Where is the Gamma Radiation? (Profchuck)

This comment was originally posted by ECW reader profchuck here.

There is a fundamental question: “where is the gamma radiation?” Nucleosynthisis on this scale releases energy that is proportional to the mass difference between the “fuel” and the end product. No known fusion process takes place without releasing some of this energy in the form of high energy radiation and particles. To produce the 1.5 megawatt hours over the 30 odd days of the experiment should have produced gamma radiation that was not only detectable but lethal. For this to be real one must accept that not only is low temperature fusion possible but it can occur without producing detectable radiation. That would be two major discoveries that can only be called “New Physics”.

  • Josh G

    Finding an answer to this question is key. Seems to me that we’ve got a decent (but far from complete) understanding of how fusion is possible at low temperatures in condensed matter through boson condensation and electron capture. But the lack of gamma radiation is still a mystery, as far as I can tell. Everything we (currently) know about nuclear physics tells us there must be gamma radiation produced. If someone could develop a rigorous, plausible theoretical explanation for the lack of gamma radiation, this could go a long way towards overcoming scientific skepticism and developing an understanding of the LENR process so as to better harness it (and winning a Nobel).
    It may very well be that “the energy is released as phonons,” but you have to work out the mathematical details and physical properties to explain how this happens before people will take the idea seriously. I know Hagelstein has been working on this recently with a former grad student (now prof. in India) having to do with fractionated gammas or something along those lines, but I don’t know how well developed or plausible their model is. Does anybody know more about this or other explanations?

  • Gerrit

    found on the internet, so disclaimers apply:

    “Heavier-than-air flying machines are impossible.” – Lord Kelvin,
    president, Royal Society, 1895.

  • Observer

    In semiconductors there are direct and indirect transitions of electrons in the conduction band going to the valance band. In direct semiconductors (e.g. GaAs) the electron does not change momentum, allowing a photon to release the excess energy. In an indirect semiconductor (e.g. Silicon) the electron needs to change momentum to get from the bottom of the conduction band to the top of the valence band, and thus phonons are needed to release the excess energy. A atom in free space has no latice to release phonons and thus must rely on photons (gamma rays) or expell excited particles. Nuclear reactions with in a lattice may have alternate means of releasing excess energy (e.g. phonons).

  • GreenWin

    Electrons in pycnonuclear condensed matter (e.g. in stars) effectively screen the Coulomb repulsion making nuclear reactions highly probable at low temperatures. Ikegami and the chemonuclear transition provides answers. It may be that Rossi’s reaction starts with the collapse of hydride H-H bonds followed by Li7+Li7 leveraging the Gibbs temp energy change. Ikegami shows high fusion rates in liquid Li surface embedded with H, D, D2. http://www.roxit.ax/CN.pdf

    The heavy hydrogen Li-system is patented and assigned to Japanese energy system maker Sakaguchi E. H Voc. It seems reasonable to expect a commercial heat product from Sakaguchi around the time other LENR products arrive.

  • Where is the Gamma Radiation?

    The gamma radiation is in your perhaps faulty assumption that what’s happening has to be something we already understand and have cataloged.

    Of course the ash sample or report data could be incorrect or even faked. We all have to make assumptions in this situation and nobody knows exactly what’s happening or who to believe.

    But observation trumps theory. Let’s see where the data takes us and not get too hung up on how it’s different than anything we’ve seen before or what was expected. if we can work out a good model for it then we’ll be able to test it to see if it’s real or BS.

    • Ophelia Rump

      The Real or BS argument is settled. Please accept that and move on.
      If you would continue to replay that loop, this way lies madness.

      There is something producing heat. Heat is real.
      If you do not believe that heat is real, for you my friend there is waste, waste is real.
      A little something for the never believer.

      • At this juncture I am persuaded. However I remain doggedly open-minded and will pursue all leads, fantastic or ugly.

        I’m on team Truth, not team Rossi or team Dogma.

  • Gerrit

    The authors say basically the same thing in the abstract of the paper: “Nuclear reactions are therefore indicated to be present in the run process, which however is hard to reconcile with the fact that no radioactivity was detected outside the reactor during the run.”

    • Ophelia Rump

      By definition all factors are irreconcilable for the production of the impossible.

      Throw that logic out the window. It proves only that what you believe to be impossible becomes immune to reason.

  • Pekka Janhunen

    That Bianchini measured no radiation above background from the reactor and from the ash is an extremely stringent constraint. One gram of nickel contains about 1e22 Ni atoms. If one in million of them would be Ni(59), for example, the number of Ni(59) atoms would be 1e16. Ni(59) half-life is 76000 years (2.4e12 s), so every second there would be 4000 decays. Ni(59) decays by beta+, that is, positron emission. The positrons are annihilated with electrons which produces a pair of 511 keV gammas for each decay. Such gammas are already quite penetrating and thus easily detected. Bianchini used several measurement techniques based on different principles. It is inconceivable that all of them would have malfunctioned.

