1995.09.23 / Mitchell Jones / Re: The Protoneutron Theory of "Cold Fusion" Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: The Protoneutron Theory of "Cold Fusion" Date: Sat, 23 Sep 1995 11:00:38 -0500 Organization: 21st Century Logic The second post on which this thread is based is the following: ----------------------------------------------- OK, you guys, I've been waiting for two days to see if somebody was going to earn three ataboys by noticing that the protoneutron theory explains the Griggs result! Time's up! The ataboy window is closed! Here's the way Griggs' machine generates excess heat: (1) The 3500 rpm rotation of the "cylinder full of holes" produces massive turbulence in the water, and generates millions of tiny cavitation voids. (2) Since "there ain't nothin in them voids," they collapse as soon as they move to a less turbulent point in the flow. (3) When a void collapses, the walls crash together at the center, producing momentary giant overpressures. (This is a phenomenon well known to plumbers. When, for example, a long column of water running in an empty pipe slams into a closed valve, the giant overpressure that results is known as a "surge," and a blown valve is the frequent result.) (4) In the moment of final collapse, there is a tiny sphere at the center of the former void, where the situation is sufficiently crowded at the microscopic level so that electrons in the lowermost (K) shells of hydrogen atoms do not have room to orbit. While the duration of this moment is short in human terms, it is an eternity on the time scale of a hydrogen atom, whose electron revolves around its nucleus at a frequency of 10 quadrillion revolutions per second. (5) For an instant, therefore, such spheres are packed with protoneutrons. (6) Since protoneutrons are rapaciously hungry for the energy of transformation into neutrons, and since there are occasional gammas in the background radiation to which we are all exposed, a tiny percentage of these protoneutrons will capture gammas and transform into thermal neutrons. (7) Those thermal neutrons will quickly enter the nuclei of neighboring H or O atoms, and kick off gammas of their own when the product nuclei drop to the ground state. In the case of neutron absorption by a hydrogen, for example, we have: p + n --> d + 2.22 Mev (gamma), as previously discussed in the explanation of CF electrolysis. (8) Since the surrounding solution contains millions of cavitation bubbles in all stages of collapse, it follows that at any given moment the solution contains billions of protoneutrons. Which means: the 2.22 Mev gamma is surrounded by protoneutrons and isn't going anywhere: they are going to suck that puppy dry--which means: it is going to produce 2 more thermal neutrons, plus dump some heat into the fluid. (9) So here, again, we have the same old chain reaction that we discussed the other day, in which one thermal neutron begets two plus heat, which beget four plus more heat, which beget 8 plus heat, and on and on. This time, however, the process occurs in the dynamic environment of a turbulent fluid, rather than in the static environment of a palladium lattice. Bottom line: the Griggs device would be "cold fusion," except that "cold fusion" itself isn't fusion. What we have here, instead, is a protoneutron heat engine: an odd device that burns hydrogen so completely that even the nuclear energy is extracted, and yet it produces no pollution and no radiation. It is a "soft" energy source, as benign and non-threatening as flowers on a sunny day, and what it means is simple: the age of pollution, of centralized energy sources, and of centralized political power, is over. --Mitchell Jones =========================================================== =========================================================== cudkeys: cuddy23 cuden21cenlogic cudfnMitchell cudlnJones cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / A Plutonium / Re: First experimental predictions of Hydrogen Atom Systems Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Re: First experimental predictions of Hydrogen Atom Systems Date: 23 Sep 1995 17:30:09 GMT Organization: Plutonium College In article <43sua4$d65@news.utdallas.edu> iblum@utdallas.edu (Ira K Blum) writes: > You are forgetting pions and kaons and muons, particles which are not > protons, electrons, photons, or neutrinos, yet are smaller than 939 MeV, > which is the mass of your HYASYS. Tell me Ira, what does a muon decay into since its lifetime is 2.2 x 10^-6 sec yet its mass is 105.66 MEV as compared with electron mass of what about .5 MEV. Please, if you could or some other reader, list the lifetimes and the MEVs of all of these particles and what they decay into. Physicists for the past 40 years have not been truthful to themselves. They should realize that when a particle exists for such a fleeting time as 10^-6 sec that such a particle is really not on "par" to the existence of say the proton or electron etc. To put it on terms that all readers can understand. If we were to give a comprehensive science study of Homo sapiens, should we spend any time whatsoever or some little time on the fact that a worm lives in the eyelashes of humans, considering that we are really talking about Homo sapiens and not an organism that lives on Homo sapiens? In the same vain, is the muon a particle on par with the electron, or is it a energy exchange, that is, the muon is merely an electron with stored up extra energy and not a true particle. Each particle should be looked at as to "how long it exists", "what it decays into". In this manner, only hydrogen atoms exist and everything else is a hydrogen atom system with more energy than the 938 MEV hydrogen. What is the muon but an electron with a lot of extra energy? What needs explanation therefrom is why the quantization of 105 MEV? Why will the first quantization of an electron be 105 MEV or are there electrons between the .5 MEV and the 105 MEV? These are the questions to ask. And never to think that the muon is as basic and fundamental as an electron. Why does it take 9 muons to make a Hydrogen Atom System? All particles be they elements, isotopes, or this fancy schmancy things out of CERN are nothing but Hydrogen Atom Systems. This stuff is fun and it must be correct since it is too beautiful not to. cudkeys: cuddy23 cudenPlutonium cudfnArchimedes cudlnPlutonium cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / ZoltanCCC / Re: ? Singularity Technology ? Originally-From: zoltanccc@aol.com (ZoltanCCC) Newsgroups: sci.physics.fusion Subject: Re: ? Singularity Technology ? Date: 23 Sep 1995 16:29:45 -0400 Organization: America Online, Inc. (1-800-827-6364) Would you tell us where the RGeorge Home Page is. Thanks, Zoltan Szakaly cudkeys: cuddy23 cudenzoltanccc cudlnZoltanCCC cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / ZoltanCCC / Re: The electron capture theory of cold fusion Originally-From: zoltanccc@aol.com (ZoltanCCC) Newsgroups: sci.physics.fusion Subject: Re: The electron capture theory of cold fusion Date: 23 Sep 1995 18:39:24 -0400 Organization: America Online, Inc. (1-800-827-6364) In article <4409ec$nlq@newsbf02.news.aol.com>, mrichar353@aol.com (MRichar353) writes: > >Oh, come on. I have refuted Robin's idea conclusively, at least for people >who know physics at an undergraduate level. What you and he are talking >about violates energy conservation, it's as simple as that. > > Mark, unfortunately your half cocked reference to the conservation of energy did not cut it with me. Up to this point nobody has described the situation as the way I see it now. Here it follows: Before the electron falls into the potential well of the deuteron there is energy associated with the field of the proton as well as the field of the electron. In case of the electron this field is a major contributor to the electron's mass. If the electron were pointlike, the field would be infinitely massive. In fact we can calculate the radius of the electron by assuming that all of the electron's mass comes from the field. This radius is a lower limit. The electron cannot be smaller than that. I believe Einstein made a theory that explained the electron's inertia based on its electric field. In any case, the field of the proton and also the field of the electron disappear as the electron closely approaches the proton because their charges cancel each other. The mass of the two fields combined disappears and accelerates the electron which becomes heavier by the same mass. Its kinetic energy is already counted when we add up the initial masses of the electron and deuteron and so the electron does not have enough energy to initiate the electron capture I described earlier. I retract my claim now that the electron could simply fall into a proton and cause electron capture. The energy has to come from somewhere else. On the other hand it is interesting to see that the electron capture requires as much energy as there is in both fields, it seems to me like strange coincidence. From Robin's calculation it seems that the mass of the electron's field is about 0.782/2 =0.391 MeV, while the rest mass of the electron is 0.511 MeV This gives us an idea of how much energy we need for electron capture to happen. If there is a mechanism in the lattice that produces excited states of the deuteron, electron capture may become feasible. Zoltan Szakaly Zoltan Szakaly cudkeys: cuddy23 cudenzoltanccc cudlnZoltanCCC cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.22 / Henry Spencer / Re: Making He3 and related questions Originally-From: henry@zoo.toronto.edu (Henry Spencer) Newsgroups: sci.space.tech,sci.physics,sci.physics.fusion Subject: Re: Making He3 and related questions Date: Fri, 22 Sep 1995 20:33:10 GMT Organization: U of Toronto Zoology In article <43nb29$mnm@morgoth.sfu.ca> gay@sfu.ca (Ian Gay) writes: >>Hence, He-3 is a byproduct of our nuclear weapons stockpile. So it is >>being manufactured "earthside". Quite expensive, but I don't have a >>figure. Off the top of my head I seem to remember $1000/STP-liter. > >Just tried to look it up. All my current isotope catalogs fail to give a >price (Why? wildly fluctuating?) ... Probably very limited availability. The government is the only supplier, and they're not noted for rational decision making. In this case, "please call" probably means "we're not sure whether we can get it or how much it would cost -- it might depend on who you are and what you want to do with it". -- The problem is, every time something goes wrong, | Henry Spencer the paperwork is found in order... -Walker on NASA | henry@zoo.toronto.edu cudkeys: cuddy22 cudenhenry cudfnHenry cudlnSpencer cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / A Plutonium / Mark Richardson, Mr. Dupree, and Mr. Esch Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Mark Richardson, Mr. Dupree, and Mr. Esch Date: 23 Sep 1995 22:55:54 GMT Organization: Plutonium College In article <43tc0g$q4s@newsbf02.news.aol.com> mrichar353@aol.com (MRichar353) writes: > > God how idiotic! > > Superposition has absolutely nothing to do with building up a > multi-particle system from individual particles. It is the principle that > a linear combination of two wave functions is another wave function. These > two (or more) wave functions represent *alternatives*; they cannot both be > true at the same time (at least when a measurement occurs). > > Then again, considering the source, what should I expect? > > I *must* get a newsreader with a killfile capability! > > Mark Richardson I was updating the go-to-hell file which I keep and saw that Mr.Craig Dupree and Mr. Patrick P. E. Esch were on it. Reconsider. PU, PLuto revert both Craig Dupree and Patrick P. E. Esch back to point of neutrality, away from hell. Atom As to Mark Richardson who I told to go to hell, his spite and hatred still spews forth in greater amounts, I say go to hell, Styx this time. cudkeys: cuddy23 cudenPlutonium cudfnArchimedes cudlnPlutonium cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / A Plutonium / Re: All atoms are Hydrogen Atom Systems -> Superposition Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Re: All atoms are Hydrogen Atom Systems -> Superposition Date: 23 Sep 1995 23:13:14 GMT Organization: Plutonium College In article jeffo@rayleigh.lanl.gov (Jeff Olson) writes: > In article <43q0va$9ja@news.utdallas.edu> iblum@utdallas.edu (Ira K Blum) writes: > > >In article , jeffo@rayleigh.lanl.gov (Jeff > >Olson) writes: > > >|> According to the ancient periodic table on my wall here (back when O was > >|> exactly 16 au), the neutron has a rest mass of 1.0087 au and an H atom of > >|> 1.008 au. If the neutron mass were less, why would it decay? > >|> > > >M(p) = 938.27231 +/- 0.00028 MeV > >M(n) = 939.56563 +/- 0.00028 MeV > >M(e) = 0.51099906 +/- 0.00000015 MeV > >M(p+e) = 938.78331 +/- 0.00028 MeV > > >I guess you are right, I haven't looked at it in quite some time, so > >there is more energy in a neutron than a proton and an electron. (also > >the dominant decay is n-> p e nu (note that the neutrino is present and > >carries away momentum.) > > Actually, it's an anti-neutrino, I think. Someone please mind posting a representative list of elementary particles, just the most important members of each family showing the MEV, all in MEV. Plus indicate the lifetime of the particle all in seconds. Then what the particle decays into. Please post with a name of the particle not the Greek symbol. Thanks cudkeys: cuddy23 cudenPlutonium cudfnArchimedes cudlnPlutonium cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / Mitchell Jones / Re: The Protoneutron Theory of "Cold Fusion" Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: The Protoneutron Theory of "Cold Fusion" Date: Sat, 23 Sep 1995 10:56:55 -0500 Organization: 21st Century Logic I am getting more and more stuff, both in the form of posts or via e-mail, that reveals a massive misunderstanding of the protoneutron theory. Apparently many people have become interested in this thread after the original posts rolled off of their servers. Therefore, to preserve bandwidth by squelching the uninformed posts, I am going to re-post the two main articles on which this thread is based. The first follows: ---------------------------------------------------------- In article , hheffner@matsu.ak.net (Horace Heffner) wrote: > In article <21cenlogic-0509951113520001@austin-2-12.i-link.net>, > 21cenlogic@i-link.net (Mitchell Jones) wrote: > > [snip] > > > ***{The point here is to accurately visualize what is going on in the > > palladium cathode. To do that, we must accept the fact that the hydrogen > > ions (protons and/or deuterons) retain their positive charges until > > *after* they enter the cathode. Think about it: if, when a hydrogen ion > > appeared at the surface of the cathode, it were to be handed an electron, > > it would immediately become electrically neutral, and would cease to be > > attracted to the cathode. In that case, it would bubble away as a gas and > > loading would be impossible. Here, instead, is what must happen: (1) When > > an H+ reaches the surface of the cathode, it retains its charge, and > > enters the lattice structure. (2) The presence of H+ ions within the outer > > layers of the lattice gives those layers a positive charge, while the > > inner layers retain their negative charges. (3) When enough H+ ions have > > packed into the outer layers of the cathode to neutralize the charge of > > the cathode, the voltage gradient between the anode and the surface of the > > cathode disappears, and the H+ ions in solution are no longer attracted to > > the surface of the cathode. Thus the + charge in the outer layer of the > > cathode does not build up to "millions of volts," but only to the level > > necessary to eliminate the voltage gradient from anode to the surface of > > the cathode. In effect, the anode charge migrates into the outer surface > > of the palladium electrode, and the cathode charge retreats into the inner > > portion of the palladium electrode. (4)) Once the anode's charge has > > migrated to the surface of the cathode via this mechanism, loading of the > > palladium electrode does *not* cease, because H+ ions on the inner surface > > of the positive region are constantly having electrons passed to them from > > the negatively charged central region. Each time this happens, the > > affected H+ ion becomes a neutral H atom. However, it can't bubble out of > > the solution, because it is trapped inside the palladium electrode. (5) > > Each time a new, neutralized hydrogen atom is added to the pool of those > > trapped inside, the positive charge of the surface region drops slightly, > > and a new H+ ion enters the surface region from the outside, again > > eliminating the voltage gradient between the anode and the surface of the > > cathode. (6) Then, another H+ ion inside the palladium electrode is handed > > yet another electron and neutralized. (7) Then, another H+ ion enters the > > surface. And on and on it goes, until saturation loading is approached, > > and, hopefully, "cold fusion" kicks in. Bottom line: it isn't necessary > > that the cathode achieve a charge of "millions of volts;" but it *is* > > necessary that it build up a positively charged region on its outer > > surface, in order for the loading process to take place at all. --Mitchell > > Jones}*** > > > > > > In article , > > hheffner@matsu.ak.net (Horace Heffner) wrote: > > > > Requoted here with minor typo's corrected to aid in voltage calculation to > follow: > > > > I would like to take this opportunity to clear up some mistaken thoughts I > > > previously posted here. First, some data from the Handbook of Chemistry > > > and Physics (HCP) and a Sargent-Welch periodic table. All sizes in A > > > (1e-10 m). > > > > > > Radius of Pd atom: (Pd) 1.79, (Pd covalent) 1.28, (Pd+2) .80, (Pd+4) .65 > > > Radius of H atom: (H) .79, (H-1) 1.54, H (covalent) .32 ***{Note: the radius of the innermost Bohr orbit is .53 Å, and that of the H+ ion (a proton) is .0000137 Å. --Mitchell Jones}*** > > > Bond lengths: (H-H) .746, (Pd-Pd) 2.751 > > > Density of Pd: 12.0 g/cm^3 > > > At. Wt. of Pd: 106.4 > > > > > > Now, the Pd lattice is cubic, face centered. ***{It is misleading to refer to this lattice as cubic. The smallest unit cell consists of two four-sided pyramids (e.g., like the Egyptian pyramids) joined base-to-base. The faces of each pyramid are equilateral triangles with a side length of about 2.67 Å. The cubic unit cell to which you refer is three times as large, and is not a satisfactory conceptual unit for present purposes. The reason: when loading takes place, it is the double-pyramid unit cells that are loaded--i.e., that have H atoms placed in their centers. --Mitchell Jones}*** You could imagine building > > > this lattice by taking 1 atom thick sheets of spheres arranged in square > > > patterns and laying them one on top of each other, but the top layer atoms > > > resting naturally in the spaces between the bottom layer atoms. The bonds > > > formed are between atoms in seperate layers only. If we call the distance > > > between alternate sheets S, then the bond length D will be 1/2 the length > > > of the diagonal of the cube S^3, or .866 S. I seem to recall the jist of > > > this being posted before. > > > > > > For simplicity sake let's assume all atoms are spheres. Now, applying > > > common sense, suppose we considered an imaginary Pd made up of only > > > alternate sheets. It would have half the density, or 6.0 g/cm^3. Also, > > > each atom would occupy a cube (S^3) of volume V. Applying Avogadro's > > > number we get 6.79E22 atoms/cm^3 for Pd. Our half density sheets would > > > therefore have 3.40E22 atoms/cm^3, so V = 2.94E-23 cm^3, and S = 3.09E-8 > > > cm = 3.09 A. This gives D = 2.67 A, which corresponds within about 3 > > > percent with the published bond length of 2.751. So this is good, common > > > sense and published values agree. > > > > > > Now this means we have a radius R for Pd in the lattice because R =(0.5)D = > > > 1.38 A. This radius corresponds best with a covalent radius. > > > > > > Now, the largest sphere that can be placed between the Pd spheres is > > > diameter d = S - R - R = S - D = .134 S. Since S = 3.09 A, d = (.134 * > > > 3.09) A = .414 A. This means the H atom will have a radius r = .5 d = .207 > > > A. ***{Using the calculated bond length of 2.67 Å rather than the published one, the size of the entry hole in one of the triangular pyramid faces is large enough to permit the passage of a sphere with a radius of .206 Å. This means entry is permitted only to H+ (protons) or D+ (deuterons). A neutral hydrogen atom is too large to pass through the face and into the unit cell. Once inside the cell, of course, there is room in the center for a sphere of radius .55 Å, which will accomodate one neutral H atom with its electron orbiting at the innermost Bohr radius of .53 Å. To get there, however, its parts must enter as separate charged particles--i.e., as a proton and an electron---and combine after they are inside. At this point, it is appropriate to ask ourselves what sort of loading ratios result from such a process. For example, if there is one H atom inside each unit cell of the lattice, what is the loading ratio? Well, for an infinite lattice the count of unit cells is simply three times the count of the number of Pd atoms in the lattice. (Since each Pd atom in a layer is the top of a vertical unit cell in the two layers below it, the number of vertical unit cells equals the number of Pd atoms in the lattice. The same is true of unit cells oriented right to left and front to back. Hence there are 3 times as many unit cells as Pd atoms in the lattice.) If we stuff a single, neutral H atom into each unit cell, therefore, we would have a ratio of 3 H to every Pd! It therefore follows that, for the loading ratios that are actually seen (not much above .8), more than two thirds of the unit cells are unoccupied. --Mitchell Jones}*** This appears to be too small to be covalent (i.e. .32), so we are led > > > to think the bond must be ionic. This agrees with other chemistry if I > > > recall correctly, and is the basis for Marshall Dudley's Hypothesis. > > > Loading above 1-1 leaves no room for H orbitals. > > > > > > Anyone, please comment on flaws in this line of reasoning. > > > > > [snip] > > Now, to check your model of the electrode surface, let's assume a single > H+ ion blocks each entry hole to the lattice (i.e. the ion layer is only > one atom thick.) ***{I never said they would block entry. They are, in fact, too small to block entry. They do, however, carry their positive charges with them into the lattice until such time as they meet up with outgoing electrons. Since protons are more than 1837 times as massive as electrons, their drift velocities in the lattice are vastly reduced, and they tend to not penetrate very deeply before being neutralized. Thus the positive charge that migrates to the cathode will be concentrated in the outer regions of the lattice. --Mitchell Jones}*** From the above calculations we have surface atoms in a > square array, with the sides of each square rougly S = 3.09E-8 cm. > Therefore each Pd atom presents an area of roughly S^2 = 9.55E-16 cm^2. > So we have roughly 1/(S^2) = 1.05E15 ions in a 1 cm^2 area. For > simplicity, let's assume we area talking about a 1 cm^2 surface area > cathode. This cathode would then have a charge Q=(1.06E-19C/ion)*(1.05E15 > ions) =1.11E-4 Coulombs. > > Now, what we are talking about is essentially a capacitor with area = 1 > cm^2 and a plate separation of S. This let's us use the capacitance > formula: > > C = e_k * A / S = (8.85E-12 F/m) (.01 m)^2 / (3.09E-10 m) > C = (8.85E-12 * 1E-4 / 3.09E-10 ) F > C = 2.86E-6 F (about 3 uF) > > Now volts = Q/C = (1.11E-4)/(2.86E-6) = 38 volts. > > This is still a pretty big barrier. This implies that to just balance an > electrode voltage of 2 volts, thus terminating electrolysis, that only > about one site in 19 could be occupied by a + ion. ***{The fact that a minority of the sites are occupied by H+ ions at a given instant is not a problem. The absolutely crucial point here is not that the surface of the cathode acquires the full positive charge of the anode, but rather that H+ ions *must* penetrate into the lattice before being neutralized in order for loading to take place. The reason: electrons that are given over to H+ ions before they enter the unit cells of the lattice result in gas production, not loading. Once a particular unit cell is occupied by an H+ ion, the next step is to hand it an electron, converting it into a neutral H atom which, because the radius of the innermost Bohr orbit is .53 Å, is now trapped inside the unit cell. This is the way the lattice is loaded, *because it has to be.* To repeat: electrons that are given over to H+ ions before they enter the unit cells of the lattice result in gas production, not loading. --Mitchell Jones}*** > > I have a different mental model I would like to share. I don't know if it > is correct, but I'll spell it out, as if it's a fact, so it can be > dissected. ***{I am not sure why you are presenting an alternative model here. None of the statements that you made above revealed any reason for rejecting the model that I have proposed. If your difficulty lies solely in the implication, admittedly conveyed by my early description, that the full anode charge migrates to the surface of the cathode, then let me emphasize again that this is not necessary. What is necessary--absolutely necessary--is that H+ ions enter the lattice cells before they receive electrons. This fact implies that there is going to be some degree of migration of the anode charge into the outer portion of the cathode, but does not necessitate that the full anode charge do so. --Mitchell Jones}*** > > Water molecules are bipolar, electrostatically speaking. So, if a + ion is > in solution, the adjacent water molecules align their - sides toward the + > ion. Now we have a sphere with all + sides of the water molecules on the > surface. The water molecules adjacent to the sphere align their - sides > toward that layer around the sphere, and so on. There is in effect a hugh > ionically bonded +1 charge molecule (clump) that migrates through the > electrolyte. When the clump reaches the cathode, the elctrostatic force > of the cathode, if sufficient voltage, can separate the + ion from the > clump. > > It seems like, if this is true, H atoms preceeding a clump to the cathode, > and at the surface of the cathode, would have a very high probability of > being forced into a site by a clump. Also, some would H atoms would > escape, combining with other H atoms to form H2 and bubble out. > > What do you think? ***{To repeat: I do not see a need for an alternative model. A voltage gradient is all that is required to attract an H+ ion into the unit cells of the cathode. It doesn't have to be "forced" in: the openings in the triangular faces of the unit cells are large enough for it to move in *very* easily--as easily as a fly into an amphitheater, in fact. Remember: an H+ is simply a proton--i.e., the nucleus of a hydrogen atom. The standard formula, curve-fitted to the experimental data, for computing the radius of a nucleus is simply r = (1.37E-13)A^(1/3), where r is the radius in cm and A is the mass number. For H+ the mass number (number of protons plus number of neutrons) is 1, so r = 1.37E-13 cm = .0000137 A. Once inside the cathode, such an H+ ion will drift toward the electron source until it acquires an orbital electron. At that point, it will become an electrically neutral H atom, with an effective radius as per the innermost Bohr orbit (.53 Å or slightly larger). This means, roughly speaking, that its radius will increase by four orders of magnitude when it captures an electron, and it will become trapped in the unit cell which it is occupying at that time. --Mitchell Jones}*** > > > > > Horace, your posts are tightly focused, high on content and relevant > > logic, and are excellent food for thought. I, for one, appreciate them > > very much. Thank you! > > > > --Mitchell Jones > > > > =========================================================== > > Thanks for the support much needed by this newbie. > > Horace > > -- > Horace Heffner 907-746-0820 > PO Box 325 Palmer, AK 99645 ***{At this point, I would like to toss out what I call the protoneutron theory of "cold fusion." If this theory is correct, then "cold fusion" ain't fusion! Note that the drift of the H+ ions into the cathode does not require a free path through unoccupied unit cells. In an occupied unit cell, the occupant is a neutral H atom. Thus there is no voltage gradient barring an H+ ion from drifting into an occupied unit cell, and there is no voltage gradient preventing an outward drifting electron from passing into an occupied unit cell, either. This means it is purely a matter of statistical probability, regarding the question of whether a drifting H+ meets an electron in an occupied unit cell. If it does, it will attempt to "pop" into a neutral hydrogen atom. Unfortunately, because the unit cell is already occupied, when the second H atom attempts to form it will not have sufficient room for its electron at even its innermost Bohr orbit. Result: the electron will spiral down toward the nucleus, where it will linger at grazing altitude in a particle form which we may term a *protoneutron.