Sunday, February 10, 2013

Ron Maimon's theory (2)

In a previous post, I attempted to provide a layman's overview of a theory put forward by Ron Maimon concerning what is generating anomalous heat in the palladium deuteride LENR experiments.  The theory, which I have nicknamed a theory of "Auger deuterons," nicely incorporates the primary elements of the signatures seen in many of the Pd/D experiments:
  • Heat
  • Broadband x-ray spectra
  • Fast alpha particles and protons
  • 4He off-gas
  • Transmutations of various kinds
Part of the motivation for the earlier blog post was to get feedback from people on vortex-l, and as I hoped would happen, Robin, who is on that list, pinpointed several difficulties that needed to be addressed.  Here are questions from him (1 and 2) and me (3 and 4) that resulted from the thread:
  1. How do two deuterons approach the Pd nucleus concurrently, as this seems a very unlikely thing to occur?
  2. The d+d fusion cross section becomes negligible below 5 keV.  Assuming a loss of 400 eV per palladium atom that a 20 keV deuteron passes through as a result of interactions with its electrons, the energy of the deuteron will drop below the 5 keV threshold after passing through 38 palladium atoms.  In the unlikely event that it hits another deuteron head-on before that, even then a fusion is not assured.  So all-in-all the likelihood of a self-sustaining reaction seems small.  How can one be obtained under these circumstances?
  3. The regular branches for d+d fusion are (a) d+d→t+p (50 percent), (b) d+d→3He+n (50 percent) and (c) d+d→4He+ɣ (almost negligible).  What causes branches (a) and (b) to be suppressed and branch (c) to become dominant?
  4. When you have fast particles flying through a deuterated metal lattice, a particle is likely to bump into a deuteron, and it in turn will hit another deuteron. Occasionally a side reaction of branch (b), above, will occur, yielding a significant number of neutrons which would then exit the system.  But neutrons are only rarely seen and at levels barely above the sensitivity of the neutron counters.  For this reason Peter Hagelstein places a 20 keV upper limit on the energy of the particles in the system.  Ron's account involves alpha particles with energies of tens of MeV, so the lack of neutrons from side reactions on an order above that currently seen could be expected, presenting a challenge to be addressed.
The full vortex-l thread can be found here.  I am sure that the difficulties go back to my own understanding and have been anticipated by Ron.  I will be interested to hear how he addresses them, especially (3).

EDIT: Concerning items (1) and (3), above, Ron addresses these questions in his original physics.SE post:
The fusion of deuterons always happens through unstable intermediate states, and the cross section to alpha particle is only small because of the same non-relativistic issue. To get an alpha, you need to emit a gamma-ray photon, and emissions of photons are suppressed by 1/c factors. When there is a nucleus nearby, it can be kicked electrostatically, and this process is easier than kicking out a photon, because it is nonrelativistic (the same holds for an electron, but with much smaller cross section due to the smaller charge, and there is no reason to suspect concentration of wavefunction around electron density, as there is for a nucleus). 
The time-scale for kicking a nucleus is the lifetime of the two-deuteron resonance, which is not very long, in terms of distance, it is about 100 fermis, this is about the same size as the inner shell. If the deuterons are kicking about at random, this coincidence is not significant, but if the deuteron-hole excitations are banded, it is plausible that nearly all the energetic deuteron-deuteron collisions take place very close to a nucleus, as explained above. 
There are conservation laws broken when a nucleus is nearby. The nucleus breaks parity, so it might open up a fusion channel, by allowing deuteron pairs to decay to an alpha from a parity odd state. Such a transition would never be observed in a dilute beam fusion, because these fusions happen far away from anything else. This hypothesis is not excluded by alpha particle spectroscopy (there are a lot of relevant levels of different parities), but it is not predicted either.
This only hints at an answer to question (1), by saying that the banded state makes it "plausible" that the energetic deuterons will encounter one another near a palladium nucleus.


