Article 4739 of sci.physics: Subject: New theory of room temperature fusion Date: 31 Mar 8

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Article 4739 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!rutgers!ucsd!ames!lll-winken!uunet!mcvax!ukc!etive!bjp From: bjp@etive.ed.ac.uk (B Pendleton) Newsgroups: sci.physics Subject: New theory of room temperature fusion Message-ID: <1672@etive.ed.ac.uk> Date: 31 Mar 89 23:12:49 GMT Reply-To: bjp@etive.ed.ac.uk (B Pendleton) Organization: Edinburgh University Computer Services Lines: 43 Posted: Fri Mar 31 18:12:49 1989 Posted for a friend without access to the net -- *please* do *not* reply to me. Responses to J.Butcher@edinburgh.ac.uk Theory of "Kitchen Table Fusion" -------------------------------- Most physicists agree that the most puzzling part of the recently announced room temperature fusion is how two deuterons can overcome the intense coulomb repulsion and get sufficiently close for the short range attractive nuclear force to take over and produce a triton plus emitted neutron. It has been suggested that there is a catalysis or tunneling process that allows the two deuterons to get very close. I shall argue that this is not necessary. What may actually happen is the following. Deuterons are absorbed into the regular crystal lattice of the Palladium electrode in such a way that the deuterons are aligned in their internal isospin space. Now, since isospin is a continuous symmetry, and this alignment picks out a particular direction in isospace, then Goldstone's theorem tells us that there must be a massless Goldstone boson associated with this symmetry breaking. For the proton and neutron, this isospin symmetry is not exact -- the u and d quarks are not exactly massless and the Goldstone boson corresponding to axial symmetry breaking (namely the isotriplet pion) is not exactly massless -- it has a mass of 139 Mev/C^2. However, the deuteron contains equal numbers of u and d quarks, moreover, it is the *vector* isospin symmetry that is broken in the Palladium crystal, and the symmetry breaking factor cancels to some extent. A back-of-the-envelope calculation reveals that the Goldstone of the new symmetry breaking would have a mass of the order of tens of Kev/C^2 -- or about the same as the energy in the Coulomb barrier!! This balance of energies allows fusion to occur in a controlled fashion. In other words, there is little chance of an "accidental H-bomb". Another way of looking at it is to notice that this new Goldstone gives the nuclear force a longer range than is usual and so the tunneling or catalysis mechanism is not necessary -- the two deuterons do not need to get close together for fusion to occur. In addition, the energies of the emitted neutrons will be *small* -- one does not need to "get back" the coulomb energy. Of course, the theory has to be worked out in more detail, but it does seem more promising than any other I have seen. John Butcher -- Hello! Article 35 of sci.chem: Path: dasys1!cucard!rocky8!cmcl2!husc6!ukma!gatech!bbn!rochester!dietz From: dietz@cs.rochester.edu (Paul Dietz) Newsgroups: sci.physics,sci.chem,sci.research,sci.space Subject: Re: Reactions described in the Pons seminar summary Message-ID: <1989Apr3.152949.23607@cs.rochester.edu> Date: 3 Apr 89 19:29:49 GMT References: <1495@wasatch.UUCP> <3604@silver.bacs.indiana.edu> <24015@beta.lanl.gov> Reply-To: dietz@cs.rochester.edu (Paul Dietz) Organization: U of Rochester, CS Dept, Rochester, NY Lines: 34 Xref: dasys1 sci.physics:4771 sci.chem:35 sci.research:684 sci.space:8441 Posted: Mon Apr 3 14:29:49 1989 In article <24015@beta.lanl.gov> mwj@beta.lanl.gov (William Johnson) writes: >However, I would like to point out that the most mystifying thing about the >Fleischmann-Pons experiment -- and many things about it are mystifying -- is >that *none* of the nuclear physics makes sense. I say this not implying that >F&P don't know what they are talking about, but rather that many things about >the experiment -- notably the enormous dearth of neutrons observed relative to >the energy allegedly released -- fly in the face of what we *think* we know >about (d,d) reactions. Everyone has been assuming that the neutrons are coming from catalyzed dd reactions. If, instead, some exotic fusion reaction was occuring -- say, Li6 + d -- we'd expect some neutrons anyway. First, a fast charged fusion product would occasionally break up a deuteron before stopping. Second, deuterons would occasionally be scattered and fuse with another deuteron. Some proposed experiments: (1) Measure the ratio of neutron rate/power density as the density of d atoms increases. It should increase if this model is true. (2) Measure the neutron spectrum -- it should differ considerably from that of cold dd