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Andrea Rossi and cold fusion

There are lots of interesting articles on Anthony Watts' blog but when I saw the text by Ric Werme

Andrea Rossi’s E-cat fusion device on target
and noticed that it has a perfect, 5-star rating, I had to tell myself: holy cow! Do the skeptics really abandon 100% of their skepticism in order to believe this self-evident garbage? Is it just because there's not a lot of publicity about this "technology" (so far)? Well, it may change.

It's being claimed that you may buy a cold fusion power plant sometime next year. You will receive a simple low-energy gadget. It has some hydrogen and nickel in it, and some electricity. So it doesn't differ from an NiMH battery so much.

Here a miracle occurs. You surely don't want to be annoyed by some technical details, do you? It's enough to trust the ingenious experts who invented - despite the disbelief of the physicists - a free energy source for you. Just for most technically inclined readers, here is the detailed engineering scheme of the new device:

A schematic diagram of the 2011 commercial perpetual motion cold fusion thermonuclear reactor.

And now, while emitting some - surely harmless - beta rays, neutrinos, and gamma rays, you have turned your nickel into copper (unfortunately not gold) while getting 1 megawatt of energy along the way. There's apparently nothing problematic about it and WUWT readers just vote that it is a 5 stars project. Well, it's plausible that the results wouldn't be much more sensible here but I still think that they would be at least slightly more sensible.

Werme's article contains no science - it's purely about marketing: it tries to answer questions "why should I care, what about me" etc. but never pays any attention how this could possibly work. You have to go to other sources to see what they're claiming to be achieving in their gadgets.

By a Google search, you may instantly find Journal of Nuclear Physics which is actually no journal of nuclear physics but a crackpot blog promoting this particular "fusion" (not to speak about similar crackpot theories of discrete spacetime - indeed, it does belong there, too). An even more official Free Energy Truth blog just announced the beginning of the World War III. :-)

One of the pages tries to answer the question how can 30% of nickel become copper (note that \(Ni,Cu\) have \(Z=28,29\), respectively) in those batteries. A good question, indeed. ;-) Well, the first reaction that should occur is
\[Ni_{58} + p \to Cu_{59} \]
By comparing the \(E=mc^2\) energies stored in the rest masses, you will may see that this produces about \(4 {\rm MeV}\) of energy. Nice and innocent. The lifetime of copper 58 is 118 seconds and then your healthy reactor is supposed to \(\beta^+\)-decay:
\[Cu_{59} \to Ni_{59}+\nu + e^+ \]
The positron annihilates with an electron to two gamma rays. Everyone is happy and your clean thermonuclear reactor may be used in your infant's bedroom. (This is of course also rubbish. Even if your substance were beta-decaying at this huge rate, it would produce lots of harmful ionizing radiation.)

An attempt to stabilize WUWT readers' brain

But wait a minute. Does it make any sense? Can the nuclei get close to each other for the fusion to occur? The proton and the initial nickel nucleus are positively charged, so they repel by the electrostatic (Coulomb) interaction. In fact, the maximum of the potential energy - when the distance is such that they nearly touch (but not quite, so that the attractive strong force is still essentially zero) - is a few \({\rm MeV}\) once again. You surely won't be able to subsidize the neutron - or the nickel nucleus - by your NiMH battery whose voltage is \(1.5\, {\rm V}\).

(It's not hard to see why the potential barrier is a few \({\rm MeV}\). The proton and the nucleus are roughly \(5,000+\) times closer than the proton and the electron in a hydrogen atom; the potential energy goes like \(1/r\). And the charge of the nucleus is \(Z=20\); the potential energy goes like \(Z\). So multiply these two numbers as well as \(10\,{eV}\) from the electrostatic energy in the hydrogen atom to get your estimate, \(5\,000\times 20\times 10\,{\rm eV} = 1\,{\rm MeV}\), for the electrostatic repulsive energy of the proton and the nucleus.)

Note that if a charged particle with charge \(+e\) - such as the proton - is accelerated in between the two poles of such a \(1.5\, {\rm V}\) battery, it will acquire energy \(1.5\, {\rm eV}\). You know, this is why the unit of energy is called an electronvolt: the unit is a product of an electron and a Volt. Because "mega" is "one million", you need millions of batteries to add up their voltages. You need to create a collider of this kind that will shoot the protons against the nickel nuclei. If it works, it will surely not resemble a NiMH battery.

If you want a proton to carry an extra kinetic energy of just \(1\,{\rm MeV}\), then \(\sqrt{1-\beta^2}\) has to be \(0.999\) which means that \(1-\beta^2\) is \(0.9995\) and \(\beta^2 = 0.0005\). It follows that the speed \(\beta\) is \(\sqrt{0.0005}=0.022\) times the speed of light - about seven thousand kilometers per second. That's the speed you would have to give to the protons (or the nickel nuclei!) for them to have a remote chance to get through the Coulomb barrier.

But the creative guys around Rossi have a solution. In the case that you're going to laugh out loud, be aware that the picture is not mine; it is taken from the pages promoting this stuff I linked above.

Picture via Journal of Nuclear Physics

You see that they "reject" a proton: a nucleus decides that a proton is a heretic who has to be excommunicated (I guess they meant "eject", but this is just the beginning). The angry proton borrows sunglasses from a carnival, masks himself to look like a neutron, and by this method borrowed from Agent 007 comfortably (and without any electrostatic repulsion) marches into the initial nickel atom, in order to revolutionize it and turn it into copper.

So Rossi's apparatus is mainly a device that produces these sunglasses for the proton and trains him to act as Agent 007. I mean James Bond.

Another page on the server tries to tell you additional details about the "probable" mechanism of the hydrogen/nickel cold fusion. (Note that they first managed to create a tabletop fusion reactor, and then they were looking how it's possible that they were so ingenious to guess it.) The page clarifies the training camp for the proton agents. How do they pretend to be neutrons? Well, they borrow an electron and produce a "mini-atom". That's just like an atom but it is as small as a nucleus. ;-)

Experienced TRF readers may recognize that at this moment, the cold fusion industry has incorporated another revolutionary discipline of physics, one about the hydrinos coined by Randell Mills who has earned tens of millions of dollars with this utterly stupid scam. The mini-atom cold fusion explanations are pretty much copied from his hydrino papers. Maybe these two groups even pay dividends to each other.

Quantum mechanics is evil, the uncertainty principle doesn't hold, and if you kindly ask the electrons to help you, their average distance from the proton may be 10,000 smaller than what they naturally like. If you carefully explain the electrons that they will help to make Mr Rossi quite some profit, the electrons won't hesitate and they will help to mask the protons so that they may pretend that they're neutrons.

Don't get me wrong. I find it plausible (yet unlikely) that a clever engineering setup will be found that will allow the protons in similar reactions to overcome the Coulomb barrier in the future, despite a "cold environment". Maybe a few protons will be exposed to some huge combined electromagnetic force from many electrons squeezed into a small volume. (Similar strategies exist in which laser beams are pushing the protons or nuclei.)

One by one, the protons will be fused, producing lots of energy. The only thing I am sure about is that these folks haven't managed to do anything of the sort. There is absolutely nothing nontrivial happening at the nuclear level in their devices. It's a NiMH battery claimed to become copper. Moreover, they claim that a nuclear transmutation of grams or kilograms of rather heavy elements is healthy enough for you to have it at home.

They have no idea what they're talking about and the actual mechanism behind their profits is that their customers are even more ignorant about basic physics than the authors of this "technology" themselves. Too bad to find out that this appraisal holds for most WUWT readers as well.

You can't turn macroscopic amounts of nickel into copper in a safe tabletop experiment. Fusion is physically doable but at least for a short period of time, the relevant nuclei simply have to have the energy corresponding to the temperature of millions of degrees. If you wish, the fusion reactor must inevitably "borrow" a huge energy comparable to the energy produced at the end. This follows from the shape of the potential. In this sense, the very term "cold fusion" is an oxymoron.

You can't get cold fusion, OK? This is just like the perpetuum mobile devices. Mr Rossi probably is a scammer but he may also be an honest re-discoverer of an NiMH battery: he just incorrectly measured how much energy such a battery may produce.

See Cbullitt for some fresh yet historical memories about cold fusion - and a nice compliment. ;-)

Wireless power transfer

By the way, if some people were observing some experiment where 50 cubic centimeters of a black box generated a very high amount of energy - kilowatthours - which can't occur chemically, well, then it is scam and there was either a secret extra conductor bringing extra energy or the energy was transmitted wirelessly. Some infrared radiation directed at the "reactor" would do.

Intel shows a light bulb shining in the air: the power is transmitted via electromagnetic waves, wirelessly. To transmit ordinary heat "wirelessly" is of course much easier.

I have made the same comment - either hidden wires or wireless power transmission - when I saw a video in which two burning candles power a light bulb or a motor etc.

I would never forget about the wireless possibility since ten years ago or so when I bought a rechargable toothbrush via The recharging occurs without any wires (which could be dangerous in a wet bathroom) - by electromagnetic waves.

It wasn't explained on the product so I was stunned - how it could recharge? :-) Of course, after some time - which wasn't limited, maybe hour, maybe even longer - I figured out what had to be going on. So then I could show the same trick e.g. to Prof Shiraz Minwalla who was equally shocked. How do the electrons get there through the plastic? :-)

Well, one doesn't need electrons to get through the plastic. One only needs the electrons behind the plastic to move and coils are enough to achieve it through the plastic. People, including smart and educated ones, may be easily fooled.

Algorithm of the scam

In the likely case that this is a scam and the people behind it know what they're doing, I guess that both the Greek and the U.S. companies are parts of the scam - they are meant to be the role models that lure other investors - other companies ready to produce this stuff and pay license fees.

The money for licenses will only come from other, new companies. The license contracts will be constructed in such a way that the dysfunction of the full engines won't make the contracts invalid. Moreover, I believe that the October deadline will be ignored and new "tight schedules" will be promoted to new investors without admitting that the previous ones were missed.

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reader Greg said...

Currently it's 3.5 stars on WUWT so the the skeptics are downgrading it. I also found your comments, Lubos.

I don't know the physics well enough to comment on this stuff, but clearly you do and I've read many of your posts that are WAY over my head. I'm just one of the idiots who'd like to see something like this be real. However, reality intrudes...

So let's imagine that Bob comes to your lab and puts a black box on your table, small enough to actually fit on your table.

He claims that the box takes in input energy, say 220v, and outputs *way* more energy, for as long as you care to keep it plugged in. Let's say it's on an order of what Rossi is claiming.

What would be needed to prove to you, or any physicist, that the thing is real, with "massive excess energy" being output, and not a fancy battery, massive electrical hack, or otherwise a scam? I'm not saying this thing is real, I'm asking what would need to be done to prove that something like this IS real?


reader Renee said...

Fascinating discussion, to which I would like to add a bit of nitpicking. In the paragraph that starts "It's not hard to see why...", your estimation of the number of eV in the potential barrier is off by a factor of 10: 5,000 x 20 x 10eV = 0.1 MeV, not 1 MeV as you stated.

Other than that, I can't really check your assumptions without spending a lot of Google time; I'm afraid that most of what I need to vremember about physics can be summed up in the statement "runs to red" (which describes the movement of negatively charged biomolecules in an electric field towards the red anode).

I've run into some of Rossi's devotees before; they sure made a fuss when I pointed out that there is no way Rossi could have invented a device that does what he claims without having a basic understanding of the mechanism involved.

reader Luboš Motl said...

Thanks for your kind words, Renee, but there is no numerical mistake in that paragraph.

Five thousand times twenty is equal to one hundred thousand.

One hundred thousand times ten electronvolts is one million electronvolts, i.e. one megaelectronvolt.

I agree such a thing is unlikely to be discovered "by accident". Combining some random ordinary materials and hoping that they will work as a perpetuum mobile is just not how physics may make progress.

reader Renee said...

You are correct. That just proves I shouldn't be trying to do math early in the morning.

reader Luboš Motl said...

Dear Ron, pretty much all experiments that tried to reproduce Pons-Fleischmann concluded that the effect doesn't exist.

Your numbers and words make no sense and they're totally incompatible with each other. You are talking about cold fusion but at the same moment, you are admitting that all the relevant energies are at least tens of keVs and more typically many MeVs.

Do you understand what is the temperature if just one modest keV is given to a degree of freedom? If kT is equal to 1 keV, i.e. 1.602 x 10^{-16} joules, the temperature is already 1.16 million kelvins.

This is not just a proof of an innocent adjective, "hot", which is associated with any fusion reaction. It's also a proof that any conventional vessel that would try to "contain" any thermonuclear reaction would immediately melt.

Even when divided to thousands of nearby atoms, you get the temperature of thousands of degrees, so even if the reaction only occurred "slowly", it would gradually melt the inner surface of the vessel. It's inevitable. You can't "contain" a fusing fuel in a cold vessel.

Sometimes you look very smart but this fusion stuff speaks an immensely different language. You speak like some of the most uneducated cranks who are out there.

reader Ron Maimon said...

Yes, Lubos, you are absolutely right--- the KeV's would be impossible if they would be thermal energy. This is how I am sure they aren't thermal.

You are wrong about reproductions of Pons/Fleischamann, many reproductions succeeded in 1989, at least 3 immediately, and all of them would have succeeded if they had patience to run for a few weeks on good samples at high deuteron loading, as Pons and Fleischmann had patience. Further, the lack of reproducibility is a scientific fraud propagated by hot-fusion folks at MIT. But people bought it because of the thermalization argument you give.

The whole cold fusion thing is extraordinarily out of equilibrium--- the energy is in ridiculous modes far far above any thermal modes in energy: K-shell holes and KeV deuterons. These take many steps to turn into thermal energy. Usually, they don't turn thermal at all, they just emit X-rays which escape the material. You might be puzzled by this. Why don't they turn thermal?

High energy particles don't have a lot of thermal friction. This is the paradox of high-energy particles. You would naively think they would quickly thermalize their energy into thermal modes. But they don't. They don't heat up the solid, they ionize it, and preferentially in inner shells. This is well known since the 1910s experimentally, understanding this was a major motivation for Bohr in 1913, and it was completely explained by QED in the 1940s, by Hans Bethe. The Bethe theory describes what happens to fast charged ionizing radiation, it loses energy to ionizing atoms in a line, and with a completely non-thermal energy distribution--- the high energy modes are much more likely to be excited than the low-energy modes. The high energy modes then produce X-rays at KeV energies, not thermal photons at fractions of an eV.

The result stays a crystal, the solid doesn't melt locally, it just is ionized. This is even though the energy, if translated to thermal terms, would be millions or billions of degrees. These ionized atoms just can't melt the crystal, because a nucleus can't be kicked all together with it's core electrons (even though there is enough energy to do this), this is phase-space impossible. All you can do is kick out an individual electron. It's the phase-space problem that a melted solid is not immediately phase-space accessible from a crystal solid by kicking a single particle.

On the other hand, you can kick a deuteron, because a deuteron has no core. So there is no obstacle to having fast deuterons in a crystal, as there is to having fast Pd cores. Ionizing radiation can dump its energy into ionizing electrons or accelerating deuterons, but not into melting the crystal. This is why we have X-ray emissions without crystal melting, and you are right that back-of-the-envelope, these are not possible (this is why you need to understand QED to understand high energy particle energy transfer).

The cold fusion process is exploiting the long thermalization time to do many fusions before the crystal melts. The crystal eventually does melt, it explodes in little clumps a few microns across. This is what happens to cold fusion devices. Scanning electron microscopy on used cathodes shows that they are full of micron-scale pits where the Pd melted and exploded outward. The heat output is in extremely localized flashes at very high energy on the surface, it isn't uniform. This was another clue for me as to the theory.

But melting the crystal only happens after a bazillion fusions, once the KeV scale levels have time to mix with thermal levels. This is what prevents a runaway chain-reaction and an explosion, the chain reaction stops once the solid melts and the deuterons are far from ionized atoms. Still, you can have a chain-reaction in Pd/d, and there are explosions (although not of nuclear scale). The self-limiting aspects are what make this interesting.

reader Ron Maimon said...

You don't need to publish this comment if you don't want to--- it's just a correction to your remarks below: the K-shell of Pd is a little over 20KeV, around 22KeV. The next level is at 3KeV. The beam cross section for fusion on deuterated palladium(yes, this experiment is done!) is 1 d-d neutron emitting fusion in 10,000-100,000 at around 3-20KeV, (I forgot the exact numbers, it was something like some thousands of fusions in a beam of a billion particles but I can dig up the paper if you want). The K-shell is not "something like KeV", it is exactly what I give here, and the fusion rate at 20KeV is not at all negligible, it's very high (this is higher temperature than the center of the sun, if converted to thermal units). The enhancement of fusion that lowers the barrier for d-d fusion to KeV's as opposed to MeV's (which is the naive classical Coulomb barrier for deuterons to touch) is well known, it is what allows ordinary fusion in H-bombs and muon-catalyzed fusion, and all other known accepted forms--- it's ordinary quantum tunnelling during the scattering process. Please fix these errors, as these are a little bit demoralizing considering your usual scientific honesty.

reader Ron Maimon said...

Hey Lubos, What happened to my comments? If you insist on deleting my corrections to your false claims, I will permanently move you from "scientist" column to the "politician" column. That means: I will no longer take anything you say seriously, and I'll stop upvoting your stackexchange answers, dude.

In a nutshell--- high energy particles don't thermalize crystals, as was known already to Bohr, high energy charged particles deposit energy in ionization of atoms, highest energy levels first. This makes thermalization very inefficient, and the result is that you can have materials emitting X-rays at KeV energies without melting the lattice, even though X-rays have an energy corresponding to a temperature of millions of degrees. When an MeV electron or proton goes through a crystal, the crystal doesn't locally melt, it ionizes and reemits X-rays.

This is a phase-space restriction on the thermalization, it occurs because you can't locally melt a crystal by just kicking the nucleus, because this would fully ionize the nucleus. You would need to kick the entire atomic core, the nucleus plus inner electrons, and that is phase space impossible. What happens instead is that electrons are kicked out individually. This was described by Hans Bethe in the 1940s, using QED. The Bethe formula correctly predicts the stopping times of charged particles in matter of all sorts, and is well established science.

Bethe's process does not lead to melting, it leads to secondary X-ray and beta emissions. The beta emissions are from electrons falling back into the K-shell (and other shell) holes produced by the charged particle. In a deuterated metal, the deuterons don't have cores, and they can be kicked around. The deuterons accelerate from K-shell holes and charged particles, they fuse near a nucleus, and the resulting fast charged particle emissions lead to more K-shell holes.

The K-shell energy of Pd is 20KeV. The fusion rate of beams on deuterated Pd is published and available, and the neutron emission fusion rate is 1 in 100,000 atoms or so, at around 3KeV, before the deuterons stop. The rate is not negligible at 1KeV, because of known tunneling enhancements to fusion cross sections, things that are known from hot fusion and muon catalyzed fusion.

I know this is your blog, but have some freaking respect for scientific honesty. I have just demolished your retarded argument against cold fusion, so please change your mind on this, and do it quickly. Anything else is not befitting a scientist.

reader Luboš Motl said...

Dear Ron, your comments were waiting in the moderation queue. I am not available 24 hours a day, especially not on Sunday.

reader Luboš Motl said...

You're being completely ludicrous, Ron. The Coulomb barrier between pairs of nuclei is never smaller than 0.4 MeV and it is dozens of MeV for larger nuclei so even if you could subtract 20 keV, it would make the same impact as spitting on your "cold fusion apparatus". Most of the barrier remains and the reaction rate by tunneling is exponentially negligible.

reader Ron Maimon said...

Oops, sorry! I'm an idiot. Thanks Lubos, sorry for repeating myself, I really thought you were removing the comments. You can call me a crackpot all you like, I was just annoyed at what I thought was censorship. I won't make repeat comments again. Please delete the last comment of mine, it was rude, redundant, and wrong.

reader Ron Maimon said...

Hi Lubos, I won't bother you again, but I am now certain the theory is correct, as I have figured out the last major discrepancy with the observation. The fusion-near-a-nucleus story is correct, and you can see it is true because if you have a K-shell hole and you kick a deuteron, it will get just enough KE to approach the nucleus to the distance of a K-shell hole, i.e. 100 fermis. This gives a classical turning point, and a wavefunction enhancement as the inverse square-root of the classical velocity, and when two such accelerated deuterons meet, they meet right next to a Pd nucleus, leading to the 3-body fusion. The bands are real, the fusion is real, the result is excess heat at 24 MeV per deuteron pair, and if you were clever, you could have predicted it theoretically with no experimental data already before 1989.

The reason I am certain is the pattern of transmutation products in the Pd, it matches the theoretical predictions to a tee. At first I thought the transmutations would only happen _after_ the fusion, from the alpha bombardment, but that's not true. A fusion by a nucleus transferring the energy electrostatically fragments the nucleus. The theory is complete, and I am very happy. You don't need to publish this, I thought you might like to know.

reader Luboš Motl said...

Thanks, Ron, it's very good to know. Why don't you borrow a few thousand dollars and start to produce electricity almost for free?

More seriously, could you please be more specific about how large a wave function enhancement you get and why you think it changes anything?

reader Ron Maimon said...

Yes, I'll submit a few articles somewhere after a few weekends (I have an interesting bioinformatics money job I have been neglecting) . This will consist of three major parts: 1. The theory of deep holes--- the K-shell electrons. These are described by the creation-annihilation inverted nonrelativistic spinning Schrodinger field (the Pauli field equation, or the nonrelativistic Dirac equation). They behave as positive charged particles with negative mass, and this allows you to compute the X-ray transition spectra (including selection rules) from single particle QM. The results are interesting, as they allow you to simply calculate the Moseley law screening coefficient from Hartree fock type effective potentials in the electron ground state very accurately. 2. The interaction of these deep holes with deuterons in a deuterated metal leads to the banding of X-rays in hydrogenated metals (this will be uncontroversial and important), this allows X-ray refraction and easy downconversion, together with potential solid state X-ray laser, not plasma (although such a laser will be risky with deuterium, due to the risk of small atomic explosion). The band-width calculation I haven't done but it's not hard 3. The effective field theory of deuteron resonances--- there are about 6 of these. The coefficients for transitions between these are not observed in cases where you have a strong electric field. I tried to formulate this theory about 3 years ago, but there wasn't enough data. I think it might be possible to get around this with some heuristic estimates using some nuclear potentials. This requires more expertise in nuclear physics. 4. The wavefunction of the deuterons in certain positions in the X-ray band (it has a lot of states) will correspond to classical orbits which go around from nucleus to nucleus with turning points of closest approach arount 100fm. The quantitative estimates for the fusion probability when such states have a given occupation number can be given from the theory in 4. The important thing is the electrostatic transfer of energy to the nucleus, which leads to fragmentation similar to electron bombardment at 20 MeV (this involves a virtual photon of about the same energy). This leads to nuclear fragmentation and recoil which is quantitatively understood from some 1970s measurements of electron nuclear interaction. The resulting fragmentation spectrum involves a whole bunch of different species, and one will have a roughly quantitatively accurate list of emitted light isotopes during fusion. There is a lot of nuclear data for what happens when these isotopes collide with heavy nuclei, but the general estimates can be just from the Coulomb barrier to estimate absorption probability, and assuming absorption plus gamma/beta decay or one-particle or one-alpha ejection is the major tranmutation source. This allows you to give a quantitative prediction for the tranmutation spectrum for the cold fusion which is a smoking-gun signature--- you get light elements and heavy elements obeying a sum rule: (Pd-X)+X = (Pd+X) where X is a light ejecta, and (plus minus one or two in charge to account subsequent beta decay, and with some corrections for known alpha ejection probability). These will be for all transmutations in mass more than +4. For less than +4, the tranmutation spectrum is the exactly same as Pd and d bombardment with 10-20 MeV alphas in a precise distribution (which is dependent on what ejecta comes out). This is exactly consistent with Wolf and Iwamura's transmutation data (although I should read Mizuno's book, which is available).

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reader Doktor Bob said...

A international science team have made a independent report based on two tests of Cold Fusion reactors. The tests went on for about 100 hours each and the results came out very positive.

Please make a update article / blog on this

Love / Dr Bob

reader Luboš Motl said...

Dear Doktor Bob,

the "test" wasn't independent and it is completely wrong, too. You may read about it in the yesterday's article

I have nothing to add or correct in the blog entry above.


reader Doktor Bob said...

Among the testers you find a rep from Swedens Most prestigious University as well as Europes Oldest.

The tests where financed by Elforsk, a collaboration of companies and utilities.

I do not know if there is a "specific definition" of the meaning of the word independent in relation to "science".

If you ask me... I would define it as independent.


reader Luboš Motl said...

Dear Doktor Bob, the paper isn't an independent test because it describes two presentations run by Rossi on which some of the authors of the paper were witnesses - as obedient as Jehovah's Witnesses.

You may describe these people by impressively sounding names but it still changes nothing about the fact that it's a shameful pseudoscience, scam, and the people are a mixture of fraudsters and gullible idiots.

reader Doktor Bob said...

Well the conclusion of the report is that they will do another testrun, this time for 6 months.

And I did read an interview with one of the scientists explaining that Rossi was not running the tests, he was around to answer questions if needed (not word by word quote)

Im not sure if i agree with your definitions of words such as "facts", and I did know about some of these scientists before me OR they heard about Cold Fusion, thats why I referee to them with respect for their background and competence.

They have their reputations to worry about and I think they do not take lightly on this task.

But always good to express opinions.

Peace Out

reader Luboš Motl said...

Could they please follow the basic scientific standards? Ethan Siegel summarized some errors they have to avoid for their tests to have any value here:

reader Eugene S said...

In my humble opinion, this post by an electrical engineer on Dorigo's blog kills the Swedish-Italian paper dead and it has the advantage that anyone can understand it at first reading.