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265) A paper of Antonella, Antonio, Antonietta, Emilio and Guiliano

Ludwik Kowalski (10/20/05)
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043

How can cold fusion be explained? The answer to this question can be found in the appendix of a paper of Antonella De Ninno et al. “Experimental evidence of 4He production in a cold fusion experiment.” The pdf file containing this paper was sent to me (October 2005) by Jed Rothwell. Thanks, Jed. Imagine molecules of air through which sound is propagating. They oscillate with amplitude which increases with loudness. Layers of compression and layers of rarefaction follow each other as water crests in a wave propagating along the surface of a pond or pool. Oscillations of sound, however, are in the direction of propagation, not perpendicular to it, as in familiar water waves.

Looking on two nearby molecules, oscillating along the direction of sound propagation, one would see that the distance between them tends to increase and decrease periodically. Something similar might occur in palladium loaded with deuterons. The average distance between the two nearby 2D ions is too large to allow fusion but the nuclei can occasionally fuse when distances are close to the minimum. But this is only part of the story. The other part has to do with answering a different question: “why are cold fusion products,are different from the hot fusion products? Most CMNS products are 4H while most CF products are n, p, 3He and radioactive 3T.”

The authors say that, according to QED (Quantum ElectroDynamics), sudden change is expected to occur in Pd when atomic concentration of dissolved 2D ions exceeds 0.7. Deuterium ions start oscillating “inside shallow potential waves . . . [and] “the Coulomb barrier turns out to be smaller than in vacuo, because of the high concentration of the negative” electrons. That is the answer to the first question “why does cold fusion happen?” Addressing the second question the authors describe cold fusion as a three step process:

a) Two 2D ions fuse forming the excited compound nucleus 4He. The well known excitation energy is 23.8 MeV.
b) Rapid “cooling” -- decay of the excited nucleus to the ground state
c) Transfer of the released energy to the Pd crystal.

The step (a) must occur in any kind of fusion (in gas-like plasma and in condensed matter). But steps (b) and (c) are very different in these two kinds of fusion. In plasma deexcitation , most often, takes place by by fragmentation of the compound nucleus. It breaks into either n+3He or p+3H. Deexcitation by emission of gamma rays is also possible but it is much less probably. In condensed matter, on the other hand, fragmentation is prevented by close proximity of charged particles, negative electrons and positive ions. The released energy goes to charged particles, mostly to surrounding ions. Metallic grains in which fusion occurs become very hot, momentarily, and loose heat by well known mechanisms (radiation, convection and conduction). Helium gas is most often released from very hot “micro furnaces.” The appendix end with this sentence: “From our ‘prejudice’ that almost all fusion occurring in the Pd lattice should produce 4He, we derive that the easiest way to evaluate the energy output is to count 4He atoms in the cell gasses.” And that what the main part of the paper is about.

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Dear Dr. Kowalski-
      The more I examine the information about cold fusion, the more it appears to me that the most probable explanation of the phenomenon is the Palladium catalyzed conversion of  D:D to He4 along with an overlooked concurrent Pd catalyzed conversion of H:H to Deuterium atoms.  It is very possible that the electrical field is not necessary and the electrolytic cell is serving only to furnish D:D  for conversion.
      It is possible that Hydrogen gas forced through a thin sheet of Palladium would show Deuterium enrichment on the other side and a temperature rise of the Palladium barrier.  Similarly, if Hydrogen gas heavily enriched with Deuterium were forced through a Palladium barrier,  it would not be surprising to see heat, Tritium, He3 and He4 produced.   In so far as I can tell, no one has considered the molecule to "atomic isomer" transformations as a possible explanation of the phenomenon.

      If one looks at the published data re size and mass of the electron and the proton one finds an interesting situation. We usually think of the proton as being large and massive and the electron small and light.  However, if the electron were scaled up to a 1 cm. water balloon, the proton would be more like a 100 meter soap bubble!  You're undoubtedly a better mathematician than I ,  so you probably can work out a better comparison than that.  The point is that the mass and charge density of the electron  is many times that of the proton, indeed,  the disparity is so great that there seems no reason at all that electrons cannot pass freely through protons.  Similarly, two protons would  have to approach to about the diameter of an electron to experience the same repulsive forces as two electrons colliding. In other words,  there is a fair chance that a Hydrogen Molecule spends at least some time in the same configuration as a Deuterium atom, but does not stabilize to that form because of the vibrational energy content...,.(This could be considered as essentially the same reason that absolutely pure gaseous water will not condense to liquid droplets ......)   The same kind of analysis applies to the  D:D,  He4 system and the  H:D  to Tritium  possible conversion....Best Wishes,  Dean L. Sinclair, BA, MS, PhD

From previous correspondence with Dean, whose life circumstances are not favorable to conduct research, I know that he would be happy to receive comments. Here is his email address: <>

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Let me also share the URL where Bill Beaty, a friend, and a longtime observer of the CMNS field, tells us how to recognize fraudulent technological claim. I think that potential investors should benefit from reading his piece. Share this URL with them:


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