Re: [Vo]:The excimer laser
On Fri, Nov 23, 2012 at 10:33 PM, mix...@bigpond.com wrote: I think what he's trying to say is that a fast D nucleus can also knock an inner electron out of Pd, which can then in turn accelerate another D nucleus, in a train of reactions. I am reminded of a pinball machine, where the palladium atoms are the devices with the rubber bouncers that flip the ball across to the other side. Once you drop a pinball into the machine, it can go for quite a while before falling through the opening at the bottom. Another image that comes to mind is letting go of several marbles at the top of a board with nails nailed into it in a cross-hatch fashion. Maybe the cracks that Ed Storms draws our attention to facilitate something here -- in a perfect lattice, perhaps there is less occasion for movement of this kind, whereas it becomes more possible when there is a small amount of void for the D nuclei to bounce around in. In a noble gas, maybe the analogy would be that of letting a bull go in a china shop. ;) One question I have has to do with the energies. At 20 keV, would an incident D nucleus impart enough momentum to the palladium atom enough to unseat it from the surrounding lattice? If so, it does not seem like such an interaction could last for very long before the local region was altered significantly. Eric
Re: [Vo]:The excimer laser
Sent from my iPhone On Nov 24, 2012, at 13:33, mix...@bigpond.com wrote: As mentioned previously, the fast D nuclei will lose most of their energy to valence Pd electrons, this analogy doesn't work very well. It would be more like a pinball machine that was coated with glue. The ball doesn't get very far. If Ron's mechanism works with Kshell vacancies, could it also work with valence electron vacancies? I think this would mean there would be an elastic collision instead of photon emission. In that case, would there necessarily be a loss of momentum for the lower energy incident D nucleus? Eric
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Sat, 24 Nov 2012 13:56:31 -0800: Hi, [snip] Sent from my iPhone On Nov 24, 2012, at 13:33, mix...@bigpond.com wrote: As mentioned previously, the fast D nuclei will lose most of their energy to valence Pd electrons, this analogy doesn't work very well. It would be more like a pinball machine that was coated with glue. The ball doesn't get very far. If Ron's mechanism works with Kshell vacancies, could it also work with valence electron vacancies? Possibly, but it wouldn't do you any good. The only possible use for a D nucleus with a kinetic energy of a few eV is in spreading heat around. I think this would mean there would be an elastic collision instead of photon emission. In that case, would there necessarily be a loss of momentum for the lower energy incident D nucleus? Pointless line of thought, and no it wouldn't be elastic. (The ball in the pinball machine heats up the glue). ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
On Nov 24, 2012, at 14:47, mix...@bigpond.com wrote: Pointless line of thought, and no it wouldn't be elastic. (The ball in the pinball machine heats up the glue). ;) I'm sure you're right. I was just thinking that if the majority of the heat energy was confined to the D nuclei, then dissipation of energy from the system would not be too great, and there would be occasion for a large number of collisions with varying levels of energy -- I think you were saying that a large number of collisions would be needed to get a fusion. If the collisions are inelastic, then I understand there will be energy transferred to the lattice. But the size of the substrate atoms is large compared to the deuterons, so perhaps the treansfer of energy would be minimal? I will go read a physics textbook now and spare you further interrogation. ;) Eric
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Sat, 24 Nov 2012 15:21:23 -0800: Hi Eric, [snip] On Nov 24, 2012, at 14:47, mix...@bigpond.com wrote: Pointless line of thought, and no it wouldn't be elastic. (The ball in the pinball machine heats up the glue). ;) I'm sure you're right. I was just thinking that if the majority of the heat energy was confined to the D nuclei, then dissipation of energy from the system would not be too great, and there would be occasion for a large number of collisions with varying levels of energy -- I think you were saying that a large number of collisions would be needed to get a fusion. Not just a large number of collisions. Ron's theory is specifically about D nuclei that approach a Pd nucleus to within about 100 fm (i.e. the size of the K shell). So it's not just about collisions, it's about two D nuclei being at a distance of 100 fm from a Pd nucleus at the same time. (Mind you I think the chances of this are about on a par with a snowflakes hope in hell ;) (Though I suspect that a head on collision between two 20 keV D nuclei in free space would be almost as useful). Note that only the 20 keV D nuclei can get that close, and they have to get lucky and penetrate to the K shell before they lose too much energy ionizing Pd valence electrons. The question that needs to be answered is: If a 23 MeV alpha from the fusion reaction creates 100(?) 