    One must follow the evidence, and the evidence is that there is zero radiation from this version of HotCat.

    One possible difference to earlier E-cats is that the new model apparently uses hydrogen from which deuterium has been removed (I think it’s said somewhere in the report).

    • Pekka Janhunen

      It’s worth remembering that Rossi himself is puzzled by the 99% Ni-62 outcome and is investigating the issue. Since earlier Rossi has said that Ni-62 and Ni-64 perform well, it’s possible and even likely that he has made some tests also with enriched Ni-62 in his lab. Then it’s not inconceivable if the analysed grain was contamination from earlier tests, either by the same or different reactor. I’m waiting for Rossi’s and/or the testers’ further comments about the Ni-62 issue before taking it really seriously.

      • Gerard McEk

        Can it be that in the latest Ecat Lithium has taken the role of Hydrogen in the older Ecat designs? I do not believe that Hydrogen can be kept in this new Alumina Ecat. It would disappear very fast. That will cause also the different isotopic shift of the nickel ash.

        • There’s not enough Lithium for it to be one of the primary reactants. It may be an important catalyst.

          The amount of Ni-62 created can only be explained by the iron and aluminum in the fuel (as well as the other nickel isotopes in the fuel) providing the raw material.

          Said another way, the ratio of the rise in Nickel nucleons in the ash to loss of Lithium nucleons in the ash is about 50:1.

          See https://docs.google.com/spreadsheets/d/1JJjNVq_2euIwwmfOlVb4MK_UigkcoriisW5VsB7hu5c/edit#gid=0

          • Actually I’m going to walk that back a bit. Lithium could be considered on par with the other reactants but only if swallowed wholesale, not just contributing a single neutron (or proton).

        • Pekka Janhunen

          Sounds possible, unless he has some hydrogen barrier inside the reactor. I haven’t calculated hydrogen diffusion rate through alumina, but it sounds plausible that it would otherwise escape. A replication attempt would answer this question of hydrogen escape rate.

        • Gerard McEk

          I was wrong with my feeling that aluminia would be not very Hydrogen tight, I found:
          “Hydrogen permeability of high-density, high-purity sintered alumina tubes was measures is a function of temperature and pressure using tritium as a tracer. The permeability, ø, at 1200° to 1450°C and hydrogen partial pressure between 2 and 50 kPa is:

          φ(H atoms cm cm−2 s−1 kPa−0.43) = exp(48.95±0.61)

          exp[−318.2±18.8 kJ(g·atom)−1 (RT)−1]

          Diffusion coefficients and solubility values deduced from the permeation experiments are consistent with earlier independent measurements. No accelerated permeation due to micro-structural defects or changes during the experiments was observed. Comparison of hydrogen permeability of alumina with the values observed for metals shows that alumina may be a suitable coating material for use as a hydrogen permeation barrier.”

          • Pekka Janhunen

            Ok, good find. Good old hydrogen…

          • Sanjeev

            Great. That solves a big question.

    • Andrea

      Dear Pekka, I think Ni59 decays mainly by electron capture and transforms to the stable Co59. The brunching to beta+ (positron emission) is about very low: 0,00003%.
      Please correct me if I am wrong.

      • Pekka Janhunen

        Yes, I think you are correct. Nevertheless, electron capture is associated with gamma emissions. The energies of those gammas must exist in some tables, although I didn’t yet find them easily.

        • Andrea

          The Auger electron emissions should be in the linked table. They are in the X range: a few [keV].

  • fritz194

    Analyzing Rossi´s comments – it´s tuned for no gamma. In the earlier e-cats – there was mild gamma available – and some lead shielding in place. So maybe it´s a superposition of known and new physics.
    I also remember that gamma and neutron radiation was detected on blowing up some e-cats…

  • LCD

    I think to some extent you are right but remember you are thinking of hot particle physics as your basis.

    If there are n-body reactions where n is >>2 we can imagine everything working within the laws of physics.

    But it’s hard to imagine how n-body reactions are taking place.

    Also whatever enables the reactions also prefers to create stable isotopes. That’s wired. But if you think about it it has to be that way or else there would be dangerous radiation no matter what. At least it is self consistent.

  • Andrew

    Using law and rules for know reactions don’t apply to the unknown.

  • Ophelia Rump

    Very Cool, Prof Chuck.

    I want to hear more about the edges of where this new science resides.