RNBE 2016 William Collis - a heretical theory involving unusual particles.

@Wyttenbach

could a nucleus in an excited energy state or perhaps in a excited isomer be considered to be a halo nuclei for the brief time it is in the excited state?

i suppose halo nuclei are strictly speaking in their ground state though?

I recall I think that Andrea Rossi and NormanCooks theory requires the Li7 to be in its first excited state to absorbe the proton?

I wonder if the neutron or proton capture cross-section increase in size if the nuclei are in excited states.

I suppose the respective parity spin states of the particle and nucleus may also be important. But perhaps there is also a structure aspect. For example a relatively unbound neutron proton pair could maybe more easily bind to a deuteron perhaps to form a bound alpha particle? Or Inthe case of li7 as per Rossi and Cook a proton + 2 neutron triplet can more easily bind a proton perhaps?

On a separate point: Interestingly although I mentioned above Ba138 is neutron magic with82 neutrons. Ni is proton magic with 28 protons so would not really want to absorb protons. Lighter isotopes of Ni do want to absorb neutrons though. It looks like Ni is getting Protons but wants Neutrons. So if there is a process of transmutation and decay with Ni or evolution towards Ni62 perhaps it is subtly different process than the evolution of Ba to Sm but both requires similar environment with the particles approaching close enough to a nucleus for sufficient time to allow nuclear force interaction or tunneling type event or force interaction.

Whilst on the topic of magic nuclei I'm not sure if the following is relevant but it's a strange coincidence none the less.

in the Iwamura paper discussed above in one of the postings here the deuterium ion production is enhanced with a CaO layer (that is not seen with MgO for example)

It is recognized in that paper that the process is not know but It seems most likely this is a chemical effect somehow enhancing maybe the known dissociation of D2 into atomic d on the surface of the palladium.

i wonder however if there is some other aspect about CaO however that is not purely chemical.

Surprisingly the most abundant natural isotopes of both Ca and O (Ca40 and O16) are doubly magic nuclei both with equal numbers of protons and neutrons. O16 has 8 protons and 8 neutrons (both magic) Ca40 has 20 protons and 20 neutrons also both magic. I'm not sure how these could enhance atomic deuterium production but they certainly would well bound nuclei and not not be thirsty for additional deuterons, neutrons or protons I suppose.

perhaps it is the extra layers of a this relatively well bound nuclei in CaO between the layers of palladium that increases the activity of surface of palladium available for deuteron production. So it is a kind of nuclear/chemical affect that is not inhibiting the process as much as other materials such as MgO somehow. I'm not sure how and if it could work but it's an interesting coincidence that both Ca40 and O16 are both doubly magic nuclei.

Quote from Eric Walker

1991 patent,

I had a quick look at the patent. I have to say that my trust on the scientific value of patents is near to nil. The explanation given for the accelerated decay goes on the lines of your explanation, sure. The problem is the realistic magnitude of the influence of the atomic (i.e. electronic) environment on nuclear decay matters.

Quote from Eric Walker

Another relevant article that you might find interesting (dated 1993): math.ucr.edu/home/baez/physics…dNuclear/decay_rates.html.

The comment at the end of the web page says: “All told, the existence of changes in radioactive decay rates due to the environment of the decaying nuclei is on solid grounds both experimentally and theoretically. But the magnitude of the changes is nothing to get very excited about.“

I am far from an expert of these subjects, but I would trust the comment.

Your attempt Eric seems to me to share the problems of the LENR theories which predict extraordinary concentration of energy in a single very small volume of the lattice. In fact obtaining very high electron densities, which would significantly influence nuclear decay rates, would require extraordinary high energy concentration, which is thermodynamically too unlikely.

Quote from Eric Walker

on what basis do you assert that the electron has structure? Shouldn’t that be something borne out by experiment?

The ZB frequency is very high, so it is not easy measure it, however there are some experiments showing data in agreement with the ZB of the electron:

http://repository.ust.hk/ir/Record/1783.1-19361

http://repository.ust.hk/ir/Record/1783.1-22246

David Hestenes commented this effort in the reference I gave you some comments ago:

geocalc.clas.asu.edu/pdf/SnarkPaper.pp.pdf

Essentially an 80MeV electron beam crosses a thin silicon crystal and at 0[deg] (pass through) there is an 8% dip in the transmission rate at the expected energy/frequency.

You can also have a look at some of these:

https://arxiv.org/abs/0909.0674

http://arxiv.org/pdf/1606.02618.pdf

http://arxiv.org/abs/1403.7037

…

Quote from Eric Walker

Charge independence is not simply a theoretical supposition; it’s an experimental conclusion. It is a conclusion that comes from doing things like noting the isospin of various systems at different energy levels, and then looking at angular correlation in neutron-proton and proton-proton scattering. All else being equal, systems with the same isospin demonstrate the same behavior. This is a very different case from, e.g., string theory, where there’s very little experimental evidence that can be produced to sift between different possibilities. In this case there are lots of experiments to account for.

An example is the deviation from Rutherford scattering, i.e., deviation from a pure Coulomb potential as neutrons and protons are scattered at moderate energies, which is also seen as protons and protons are scattered. Once one factors out the Coulomb repulsion that arises only between two protons, the scattering data look similar. Since the strong force is 1000 times stronger than the Coulomb force, there should be a clear signal.

Quote from Eric Walker

See also ch. 4 of Krane, “Introduction to Nuclear Physics” (a textbook), as well much of Bethe and Morrison, “Elementary Nuclear Theory” (another textbook). It’s fine to take a position at variance with mainstream physics on the question of charge independence, I suppose, but be prepared to be asked to respond to some very difficult experimental questions that point in a different direction. Physicists will ignore you if you don't have good answers for those questions, so it's worth the time to really look into this.

So, charge independence. The absence of a bound state for p-p and n-n comes from the the theory of the magnetic attraction, and I can not change it. As I said, I am discussing this matters with Norman Cook and Valerio Dallacasa, who know much more than me about scattering results with nucleons. I know that p-p (corrected for the Coulomb interaction) and n-n scatterings have similar scattering lengths and effective ranges. Which I suppose is not in contrast with the features of the magnetic attraction mechanism. p-n scattering has instead slightly different parameters.

The magnetic force hypothesis predicts a dependence on reciprocal spin orientation and can depend on the momentum of the nucleons for high energies. However it is not clear if it can be responsible also for the repulsion between nucleons.

Anyway I think you are right about physicists ignoring the magnetic force proposal, and, with it the whole EMNR theory, if the model does not answer all detailed questions correctly. I am aware of this. I will try and study more in this direction.

I would like to stress however that my contribution can not be in going into the details of the magnetic attraction mechanism and compare it with a mole of scattering and other experiments and nuclear data. This could be done by professionals in the field. I do something very different for a living. My contribution instead can be in suggesting that, if mass is rotational frequency, and identical (or multiple) rotational frequency means attraction, the electron with its mass (and intrinsic frequency), can feel the nuclear force. This requires an orbital contribution for the “frequency match” which can be provided by some sub-valence electron orbitals. If one uses the ionization energies as proxy of the energy/frequency of these sub-valence orbitals, the list of the best atoms starts with Osmium, Calcium and Palladium, and Zr is not far. Osmium is rare, expensive and nasty, so it has never been used in LENR experiments (as far as I know). But Ca and Pd are the base of the device of Iwamura, who has the only device activated by diffusion only(!). And Pd is the material which started the whole story ...

At 2 fm the Coulomb force is not 1000 times smaller than the nuclear force. It is smaller, but by less than one order of magnitude.

Quote from Eric Walker

How do you differentiate between an electron trapped in the Coulomb potential of a positively charged proton (i.e., an atom) and an electron-proton system that forms a neutral (and very large) composite particle (a nucleon)? Is the neutron also composed of an electron and a proton in this scheme, with the electron orbiting within a much smaller radius? If not, why not?

