Branching ratio, collective effects, and special relativity

When you read cold fusion theories you hear a lot about collective effects, ways that atoms in a solid work together to do things individual atoms cannot. One iron atom cannot be a permanent magnet, but a trillion of them together can be.

Or here’s another example. Two deuterons bouncing around at room temperature do not have enough energy to surmount the Coulomb barrier and fuse. But what if a million deuterons pooled their energy together, spontaneously transferring it all to a single deuteron? Actually this particular example is just wishful thinking. Atoms don’t do that, it would violate the second law of thermodynamics. 😛 I chose this example to illustrate how easily one can dream up collective effects that, upon closer scrutiny, are inconsistent with the laws of physics!

Enough about the Coulomb barrier, one of the two great mysteries of cold fusion. The subject of this blog post is the other great mystery: The branching ratio mystery, a.k.a. mystery of the missing radiation. Here, too, you will hear much about collective effects if you read cold fusion theory papers.

But first a reminder: What is the branching ratio mystery? Well, mainstream nuclear physicists will tell you that when two deuterons collide and react, the result may be:

  • D+D –> helium-3 + neutron + 3.3 MeV (~50% probability)
  • D+D –> tritium + proton + 4.0 MeV (~50% probability)
  • D+D –> helium-4 + gamma-ray + 23.8 MeV (0.0001% of the time).

Basically, when D+D come together, you wind up with two protons and two neutrons and a lot of extra energy (compared to the ground state of helium-4). There is enough energy for one of the neutrons to jump out of the nucleus (the first branch), or for one of the protons to jump out (the second branch). Or, the extra energy can be emitted in an electromagnetic wave (gamma ray), but that’s a relatively slow process; normally a proton or neutron boils off long before the gamma ray photon can be formed.

So much for mainstream nuclear physics; what happens in cold fusion? The most reliable (or “least unreliable”??) cold fusion experiments, as far as I understand, leave little or no trace of neutrons, tritium, helium-3, or gamma-rays, certainly many orders of magnitude less than one would expect from the amount of fusion allegedly going on. Instead, the experiments (allegedly) create a lot of helium-4 and a lot of heat, allegedly in the ratio of ~24 MeV of heat per helium-4 atom. So this obviously contradicts the conventional branching ratio above.

To resolve the contradiction, one might suppose that the branching ratio is modified by some kind of collective effect, i.e. interactions with other particles in the solid. If it’s just one other particle we would call it a spectator. If it’s dozens or billions of other particles, we would call it a collective effect. The comments below apply to both.

A critical constraint on collective nuclear reactions

A lot of these collective-action theories are futile for a simple reason. There’s an important constraint on collective effects modifying the branching ratio, and it’s the following inequality:

d < tc \qquad  (1)

where d is the distance of a participating particle from reaction center, t is the duration of the interaction, and c is the speed of light.

You need to be careful in defining “duration of the interaction”, but if you do it right, this is about as watertight and inviolable a principle of theoretical physics as you’ll ever see. It comes straight from relativistic causality: “No information or influence can travel faster than light.”

Example: The Mössbauer effect

An instructive example is the Mössbauer effect, a rare example of a true collective nuclear reaction within mainstream physics. In the Mössbauer effect, a nucleus absorbs a gamma-ray photon—both its energy and its momentum. The momentum should cause the nucleus to recoil. But instead, the momentum kick is shared by all the atoms in the crystal together. The Mössbauer effect seems like it should be reassuring to cold-fusion proponents: If billions of atoms can share the momentum of an absorbed gamma-ray, why can’t billions of atoms share the energy of a nuclear fusion?

…I am not so reassured! Let’s look again at Eq. (1). As it happens, the gamma-absorption and gamma-emission events studied in Mössbauer spectroscopy have a very long duration! How long? For example, the famous iron-57 Mössbauer isotope has a gamma-emission (and absorption) linewidth of ~10-8eV, i.e. the gamma electromagnetic wave has a frequency spread (bandwidth) of ~2 MHz. As every electrical engineer knows, the time-domain version of this wave has to have a duration of at least (roughly) the reciprocal of that, i.e. >10-7s. (This is an example of the Energy-Time Uncertainty Principle.) So the nucleus is really absorbing that gamma ray over the course of at least 100ns. In that amount of time, light can travel 30 meters. So Eq. (1) tells us there is plenty of time for grillions of nearby atoms to push and pull on the nucleus as it is absorbing the photon, and hence take a share of the momentum.

