Discussion between Eric Walker and Ron Maimon

Eric Walker

03:50

@RonMaimon, I'm interested in better understanding the details of your CF explanation, suggested here: physics.stackexchange.com/questions/3799/…. This is the first chat room I've created -- hopefully I'm not doing something wrong or rude.

My first question has to do with the deuteration of the Pd K-shell holes. If I understand your meaning, the deuterons are in a band at 100 fermis from the Pd nuclei. There are some K-shell holes (from previous excitation) and some deuterons, and they're all mixed together. In this situation, how is it that the positively charged deuterons are not electrostatically expelled from the nucleus, given their close proximity? What is it that is keeping them there?

Is it that they're now "orbiting" the nuclei with 20keV, and it is the amount of energy they have absorbed that keeps them this close to the nuclei?

By "expelled from the nucleus," I mean expelled from the vicinity of the nucleus, not the nucleus proper.

Eric Walker

04:13

room topic changed to Discussion between Eric Walker and Ron Maimon: Trying to better understand the details of Ron's CF explanation (no tags)

Ron Maimon

04:38

Hi Eric, no, this is how you do it--- although you could link to the page in your comment to the question. If we don't chat here every day, it will get locked tho. Regarding the specific questions--- no the deuterons are not specifically 100fermis from the nucleus, they are everywhere in the metal. The "mixing" is a quantum mechanical concept--- a process which starts with a K-shell hole fills the hole with an electron, and dumps the energy into a deuteron, which then wanders around the lattice.

This deuteron has 20KeV of energy, so it can approach up to 100fermis away from a nucleus (a little more if the band is more then 20KeV--- it should be around this, because this is the naked hole energy, but if you have mixing, the energies shift a little bit, and it could be a 40% effect, I don't know how big it is. The easiest way to figure it out is experimentally). So you have deuterons in a metal wandering around with two dozen KeV, and this means that they have a non-negligible fusion.

The 100fermis is just the statement that the wavefunction of the deuterons in this band has a turnaround point within 100 fermis of a nucleus. When the band is highly occupied, you have a lot of deuterons delocalized in a patch of the metal, and with a high enough concentration, you can have two deuterons near a nucleus "at the same time" (I put it in quotes, because it's quantum mechanics, so both deuterons are smeared over many nuclei, and just have an enhanced wavefunction near a nucleus).

These deuterons fuse and fall apart into two deuterons again all the time (there are several highly unstable 2d resonances at 24MeV above the alpha-particle ground state). But when they are within 100fermis of a Pd nucleus, these alpha-particle excitations can transfer the energy to the nucleus and go the ground state. The energy transferred is about 24 MeV, and this is just enough to eject a proton, or eject a full shell from the Pd nucleus. This produces the transmutations.

Full shell ejection is observed in electromagnetic disintegration of nuclei at 10s of MeVs (there are nice LINAC expriments at 20MeV which should give the precise spectrum of ejecta, but so far, the literature I read just showed that you get ejected protons, ejected "full shells", so something like an alpha particle ejected, a Be8, O16, things like that with an integer number of alphas, which are highly stable small nuclei). The result is transmutations with a tendency for -8 -16, and light

Eric Walker

I assume the delocalized deuterons are an extension of another example of nuclei that have become delocalized? By "delocalized," I'm imagining a deuteron cloud in a region of the lattice. What other examples of this are there?

Ron Maimon

elements. The ejected Be8 (or whatever) is going pretty fast, it has about 10MeV of KE left, and depending on the charge, it can get absorbed by another Pd to give a +8 transutation, also +16, explaining Iwamura.

The phenomenon of delocalization is universal to QM. It isn't apparent for nuclei like for electrons (electrons delocalize in metals, nuclei hardly ever delocalize at all) because the nuclei are in the "Mott phase" all the time. The Mott phase means that the quantum mechanical ground state is a superposition of approximate lattice configurations in space, rather than delocalized wavefunctions in space. This is a property that is caused by the repulsion of charged things. Electrons can be Mott.

But the reason Fleischmann was puzzled by Palladium, and the motivation for cold fusion, is that Pd has highly mobile hydrogen, as if there are configurations where the hydrogen isotopes are freely mobile, or delocalized. This is why he investigated the system for so long--- it looks like a case where the delocalized nuclei idea is realized to some extent. There is a wonderful lecture by him from the 1990s where he talks about the motivations for cold fusion, and discusses these experiments.

