On Apr 12, 2011, at 11:23 PM, [email protected] wrote:
In reply to Horace Heffner's message of Tue, 12 Apr 2011 21:05:45
-0800:
Hi,
[snip]
The feasibility of such states indicates that
when QM computation means are available, they will likely show such
states to be Rydberg like, with quantum fuzziness unimportant to the
state.
Gee Horace, this sounds a lot like my model. :)
(See http://rvanspaa.freehostia.com/theory-paper.html , in
particular the
section on HUP).
Actually, it is not. You assume quantum states in which the de
Broglie wave wraps around the orbit. I assumed the orbit was small
enough and the kinetic energy high enough, due to the additional
force of magnetic binding, that the radii of the nucleus and and
electron are less than their de Broglie wavelengths - thus the
quantum nature to the orbital is removed, as it is with electrons in
the distant portion of a Rydberg orbital. I provided this
approximation of the deuterium deflated state in 2007:
http://www.mtaonline.net/~hheffner/FusionSpreadDualRel.pdf
but I doubt anyone actually has actually looked at it.
ENERGY TRAPPING
Assume the kinetic energy of a specific electron in the deflated
state is momentarily around 1 MeV, thus the potential energy in that
case is about -1 MeV. Upon fusion with Ni, the kinetic energy in that
case remains initially at 1 MeV, but, due to the suddenly present 28
extra Ni nucleus charges, the potential energy is reduced to -29 Mev,
and the net potential plus kinetic energy is reduced by 28 MeV. The
potential energy gained by the proton approach to the nucleus is
actually present, but for exogenous thermodynamic purposes lost due
to the interaction of the strong force. This loss of potential
energy does not prevent electron capture of the now energetically
trapped electron, if capture occurs very fast, because that electron
initially has the kinetic energy necessary.
It isn't clear why kinetic energy is "necessary".
Uh ... why do you say "necessary". It *is* there. As the potential
energy of the system, the field energy, is lost, the kinetic energy +
relativistic mass of the electron (and hydrogen nucleus, because in
relativistic orbitals they rotate about each other) makes up for it.
When a proton or deuteron tunnels, the ensemble of particles it
represents does not change in nature. It remains a proton or
deuteron, which entails particles and their kinetic energies. I
simply am saying that when an electron-nucleus pair tunnels, it also
retains its nature, including its breakdown of kinetic and potential
energy.
Is it necessary that the
energy be kinetic in nature, or just that a specific quantity of
energy be
available, and if the latter then how much?
The precise "how much" at all times is a stochastic variable, becuase
it depends on r, which is itself a stochastic variable. The
energetics of the deflation fusion process is described on pp 3-4 of
http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf
The ramifications of the stochastic nature of the process are
described on pp 6-11.
The amount of kinetic energy matches the loss in potential energy.
The kinetic energy depends on the radius at which the electron is
found, and the charge of the nucleus it is in. When the electron is
inside the nucleus the electron kinetic energy is quickly thermalized
and results in photon radiation.
You can look the energetics another way. When, in hot fusion, a
naked proton is forced into a Ni nucleus to the radius where 28 MeV
(for example) is released upon its departure if fissioned, then that
energy was in the form of kinetic energy a priori. Once the proton
binds to the nucleus by the strong force, that a priori kinetic
energy is gone - it is now in the form of potential energy, the field
energy of the new nucleus. It is locked in, no longer available,
lost, in either hot or cold fusion.
This is only part of the story when an electron-hydrogen-nucleus
pair, be the hydrogen nucleus protium or deuterium, enters the
nucleus. No a priori kinetic energy is required, because their net
charge is zero. The nucleus itself is not much different from the
hot state nucleus. It has the same greatly increased potential
energy supplied to its field energy by the added proton in ordinary
hot fusion, plus it has the kinetic energy supplied to it that
existed previously in the proton in the deflated state hydrogen. The
difference between hot and cold fusion is in cold fusion no huge
energy has to be supplied to the proton initially, but the resultant
nucleus has an electron trapped with a hugely negative potential
energy, and a kinetic energy insufficient to escape. In fact, the
total energy from the fusion, plus the kinetic energy existing a
priori in the deflated state, is not sufficient to escape. The net
energy of the new compound system is often negative. I show this
balance of energy, the fusion energy less the lost energy energy of
the electron, in brackets in all the equations provided in these
reports:
http://www.mtaonline.net/~hheffner/dfRpt
with exceptions noted, the radius used for the calculations is the
radius of the new compound nucleus. I think in practice that mean
radius should probably be the mean location of a thermalized electron
in the nucleus.
I am left wondering (literally) where the energy comes from during
normal beta
decay for the electron to escape the nucleus. The only conclusion I
can come to
it that conversion of a neutron to a proton results in 28 MeV more,
The amount is stochastic, it isn't a fixed "28 MeV".
of nuclear
binding energy, than we usually calculate as the beta decay energy,
which
additional 28 MeV is then carried away by the electron, and mostly
lost during
it's uphill battle to escape the Coulomb attraction of the nucleus.
This in turn
would imply that adding a "fresh" proton to the nucleus adds an
additional 28
MeV of binding energy, and since the proton didn't need to overcome
the Coulomb
barrier on the way in, that additional 28 MeV is available to the
electron to
escape.
(Sorry, I just don't buy the ZPE extraction part of your theory).
Well, then, you can't buy the trapping part either - so you are left
having to explain why heavy element LENR provides neither the
enthalpy nor the high energy signatures that would be expected, and
why branching ratios change for cold fusion D+D-->He, etc, etc., etc.
You can no longer explain LENR. It is an experimental fact that
stable nuclei have heat. See: E. Melby et al, “Observation of
Thermodynamical Properties in the 162Dy, 166Er, and 172Yb Nuclei ,”
Phys. Rev. Lett. 83/16 (October 1999): 3150 - p. 3153, the basis for
my theory in:
http://mtaonline.net/~hheffner/NuclearZPEtapping.pdf
NUCLEAR HEAT
I think it is also true that nuclear heat may prolong the ability of
the electron to both radiate and be captured. The nuclear heat to
which I refer is provided by zero point energy, in a manner as
described here:
http://mtaonline.net/~hheffner/NuclearZPEtapping.pdf
in which I estimate the nuclear temperature for Ni to be 1.02 MeV.
The energetically trapped electron thereby provides a method of
capturing zero point energy.
UNCERTAINTY PRESSURE
The electron only expands its orbital outside the nucleus if a weak
reaction does not quickly follow the strong reaction. This orbital
expansion is driven by zero point energy, uncertainty energy. The
proximity of the electron to the deflated state hydrogen nucleus, and
its kinetic energy, prior to tunneling into a heavy nucleus, are for
practical purposes random variables.
However, I think you can impose reasonable limits on the distance
between proton
and electron in the deflated state. If the distance is too great,
then the
electron won't shield the proton adequately, and tunneling won't
happen in a
reasonable time (i.e. a short enough time to be useful for LENR).
Just as everyone else does, you keep projecting your own mental
models and biases onto parts of my theory, in effect changing the
theory. You have to accept all the givens as hypothetically proposed
and then compare the predicted consequences to experimental results
to make any sense of my theory. You can't shoehorn it into your
perception of reality and expect the same results. You have to accept
the Gestalt to make any sense.
Tunneling in the case of the deflation fusion model is considered to
be precipitated by wavefunction collapse. The wavefunction of the
deflated state ensemble is already "there" with some probability.
Wavefunction collapse thus takes zero time. The wavefunction
collapse merely amounts to an "observation". What is limited by c is
the adjustment to the EM field required by the wavefunction collapse.
You can probably use the ordinary QM equation
No. Such an analytic capability does not yet exist, due to the
relativistic requirements, and the requirement that the nucleus
motion is significant. The closest thing to it was an analysis of a
hydrino state done by Naudt, wherein the nucleus was considered
fixed, but even that is controversial at this point.
of the H atom to determine the
percentage of time that the electron is within an adequately small
radius
(though this is a bit deceptive as tunneling is possible at all
distances, so
the net probability requires integration of the probabilities at
each distance).
This kind of approach yields a too small probability of the state.
The inspiration for my theory was the first of the following related
articles:
“A Water Molecule's Chemical Formula is Really Not H2O”,Physics News
Update,
Number 648 #1, July 31, 2003 by Phil Schewe, James Riordon, and Ben
Stein,
http://www.aip.org/enews/physnews/2003/split/648-1.html
C. A. Chatzidimitriou-Dreismann et al,“Anomalous Deep Inelastic Neutron
Scattering from Liquid H2O-D2O: Evidence of Nuclear Quantum
Entanglement”,Phys. Rev. Lett. 79, 2839 - 2842 (1997),
http://prola.aps.org/abstract/PRL/v79/i15/p2839_1
The fact that, when H2O is looked at fast enough, it is H1.5O,
meaning 1/4 of the time a proton "disappears", is just too much
linked to cold fusion to ignore! Even though the disappearance is
only attoseconds in duration, it must be repeated frequently i a
water molecule to have such a probability.
The resulting associated values
post tunneling are thus also random variables. The energy balance for
individual LENR reactions are therefore also random variables.
Energy does not appear to be conserved, because vacuum energy
transactions are involved. Time of electron near the composite fused
nucleus is a random variable, and one which, along with the other
random variables, affects the branching ratios.
It seems clear that (1) the electron which catalyzes the hydrogen
entry to the nucleus is necessarily trapped there in many cases,
because there is not enough energy from the strong force reaction to
expel it, (2) some heavy LENR reactions clearly do not involve weak
reactions, and (3) the trapped electron can not remain in the
nucleus, otherwise it would be indistinguishable chemically from the
product of a follow-on weak reaction, and the mass of the nucleus
would remain about 0.8 MeV too high.
The mass wouldn't be 0.8 MeV too high, because that mass was lost
to the
creation of the kinetic energy in the first place. Note your own
words:
"The deflated state is a degenerate state of the hydrogen within its
molecular or lattice environment, and has the near ground state
binding energy of that state."
...IOW there is no extra 0.8 MeV. (The combined *rest* mass of the
electron and
proton in the deflated state has to be 0.8 MeV less than that of an
H atom,
because 0.8 MeV of rest mass has been converted into kinetic energy
of the
electron, leaving the total relativistic mass constant and equal to
the rest
mass of the Hydrogen atom - which BTW is still 782 keV shy of the
energy
required to create a free neutron - which energy deficit is
precisely why free
neutrons decay into protons and electrons).
This is your assumption based on your model of reality. My
assumption is the kinetic energy and relativistic mass of the
deflated state comes from the field energy resident in the vacuum.
The state normally only exists for attosecond time frames. However,
if it is translated into a nucleus, the energetics of the ensemble
remain initially unchanged.
Experimental results thus
indicate the trapped electron has an exit. The energy for that exit
clearly can be provided by zero point energy, "uncertainty pressure",
if you will, expanding the electron wavefunction, and its de Broglie
wavelength.
[snip]
Regards,
Robin van Spaandonk
http://rvanspaa.freehostia.com/Project.html
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