Back to D+D fusion

Excited nuclear states are not like that. If two deuterons somehow come together within range of nuclear forces, then the amount of time that they are in this embrace cannot be longer than the lifetime of an excited state of 4He. After all, “excited state of 4He” is just another way of describing the same configuration. According to Tilley / Weller / Hale, if I’m reading it right (Table 4.3), the longest-lived excited state of 4He has a linewidth of 0.5MeV, which corresponds to a lifetime of 10-21s, during which time light can travel about 1 picometer (10-12m). So that’s how close something has to be to participate in the D+D fusion. For comparison, the size of a Pd unit cell is ~400pm, and the typical separation between a hydrogen nucleus and its electron is ~50pm. [1pm is roughly the separation between a palladium nucleus and its closest (1s) electrons, though I don’t see how that is relevant or helpful.]

Incidentally, I made this (to-scale!) energy plot:

He4energyLevels

Note that Tilley / Weller / Hale do not list any helium-4 excited states below the level of p+T. That explains why they are so short-lived—a proton or neutron can and will jump right out of the nucleus.

Reviewing cold-fusion theories

I’ve discussed a number of cold fusion theories on my blog. How do they stack up against Eq. (1)?

  • BEC theory: If I understand correctly—a big “if”!—this theory proposes that there is a macroscopic Bose-Einstein condensate, and the fusion energy gets shared by every atom in the condensate. So I’m inclined to say it grossly violates Eq. (1).
  • Hagelstein spin-boson model: As I understand it, this theory posits that separated D+D goes directly to ground-state helium-4 without the intermediate step of the D+D getting close to each other. What’s the duration of this interaction? It’s effectively infinite. The D+D can sit side-by-side for the lifetime of the universe without spontaneously fusing. So while I am skeptical of the Hagelstein spin-boson theory, I don’t think it violates Eq. (1). My complaints are very different—e.g., quantum decoherence, and the implausibility of the strong coupling required for the model to work.
  • Widom-Larsen theory: This theory requires, among other things, collective effects altering the final state of a weak interaction. (After electron capture, the output neutron is supposed to be moving ridiculously slowly!) With weak interactions, you may spend a long time waiting for them to start (i.e., long half-life), but once it’s happening, the duration of the interaction is extremely short. In fact it’s easy to calculate. The beginning and end of a weak interaction are marked by the blinking in and out of existence of a virtual W boson. There is not enough energy to create a real W boson, because its mass is huge, 90GeV. So we can use the energy-time uncertainty principle, with 90GeV as the energy, to guess the lifetime of the W boson, and we get ~10-26 seconds. So Eq. (1) tells us that particles within 10-18 meters may have an effect on the final state. (This back-of-the-envelope calculation agrees with nuclear physics textbooks describing the range of the weak force.) So I would say that Widom-Larsen theory grossly violates Eq. (1).

13 thoughts on “Branching ratio, collective effects, and special relativity

  1. David French

    David French writes:

    I have a premise to explain why branching need not necessarily occur in the case of the Cold Fusion formation of Helium-4 from two Deuterium. I will summon the courage to put forward a concept that I have presented to several Gurus who were not impressed.

    “Cold fusion” got condemned in 1989-90 by physicists who insisted that there had to be a branching ratio that would produce tritium and helium-3, along with energetic particles. No energetic particles were being detected, at least not significant levels in Cold Fusion experiments; therefore no fusion of deuterium was occurring, they concluded.

    My premise is based on the premise that the branching ratio, meaning the instantaneous fragmentation of a newly formed helium-4 nucleus, has been observed in high-energy environments, such as in the Sun and in high-energy accelerators. It is not, however, observed on the tabletop in LENR events. Why not?

    My proposal is that the branching effect is demanded by the requirement to conserve momentum, a requirement that must be resolved at the moment of creation of helium-4. There is also the issue of conservation of energy. But that does not have to be resolved instantaneously. A Helium nucleus can absorb 30 million electron volts in an excited state. But momentum cannot be stored. In high-energy collisions, fragmentation of the helium-4 is required in order to balance the momentum equation.

    However, in the solid-state environment of an LENR reaction, the initial momentum values of the colliding particles would be very low. Furthermore, momentum could be absorbed by the surrounding lattice. Hence, this would explain why we do not see the formation of tritium and helium-3 in association with high-energy particle emissions when they occur within condensed matter under near room temperature conditions.

    The discussion that I have just provided explains how that phenomena could be occurring without the necessary presence of a branching phenomenon.