The delocalization of nuclei is also apparent in liquid He4 (or He3) where you don't have a Mott solid (people don't usually use the word "Mott solid" for nuclei, they just say solid. I am just making the analogy with electron Mott localization, although historically it went the other way--- Mott is inspired by solid crystals to identify the phenomenon in electrons). But all these considerations are academic--- a 20KeV deuteron is definitely delocalized, because the dinky repulsion energies

Eric Walker

04:54

So "delocalization" is a mathematical property of the deuterons as elements in a QM system. For electrons, this has the result that they have a diffuse area in which they might be detected, but for nuclei this generally is not the case. The fact that a nucleon has become "delocalized" is just a statement about its susceptibility to certain processes. But in the case of hydrogen, it might actually mean something closer to what happens to an electon?

Ron Maimon

are hardly enough to localize even stationary deuterons.

delocalization just means that the wavefunction is extended over a large region. For an electron, this means a metal. For a deuterated metal, it means the deuteron wavefunctions intersperse large chunks of the metal, instead of just piling up in interstitial positions. Delocalization is hindered by the repulsion two electrons feel when they are close, if you have a lot of delocalized electrons, they have to sit on top of each other.

Eric Walker

I see -- and liquid 4He and 3He offer another example of this diffuse wavefunction, meaning that the protons and neutrons that make up the nuclei are not as easy to pin down at a specific location.

Ron Maimon

quantum mechanics default is always delocalization. It's the localization that you need to explain. In metals, localization can happen in a Mott transition. For nuclei, localization is the norm, because the tunnelling amplitude to go from one lattice position to another is so small, and the repulsion for two nuclei to be close is so big (you can see this because defect mobilities are not especially high).

Yes, He3 and He4 are delocalized nuclei.

This is not so uncommon--- delocalized nuclei are also the BEC's of recent years, although those are best thought of as delocalized entire atoms (so is He4).

Eric Walker

How straightforward or, alternatively, controversial, is the possibility of the delocalization of 2H in Pd that you're thinking is happening?

Ron Maimon

In the Pd/d system, the d's are partly delocalized, although probably not in the ground state (I don't know the physics or chemistry of the ground state so very well as Fleischmann did). If the d's were completely localized, you couldn't get them to fill the metal.

Eric Walker

05:00

When the 2H wavefunctions approach the "classical turning point," at 100 fermis from the Pd nuclei, do they momentarily become more localized?

Ron Maimon

But they do fill the metal, they pour into the solid (although not at conduction band rates, so take it with a grain of salt), over a period of hours or days, and the metal gets deuterated. This is not the delocalization I am talking about, it's just vacancy hopping and filling of interstitials with deuterons.

No, approaching the turning point is not a localization--- localization is a many-body effect due to repulsions, disorder, or coupling to phonons, none of which is relevant at 20KeV.

All of these considerations are completely moot for 20KeV deuteron excitations, you don't have to work to get those delocalized--- they are always delocalized by their nature--- the deuterons is smashing through the metal hopping from site to site nearly classically (not quite though). The wavelength of the deuteron is much shorter than the interatomic spacing under these conditions/

The delocalization of the deuterons at 20KeV is not controversial at all, as far as I can see. The mixing with the hole excitations is what is not really straightforward, I don't know how it works. It might be just pure downconversion--- holes kicking deuterons into high momentum states, that then knock around as pure deuteron states. But I doubt it, because the deuterons have friction, they can excite electrons in the next energy level and slow down some. The dynamics is complicated.

The whole point is that when there are enough of these fast deterons in a region, the fusion can reproduce enough K-shell holes to populate the band more, and replenish it faster than it decays by friction. Then you get the self-sustaining fusion reaction. I hope I am being clear, I don't know how much of this is familiar to you.

Eric Walker

None of it is familiar to me -- I'm trying to get a foothold to do some more reading.

When I looked over the Bethe formula, I saw this chart, which suggests that the mean excitation is on the order of 10 eV for large Z:

Ron Maimon

Ok. The ideas of band theory are simple, if you know the Schrodinger equation. They are covered in many introductory condensed matter books, although I never read them cover to cover, I worked it out for myself mostly (it's not too hard).

Eric Walker

for the image, see the second chart here:

Bethe formula

In nuclear physics and theoretical physics, charged particles moving through matter interact with the electrons of atoms in the material. The interaction excites or ionizes the atoms. This leads to an energy loss of the traveling particle. The Bethe formula describes the energy loss per distance travelled of swift charged particles (protons, alpha particles, atomic ions, but not electrons) traversing matter (or alternatively the stopping power of the material). The non-relativistic version was found by Hans Bethe in 1930; the relativistic version (shown below) was found by him in 1932 (Sigm...

