At 12:27 AM 12/8/2009, Horace Heffner wrote:
On Dec 7, 2009, at 6:39 PM, Abd ul-Rahman Lomax wrote:
I don't think so, because the doubt and question you raise isn't
about the triple tracks. In a way, you may be making the same
mistake as Mr. Beene, confusing the neutron inference with the
hypothesized source of the neutrons.
Not at all. A 50-50 mix of D-T will produce D+D --> T+p reactions.
The T will be hot. There should be T+T reactions from the "hot" T.
If *huge* amounts of hot T in a lattice produce barely observable
amount of neutrons, then it is indeed highly questionable that a
lattice with non-detectable amounts of T can produce hot neutrons.
Perhaps I don't know what you are saying here.
Good guess. I'm saying that if the main CF
reaction sometimes produces, instead of helium,
hot tritons, this will, with high probability,
generate neutrons through this hot triton finding
and fusing with a deuterium nucleus. It will have
the energy to overcome the coulomb barrier, from the manner of its generation.
Adding cold tritium to the mix will not increase
this. It will increase the incidence of d-t
fusion, but each one of these reactions might
simply replace a d-d fusion. (i.e., instead of 4D
-> Be8*, we get 3D + T -> Be9*, and whatever that generates.)
So even if tritium is the source of the neutrons,
adding tritium may not increase the neutrons at all, or only a little.
The tritium present in CF cells, the baseline, is
tritium generated by the CF reactions, so it will
probably be hot. Thus the neutron flux may be
giving us an immediate metric for the level of
tritium generation. Adding tritium will not
increase the tritium generation.... it may do
other things, but if the amount of tritium added
is small, those "other things" won't predominate,
even in terms of neutron output.
I'm coming to the position that I expect adding
tritium will not increase neutron output, though
all bets are off as to what will happen if D is
significantly replaced by T. That will create a
new predominant reaction, quite likely. All
tritium, what about 4T -> Be12? And what would that do?
The conclusion of the Mosier-Boss et al. article, that triple
tracks are due to the
12C(n,n)3á reaction, implies the need for repeating exactly the
same experiment
using D2O + trace T2O instead of just D2O.
I've said that tritium doping is an obvious thing to try, but it's
quite possible that even if the neutrons are from tritium fusion,
there would be no increase.
You might be surprised how much resistance from intelligent
physicists one might encounter to the suggestion of using tritium. I
have to wonder if it might be a subconscious fear of bursting a
fantasy bubble.
What I'm doing is trying to create a standard
experiment. It would then be simple to add some
tritium, and that resistance, to the extent that
it remained, would be obvious foolishness. If I
expect a result to be negative, I'm not motivated
to set up a complex experiment to test it, unless
I'm funded to do replication work, normally
that's the job of academic and pure science
institutions, that's an appropriate job for
public funding, because venture capital isn't
interested in boring replications, it wants to
see improvement of quantitative outcomes.
But if it's cheap and easy, buy a kit for under
$100, add some tritium, connect it to a power
supply, run it for a couple of weeks, then
develop the CR-39 -- and run another cell for
another $100 in series so that the current
profile, everything is identical except for that
tritium doping, that's peanuts. The only
difficult and unique thing there is having enough
tritium. However, it might only need to be a little?
From the neutron flux and the detected tritium
that doesn't fuse, it should be possible to
estimate how much tritium is being generated in
the cell, and adding that much -- which is very
little -- should have an observable effect under
some scenarios. I just don't think that the
absence of such an effect would prove much about
the source of the neutrons, because if what I'm
claiming is correct, it won't. The neutrons would
depend on the rate of generation of tritium by
the CF reactions, not on the level of tritium
existing in the cell before the reactions.
Now, if the increase were from tritium fusing as the initial
reaction, you'd be right. And that is a possibility. After all, if
there is some tritium in the cell, and it ends up in the place
where some deuterium would fuse, it would also likely fuse. But
very little of this tritium would fuse, just as very little of the
deuterium fuses. Good thing, eh?
But if the tritium is generated in the initial reaction, sometimes,
it would be hot, and very likely to fuse. Adding more tritium would
not increase this, because the new tritium would not be hot.
This is an interesting idea, but as noted above, a 50-50 D-T
experiment produced nominal high energy neutrons. For this idea to
be valid that hot T fusion works in a lattice, and cold T fusion does
not, there needs to be an explanation as to *why* lattice conditions
foster cold D+D but not cold D+T.
I'd suspect that the conditions *do* foster D+T,
to the extent that T is there. And very high
levels of tritium would create new reactions,
some of which might generate more neutrons than
the N(d) -> (N/2)(X)* reaction that predominates
with pure deuterium. But that's not relevant.
What I'm claiming is that most of the tritium
generated probably ends up as having fused with d
to generate an energetic neutron, through some
non-predominant pathway. So the neutron flux
depends on the rate of tritium generated by that
primary reaction, not on the level of tritium
already there, until that level rises so high
that it influences the primary reaction, since
the proportion of deuterium to tritium has been lowered.
In other words, hot T should not
be necessary for a reaction. If cold D fuses then cold T should also
fuse, and if anything better. It may well be there is a reason, but
on the surface it looks nonsensical.
Yes. That's correct. Hot T should not be
necessary for a reaction with tritium, where
tritium replaces d. But adding tritium simply replaces some D with T.
This is a *different effect* from generation of
neutrons that takes place in the cell when there
is very little tritium there to start. The
tritium generated and later detected would only
be the tritium that fails to fuse when generated.
Most of it, if I'm correct, i.e., if it is
generated sufficiently hot, will fuse.
Basically, the CF reaction is creating a
secondary hot fusion reaction with high
cross-section. It is a source of high tritons.
It's conceivable that this reaction could somehow
be encouraged, and it's possible that gold cathodes do this.
We know that something generates tritium in these cells, at low
levels. And it would quite likely be hot tritium formed, so it
would probably fuse with deuterium, and it would then produce the
required neutron to produce the triple tracks.
This might explain a significant decrease in T lattice half life.
However, it does not explain why T does not fuse cold in exactly the
same conditions D does.
But it probably does. You won't see any
difference, though, until T/D rises to quite high
levels, and then it would only be the difference
in outcome from (d,t) fusion instead of (d,d)
fusion. Say M(d) + N(t) -> X, where M + N = 4, or
sometimes other integer. If that's the primary reaction.
In other words, from what we already know (tritium generation),
we'd expect a low level of neutrons. So neutrons are merely a
confirmation of this. It's a coherent picture already, and, while
that doesn't prove that this picture corresponds exactly to the
reality, there isn't much need to conjure up exotic reactions.
The exotic reactions are merely a natural outgrowth of a specific
model, a natural consequence. They are not expected to be the
*prominent* channels, but are of interest because they may be
feasible channels, and of diagnostic use. Their possibility was a
natural consequence of the deflation fusion model. It struck me as
worthwhile to alert the community as to the possibilities, as they
are new and verifiable.
That's great. And to the extent that these
predictions can be reduced to simple experiments,
to that extent tests become more likely. I'd like
to look at what tritium sources are legal for me
to obtain and use, and I'd consider the test, a
few months down the road, and it is an obvious
application for the kits I'm designing and testing.
Except, of course, for whatever it is that is going on in the cell
that would generate tritium! I think the TSC
I used to know what the TSC was, but don't any more.
Tetrahedral Symmetric Condensate.
If you are
referring to Be -> 2 alpha, due to a condensate forming then there
still has to be an explanation as to how co-location is overcome and
why the bremsstrahlung, gammas, and alphas disappear even near the
surface.
You should read Takahashi more thoroughly. There
was a relatively early paper which described what
happens when Be-8* is formed. It is not likely to
be two alphas at 23.8 MeV, that was an erroneous
assumption of mine, an assumption which
apparently has also been made by others. The
actual alphas are more likely generated in the
KeV range, with most of the energy being dumped
through a whole series of photon emissions from
the excited nucleus; apparently there is time for
that before the Be-8 decays into the alphas.
I'll try to find reference to this paper.
would do it, and a lot of other things, both directly if it
encounters a nucleus before collapsing and decaying, and indirectly
as it generates a few very hot alphas. I believe that if it's
generating seriously hot alphas (these would be from early decay of
the Be-8, before the bulk of the energy of the excited nucleus has
been emitted as EUV or whatever it is), these would be generated
below the surface, there would be significant loss of energy before
they escape the cathode.
I don't think that is true. CF also occurs very near the surface.
There would have to be *huge* numbers of alphas and gammas produced
to account for excess heat.
Well, "huge" is what is detected by the helium.
The alphas *are* being produced. That's not
really questionable at this point, the only
question is what energy they have when they
emerge from the lattice. Probably not much.
Basically, photon emission from Be-8 is the
Takahashi answer to how the energy of the
reaction is communicated to the lattice. Those
photons are at low enough energy that they won't
escape at all. They could be very difficult to
detect, but you have suggested a possibility, an
etch-fabricated cathode with tiny holes in it.
Come to think of it, that's done with CR-39 and
it might be possible to use a standard
polycarbonate microfilter for the cathode, just plate the damn thing.
Gammas aren't absorbed, they are
attenuated (except in some unusual CF models).
The photons aren't in the gamma range, the
emissions take place in a series of reductions in
nuclear excitation, Takahashi lists them.
It is one thing for
gamma producing reactions to occur once every few seconds, and
another entirely to occur at large excess heat rates. Such a
reaction might be an oddball reaction, like the exotic strange matter
reactions I suggested should be feasible, but not the predominate
reaction, certainly not an explanation for CF excess heat.
It is easy to see why CF was rejected, it was due
to assumptions about the nature of the reaction.
But, contrary to what I'd said and to what others
thought when I asked about it, the Be-8 model
doesn't predict gammas at all, it transfers,
indeed, most of the energy to the lattice through photon emission.
I'd say that Takahashi needs a publicist. His
model has not been widely explained, and I'm not
sure I get it enough to be confident about what
I'm saying about it. I've seen no serious
criticism of his model that was based on the
whole body of work, nor serious and clear
explanation of it. Storms certainly mentions it
as a notable theory, but doesn't say much....
Takahashi's paper purports to predict fusion to
Be-8, 100% within a femtosecond or so, if the TSC
forms, which appears to be as simple as two
deuterium molecules being momentarily confined by
a single lattice cell. In other words, the fusion
rate may depend entirely on the frequency of that
occurrence. Some of the reaction to the Be-8
theory has been based on an assumption that this occurence would be unlikely.
But it *is* unlikely, that's the whole point. It
only occurs at a very low rate that is sensitive
to the very cathode conditions we know are
crucial to formation of the nuclear active
environment. Deuterium gas cannot exist deep in
the lattice, much less two molecules in the same
spot. But very close to the surface, where
deuterium gas is being generated -- it is about
100% outside the lattice, there will be some
occurrence of single confinements. Double
confinement would still be quite unlikely. But
"quite unlikely" might be just the right number
to explain the effect, which clearly depends on a quite unlikely reaction.
Or else a vaporized laboratory.
There is some small risk that someone, trying out
some serious variations all at once, will happen
upon something that dramatically raises the
incidence of TSC. I suggest small variations, so
that increased frequency is gradually approached.
There is such a thing as too much success, too
quickly. Always remember that Fleischmann
meltdown. It could have been much worse. I'm,
myself, sticking with known techniques, it's only
fair to my neighbors, don't you think?