    I do not have access to the mathematics that supports the branching ratio phenomenon. But if you could help validate this proposition that I’m putting forth I would appreciate this very much.

    How the fusion event occurs, and how the energy of an excited helium-4 nucleus is discharged are separate issues that I cannot address.

    David French

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    1. steve Post author

      Yes, a helium nucleus can store energy as an excited helium nucleus. But as far as I can tell, it can only store it for about 10-21s, and nearby particles can interact with this excited state only if they are within about 1 picometer (see post). If two D’s are hitting each other and nothing else is within 1 picometer, they might as well be in a vacuum, AFAICT. A relevant quantity is “energy in the COM frame” (center of momentum). If there are no other particles within 1 picometer, the branching ratio should only depend on energy in the COM frame, not whether the COM frame is the same as the laboratory frame, and not whether the deuterons are in Utah or ITER or the sun. Of course, you can argue that the COM energy is lower in a cold-fusion experiment, and that might plausibly affect the branching ratio for the three branches, but it still leaves the issue that zero of the three branches are in any obvious agreement with what is allegedly happening in cold fusion. We would need a fourth branch, or an entirely different pathway.

      Keep in mind that in a collision like A+B –> C, if you want to conserve momentum but not energy, that’s always possible to do. You don’t need a lattice or anything special to make that happen. Just assign a velocity to C which conserves the momentum.

      (OK, well in addition to “energy in COM frame”, the reaction could also depend on initial spin alignment and maybe other things too.)

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      1. Dan steinberg

        There are lots of free electrons within 1 picometer. These electrons may interact with the compound nucleus, and perhaps could acquire some of the released energy.

        There is good evidence (theoretical and experimental) suggesting that the coulomb barrier is overcome by electron screening. If this is the case, then the impact energy and angular momentum of the compound nucleus will be very low. There is evidence (theoretical and experimental) indicating that this will affect the branching ratio. For example, in muon catalyzed fusion, the angular momentum affects the branching ratio.

        Also, if electron screening is the mechanism, then the participating electron will definitely be within 1 picometer at the moment of fusion.

        Muon catalyzed dd fusion has a similar branching ratio as hot fusion. Since mcf occurs by charge screening, it has been argued that the charge screening process does not affect the branching ratio. But his is erroneous, because in dd mcf, there is actually a lot of angular momentum! The angular momentum comes from the dd-muon molecule, which is stuck in a J=1 rotation state. Almost all dd mcf fustion reactions proceed from tje j=1 state, and this momentum winds up in the compound nucleus.

        Huizenga and other cold fusion critics argue that the T/3He branching ratio in dd fusion must always be 1, because this is the result of fundamental symmetries. This is absolutely wrong. In dd mcf, the ratio is affected by temperature!

        There is solid theoretical suppprt for cold fusion, and overwhelming experimental evidence. The physics community really screwed up the cold fusion thing. Very embarrassing for the physics community.

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      2. steve Post author

        Dear Dan, thanks for your comment! You say “there are plenty of free electrons within 1 picometer”, but I calculate that there is only ~1 millionth of an electron per cubic picometer in palladium, and that the density of electrons between the two deuterium atoms in a D2 molecule is about the same, 10^-6 per cubic picometer. So the questions are: Do you agree with those figures? If so, do you know of any argument explaining why 1 millionth of an electron is enough to make a difference? And why does deuterium gas (without muons) not fuse if it has about the same electron density screening? (Or does it?) Also, my understanding was that of the three hot fusion mechanisms, zero of them are consistent with cold fusion experiments, which produce lots of He4, negligible He3, negligible T, and negligible gammas (where I use the term “negligible” not to describe absolute numbers, but rather how many appear compared to the number you would infer from the power generation rate and energy per fusion event). So we’re not just trying to explain a shift of probabilities among the three paths, but rather a whole new path. If you disagree with that, what do you think the reaction is? Sorry for the deluge of questions, you don’t have to answer them all at once 😀

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      3. Dan steinberg

        Hello

        I should have been more specific about electron density. It is orders of magnitude higher in Pd than in fusing plasma. Also, i think what matters is the density of free/conduction electrons. Pd has lots of those (about 5 per arom if i recall). Deuterium/hydrogen has no free electrons

        I think your calculation is about right. Volume of a pd atom is About 5 million cube picometers. 5 free eelctrons per atom leads to 1millionthnof an electron per pm.

        note that an electron participating in charge screening will be closer than 1pm at the moment of fusion.
        I said:
        “Also, if electron screening is the mechanism, then the participating electron will be within 1 picometer at the moment of fusion.”

        your calculation provides the electron density averaged over time. But i assert that what really matters is the presence of an electron between deuterons when fusion occurs. Fusion occurs from a coincidental collision of 3particles: two deuterons and a free electron.

