On Apr 14, 2011, at 6:35 AM, Jones Beene wrote:
Horace
I wrote:
Let’s look at 58Ni specifically which is over 2/3 of all nickel
The energy deficits for Ni are all huge. For example (energy
deficits in square brackets):
58Ni28 + p* --> 59Cu29 * + 3.419 MeV [-6.329 MeV] --> 59Ni28 +
neutrino + ~2.6 MeV
Jones writes:
Ok, as I interpret your theory, part of the large 6.3 MeV “deficit”
could conceivably function as ‘makeup’ for zero point energy
already removed from the Rossi device by another mechanism, but
let’s not go into that other mechanism for now. This is the part I
like, even if you do not interpret it this way.
My theory is exactly the opposite. The 6.3 MeV deficit that results
from the neutral entity tunneling is made up in part (but not in each
case precisely, due to the stochastic nature of the process) by
energy from the zero point field. If and when the Coulombic
potential energy of the new nucleus is released by fissioning, that
Coulombic potential energy is recovered, and a net energy will have
been produced from the vacuum. This is why I say the overall process
is not energy conservative.
However, there is a problem with converting a deflated proton into
a neutron without a neutrino,
I never implied to my knowledge that no neutrino would be produced in
the electron capture process.
even with an energy deficit.
Yes, there is an overall energy deficit, but the electron initially
retains its kinetic energy, which can be used for a weak reaction.
Not only that but, and this is an aside comment, the temperature of
the Ni nucleus is maintained at about 1 MeV by the zero point field,
so sufficient kinetic energy remains available to the electron while
it remains within the compound nucleus. The probability of a weak
reaction for an electron that remains in a nucleus is of course
vastly greater than that for orbital electrons which only
occasionally, and with very brief duration, transit the nucleus, i.e.
in a QM interpretation have a very small probability of being
observed in the nucleus.
It is almost like saying that part of this deficit takes the place
of the missing neutrino, and it is the same neutrino that shows up
on the other side of the equation, so it cannot be ‘borrowed’ in
the QM sense since the arrow of time goes the other way.
Again, I explicitly say a neutrino is produced. (Actually, due to a
consistent typo when I replaced v in my paper with the neutrino
symbol, I accidentally replaced all of them with an erroneous anti-
neutrino symbol. It was corrected in my article here I think:
http://www.journal-of-nuclear-physics.com/?p=179). I could really
use a good editor!
When a weak reaction follows the strong reaction in deflation fusion
there is very little energy released. There are no orbital x-rays or
auger electrons, the radiating time of the electron in the nucleus is
cut short, its kinetic energy is used in the formation of the
neutron, and carried off by the neutrino. This lack of high energy
signatures and significant enthalpy is highly characteristic of LENR
transmutation reactions. This probably accounts for a lack of
research into this area, because it is of little use in providing
energy, and because the effects can not even be detected unless they
are specifically looked for in the "ash". The enthalpy producing
reactions in Rossi's case are only:
62Ni28 + p* --> 63Cu29 + 6.122 MeV [-3.415 MeV]
64Ni28 + p* --> 65Cu29 + 7.453 MeV [-1.985 MeV]
Anyway, even if we can get past that one, the next problem resolves
to the 59Ni and that large amount of ‘real’ energy 2.6 MeV. Even if
most of the energy were carried away by the neutrino, most of the
time – in practice there is always secondary gamma or
bremsstrahlung from weak force reactions, which should have shown up.
You refer here to ordinary nuclear reactions, which are irrelevant.
Is there an example in nature of a radiation-free weak force reaction?
Yes, heavy element LENR. This fact is one of the strong validating
points for my theory. Among others stated in my article: "There is
thus (1) no need to explain how a sub-ground state hydrogen is formed
in a lattice, (2) no need for large sub-ground state binding energies
to overcome the Coulomb barrier, (3) no need to explain how neutrons
in the lattice are generated en mass and yet not readily detectable
directly or through neutron activation of materials in the lattice,
or (4) to explain how the high energy barriers of high mass lattice
elements are also defeated. There is also (5) no further need to
explain why there is a lack of high energy gamma signatures or to (6)
explain how MeV magnitudes of reaction energy is carried off by
phonons or through simultaneous action of large bodies of lattice
atoms, or why (7) large numbers of high kinetic energy particles are
not detected or why (8) readily observable high energy bremsstrahlung
are not seen." There are various others.
