Ed--Bob Cook here
Spin states of a quantum system reflect the angular momentum of the system and
hence the energy associated with that angular momentum. High spin quantum
numbers reflect the higher energy of the system. The allowable states are
quantized. In magnetic fields the direction of the spin is controlled more or
less depending upon the field strength. The allowable number of states is
reduced from the situation where there is no magnetic field. Resonant magnetic
oscillating fields input to a nucleus with a magnetic moment and non-zero spin
state for its ground state, can add energy to the quantum system by changing
the spin number of the quantum system. This is the basis for the MRI
technology which is an accounting of the energy absorption at a given
resonance frequency at well determined locations, identifying the nucleus with
the specific resonance frequency absorption .
If there is spin coupling, (a basic assumption is that spin is conserved in
any nuclear reaction at the end of the reaction) a coupling between various
particles subject to integer, J, quantum seems probable. Thus, any He-4* with
a high spin integer J quantum number and excess energy--say 10 mev--would
distribute this high angular momentum to electrons or other particles in the
quantum system--all the many electrons and particles at the same time. The
electrons (and other particles) in turn would distribute their excess spin
energy (angular momentum) to the lattice as electromagnetic field oscillations
or radiation and hence lattice heat. In the end the net spin would be what it
was to start with. The reaction would be fast and cause results of the
distribution of quantum angular momentum and lattice motion instantaneously.
No energetic (kinetic energy) particles are involved, only angular momentum
with its corresponding rotational energy. The rotational energy may actually
be rotating electric and or magnetic fields associated with the particle with
the high spin quantum state.
Again I do not understand the details of spin coupling, the actual timing nor
the most likely fractionation of the spin/angular momentum among the particles
of the quantum system. The basic idea is that the energy associated with the
mass loss first shows up as angular momentum or spin of the newly found He-4*
and this spin is distributed to the rest of the system.
Bob Cook
----- Original Message -----
From: Edmund Storms
To: vortex-l@eskimo.com
Cc: Edmund Storms
Sent: Friday, February 07, 2014 2:34 PM
Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 7, 2014, at 1:42 PM, Bob Cook wrote:
Ed--Bob Cook here--
Another question is if D is formed in the Ni-H system as you propose, why
not the generation of He-4 as in the Pd system without the nasty fragmentation
or fission of the Ni?
That answer is too complicated to explain here. That is why the book is
required. Take my word for the present that I have a good reason for this
model.
The key to controlling the Rossi process maybe controlling the formation
of D. The energy transfer process would be coupled by spin and distribution of
angular momentum between the initially excited He-4* at a high spin state and
the electrons of the system and maybe the various Ni nuclei in the system.
Nuclear-magnetic spin of nuclei is of course coupled to electromagnetic
irradiation signals in MRI technology. The math must be well established.
These fragmentation products release about 11 MeV/fragment. Please tell me
how spin state coupling can transfer this amount of energy. Even adding a p to
Ni requires about 6 MeV be dissipated. I know of no example of this much energy
being transferred by any kind of coupling mechanism. Do you?
Ed Storms
Again spin state coupling with appropriate energy transfer should be
explored in theory--this is above my head. Do you know if anyone has looked
at this?
Bob
----- Original Message -----
From: Edmund Storms
To: vortex-l@eskimo.com
Cc: Edmund Storms
Sent: Friday, February 07, 2014 10:32 AM
Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems
Bob and Eric, the issue of transmutation is basic to understanding LENR.
First of all, transmutation has a very high barrier requiring an explanation of
how this can be overcome. Second, the resulting energy has to be dissipated in
ways known to be possible. I propose the hydrogen fusion process provides the
required energy and dissipates much of the excess mass-energy. In other words,
transmutation can not occur unless fusion is taking place at the same time and
place in the material.
We now know that two kinds of transmutation occur. Iwamura shows that D
can be added to a target resulting in a stable heavier product. Most other
claims for transmutation are based on fragments of Pd being found. Explaining
these two different results is the challenge.