  1. To answer the concerns:

    1+2. The dynamics of charged particles in materials is not as "frictional" as you imagine. Charged particles don't slow down through bulk dynamics like a baseball in water, they have to transfer their energy through ionization of atoms, and in the case of 20KeV deuterons, they need to ionize other inner shells. They will eventually slow down, but it isn't clear how quickly, because it depends the precise density of states. If you have banded excitations, there are sometimes band-gaps where you have no states, so you can sometimes have extremely long-lived high energy states just because there aren't any available states to allow a simple de-excitation process. So if there are no available states at 17 KeV to transition to when you kick out 3KeV in exciting the n=1 p-wave electrons, these things could live a long time. I am not saying that this is necessarily so, I have no idea how long lived a 20 KeV deuteron is, but it can have a lifetime from a fraction of a second to minutes. But in any case, there is certainly not a barrier to having a 20KeV particle go microns or a millimeter through a metal before slowing down to 5KeV. The penetration depth of 20 KeV charged particles is not 10 atoms.

    In order to have the cold-fusion proceed, you do need a region with high density of deuterons, high enough for 2 deuterons to fuse. When this is so, this first happens right near a nucleus, because this is where the wavefunction of 20KeV deuterons are most concentrated. The wavefunction of such a deuteron has turning points near the nucleus, and you can describe it semi-classically, and the thing is peaked at the K-shell radius, about 100 fermis from the nucleus.

    The concentration of deuterons can plausibly happen in many ways, just from local electric fields concentrating the positively charged fluid into some region, it doesn't require conspiracy. The deuterons are charged and macroscopically flowing through the metal, like electrons, except at much higher energy. The result is that they can concentrate in regions where the electric field is big. Whether they flow in the direction of E or paradoxically against the direction of E depends on details of the energy as a function of wavenumber in the high-energy band, but my intuition is that they will flow like positively charged objects, in the direction of E, and opposite the direction of electron flow. This is a bit confusing, because this means they flow away from the surface of the cathode toward the interior. But they could also flow the other way, if their effective mass is negative, meaning if you produce them near a maximum of the band dispersion relation.

    3. The whole point is that the process is d+d fusion right next to a Pd nucleus, so that the process completes with electrostatic energy transfer to the nucleus, not with the standard transfer of energy to an ejected proton or neutron. Then the result is an ejected He4 and a fragmented nucleus. We have no data on three-body fusion collisions, they never occur in dilute plasmas, so we don't even have an estimate for the cross section (except order of magnitude it should be not too many orders of magnitude larger than usual fusion, and this is ok for the theory to work). The point here is that due to the concentration of the wavefunction of the deuterons near the Pd nuclei, from the turning-point, you get a larger density of deuterium near the Pd, and you can have fusion when the local density of deuterium exceeds a certain threshhold.

    4. The random bumping around fusion will occur in this theory, you will get occasional sporadic neutrons from the system. I don't consider this a problem, as it explains the neutron signal people see. Neutrons are not easy to detect anyway.

    1. Thank you, Ron, for the clarifications. I'll follow up with further questions before too long.

  2. Hi again, Robin Van Spaandonk says some silly things, so I want to correct them: first, the penetration depth of 20KeV deuterons in a metal is not 30 atoms, it's measured, and it's like microns or a mm (I didn't look it up, but it's the same as alpha particle penetration in order of magnitude), it's just not 30 atoms. The ionization is not as efficient as he imagines. I am tired of repeating this, it is well known experimental fact, no matter how counterintuitive it is that you can have charged particles zipping through matter. This was the original stopping power mystery that motivated Bohr and Bethe.

    Regarding the "entering the Pd nucleus to fuse" OF COURSE NOT, the deuterons do not enter the Pd nucleus, they have nowhere near enough energy. They fuse about 100 Fermis away from the Pd nucleus, a K-shell radius away, and then they transfer the energy by virtual photon (electrostatic interaction) to the Pd nucleus. This happens at a distance, it doesn't require contact. Electrostatics is nonlocal. The matrix elements are not small, because the charge on the Pd nucleus is 46, that's a lot of charge.