        “And why does deuterium gas (without muons) not fuse if it has about the same electron density screening”

        The explanation is simple: the electrons are held in orbits too far from the nuclei to provide charge screening at small nuclei separation distance. The electrons are not free.

        In DD loaded Pd, the free electrons are conjectured to by chance pass between the deuterons at the moment they collide. The electron screens the coulomb repulsion, allowing an approach close enough to facilitate fusion. Just like in muon catalyzed fusion.

        Hot fusion almost always produces a compound nucleus with high angular momentum. I belive this explains the branching rafio difference.

        The next question is: where is the gamma? I diont know. I expect the close presence of the screening electron does something to interfere with the gamma emission. A critic can argue: but then why isnt the gamma also inhibited in muon fusion? I dont know the answer to that.

        I think electron charge screening and low angular momentum may explain 2 of the objections to cold fusion: traversing the coulomb barrier and the branching ratio.

        In studies of low energy D bombardment kf metals, a remarkable “coincidence” has been found. Palladium enhances DD fusion rate, by a charge screening mechanism, better than any other metal tested. It seems to work even better if lithium is present. Palladium is also the best metal for cold fusion. The most parsimonious explanation is that Cold fusion occurs in Pd becauss of charge acreening.

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      4. steve Post author

        Dear Dan, thanks for your comment!

        The branching ratio issue is key. (If you tell me a story about how the Coulomb barrier is overcome, but leave out how the energy turns into helium-4 and heat, then why bother? It’s a moot point!) So a few comments on the branching ratio. First, the gamma branch is infrequent because the process whereby a proton or neutron jumps out of the unstable excited He4 intermediate state is both energetically allowed and fast, whereas electromagnetic emission is slow because the nucleus is very small compared to the wavelength (it makes a bad antenna, loosely speaking). I don’t see how these considerations are very sensitive to angular momentum or energy or anything else I can think of. I mean, even at zero angular momentum and zero D-D kinetic energy, it is legal and likely for a proton or neutron to jump out of the excited He4 nucleus. Second, if the gamma branch is relatively more likely at lower energy, why isn’t that fact visible in the tabulated curves as a difference between 100keV and 50keV and 10keV? Third, why isn’t gamma branch more common in muon-catalyzed fusion? Fourth, and most importantly, if it’s the gamma branch, where is the gamma? Something has to take the extra energy. It’s not insane to propose that the electron is a spectator and winds up as a 24MeV beta particle. But even if the nuclear physics transition probability math worked out—and I don’t think it does—24MeV betas are experimentally ruled out just like 24MeV gammas are.

        Still, I’m happy to talk about the Coulomb barrier too. We were talking about the fact that D2 gas does not fuse, despite having a similar overall electron density as Pd-D, if electron screening were the key. You explain that as “the electrons are held in orbits too far from the nuclei to provide charge screening at small nuclei separation distance. The electrons are not free.” If we look at, for example, here, we see that in D2 gas, electrons do have most of their probability amplitude in between the two D’s, at least when the D’s are at their usual separation. Bring them closer, and sure, the electron weight between them goes down. The same is true for Pd-D, for the same reason.

        In the Pd-D case (but not the D-D case apparently), you seem to be treating the electron wavefunction as a probability distribution for an electron to be in different positions. That’s sorta OK if we were talking about one instant in time, but we’re not. The DD approach is gradual (relatively speaking), and the electron won’t sit around between the D’s and wait for the whole process to happen. I suggest that your intuition here should be based on the Born-Oppenheimer approximation. Visualize the electrons as fast—so fast that they’re instantaneous—and ions as slow-moving. As the ions gradually approach each other, the electrons move around super fast and sample every part of that probability distribution at every moment. So if, for example, 1% of the electron probability weight is between the two D’s, you shouldn’t think of it as 1 electron being present during 1% of the approaches, it really is much more like 1/100th of an electron is present during during 100% of the approaches, or if you prefer, 1 electron making sporadic cameo appearances between the D’s that add up to 1% of the D approach time.