20 keV deuterons, then what is the chance that two of them will arrive at the K shell of the same Pd atom concurrently? (They don't hang around for very long.) If the collisions are inelastic, then I understand there will be energy transferred to the lattice. But the size of the substrate atoms is large compared to the deuterons, so perhaps the treansfer of energy would be minimal? A deuteron is a charged particle, and as it passes through another atom, it disrupts the (mostly valence) electrons of the atom it's passing through, and ionizes the atom to varying extents. This costs energy which comes from the kinetic energy of the passing charged particle (in this case a deuteron). Consequently, every atom along the way that gets ionized decreases the energy of the passing deuteron. BTW the same thing happens to the fast alpha from the fusion reaction. Later these ionized atoms will retrieve free electrons and release photons of various wavelengths. Thus the energy of the reaction is spread across many atoms in the lattice as chemical energy which is ultimately converted (degraded) into heat. BTW2 This may also tie in with Paul Brown's device. If the ionized electrons have kinetic energy of their own that is well in excess of the ionization energy, and a strong magnetic field is in place, then you may get cyclotron radiation as the electrons head for home, i.e. back toward a positively charged ion. If you can find a way to synchronize this cyclotron radiation and capture it, then you have a means of converting fast particle energy into electric power, without using heat as an intermediary. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
If a pinball machine were to follow the rules of quantum mechanics, this is how it would work. The pinballs would be strongly attracted to the rubber bouncers like they were magnets…they would gain more energy if the obstruction was large. For a large obstruction, the pinballs would stick like glue to the rubber bouncer and gain loads of speed and momentum. This is called Anderson Localization. In our discussions to date, the question that has not yet been addressed in detail is how fatigue cracks in cold fusion electrodes, nano-hairs on the surface of micro-grains and pitting in the wire that Celani uses all contribute to the cold fusion process. This question revolves around the wave based quantum mechanical property called Anderson localization. http://en.wikipedia.org/wiki/Anderson_localization What nature does in one instance, she can act in an opposite way in another. For instants, the wave nature of a quantum particle can cause a quantum mechanical phenomenon where a particle tunnels through a barrier that classically it could not surmount. Anderson localization is the opposite quantum mechanical phenomenon where a particle is fixed at a location that it classically should have no problem in surmounting. Think of it this way: in our classical world, a helicopter can fly over a mountain range without being disturbed by the underlying landscape, provided that it flies higher than the highest mountain, or provided that for the height at which it flies, there is a labyrinth of valleys allowing it to cross the mountain chain. But in the quantum world, a quantum helicopter has a very good chance of being unable to cross the chain, even if there is a percolating path of valleys, and, in some situations, even if it has enough energy to fly over the highest peak. And even more perplexing, the higher this quantum helicopter flies the less chance it has to get over the mountain. What happens instead is that its quantum wave remains trapped, due to the interference of the multiply reflected wave at the various mountain peaks. And the lager the electron is, that is, the more energy it has, the more likely the obstacles in its path will nail it to its original position; this strange behavior gives rise to a phenomenon known as Anderson localization. Read more at: http://phys.org/news/2012-09-broadens-quantum-mechanics.html#jCp When high energy electrons flow over a cracked, hairy, or pitted surface, these electrons will pile up and accumulate because their large wave forms are snagged by these surface imperfections. The bigger these quantum particle wave forms are, the more likely that these particles will be impaled and imprisoned by these surface imperfections. The same is true for proton cooper pairs that these imprisoned high energy electrons produce via the Shukla-Eliasson effect. These cooper