    The details of banding of deuterons, if it happens at all, are difficult to work out, I don't know if they behave more like particles or like waves, and it doesn't really make much difference. That's simply a question of their coherence time, the time to do something to a phonon or an electron. In any case, the deuterons bump around the atoms, focusing near the nuclei, and the first place where you would see fusion is near a nucleus, at a range where you could potentially transfer energy to the nucleus. This doesn't require massive conspiracy, or some weird violation of physical law. All you need is a concentration of deuterons high enough to lead to fusion, as this fusion will naturally happen unnaturally close to a Pd nucleus, simply from the mechanism of acceleration of the deuterons, through the K-shell holes. It's not a crazy idea, it's just something that needs to be tested in detail, through experiment.

    The easiest experiment is simply to shine X-rays on a deuterated metal, and see the broadening of the K-shell line into a band (if it happens). This is an indication that the K-shell and the deuteron states are mixed up. You can also see this from enhancements of beam fusion on deuterated targets, and this has been observed in Russia, in the beam experiments, and the enhancement for the process is by a small factor, but the detection is through detecting neutrons, so it is blind to any three-body reaction that might be taking place.

    It is also possible to test this by calculations, since the entire theory is known. For this, however, one needs the detailed alpha particle spectroscopy, all the unstable intermediate states, and there isn't good data on this (I looked), but it might be possible to make something qualitative using effective proton-neutron potentials. I don't know the state of the art on these models, or how well they reproduce alpha particle spectroscopy, but I am sure that it doesn't rule this out back-of-the-envelope, and if this theory fails, it fails only because of hard-to-calculate detailed effects.

    1. Ron -- Robin has replied to your points here:

    2. The stopping distance is around 600 atoms according to Spaandonk, and I don't dispute the theoretical value, but I don't know if this is accurate, it can be larger or smaller, because the calculations are theoretical, and this energy regime is comparable to the energy levels involved, so experiment is the best way.

      The reason that the deuterons have unnatural concentration near the nucleus is precisely because of the repulsion--- they turn around at a distance of 100 fermis from the nucleus. This is because their kinetic energy is equal to the potential energy of an electron-hole in a K-shell, so that you get stopping at exactly the K-shell radius, give or take a factor of 2.

      The turn-around region is where the wavefunction is most concentrated, because semi-classical wavefunctions go as 1 over the square root of the speed. This enhancement is like a beam-focusing device.

      Robin is right that there are two parts to test--- the fusion and the Auger transfer. The Auger transfer one knows how to test, but this is likely to work, but the details of the banding are still mysterious. The other part is the fusion rates. This is confounding to me.

      The beam experiments did not see excess heat, because the number of fusions is ridiculously miniscule, it's a beam with order 10^10 particles, not a macroscopic amount of stuff. They see fusion products only.

      The way I was hoping to investigate the 3-body fusion is through a theoretical model of the proton-neutron force, and the experimental data on the alpha-4 resonance. This should allow one to theoretically calculate the electrostatic matrix element between the high energy states and the ground state, although it's hard, because the resonance data is so crappy.

      Other than that, I'm stuck, because I can't think of any circumstance where you would see 3-body events other than in cold fusion. So the actual rate needs to be attacked theoretically, and it's annoyingly hard.

      I'll think about it again soon, I have been doing biology for the past months. Thanks to Robin for the considered comments.

    3. In Robin's reply to your comments above, he has two ideas -- (1) adapting a Mathcad file he has to get an upper bound on the d+d fusion rates as a function of separation distance; and (2) using muon catalyzed fusion to get a sense of what a 3-body event would look like in the context of d+d in Pd.

    4. Fusion doesn't have a rate as a function of distance, it's a coherent tunneling effect in deuterons and it depends on the whole wavefunction shape not on the classical "distance between nucleons".