* Why doesn't it become an actual neutron? The answer: in the reaction p + e --> n, the mass of the proton is 1.67239E-24 grams while that of the electron is .00091E-24 grams. Thus the sum of p + e on the left side of the equation is 1.6733E-24 grams. On the right side of the equation, the mass of the neutron is 1.6747E-24 grams. Since the neutron outweighs the proton plus the electron by .0014E-24 grams, which equates to .7875 Mev, this amount of energy must be available from somewhere in order for the protoneutron to convert into a real neutron. [Note: to convert .0014E-24 grams to ergs, we multiply by the square of the velocity of light in cm per sec, giving: (.0014E-24)(9E+20) = .0126E-4 ergs. Since 1 electron-volt equals 1.6E-12 ergs, it follows that this reaction absorbs .0126E-4/1.6E-12 = .7875E+6 electron-volts, which is .7875 Mev per neutron created.] Since this amount of energy is unlikely to be available, the bizarre and unstable protoneutron will only endure for a tiny fraction of a second, until it bounces into an unoccupied unit cell. At that point, it will expand its radius like an exploding piece of popcorn, and become a hydrogen atom trapped in that unit cell. Note, however, that such events have a cumulative effect: if electrons tend to meet protons in a particular region of the lattice, all of the adjacent unit cells in that region are quickly going to become occupied, and the size of that continuously occupied region is going to grow inexorably. Result: protoneutrons that are formed in that region will endure for longer and longer, before finally bouncing into an unoccupied unit cell adjacent to the region and "popping" into hydrogen atoms. Eventually the size of the "loaded" region will become so immense, relative to the slow moving, uncharged protoneutrons, that they will begin to accumulate in the region. Why do they persist in this bizarre netherworld state, as "protoneutrons?" For two reasons: (1) Enough thermal energy is available for them to "pop" into hydrogen atoms, but that low energy state requires a lot of space--more space than is available in the loaded region of the lattice which they occupy. (2) Enough space is available for them to collapse into neutrons, but that state requires more energy than is available. Result: thermal protoneutrons inside large, continuously loaded regions of the lattice will simply endure, desperately hungering for the energy or the space they need to enter a more stable state. This means that protoneutron loading will take place in subregions of the lattice that are fully packed with hydrogen atoms. First there will be one, then two, then ten, then hundreds, then thousands, then millions of protoneutrons trapped in the region, *and all will be desperately awaiting some event capable of supplying a .7875 Mev jolt in the right place.* They have no prospects of being transformed into hydrogen atoms, because that is a low energy-high space transformation, and the space is not available. Thus their only allowed transformation is the low space-high energy transformation into neutrons. When such a transformation finally happens, the loaded palladium electrode will "turn on," and begin to produce "excess heat." To understand the production of "excess heat," assume that in a subregion of the lattice in which there are millions of trapped protoneutrons, one of them finds the energy it needs to become a neutron. Result: a thermal neutron will be created, and will quickly drift into a nearby nucleus. If, for example, it drifts into an H nucleus, the result is D plus a 2.22 Mev gamma. The gamma, however, will not exit the lattice, because it is emitted in a region packed with protoneutrons, *and it is my conjecture that protoneutrons are so rapacious in terms of their demand for energy of transformation that they will absorb all of the gamma's energy before it can exit from the lattice.* This means that the 2.22 Mev gamma will transform 2 protoneutrons into neutrons, and, in addition, will give a third protoneutron about .645 Mev of kinetic energy. Naturally, the resulting fast protoneutron will fly through the lattice until it encounters an empty unit cell, where it will virtually instantly "pop" into a hydrogen atom. But remember: the hydrogen atom is too large to move freely in the lattice. It will thus slam into the opposite side of the unit cell in which it forms, and deliver its .645 Mev to the lattice as heat. As for the two thermal neutrons that were created, they will quickly drift into nearby nuclei, producing isotope shifts and more electromagnetic emissions. When those emissions are insufficient to transform a protoneutron into a neutron, they give it kinetic energy, and cause it to fly through the lattice until it comes to an unoccupied unit cell, where it pops into a hydrogen atom, slams to a stop, and delivers its energy to the lattice in the form of heat. When the emissions are energetic enough to transform protoneutrons into neutrons, on the other hand, they will trigger additional neutron creation, which in turn will produce still more isotope shifts, and so on. In short, a kind of chain reaction will begin within the subregion of the electrode that is packed with protoneutrons. That reaction will have three odd characteristics: (1) It will be self-controlling: if the lattice begins to overheat, it will deload hydrogen, which will reduce the population of protoneutrons, which in turn will slow the reaction back down. (2) The reaction will be self sustaining: once large numbers of thermal neutrons are afoot in the lattice, a constant sprinkling of gamma emissions will be assured, thereby continuing the process. (3) There will be no radiation hazard: radiation, for practical purposes, cannot escape from a lattice that contains a substantial population of protoneutrons. Anyway, that's enough for now. I am not going to go into the various isotope shifts which can result from this process, nor into the energies of the associated gammas, betas, etc., that can result. Instead, I am going to offer a simple prediction which can be verified or falsified: in order for the protoneutron theory to be correct, a palladium lattice loaded with protoneutrons must have an abnormal hunger for energy, and an abnormal ability to suppress internally emitted gammas. Otherwise, the experimenters who have touched off "cold fusion" would all be dead. This can be explained in only two ways: either the presence of nearby protoneutrons absorbs the energy before the gammas can form--i.e., before the photons can attain the speed of light--or else they absorb it after the gammas have formed but before they exit the lattice. To test this, here is what I propose: (1)To determine whether the energy is absorbed *after* gammas form, load a palladium cathode to the point where "cold fusion" begins, and then bombard it with gamma radiation. In that case: (a) a bizarre and extreme increase in gamma absorption will be noted, by comparison to measurements taken with the same cathode prior to loading; (b) it will be possible to "turn on" loaded cathodes using gammas that have enough energy to turn protoneutrons into neutrons. (Note: lots of other predictions could be made here--e.g., that it should be possible to turn on loaded cathodes by supplying them with thermal neutrons--but I am deliberately trying to restrain myself.) (2) To determine if the energy is absorbed *before* gammas form, supply the cathode with thermal neutrons before loading, measure the gammas produced, and then repeat the procedure with a loaded cathode that is producing anomalous heat. Far more gammas should be emitted from the cathode in the first case than in the second. Now, to a different point: earlier I said that if the protoneutron theory is correct, then "cold fusion" isn't fusion. The reason is simple: fusion refers to the combination of the nuclei of different atoms to form atoms of yet another kind. Since an electron isn't the nucleus of an atom, the combination of an electron and a proton to form a neutron isn't fusion. And, since a neutron isn't a nucleus, when a neutron combines with a preexisting atomic nucleus to form a different isotope, that, too, is not fusion. Bottom line: if the protoneutron theory is correct, then "cold fusion" ain't fusion! I would add that I