The hyd formation is actually exothermic. I could be wrong.
This is a very clear and easy account of the various types of metal hydrides, including information on exothermy.
RNBE 2016 William Collis - a heretical theory involving unusual particles.
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This is a very clear and easy account of the various types of metal hydrides, including information on exothermy.
Hi Alan: This is a terminological overlay of hyd... Deep orbit H* etc. was meant...
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Above you’re merely echoing the common wisdom that the atomic environment has little effect on nuclear decays;
Yes, I am echoing something that comes from the expertise of most physicist who worked in the field, I guess. This matter has been explored by hundreds of carefully checked experiments, and non-equilibrium-thermodynamics does not go that far in increasing electron density.“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!)Let me try to reply to these unlucky forecasts:
- Rutherford was commenting on the work of two of his students who had bombarded lithium with ACCELERATED protons, and goes back to 1932. From Wikipedia: “accelerator-induced fission of light elements remains too inefficient to be used in this way, even today”. The phrase of Rutherford sounds to apply to any nuclear reaction, but it was based only on that experience with accelerated protons. Neutrons were discovered by James Chadwick the same year of the unlucky phrase of Rutherford.
- The phrase of Einstein is about a technology, which at that time would have had to be based on an unknown mechanism. In fact he said “indication”, not that it was impossible.
- Lord Kelvin was simply an arrogant chap.
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
My experience is that thermodynamics has a lot to say about non-equilibrium. Going outside the domain of non-equilibrium thermodynamics inside condensed matter requires extraordinary conditions … possibly those generated by well orchestrated nuclear reactions. The concentration of energy near to nuclei for very high electron densities based on chemical processes remains utterly unlikely in my opinion. To obtain high electron densitites the electron orbitals near to the nucleus of relatively heavy atoms should be squeezed inside. But chemistry can hardly touch orbitals beyond valence ones.
In my opinion you are following the impossible path of energy concentration for your theory. Sharp protrusions, z-pinch, … what about non-linear phonos? I think you are betting on the wrong horse. Chemical processes (in their widest definition) can not access the nuclear well without specific mechanisms outside the standard model. I am repeating myself …
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?
Here I do not exactly understand this question.
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?
Yes, absolutely, definitively, … yes.
A fact is however that p-p and n-n are not bound, while deuterium, although weakly bound in its triplet state is stable. Moreover if nucleons arrange in nuclei in the fcc isospin - antiferromagnetic lattice, as suggested by Cook, the data about scattering from nuclei (above hydrogen and neutrons) could well be very similar to that of a charge-independent nuclear force, despite having only p-n attractions, I guess.
I just noticed that another word that can be used here is “rosetta”.
I prefer spirograph. Words reminding of childhood are always better for our brains, I think.
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:
The electron in Hyds cannot exist into orbitals because the very existence of an electron IS its ZB (the ZB gives it mass, inertia, magnetic moment, …), so changing it would mean to destroy the electron. Therefore the only thing that could happen is the increase in the (spirograph) orbital component. The orbital momentum of the Hyd must be quantized, I agree. I am not sure however if it can change value. It should instead remain equal to the sum of the spin angular momentum of the electron plus that of the hydrogen nucleus, like in any nucleus. The orbital component (due to the spirograph-like movement) is very tiny and should in some way be “eaten” by the electron spin, analogously to what happens to the ZB radius seen by an external observer which shrinks when the electron reaches relativistic speeds relative to that observer. I should investigate better these matters.
I think the orbital angular momentum will also be relevant in obtaining the magnetic moment?
If you actually calculate its contribution it is a minuscule component. In fact it corresponds to the unitary charge (e) travelling at the electron ZB radius (193[fm]) with a frequency which is much lower than the ZB electron frequency, and so it generates a correspondingly lower magnetic moment (the ZB motion generates the magnetic moment of the electron). The frequency ratio (here I write the circular frequencies) is about 2E16/2.5E20=8E-5, so the orbital component is this tiny fraction of the electron magnetic moment.
