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To: Kevmo
I followed the authors' logic pretty well (and detected relatively little double-talk BS) -- including the premise that the lochon's lowering effect on the Coulomb barrier supposedly enables not only deuterium-deuterium, but also proton-proton interaction. And, I follow their claim that, since no resulting gamma radiation is detected, the released energy is emitted as heat (as observed).

But, they lost me at their "one equals zero" statement, below:

"The key to the mechanism is the lochon, which during the collision process attains significant energy (keV to MeV range); but,

being tightly bound in an l = 0 ground state,

it does not radiate. "

Either that is some "quantum believability" bit that I have not encountered, or these guys are -- very skillfully -- pulling our collective legs...

If, however, they are correct, it is encouraging, since protons (H nuclei) are far more abundant than Deuterium nuclei...

~~~~~

<DISCLAIMER> It was well after I left academia for industry that quantum physics knowledge really flourished -- so, like most of you fellow FReepers -- I find this stuff to be "fairly deep wading"...

24 posted on 07/02/2011 8:06:37 AM PDT by TXnMA (There is no Constitutional right to NOT be offended.)
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To: Kevmo
Hmmm -- been having problems with my satellite ISP, so, when I got a brower timeout on my first post attempt I blamed it on Wild Blue. Apparently, though, it was FR's delay in acknowledging receipt that was the problem -- hence the double post.

FWIW, I perceive this publication as nothing more than a bunch of "blackboard theorizing" -- with no experimental attempt by the authors to confirm their conjectures. IMH(engineer's)O, that makes it only slightly more valuable than the chalk dust now on the erasers... '-)

26 posted on 07/02/2011 8:18:11 AM PDT by TXnMA (There is no Constitutional right to NOT be offended.)
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To: TXnMA

The key to the mechanism is the lochon, which during the collision process attains significant energy (keV to MeV
range); but, being tightly bound in an l = 0 ground state, it does not radiate.
***That’s not 1=0, it is “L”, l=0, when you copy from PDF there are tons of artifacts.


28 posted on 07/02/2011 8:31:02 AM PDT by Kevmo (Turning the Party over to the so-called moderates wouldn't make any sense at all. ~Ronald Reagan)
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To: TXnMA
being tightly bound in an l = 0 ground state,

"l" is not the number one but the letter lower case "l" attempting to refer to quantum mechanical angular momentum state. Now this is nonsense because l is not a quantum mechanical observable. The QM observables are Jz or J^2=j*(j+1), and since we are talking about an electron with spin 1/2 there is no l=0 state. The lowest angular momentum state can be j=1/2 corresponding to a linear combination of l=0 and l=1 states with the spin 1/2 of the electron.

So even where they lost you this is nonsense, but that is not really the point, here.

the lochon's lowering effect on the Coulomb barrier supposedly enables not only deuterium-deuterium, but also proton-proton interaction. And, I follow their claim that, since no resulting gamma radiation is detected, the released energy is emitted as heat

But this is all BS. The lochon cannot exist because its properties (tightly localized electron density providing effective charge screening) violates the Heisenburg uncertainty principle. It also violates the Pauli principle (Fermi-Dirac statistics) since only electrons with energies above the Fermi energy can participate in this kind of screening. We already have an example of electron phonon interactions producing localized states. It is called the cooper pairing of superconductors. Properly taking account of the strength of electron phonon coupling and the fact that the pairing can only occur among electrons with energies near the Fermi-edge produces coherence lengths of the order of 3000 Angstroms (1000 lattice spaces not 1/10th of a lattice space as this lochon theory requires). [See Mourakchine http://arxiv.org/PS_cache/cond-mat/pdf/0405/0405602v1.pdf].

Now you might argue well who is to be believed. The theory of superconductivity is well established, experimentally well studied and theoretically well explained. It has all been worked out in very great detail using physical laws and principles that are broadly accepted.

None of the gibberish that he writes is accepted by any competent physicist.

And there was no peer review, since any competent solid state physicist, and I am not a solid state physicist, would have pointed out these issues and many many more.

Then there is all the nuclear physics nonsense,since no resulting gamma radiation is detected, the released energy is emitted as heat

But there is no way that nuclear processes (scale lengths of a fermi - 10-15cm couples to lattice phonons directly - scale lengths of 10-8 cm). It is like rocking a battleship with pond ripples. The frequencies are incommensurate by the same number of orders of magnitude, and the energy does not work either. You cannot put 2Mev of energy into a phonon or any finite number of phonons.

And then there is the fact that you cannot even get such a nuclear transition in the first place because of the laws of nuclear physics.

The simple underlying problem is that these guys are trying to explain a phenomenon that have never been consistently and repeatedly demonstrated with a claimed set of processes that violate every principle of physics known. These are, as another critic stated, extraordinary claims. Acceptance will require extraordinary proof. Such extraordinary, convincing proof has not been forthcoming.

Naysayers are allowed to remain skeptical until there is overwhelming evidence to the contrary.

41 posted on 07/02/2011 10:52:12 AM PDT by AndyJackson
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