Spectators are not the answer

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)

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

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.)

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.

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 10⁴ n/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.

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.

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.

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s