        In this picture, I don’t see any difference between free and non-free electrons. Neither will be found in an orbital with a high probability amplitude to be in between two D’s as they get very close. A free electron is usually defined as one that is delocalized, and/or which has similar-energy empty states that it can move into. Neither of those things is helpful for screening the Coulomb repulsion, as far as I see. So I don’t follow why the purported difference between Pd-D and D2 gas can be explained by whether or not the electrons are free.

        My belief is still that electron screening will reduce the Coulomb barrier in the early part of the D-D approach, but not the later part where most of the energy requirement lies. If you reduce the energy barrier by 1keV or whatever, I’m sure that makes the transition probability go up by many orders of magnitude, but it still seems like it should be too low to actually happen with the frequency necessary to explain cold fusion. (Or have you seen anyone go through the math to argue to the contrary?)

        If I’m misunderstanding your ideas, sorry and please do try again. 😀

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      5. dan s

        “First, the gamma branch is infrequent because the process whereby a proton or neutron jumps out of the unstable excited He4 intermediate state is both energetically allowed and fast, whereas electromagnetic emission is slow because the nucleus is very small compared to the wavelength (it makes a bad antenna, loosely speaking).”

        I read that neutron/proton emission is affected by angular momentum of the compound nucleus, at least for larger nuclei. There is lots of experimental work on high angular momentum states in heavy nuclei.

        I searched and searched for experimental evidence of the impact of angular momentum for small nuclei, and it appears very little work as been done on this. So I dont think there is empirical evidence supporting the theoretical reasoning for the D+D branching ratio. Seems to be based on conjecture/hypothesis.

        My impression from the literature is that high angular momentum in 4He may be responsible for the neutron emission.

        Your comment gave me a thought: what happens to the “antenna” characteristics of the 4He nucleus if there is an electron very close by? Seems that arrangement could make the nucleus a much better antenna, no? The electric field from the electron may affect the dipole moment of the 4He nucleus such that it can radiate better. Just a thought.

        “I mean, even at zero angular momentum and zero D-D kinetic energy, it is legal and likely for a proton or neutron to jump out of the excited He4 nucleus”

        I looked for experimental evidence for this and found none. But maybe I was looking in the wrong place?

        “Second, if the gamma branch is relatively more likely at lower energy, why isn’t that fact visible in the tabulated curves as a difference between 100keV and 50keV and 10keV?”

        Such impact energies will almost always produce a compound nucleus with very high angular momentum.

        “Third, why isn’t gamma branch more common in muon-catalyzed fusion? ”

        Angular momentum. I looked into this question closely. Muonic D2 is almost entirely in the J=1 rotation state, and this angular momentum winds up in the compound nucleus after D+D fusion.

        “Fourth, and most importantly, if it’s the gamma branch, where is the gamma?”

        That I dont know! But hey, two out of three isn’t bad, right? lol I suggest its got something to do with the close proximity of the electron to the excited compound nucleus. See my comment about the proximate (screening) electron affecting the radiative properties of the compound nucleus.

        So, in other words, the D+D cold fusion process has a unique combination of features not found in hot fusion or muon fusion:
        zero angular momentum
        zero or very low impact kinetic energy (e.g. thermal)
        high electron density of cool electrons (room temp or <600K or so)
        close proximity of electron (the one that does the screening)

        I believe this particular combination of features may explain the mysteries of D+D cold fusion in metals.

        "You explain that as “the electrons are held in orbits too far from the nuclei to provide charge screening at small nuclei separation distance. The electrons are not free.”"

        I should have also mentioned that the electron bonding keeps the D nuclei separated.

        "So if, for example, 1% of the electron probability weight is between the two D’s, you shouldn’t think of it as 1 electron being present during 1% of the approaches, it really is much more like 1/100th of an electron is present during during 100% of the approaches, or if you prefer, 1 electron making sporadic cameo appearances between the D’s that add up to 1% of the D approach time."

        I dont think that is the correct way to think about this situation. The electron has a definite location (I do not subscribe to the Copenhagen interpretation. I prefer Bohms formulation). Looking at the average e-density makes sense for answering some problems, but not D+D catalyzed by charge screening, because the charge screening catalysis depends on the specific location of each electron, not the average electron density. Considering only the average density gives the wrong answer, IMO.

        I think the D+D fusion likely happens in dislocations containing two D+ ions. These ions vibrate and bounce off each other millions of times per second. I have a few papers calculating this impact rate. If an electron happens to wander between the D+ ions at the time and location of their closest approach, fusion probability is dramatically increased.