Ron Maimon

The mean excitation is 10eV because there are so many low energy excitations. You have to take a weighted average by energy--- if you have a 10MeV particles, what percentage of the energy ends up in 1eV,10eV excitations and what percentage at 20KeV. It's not a negligible percentage of the energy. The Bethe formula is the most important thing to understand, this is the key to the theory.

Eric Walker

05:08

I see

Cool -- I'll read up on the Schrodinger equation, then.

Ron Maimon

Oh, I see you are linking to the stopping power formula. Bethe does more--- he calculates the fraction of the time you excite each level, so he gets the complete spectrum of excitations as the charged particle is ripping through. But for a fast particle, you can use this cheat: the interaction with each electron is essentially independent of all the others, so each electron gets knocked out independently with some energy that depends on the incoming particle energy.

The result is that something like 10% of the energy gets dumped into K-shell holes (I only did qualitative futzing around using the intuition from reading Bethe's calculation), and something like 90% gets dumped into other holes, or into the valence band. The 10% is important though--- it means you get 10% of the particle energy converted to secondary x-rays as it passes through the material.

Eric Walker

Right -- and your point in reference to Robin van Spaandonk's objection is that there's a lot of energy to go around, and a significant percentage of it will go into exciting k-holes.

Ron Maimon

Yes on the comment above. The auger electrons (secondary ejected electrons from decay of k-shell holes) have an analog in auger deuterons in the theory I am pushing. The auger deuterons are going very fast through the Pd lattice, and are somewhat delocalized. They eventually decay to interstitials by friction (although how long this takes, your guess is as good as mine), but in the meanwhile, they are focused by nuclei into conc

It's actually quite paradoxical, the energy loss of fast charged particles (continuing on the line of your previous comment): this was what mystified Bohr in the 1910s--- why do charged particles lose energy in such a strange way in lumps of discrete steps as they pass through matter? He created the shell model in part to answer this, but the work wasn't completed until the Bethe formula in the 1940s. It's a beautiful neglected part of physics history.

Eric Walker

Another question I had related to the Augur electrons -- in your explanation, you mentioned that occasionally a k-hole decays by ejecting an electron in another atom. I'm used to thinking of Augur electrons as being ejected from the same atom.

Ron Maimon

Anyway, the 20KeV deuterons are flying around delocalized, and you can stuff as many as you like in a region, their repulsion is negligible at this scale. They are mixed with the electronic fluid in some way that can involve reexciting the K-shell as they pass (they are resonant with the K-shell by construction, although this might be true only at one point in the band, depending in the details of the k-shell band structure). You really need simulations to answer the detailed questions.

Oh, it could be the same atom too (in fact it's more likely, because the electrons in other atoms are further away)

I just meant "any old other electron, on itself or on another atom".

Eric Walker

05:16

gotcha

another question -- in the Augur process, a positively charged k-hole dumps its energy into a negatively charged electron.

I would have figured the charge would be important here, but it seems like it isn't, and that the k-hole can transfer its energy to a positively charged deuteron just as easily.

Ron Maimon

The deuteron is better for ejecting because of density of states--- deuterons are heavier, and the quantum mechanical probability of transitioning to a state depends on how many other states like it are nearby. This is enhanced by a higher mass.

Eric Walker

So I take it the electrostatic charge is not a factor, here?

Ron Maimon

The sign of the charge is not important, just that the particle is charged. You get an opposite sign amplitude for opposite charge, but it's the same process.

Eric Walker

So there would be no "augur neutrons," for thermal neutrons in the area, but there could be Augur protons and Augur deuterons, because they have charge.

Ron Maimon

This is from the Feynman diagrams of condensed matter physics--- you can learn that in a little book called "Feynman diagrams in condensed matter physics", but again, I think it's easier to work it out for yourself if you understand the Feynman diagrams in high energy.

The key difference is that there is an instantaneous Coulomb Feynman diagram for a particle at momentum k to get deflected, and transfer the momentum instantaneously by Coulomb repulsion to another particle. This doesn't just work for free particles, for any process, you can transfer the momentum energy to another charged particle electrostatically. This is the magic of electrostatics--- it's instantaneous for nonrelativistic particles (and for relativistic particles too in Dirac gauge)

Eric Walker

05:20

Another question -- I would have assumed that the phenomenon of Augur deuterons would have been noticed by now, since Pd/D is a well-studied system. Is it just something that has been overlooked?

Ron Maimon

The Feynman diagrams for condensed matter you can work out from the nonrelativistic Schrodinger field (I wrote Schrodinger field on Wikipedia to help people out with this). Once you have the nonrelativistic Schrodinger field Lagrangian, it's straightforward to work out).

Eric Walker

By "instantaneous," do you mean by some approximation, or do you mean exactly instantanous?

Ron Maimon

I don't think anyone has studied radioactivity in Pd/d. Auger deuterons have been overlooked--- if you ask me why, it's just because nobody had a reason to look for them. There is zero chance they don't exist--- this is completely uncontroversial. If you can kick an electron, you can kick a deuteron, the only question is the branching fraction, and I would guess.

Eric Walker

I'll check out the Wikipedia article.

Ron Maimon

I mean instantaneous in the nonrelativistic approximation, although you can treat it as instantaneous in the relativistic theory too, but then there is a miracle of cancellation between this instantaneous interaction and the photon part of the interaction that enforces proper relativistic speed-of-light causality. This is a well known thing, it's Dirac gauge, and it's described in several places (I think I read it in Bogoliubov and Shirkov, but that's outdated, don't read. Also Sidney Coleman's

QFT lectures cover Dirac gauge). Just google "Dirac gauge" and you can find it. I might have read it in Dirac's original paper too, I am not sure, it was many years ago. The instantaneous is a perfect approximation in nonrelativistic systems, where the speed of light is not an issue.

Eric Walker

05:24

Cool -- that gives me some places to start!

Ron Maimon

That's borderline for KeV excitations, because the wavelength of the emitted X-ray is about 1 Angstrom, and that's also the distance to the nearest deuteron, but good enough for day-to-day work to pretend the nonrelativistic approximation is exact (the lifetime of the K-shell is relatively long, so I think this improves the nonrelativistic approximation, but I am really not worried about the speed of light here, I don't think the retardation makes any difference to the auger deuterons).

Eric Walker

And the reason we're working in a nonrelativistic approximation is because the energy of the deuterons is low, relatively speaking?

Ron Maimon

Regarding Auger deuterons, this is an interesting prediction you can test without reference to the other theory, simply by irradiating hydrogenated Pd (don't irradiate deuterated Pd, because you can make a small atomic explosion, you might get so many deuterons that you get fusion in many spots).

The nonrelativistic approximation is because everything is moving much slower than light--- the ratio of energy to mass is small. The mass of a deuteron is 3600KeV, the mass of an electron is 511KeV, the mass of a proton is 1800KeV (more, but whatever), and the energies involved in the pre-fusion part are all about 1% of the mass (more for the electron, but even these electrons are nonrelativistic).

Eric Walker

so there's the possibility for weaponization that Fleischmann was worried about -- that can be triggered by irradiation. (X-ray, neutron, gamma?)

Ron Maimon

Not really--- it won't be a good weapon, but it might kill the experimentalist! the explosions will be tiny, even with X-rays, because if you actually release a large amount of energy, you'll blast the solid apart, and the reaction will stop. This is very self-limiting, but it can explode in a meltdown, or produce a large number of neutrons, so one has to be careful.

The reason I say it's not a good weapon is because the fusion is not inertially confined or hot, you don't have the kinds of pressure you find in an Ulam-Teller device. It requires the lattice to be around, so you need the Pd at below it's melting temperature, and the disintegration of the deuteron bands happens in coordination with the disintegration of the metal, so you just wouldn't be able to make a huge explosion. But you can make a chemical scale explosion with a small sample, sure

Eric Walker

05:31

Gotcha

Eugene Seidel

Hello Ron, may I ask you a personal question?

Ron Maimon

but this was already observed by Mizuno--- he had a runaway cathode the produced heat for days, evaporating water off a bucket for 3 days straight

go for it.

Also, Pons and Fleischmann had a runaway cell.

It melted their table.

and blew a hole in the concrete (probably not an explosion, more of a meltdown).

I don't think you can sustain the reaction in a melting cell--- Pd ejects hydrogen past 300 degrees, it's actually mysterious what exactly happened in the PF runaway experiment (although not in the Mizuno one--- he told you).

what's the question

?

Eugene Seidel

On your personal info page you write that you are not a physics Ph.D. but does that mean you were a physics undergrad in college then went to grad school and finished ABD... or are you entirely self taught?

Ron Maimon

ABD. I am self- taught though, I only went to school for accreditation.

I had a thesis worth of work at the time I left grad-school,

Eugene Seidel

ok thanks

Ron Maimon

05:34

I was just kind of sickened by academic stuff that was going on--- large extra dimensions were popular then.

Eric Walker

Anyway, thanks Ron -- I'll get back to you with more questions soon, I'm sure.

Ron Maimon

Also I was at Cornell, my advisor left for Cincinnatti, and I was not in very good standing there (I was kind of a jerk, as I still am). Some friends wanted to start a biotech company called "Gene Network Sciences", and I joined them.

