On Jun 7, 2007, at 2:10 PM, Jones Beene wrote:
Horace,
This is a nonsensical model of the process and certainly *not* one
implied by me.
Well - playing devil's advocate once again,
I think my interest here is fast ending. I have a lot of mundane
things I have to do before winter, and it looks like I won't be able
to get to the science things that are of most interest to me.
if tritium were coming off in the vacuum exhaust in well-equipped
labs, it would set off a warning - but maybe they did not have any
such precaution... nevertheless ... in trying to get a better
protocol pinned-down, in case anyone (such as a Mills proponent)
might wish to whittle down the open possibilities, it would seem
that tritium has such a unique signature that it would not be hard
to find it, especially with a dedicated tritium detector, unless it
is ALL at the much lower energy level (and how could that be?) ....
That is, if one looks in the right place like the vacuum exhaust,
or turning the pump off, tritium detectors should spot it like a
sore thumb.... and also - another factor weighing against tritium
is that one can doubt that helium would have much of an effect on
tritium release, in the situation where there was only helium, but
less fogging.
The above is a red herring. I never said anything about normal tritium.
You would agree that if a vacuum is drawn on a tight seal, then
pump turned off for the multi-hour exposure, and there is still
only minimal fogging - then the exposure is not due to the release
of tritium ?
Of course I would not agree! I just went to a lot of trouble to show
why. This is utterly frustrating.
Ef = p(T) * (density of T at P) = R * P * (density of T at P)
If the pressure P is zero then partial pressure p(T) of T is zero so
the film exposure rate Ef is zero. If the p remains near zero then
the exposure rate remains near zero. To the extent there is a
vacuum, the exposure rate is diminished.
If tritium can be eliminated, then beta decay of the neutron is
still an open possibility
Except for the fact the neutron decay energy (782,350 V) is about
2600 times too high.
- but a vacuum would not have eliminated that before (in the
original) - and the crux of this puzzle is that the effect goes
away with a either a vacuum or with an unreactive gas (He, Ar)...
and also - the other factor weighing against tritium is that helium
should not have much of an effect.
If I am understanding this, with a reactive gas present - O2 or N2
Just O2.
"Some samples were also kept in atmospheres of nitrogen,
helium and argon gases. The gas pressure was retained slightly (~ 50
mbar) above one
atmosphere. The exposure time in all the cases was 96 h. No
radiation, above threshold, was
observed on any of these autoradiographs."
Maybe would get it with N2 also if an ammonia forming catalyst were
present, but ammonia would not respond to an E field. Just because
the E field experiment is not conclusive, due to a lack of control,
does not mean there is reason to completely throw it out yet. There
were runs with similar width gaps at 0 potential, so the results may
be valid, and if so it will still require both positive and negative
ions of the species to explain.
there is an fairly large signal and it is not due to photons. If
tritium, photons and neutron decay are eliminated and 300 volt
electrons are the culprit, then my original take on this was to
look for a species that would displace an inner electron of O2 or
N2. Auger electron spectroscopy is where you usually see electrons
of this energy.
But you see x-rays from auger electrons, lots of photons from the
cascades, and where's the needed 300 eV particles?
Regards,
Horace Heffner