Is the binding energy of the Hyd a smoothly varying function, or is it quantized into energy levels? Or is it constant?

If the force that binds the electron to the proton in the Hyd is magnetism, does the Hyd have a large magnetic moment?

An electron in an atomic orbital cannot feel the nuclear force because it doesn’t have the correct orbital frequency and, when it is near it, it is shielded by the other core electrons. So it resides in an orbital, which is defined by an Hamiltonian contemplating only kinetic and coulomb (plus spin, …) energies. The electron in the Hyd instead feels a stronger force and its Hamiltonian has a “magnetic attraction term” which sort of “prevails” on the kinetic energy operator (the Coulomb energy is anyway attractive). It is this term which I guess could be the additional term missing in the Hamiltonian for high temperature superconductivity. This term should be responsible for the otherwise inexplicably strong electron-phonon necessary for the formation of Cooper pairs at high temperature. If sub-valence orbitals, which are the “floor” for metallic (and superconductive) orbitals, react strongly to phonons, the (metallic)electron-phonon coupling should appear.

I regard the Hyd as a neutral nucleus, because the mechanism which keeps the electron and the hydrogen nucleus together is the same which keeps nucleons together in ordinary nuclei.

A neutron is a totally different thing. It is almost as a proton, small, massive … and differs from the proton not only by an electron, but also by an anti-neutrino. I guess the neutrino component is not electromagnetic, so Hyd and neutron are two very different objects.

Energy levels in the Hyd. Excellent question! You are the first to ask this. The equations say that the Hyd could have energy levels, since the frequencies of the electron and of the hydrogen nucleus have only to be multiple of each other, but the multiplier is free.

Attached you can find the table of the first “possible” energy levels.

Condition I is when the frequency of the electron plus the orbital contribution is equal to the frequency of the proton divided by the integer in the first column. Condition II is when the frequency of the proton plus the contribution divided by the integer is equal to the electron frequency.

The energy for the formation of the Hyd at its lowest energy level is 85 eV. But then the Hyd could absorb a photon of 557[eV] (or 556.4 when inverting the rotation direction) and reach the next energy level (one upper and one lower in the multiplier scale) which has a higher spirograph-like component.

Condition II in chemical system is clearly unreachable.

It is interesting that the first excitation level in Condition I and II could have been seen respectively by Holmlid and by the people of the p+Li7->Be8 internal pair production anomaly.

Still quite an open question, but different energy levels would mean a simple way of detection of the Hyds!

The magnetic moment of the Hyd should be practically the sum of the magnetic moments of its components, like any nucleus. So its magnetic moment should be very similar to that of the electron, which is very large in nuclear terms. You can find this on slide 14 of my presentation. The large magnetic moment should be responsible for the significant deviation in the angular correlation in the internal pair creation from protons against Li7. Plus the large magnetic moment should also allow to trap Hyds in solids or magnetic traps. Explosions in LERN experiments should be essentially due to the accumulation of Hyd in condensed matter.

Quote from StephenC

perhaps it is the extra layers of a this relatively well bound nuclei in CaO between the layers of palladium that increases the activity of surface of palladium available for deuteron production.

Some years ago the Japanese calculated the 2x2x2 grid-cell of Pd(Zr)LiH lattice with the addition of Ca. The result was an increase of about 200eV Cell energy. That allows the conclusion that if Ca moves out of the grid a potential of 200eV most be redistributed! This 200eV could be used to activate a LENR reaction.

Quote

I had a quick look at the patent. I have to say that my trust on the scientific value of patents is near to nil. The explanation given for the accelerated decay goes on the lines of your explanation, sure. The problem is the realistic magnitude of the influence of the atomic (i.e. electronic) environment on nuclear decay matters.

Yes, the patent provides little scientific value; it is one of the initial sources for my suggestion, so I just want to give Barker credit and let him elaborate on the general idea. Above you’re merely echoing the common wisdom that the atomic environment has little effect on nuclear decays; the question at issue is whether that wisdom is correct.

