Cold Fusion Proven True by U.S. Navy Researchers

[TallDave]

Man you guys are willing to be open to cold fusion, which has a sordid history and next to no evidence... but not other things which have copious amounts of evidence... it never ceases to amaze me.

It's a question of epistemology and the scientific method it underlies. Things we can replicate in a lab enjoy a different class of evidence than things we can't. That's why the consensus on Earth's gravity remained 10 m/s/s during a period when Steady State was completely overturned, and why we are so dead-certain about QM even though it runs totally counter to classical physics but quite uncertain about dark matter.

Things like AGW and cosmology are very very hard to bring to the level of certainty we have for things like QM or the force of gravity, where there are millions of measurements of the phenomenae in question. For AGW, the debate is mostly about what level of certainty we can apply to computer models. The IPCC politicians say 90%, actual scientists in the field of forecasting say 10-20%.

If the LENR guys are producing results that can be replicated, they deserve to be taken seriously. If not, the whole episode will soon be forgotten. I'm certainly open to the idea of attempting replication.

One guy told me he thought it was a new quantum mechanical process.

Was his name Mills? (Sorry, couldn't resist.)

W-L is interesting. http://www.i-sis.org.uk/Widom-Larsen.php

Lithium-6 + 2 neutrons -> 2 helium-4 + beta particle + neutrino + 26.9 MeV

[BenTC]

But I absolutely stand by what I say, and I say it for objective reasons, there is no way an unguarded nuclear process of D+D->4He can be observed to generate heat by anyone alive.

Perhaps it is not D+D. Reading up on kurt9's reference to Widom-Larson Theory leads to this proposal http://www.i-sis.org.uk/Widom-Larsen.php

e- + d+ -> 2 neutrons + neutrino

This type of neutron production due to weak interactions in very high surface electric fields is well-described by the generally accepted electroweak theory [3] on which the W-L theory of LENRs is based.

An isolated ‘normal’ thermal neutron outside a nucleus travelling through a solid has a quantum mechanical wavelength of about 0.2 nanometre (1 nanometre is 10-9m) and a speed of about 2 200 metres per second, which is faster than a rifle bullet. Interestingly, the ‘size’ of a neutron confined inside an atomic nucleus is even smaller, at several femtometres (10-12 m). By contrast, the neutron formed on a metallic hydride surface in a LENR is more-or-less standing still. Being formed collectively, these Ultra Low Momentum neutrons have almost no kinetic energy at the instant of their creation, effectively zero. This gives them huge quantum mechanical wavelengths compared to ‘normal’ neutrons.
...
ULM neutrons’ huge size is exactly why biologically dangerous energetic (‘hot’) neutrons are not released by LENR systems. ULM neutrons are extraordinarily ‘cold’ to begin with; and virtually all are absorbed locally; they never get a chance to escape and go anywhere. It is the first reason why LENRs are safe and environmentally friendly in comparison with heavy element neutron-triggered fission and light element hot fusion.

After being created, ULM neutrons are efficiently absorbed by nearby target atoms, resulting in nuclear transmutations into different elements or isotopes [5]. Unstable transmutation products undergo subsequent weak interaction beta decays that, depending upon exactly which nearby target elements were used as ‘fuel,’ can then release large amounts of nuclear binding energy

As I understand it from my brief reading, experimental results do produce a variety of unexpected elements.

I am still working through these:
+ http://www.newenergytimes.com/v2/sr/WL/ ... Slides.pdf
+ http://www.newenergytimes.com/v2/sr/WL/ ... Slides.pdf

[chrismb]

Perhaps it is not D+D. Reading up on kurt9's reference to Widom-Larson Theory leads to this proposal http://www.i-sis.org.uk/Widom-Larsen.php

e- + d+ -> 2 neutrons + neutrino

mass of two neutron = 1879.131 MeV/c2

mass of deuterium atom = 1876.123 MeV/c2

More mass after reaction = endothermic.

oops... only cooling from this reaction, no heating.

Now I don't necessarily state that heat cannot be evolved from this, but to do so it *must* react by neutron absorption and transmutation of the surrounding material, so odd isotopes must be in abundance, and many of those reactions would surely release easily identified gammas.

I will add aswell, though I am sure you will find it nit-picking, but this isn't a nuclear reaction. Electrons aren't nucleons, just as the emission of light from an excited atom isn't *a nuclear process*.

