Spectators are not the answer

Subpage of Steven Byrnes

May 6, 2014.

In ordinary “hot” deuterium-deuterium fusion, you get:

  • D+D → neutron + helium-3 (~50% of the time),
  • D+D → hydrogen + tritium (~50% of the time),
  • D+D → helium-4 + a gamma-ray (0.0001% of the time)

Yes. That is “ordinary d-d fusion,” and it doesn’t actually matter if it is “hot,” i.e., muon-catalyzed fusion, very not hot, still shows the same branching, I understand.

In palladium-deuteride cold fusion, you allegedly get more-or-less only helium-4, plus energy that winds up as heat. Very strange!

It is only strange if we think we are looking at ordinary d-d fusion. We are not. We are probably not looking at d-d fusion at all, but something else, which includes the possibility of multibody fusion, which seems at first glance to be ridiculously unlikely, but that ridiculousness comes from thinking about fusion probabilities in a plasma. Condensed matter might be quite different, and, in fact, it’s reasonably established by experimental evidence — not well enough confirmed for my taste, but more than just a speculation — that the fusion rate for three-deuteron fusion in PdD under deuteron bombardment is hugely enhanced , 1026 being reported. (See Takahashi, A., et al., Detection of three-body deuteron fusion in titanium deuteride under the stimulation by a deuteron beam. Phys. Lett. A, 1999. 255: p. 89. ResearchGate. )

A reasonable guess is that the reaction is different because there is a third particle, besides the D+D, involved in the fusion reaction as a “spectator”

    • D + D + spectator → helium-4 + spectator

It’s a reasonable guess, but alas, the experiments show that this is apparently not the case. (Something like that could occur occasionally on the side, but it is not the main event producing all the heat.)

I agree, and Byrnes properly hedges this before he is done.

The reason we know this was lucidly explained by Peter Hagelstein in Constraints on energetic particles in the Fleischmann–Pons experiment, relying on the complete (or almost-complete) lack of neutrons in experimental measurements, along with other measurements.

That article is crucial to understanding what cold fusion is not. Basically, the spectator idea has the spectator, if not too massive, carry away the energy of fusion, or the helium product carries it. That would be hot helium, almost 24 MeV minus the energy of the other particle. This would be very visible. If a difficult-to-detect particle carries away the energy, it would not show up as heat. Simple.

I actually prefer his succinct summary in a different paper (“Energy exchange in the lossy spin-boson model“). Here he explains why d + d + (something) → 4He + (something) does not work, regardless of what the “something” is. He goes through the possibilities one-by-one:

  • 4He + Pd (an example where the alpha energy is maximized), with the alpha particle ending up with about 23 MeV. Although fast alphas are not penetrating, they cause α(d,n+p)α deuteron break-up reactions with a high yield, with fast neutrons that are penetrating. We calculated an expected yield of 107 n/J, which is nine orders of magnitude above the neutron per unit energy upper limit from experiment.
  • 4He + d (since there are deuterons in the system), so that the alpha particle ends up with about 8 MeV. We would expect about 10n/J from the same alpha-induced deuteron break up reaction, which is now six orders of magnitude above experiment. However, the deuteron will have 16 MeV, which would make dd-fusion neutrons with a yield of just under 108 n/J, which is a bit less than 10 orders of magnitude above the upper limit from experiment.
  • 4He + p, so that we get the minimum alpha particle recoil for any nucleus, and the alpha ends up with 4.8 MeV. The number of secondary neutrons produced as a result of primary collisions between the alphas and deuterons in the lattice now is reduced to about 200 n/J, which is about four orders of magnitude above the experimental limit. The energetic protons in this case would cause deuteron break up reactions with a yield near 107 n/J, which is nine orders of magnitude above the experimental limit.
  • 4He + e, which gives close to the minimum alpha recoil for any single particle, and the alpha ends up with about 76 keV. Now the secondary neutron emission due to the alphas is down to 10 n/J, only three orders of magnitude above experiment. However, penetrating 24 MeV electrons produced at the watt level would again constitute a significant health hazard for any experimentalists nearby. For an experimentalist within a meter of an experiment producing a watt of 24 MeV betas, the radiation dose would be on the order of 1 rem/s (assuming a 10 cm range) which would be lethal in about 1 min.
  • 4He + γ, again giving 76 keV recoil energy for the alpha, and again 10 n/J which is again three orders of magnitude above experiment. Penetrating 24 MeV gammas at the watt level would be a major health hazard for any human beings in the general vicinity. As in the case of fast electrons, 24 MeV gammas at the watt level would be lethal for an exposure of about 1 min at a meter distance.
  • 4He + neutrino (as advocated by Li), also gives 76 keV recoil energy for the alpha, so we would expect three orders of magnitude more neutrons than the experimental upper limit. The neutrinos in this case are not a health hazard, and we would not know from direct measurements if they were there. However, most of the reaction energy would go into the neutrinos, so that the observed reaction Q-value [i.e., heat generated per D+D fusion] would be about 76 keV, which differs from the experimental value by about 300.

Hagelstein can be very clear, and he was, here.

If you’re not sure what’s going on: α(d,n+p)α is another way to write α + d → α + n + p, i.e. a deuteron can be cracked in half if you knock it hard enough, creating a proton and a neutron, and the latter may exit the system and get detected. The energy figures are computed from the total energy released as kinetic energy (24 MeV) and the masses of the two final particles, assuming that the fusion happens more-or-less at rest in the laboratory frame-of-reference. That calculation is basic special relativity, using conservation of energy and momentum. The lighter particle always winds up with a greater share of the kinetic energy.

And great minds think alike.

Are these arguments airtight, or might there be “loopholes”? For example, might there be something special about the material that makes the deuteron breakup reaction very very unlikely, in comparison to normal expectations? Well, I don’t know enough about this topic (SRIM-related physics) to say for sure. But I have the impression that the arguments are airtight.

Let’s just agree that they are strong, and move on. Reality will judge between us in that wherein we differ. (– Qur’an, my gloss).

(Acknowledgements: I learned about this topic from Ron Maimon. But all mistakes are my own.)

Ah, Ron Maimon. Nice to see him acknowledged. He popped into the Wikiversity resource before a Wikiversity bureaucrat decided to ban me and censor all fringe topics. Long story, but here is the discussion:


and his theory page: Wikiversity/Cold_fusion/Theory/Ron_Maimon_Theory

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