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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
The Hyd (if it exists) must be defined by an equation analogous to the Shrödinger/Pauli/Dirac equation, but with an additional energy term which prevails over the electrostatic energy. The Hyd is not the Hydrino. You keep considering the Hyd an Hydrino, but it is not so.
These facts are in line with many LENR experiments which show that the ignition is primarily of kinetic (Delta function at nuclear level?) nature.
Ignition of kinetic nature? Could you please be more specific? Delta function at nuclear level? What do you mean?The question was how you can add 85 eV and lower the electron trajectory to get H*? - That's phantasy.
What definitely is impossible: A low orbit hyd state can not be reached by absorbing a photon. In contrary the enrgy must be released!
You do not add 85 eV. The initial 85 eV necessary to extract the electron from its orbital are given by the nuclear force (magnetic attraction) between the electron and the hydrogen nucleus. Then the rest of the binding energy of the Hyd is liberated in chunks with energy near to 85 eV. The binding energy of the Hyd should be in the hundreds of keV range.You must remove (re-emit!) 85 eV, which only could work for the transversal kinetic Mills process.
We've understood you favour Mills' theory ...
Transversal kinetic? If you think Mills' is the correct theory for LENR, why bother analysing other theories and try to reinterpret them through Mills'?But up to now absorption of energy always leads to an increase of charge radius!
Increase in charge radius? What do you mean? Anything seen travelling at relativistic speeds contracts, but does not increase. -
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!Please read my presentation before commenting. I would appreciate it.
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Please read my presentation before commenting. I would appreciate it.
I have read it some time ago.
If an electron accepts energy its "charge radius expands". You assume that the 85eV are perturbative energy. But to reach a charge radius of 193fm, where the magnetic energy is large enough, the electron first must release a huge amount of energy, to get down to a deep level.
This is the process - how to reach the deep Dirac level - you must define first!
Regarding Mill's. , I already said that there is no simple statistics suporting hydrino-levels "below" the first 13 states and thus his theory is to simple. But the upper level resonances he postulates exist, as different experiments have shown.
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Increase in charge radius? What do you mean? Anything seen travelling at relativistic speeds contracts, but does not increase.
When a low energy photon is absorbed by an electron orbiting a hydrogen nucleus with energy insufficient to ionize the electron, the electron is kicked out into a highly eccentric Rydberg orbital, barely bound, with a radius possibly 100,000's of thousands of times that of a normal hydrogen atom. When a photon in the keV is absorbed by an inner shell electron in a heavy nucleus, which is orbiting at relativistic velocities, the electron is either ionized or is kicked out into a shell with large n, after which the hole that was formed is rapidly filled. When the electron orbiting a Hyd absorbs a photon in the ~ 90 eV, the electron orbital assumes a smaller volume. Why do electrons in Hyd's play by different rules than normal electrons, even relativistic ones? Is this not a hint that the system is unphysical? -
@Andrea Calaon. I'm wondering why you need to "abandon slightly" the standard model with regard the strong force to explain your theory?
i appreciate that you consider the strong force as an EM force but is this necessary to explain your ideas?
Could the magnetic field in your ideas be equally explained as originating in the residual spin of the nucleus?
the strong force normally has a very range limited to a few fm and is not really significant at the kinds of ranges in the hyd. It also explains well the relative sizes of nuclei made up of neutrons and protons and why an alpha particle is more stable than a Deutron say. That's not to say there is not some underlying EM force behind the strong force and its short range behaviour but I wonder if it is necessary for your ideas.
it is well known that the residual spin and angular momentum of the nucleus can affect the dipole and quadropole nature of the Coulomb barrier effectively its electric field. But could it have other longer range magnetic field effects?
If I remember right the magnetic quadropole field strength drops off faster with distance than the electric field 1/r3 instead of 1/r2? So would be come much stronger closer to the nucleus i suppose for nuclei with very large neutron cross section this could be very high close to the nucleus?
curiously neutron capture and scattering cross-sections can be very large up to several 10,000 barns in some cases this is much larger than the nuclei size the neutron wave function size and the strong force range. I think that they are also large for nuclei with large residual spin I maybe wrong and looking at it too simply but It seems to me in fact the active force catching the neutron in this case is the magnetic field.