No-one has yet suggested the alternative to my *where's the momentum gone* conundrum. Just to show I am thinking about it, momentum would be conserved if there was a 4 body interaction, 4*D->8Be->2He. The momentum coul dbe balanced then. But I cannot believe this unless an experiment is done in which 6Li is used in the cell and it all disappears (because the implication of this 4 body reaction would be that 6Li+D is so easy as to *always* move on to the 4 body reaction before the 6Li drifts off).

[DeltaV]

No-one has yet suggested the alternative to my *where's the momentum gone* conundrum.

Somebody upstream on the experiment's worldline is operating a Mach-Lorentz Thruster ("Tomorrow's momentum today!").

[chrismb]

No-one has yet suggested the alternative to my *where's the momentum gone* conundrum.

Somebody upstream on the experiment's worldline is operating a Mach-Lorentz Thruster ("Tomorrow's momentum today!").

Maybe in someone else's universe, not in the one I live in....

[Tom Ligon]

Chris,

At the risk of proving I have no idea what your real point is (it could be true), here is my simpleminded answer about where the momentum goes.

You presume an alpha at 22 MeV, released into water at, lets's say, 300 K, or 0.026 eV. I presume you are saying the energy is actually split between the alpha and the other resulting particle, the tattered nucleus from which it came.

So these particles go crashing thru the surrounding water molecules. The initial collisions are way higher in energy than the chemical bond energy of the water, so water is broken down. The energy transfer in the first few collisions is, in fact, high enough you might well get some secondary nuclear reactions, fusing whatever comes off those collisions with whatever takes a direct enough hit. But for the most part I would expect momentum to transfer to more and more molecules.

In about 9e8 collisions the energy is spread over enough molecules that you are essentially at room temperature. Actually, the stopping power calculation is the way to go, but this simpleminded approximation at least conveys the idea. The initial particle transfers large amounts of momentum to the first few , uh ... thousand maybe ... molecules it hits, which in turn collide with their neighbors. We wind up with some OH radicals, H2O2, hydrogen, and oxygen, but mostly buzzed water molecules.

This sounds like a lot of collisions, but a milliliter of water contains about 3.3e22 molecules.

And that 23 MeV is 3.7e-12 joules, so the water temperature rise from a single event is trivial.

If all that were happening in these reactions was release of charged particles, I would not expect much of a hazard outside apparatus containing a teacup of water. I would expect some localized high energy events close to the source of the high energy particles. Somebody posted a link to a PowerPoint show (I do not wish to imply a PowerPoint presentation is, in itself, proof of anything beyond the existence of MicroSoft). It does show localized damage that would seem consistent with something of this sort, plus reports some x-ray emissions and neutron tracks in a detector material. I would expect light emissions as well.

You say you would not worry about standing over an operating swimming pool reactor. I think we're talking degree of comfort here, just how much water you think is enough.

[Tom Ligon]

All,

If I recall correctly, about the first page or so of Dr. Bussard's CF paper, where he discusses Gamow barrier penetration dominance at low collision energies, he notes that the branching we normally expect of DD is not the approximate 50-50 split, half to making T and a proton, half to making He3 and a neutron. At the low energies the reaction takes a decidedly aneutronic bias.

I'm not sure it is relevant in the long run, as the virtual reactions that are actually suspected are not straight DD fusion.

[chrismb]

At the risk of proving I have no idea what your real point is (it could be true), here is my simpleminded answer about where the momentum goes.

You presume an alpha at 22 MeV, released into water at, lets's say, 300 K, or 0.026 eV. I presume you are saying the energy is actually split between the alpha and the other resulting particle, the tattered nucleus from which it came.

...hmmm. I think you hit the nail in your first phrase, I'm afraid

My issue is about how the alpha gets to be projected 'out' from the point of the reaction. Where is the 'recoil'? When you say 'the tattered remains', this is the point - there is no remains from two deuterium nucleii, so there is nothing to recoil against.

How can you shoot a bullet if you don't have a gun to fire it in? It just goes 'bang' if you heat up a bullet until it expoldes, it doesn't go anywhere. Only the bits os shrapnel will go somewhere, but you have to have an equal and opposite amounts of shrapnel [momentum] going in every direction, you can't get all the shrapnel exploding only to one side, it's not possible. Just as it isn't with CF.

