On Oct 11, 2009, at 6:54 PM, [email protected] wrote:
In reply to Horace Heffner's message of Fri, 9 Oct 2009 10:55:10
-0800:
Hi Horace,
I'm afraid I can't make any sense of this at all. Perhaps a
specific example
with calculations would make it clearer?
[snip]
I forgot to note another very strong indication that the ability of
the electron to radiate photonic energy is not the primary reason for
the change in branching ratios, but rather the deflated energy of the
initial result of the wavefunction collapse. In other words there is
another indication there is a vacuum exchange of energy upon fusion
resulting in an apparent reduced Q of the cold fusion reactions. That
indication is the nearly complete lack of evidence of any excess heat
or particle emissions from the heavy nucleus fusions, despite large
nuclear mass changes. There is plenty of evidence of lattice element
transmutation, but little evidence of excess heat.
Hopefully we agree to here.
The reason this
is so is that hydrogen fusion adds one positive charge to the
nucleus, increasing the bonding of the electron by a factor of two,
one unit of which is offset by the electron's kinetic energy, and
other unit of which is lost energy due to the added proton.
Numerical examples of this given in immediately prior post.
In the
case of fusion of deflated hydrogen with an A proton nucleus, fusion
adds A positive charges to the nucleus, increasing the bonding of the
electron by a factor of A+1, one unit of which is offset by the
electron's kinetic energy, and the other A units of which is lost
energy due to the added protons.
BTW, I should have used Z instead of A, but that changes nothing.
When the neutral deflated state hydrogen (D e) with a single charge
shares a wavefunction collapse with a nucleus having A charges the
result is a nucleus with A+1 charges. In the case of fusion with
another deuteron we have A = 1 and the calculation is the same as
before, with the binding energy of the electron doubling, but its
kinetic energy taking away half of that, which in that arbitrary
example was 11.9 MeV, leaving 11.9 MeV, which was not enough to pull
off a fission of the He4*. Suppose A = 3, as in the case of
lithium. We then have for the same initial nucleus size:
Pe = k (3q)(-q)(1/r) = (-4.32x10^-9 eV m) (1/(1.2x10^-16 m)) = 36
MeV
with the electron kinetic energy still initially 11.9 MeV, for a net
deficit of 24.1 MeV. This exceeds the 22.4 MeV available from the D +
Li6 reaction. Of course in the case of lithium the initial
wavefunction might be larger, the deficit less and the fission may
produce net energy. Here we have:
D + 6Li -> 2 4He + 22.4 MeV
As the z of lattice element increases, however, the energy of fusion
drops eventually to a negative, while the energy deficit of wavefrom
collapse increases. Note, however, that this still does not preclude
the fusion under this model. The tunneling is still energetically
favorable. There simply is no net release of energy. There may also
be no fusion. There may still be release of low energy photons as
the electron rebuilds it waveform. There may be greatly increased
probability of weak reactions due to the long residence time of the
electron and the high energy virtual photon exchanges with the nucleus.
The lack of appropriate net energy
emerging from the new nucleus can not be due to the photon radiation
of the trapped electron. It has to have been carried off by vacuum
transactions, possibly electroweak reactions involving neutral
species. This is not a large leap of intuition when you consider the
fact that much of the mass of hadrons is not really there, but
results from vacuum transactions in which particle pairs, including
strange quarks, pop in and out of existence within Heisenberg limits.
The nucleus is a hotbed of vacuum transactions, so it requires no
stretch of the imagination to expect an internuclear electron to be
involved in such transactions.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
