1995.09.23 / Mitchell Jones /  Re: The Protoneutron Theory of "Cold Fusion"
     
Originally-From: 21cenlogic@i-link.net (Mitchell Jones)
Newsgroups: sci.physics.fusion
Subject: Re: The Protoneutron Theory of "Cold Fusion"
Date: Sat, 23 Sep 1995 11:00:38 -0500
Organization: 21st Century Logic

The second post on which this thread is based is the following:

 -----------------------------------------------
OK, you guys, I've been waiting for two days to see if somebody was going
to earn three ataboys by noticing that the protoneutron theory explains
the Griggs result! Time's up! The ataboy window is closed!

Here's the way Griggs' machine generates excess heat: 

(1) The 3500 rpm rotation of the "cylinder full of holes" produces massive
turbulence in the water, and generates millions of tiny cavitation voids. 

(2) Since "there ain't nothin in them voids," they collapse as soon as
they move to a less turbulent point in the flow. 

(3) When a void collapses, the walls crash together at the center,
producing momentary giant overpressures. (This is a phenomenon well known
to plumbers. When, for example, a long column of water running in an empty
pipe slams into a closed valve, the giant overpressure that results is
known as a "surge," and a blown valve is the frequent result.)

(4) In the moment of final collapse, there is a tiny sphere at the center
of the former void, where the situation is sufficiently crowded at the
microscopic level so that electrons in the lowermost (K) shells of
hydrogen atoms do not have room to orbit. While the duration of this
moment is short in human terms, it is an eternity on the time scale of a
hydrogen atom, whose electron revolves around its nucleus at a frequency
of 10 quadrillion revolutions per second.  

(5) For an instant, therefore, such spheres are packed with protoneutrons. 

(6) Since protoneutrons are rapaciously hungry for the energy of
transformation into neutrons, and since there are occasional gammas in the
background radiation to which we are all exposed, a tiny percentage of
these protoneutrons will capture gammas and transform into thermal
neutrons. 

(7) Those thermal neutrons will quickly enter the nuclei of neighboring  H
or O atoms, and kick off gammas of their own when the product nuclei drop
to the ground state. In the case of neutron absorption by a hydrogen, for
example, we have: p + n --> d + 2.22 Mev (gamma), as previously discussed
in the explanation of CF electrolysis.

(8) Since the surrounding solution contains millions of cavitation bubbles
in all stages of collapse, it follows that at any given moment the
solution contains billions of protoneutrons. Which means: the 2.22 Mev
gamma is surrounded by protoneutrons and isn't going anywhere: they are
going to suck that puppy dry--which means: it is going to produce 2 more
thermal neutrons, plus dump some heat into the fluid. 

(9) So here, again, we have the same old chain reaction that we discussed
the other day, in which one thermal neutron begets two plus heat, which
beget four plus more heat, which beget 8 plus heat, and on and on. This
time, however, the process occurs in the dynamic environment of a
turbulent fluid, rather than in the static environment of a palladium
lattice.  

Bottom line: the Griggs device would be "cold fusion," except that "cold
fusion" itself isn't fusion. What we have here, instead, is a protoneutron
heat engine: an odd device that burns hydrogen so completely that even the
nuclear energy is extracted, and yet it produces no pollution and no
radiation. It is a "soft" energy source, as benign and non-threatening as
flowers on a sunny day, and what it means is simple: the age of pollution,
of centralized energy sources, and of centralized political power, is
over.

--Mitchell Jones

===========================================================

===========================================================
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1995.09.23 / A Plutonium /  Re: First experimental predictions of Hydrogen Atom Systems
     
Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Re: First experimental predictions of Hydrogen Atom Systems
Date: 23 Sep 1995 17:30:09 GMT
Organization: Plutonium College

In article <43sua4$d65@news.utdallas.edu>
iblum@utdallas.edu (Ira K Blum) writes:

> You are forgetting pions and kaons and muons, particles which are not 
> protons, electrons, photons, or neutrinos, yet are smaller than 939 MeV, 
> which is the mass of your HYASYS.

  Tell me Ira, what does a muon decay into since its lifetime is 2.2 x
10^-6 sec yet its mass is 105.66 MEV as compared with electron mass of
what about .5 MEV.

   Please, if you could or some other reader, list the lifetimes and
the MEVs of all of these particles and what they decay into.

  Physicists for the past 40 years have not been truthful to
themselves. They should realize that when a particle exists for such a
fleeting time as 10^-6 sec that such a particle is really not on "par"
to the existence of say the proton or electron etc.
   To put it on terms that all readers can understand. If we were to
give a comprehensive science study of Homo sapiens, should we spend any
time whatsoever or some little time on the fact that a worm lives in
the eyelashes of humans, considering that we are really talking about
Homo sapiens and not an organism that lives on Homo sapiens? In the
same vain, is the muon a particle on par with the electron, or is it a
energy exchange, that is, the muon is merely an electron with stored up
extra energy and not a true particle. 
  Each particle should be looked at as to "how long it exists",  "what
it decays into". In this manner, only hydrogen atoms exist and
everything else is a hydrogen atom system with more energy than the 938
MEV hydrogen.

  What is the muon but an electron with a lot of extra energy? What
needs explanation therefrom is why the quantization of 105 MEV? Why
will the first quantization of an electron be 105 MEV or are there
electrons between the .5 MEV and the 105 MEV? These are the questions
to ask. And never to think that the muon is as basic and fundamental as
an electron. Why does it take 9 muons to make a Hydrogen Atom System?
All particles be they elements, isotopes, or this fancy schmancy things
out of CERN are nothing but Hydrogen Atom Systems.

  This stuff is fun and it must be correct since it is too beautiful
not to.
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1995.09.23 /  ZoltanCCC /  Re: ? Singularity Technology ?
     
Originally-From: zoltanccc@aol.com (ZoltanCCC)
Newsgroups: sci.physics.fusion
Subject: Re: ? Singularity Technology ?
Date: 23 Sep 1995 16:29:45 -0400
Organization: America Online, Inc. (1-800-827-6364)

Would you tell us where the RGeorge Home Page is.


Thanks,


Zoltan Szakaly
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1995.09.23 /  ZoltanCCC /  Re: The electron capture theory of cold fusion
     
Originally-From: zoltanccc@aol.com (ZoltanCCC)
Newsgroups: sci.physics.fusion
Subject: Re: The electron capture theory of cold fusion
Date: 23 Sep 1995 18:39:24 -0400
Organization: America Online, Inc. (1-800-827-6364)

In article <4409ec$nlq@newsbf02.news.aol.com>, mrichar353@aol.com
(MRichar353) writes:

>
>Oh, come on. I have refuted Robin's idea conclusively, at least for
people
>who know physics at an undergraduate level. What you and he are talking
>about violates energy conservation, it's as simple as that.
>
>

Mark, unfortunately your half cocked reference to the conservation of
energy did not cut it with me. Up to this point nobody has described the
situation as the way I see it now. Here it follows:

Before the electron falls into the potential well of the deuteron there is
energy associated with the field of the proton as well as the field of the
electron. In case of the electron this field is a major contributor to the
electron's mass. If the electron were pointlike, the field would be
infinitely massive. In fact we can calculate the radius of the electron by
assuming that all of the electron's mass comes from the field. This radius
is a lower limit. The electron cannot be smaller than that. I believe
Einstein made a theory that explained the electron's inertia based on its
electric field. In any case, the field of the proton and also the field of
the electron disappear as the electron closely approaches the proton
because their charges cancel each other. The mass of the two fields
combined disappears and accelerates the electron which becomes heavier by
the same mass. Its kinetic energy is already counted when we add up the
initial masses of the electron and deuteron and so the electron does not
have enough energy to initiate the electron capture I described earlier.

I retract my claim now that the electron could simply fall into a proton
and cause electron capture. The energy has to come from somewhere else. 

On the other hand it is interesting to see that the electron capture
requires as much energy as there is in both fields, it seems to me like
strange coincidence. From Robin's calculation it seems that the mass of
the electron's field is about 

0.782/2 =0.391 MeV, while the rest mass of the electron is 0.511 MeV 

This gives us an idea of how much energy we need for electron capture to
happen. If there is a mechanism in the lattice that produces excited
states of the deuteron, electron capture may become feasible.

Zoltan Szakaly



Zoltan Szakaly
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------------------------------
1995.09.22 / Henry Spencer /  Re: Making He3 and related questions
     
Originally-From: henry@zoo.toronto.edu (Henry Spencer)
Newsgroups: sci.space.tech,sci.physics,sci.physics.fusion
Subject: Re: Making He3 and related questions
Date: Fri, 22 Sep 1995 20:33:10 GMT
Organization: U of Toronto Zoology

In article <43nb29$mnm@morgoth.sfu.ca> gay@sfu.ca (Ian Gay) writes:
>>Hence, He-3 is a byproduct of our nuclear weapons stockpile. So it is 
>>being manufactured "earthside". Quite expensive, but I don't have a 
>>figure. Off the top of my head I seem to remember $1000/STP-liter.
>
>Just tried to look it up. All my current isotope catalogs fail to give a 
>price (Why? wildly fluctuating?) ...

Probably very limited availability.  The government is the only supplier,
and they're not noted for rational decision making.  In this case, "please
call" probably means "we're not sure whether we can get it or how much it
would cost -- it might depend on who you are and what you want to do with
it". 
-- 
The problem is, every time something goes wrong,   |       Henry Spencer
the paperwork is found in order... -Walker on NASA |   henry@zoo.toronto.edu
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1995.09.23 / A Plutonium /  Mark Richardson, Mr. Dupree, and Mr. Esch
     
Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Mark Richardson, Mr. Dupree, and Mr. Esch
Date: 23 Sep 1995 22:55:54 GMT
Organization: Plutonium College

In article <43tc0g$q4s@newsbf02.news.aol.com>
mrichar353@aol.com (MRichar353) writes:

> 
> God how idiotic!
> 
> Superposition has absolutely nothing to do with building up a
> multi-particle system from individual particles. It is the principle that
> a linear combination of two wave functions is another wave function. These
> two (or more) wave functions represent *alternatives*; they cannot both be
> true at the same time (at least when a measurement occurs).
> 
> Then again, considering the source, what should I expect?
> 
> I *must* get a newsreader with a killfile capability!
> 
> Mark Richardson

  I was updating the go-to-hell file which I keep and saw that Mr.Craig
Dupree and Mr. Patrick P. E. Esch were on it. Reconsider.
  PU, PLuto revert both Craig Dupree and Patrick P. E. Esch back to
point of neutrality, away from hell. Atom

  As to Mark Richardson who I told to go to hell, his spite and hatred
still spews forth in greater amounts, I say go to hell, Styx this time.
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1995.09.23 / A Plutonium /  Re: All atoms are Hydrogen Atom Systems -> Superposition 
     
Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Re: All atoms are Hydrogen Atom Systems -> Superposition 
Date: 23 Sep 1995 23:13:14 GMT
Organization: Plutonium College

In article <jeffo.338.3060A8E4@rayleigh.lanl.gov>
jeffo@rayleigh.lanl.gov (Jeff Olson) writes:

> In article <43q0va$9ja@news.utdallas.edu> iblum@utdallas.edu (Ira K Blum) writes:
> 
> >In article <jeffo.326.306040FF@rayleigh.lanl.gov>, jeffo@rayleigh.lanl.gov (Jeff
> >Olson) writes:
> 
> >|> According to the ancient periodic table on my wall here (back when O was 
> >|> exactly 16 au), the neutron has a rest mass of 1.0087 au and an H atom of 
> >|> 1.008 au.  If the neutron mass were less, why would it decay?
> >|> 
> 
> >M(p) = 938.27231 +/- 0.00028 MeV
> >M(n) = 939.56563 +/- 0.00028 MeV
> >M(e) = 0.51099906 +/- 0.00000015 MeV
> >M(p+e) = 938.78331 +/- 0.00028 MeV
> 
> >I guess you are right, I haven't looked at it in quite some time, so 
> >there is more energy in a neutron than a proton and an electron.  (also 
> >the dominant decay is n-> p e nu  (note that the neutrino is present and 
> >carries away momentum.)
> 
> Actually, it's an anti-neutrino, I think.

  Someone please mind posting a representative list of elementary
particles, just the most important members of each family showing the
MEV, all in MEV. Plus indicate the lifetime of the particle all in
seconds. Then what the particle decays into. Please post with a name of
the particle not the Greek symbol.  Thanks
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------------------------------
1995.09.23 / Mitchell Jones /  Re: The Protoneutron Theory of "Cold Fusion"
     
Originally-From: 21cenlogic@i-link.net (Mitchell Jones)
Newsgroups: sci.physics.fusion
Subject: Re: The Protoneutron Theory of "Cold Fusion"
Date: Sat, 23 Sep 1995 10:56:55 -0500
Organization: 21st Century Logic

I am getting more and more stuff, both in the form of posts or via e-mail,
that reveals a massive misunderstanding of the protoneutron theory.
Apparently many people have become interested in this thread after the
original posts rolled off of their servers. Therefore, to preserve
bandwidth by squelching the uninformed posts, I am going to re-post the
two main articles on which this thread is based. The first follows:

 ----------------------------------------------------------
In article <hheffner-0609951722310001@204.57.193.72>,
hheffner@matsu.ak.net (Horace Heffner) wrote:

> In article <21cenlogic-0509951113520001@austin-2-12.i-link.net>,
> 21cenlogic@i-link.net (Mitchell Jones) wrote:
> 
> [snip]
> 
> > ***{The point here is to accurately visualize what is going on in the
> > palladium cathode. To do that, we must accept the fact that the hydrogen
> > ions (protons and/or deuterons) retain their positive charges until
> > *after* they enter the cathode. Think about it: if, when a hydrogen ion
> > appeared at the surface of the cathode, it were to be handed an electron,
> > it would immediately become electrically neutral, and would cease to be
> > attracted to the cathode. In that case, it would bubble away as a gas and
> > loading would be impossible. Here, instead, is what must happen: (1) When
> > an H+ reaches the surface of the cathode, it retains its charge, and
> > enters the lattice structure. (2) The presence of H+ ions within the outer
> > layers of the lattice gives those layers a positive charge, while the
> > inner layers retain their negative charges. (3) When enough H+ ions have
> > packed into the outer layers of the cathode to neutralize the charge of
> > the cathode, the voltage gradient between the anode and the surface of the
> > cathode disappears, and the H+ ions in solution are no longer attracted to
> > the surface of the cathode. Thus the + charge in the outer layer of the
> > cathode does not build up to "millions of volts," but only to the level
> > necessary to eliminate the voltage gradient from anode to the surface of
> > the cathode. In effect, the anode charge migrates into the outer surface
> > of the palladium electrode, and the cathode charge retreats into the inner
> > portion of the palladium electrode. (4)) Once the anode's charge has
> > migrated to the surface of the cathode via this mechanism, loading of the
> > palladium electrode does *not* cease, because H+ ions on the inner surface
> > of the positive region are constantly having electrons passed to them from
> > the negatively charged central region. Each time this happens, the
> > affected H+ ion becomes a neutral H atom. However, it can't bubble out of
> > the solution, because it is trapped inside the palladium electrode. (5)
> > Each time a new, neutralized hydrogen atom is added to the pool of those
> > trapped inside, the positive charge of the surface region drops slightly,
> > and a new H+ ion enters the surface region from the outside, again
> > eliminating the voltage gradient between the anode and the surface of the
> > cathode. (6) Then, another H+ ion inside the palladium electrode is handed
> > yet another electron and neutralized. (7) Then, another H+ ion enters the
> > surface. And on and on it goes, until saturation loading is approached,
> > and, hopefully, "cold fusion" kicks in. Bottom line: it isn't necessary
> > that the cathode achieve a charge of "millions of volts;" but it *is*
> > necessary that it build up a positively charged region on its outer
> > surface, in order for the loading process to take place at all. --Mitchell
> > Jones}***  
> > > 
> 
> > In article <hheffner-0209950613530001@204.57.193.68>,
> > hheffner@matsu.ak.net (Horace Heffner) wrote:
> > 
> 
> Requoted here with minor typo's corrected to aid in voltage calculation to
> follow:
> 
> > > I would like to take this opportunity to clear up some mistaken thoughts I
> > > previously posted here. First, some data from the Handbook of Chemistry
> > > and Physics (HCP) and a Sargent-Welch periodic table. All sizes in A
> > > (1e-10 m).
> > > 
> > > Radius of Pd atom: (Pd) 1.79, (Pd covalent) 1.28, (Pd+2) .80, (Pd+4) .65
> > > Radius of H atom: (H) .79, (H-1) 1.54, H (covalent) .32

***{Note: the radius of the innermost Bohr orbit is .53 Å, and that of the
H+ ion (a proton) is .0000137 Å. --Mitchell Jones}***

> > > Bond lengths: (H-H) .746, (Pd-Pd) 2.751
> > > Density of Pd: 12.0 g/cm^3
> > > At. Wt. of Pd: 106.4
> > > 
> > > Now, the Pd lattice is cubic, face centered.
 
***{It is misleading to refer to this lattice as cubic. The smallest unit
cell consists of two four-sided pyramids (e.g., like the Egyptian
pyramids) joined base-to-base. The faces of each pyramid are equilateral
triangles with a side length of about 2.67 Å. The cubic unit cell to which
you refer is three times as large, and is not a satisfactory conceptual
unit for present purposes. The reason: when loading takes place, it is the
double-pyramid unit cells that are loaded--i.e., that have H atoms placed
in their centers. --Mitchell Jones}*** 

You could imagine building
> > > this lattice by taking 1 atom thick sheets of spheres arranged in square
> > > patterns and laying them one on top of each other, but the top layer atoms
> > > resting naturally in the spaces between the bottom layer atoms.  The bonds
> > > formed are between atoms in seperate layers only.  If we call the distance
> > > between alternate sheets S, then the bond length D will be 1/2 the length
> > > of the diagonal of the cube S^3, or .866 S. I seem to recall the jist of
> > > this being posted before.
> > > 
> > > For simplicity sake let's assume all atoms are spheres.  Now, applying
> > > common sense, suppose we considered an imaginary Pd made up of only
> > > alternate sheets. It would have half the density, or 6.0 g/cm^3.  Also,
> > > each atom would occupy a cube (S^3) of volume V. Applying Avogadro's
> > > number we get 6.79E22 atoms/cm^3 for Pd. Our half density sheets would
> > > therefore have 3.40E22 atoms/cm^3, so V = 2.94E-23 cm^3, and S = 3.09E-8
> > > cm = 3.09 A.  This gives D = 2.67 A, which corresponds within about 3
> > > percent with the published bond length of 2.751. So this is good, common
> > > sense and published values agree.
> > > 
> > > Now this means we have a radius R for Pd in the lattice because R
=(0.5)D =
> > > 1.38 A.  This radius corresponds best with a covalent radius. 
> > > 
> > > Now, the largest sphere that can be placed between the Pd spheres is
> > > diameter d = S - R - R = S - D = .134 S.  Since S = 3.09 A, d = (.134 *
> > > 3.09) A = .414 A. This means the H atom will have a radius r = .5 d = .207
> > > A. 
 
***{Using the calculated bond length of 2.67 Å rather than the published
one, the size of the entry hole in one of the triangular pyramid faces is
large enough to permit the passage of a sphere with a radius of .206 Å.
This means entry is permitted only to H+ (protons) or D+ (deuterons). A
neutral hydrogen atom is too large to pass through the face and into the
unit cell. Once inside the cell, of course, there is room in the center
for a sphere of radius .55 Å, which will accomodate one neutral H atom
with its electron orbiting at the innermost Bohr radius of .53 Å. To get
there, however, its parts must enter as separate charged particles--i.e.,
as a proton and an electron---and combine after they are inside. At this
point, it is appropriate to ask ourselves what sort of loading ratios
result from such a process. For example, if there is one H atom inside
each unit cell of the lattice, what is the loading ratio? Well, for an
infinite lattice the count of unit cells is simply three times the count
of the number of Pd atoms in the lattice. (Since each Pd atom in a layer
is the top of a vertical unit cell in the two layers below it, the number
of vertical unit cells equals the number of Pd atoms in the lattice. The
same is true of unit cells oriented right to left and front to back. Hence
there are 3 times as many unit cells as Pd atoms in the lattice.) If we
stuff a single, neutral H atom into each unit cell, therefore, we would
have a ratio of 3 H to every Pd! It therefore follows that, for the
loading ratios that are actually seen (not much above .8), more than two
thirds of the unit cells are unoccupied.  --Mitchell Jones}***

This appears to be too small to be covalent (i.e. .32), so we are led
> > > to think the bond must be ionic. This agrees with other chemistry if I
> > > recall correctly, and is the basis for Marshall Dudley's Hypothesis.
> > > Loading above 1-1 leaves no room for H orbitals.
> > > 
> > > Anyone, please comment on flaws in this line of reasoning.
> > > 
> 
> [snip]
> 
> Now, to check your model of the electrode surface, let's assume a single
> H+ ion blocks each entry hole to the lattice (i.e. the ion layer is only
> one atom thick.) 

***{I never said they would block entry. They are, in fact, too small to
block entry. They do, however, carry their positive charges with them into
the lattice until such time as they meet up with outgoing electrons. Since
protons are more than 1837 times as massive as electrons, their drift
velocities in the lattice are vastly reduced, and they tend to not
penetrate very deeply before being neutralized. Thus the positive charge
that migrates to the cathode will be concentrated in the outer regions of
the lattice. --Mitchell Jones}***

 From the above calculations we have surface atoms in a
> square array, with the sides of each square rougly S = 3.09E-8 cm. 
> Therefore each Pd atom presents an area of roughly S^2 = 9.55E-16 cm^2. 
> So we have roughly 1/(S^2) = 1.05E15 ions in a 1 cm^2 area. For
> simplicity, let's assume we area talking about a 1 cm^2 surface area
> cathode. This cathode would then have a charge Q=(1.06E-19C/ion)*(1.05E15
> ions) =1.11E-4 Coulombs.
> 
> Now, what we are talking about is essentially a  capacitor with area = 1
> cm^2 and a plate separation of S. This let's us use the capacitance
> formula:
> 
> C = e_k * A / S =  (8.85E-12 F/m) (.01 m)^2 / (3.09E-10 m)
> C = (8.85E-12 * 1E-4 / 3.09E-10 ) F
> C = 2.86E-6 F (about 3 uF)
> 
> Now volts = Q/C = (1.11E-4)/(2.86E-6) = 38 volts. 
> 
> This is still a pretty big barrier.  This implies that to just balance an
> electrode voltage of 2 volts, thus terminating electrolysis, that only
> about one site in 19 could be occupied by a + ion.

***{The fact that a minority of the sites are occupied by H+ ions at a
given instant is not a problem. The absolutely crucial point here is not
that the surface of the cathode acquires the full positive charge of the
anode, but rather that H+ ions *must* penetrate into the lattice before
being neutralized in order for loading to take place. The reason:
electrons that are given over to H+ ions before they enter the unit cells
of the lattice result in gas production, not loading. Once a particular
unit cell is occupied by an H+ ion, the next step is to hand it an
electron, converting it into a neutral H atom which, because the radius of
the innermost Bohr orbit is .53 Å, is now trapped inside the unit cell.
This is the way the lattice is loaded, *because it has to be.* To repeat:
electrons that are given over to H+ ions before they enter the unit cells
of the lattice result in gas production, not loading. --Mitchell Jones}***

> 
> I have a different mental model I would like to share. I don't know if it
> is correct, but I'll spell it out, as if it's a fact, so it can be
> dissected.

***{I am not sure why you are presenting an alternative model here. None
of the statements that you made above revealed any reason for rejecting
the model that I have proposed. If your difficulty lies solely in the
implication, admittedly conveyed by my early description, that the full
anode charge migrates to the surface of the cathode, then let me emphasize
again that this is not necessary. What is necessary--absolutely
necessary--is that H+ ions enter the lattice cells before they receive
electrons. This fact implies that there is going to be some degree of
migration of the anode charge into the outer portion of the cathode, but
does not necessitate that the full anode charge do so. --Mitchell
Jones}***  
> 
> Water molecules are bipolar, electrostatically speaking. So, if a + ion is
> in solution, the adjacent water molecules align their - sides toward the +
> ion. Now we have a sphere with all + sides of the water molecules on the
> surface. The water molecules adjacent to the sphere align their - sides
> toward that layer around the sphere, and so on. There is in effect a hugh
> ionically bonded +1 charge molecule (clump) that migrates through the
> electrolyte.  When the clump reaches the cathode, the elctrostatic force
> of the cathode, if sufficient voltage, can separate the + ion from the
> clump.
> 
> It seems like, if this is true, H atoms preceeding a clump to the cathode,
> and at the surface of the cathode, would have a very high probability of
> being forced into a site by a clump. Also, some would H atoms would
> escape, combining with other H atoms to form H2 and bubble out.
> 
> What do you think?

***{To repeat: I do not see a need for an alternative model. A voltage
gradient is all that is required to attract an H+ ion into the unit cells
of the cathode. It doesn't have to be "forced" in: the openings in the
triangular faces of the unit cells are large enough for it to move in
*very* easily--as easily as a fly into an amphitheater, in fact. Remember:
an H+ is simply a proton--i.e., the nucleus of a hydrogen atom. The
standard formula, curve-fitted to the experimental data, for computing the
radius of a nucleus is simply r = (1.37E-13)A^(1/3), where r is the radius
in cm and A is the mass number. For H+ the mass number (number of protons
plus number of neutrons) is 1, so r = 1.37E-13 cm = .0000137 A.  Once
inside the cathode, such an H+ ion will drift toward the electron source
until it acquires an orbital electron. At that point, it will become an
electrically neutral H atom, with an effective radius as per the innermost
Bohr orbit (.53 Å or slightly larger). This means, roughly speaking, that
its radius will increase by four orders of magnitude when it captures an
electron, and it will become trapped in the unit cell which it is
occupying at that time. --Mitchell Jones}*** 
> 
> > 
> > Horace, your posts are tightly focused, high on content and relevant
> > logic, and are excellent food for thought. I, for one, appreciate them
> > very much. Thank you! 
> > 
> > --Mitchell Jones
> > 
> > ===========================================================
> 
> Thanks for the support much needed by this newbie.
> 
> Horace
> 
> -- 
> Horace Heffner 907-746-0820    <hheffner@matsu.ak.net>
> PO Box 325 Palmer, AK 99645

***{At this point, I would like to toss out what I call the protoneutron
theory of "cold fusion." If this theory is correct, then "cold fusion"
ain't fusion! 

Note that the drift of the H+ ions into the cathode does not require a
free path through unoccupied unit cells. In an occupied unit cell, the
occupant is a neutral H atom. Thus there is no voltage gradient barring an
H+ ion from drifting into an occupied unit cell, and there is no voltage
gradient preventing an outward drifting electron from passing into an
occupied unit cell, either. This means it is purely a matter of
statistical probability, regarding the question of whether a drifting H+
meets an electron in an occupied unit cell. If it does, it will attempt to
"pop" into a neutral hydrogen atom. Unfortunately, because the unit cell
is already occupied, when the second H atom attempts to form it will not
have sufficient room for its electron at even its innermost Bohr orbit.
Result: the electron will spiral down toward the nucleus, where it will
linger at grazing altitude in a particle form which we may term a
*protoneutron.* Why doesn't it become an actual neutron? The answer: in
the reaction p + e --> n, the mass of the proton is 1.67239E-24 grams
while that of the electron is .00091E-24 grams. Thus the sum of p + e on
the left side of the equation is 1.6733E-24 grams. On the right side of
the equation, the mass of the neutron is 1.6747E-24 grams. Since the
neutron outweighs the proton plus the electron by .0014E-24 grams, which
equates to .7875 Mev, this amount of energy must be available from
somewhere in order for the protoneutron to convert into a real neutron.
[Note: to convert .0014E-24 grams to ergs, we multiply by the square of
the velocity of light in cm per sec, giving: (.0014E-24)(9E+20) = 
.0126E-4 ergs. Since 1 electron-volt equals 1.6E-12 ergs, it follows that
this reaction absorbs  .0126E-4/1.6E-12 = .7875E+6 electron-volts, which
is .7875 Mev per neutron created.] Since this amount of energy is unlikely
to be available, the bizarre and unstable protoneutron will only endure
for a tiny fraction of a second, until it bounces into an unoccupied unit
cell. At that point, it will expand its radius like an exploding piece of
popcorn, and become a hydrogen atom trapped in that unit cell. 

Note, however, that such events have a cumulative effect: if electrons
tend to meet protons in a particular region of the lattice, all of the
adjacent unit cells in that region are quickly going to become occupied,
and the size of that continuously occupied region is going to grow
inexorably. Result: protoneutrons that are formed in that region will
endure for longer and longer, before finally bouncing into an unoccupied
unit cell adjacent to the region and "popping" into hydrogen atoms.
Eventually the size of the "loaded" region will become so immense,
relative to the slow moving, uncharged protoneutrons, that they will begin
to accumulate in the region. Why do they persist in this bizarre
netherworld state, as "protoneutrons?" For two reasons: (1) Enough thermal
energy is available for them to "pop" into hydrogen atoms, but that low
energy state requires a lot of space--more space than is available in the
loaded region of the lattice which they occupy. (2) Enough space is
available for them to collapse into neutrons, but that state requires more
energy than is available. Result: thermal protoneutrons inside large,
continuously loaded regions of the lattice will simply endure, desperately
hungering for the energy or the space they need to enter a more stable
state. 

This means that protoneutron loading will take place in subregions of the
lattice that are fully packed with hydrogen atoms. First there will be
one, then two, then ten, then hundreds, then thousands, then millions of
protoneutrons trapped in the region, *and all will be desperately awaiting
some event capable of supplying a .7875 Mev jolt in the right place.* They
have no prospects of being transformed into hydrogen atoms, because that
is a low energy-high space transformation, and the space is not available.
Thus their only allowed transformation is the low space-high energy
transformation into neutrons. When such a transformation finally happens,
the loaded palladium electrode will "turn on," and begin to produce
"excess heat."

To understand the production of "excess heat," assume that in a subregion
of the lattice in which there are millions of trapped protoneutrons, one
of them finds the energy it needs to become a neutron. Result: a thermal
neutron will be created, and will quickly drift into a nearby nucleus. If,
for example, it drifts into an H nucleus, the result is D plus a 2.22 Mev
gamma. The gamma, however, will not exit the lattice, because it is
emitted in a region packed with protoneutrons, *and it is my conjecture
that protoneutrons are so rapacious in terms of their demand for energy of
transformation that they will absorb all of the gamma's energy before it
can exit from the lattice.* This means that the 2.22 Mev gamma will
transform 2 protoneutrons into neutrons, and, in addition, will give a
third protoneutron about .645 Mev of kinetic energy. Naturally, the
resulting fast protoneutron will fly through the lattice until it
encounters an empty unit cell, where it will virtually instantly "pop"
into a hydrogen atom. But remember: the hydrogen atom is too large to move
freely in the lattice. It will thus slam into the opposite side of the
unit cell in which it forms,  and deliver its .645 Mev to the lattice as
heat. As for the two thermal neutrons that were created, they will quickly
drift into nearby nuclei, producing isotope shifts and more
electromagnetic emissions. When those emissions are insufficient to
transform a protoneutron into a neutron, they give it kinetic energy, and
cause it to fly through the lattice until it comes to an unoccupied unit
cell, where it pops into a hydrogen atom, slams to a stop, and delivers
its energy to the lattice in the form of heat. When the emissions are
energetic enough to transform protoneutrons into neutrons, on the other
hand, they will trigger additional neutron creation, which in turn will
produce still more isotope shifts, and so on. In short, a kind of chain
reaction will begin within the subregion of the electrode that is packed
with protoneutrons. That reaction will have three odd characteristics: (1)
It will be self-controlling: if the lattice begins to overheat, it will
deload hydrogen, which will reduce the population of protoneutrons, which
in turn will slow the reaction back down. (2) The reaction will be self
sustaining: once large numbers of thermal neutrons are afoot in the
lattice, a constant sprinkling of gamma emissions will be assured, thereby
continuing the process. (3) There will be no radiation hazard: radiation,
for practical purposes, cannot escape from a lattice that contains a
substantial population of protoneutrons.        

Anyway, that's enough for now. I am not going to go into the various
isotope shifts which can result from this process, nor into the energies
of the associated gammas, betas, etc., that can result. Instead, I am
going to offer a simple prediction which can be verified or falsified: in
order for the protoneutron theory to be correct, a palladium lattice
loaded with protoneutrons must have an abnormal hunger for energy, and an
abnormal ability to suppress internally emitted gammas. Otherwise, the
experimenters who have touched off "cold fusion" would all be dead. This
can be explained in only two ways: either the presence of nearby
protoneutrons absorbs the energy before the gammas can form--i.e., before
the photons can attain the speed of light--or else they absorb it after
the gammas have formed but before they exit the lattice. To test this,
here is what I propose: 

(1)To determine whether the energy is absorbed *after* gammas form, load a
palladium cathode to the point where "cold fusion" begins, and then
bombard it with gamma radiation. In that case: (a) a bizarre and extreme
increase in gamma absorption will be noted, by comparison to measurements
taken with the same cathode prior to loading; (b) it will be possible to
"turn on" loaded cathodes using gammas that have enough energy to turn
protoneutrons into neutrons. (Note: lots of other predictions could be
made here--e.g., that it should be possible to turn on loaded cathodes by
supplying them with thermal neutrons--but I am deliberately trying to
restrain myself.)

(2) To determine if the energy is absorbed *before* gammas form, supply
the cathode with thermal neutrons before loading, measure the gammas
produced, and then repeat the procedure with a loaded cathode that is
producing anomalous heat. Far more gammas should be emitted from the
cathode in the first case than in the second.   

Now, to a different point: earlier I said that if the protoneutron theory
is correct, then "cold fusion" isn't fusion. The reason is simple: fusion
refers to the combination of the nuclei of different atoms to form atoms
of yet another kind. Since an electron isn't the nucleus of an atom, the
combination of an electron and a proton to form a neutron isn't fusion.
And, since a neutron isn't a nucleus, when a neutron combines with a
preexisting atomic nucleus to form a different isotope, that, too, is not
fusion. Bottom line: if the protoneutron theory is correct, then "cold
fusion" ain't fusion!

I would add that I am perfectly aware of the conflict between the
protoneutron theory and "quantum mechanics." This is a classical
mechanical theory, and classical mechanics is based on the principle of
continuity--i.e., the principle that no entity may come into existence out
of nothing or vanish into nothing. The principle of continuity indicates
that an entity can arrive at, or leave, a location in one way only: by
successively occupying each position in a continuous spatial pathway to,
or away from, that location. The implication is that motion is continuous,
not a series of quantized "jumps." This means, for example, that an
electron moving from one stable orbit to another does not "jump" in the
quantum mechanical sense: it follows a continuous spatial pathway from the
former orbit to the latter, and it exists just as surely during those
instants when it is between the "preferred" orbits, as when it is in one
of them. In the case presently under discussion, the principle of
continuity means that an electron exists and follows a continuous spatial
pathway when it is transiting from the lower Bohr orbit to a position in
the nucleus. It does not merely exist "in" the nucleus and "in" the
innermost Bohr orbit, but also in between. The protoneutron, in short, is
a classical mechanical phenomenon. It violates the most fundamental
precept of "quantum mechanics" and, if it exists, it constitutes one more
piece of evidence, in a pile that is already mountainous, indicating that
motion in the microcosm is not "quantized" and, hence, that "quantum
mechanics" is wrong.    

Needless to say, I present the protoneutron theory as a hypothesis subject
to verification or falsification. If it proves to be indefensible, then I
will abandon it. On that, you have my absolute guarantee. --Mitchell
Jones}***

===========================================================

===========================================================
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------------------------------
1995.09.23 / Darin Vrancic /  Re: ? Singularity Technology ?
     
Originally-From: darin@iglou.com (Darin Vrancic)
Newsgroups: sci.physics.fusion
Subject: Re: ? Singularity Technology ?
Date: Sat, 23 Sep 1995 10:09:31 GMT
Organization: IgLou Internet Services (1-800-436-4456)

zoltanccc@aol.com (ZoltanCCC) wrote:



>I have an idea. Since the most common isotope of Lithium is Li7, this
>substance is cheap. If you add a neutron to Li7 you get Li8 which is
>exactly two alphas. 
We see that in your later posts you noticed your isotopic error. I
only included this part for reference.

>I don't know if the energy losses in hot fusion reactors come mainly from
>neutrons leaving the plasma core, but I suppose you could try blanketing
>the equipment with massive volumes of Li7. The neutrons that escape from
>the plasma core would be absorbed by the Li blanket and the reaction:

>n + Li7 --> He4  +   He4                       (1)

>would produce plenty of heat. The key is to suppress the alternative
>reaction:

>n + Li7  -> Li6 + n + n (endothermic)     (2)

>which would absorb the heat. The He4 produced in the first reaction would
>be fast and the equipment would have to be designed to slow it down to
>convert the energy to heat.

>Now this would not be strictly a hot fusion reactor that breaks even but
>it would sure help the total energy output of the system. In fact in terms
>of total energy in minus total energy out it may break even due to the
>Lithium fission.

>Of course this may not be my original idea, it is kind of obvious.

	No, this is not original to you, it goes back at least 15 years,
though the most common purpose quoted for a lithium blanket is to
absorb the neutrons to keep them from exiting the reactor and becoming
a radiation hazard. These designs usually use the Lithium in liquid
form to capture the gammas as well as the neutrons in order to
transfer the released energy from the core to a boiler to produce
electricity. The problem with expecting this reaction to increase the
energy efficiency of a fusion reactor, however, is that for it to do
so, the fusion reaction must be generating enormous numbers of
neutrons, which of course means it is generating large amounts of
power already. Not that it doesn't make sense to increase efficiency
any way one can, but reactions currently need to be increased by 2-4
orders of magnitude before the technology even begins to be
commercially feasible.

	Now, these cold fusion claims, if validated once and for all, would
change the energy landscape even more than hot fusion would, since
they hold the promise of small energy providers (companies, even
individual households, eventually providing their own power, with
power companies relegated to being no more than maintenance
providers).

	Some of you might wish to check out the RGeorge Home Page, as he
discusses a "cold" fusion technique which may have been discussed
here, but if it was I missed it. It also supposedly produces no free
neutrons and no high-energy photons. He is well spoken and is working
on patents of his own, and you may be surprised at what you find
there.


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------------------------------
1995.09.23 / Mitchell Jones /  Re: Testing the Mitchell Jones Hypothesis
     
Originally-From: 21cenlogic@i-link.net (Mitchell Jones)
Newsgroups: sci.physics.fusion
Subject: Re: Testing the Mitchell Jones Hypothesis
Date: Sat, 23 Sep 1995 02:19:36 -0500
Organization: 21st Century Logic

In article <43srd2$3kko@lamar.ColoState.EDU>, gelfand@lamar.ColoState.EDU
(Martin Gelfand) wrote:

> I feel compelled to comment on just one of the ravings
> in the long post by M.J....
> In article <21cenlogic-2009951048240001@austin-1-4.i-link.net>
21cenlogic@i-link.net (Mitchell Jones) writes:
> >
> >By the way, speaking of ether theory, are you aware that the existence of
> >the ether was proven more than 60 years ago, and is generally accepted by
> >top physicists today? I'll bet not. That's because you buy into the
> >dissembling and the crapola. You don't notice that every knowledgeable
> >physicist today accepts the existence of "Dirac's ocean"--a particulate
> >medium that pervades all of space and accounts for most of the phenomena
> >of electromagnetism. Of course, they have all agreed to not *call* it the
> >ether, because that would have required them to clash head-on with another
> >dogma: that of Einstein's hare-brained "relativity" theory. But behind the
> >political dissembling lies the reality of classical mechanics triumphant,
> >in all areas of physics. The myth of its demise is a politically expedient
> >fantasy, nothing more. And soon even that will be gone.  
> >
> >--Mitchell Jones
> 
> The statement 'every knowledgeable physicist today accepts 
> the existence of "Dirac's ocean"' is about forty to fifty years out
> of date, now.  The "Dirac sea" of negative energy electrons was
> a necessary construct back in the days when the Dirac equation
> was the leading theory of relativistic quantum mechanics for
> electrons.  However, with the advent of relativistic quantum field 
> theory a Dirac sea is no longer required, and nobody believes
> in its existence.

***{Martin, the issue here is not what present day physicists will *say*
if asked whether they believe in an "ether" or in "Dirac's ocean." What
they will say is influenced by non-rational considerations (e.g., the
desire not to be laughed at or labeled a crackpot). The question, instead,
has to do with whether an observer who is not concerned about peer group
acceptance might reasonably say that they believe in such things as an
ether or Dirac's ocean.

The problem that Dirac solved was simply that during the period from
roughly 1900 to 1930, the rejection of the classical ether theories
hardened into an enshrined, sacred dogma. A climate of fear existed in
physics--fear of being laughed at--and physical thought was crippled by
the determination to not treat "space" as an active agent in problem
solving. