>
> I believe as follows:
>
> The magnetic fields produced by SPP solitons catalyze nuclear reactions in
> matter that this field falls upon. The soliton is PUMPED by heat photons.
>
> The soliton produces two kinds of magnetic photons: real and virtual. The
> power of the anapole magnetic field is
I wrote:
it might be good to work backwards from known and plausible reactions;
> e.g., a proton being stripped off of a deuterium nucleus and hopping over
> to the lattice site.
>
Sorry, that should have been "a neutron being stripped off of a deuterium
nucleus," which would lead to the proton t
On Fri, Apr 10, 2015 at 8:04 AM, Bob Higgins
wrote:
This suggests that something nuclear is happening in the branch of the
> reaction that results in the ejection of the 6 MeV proton to supply the
> proton with its 6 MeV of energy.
>
The impression I've taken away from what I've read of Piantell
http://phys.org/news/2015-04-gold-atoms-metal.html
Small nanoparticles store energy and are magnetic, while big nanoparticles
do not store energy and are not magnetic.
In reply to Jones Beene's message of Fri, 10 Apr 2015 17:45:06 -0700:
Hi,
[snip]
>A sheet of paper makes no difference whatsoever. A sheet of aluminum foil
>doesn't make much difference but a nickel coin blocks most of this small
>sample. The Gamma Scout is picking up x-rays which are not stopped
A sheet of paper makes no difference whatsoever. A sheet of aluminum foil
doesn't make much difference but a nickel coin blocks most of this small
sample. The Gamma Scout is picking up x-rays which are not stopped by paper
(even though alpha particles themselves would be). However, to be fair - a
Some pitchblende contains radium how emits gammas.
On Sat, 11 Apr 2015 09:31:35 +1000, mix...@bigpond.com wrote:
> In reply to Jones Beene's message of Fri, 10 Apr 2015 12:48:35 -0700:
> Hi,
> [snip]
>>which is why even small pitchblende samples make the Geiger
>>counter go wild.
>
> Try puttin
In reply to Jones Beene's message of Fri, 10 Apr 2015 15:54:53 -0700:
Hi,
[snip]
>-Original Message-
>From: mix...@bigpond.com
>
>> He calls them "Hydrinohydride". The smallest is for p = 24. I.e. 24 times
>smaller than normal H-. For greater values of p (i.e. further shrunken), the
>seco
In reply to Jones Beene's message of Fri, 10 Apr 2015 13:53:13 -0700:
Hi,
[snip]
>If neutrons were involved there would be neutron activation, a widely known
>phenomenon - not seen at Lugano.
[snip]
Indeed, the reaction:
Al27 + Li7 => Al28 + Li6 + 0.475 MeV would produce radioactive Al28 with a
In reply to Jones Beene's message of Fri, 10 Apr 2015 12:48:35 -0700:
Hi,
[snip]
>which is why even small pitchblende samples make the Geiger
>counter go wild.
Try putting a sheet of paper between the Geiger counter and the pitchblende. I
think you will find that it makes a huge difference. Most
In reply to Bob Higgins's message of Fri, 10 Apr 2015 09:04:29 -0600:
Hi,
[snip]
>I cannot pretend to be a spokesman for Dr. Piantelli's theory. I have a
>couple of observations from this theory that I still cannot internally
>justify (from my own limited understanding of nuclear physics). The fi
-Original Message-
From: mix...@bigpond.com
> He calls them "Hydrinohydride". The smallest is for p = 24. I.e. 24 times
smaller than normal H-. For greater values of p (i.e. further shrunken), the
second electron is unbound, according to his formula, so there is no
Hydrinohydride for larg
In reply to Bob Higgins's message of Fri, 10 Apr 2015 08:45:31 -0600:
Hi,
[snip]
>Well, Piantelli may not be saying it is a hydrino because he doesn't look
>at it that way. He has black box evidence that the Ni and the H- anion
>nuclei coalesce producing a specific set of branched outcomes, one o
This is not what Cook and Rossi are now saying. They are claiming in this
so-called “mainstream physics” paper, that lithium + proton fusion to helium
accounts for the gain.
If neutrons were involved there would be neutron activation, a widely known
phenomenon - not seen at Lugano.
It l
The amount of nickel Ni62 in the fuel load doubled from some unknown
combination of lighter elements. This fusion process should have released a
huge amount of nuclear binding energy. It is possible that the only thing
that lithium did was donate its neutron to the Nickel 58 to turn it into
Nickel
-Original Message-
> The argument can be made that there was NEVER enough lithium present in
the Lugano reactor to provide the reported net energy gain (1.5 MW-hrs) over
32 hours- even if 100% of the lithium was consumed and converted into
helium.
For the record - The total Lugano Fuel sa
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