fusion. (3) Try to detect energetic fusion product nuclei by mixing the Pd with beryllium and observing the neutron flux. (4) Try to observe fusion products directly by using a low energy deuterium ion beam to saturate a very thin target of Pd. Turn off the beam and observe any charged particles emitted. Paul F. Dietz dietz@cs.rochester.edu Article 4761 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!husc6!bloom-beacon!tut.cis.ohio-state.edu!ucbvax!decwrl!shelby!glacier!jbn From: jbn@glacier.STANFORD.EDU (John B. Nagle) Newsgroups: sci.research,sci.physics Subject: Energy efficiency of Pons-type cold fusion Keywords: fusion Pons Message-ID: <18244@glacier.STANFORD.EDU> Date: 2 Apr 89 18:45:16 GMT Sender: John B. Nagle Organization: Stanford University Lines: 23 Xref: dasys1 sci.research:678 sci.physics:4761 Posted: Sun Apr 2 13:45:16 1989 Does this thing really result in a net gain in energy? Pons reports that one must pump energy in for weeks to months before getting energy out. Further, there seems to be an implication that the energy output doesn't go on indefinitely once output is achieved. But Pons's numbers on output only seem to refer to the situation that exists during the output phase. He doesn't integrate the total energy input during the startup phase into his "gain" figures. The obvious question to ask is whether this is some kind of energy storage phenomenon. Suppose that there are actually two phenomena taking place here. One is tunnelling-type fusion as previously reported, which would account for a modest output of neutrons. The other is some kind of energy storage, which may be chemical in nature, as in a battery. This would explain the low output of neutrons along with the high energy output. From this standpoint, a reasonable question to ask is whether the energy density observed is totally out of reach for an electrochemical system. Is a battery hypothesis totally untenable? John Nagle Article 4750 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!rutgers!cs.utexas.edu!ut-emx!ethan From: ethan@ut-emx.UUCP (Ethan Tecumseh Vishniac) Newsgroups: sci.physics Subject: Re: Success with cold fusion reported Summary: making Helium4 Message-ID: <11655@ut-emx.UUCP> Date: 1 Apr 89 22:12:09 GMT References: <13268@sequent.UUCP> <8904012009.AA06967@cunixd.cc.columbia.edu> Organization: The University of Texas at Austin, Austin, Texas Lines: 40 Posted: Sat Apr 1 17:12:09 1989 As I understand it, from summaries of talks by Pons and Fleischmann (not from their preprint) they have detected Helium4 as a byproduct of their reaction. Their estimation of production rates for Helium3 and tritium are consistent with each other and much lower (by about 10^9) than the rates required by the observed (or claimed) energy production. It therefore seems impossible to invoke any reactions that use He3 or tritium as fuel to explain the energy production. This leaves 2D into He4, but under normal circumstances this is down from the production rates for He3 and tritium by one power of the fine structure constant since it involves emitting another photon. So we can either discard this possibility or choose to believe in an enhancement of the rate due to the emission of phonons with energies of several MeV. (Incidentally, this makes the fusion experiment the shrillest noise on Earth :-)). I find the idea of these "super-phonons" unlikely, but I'm hardly an expert. The only alternative that comes to mind is Li6 + D into 2He4. This has the virtue of avoiding nucleon emission in a reaction whose rate is not obviously tied to the measured reaction rates for 2D into something. (Actually, I assume that this would make Be8 which would decay into 2He4 on a time scale of 10^(-8?) seconds.) This means the addition of LiOD to the solution is important for physical, as opposed to chemical, reasons. However, is it really true that the lithium would dissolve into the palladium? If not, it is difficult to see how it could be involved. The results of P and F still don't make much sense to me, but if they are correct then this would seem like the most likely hypothesis (which may be a dubious honor). If this is right then it suggests that lithium is about to become a little more expensive. :-) (sort of). -- I'm not afraid of dying Ethan Vishniac, Dept of Astronomy, Univ. of Texas I just don't want to be {charm,ut-sally,ut-emx,noao}!utastro!ethan there when it happens. (arpanet) ethan@astro.AS.UTEXAS.EDU - Woody Allen (bitnet) ethan%astro.as.utexas.edu@CUNYVM.CUNY.EDU These must be my opinions. Who else would bother? Article 4759 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!ames!amdahl!pyramid!prls!philabs!linus!mbunix!eachus From: eachus@mbunix.mitre.org (Robert Eachus) Newsgroups: sci.physics Subject: Re: UU Power Requirements