On Oct 11, 2009, at 6:31 PM, [email protected] wrote:
In reply to Horace Heffner's message of Fri, 9 Oct 2009 04:26:08
-0800:
Hi Horace,
[snip]
It simply is very difficult to establish with high sigma results
whether energy is produced at 23.8 MeV or 18.8 MeV per He4, or
something in between.
In the case of the fusion reaction:
D + T -> He4 (3.5 MeV) + n (14.1 MeV)
it is much easier to determine if the energy spectrum of the
neutrons, which result from every such reaction, is consistent with a
mean of 14.1 MeV.
Even if it isn't, there are possibilities other than a ZPE
interaction that
might explain the deviation. e.g. some of the neutrons could come
from D-D
reactions according to the usual reactions (since D must be present
in a D-T
experiment).
I don't think so, not unless even more energy is given up to the
vacuum transactions, at the moment of branch selection, than one
might expect. The hot fusion D + D reaction produces 2.45 MeV
neutrons. An ordinary fusion reaction produces 14.1 MeV neutrons.
If about 5 MeV or so is given up to vacuum transactions that still
leaves 9.1 MeV which is about what is needed to pull off the triple
alpha track trace observed by SPAWAR.
Furthermore, even in the D-T reaction, if/when the electron is
expelled along
with the neutron,
Under the deflation fusion scenario the electron can barely make it
to ground state. Initially id doesn't have enough energy to get
there. That is why it is trapped.
one might logically expect all three particles to share the
energy resulting in a broad spectrum of neutron energies. (Three
particles
because the He4 also counts as one). Qua energy distribution, this
should result
in a spectrum analogous to that of beta-decay.
No readily observable beta decay is likely under the proposed
deflation fusion scenario. Of course the scenario could be wrong, but
that is the already stated case.
I think there is in fact some indication that the
neutrons resulting from cold fusion are considerably less than 14.1
MeV on average. This is because only a few percent are above 9.4
MeV. It remains for the spectrum to be nailed down. I think this is
the most important experimental work at hand.
There is also the theoretical question of whether the electron is
capable of radiating away the energy released by the strong force
when the helium nucleus is formed.
The electron doesn't need to radiate the energy away. It can simply
absorb it as
kinetic energy, and be expelled from the nucleus at near light speed.
This clearly doesn't happen. Therefore any model which suggests it
does is wrong. The deflation fusion model does not predict this.
This then
leaves the nucleus itself behind as an ion, which later reacquires
an electron
from the environment to become a neutral atom. The energetic
electron loses
energy to the environment by ionizing other atoms.
This is some other model, not mine.
It strikes me as likely in
ordinary D-D hot fusion that an energetic helium nucleus is
momentarily formed prior to the release of either a proton or
neutron. That means the nucleus momentarily contains the full 23.9
MeV of a D-D fusion heat, of strong force binding of the deuterons.
Agreed.
The breaking of strong force bonds, fission of the He4*, to release
an n or p then saps away some of that heat. In the case of D(D,p)T
the energy required to fission off a proton is 23.9 MeV - 4.03 MeV =
19.87 MeV. In the case of the D(D,n)He3 reaction the fission energy
to produce the neutron is 23.9 MeV - 3.27 MeV = 20.63 MeV. This
results in the classic hot fusion energies:
D(D,p)T 4.03 MeV
D(D,n)He3 3.27 MeV
D(D,gamma)He4 23.9 MeV
Agreed.
Now, if a small electron is present in the fused nucleus, the nucleus
kinetic energy is reduced a priori,
Why?
This was explained on pages 9-11 of:
http://www.mtaonline.net/%7Ehheffner/DeflationFusion2.pdf
but apparently I did a very bad job of it because it seems no one
thinks it is credible enough to ask question like you are.
Let's start with the idea of 3 point charges, two deutrium nucleii
(+) and and electron (-), all in
a line in the x axis separated by (an initial) distance d1 = d2 = of
10^-11 m:
(+)........d1.......(-)........d2.......(+)
The force between the left deuterium nucleus and the electron and is
given by:
F1 = q^2/(4 Pi e0 r^2) = 8.98 N
and is to the right towards the electron. The force the two deuterium
nucleii is repulsive and is 1/4 the magnitude of the force between
the deuterium and the electron because the distance is doubled, i.e.
d1 + d2 = 2 d1. So the net force on the left deuteron is 3/4 * 8.98 N
= 6.74 N and is to the right. Similarly, the net force on the right
deuteron is 6.74 N and is to the left. The net force on the electron
balances out to zero.