In the case of Ni+H, I propose the p-e-p fusion process deposits the
resulting d in the Ni nucleus, resulting in fragmentation of the product in
order to dissipate the excess mass-energy. I believe 2d enter all isotopes of
Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained
from the p-e-p reaction is added to any energy resulting from occasional
transmutation. When the Ni fissions, it must conserve n and p, which produces a
distribution of products that can be calculated. This calculation shows a
distribution that is consistent with what is reported and reveals Ni-58 to be
the most active isotope for energy production. I will provide much more detail
and justification in my book. Meanwhile, you might consider this proposed
process.
I propose transmutation takes place in the Rossi cell, but he has
incorrectly identified its source and incorrectly attributed the energy to
transmutation. I propose most energy results from p-e-p=d fusion, with
transmutation resulting from fission of Ni adding only a minor amount of
energy. If this is the case, focus on Ni is a waste of time.
Ed Storms
On Feb 7, 2014, at 10:39 AM, Bob Cook wrote:
Eric--
Your bring up some interesting questions about the Rossi reactor. The
information I have included come from Rossi and Focardi's international patent
application noted below.
1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor?
This would be expensive. The natural isotopic abundances are:
Ni-58, 68.08%;
Ni-59, 0%--its radioactive with 1/2 life of 80,000 years;
Ni 60, 26.22%;
Ni-61, 1.14%;
Ni-62, 3.63%;
Ni-63, 0%--its radioactive with 1/2 life of 92 years;
Ni-64, 0.93%.
I would pick Ni-60 because it is more than one transmutation (Ni-proton
fusion) away from a radioactive residue.
2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors?
Rossi and Focardi seem to contradict themselves with the statements
below:
"...we believe that form of energy involved is nuclear, and
more specifically, due to fusion processes between protons and Nickel nuclei.
They are exothermic with an energy release in the range 3-7,5
MeV, depending on the Nickel isotope involved."
"No radioactivity has been found also in the Nickel residual
from the process."
This information attributed to Focardi and Rossi comes from their
instructive statements, which suggest the nuclear Ni-proton fusion, in the
following paper:
A new energy source from nuclear fusion
S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna University
and INFN Bologna Section, (2)Leonardo Corp. (USA) - Inventor of the
Patent, March 22, 2010 (international patent publication N. WO 2009/125444 A1)
My final observation is that the Rossi-Focardi comment that there is no
radioactivity in the residue needs to be checked. Other Ni-hydrogen materials
that have been produced by other experimenters should be carefully checked for
both the potential radioactive Ni isotopes---Ni-59 and Ni-63. They should be
easy to detect given their well known decay modes and probable gamma emissions.
(I will look up this information and put it in a subsequent comment.) I know
that both Ni-59 and Ni-63 are problems when it comes to nuclear waste disposal
of activated metals.) A null radioactivity essay would be revealing as to the
process actually occurring in the Ni-hydrogen reactions.
Bob
----- Original Message -----
From: Eric Walker
To: vortex-l@eskimo.com
Sent: Thursday, February 06, 2014 7:45 PM
Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Thu, Feb 6, 2014 at 2:26 PM, Bob Cook <frobertc...@hotmail.com>
wrote:
Also I suspect that the nano Ni that is produced is pretty pure.
That may be why Rossi uses it and may be the reason other researchers do not
have very good luck at getting a good reaction.
I'm guessing that the purity of Rossi's nickel (in terms of 62Ni and
64Ni) is related to avoiding beta-plus and beta-minus decay, and, with
beta-plus decay, the 511 keV positron-electron annihilation photons.
Some vorts may enjoy this video of a small cloud chamber [1]. It's
remarkable that such a small event can have macroscopic effects.
Eric
[1] http://www.youtube.com/watch?v=xQVMrkJYShc