am perfectly aware of the conflict between the protoneutron theory and "quantum mechanics." This is a classical mechanical theory, and classical mechanics is based on the principle of continuity--i.e., the principle that no entity may come into existence out of nothing or vanish into nothing. The principle of continuity indicates that an entity can arrive at, or leave, a location in one way only: by successively occupying each position in a continuous spatial pathway to, or away from, that location. The implication is that motion is continuous, not a series of quantized "jumps." This means, for example, that an electron moving from one stable orbit to another does not "jump" in the quantum mechanical sense: it follows a continuous spatial pathway from the former orbit to the latter, and it exists just as surely during those instants when it is between the "preferred" orbits, as when it is in one of them. In the case presently under discussion, the principle of continuity means that an electron exists and follows a continuous spatial pathway when it is transiting from the lower Bohr orbit to a position in the nucleus. It does not merely exist "in" the nucleus and "in" the innermost Bohr orbit, but also in between. The protoneutron, in short, is a classical mechanical phenomenon. It violates the most fundamental precept of "quantum mechanics" and, if it exists, it constitutes one more piece of evidence, in a pile that is already mountainous, indicating that motion in the microcosm is not "quantized" and, hence, that "quantum mechanics" is wrong. Needless to say, I present the protoneutron theory as a hypothesis subject to verification or falsification. If it proves to be indefensible, then I will abandon it. On that, you have my absolute guarantee. --Mitchell Jones}*** =========================================================== =========================================================== cudkeys: cuddy23 cuden21cenlogic cudfnMitchell cudlnJones cudmo9 cudqt3 cudszL cudyr1995 ------------------------------ 1995.09.23 / Darin Vrancic / Re: ? Singularity Technology ? Originally-From: darin@iglou.com (Darin Vrancic) Newsgroups: sci.physics.fusion Subject: Re: ? Singularity Technology ? Date: Sat, 23 Sep 1995 10:09:31 GMT Organization: IgLou Internet Services (1-800-436-4456) zoltanccc@aol.com (ZoltanCCC) wrote: >I have an idea. Since the most common isotope of Lithium is Li7, this >substance is cheap. If you add a neutron to Li7 you get Li8 which is >exactly two alphas. We see that in your later posts you noticed your isotopic error. I only included this part for reference. >I don't know if the energy losses in hot fusion reactors come mainly from >neutrons leaving the plasma core, but I suppose you could try blanketing >the equipment with massive volumes of Li7. The neutrons that escape from >the plasma core would be absorbed by the Li blanket and the reaction: >n + Li7 --> He4 + He4 (1) >would produce plenty of heat. The key is to suppress the alternative >reaction: >n + Li7 -> Li6 + n + n (endothermic) (2) >which would absorb the heat. The He4 produced in the first reaction would >be fast and the equipment would have to be designed to slow it down to >convert the energy to heat. >Now this would not be strictly a hot fusion reactor that breaks even but >it would sure help the total energy output of the system. In fact in terms >of total energy in minus total energy out it may break even due to the >Lithium fission. >Of course this may not be my original idea, it is kind of obvious. No, this is not original to you, it goes back at least 15 years, though the most common purpose quoted for a lithium blanket is to absorb the neutrons to keep them from exiting the reactor and becoming a radiation hazard. These designs usually use the Lithium in liquid form to capture the gammas as well as the neutrons in order to transfer the released energy from the core to a boiler to produce electricity. The problem with expecting this reaction to increase the energy efficiency of a fusion reactor, however, is that for it to do so, the fusion reaction must be generating enormous numbers of neutrons, which of course means it is generating large amounts of power already. Not that it doesn't make sense to increase efficiency any way one can, but reactions currently need to be increased by 2-4 orders of magnitude before the technology even begins to be commercially feasible. Now, these cold fusion claims, if validated once and for all, would change the energy landscape even more than hot fusion would, since they hold the promise of small energy providers (companies, even individual households, eventually providing their own power, with power companies relegated to being no more than maintenance providers). Some of you might wish to check out the RGeorge Home Page, as he discusses a "cold" fusion technique which may have been discussed here, but if it was I missed it. It also supposedly produces no free neutrons and no high-energy photons. He is well spoken and is working on patents of his own, and you may be surprised at what you find there. cudkeys: cuddy23 cudendarin cudfnDarin cudlnVrancic cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / Mitchell Jones / Re: Testing the Mitchell Jones Hypothesis Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: Testing the Mitchell Jones Hypothesis Date: Sat, 23 Sep 1995 02:19:36 -0500 Organization: 21st Century Logic In article <43srd2$3kko@lamar.ColoState.EDU>, gelfand@lamar.ColoState.EDU (Martin Gelfand) wrote: > I feel compelled to comment on just one of the ravings > in the long post by M.J.... > In article <21cenlogic-2009951048240001@austin-1-4.i-link.net> 21cenlogic@i-link.net (Mitchell Jones) writes: > > > >By the way, speaking of ether theory, are you aware that the existence of > >the ether was proven more than 60 years ago, and is generally accepted by > >top physicists today? I'll bet not. That's because you buy into the > >dissembling and the crapola. You don't notice that every knowledgeable > >physicist today accepts the existence of "Dirac's ocean"--a particulate > >medium that pervades all of space and accounts for most of the phenomena > >of electromagnetism. Of course, they have all agreed to not *call* it the > >ether, because that would have required them to clash head-on with another > >dogma: that of Einstein's hare-brained "relativity" theory. But behind the > >political dissembling lies the reality of classical mechanics triumphant, > >in all areas of physics. The myth of its demise is a politically expedient > >fantasy, nothing more. And soon even that will be gone. > > > >--Mitchell Jones > > The statement 'every knowledgeable physicist today accepts > the existence of "Dirac's ocean"' is about forty to fifty years out > of date, now. The "Dirac sea" of negative energy electrons was > a necessary construct back in the days when the Dirac equation > was the leading theory of relativistic quantum mechanics for > electrons. However, with the advent of relativistic quantum field > theory a Dirac sea is no longer required, and nobody believes > in its existence. ***{Martin, the issue here is not what present day physicists will *say* if asked whether they believe in an "ether" or in "Dirac's ocean." What they will say is influenced by non-rational considerations (e.g., the desire not to be laughed at or labeled a crackpot). The question, instead, has to do with whether an observer who is not concerned about peer group acceptance might reasonably say that they believe in such things as an ether or Dirac's ocean. The problem that Dirac solved was simply that during the period from roughly 1900 to 1930, the rejection of the classical ether theories hardened into an enshrined, sacred dogma. A climate of fear existed in physics--fear of being laughed at--and physical thought was crippled by the determination to not treat "space" as an active agent in problem solving. Dirac ended that by putting forth his notion of an "ocean of extraordinary electrons" which filled all of space, and by making accurate predictions (e.g., the existence of the positron) based on that view. Of course, he denied that his theory was an ether theory. Not even he dared to use such a loaded term. But he did take on, and overturn, the associated taboo against imagining that a substance filled all of "space." Because of him, it became OK once again, as it had been in the 19th century, to imagine that the vacuum was not empty. What had been a taboo against a certain kind of thought was, by him, reduced to a taboo about the use of a word. That, in my view, was a significant accomplishment. In summary, you are right to say that nobody believes in "Dirac's ocean," if by that you mean that few physicists today would use those words. However, every person today who has thought deeply about the problems of electromagnetism recognizes that "the vacuum" is a complex substance with myriad properties. Dirac, by removing a taboo, made that possible. Even you, I dare say, would not claim that space is "empty" in the sense in which the term was used in the early years of this century! --Mitchell Jones}*** > > And even worse, from Mitchell's viewpoint, is that even if > you wanted to believe in the existence of the Dirac sea it > wouldn't serve as a classical ether. After all, the Dirac > equation _is_ a relativistic theory. ***{You fail to understand that the curve fitted mathematics associated with "relativity theory" can be interpreted classically. One need not, for example, treat Einstein's t' as indicating time distortion. An accelerated clock may simply run slower than a clock that is not accelerated. Why? *Perhaps because the pressure exerted by particles of "Dirac's ocean" (i.e., "the ether") moving between the atoms of the clock causes it to run slowly!