Could it be this effect that is behind the hyd and also attracting nucleons and the hyds closer to the Coulomb barrier? Or is it really necessary to replace the strong force with an EM force to explain the effect?
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Yes, I am echoing something that comes from the expertise of most physicist who worked in the field, I guess. This matter has been explored by hundreds of carefully checked experiments, and non-equilibrium-thermodynamics does not go that far in increasing electron density.
Are you basing your impression that the scenarios we were discussing, non-equilibrium electron densities under strong external perturbations at surface defects and in currents, have been thoroughly explored and found uninteresting on the basis of the investigations of people you know? Perhaps people one step removed? Are the people with those negative impressions two steps removed? You probably know the name of one of these investigators that has shown these two specific scenarios to be uninteresting. Or is this instead a vague, general impression that physicists on the Internet have looked into all of this stuff and it’s not really interesting, but their work was rigorous because it’s all physics anyway, and it doesn’t matter if the experiments were not exactly pertinent, because we have a hard time modifying the density distribution of valence electrons?
Hopefully those hundreds of physicists, who have shown that nonequilibrium electron densities under strong external perturbations at surface defects and in currents are very uninteresting in the present context, will have done a better job cross-checking one another’s work than the engineers who built this bridge, or the space shuttle Challenger.
QuoteEric Walker wrote:
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?Here I do not exactly understand this question.
The idea was that in the system you’re describing, where the residual nuclear force is actually a magnetic force arising from the movement of point charges in the nucleons, you have, then, neutrons and protons. The circulating point charges in the protons have a net electrostatic charge of 1e. The point charges in the neutrons have a net electrostatic charge of 0e. Does this not have an important effect on the magnetic fields that are generated by those circulating point charges? Wouldn’t the magnetic field set up by the point charges whose sum is 1e (a proton) be larger than that set up by the point charges whose sum is 0e (a neutron)? But if so, that would seem to work against the experimental conclusion of the charge independence of the residual nuclear force.QuoteThe orbital momentum of the Hyd must be quantized, I agree. I am not sure however if it can change value. It should instead remain equal to the sum of the spin angular momentum of the electron plus that of the hydrogen nucleus, like in any nucleus. The orbital component (due to the spirograph-like movement) is very tiny and should in some way be “eaten” by the electron spin, analogously to what happens to the ZB radius seen by an external observer which shrinks when the electron reaches relativistic speeds relative to that observer. I should investigate better these matters.
In nuclei, there is the spin-orbit interaction, which gives rise to a noncentral force on the nucleons, causing some of them to move about in a way that deforms the spherical symmetry of the nucleus. In heavy nuclei this is readily apparent in the very deformed, ellipsoid shapes of the nuclei. Similarly, the orbital angular momentum of an electron adds to the intrinsic angular momentum a non-central tendency which deforms (radically) a movement that otherwise could be characterized solely in terms of the distance of the electron from the nucleus (i.e., a spherical orbit).In the case of the ZB, where the electron is proposed to have a spirograph-like movement, why would the addition of orbital angular momentum not similarly twist the spirograph-like movement into something radically different, following the analogy of the spin-orbit interaction of nucleons or the non-radial orbits of electrons in normal shells? Is the orbital angular momentum in the case of ZB so tiny that the spirograph-like movement does not change appreciably? Why is there such a large deformation of the orbit in the case of the nucleon and the normal electron orbits, but not in the case of the ZB?
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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.
Collis explains the lack of radiation through these three properties of LENR reactions:
- There should be two reactants (no decay and no multi-body reactions) and more than one product, in order to have the chance to conserve linear momentum and avoid prompt gammas. In fact whenever momentum is not conserved the excess energy of the daughter is liberated in single large quanta (prompt gammas).
- Spin should be conserved, again to avoid prompt gammas, which would be needed to take away the excess of spin angular momentum.
