At 04:32 PM 10/20/2010, [email protected] wrote:
Of course, but IMO it still qualifies as DD fusion, in as much as D
is the fuel,
nothing else is involved, and He4 is the ash.
"DD fusion" strongly implies a reaction between two deuterons. Even
though you can justify another meaning for it, that's not what is
associated with it. Again and again, you will encounter in the
skeptical literature an assertion that the characteristics of "DD
fusion" are well known.
I've been simply calling it deuterium fusion, and making the point
that it is probably not "d-d fusion." It has the heat/helium ration
of d-d fusion, but any fusion process that starts with deuterium and
ends with helium will, no matter what the pathway (unless, of course,
there are significant other products.
I've identified this semantic error as a major part of how cold
fusion was assassinated. I don't think they intended to be deceptive,
the skeptics, but they simply imagined that everyone who was claiming
"fusion" was claiming d-d fusion, and the attempts to rationalize d-d
fusion didn't help, i.e., the attempt to, say, postulate a
Mossbauer-effect-like transfer of energy to the lattice instead of
the emission of a gamma ray.
There were so many problems with this that, politically, it would
have been better, far better, to temporarily and provisionally
abandon any idea that this was d-d fusion. It was an "unknown nuclear
reaction," and it still is, but because we now know that the ash is
helium and can rather safely assume, at this point, that the fuel is
deuterium (not ruling out the possiblity of other kinds of fusion
under different conditions), so it's "deuterium fusion" with an
unknown mechanism. but some idea that it might be cluster fusion, a
simple form which explains very much of the experimental evidence is 4D fusion.
By the way, the fast electrons you mentioned might possibly be an
answer. I don't know enough to speculate much. If we get a TSC
condensate, and if in that state the Be-8 lasts longer than it would
otherwise -- it might -- then the excited nucleus could have time to
radiate the excitation energy to the lattice as photons (euv?). These
would be difficult to detect.
What happens if the Be-8 then fissions while still in the BEC? The
BEC, now consisting of two helium atoms, would fly apart, I
understand, and the electrons would share in that expansion. There
are a number of possibilities, Storms mentions that we might see, for
example, fast atoms. Neutrally charged! That is not a common
phenomenon. A helium nucleus with its electrons at 23.8 MeV? Those
electrons would not stay attached for long, and most of the energy
would be with the nucleus. That doesn't work. But if the electrons
and the helium expand in six different (but balanced, of course,
conservation of momentum) directions, how much energy would each carry?
I don't know. I only pretend I learned a little physics so many years
ago. Shallow, it was. Just enough to be able to recognise light from dark.
The total energy would be about 180 KeV, as I recall, from the ground
state decay of Be-8, average 30 KeV per particle, or perhaps it would
be 45 KeV for the electrons. I'd think that would generate
Bremsstrahlung that would be observed.
But larger clusters could have unfused atoms that would also carry away energy.