I've no particular contest on your other points, only to say that, exactly as you say, the total flux of these 23MeV alphas must be very high to generate much heat and you *can be sure* that bags of gammas will come out of a high 23MeV flux of anything hitting anything.

[Tom Ligon]

Chris,

What if the alpha or other small charged particle is ejected from a heavier nucleus? Palladium or nickel electrode metal, for example, in which a "virtual" reaction has taken place?

I was about to say "This is not your grandfather's fusion," but actually it might be. Or maybe great-grandfather's. Old-fashioned linear accelerators hitting solid targets, run back in the 1930's. Normal thinking is that anything heavier than iron has no energy to offer via fusion, but Bussard's paper argues otherwise ... the energy should be much lower than from DD. I would have aimed for lower Z on the periodic table.

I will not admit to have actually done the experiment, but, ah, I can say with some certainty that when a .22 caliber cartridge is heated sufficiently, the bullet does not do much but the cartridge casing really flies. And there is surprisingly little bang.

[MSimon]

there is nothing to recoil against.

The lattice in which it is formed. Or did you miss my reference to Mossbauer above?

[Josh Cryer]

The IPCC politicians say 90%, actual scientists in the field of forecasting say 10-20%.

Just don't forget that the "guys in forecasting" use models built for free by the guys whom the IPCC "politicians" represent.

eg, NOAA makes the models that the forecasters plug into computers, and then just say what the NOAA tells them to say (because when it comes to weather, the NOAA is very very* good).

* compared to when I was a kid when almanacs were used for weather prediction, which btw, is the farmers astrology.

[MSimon]

The IPCC politicians say 90%, actual scientists in the field of forecasting say 10-20%.

Just don't forget that the "guys in forecasting" use models built for free by the guys whom the IPCC "politicians" represent.

eg, NOAA makes the models that the forecasters plug into computers, and then just say what the NOAA tells them to say (because when it comes to weather, the NOAA is very very* good).

* compared to when I was a kid when almanacs were used for weather prediction, which btw, is the farmers astrology.

The Met Office in Britain is in trouble because they got two winters in a row wrong. Beaten twice by Piers Corbyn who uses a solar model.

[chrismb]

Chris,

What if the alpha or other small charged particle is ejected from a heavier nucleus? Palladium or nickel electrode metal, for example, in which a "virtual" reaction has taken place?

there is nothing to recoil against.

The lattice in which it is formed. Or did you miss my reference to Mossbauer above?

Are you trying for me to feel yet more offended?

My objection above was that there was an implication my comments are driven by a belief system which was unfair because I have specific objective analyses not at all based on my personal interpretation.

Yet you two are banging on with your own personal preferences as to what might be going on (without the very slightest wiff of evidence) as some sort of defence against my objective critique.

If you are saying that in some way the palladium fuses with two deuterium nucleii to form a ?.. cadmium atom and that then decays by an alpha decay, well, possibly. I would accept that as *physically possibility*, though there are no cadmiun isotopes that are alpha emitters, so why would it decay with an alpha?

If you are saying that these chemically bound deuterium atoms can be nuclear-ejected, you're talking crap. How the heck can a chemical bond of the order of eV have any possible effect on a MeV-energy scaled particle? You can only react momentum of chemical reactions through chemical bonds.

It's a bit like you going back to the bullet example I gave above and saying "sure a bullet can react against something if it is just heated up - it will push against the air around it". The fact that you do not really appear to have thought about it, or just understand so little about conventional mechanics that you don't realise you're simply not even wrong, suggests to me you know you're just winding me up after I said I was offended by faith-based people saying *my* objective analyses were faith-based. I'm sure it is one of those methods to try to win an argument when you've got nothing to argue with (accuse the other person of your own weaknesses) but you've got me quite annoyed on this one.

[Josh Cryer]

The Met Office in Britain is in trouble because they got two winters in a row wrong. Beaten twice by Piers Corbyn who uses a solar model.

He's been as good as a coin flip.

[MSimon]

The Met Office in Britain is in trouble because they got two winters in a row wrong. Beaten twice by Piers Corbyn who uses a solar model.

He's been as good as a coin flip.

That is not too bad compared to the Met Office.