* Moreover, similar alternative interpretations are available for all aspects of the mathematics used by Einstein. Just as there is nothing in the mathematics of "quantum mechanics" that requires the "Copenhagen interpretation," so there is nothing in the mathematics of "relativity" that requires Einstein's interpretation. (The math, as I have said repeatedly, doesn't give a hoot in hell how it is "interpreted.") Bottom line: since Einstein's mathematics can be interpreted classically, the fact that Dirac incorporated it into his own theories does not imply that his "ocean" cannot serve as a classical ether. --Mitchell Jones}*** > > > Martin Gelfand > Department of Physics, > Colorado State University =========================================================== cudkeys: cuddy23 cuden21cenlogic cudfnMitchell cudlnJones cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / Politics@usa.c / Proposed State & Federal Regulations for the INTERNET! Originally-From: Politics@usa.com Newsgroups: sci.physics.fusion Subject: Proposed State & Federal Regulations for the INTERNET! Date: 23 Sep 1995 07:31:58 GMT Organization: Wisconsin Internet Services My name is Scott Glasrud, and I am running for the New Mexico State Senate during the 1996 elections. One of the reasons I have chosen to run is to combat the proposed state and federal regulations of the Internet. As you know, the Internet was never designed to be regulated! It was designed to allow communications in the event of anuclear war or a major catastrophe. I OPPOSE REGULATION, and if elected will fight to preserve your constitutional rights. HOWEVER, I NEED YOUR HELP! I am asking each person who reeives this message to send $5.00 to the Scott Glasrud Campaign Committee. If we pull together, we CAN protect our first amendment rights! HELP ME show the politicians the POWER behind this important NETWORK. Please send contributions to: The Scott Glasrud Campaign Committee 11024 Montgomery Blvd. NE, Suite 179 Albuquerque, New Mexico 87111 Thank you! cudkeys: cuddy23 cudenPolitics cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.22 / MARSHALL DUDLEY / Re: Pathological Skepticism! Originally-From: mdudley@brbbs.brbbs.com (MARSHALL DUDLEY) Newsgroups: sci.physics.fusion Subject: Re: Pathological Skepticism! Date: Fri, 22 Sep 1995 16:20 -0500 (EST) "Alan M. Dunsmuir" writes: -> Hydrogen - created by the solar-powered hydrolysis of sea water - will ^^^^^^^^^^ Huh? I show hydrolysis to mean a chemical reaction in which a substance reacts with water to produce another substance, such as starch & water to make glucose. I am not sure if you are misusing a term here, or if there is another definition of which I am unaware. Please explain exactly what solar-powered hydrolysis is. If you are talking about a process that converts water to hydrogen and oxygen directly from sunlight I would be very interested in knowing how to do that. If you can cite any papers that would even be better. Marshall cudkeys: cuddy22 cudenmdudley cudfnMARSHALL cudlnDUDLEY cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / Alan M / Re: Pathological Skepticism! Originally-From: "Alan M. Dunsmuir" Newsgroups: sci.physics.fusion Subject: Re: Pathological Skepticism! Date: Sat, 23 Sep 1995 07:48:26 +0100 Organization: Home In article: mdudley@brbbs.brbbs.com (MARSHALL DUDLEY) writes: > Huh? I show hydrolysis to mean a chemical reaction in which a substance reacts > with water to produce another substance, such as starch & water to make > glucose. I am not sure if you are misusing a term here, or if there is another > definition of which I am unaware. > Sorry - finger/brain trouble. For 'hydrolysis' read 'electrolysis'. -- Alan M. Dunsmuir [@ his wits end] (Can't even quote poetry right) I am his Highness' dog at Kew Pray tell me sir, whose dog are you? [Alexander Pope] PGP Public Key available on request. cudkeys: cuddy23 cudenAlan cudfnAlan cudlnM cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / Alan M / Re: Testing the Mitchell Jones Hypothesis Originally-From: "Alan M. Dunsmuir" Newsgroups: sci.physics.fusion Subject: Re: Testing the Mitchell Jones Hypothesis Date: Sat, 23 Sep 1995 07:50:01 +0100 Organization: Home Mitchell Jones seems to have developed paralysis of the fingers regarding his 'slower than light' photons. Whatever can this mean? -- Alan M. Dunsmuir [@ his wits end] (Can't even quote poetry right) I am his Highness' dog at Kew Pray tell me sir, whose dog are you? [Alexander Pope] PGP Public Key available on request. cudkeys: cuddy23 cudenAlan cudfnAlan cudlnM cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.24 / Bruce Simpson / Re: French nuclear test agenda Originally-From: bruce@faxmail.co.nz (Bruce Simpson) Newsgroups: sci.physics.fusion Subject: Re: French nuclear test agenda Date: Sun, 24 Sep 1995 06:17:49 GMT Organization: FaxMail Technologies Mario Pain wrote: >bruce@faxmail.co.nz (Bruce Simpson) wrote: >>But France has repeatedly claimed that there is *NO* risk associated >>with these tests. If this is indeed the case then it matters not >>whether the devices are tested in downtown Paris or anywhere else for >>that matter. You can't claim total safety on one hand and "reduced >>risk" on the other. >> > I do not think France has ever claimed there was NO risk associated >with this tests. Such a claim will of course be absurd, as there is >no human act which has "zero risk". What is claimed is that this risk >is "in the noise", that is, the probability of an accident happening is >negligible when compared with any other possible accident. To be schematic, >a polynesian has more chance of being eaten by an alligator when getting >out of bed than to be affected by this tests. But that is a risk that the polynesian is well aware of and accepts as part of his daily life. French nuclear testing in the South Pacific is somewhat akin to New Zealand placing a nuclear device 100Km off the French coast and then connecting the trigger mechanism to a huge "wheel of fortune" device which gave a one in ten million chance of detonation. This is a *very* low risk - but would it be acceptable? The French would surely say "why not do this in your own country?" or perhaps "why endanger the lives of Frenchmen when there is no benefit to us associated with the risk?". This is all we are saying. With the nuclear testing at Mururoa the benefit is only to France but the risk is shared by many other countries. I do not accept that France has any right to endanger the lives of myself or any of my friends and familly simply to ensure its own sovereignty. > That does not mean of course that France should not take any measure >which will reduce both the probability of a nuclear accident, and the >potential effect of it. In that sense, the Mururoa site is the best of >the available ones. *best* for who? Certainly not the inhabitants of the South Pacific. France's attitude is not only arrogant but selfish. *----[Fixed-price software development over the net ]----* | bsimpson@iprolink.co.nz or bruce@faxmail.co.nz | *--[C/C++, Win, OS/2, POSIX, device-drivers, fax, comms]-* cudkeys: cuddy24 cudenbruce cudfnBruce cudlnSimpson cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / beefjerk / cmsg cancel <440d5e$a91@newsie.wis.com> Originally-From: Politics@usa.com (beefjerk) Newsgroups: sci.physics.fusion Subject: cmsg cancel <440d5e$a91@newsie.wis.com> Date: Sat, 23 Sep 1995 05:40:43 GMT Organization: Wisconsin Internet Services Cancelled-This post is inappropriate for this group. cudkeys: cuddy23 cudenPolitics cudlnbeefjerk cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / John Logajan / Re: Seeking explanation for observed events (revisited) Originally-From: jlogajan@skypoint.com (John Logajan) Newsgroups: sci.physics.fusion Subject: Re: Seeking explanation for observed events (revisited) Date: 23 Sep 1995 14:02:40 GMT Organization: SkyPoint Communications, Inc. Marcus Leech (mleech@bnr.ca) wrote: : > After a few experiments, we concluded that it : > had to do with the compression of the packed-granular snow at the site. : > Every time the snow made a 'crunching' noise, we observed broad-spectrum : > RF noise on the spectrum analyser. : > : > Is this actually related to the compression of the snow? And if so, how? : > I can concieve of both a small-scale piezoelectric mechanism and : > a static-discharge mechanism. On the day we did this, it was very : > cold (-15C), and dry. Airborne snow, of course, carries static charge -- we often see lightning strikes at a nearby TV antenna tower in the dead of winter just after a snowfall starts. I've also seen a snow fall generate high static charge in a small antenna I had mounted on the roof. The coaxial cable coming into the house would "twitch" from the changing high potential and sparks would periodically jump from the connector tip to the connector ground sheath. I ended up installing a permanent bleeder resistor to prevent possible damage to my radio equipment. I don't know if snow packed on the ground would be as effective at isolating charge. Another mechanism to consider is the material of the