pairs first form a pile of stuff called a Bose glass. A Bose glass is a disordered form of a Bose-Einstein condensate. When the conditions become favorable, these localize pairs form a Bose-Einstein condensate. In QM speak, these nonlinear bosonic matter waves can undergo a localization-delocalization quantum phase transition in any spatial dimension when the interaction strength is varied; the transition brings the system from a non-interacting Anderson insulator to an interacting superfluid. For the research that supports this new quantum mechanical interpretation see http://www.google.com/url?sa=trct=jq=esrc=sfrm=1source=webcd=1cad=rjasqi=2ved=0CB8QFjAAurl=http%3A%2F%2Fwww.nature.com%2Fnature%2Fjournal%2Fv489%2Fn7416%2Ffull%2Fnature11406.htmlei=635iULfnNYTO0QHU8YDoDQusg=AFQjCNEFWcWRYj5-jhRJNdgy7xEmcrTgRQsig2=_-S22pviwufHLkkd99P9iA Also see. http://arxiv.org/pdf/1109.4403 This reference is the underlying paper called Bose glass and Mott glass of quasiparticles in a doped quantum magnet This concept is one of the important concepts in LENR and I will not tire in explaining it until you understand quantum mechanics enables LENR. Cheers:Axil On Sat, Nov 24, 2012 at 2:16 PM, Eric Walker eric.wal...@gmail.com wrote: On Fri, Nov 23, 2012 at 10:33 PM, mix...@bigpond.com wrote: I think what he's trying to say is that a fast D nucleus can also knock an inner electron out of Pd, which can then in turn accelerate another D nucleus, in a train of reactions. I am reminded of a pinball machine, where the palladium atoms are the devices with the rubber bouncers that flip the ball across to the other side. Once you drop a pinball into the machine, it can go for quite a while before falling through the opening at the bottom. Another image that comes to mind is letting go of several marbles at the top of a board with nails nailed into it in a cross-hatch fashion. Maybe the cracks that Ed Storms draws our attention to facilitate something here -- in a perfect lattice, perhaps there is less occasion for movement of this kind, whereas it becomes more possible when there is a small amount
Re: [Vo]:The excimer laser
The fact that there is photon emission in the soft x-ray range for the heavier elements is interesting. I am reminded of Ron Maimon's suggestion, assuming I have understood it: if you kick out an inner shell electron in one of the heavier elements (Ar, Kr and Xe, below, but also Pd and perhaps Ni), the resulting vacancy will have potential energy in the keV -- he mentioned 20 keV for palladium. The two ways that are commonly understood to dissipate this energy are characteristic photons, where a higher shell electron falls into the vacancy and emits a photon in the process, and Augur electrons. The crux of Ron Maimon's proposal is that there is a third way to deal with the resulting potential energy -- it could end up being transferred to a deuteron in the area in the form of kinetic energy (if I have understood him). So instead of a characteristic photon or an Augur electron you would have a deuteron with ~20 keV energy. According to Wikipedia, the optimum temperature for D-D fusion is 15 keV [1], so all else being equal, the energies appear to be within the realm of possibility. Maimon proposes that transmutations are a combination of (1) the shattering of the spectator nucleus (e.g., the palladium atom) by the energy of the reaction and (2) the absorption of daughters of the fusion reaction into the spectator nucleus. Alphas and other fragments that are formed and not absorbed in this way race through the local system, ionizing nuclei as they go and carrying the reaction forward. Note that although Ron Maimon has been talking about the Pd/D solid-phase system, there is nothing obvious that would restrict this description to that system. Perhaps you could see something similar going on in a gas phase system with species entirely different from palladium (although I suspect the presence of deuterium would be necessary). I think you would need heavy gas atoms, though -- perhaps Ar, Kr and Xe, for example. Eric [1] http://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions On Fri, Nov 23, 2012 at 12:36 PM, Axil Axil janap...@gmail.com wrote: Chlorine/noble gas combo produces the most powerful laser effect in the 150 and 173 nm wave length range, were the shorter the wavelength is, the closer the laser is to the soft x-ray range. Floride/noble gas produces a less powerful laser emination The wavelength of an excimer laser depends on the molecules used, and is usually in the ultraviolet: Excimer /Wavelength /Relative Power mW Ar2* /126 nm Kr2* /146 nm Xe2* /172 175 nm ArF /193 nm /60 KrF /248 nm /100 XeBr /282 nm XeCl /308 nm /50 XeF /351 nm /45 KrCl /222 nm / 25 See http://en.wikipedia.org/wiki/Excimer_laser * Notice that Ar, Kr, and Xe can produce powerful soft x-ray laser radiation on their own. Cheers: Axil