Quote

The comment at the end of the web page says: “All told, the existence of changes in radioactive decay rates due to the environment of the decaying nuclei is on solid grounds both experimentally and theoretically. But the magnitude of the changes is nothing to get very excited about.“

I am far from an expert of these subjects, but I would trust the comment.

Again, I’m just giving credit where credit is due, in this case to Bill Johnson. I assume Johnson is incorrect about that last suggestion. But perhaps if you like that suggestion you will also trust the wisdom in these ones?

• “We might in these [nuclear] processes obtain very much more energy than the proton supplied, but on the average we could not expect to obtain energy in this way. It was a very poor and inefficient way of producing energy, and anyone who looked for a source of power in the transformation of the atoms was talking moonshine.” (Rutherford.)
• “There is not the slightest indication that [nuclear energy] will ever be obtainable. It would mean that the atom would have to be shattered at will.” (Einstein.)
• “X-rays will prove to be a hoax.” (Lord Kelvin.)
• “There is nothing new to be discovered in physics now, All that remains is more and more precise measurement.” (Also Lord Kelvin!)

(Quotes stolen from this link.)

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Your attempt Eric seems to me to share the problems of the LENR theories which predict extraordinary concentration of energy in a single very small volume of the lattice. In fact obtaining very high electron densities, which would significantly influence nuclear decay rates, would require extraordinary high energy concentration, which is thermodynamically too unlikely.

Thermodynamics tells us that concentration of electron density would not happen spontaneously. It has little to offer us as a guide as to what would happen under dynamic perturbations from external forces. I’m suggesting that dynamic conditions brought about in the external environment might produce far more significant changes in electron density than thought up to now. Here we’re talking not about static systems but instead dynamic ones. The energy for concentrating the electrons would presumably come from external stimuli of various kinds, such as varying magnetic fields and arc discharge. What would be important would be that they induce a movement of electrons from one location to another. Forces like z-pinch might constrain the current to a narrow waist. Anything at that waist would be subject to higher electron density. Another possibility is for electrons to pool in great numbers at sharp protrusions and other surface defects. Again we’re considering the dynamic case and not the static, equilibrium case.

The assumption at the present time seems to be that electron density is very static and hard to change. But perhaps that is just laziness going back to the fact that it’s hard to model solid state systems. There might be far larger swings in electron density than understood up to now, for very brief moments. As long as the timespan for those swings is long in comparison to the timespan of the strong interaction, I think things could get interesting.

Quote

The ZB frequency is very high, so it is not easy measure it, however there are some experiments showing data in agreement with the ZB of the electron:

I see what you’re referring to as “structure” — the language Hestenes and possibly others use is along the lines of “particle substructure in quantum mechanics.” I misunderstood “structure” to mean internal structure, within the electron, normally considered an elementary (simple) particle. I have no opinion on either question, but it’s good to be clearer on what you have in mind.

Quote

So, charge independence. The absence of a bound state for p-p and n-n comes from the the theory of the magnetic attraction, and I can not change it. As I said, I am discussing this matters with Norman Cook and Valerio Dallacasa, who know much more than me about scattering results with nucleons. I know that p-p (corrected for the Coulomb interaction) and n-n scatterings have similar scattering lengths and effective ranges. Which I suppose is not in contrast with the features of the magnetic attraction mechanism. p-n scattering has instead slightly different parameters.

Wouldn’t similar scattering lengths and effective ranges for p-p and n-n scatterings suggest that protons (with charge) and neutrons (with any internal charge canceling out) are interacting with one another in a way that is independent of charge, rather than an outcome of it, since the net charge is dramatically different in each case?

The p-n scattering parameters must be similar enough to the p-p and n-n scatterings for physicists to arrive at the overwhelming conclusion that the nuclear force is an attractive (and at close ranges repulsive) force independent of the electromagnetic interaction at energies below that dictated by a unified field theory that has yet to be discovered. You might disagree with the conclusion; but perhaps you’ll agree that physicists are strongly convinced of it?