        "In this picture, I don’t see any difference between free and non-free electrons. Neither will be found in an orbital with a high probability amplitude to be in between two D’s as they get very close."

        I think the location of D+ impacts is at (or near) the center of the octahedral locations. I think the bound electrons dont go there very much. The electrons that pass through this location are primarily the conduction/free electrons.

        "So I don’t follow why the purported difference between Pd-D and D2 gas can be explained by whether or not the electrons are free."

        Another way to think about this question: in a D2 molecule, the 2 electrons are bound to the D nuclei (and to eachother). All this binding holds the nuclei apart. This is because the electrons are locked in orbits around the nucleus, and the nucleus is locked in the center of the electron orbits. The nucleus cannot go anywhere without dragging the electron orbits along. The D nuclei cannot get close to eachother because the bound electron orbits would overlap more than they want to. Thermal energy does not come close to overcoming this.

        In Pd-D, the D exist as ions (D+) with no bound electrons. The D+ ion in Pd-D can move around without bringing a bound electron with it. The D+ ions can bounce into eachother. their separation is not maintaining by associated bound electrons.

        Make sense? This is how I think about it anyway.

        This makes sense in view of muon catalyzed fusion. When the muon enters a D atom, the bound electron is ejected, and the D nucleus becomes mobile. it can enter a regular D atom (as a neutral muonic atom) and get very close to the other D nucleus.

        "(Or have you seen anyone go through the math to argue to the contrary?)"

        I have read a couple papers (from the early 90s) with this calulation, but 1) they dont quite get a high enough reaction rate to explain cold fusion, and 2) there are lots of simplifying assumptions.

        Much is unknown about the the conditions inside highly loaded Pd-D, especially in lattice defects, and the effects of contaminants (like Lithium, boron or other metals). So, I dont think its reasonable to say its impossible based on theoretical calculations. The calculations and experiments show very dramatic increases in fusion reaction rate. I think its very plausible that complications (e.g. contaminants, lattice defects) could explain the reaction rates.

        I agree with you about WL theory. Does not make sense to me. Especially the requirement that the neutrons have EXTREMELY low energy. Why would they have such low energy when there is so much energy available. Seems to me the neutrons would be thermalized very quickly.

        Please email me and I will send you some interesting papers to discuss.

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  2. Dan steinberg

    The coulomb barrier in cold fusion is not a mystery at all. Cold fusion occurs in metals, which have high electron densities. The electrons screen the coulomb repulsion. In other words when an electron is present between the deuteron nuclei, the coulomb repulsion vanishes. In fact, a slight coulomb attractin is present, pulling the deuterons together.

    Charge screening is highly effective for inducing fusion. The fusion reaction rate can be enhanced by 50-70 orders of magnitude by charge screening.

    Muons catalyze fusion by charge screening. And muons can do this at cryogenic temperatures, in liquid hydrogen. Cant get much colder than that. So cold fusion is not so surprising.

    I am always amazed when physicists are mystified by cold fusion. Muon catalyzed fusion has been studied to death since the 1950s.

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    1. steve Post author

      Dear Dan, thanks for your comment! I’m very interested to see the calculation that concludes a 50-70 order-of-magnitude enhancement. Can you provide a link or reference? Thanks, Steve

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  3. Christian Teske

    Dear Steve, there are quite a few interesting publications (peer reviewed) regarding the measured screening potential for deuterated Palladium (800eV) or Platinium (675eV) (for this see for example F. Raiola; Physics Letters B547 (2002) 193-199 or F. Raiola; J. Phys. G: Nucl. Part. Phys. 31 (2005) 1141-1149). Based on this experimental evidence the fusion reaction rate is indeed increased more than 70 orders of magnitude. This has been shown in F.M. Prados-Estevez; Nuclear Instr. and Methods in Physics Research, B: 407 (2017) 67-72. Further readings regarding the topic of thermonuclear burning in with strong plasma screening are A. I. Chugunov; Physical Review D 76, 025028 (2007) and A. I. Chugunov; Physical Review C 80, 014611 (2009).
    Regards, Christian

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    1. steve Post author

      Dear Christian, thank you for the links. I don’t think they contradict anything I wrote. I agree that electron screening exists and can increase reaction rates by a large ratio. But the ratio, large though it is, is nevertheless increasing the reaction rate from “effectively 0” to “effectively 0”, when everything is calculated correctly. Screening is relevant as deuterons go from “far” to “close”, but screening does not appreciably help deuterons go from “close” to “very very close”, and the latter is where virtually all the energy and improbability are concentrated. If you disagree, please let me know exactly where and why. Thanks, Steve

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  4. Christian Teske

    Dear Steve, first of all, thanks for the response and also for your interesting blog.