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Fri, 23 Nov 2012 13:56:15 -0800: Hi, [snip] The crux of Ron Maimon's proposal is that there is a third way to deal with the resulting potential energy -- it could end up being transferred to a deuteron in the area in the form of kinetic energy (if I have understood him). So instead of a characteristic photon or an Augur electron you would have a deuteron with ~20 keV energy. [snip] The problem with this approach is lack of ROI. To start with only a fraction of the incident x-rays are going to kick an electron out of a lower orbital. When it does happen, only a fraction of the time would this produce an energetic D nucleus. Then only a fraction of those energetic D nuclei would actually undergo fusion. All in all, I fear that all those fractions multiplied together are going to result in a COP 1. Besides it's a very indirect approach. It's much more efficient to just use an RF source to ionize D atoms, then accelerate the resulting nuclei in an electric field of 20 kV of so. This is in fact how the first fusion reactions were created, yet even this direct approach has a COP 1. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
On Fri, Nov 23, 2012 at 2:29 PM, mix...@bigpond.com wrote: The problem with this approach is lack of ROI. To start with only a fraction of the incident x-rays are going to kick an electron out of a lower orbital. When it does happen, only a fraction of the time would this produce an energetic D nucleus. Then only a fraction of those energetic D nuclei would actually undergo fusion. All in all, I fear that all those fractions multiplied together are going to result in a COP 1. This is my way of learning nuclear physics on the sly -- I say things, and then Robin sets the record straight. ;) There was one detail I left out, because I didn't understand it -- Ron referred to the classical turning point. It almost sounded like he envisioned two (and not just one) dueterons being pulled in together (or pushed out, together?) and then meeting at a specific location; i.e., the movement of the dueterons seemed to be directed rather than thermal. If true, perhaps this would take care of some of the loss of COP through fractions being multiplied together. I don't have a sophisticated enough understanding of the forces involved to see how this is supposed to work; perhaps it is either of: (1) A higher-shell electron moves in to fill the vacancy, pulling in a deuteron as it does, until it reaches the classical turning point -- maybe the point at which coulomb repulsion stops the deuteron from going any further; presumably it will not be moving for the brief moment that it is at that point, but perhaps Ron Maimon only intends that the fusion event occur before or after this point. (2) The original, ejected electron pulls the deuteron outwards. This would seem to have the disadvantage of not resulting in an especially directed focusing of deuterons at one another. I think Ron Maimon was proposing that there will have been two inner shell electrons that will have been ejected, but I'm not sure. Eric
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Fri, 23 Nov 2012 14:57:02 -0800: Hi, [snip] There was one detail I left out, because I didn't understand it -- Ron referred to the classical turning point. It almost sounded like he I suspect that the classical turning point refers the distance from the nucleus where an approaching positively charged particle has used up all it's kinetic energy, and momentarily comes to a halt, before being ejected. IOW the bounce distance. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Fri, 23 Nov 2012 14:57:02 -0800: Hi, [snip] There was one detail I left out, because I didn't understand it -- Ron referred to the classical turning point. It almost sounded like he envisioned two (and not just one) dueterons being pulled in together (or pushed out, together?) and then meeting at a specific location; i.e., the movement of the dueterons seemed to be directed rather than thermal. If true, perhaps this would take care of some of the loss of COP through fractions being multiplied together. [snip] Do you have a URL for Ron's work? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
On Fri, Nov 23, 2012 at 7:45 PM, mix...@bigpond.com wrote: Do you have a URL for Ron's work? See the section titled My Personal Theory and what follows it in Ron's response to this physics.SE question: http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus/13734 The explanation he proposes is part of a longer response to the question, Why is cold fusion considered bogus? Eric
Re: [Vo]:The excimer laser
In reply to Eric Walker's message of Fri, 23 Nov 2012 19:49:15 -0800:
Hi,
[snip]
On Fri, Nov 23, 2012 at 7:45 PM, mix...@bigpond.com wrote:
Do you have a URL for Ron's work?
