasiparticle is the EVO coined by Ken Shoulders. Ken said that this particle was both a monopole and a micro black hole.

What I beleive that this LENR active quasiparticle consists of is a metal nanoparticle with a covering of a Bose condinsate of polaritons. The Ultra Dense Hydrogen nanoparticle is one instance of this type of quasiparticle. Other kinds of the like particles are comprised of any transition metal nanoparticle excited by light and/or heat based EMF.

Ken Shoulders did not imagine that electrons cannot aggregate into a dense cluster because they are fermions. But when electrons on the surface of a transition metal nanoparticle are confined along with photons, electrons become bosons and they can aggregate and condense in uncountable numbers.

Ken built these quasiparticles by first vaporizing metal with a spark then when the metal vapor solidified into a nanoparticle and its surface was coated with polaritons generated by the energy from the spark, the LENR active particle (LAP) was born.

Holmlid, Shoulders, Rossi, Mills, Pons and Fleischmann, all of them... together with many others LENR experimenters have all built the LAP and got the LENR reaction to come alive. Like UDH, once created the LAP can persist for days, they are self sustaining as long as they can feed off the energy that they extract from the matter around them. They store energy in the polariton condensate coating and because they are superconducting there is almost no energy loss from diffusion and their energy stores lasts for a very long time.

Shoulders states:

“They can be just about anywhere, it is handy for me because I can analyse them in a vacuum - they can exist in a solid, literally in some of John’s [John Hutchinson] work, in the middle of the middle I’ve seen a paper that showed things like that and many other cold fusion guys. I’ve created them and kept them overnight and when I come in next morning they blow to smithereens - but I think they did some wrong things to make them do those things.”

“They are extremely ubiquitous things, you can shuffle across this road and you will have created them those marks - little marks on that door knob are the witness marks - they are just everywhere, you get out of your car, rub across your seat and snap… you just made one."

"Their physically large enough to see but they are in the diameter of a hair”

…

“I have seen a 5 and 20 um one and my present job is to get them bigger and bigger and bigger”

“I have been able to use these little biddy machines I make to get them up to 100um so far.”

“they are stable unless I intentionally blow them up”

Quote from Can

As H(0) forms? I think that is simply a matter that the closer the electrons are drawn in to the core, the tighter the atom is bound together. The change of energy level causes heat to be released.

Muon (and meson) production is a different process that can occur later on when more energy is added to the system. I personally suspect that when deuterium is used some of this energy can also come from the D-D fusion events occurring spontaneously due to the short interatomic distance of this ultra-dense hydrogen matter. Hence, the spontaneous muon signal. There is a brief description of the laser-induced process in the latest paper that Holmlid has submitted (open access; pdf available, excerpt below), but also on another recent one (also open access):

It's certainly interesting from the excerpt from the paper you've cited that just the diffuse fluorescent light of Holmlid's laboratory when turned on (the cell has a transparent window for the laser beam, when used) can reinvigorate the muon signal. However it's not clear if it only enhances a signal that has already started or if it's on its own able to start an inactive cell that has not been triggered yet for muon emission. I think the wording appears to imply the former; if it was actually the latter it would mean that H(0) (generally speaking) is not as stable as sometimes suggested.

On a loosely related note, in https://arxiv.org/abs/1302.2781 it's suggested that D(0) - at the time called D(-1) - and by extension H(0) will react with oxygen and form water when exposed to air. Perhaps this could be a good way to "deal with it" if one doesn't want to have anything to do with possible nuclear reactions/emissions and only take advantage of the heat of formation of UDH which could almost be considered a supra-chemical effect (although I guess that water formation from H(0) would be an equally strong endothermic reaction and therefore would have to be done in a different location).

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It is not possible for D(-1)/D(0) etc to react with oxygen and form water.