Quote

The energy for the formation of the Hyd at its lowest energy level is 85 eV. But then the Hyd could absorb a photon of 557[eV] (or 556.4 when inverting the rotation direction) and reach the next energy level (one upper and one lower in the multiplier scale) which has a higher spirograph-like component.

I just noticed that another word that can be used here is “rosetta”.

When the electron bound in a Hyd absorbs a photon, will it will acquire one unit of angular momentum? In atomic orbitals, the additional orbital angular momentum leads to a transition from, e.g., an s-orbital (no angular momentum) to a p-orbital. The orbital angular momentum is separate from the intrinsic angular momentum, and it forces in the electron into occupying the strange and remarkable shapes described by the spherical harmonics:

When the electron partner in the Hyd picks up a unit of angular momentum by absorbing a photon, why would it continue on as before in its spirograph-like pattern and not change to something else as a result of the additional orbital angular momentum?

Quote

The magnetic moment of the Hyd should be practically the sum of the magnetic moments of its components, like any nucleus. So its magnetic moment should be very similar to that of the electron, which is very large in nuclear terms.

I think the orbital angular momentum will also be relevant in obtaining the magnetic moment?

Quote from Andrea Calaon

The path from a stable nuclear state/nucleon/set of nuclei to another eventually stable state/... with lower energy is almost always full of intermediate states which are not stable and emit "radiation".

Most nuclear reactions "almost always" fail to conserve linear momentum and spin without gamma emission. The fact that there is an unexpected class of reactions which do not predict radiation is intriguing. I feel you still haven't explained why energy considerations cannot explain lack of radiation. In order for Collis' ENP reactions to produce gammas, there would need to populate a low lying excited state of the appropriate spin. These are unlikely, so cases probably haven't been discovered yet. Incidentlly your comment would apply to your own theory too.

Quote from Eric Walker

When the electron bound in a Hyd absorbs a photon, will it will acquire one unit of angular momentum? In atomic orbitals, the additional orbital angular momentum leads to a transition from, e.g., an s-orbital (no angular momentum) to a p-orbital.

As far as I understand the theory, hyd level electron orbits are below QM minimum levels. R. Mills, with the help of semi-classical analysis, showed that hyd states posses no possibility to radiate:

Citation: The current density function possesses no spacetime Fourier components synchronous with waves traveling at the speed of light; thus it is nonradiative.
This is for gammas emitted by electrons only!!

These facts are in line with many LENR experiments which show that the ignition is primarily of kinetic (Delta function at nuclear level?) nature.

Quote from Hermes

Most nuclear reactions "almost always" fail to conserve linear momentum and spin without gamma emission.

What about the neutrinos?
Halo nuclei states are possible able to couple to the surrounding lattice and may "dicipate/mediate" momentum to the sea.

Quote from Wyttenbach

As far as I understand the theory, hyd level electron orbits are below QM minimum levels. R. Mills, with the help of semi-classical analysis, showed that hyd states posses no possibility to radiate:

Then you should let Andrea know, because he's suggesting that the electron partner in the Hyd might be able to absorb a photon and transition to a higher energy level. If the electron partner cannot radiate a photon, then presumably it cannot absorb one either? (This is one of several things I find most implausible about Mills' theory.)

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In fact obtaining very high electron densities, which would significantly influence nuclear decay rates, would require extraordinary high energy concentration, which is thermodynamically too unlikely.

Polaritons charge electrons into bosons when electrons entangle with photons. Any amount of polaritons can produce very high densities when pumped vigorously.

When electrons are in the form of polaritons, electrons don;t need to orbit the nucleus. Polaritons combine together in a condensate call a soliton.

Quote from Eric Walker

(This is one of several things I find most implausible about Mills' theory.)