See the section titled My Personal Theory and what follows it in Ron's
response to this physics.SE question:
http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus/13734
The explanation he proposes is part of a longer response to the question,
Why is cold fusion considered bogus?
Eric
I think what he's trying to say is that a fast D nucleus can also knock an inner
electron out of Pd, which can then in turn accelerate another D nucleus, in a
train of reactions.
However the problem with that is that a fast D will lose most of its energy
ionizing valence electrons rather than K shell electrons, so the process
actually dies almost instantly. Hence I don't really see a band state being
reasonably populated.
A great many concurrent reactions would need to occur for two Ds to find
themselves in the neighbourhood of the same nucleus, at the same time. I guess
the question is whether or not enough Kshell holes are created by the fusion
reaction products to make the whole process OU.
There is also the question of what gets the process kicked off initially, though
natural background radiation might act as a trigger.
Essentially what this theory does is provide a means of temporarily storing the
energy of a fusion reaction, and parceling it out to lots of other D nuclei so
that hopefully at least one two of them can fuse too.
Note also that as the Z of the host nucleus rises (with consequent increase in K
shell energy) two things happen:
1) The D's get closer which should enhance the individual reaction rate.
2) The number of K shell vacancies created by fast particles decreases, which
should decrease the overall reaction rate.
Since 1 and 2 work in opposite directions, there may be an optimal Z value for
the host lattice. Of course the host lattice also has to absorb H/D.
I note that Ron doesn't try to apply this explanation to the Ni-H results. The K
shell electron of Ni only has an ionization energy of about 7-8 keV, which is
rather on the low side.
All that having been said however I still rather like this theory. It seems to
have much going for it.
{BTW quote:
Now suppose that two of these accelerated deuterons happen to come close to the
same Pd nucleus. This can easily produce a fusion event at the turning point,
the deuterons have around 20KeV after all, and the fusion rates at 20 KeV in
beams is not that small, let alone in cases where the wave function is
concentrated near a nucleus with a classical turning point (where the wave
function is enhanced).
I think what he's trying to say here is a that when the deuterons come to a halt
near another nucleus, they are also relatively motionless with regard to one
another, but at close range. That means that there is a large overlap in their
De Broglie waves, and the tunneling probability is consequently enhanced.}
Regards,
Robin van Spaandonk
http://rvanspaa.freehostia.com/project.html
Re: [Vo]:The excimer laser
In reply to mix...@bigpond.com's message of Sat, 24 Nov 2012 17:33:49 +1100: Hi, [snip] I note that Ron doesn't try to apply this explanation to the Ni-H results. The K shell electron of Ni only has an ionization energy of about 7-8 keV, which is rather on the low side. [snip] BTW with regard to the Ni-H results, if the momentum of the reaction is shared by the newly formed nucleus and the nucleus of the host atom, then the reaction H + D = 3He + 5.49 MeV might play a role. The cross section for the reaction would likely increase since the energy would largely appear as kinetic energy of the 3He nucleus, rather than requiring a slow gamma ray emission process. H also tunnels more readily than D because it has only half the mass. So 7-8 keV might be enough, given that there is even *some* D-D fusion at about 5 keV. Even though only about 1 in every 6400 H atoms in natural H is a D atom, if you divide 5.49 MeV by 6400, you still get an average of 857 eV / H atom, which is still about 580 times more than you get from burning Hydrogen in Oxygen. IOW it would still readily explain Rossi's results. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