And even if there is one which can be tailored for this, the third
problem is the 59Ni remaining in the ash. This isotope is commonly
used in medicine IIRC, with a well-known Auger emission cascade on
EC which Levi would have immediately recognized.
AFAIK, 59Ni has a 76000 y half life. This is only a few
disintegrations per minute per gram (I have not taken the time to do
the calculation here, but feel sure this is accurate enough for
discussion purposes). This is detectable with careful looking, but
not of practical significance with regard to energetics or harmful
effects. I expect a banana is more radioactive per gram due to
potassium. How is 59Ni used in medicine?
This is the most problematic of all, given Rossi’s lack of
radioactivity in the ash.
I did suggest a reaction which does not produce 59Ni:
58Ni28 + p* --> 59Cu29 * + 3.419 MeV [-6.329 MeV] --> 59Ni28 +
neutrino + ~2.6 MeV
58Ni28 + 2 p* --> 60Zn30 * + 8.538 MeV [-11.541 MeV] --> 60Ni28 + 2
neutrinos + ~7 MeV
60Ni28 + p* --> 61Cu29 * + 4.801 MeV [-4.840 MeV] --> 61Ni28 +
neutrino + ~4.0 MeV
61Ni28 + p* --> 62Cu29 * + 5.866 MeV [-3.722 MeV] --> 62Ni28 +
neutrino + ~5.1 MeV
62Ni28 + p* --> 63Cu29 + 6.122 MeV [-3.415 MeV]
64Ni28 + p* --> 65Cu29 + 7.453 MeV [-1.985 MeV]
How much 59Ni that might be expected to remain post run depends on
the probable reaction, the lattice half-life of 59Ni, and the
duration of the run.
I think this boils down to Rossi's credibility and tendency to make
misstatements. If you put a geiger counter up to a few gram sample
of pure 59Ni there will be no significant radiation. From Rossi's
viewpoint this is good, and an adequate representation. If competent
scientists want to look for a Ni59 signature, they will find it if
Ni59 is present, but it might involve a long counting time to get a
high sigma for a small sample.
I would indeed expect 59Ni to be present if there is significant
copper. I don't know what to expect in the way of accuracy of data
or statements with regards to the Rossi experiments, or even have any
way of being sure they are not fraudulent. I merely have stated my
theory provides some explanation of the primary results. I have also
posted here methods my theory predict should greatly enhance LENR in
the regime in which Rossi is operating, the Ni-H regime.
http://www.mail-archive.com/vortex-l@eskimo.com/msg44662.html
This I consider to possibly be a very significant gift. I think it
curious that the only response has been to criticize the wrapping and
ignore the gift.
What am I missing to tie up these lose ends ?
Jones
BTW – did you ever have a look at the Nyman paper ?
http://dipole.se/
Go down to “Strong Force between Two Protons”. I think it has
relevance to deflated protons in a reaction that does not involve
nickel. Simulations made with two different kinds of physics
software both show the following:
1) Two protons placed closely together [IRH] will repel each other
most of the time.
2) Two protons shot at each other will repel each other most of the
time.
3) However, it is occasionally possible to shoot protons at each
other with the right speed and quark positions so that they latch
on to each other - held in place for an indeterminate time by the
Strong Force.
Added to Nyman’s work is this:
4) The two protons have negative binding energy, so many things
could happen.
5) This is where the ‘quark soup’ metaphor may come into play
At any rate – everyone can probably guess that what I am struggling
with is to find any possible nuclear reaction of protons,
especially a deficit energy reaction of deflated protons, that can
never result in gammas, but can operate to level a zero point field
imbalance.
This probably means the ash must be “dark matter” of some type.
I don't know why people (and other people do) think I suggest p-p or
p-e-p reactions have any significant rate of occurrence by my
theory. The strong force reaction post tunneling is key, and is also
highly improbable in the case of two protons. I explicitly state on
page 12: "There have been indications of the possibility of proton
cold fusion reactions in various experiments.35 36 It is suggested
here that neither the conventional hot p-p nor the conventional hot p-
e-p reactions could be expected to have reaction rates that explain
LENR excess heat, because they are weak reactions and have clear
signatures. It is expected strong force mediated lattice element x
transmutations of the form p-e-x to be many orders of magnitude more
probable, and that such transmutations may produce far less excess
heat than the nuclear reactions and mass loss would normally indicate."
I expect p-e-p reactions to be more probable by deflation fusion
means than in high energy situations, like the sun, but still of no
practical or maybe even observable probability in the lattice.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