Summary: I don't agree Keywords: power fusion Message-ID: <47059@linus.UUCP> Date: 30 Mar 89 00:53:46 GMT References: <10232@nsc.nsc.com> <10233@nsc.nsc.com> Sender: news@linus.UUCP Reply-To: eachus@mbunix.mitre.arpa (Robert I. Eachus) Organization: The MITRE Corporation, Bedford, Mass. Lines: 32 Posted: Wed Mar 29 19:53:46 1989 In article <10233@nsc.nsc.com> andrew@nsc.nsc.com (andrew) writes: >A check on the claims made here. Pons has reported : > Pd wire 0.25" diam, 1" long > 26 W/cc generated @ 100 degC after "a few minutes" > generated energy 4.5 times input energy (stuff deleted) >The output would be at 20mV with 4.6W supplied. I doubt whether this >current could be sustained for more than a minute or two, however. >In this configuration, a 12V source is dissipating 2900 W internally. >Maybe the batteries melted the concrete and exploded?! (just kidding). >Check my figures - I hope they're OK. I agree with your 4.6W figure, however as I understand the apparatus the current is not passing directly through the palladium, but is being used to electrolyze D2O. Depending on your asumptions about drop across the cell you get on the order of 2 Amperes, which is much more reasonable. I assumed that batteries were used for the demo (and the experiment) because for this type of setup a battery, electrolysis cell, and resistor loop allows you to acurately measure current and power without worrying about imposed AC. The four-inch deep hole in the concrete (assuming that it is true) would result either from spalling of the concrete or from hot battery acid removing the water. Concrete, as such, cannot melt, since once the cement has given up its water of crystalization all you have is a mix of dry powders, most of which will break down chemically before melting at one atmosphere. (CaSO4.2H20 -> 2CaSO4.H2O -> Ca(OH)2 + CaO + 2S03, etc.) Robert I. Eachus Article 4780 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!xanth!nic.MR.NET!umn-d-ub!umn-cs!ns!logajan From: logajan@ns.network.com (John Logajan) Newsgroups: sci.physics Subject: Fusion: Increasing energy gain. Message-ID: <1238@ns.network.com> Date: 3 Apr 89 17:30:39 GMT Organization: Network Systems Corp. Mpls MN Lines: 19 Posted: Mon Apr 3 12:30:39 1989 The energy-in/energy-out ratio was, I think, very carefully worded when the 10:1 figure was mentioned. I think they said the 10:1 ratio EXCLUDED the losses associated with I*I*R in the heavy-water, and possibly in the 0.8mv drop in the electrode itself. In fact, it is not clear to me if they include the several weeks worth of "pumping up" electrolysis loss. Which brings me to my next point: Even if a voltage gradient is needed during fusion (instead of just a pressurized vessel of deutrium gas), there is no reason to assume that the "pumping up" process couldn't be done in a pressurized mode. Maintaining pressure (even if it takes longer) is much more energy efficient than producing pressure via electrolysis. So, saturate the electodes in a pressure cooker first, then run them in fusion mode in the electrolytic bath. -- - John M. Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428 - - ...rutgers!umn-cs!ns!logajan / logajan@ns.network.com / john@logajan.mn.org - Article 4781 of sci.physics: Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!xanth!nic.MR.NET!umn-d-ub!umn-cs!ns!logajan From: logajan@ns.network.com (John Logajan) Newsgroups: sci.physics Subject: Notes on fusion Message-ID: <1237@ns.network.com> Date: 3 Apr 89 17:29:40 GMT Organization: Network Systems Corp. Mpls MN Lines: 24 Posted: Mon Apr 3 12:29:40 1989 I just read about the Mossbaur effect in the book Elementry Modern Physics. Seems that when a gamma ray is emitted from a free floating nucleus the nucleus recoils in the opposite direction -- the nucleus gains X kinetic energy and the gamma ray departs with Y energy. BUT!!! if the atom of the nucleus is bound in a lattice, then the recoil energy is smaller and the gamma ray energy is LARGER. In other words the fact that the atom was bound in a physical structure caused the gamma to exit with an increased frequency!!!!!!! The point is that the lattice chemical forces can affect the results of nuclear reactions!!!! Also in the book they say that typical spontaneous alpha fissions of heavy nuclei are of less energy than the coulomb gradient energy -- meaning that the alpha particles have tunneled out. They cite 8Mev as the coulomb force and 4.5Mev as the alpha particle energy -- in one instance. They also said that fusions can occur through tunneling, but they did not give a memorable example. -- - John M. Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428 - - ...rutgers!umn-cs!ns!logajan / logajan@ns.network.com / john@logajan.mn.org -

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