    I disagree with the “effectively 0” and “far”, “close” and “very very close” are ill defined. Even if I assume “very very close” means something in the range of a few femtometers. The point here is the probability for a Deuteron to tunnel through the potential barrier which is dictated mostly by the “thickness” of the barrier. And this “thickness” of the barrier is dependent on the screening potential, which can make a huge difference between “close” and “closer but still not very very close”. In highly screened plasmas (which is the case for Hydrogen in a metal lattice) the reactivity is given by (0)*exp(h) (see also (Chug2009) or (Ich1993)) were (0) is the temperature averaged reactivity without screening (equations for plotting the DD reactivity without screening can be found in (Pra2017) or in any good book/publication on fusion reactions) and exp(h) is the enhancement factor due to screening. In general h is a function of the Plasma parameter G=U(screening)/kT which includes the screening potential U(screening). It has been shown by functional integral calculations (Chug2009, etc.) that h can be approximated by h~G*(1-c*G^(4/3)) for G<170 (in (Chug2009) there are actually a large variety of functions h(G) for various values of G depending on the associated astrophysical problem), were c is a constant dependent on the material (1.8*10^(-4) for Pd according to (Pra2017)). Now, let's do a very basic calculation of the enhancement factor between "close" (screening potential of a hydrogen atom of 27eV) and "closer but still not very very close" (screening potential of Palladium at 800eV). The enhancement factor at an ion temperature of lets say kT=5eV is ~exp(27/5)=221 for the hydrogen atom screening and ~exp(800/5)=3×10^(69) (!) for Palladium. So, this answers your question above to Dan Steinberg. When everything is calculated correctly as you say, there is actually quite a large difference between "close" and "closer than close but still not very very close" (the probability for a single fusion reaction is still "very very low" compared to the probability of two deuterons fusing at a distance "very very close", however the reaction rate is relevant due to the large amount of particles involved).

    According to (Pra2017) the reactivity for the DD-reaction in Pd (Palladium) reaches 1.2×10(-9)cm^3/s at an ion temperature of approx. 5eV. This can also be verified by Fig. 2 and equation (14) and (17) in (Pra2017). This boils down to 7.44×10^(13)/s for a density of 6.2×10^(22)/cm^3 in Palladium. Even if we assume a surface effect in the Palladium just several atom layers thick, such reaction rates are far from "effectively 0". Then again, the equations for the reaction rates in (Pra2017) are based on the "standard model" for nuclear fusion reactions (for this see (Chug2007) , (Chug2009) and (Ich1993)). Though, it should be noted here that the reactivity for DD reactions in Fig. 2 of (Pra2017) are plotted as a function of U(screening)/kT which is a little misleading. However, the reaction rates can be plotted quite easily with Maple as a function of the ion temperature (the ion temperature I am referring to would be the temperature of the D-ions in the lattice). For relative high ion temperatures (keV range) the reaction rate is in the ballpark of the unscreened fusion reaction rate (as one would expect due to the fact, that the ion temperature is then well above the screening potential of 800eV or G=U(sreening)/kT5eV), which has a relevant impact on the reaction rate)). All of those publications estimated the screening potential in Pd to be in the range of several 10eV. Hence, the calculated reaction rates were several 10s of orders of magnitude off or “effectively 0”, which led the scientific community to disregard any “cold fusion” claims. Measured values for the screening potential were published much later after 2002, which actually showed a discrepancy of more than an order of magnitude between theoretical predictions and measured values of the screening potential (as has been shown by the calculation above, an order of magnitude difference in the screening potential will lead to an error of several 10s of orders of magnitude in the fusion reactivity). Very extensive measurements of the screening potentials in various metals were performed by the Raiola-Rolfs-Group (Rai2002) and (Rai2005) and others. However, those measurements were performed long after the initial “cold fusion hype”, so not until recently no one has actually published any serious papers (peer reviewed high impact factor journals) on the fusion reaction rates inside metal lattices including the revised screening potentials.