From a QM perspective you are completely right. It's is more than doubtable that most of the intermediate states (lower orbits) he describes are stable orbits. Further on I tried to developed a statistics, which is only possible if you assume many-body reactions. Nevertheless a kind of statistics for the first 13 states is possible.

But interestingly enough (like halo nuclei orbits) these lower orbit resonances do exist. (Confirmed by many secondary papers.)

R.Mills model is semi-classical and assumes the electron is traveling around a nucleus as a plain wave. Which proves to be correct with all QM consequences.
(Possibly the QM probability orbits can be deduced by applying Thomas precision.)

Finally he assumes a classical non elastic contact as a mechanism for energy-transfer. This "works" for the donor H* quite well, but for the receiver of the the energy we must find a way how this additional energy is disposed of. He usually measured soft X-rays - but the rest of the story is?

For me these points are all open questions, which must be explored in deeper tests.

From a QM point of view "really deep" orbits are allowed (Paillet, Meulenberg) and my current best guess is, that most of the intermediate states end up in the deep ones causing a part of the known LENR phenomens.
What definitely is impossible: A low orbit hyd state can not be reached by absorbing a photon. In contrary the enrgy must be released!

Quote from Wyttenbach

What definitely is impossible: A low orbit hyd state can not be reached by absorbing a photon. In contrary the enrgy must be released!

This seems ordinarily intuitive on conservation of energy grounds alone. However, if coherence is preserved then a specific phase relationship might allow photonically induced down-conversion. Where is the energy going? What about change of mass of one or another participant?

Quote from Longview

However, if coherence is preserved then a specific phase relationship might allow photonically induced down-conversion. Where is the energy going? What about change of mass of one or another participant?

Then a part (the greater) of the energy must go into the potential energy, which is only possible if e.g. the charge (central field) slightly increases. The other problem is momentum conservation...

Mills is arguing in this direction. What would happen if e.g. for relativistic H* electrons the charges increase? The logic behind this is, that a faster (fatter ?) electron does better screen the central field. Thus from the exterior we couldn't see any difference. (Of course we could, with the right experiment.)

But up to now absorption of energy always leads to an increase of charge radius!

Quote from Longview

This seems ordinarily intuitive on conservation of energy grounds alone. However, if coherence is preserved then a specific phase relationship might allow photonically induced down-conversion. Where is the energy going? What about change of mass of one or another participant?

And another possibility, from laser mediated refrigeration--- dear to "classical" rydberg matter work: Photon absorption then re-emission at a higher energy, thus cooling the recipient ion. Photonic up-conversion, if you will.

Quote from Longview

Photon absorption then re-emission at a higher energy

This is ok.
The question was how you can add 85 eV and lower the electron trajectory to get H*? - That's phantasy.

You must remove (re-emit!) 85 eV, which only could work for the transversal kinetic Mills process.

If You look at Holmlids inverse Rydberg matter, with a distance ( radius?) of 2.3 pm then the energy loss of one H* - using Mills classical formulas - is in the range fo 850 eV. What a coincidense with other findings...

Quote from Wyttenbach

This is ok.
The question was how you can add 85 eV and lower the electron trajectory to get H*? - That's phantasy.

You must remove (re-emit!) 85 eV, which only could work for the transversal kinetic Mills process.

If You look at Holmlids inverse Rydberg matter, with a distance ( radius?) of 2.3 pm then the energy loss of one H* - using Mills classical formulas - is in the range fo 850 eV. What a coincidense with other findings...

Gamma energy is stored on the SPP spin waves that accumulate on the surface of the Rydberg matter. This energy is released in two ways, heat( hackings radiation) and SPP radiation whose wavelength is proportional to the density of SPP accumulation. As more energy is accumulated, the SPP population increases and the frequencies of the stored EMF also increases in proportion. The frequency of the stored energy does not exceed soft x-rays.

Wyttenbach said " 85Ev ..is phantasy".

My understanding of the EMNR theory is poor.
this is my take on it. I see things more in chemical terms.