I did some bio work I'm very happy with (I think it'll be on my tombstone), and then I was fired, and went back to Cornell to try to finish my dissertation.

Eugene Seidel

why are you so dismissive of your fellow physicists?

Ron Maimon

I was always stuck on something stupid (that I still haven't sorted out), namely the relation between SUSY in condensed matter (Parisi Sourlas SUSY) and N=1 SUSY in 4d high energy models.

I'm not dismissive of the good ones, and there are plenty. I am annoyed by the ones who do it for career rather than as an art-form.

Eugene Seidel

ok

Ron Maimon

05:37

Mostly it's just the song and dance one has to go through to get into academic discourse that bothers me. It wasn't so at a certain point--- in the 1950s-1960s, it was pretty good in the literature (although I wasn't there).

Eugene Seidel

well, thanks for the many good answers you contribute here... more than 90 percent goes over the head but the remainder is quite instructive to me

Ron Maimon

There is also the 1980s, when strings were doing well.

I hope I can explain it better--- I really think this is right. But it's a total bitch to calculate anything. The only thing I am sure one can get quantitative predictions for are the nuclear fragmentation products,

but here one has the issue that the only experimental data of value on transmutations is Mizuno and Iwamura, and these have two problems: the transmutations are detected in exploded spots in the cathode, and you have to scrape the transmuted crud away from these spots.

There is the issue of contamination (these experiments run a long time, and every possible thing is deposited from solution onto the surface--- but this is not sufficient to explain the radioactive transmutation, despite what Kirk Shanahan pretends to believe). The crud is worrisome.

I am not sure about any of the mass-spec data, and Mizuno's book is not available online (although you can order a dead-trees copy).

Eric Walker

You should also check out Hioki's work. He strikes me as a solid researcher.

Ron Maimon

Ok--- I didn't know. Does he do transmutations quantitatively?

Eric Walker

I get the impression that the mass spec data are perilous -- you could have any random ions with the given masses.

I don't remember for sure.

Ron Maimon

05:41

But the gamma-ray spectra are unambiguous about the new elements near Pd.

kηives

Hi everyone!

Ron Maimon

I also remember that the peaks in the transmutations had regularities that are consistent across setups (with McKubre doing some work reproducing this too) hi Knives! They all get the same +8 +12 +16 mass peaks, which I think is just eject/capture of stable alpha-cluster nuclei (these are very stable nuclei, they are described by Skyrme theory).

Eric Walker

hi kηives

there's another important detail about the transmutations --

i think they don't account for more than a small amount of the total energy evolved

by "small," I mean possibly tiny.

that's just anecdotal, though.

Ron Maimon

Oh-- sure--- they are tiny. Everyone agrees. But the mystery for everyone is how they happen at all, since you can't get that kind of energy into a nucleus in any reasonable way. It only happens naturally in 3-body fusion of the type I am suggesting.

anyway, I have money work I need to do unfortunately (I am doing some bio-for-hire right now, and it's very interesting, but completely unrelated.)

kηives

Ron, I found an article on arxiv under someone of your name having to do with bio

Eric Walker

05:54

sure -- np. ttyl.

kηives

is that you?

Ron Maimon

Yeah, "Computational Theory of Biological Function I". I have parts II and III in book form lying around, but I never put it on arxiv. They are written crappy, and need a total rewrite to go on arxiv. It's a language for proteins that works, and it was pretty popular in 2001. But it's got a patent slapped on it by Gene Network Sciences, which serves only one purpose, to prevent anyone from learning or using it. I wrote the patent on assurances that we would make a free-patent license, like a GPL

But of course, I was totally naive about corporate world, and the friends of mine who founded the company turned around and made it proprietary. Then they asked me for specs and implementation, and I did a little, but I was really stalling them (in hindsight) until they give me the free viral patent license I wanted. They never did, and then they fired me. It was reasonable, I couldn't work with them anymore after what they did with the patent. I hate IP.

hate hate hate IP. The idea of a software patent (like mine) disgusts me, and having my name on one is punishment enough. It would have been nice to create the GPPL (GNU Public Patent License) but it was not to be. I even contacted the FSF asking about this, but never got a reply. It was wishful thinking--- the forces in capitalist America hardly can tolerate the GPL right now.

By written crappy, I mean parts II and III. Part I is fine.

Ron Maimon

09:24

I don't know why I said the mass of a proton is 1800 KeV, it's 931 KeV. I am used to deuterons, mass 1860 KeV, not 3800 KeV. I don't know why I had this memory block.

It doesn't change anything except reduce the density of states by 40%.

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Discussion between Eric Walker and Ro…