      You can only calculate it using a two-body -> resonance -> final product effective (nonrelativisitc) field theory. This requires the two-body -> resonance coupling constant and the resonance-> final product coupling constant, neither of which do we know for each of the ~5 relevant alpha-particule resonances, and also the dependence on an external strong electric field (provided by the nucleus). With good estimates for these coupling constants, you can simulate the process.

      Muon catalyzed fusion as an estimate is a good idea, I looked at muon catalyzed fusion, but it's not an easy estimator, because the deuterons are stationary, so the process is different from colliding deuterons, and the muon is far away, and has a small charge, so it doesn't introduce a strong electric field.

  3. Thanks for the clarification -- I would be very interested in seeing those experimental leads pursued. Also, I'm running into the limits of my knowledge of physics in advocating for your theory -- you should really start putting together your thoughts into a paper, one that does not make too many assumptions about the knowledge of the reader. This would be quite helpful as well.

    1. Thanks for your efforts--- you are very nice. One can't be sure the theory is correct, but if it's not what I am suggesting, I am totally at a loss--- it would then have to be the tooth fairy putting energy in the experiment, in collusion with Santa Clause.

  4. Hi Eric, I noticed you were now going on the path of electron density being the main absorber of momentum, just letting you know that this is also plausible, I agree with you, and it is hard to discriminate between the two processes, you need the quantity of transmutations per event compared to the total amount of He produced. It is more plausible for a Ni-H reactor to proceed through the p-d process with the electronic version, however, since it only requires one fast proton per event, the d can be stationary, but it is still extraordinary if it works, because the Nickel K-shell is so low energy. I'll believe it when I see it.

  5. Hi Ron -- good to see you around. Sorry about the whole physics.SE thing.

    As my physical intuition is gradually honed, the electron charge density lead seems less and less likely, as I had it set up. But please see this more recent post and add any comments you have:

  6. You should know that there is a great paper by Hagelstein ruling out any of the processes which involve incoherent transfer of energy. He was able to show that fast particle secondary neutrons (from alpha colliding with deuterons, for example) require that the products are born at low energies. He also gave a theoretical model that can produce low-energy products, but the modes he used were ordinary phonon modes at extremely low energy, and he was never able to complete the story.

    After reading Hagelstein's work, I am on board with his stuff, he is just right. The process I described is theoretically acceptable, but it can't be an incoherent reaction. The true story has to be a coherent version of the same thing, and this makes makes it much harder to calculate or even estimate rates, or to convince oneself that there is a good theory. I am trying to sort it out, but it is important to understand the Hagelstein bound. There are references on the talk page of the Wikiversity article I wrote on the theory, a fellow named Abd, who knows Hagelstein, pointed it out to me.

    The theory I believe is ok, I always worried about coherence, but I figured I didn't need it. In a coherent version of the theory, the predictions change, and the calculations become a quantum quagmire. It's really, really, difficult to figure out how to get a rate for the fusion, it's a mix of the hardest classical problems (nonequilibrium mode-sharing, like in turbulence) and the hardest quantum problems (nuclear couplings, high-energy solid modes). The mess is much more formidable than with a simple incoherent reaction like what I proposed, but it is required to understand, because incoherent reactions simply are not in accordance with the experimental data, whether it is electrons or Pd nuclei that serve as the third body, this is certain.

    The coherent version is the same thing as the process I said, except where the dumping of energy is to already pre-excited modes. This means there is already a pre-excited environment of shaking d's and holes draining energy continuously, and these modes are pumped by constant fusion, occuring coherently over many atoms, and dumping energy to the modes coherently. This is what I am focused on. The key is that there is still an x-ray signature, although calculating its magnitude is not as straightforward as in the incoherent version of the theory. The coherence can remove the parts of the theory that are incompatible with experiment, but at the cost of theoretical nightmares. Ok, it's a tough problem, you expect some nightmares. Thanks for you patience and advocacy, I'll keep you informed when I do something.