    Now, my comment in a nutshell referring to Huizengas three mysteries of cold fusion:
    1. Penetration of the coulomb barrier: Well, this can indeed be explained by the latest experimentally verified values of the screening potential and the application of the “standard model” of fusion reactions in highly screened media.
    2. The lack of neutron emissions: According to Robert Bussard (Inventor of the Polywell Fusion Device) the branching ratio for the DD reaction is Sigma(He,n)/Sigma(T,p)=1.352*exp(-0.897/sqrt(E)) , with the equivalent energy E in keV (Bus1989). As Bussard has shown in (Bus1989) the (3He,n) branch of the DD reaction will be frozen out at normal lattice energies (Bus1989). Which basically means that one would expect predominantly tritium production. I do admit that this is a rather flimsy explanation because either the tritium somehow escapes from the lattice or the tritium should then react with the deuterium leading to MeV neutrons, so I admit, I am grasping at straws here.
    3. The lack of high energy gammas: I don’t have the foggiest idea.

    Please note that the publications listed are not the “usual suspects” in cold fusion or LENR research but peer reviewed and high impact factor journal publications.

    Chug2007 A. I. Chugunov; Physical Review D 76, 025028 (2007)
    Chug2009 A. I. Chugunov; Physical Review C 80, 014611 (2009)
    Ich1993 S. Ichimaru; Rev. Mod. Phys. 65, 2, (1993)
    Pra2017 F.M. Prados-Estevez; Nuclear Instr. and Methods in Physics Research, B: 407 (2017) 67-72
    Leg1989 A. J. Leggett and G. Baym; Phys. Rev. Lett. 63, 191, (1989)
    Tes2010a C. Teske; Phys. Plasmas 17, 043501 (2010)
    Tes2010b C. Teske; Rev. Sci. Instr. 81, 046101 (2010)
    Tes2012 C. Teske; Phys. Plasmas 19, 033505 (2012)
    Rai2002 F. Raiola; Physics Letters B547 (2002) 193-199
    Rai2005 F. Raiola; J. Phys. G: Nucl. Part. Phys. 31 (2005) 1141-1149
    Bus1989 R. Bussard; Fus. Tech. 16 (1989)

    Kind regards
    Christian Teske

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  5. Christian Teske

    My comment above has been shortened for reasons I don’t know, however i@ll include the missing part of my comment above here:

    The point being made here is that if one includes the latest results for screening potentials in certain metals published in (Rai2002) and (Rai2005) then the calculated reaction rates (or as you state “when everything is calculated correctly”) can become relevant in a temperature range achievable by relative simple technology discharge plasmas. I would suggest using a simple theta pinch plasma which can be driven by semiconductor technology (this has been shown in (Tes2010a), (Tes2010b) and (Tes2012)) and let it implode on Palladium. The deuteron ion density and temperature inside a few surface layers should still be high enough to initiate the reaction rates calculated in (Pra2017).

    In my opinion one of the reasons (beside the lack of lethal radiation) why the scientific community has ignored the lattice fusion claims is the fact that the screening potential in Pd and other metals has been underestimated for quite a long time. Most of the serious papers on fusion reactions in metal lattice surroundings were published shortly after the Pons/Fleischmann announcement in 1989/1990 like the Leggett and Bayms publication (Legg1989) and the very extensive paper by Ichimaru (Ich1993) (Ichimaru for example states a screening potential of approx. 76eV in Pd leading to a reaction rate of 10(-37)/s at 300K or 0.025eV, compare that to the publication of Prados-Estevez and it becomes very clear what a huge difference an order of magnitude in the screening potential makes (it should be noted here that the estimation of the achievable ion temperature inside the lattice is also different in the calculations made in the two publications (0.025eV vs >5eV), which has a relevant impact on the reaction rate)). All of those publications estimated the screening potential in Pd to be in the range of several 10eV. Hence, the calculated reaction rates were several 10s of orders of magnitude off or “effectively 0”, which led the scientific community to disregard any “cold fusion” claims. Measured values for the screening potential were published much later after 2002, which actually showed a discrepancy of more than an order of magnitude between theoretical predictions and measured values of the screening potential (as has been shown by the calculation above, an order of magnitude difference in the screening potential will lead to an error of several 10s of orders of magnitude in the fusion reactivity). Very extensive measurements of the screening potentials in various metals were performed by the Raiola-Rolfs-Group (Rai2002) and (Rai2005) and others. However, those measurements were performed long after the initial “cold fusion hype”, so not until recently no one has actually published any serious papers (peer reviewed high impact factor journals) on the fusion reaction rates inside metal lattices including the revised screening potentials.

    Regards
    Christian Teske

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