The hyd formation is actually exothermic. I could be wrong. Perhaps Andrea can correct me.
The Bohr atom is based on Coulomb force modified by the DeBroglie standing wave model
The idea is that as you get closer to the nucleus the energy well gets deeper. But the Bohr atom model ignored the magnetic force.
As you get close to the nucleus the magnetic plus electrostatic force can give repulsive or attractive effects.
See Paillet www.iscmns.org/work11/12 Paillet-EDOH-Full1.pdf . Fig 1.
For the zitterbewegung this magnetic force of a charge spinning at velocity c is has a large effect and is phase dependent.
At large distances it approximates to the Coulomb force.
See Lush. 2016 http://arxiv.org/abs/1409.8271
What Andrea is suggesting, I think, is that close to the nucleus at 193 fm there is an energy well or racetrack at 85 EV energy level.due to the magnetic phase harmony between the two zitters.. the proton and the electron.
This 85 eV level is lower than the 13.6 eV Bohr ground level
De Broglie model and possibly zitter phase considerations normally prevent the electron from sliding down
from the ‘ground ‘ level to this racetrack.
However the calculations of even approximate energy profile due to electrodynamic effects at the speed of light close to the nucleus is super difficult.

Global Energy calculations

For Zirconium mediating the hyd formation
E1. Photon (80.35ev) +Zr(+4) --> Zr(+5) +e- = ionisationenergy for Zr(+4) endothermic.
E2. Photon (85eV) + H-e --> --> proton + e- = ‘ionisationenergy’ for hyd, endothermic.
E3. proton + e- --> Photon (85eV) + H-e = reverse of E2 = Hyd formation energy , exothermic
Andrea postulates the formation of H-e or hyd via Zr (+4)
What is the Energy change? Calculate by E1 +E3..

Proton + Zr(+4)--> H-e + Zr(+5) + photon (4.65eV) ,corresponds to 270 nm... deep UV emission.

For Nickel mediating the hyd formation:For Ni(+4), a similar treatment yields a photon of 139 nm..... even deeper UV.

I conclude that,as I think Andrea has mentioned somewhere during hyd formation from there should be emission of deep UV . this is due in my opinion to
the excess of Hyd formation energy released over the energy required for the removal of electrons from the transition metal ions, Zr and Ni.

HUP means heisenberg uncertainty principle.

Andrea Rossi
August 29, 2016 at 7:25 AM
JP Renoir:
Absolutely not. Electron capture in LENR is impossible. It can happen only with atoms overweighted with protons, and this is not our case.
Warm Regards,
A.R.

/* Electron capture in LENR is impossible. It can happen only with atoms overweighted with protons, and this is not our case. */

There are many other LENR & transmutation systems, not just E-Cat. For example Patterson/Niedra/Notoya have dealt long time with potassium carbonate electrolysis, where the electron capture mechanism is quite possible (potassium itself decays with electron capture mechanism). Electron capture would lead into much smaller energy yield than the actual cold fusion though.

Quote from robert bryant

What Andrea is suggesting, I think, is that close to the nucleus at 193 fm there is an energy well or racetrack at 85 EV energy level.due to the magnetic phase harmony between the two zitters.. the proton and the electron.

R.Mill extensively describes the energy transfer between a catalyst and hyd (H*, H#). Andreas proposal is theroef just a cut and paste!

The general problem with an energy quantum of 85eV (81.6eV + band gap as Mills) is, that a catalyst (Ca, Zr,Pd,Fe..) is able to accept this energy as a classical quantum, but hyd is not allow to release it!
Mill's proposal is that hyd, H* releases it's energy in a so called "catalytic reaction" (momentum transfer) based on classical physical laws. (May be a Feynam description would fit better)

Further: At 193fm the energy level of an electron is some keV not 85 eV. Deep electrons travel just along the proton core and are much closer than 193fm.
The speed of light can not be reached by any particle with a finite restmass. But a captured photon is travelling at the speed of light.