At 1:24 PM 12/3/4, Robin van Spaandonk wrote:
>
>The thing that bothers me most about this is that if hydrinos are
>producing nuclear reactions, then I would expect to see at least the
>occasional gamma ray.
>(Though there may be possible particle reactions that are far more
>probable than gamma ray production).
Maybe we can expect no gammas, at least when it comes to D + D reactions.
Consider the ordinary branching ratios:
D + D -> T (1.01 MeV) + p (3.02 MeV) (50%)
-> He3 (0.82 MeV) + n (2.45 MeV) (50%) <- most abundant fuel
-> He4 + about 20 MeV of gamma rays (about 0.0001%; depends
somewhat on temperature.)
One argument against the possiblity of CF is "We know what the branching
ratios of He* is, so CF can't be real because" there is no signature
radiation. This argument is bous. In electron catalysed fusion, or
hydrino fusion, there is an electron present in the reaction which does not
necessarily gain any momentum from the fusion which it catalyses. The CF
reaction is thus:
D + D + e -> ? + e
That catalysing electron does not "fall into the Coulomb well" and thus the
resulting excited fused nucleus is not the conventional He* at all. There
is much less excitement! How much less depends on four things: (1) the
size of the electron wavefunction at the moment of fusion, (2) the amount
of energy supplied to the catalytic electron from the ZPE sea to expand its
wavefuntion out of the nucleus, (3) the amount of energy the electron
radiates while captured by the nucleus and (4) whether or not an electron
capture occurs.
I feel it might be argued that at the moment of fusion the wavefunction of
the electron, or the entire fused body for that matter, exists at a point.
The energy of the combined wavefunctions is momentarily returned to the
vacuum. The quantum wavefuntions of the participating bodies collapse into
a single point. The ensuing wavefunction inflation depends upon energy
exchanged with the vacuum. There is no convenient formulation to determine
exactly what signature energy will be available! The potential energy U of
an electron and nucleus pair depends upon the separation of that pair. If
that separation is momentarily zero, then, by definition, a singularity
exists. It is entirely possible the electron could require the entire 20
MeV to escape. It is further even likely that the branching ratios for D +
D + e differ entirely from those for D + D. It makes sense that the most
likely branch might look like:
D + D + e + energy -> He + e - 13.59844 eV
In other words, as I posted in "THE ATOMIC EXPANSION HYPOTHESIS" thread
here some years ago, the catalysing electron may only be able to dig itself
out of the hole by borrowing (back) from the vacuum enough energy to reach
equilibrium, i.e. to become a ground state orbital electron. To that
extent, this concept is compatible with Puthoff's theory that orbital
stability, i.e. the failure of the orbit to collapse, depends on the
orbital electron reaching equilibriumn with the zero point field. If the
electron radiates during its catalysis, or the expanding orbital radiates
during its expansion, or dislocates neighboring orbitals that radiate or
produce phonons, then that accounts for the modicum of free energy
observed. Increasing the free energy then amounts to additional
confinement of the expanding orbitals.
Regards,
Horace Heffner