- One of the two reactants must be neutral.
- One of the products must be neutral.
- The products must not be unstable decaying nuclei, i.e. the reactions should go straight to the final and stable nuclei, “jumping” the formation of unstable isotopes on the reaction path.
Neutrons cannot be the neutral agents for LENR because they do not cause only reactions following rules 1 and 2, they do not select only little energy differences and they cannot select reaction paths with no radioactive nuclides. So there must be a new neutral particle with some properties which cause the 5 special features.About the preference for stable nuclides, the presentation of Collis says:
“We start from stable isotopes so if reaction energies are intrinsically low (not the case for nuclear captures) then there is a chance the products will be stable also.
Most, (but not all) beta radioactive products produce gammas which would be lethal at a few Watts.”Collis speaks about a process in which a nucleus does not capture a proton, nor a neutron. He says “we are doing something much less energetic”, and this is where his tale becomes mysterious (at least to me). My theory explains qualitatively the energy fractionation (as due to the neutral particle, but Collis does not mention my theory in his talk. So what are the less energetic reactions he is talking about?
Nuclear reactions essentially consist in changes to the number of neutrons and/or protons inside nuclei. Most of these changes liberate large quanta of energy which should be clearly “visible” in LENR experiments (proton or neutron capture/emission, beta decay (plus or minus), alpha emission and fission, all well dressed with gammas). Collis is thinking instead about nuclear reactions which:
- by following rules 1 and 2, do not emit gammas;
- produce stable nuclei directly.
He says that with “gentle” changes stable nuclei should transmute into other stable nuclei. However nuclear reactions are never “gentle”, and, apart from the LENR exception, nothing prevents the generation of radioactive nuclides if these happen to be on the reaction path, even from processes liberating relatively small quantities of energy. The only escape from this dead-end road for me is the existence of a mechanism for energy fractionation, which also (in addition) favours stable nuclides.Rules 1 and 2 are necessary to prevent prompt gamma emissions IF the nuclear reactions of LENR are identical to the known ones. However there could be a novel mechanism which does not need to satisfy rules 1 and 2 because it:
- Fractionates the nuclear quanta into much less energetic bits while not changing the intermediate nuclear “steps” (nuclides and nuclear states). In fact the intermediate steps MUST be the ones we know, otherwise the nuclei produced by the LERN would need to be DIFFERENT from those we know. And this is not possible.
- Can prevent the actual formation of unstable nuclides.
Back to my theory:My theory suggests a mechanism which should allow to break rules 1 and 2 without gamma emissions, by fractionating the large nuclear quanta into smaller EUV quanta. This thanks to the fact that the nucleus remains “trapped” inside the electron ZB, so that additional “intermediate” energy levels at distances of tens of eV appear between the known “gamma-spaced” levels.
The preference for stable nuclides should be due to the continuous perturbation of the electron which does not allow the formation of unstable nuclei. However this perturbation does not interfere with the weak interaction, so that purely beta decaying nuclei can actually be produced (and in fact tritium is definitely produced).
So the precise and “unavoidable” criteria which William Collis suggests in his presentation, with the caveats explained, seem to match what my theory predicts.
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Why do electrons in Hyd's play by different rules than normal electrons, even relativistic ones? Is this not a hint that the system is unphysical?
Eric, the Hyd may well be physical, but I think in this case we are talking about two different things.
When I say that the electron size shrinks I speak about a free electron. When its speed is relativistic relative to the observer, the observer will see a smaller electron.
The electron inside the Hyd does not "travel" at relativistic speeds. The Hyd could absorb energy and the electron would simply spirograph-like rotate more quickly "around" the hydrogen nucleus. I am not sure about the possibility of the photon absorption. However, in the case of the absorption, the size of the Hyd would not change, at least not much. In fact it would do it in proportion to the added energy, which is in the 100[eV] range, while the electron energy (that "defines" its size) is 511[keV]. The Hyd is not defined by the Hamiltonian for the electron in an atom. It is defined by a Hamiltonian in which the main component is the magnetic attraction.