boot contacting and seperating from the snow. Just last week I was holding a 0-scope probe when I got up off my cloth chair -- that created a voltage spike. Then I noticed that stomping my feet caused a voltage spike. Lifting my foot off the carpet caused a polarity spike, and placing my foot down caused the opposite polarity spike. -- - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - - WWW URL = http://www.skypoint.com/members/jlogajan - cudkeys: cuddy23 cudenjlogajan cudfnJohn cudlnLogajan cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / ZoltanCCC / Re: I have a fusion reactor in my garage!!!!!!! Originally-From: zoltanccc@aol.com (ZoltanCCC) Newsgroups: sci.physics.fusion Subject: Re: I have a fusion reactor in my garage!!!!!!! Date: 23 Sep 1995 11:43:02 -0400 Organization: America Online, Inc. (1-800-827-6364) I would appretiate it if you could email me a description of your reactor. If you want to use regular mail I provide my address: Zoltan Szakaly 11517 Dellmont dr Tujunga CA 91042 cudkeys: cuddy23 cudenzoltanccc cudlnZoltanCCC cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / ZoltanCCC / Re: I have a fusion reactor in my garage!!!!!!! Originally-From: zoltanccc@aol.com (ZoltanCCC) Newsgroups: sci.physics.fusion Subject: Re: I have a fusion reactor in my garage!!!!!!! Date: 23 Sep 1995 11:43:36 -0400 Organization: America Online, Inc. (1-800-827-6364) We would all appretiate a brief description here in this newsgroup. Zoltan Szakaly cudkeys: cuddy23 cudenzoltanccc cudlnZoltanCCC cudmo9 cudqt3 cudszS cudyr1995 ------------------------------ 1995.09.23 / Mitchell Jones / Re: Testing the Mitchell Jones Hypothesis Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: Testing the Mitchell Jones Hypothesis Date: Sat, 23 Sep 1995 10:50:29 -0500 Organization: 21st Century Logic In article <402216017wnr@moonrake.demon.co.uk>, Alan@moonrake.demon.co.uk wrote: > Mitchell Jones seems to have developed paralysis of the fingers > regarding his 'slower than light' photons. Whatever can this mean? > -- > Alan M. Dunsmuir [@ his wits end] (Can't even quote poetry right) > > I am his Highness' dog at Kew > Pray tell me sir, whose dog are you? > [Alexander Pope] > > PGP Public Key available on request. Alan, you are a glutton for punishment. The fact is that I had already replied to your post before you made it, in a reply to Bill Rowe which you either didn't read or else conveniently forgot. Frankly, re-posting material for the benefit of people who didn't read it when I posted it the first time gets a low priority with me. I prefer to respond to new comments. However, for your benefit, the relevant excerpts from my reply to Bill Rowe follow: ------------------------------------------- "At this point, we have to ask ourselves how a gamma photon gets spat out as the isotope shifted nucleus drops to ground state. The question is, do we visualize a series of transitional states in which the photon accelerates to the speed of light, or not? By the "quantum mechanical" mode of thinking, there is not merely no problem in imagining an entity leaping from zero velocity to the speed of light without passing through all of the intervening transitional states, but such images are positively to be encouraged! However, as I have noted elsewhere, such imagined gaps in motion violate the principle of continuity. However brief the velocity rise time may be, the principle of continuity requires that the photon pass through all the intervening states of motion. Period. The fact that we have not measured such states does not mean they do not exist. Indeed, they must exist because the principle that requires their existence (the principle of continuity) supports the entire structure of human knowledge. To question its validity self-reduces to an absurdity. By the way, please do not assume that I am alone in visualizing such transitional states. Even physics gurus routinely do it. Unlike me, however, they are careful to kiss the behinds of the powers that be by pretending to believe in the sacred mumbo-jumbo of "quantum mechanics." For example--and it is one among many--check out the various "dressing transformations" that have been applied to the original Hamiltonian to curve fit it to newly emerging experimental data. If memory serves, Pauli applied such a transformation to explain the Biot-Savart law. His interpretation was that the magnetic field which accompanies a moving charged particle consists of a cloud of "bound photons" which move along with it. (Yes, Virginia, photons don't always move at the speed of light!)" ------------------------------------------- [End of quoted material] So, Alan, are you still laughing? If so, then I must note that you are not merely laughing at me, but also at Wolfgang Pauli! Pretty funny, huh? --Mitchell Jones =========================================================== cudkeys: cuddy23 cuden21cenlogic cudfnMitchell cudlnJones cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / A Plutonium / Re: First experimental predictions of Hydrogen Atom Systems Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Re: First experimental predictions of Hydrogen Atom Systems Date: 23 Sep 1995 23:27:02 GMT Organization: Plutonium College In article <441g71$842@dartvax.dartmouth.edu> Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes: > Physicists for the past 40 years have not been truthful to > themselves. They should realize that when a particle exists for such a > fleeting time as 10^-6 sec that such a particle is really not on "par" > to the existence of say the proton or electron etc. > To put it on terms that all readers can understand. If we were to > give a comprehensive science study of Homo sapiens, should we spend any > time whatsoever or some little time on the fact that a worm lives in > the eyelashes of humans, considering that we are really talking about > Homo sapiens and not an organism that lives on Homo sapiens? In the > same vain, is the muon a particle on par with the electron, or is it a > energy exchange, that is, the muon is merely an electron with stored up > extra energy and not a true particle. Here is a better analogy of the present day particle physics zoo. Suppose particle physics was the start of the classification of life, a la Linneas. How would present day particle physicists have done a job on the starting out of the classification of life and models thereof. Well we have the Plant Kingdom and the Animal Kingdom and we have things like the archopods and mammalia and invertibrates etc. That was just a passing trespass of the biology classification so some biologist do not holler at me for it. BTW I need a complete biology classification scheme so if someone has it please post. But getting to the analogy of transporting today's particle zoo physicists back to the time of Linneas and ask them to work up a classification. Then I suspect what they would include is bread loaves in grocery stores because some particle zoo physicist noticed that people eat bread and so concluded that bread was alive. And so the classification of life by our present day particle zoo physicists would have been a far larger scheme to include every imaginable brand of bread and lollipops and suckers would have their space I am sure. cudkeys: cuddy23 cudenPlutonium cudfnArchimedes cudlnPlutonium cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.22 / James Panetta / Re: Testing the Mitchell Jones Hypothesis Originally-From: panetta@finch.SLAC.Stanford.EDU (James H. Panetta) Newsgroups: sci.physics.fusion Subject: Re: Testing the Mitchell Jones Hypothesis Date: Fri, 22 Sep 1995 02:56:00 GMT Organization: Stanford Linear Accelerator Center Hi All, Mitchell sent me a copy of the original protoneutron post. I replied to him via email, and he has encouraged me to post it to sci.physics.fusion. /**************start included message*********************/ Hi Mitchell, > You mentioned that you had not seen the original post. It follows. It > should answer most of the questions in your recent post. The one about > photons below lightspeed is addressed in a reply to Bill Rowe, already > posted. If you have further questions, feel free to ask. Thanks for the prompt reply. I have a few comments on some of your physics. > ***{To repeat: I do not see a need for an alternative model. A voltage > gradient is all that is required to attract an H+ ion into the unit cells > of the cathode. It doesn't have to be "forced" in: the openings in the > triangular faces of the unit cells are large enough for it to move in > *very* easily--as easily as a fly into an amphitheater, in fact. Remember: > an H+ is simply a proton--i.e., the nucleus of a hydrogen atom. The > standard formula, curve-fitted to the experimental data, for computing the > radius of a nucleus is simply r =3D (1.37E-13)A^(1/3), where r is the radius > in cm and A is the mass number. For H+ the mass number (number of protons > plus number of neutrons) is 1, so r =3D 1.37E-13 cm =3D .0000137 A. Once > inside the cathode, such an H+ ion will drift toward the electron source > until it acquires an orbital electron. At that point, it will become an > electrically neutral H atom, with an effective radius as per the innermost > Bohr orbit (.53 =C5 or slightly larger). This means, roughly speaking, that > its radius will increase by four orders of magnitude when it captures an > electron, and it will become trapped in the unit cell which it is > occupying at that time. --Mitchell Jones}*** You really can't use the proton radius to determine the size of a H+ ion, you should determine the radius at which the coulomb force is powerful enough to remove an electron from a neighboring atom. This distance is probably not less than half of the Bohr radius. > ***{At this point, I would like to toss out what I call the protoneutron > theory of "cold fusion." If this theory is correct, then "cold fusion" > ain't fusion! > > Note that the drift of the H+ ions into the cathode does not require a > free path through unoccupied unit cells. In an occupied unit cell, the > occupant is a neutral H atom. Thus there is no voltage gradient barring an > H+ ion from drifting into an occupied unit cell, and there is no voltage > gradient preventing an outward drifting electron from passing into an > occupied unit cell, either. This means it is purely a matter of > statistical probability, regarding the question of whether a drifting H+ > meets an electron in an occupied unit cell. If it does, it will attempt to > "pop" into a neutral hydrogen atom. Unfortunately, because the unit cell > is already occupied, when the second H atom attempts to form it will not > have sufficient room for its electron at even its innermost Bohr orbit. > Result: the electron will spiral down toward the nucleus, where it will > linger at grazing altitude in a particle form which we may term a > *protoneutron.* First problem, electrons do not *spiral* in to the nucleus. > Why doesn't it become an actual neutron? The answer: in > the reaction p + e --> n, the mass of the proton is 1.67239E-24 grams > while that of the electron is .00091E-24 grams. Second problem, p+e-->n violates lepton number conservation. Where's the neutrino? Third problem, electron capture is known to proceed via W exchange. This process is well known, and calculable. The theory fits the known data. What prediction would your protoneutron theory give for sodium-22 EC? Iron-55? Fourth problem, if protoneutrons exist, they will cause an effect in the endpoint energy of tritium beta decay. The endpoint curve is extremely well known due to the search for massive neutrinos. What is this effect? > I would add that I am perfectly aware of the conflict between the > protoneutron theory and "quantum mechanics." This is a classical > mechanical theory, and classical mechanics is based on the principle of > continuity--i.e., the principle that no entity may come into existence out > of nothing or vanish into nothing. The principle of continuity indicates > that an entity can arrive at, or leave, a location in one way only: by > successively occupying each position in a continuous spatial pathway to, > or away from, that location. The implication is that motion is continuous, > not a series of quantized "jumps." Then explain the photoelectric effect. > This means, for example, that an > electron moving from one stable orbit to another does not "jump" in the > quantum mechanical sense: it follows a continuous spatial pathway from the > former orbit to the latter, and it exists just as surely during those > instants when it is between the "preferred" orbits, as when it is in one > of them. In the case presently under discussion, the principle of > continuity means that an electron exists and follows a continuous spatial > pathway when it is transiting from the lower Bohr orbit to a position in > the nucleus. It does not merely exist "in" the nucleus and "in" the > innermost Bohr orbit, but also in between. The protoneutron, in short, is > a classical mechanical phenomenon. It violates the most fundamental > precept of "quantum mechanics" and, if it exists, it constitutes one more > piece of evidence, in a pile that is already mountainous, indicating that > motion in the microcosm is not "quantized" and, hence, that "quantum > mechanics" is wrong. What evidence is there that quantum mechanics is wrong? All evidence I have seen and *performed* experiments to examine has shown that quantum mechanics fits the data. > Needless to say, I present the protoneutron theory as a hypothesis subject > to verification or falsification. If it proves to be indefensible, then I > will abandon it. On that, you have my absolute guarantee. --Mitchell > Jones}*** I think that protoneutrons are indefensible, but that's the opinion of a particle physicist at Caltech and you may take it how you wish. Thanks again for responding quickly. --Jim Panetta -- -- My opinions are mine...not SLAC's...not Caltech's...not DOE's...mine. (except by random, unforseeable coincidences) panetta@cithex.caltech.edu panetta@slacvm.slac.stanford.edu cudkeys: cuddy22 cudenpanetta cudfnJames cudlnPanetta cudmo9 cudqt3 cudszL cudyr1995 ------------------------------ 1995.09.23 / / Re: The electron capture theory of cold fusion Originally-From: mrichar353@aol.com (MRichar353) Newsgroups: sci.physics.fusion Subject: Re: The electron capture theory of cold fusion Date: 23 Sep 1995 21:37:10 -0400 Organization: America Online, Inc. (1-800-827-6364) zoltanccc@aol.com (ZoltanCCC) wrote: >In article <4409ec$nlq@newsbf02.news.aol.com>, mrichar353@aol.com >(MRichar353) writes: > >> >>Oh, come on. I have refuted Robin's idea conclusively, at least for >people >>who know physics at an undergraduate level. What you and he are talking >>about violates energy conservation, it's as simple as that. >> >> > >Mark, unfortunately your half cocked reference to the conservation of >energy did not cut it with me. Up to this point nobody has described the >situation as the way I see it now. Here it follows: > [snip of an attempt to classically describe electron capture, followed by the statement that the previously-proposed method won't work] Any classical description of the process is perforce intrinsically inconsistent, and this is probably the source of the confusion in your theory. I'm glad you were still able to derive the fact that energy can't come from nowhere. The nice thing about conservation laws is that they are independent of (although consistent with) the underlying details. Sorry about my tone in my previous post, I was in a foul mood. Although you are a "creative thinker" you are not a crank, as you are able to profit from constructive criticism. Mark Richardson cudkeys: cuddy23 cudenmrichar353 cudln cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.23 / / Re: Mark Richardson, Mr. Dupree, and Mr. Esch Originally-From: mrichar353@aol.com (MRichar353) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Re: Mark Richardson, Mr. Dupree, and Mr. Esch Date: 23 Sep 1995 21:44:17 -0400 Organization: America Online, Inc. (1-800-827-6364) Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote: >In article <43tc0g$q4s@newsbf02.news.aol.com> >mrichar353@aol.com (MRichar353) writes: > >> >> God how idiotic! >> >> Superposition has absolutely nothing to do with building up a >> multi-particle system from individual particles. It is the principle that >> a linear combination of two wave functions is another wave function. These >> two (or more) wave functions represent *alternatives*; they cannot both be >> true at the same time (at least when a measurement occurs). >> >> Then again, considering the source, what should I expect? >> >> I *must* get a newsreader with a killfile capability! >> >> Mark Richardson > > I was updating the go-to-hell file which I keep and saw that Mr.Craig >Dupree and Mr. Patrick P. E. Esch were on it. Reconsider. > PU, PLuto revert both Craig Dupree and Patrick P. E. Esch back to >point of neutrality, away from hell. Atom > > As to Mark Richardson who I told to go to hell, his spite and hatred >still spews forth in greater amounts, I say go to hell, Styx this time. That was not hatred, Archie, it was frustration mixed with exasperation, tucked in the middle of which was some constructive criticism. If you want to talk about superposition you might as well know what the word means. However you don't seem to want to respond to the constructive part. Mark Richardson cudkeys: cuddy23 cudenmrichar353 cudln cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.24 / Jacques Distler / Re: All atoms are Hydrogen Atom Systems -> Superposition Principle Originally-From: distler@golem.ph.utexas.edu (Jacques Distler) Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag sci.physics.fusion,sci.physics.accelerators,sci.physics.particle Subject: Re: All atoms are Hydrogen Atom Systems -> Superposition Principle Date: Sun, 24 Sep 1995 00:05:30 -0600 Organization: Physics Department, University of Texas at Austin In article <4424aa$jku@dartvax.dartmouth.edu>, Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote: > > Someone please mind posting a representative list of elementary >particles, just the most important members of each family showing the >MEV, all in MEV. Plus indicate the lifetime of the particle all in >seconds. Then what the particle decays into. Please post with a name of >the particle not the Greek symbol. Thanks Hey, so you'd actually like to learn some physics? Very commendable. I'll do better than give you a little table. The entire Particle Data Book is online: http://www-pdg.lbl.gov/rpp/book/contents.html Check it out. More than enough information for a 100 of your posts. Jacques cudkeys: cuddy24 cudendistler cudfnJacques cudlnDistler cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ processed by cud.pl ver. 0.5 Sun Sep 24 04:37:04 EDT 1995 ------------------------------