Suppose now that, on an instantaneous basis, just after tunneling,
these particles are for all practical purposes point particles,
string size. The tunneling event is unlikely unless the final wave-
function collapse is produced in a lower energy configuration. Given
this, we can see that (1) the tunneling event can not occur unless
the catalytic electron is located between the two deuterons. If it is
located to the opposing side of one of the deuterons then a massive
repulsion exists preventing the tunneling. Note that the smaller D1
and D2 are the higher the Coulomb binding energy that results from
the tunneling. However, that does not necessarily mean that kinetic
energy was gained from the tunneling event. No acceleration was
involved in the tunneling event. The wavefunctions overlapped at a
point, and the wavefunction collapse transports the particles to the
same point. The particles don't really move per se, they were
already there with some probability. That is not to say, however,
that vacuum transactions can not occur, as a result, that in the end
produce neutrinos or weak reactions, but that is another subject.
Let's look at this another way, closer that in the paper. A deflated
state hydrogen exists as the degenerate state of a deuterium nucleus,
shared with a partial orbital state in a tetrahedral site. In this
case the Hamiltonian remains unchanged. The electron gains the
kinetic energy in the close up state to match the lost potential
energy. It can therefore hop into and out of this state rapidly.
Then a wavefunction collapse occurs in which a deuteron from an
adjacent site tunnels into the locus of a deflated state deuteron.
Again this cannot happen unless the electron is located in an
advantageous position, because it would then be energetically
denied. A similar situation arises. The nucleus now has two
positive charges and the electron can no longer escape - except
possibly through energy released by the strong force. The question
now is, just how large a deficit can the wavefunction collapse
produce, by instantly placing an additional nucleus near the electron?
From the electric potential energy Pe for separating an electron
from two newly fused deuterons at radius r we have:
Pe = k (2q)(-q)(1/r) = (-2.88x10-9 eV m) (1/r)
which we can rearrange to obtain r for a given potential energy,
r = (-2.88x10-9 eV m) (1/Pe)
and we have for -23.9 MeV:
r = (-2.88x10-9 eV m) (1/(-23.9x106 eV))
r = 1.2x10-16 m
which is about 10 times the diameter of a quark, and approximately
that of the deflated hydrogen state, and thus in the realm of
credibility. The entire energy of fusion can be momentarily
contained if the wavefunction collapse is to a small enough size, and
in fact for point sized particles (or string sized particles?) a lot
more energy can be absorbed.
Now, given the electron was energy neutral at this size, about half
of the 23.9 MeV is in electron kinetic energy at this point, leaving
an energy deficit of about 11.9 MeV. The total amount of energy to
pull off the fission of an He* at this point is 11.9 MeV. This is
way less than the D(D,p)T energy of 19.87 MeV or the the D(D,n)He3
energy of 20.63 MeV required for the fissions to create those
respective reactions.
the nuclear temperature is
reduced. The energy of fusion does not have to be radiated away. No
time is required for photon creation, or for that matter neutrino
creation or any other weak reaction, prior to the He4* nucleus
fission. The heat is simply not there to begin with.
The energy of the fusion reaction due to a change in mass has to go
somewhere.
I goes to make up the deficit.
If it isn't in the He4* nucleus, then the electron has to have it
as kinetic
energy, which is what I have been saying for some time now.
The electron is stuck in the hole until the vacuum (uncertainty
energy if you prefer) allows it to regain orbital status.
The amount of
energy so removed from the He* depends on the wavelength of the
electron at the moment of wavefunction collapse, so is a stochastic
variable.
It depends on the momentum distribution between He4, electron, and
neutron.
Since there are three particles, a broad spectrum is to be expected.
There are only two of those particles in the typical D-D cold fusion
scenario, the electron and the He4*.
For DD fusion this isn't so. If Deflation Fusion is a valid
mechanism, then the
electron should always get the overwhelming majority, since the
energy (and
momentum) is only distributed over two particles, the He4 and the
electron.
Even given the electron has *all* the kinetic energy initially, it
does not have enough to escape. It is only in the D-T reaction that
the neutron is likely to have enough energy to escape, and its
spectrum can tell what the size of the initial wavefunction collapse
product is. The total enthalpy of the reaction can not, because the
electron wavefunction expansion process can tap nuclear heat and also
radiate. Some or all of the energy lost in the collapse can be
returned.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