Chemical reactions relate to binding between outer electron clouds - in the case there is only one electron. The hypothesised compact nature of the D(-1) atom means that it cannot bind with the 100's of times larger O2 outer orbitals. If any binding were to happen it would have very very different chemical properties - and no reason to suppose oxygen a good target. And the energy needed to lift and electron back to a normal S1 orbital 9if it is supposed that happens) would be far far larger than any binding energy available from H2O.

This one sentence is absurd - so maybe something is not understood.

The original justification for D(-1) included the neat idea that D nuclei specifically could do this inverse transform, and H not, due to spins. So I'm not much less impressed with Holmlid given he has abandoned this one neat and vaguely justified plank of his work and supposes both D and h can do this. It weakens the entire hypothesis.

THHuxleynew

Perhaps not directly, but "eventually" from the less dense H(1) form (to which the H(0) seems to convert back and forth, to some extent), which would decay to regular H after a few collisions with ions and molecules. But indeed the wording here is not clear and appears to imply a direct conversion.

Quote from can

THHuxleynew

Perhaps not directly, but "eventually" from the less dense H(1) form (to which the H(0) seems to convert back and forth, to some extent), which would decay to regular H after a few collisions with ions and molecules. But indeed the wording here is not clear and appears to imply a direct conversion.

The enthalpy released by the change to H(0) is very high - far far above chemical or thermal energy levels - and therefore I don't see how a few collisions with ions and molecules could provide the necessary energy to reverse the H(1) -> H(0) transition?

Highly exothermic transitions are not easily reversed, which is why explosive reaction products do not tend spontaneously to change back to explosives. The energy here is much larger (10s or 100s of times) than the combustion energy of TNT.

THHuxleynew

Earlier in this discussion a short passage was quoted from one of Holmlid's papers published in 2012 where it was suggested that the H(0) (at the time called H(-1) as it was still proposed to be inverted matter) due to its properties absorbs energy (e.g. light) efficiently and this can cause it to temporarily convert back to the H(1) form.

The H(1) form (aka Rydberg matter) has a much larger cross section for collision with ions and molecules and is much less tightly bound together than the H(0) form and is what I suggested would be reverting to regular hydrogen atoms by that in the previous comment.

How these transitions proceed exactly has not been studied in detail yet by Holmlid's group. Such a study would likely also focus on the large energy that is supposed to be released by the H(0) formation, which has been mentioned just a few times or only implied in the so-far published related literature.

Quote from can

THHuxleynew

Earlier in this discussion a short passage was quoted from one of Holmlid's papers published in 2012 where it was suggested that the H(0) (at the time called H(-1) as it was still proposed to be inverted matter) due to its properties absorbs energy (e.g. light) efficiently and this can cause it to temporarily convert back to the H(1) form.

The H(1) form (aka Rydberg matter) has a much larger cross section for collision with ions and molecules and is much less tightly bound together than the H(0) form and is what I suggested would be reverting to regular hydrogen atoms by that in the previous comment.

How these transitions proceed exactly has not been studied in detail yet by Holmlid's group. Such a study would likely also focus on the large energy that is supposed to be released by the H(0) formation, which has been mentioned just a few times or only implied in the so-far published related literature.

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OK - sorry we have a terminological confusion. If H(1) is 1st Rydberg level than the comparison is of course with normal H. The difference in energy (normal -> H(0) versus H(1) -> H(0) ) is not significant. So could you remind me what is the binding energy or H(0) relative to normal S1 H meant to be? We can then investigate the light hypothesis further.

It does not answer the conversion through collision idea, which remains as far as I can see not even remotely plausible.

THH

You might want to refer to what JulianBianchi reported earlier in the discussion in the very same thread you're writing in: Post #8

JulianBianchi reports that the value of the energy released by the transition from RM to UDH is hard to pin down (sanity check: is this the transition we're discussing here?), but it might be upwards of 600 eV to something in the keV. Considering that EUV photos carry on the order of 10-100 eV, that is a lot of energy that would be required to reverse the process. The "light" would need to be in the soft x-rays.