If you have H+ and H-, wouldn't they have an attraction force between
themselves?  Sounds like Lochons.

Lochon Catalyzed D-D Fusion in Deuterated Palladium in the Solid State
K. P. Sinha <http://arxiv.org/find/physics/1/au:+Sinha_K/0/1/0/all/0/1>, A.
Meulenberg <http://arxiv.org/find/physics/1/au:+Meulenberg_A/0/1/0/all/0/1>
(Submitted on 4 May 2007)

Lochons (local charged bosons or local electron pairs) can form on D+ to
give D- (bosonic ions) in Palladium Deuteride in the solid state. Such
entities will occur at special sites or in linear channel owing to strong
electron-phonon interaction or due to potential inversion on metallic
electrodes. These lochons can catalyze D- - D+ fusion as a consequence of
internal conversion leading to the formation of He-4 plus production of
energy (Q=23.8 MeV) which is carried by the alpha particle and the ejected
electron-pair. The reaction rate for this fusion process is calculated.

  Comments: 3 pages: see also arXiv:cond-mat/0603213
<http://arxiv.org/abs/cond-mat/0603213> (Current Science, Vol. 91, No. 7,
pp. 907-912, 10/10/06) Accepted for publication: National Academy of
Science (India) Letters  Subjects: General Physics (physics.gen-ph)
Journal reference:
National Academy of Science (India) Letters, Vol.30, No. 7&8, 2007  Cite as:
arXiv:0705.0595 <http://arxiv.org/abs/0705.0595> [physics.gen-ph]    (or
arXiv:0705.0595v1 <http://arxiv.org/abs/0705.0595v1> [physics.gen-ph] for
this version)
 Submission history From: Andrew Meulenberg Jr. [view email
<https://arxiv.org/show-email/b63e0b65/0705.0595>]
*[v1]* Fri, 4 May 2007 11:02:23 GMT (123kb)



On Tue, Aug 5, 2014 at 8:23 PM, Kevin O'Malley <[email protected]> wrote:

>
>
>
> On Tue, Aug 5, 2014 at 8:16 PM, Axil Axil <[email protected]> wrote:
>
>> Regarding the excerpt from the patent.
>>
>> *More in detail, during the process of orbital capture, H- ions can lose
>> its own couple of electrons and form protons 1H+. A first fraction of the
>> protons 1H+ is subjected to direct nuclear capture reactions by the nuclei
>> of the same atoms of the clusters in which the orbital capture has
>> occurred, while a second fraction of the protons 1H+ can be expelled by
>> Coulomb repulsion from the nucleus of the metal atom where the orbital
>> capture has taken place. The expelled protons have an energy that can be
>> determined and characterised. For instance, in the case of Nickel, this
>> energy is about 6.7 MeV, as detected by a Wilson chamber, on the basis of
>> Bethe's equation. A part of the protons of the second portion, which does
>> not react with other nuclei of the primary material, can leave the latter
>> and interact with a material adapted to give rise to proton- dependent
>> reactions, if this is present. *
>>
>> A cooper pair of protons enters the transition metal nucleus, one
>> is captured an the other is expelled carrying 6.7 MeV which is the excess
>> binding energy.
>>
>> The proton pair has a spin of zero. which is consistent with magnetic
>> interaction with protons. The same magnetic glue has just be found to
>> produce cooper pairs of electrons in superconductors is found in Ni/H. The
>> Magnetic field in the Ni/H reaction is far stronger than can be found in
>> superconductors so there is no high temperature falloff.
>>
>> I would strongly suspect that the expelled proton is not seen when the
>> reactor is in operation. Such an experimental  detail should not appear in
>> the patent because it is not part of the reaction mechanism of an
>> operational reactor. The 6.7 MeV would be thermalized by BEC formation in
>> the operational reactor, IMHO.
>>
>> To see the 6.7 MeV, the nickel must be removed from the influence of the
>> BEC and placed in a cloud chamber after a considerable amount of time.
>>
>>
>> On Tue, Aug 5, 2014 at 7:11 PM, Ron Kita <[email protected]> wrote:
>>
>>> Greetings All,
>>>
>>> I just saw this on Alain Coetmeur s Scoop.it website.
>>>
>>> Not sure IF it made it to Vortex.....yet:
>>> http://www.scoop.it/t/lenr-revolution-in-process-cold-fusion
>>>
>>> Ad astra,
>>> Ron
>>>
>>>
>>>
>>
>

Reply via email to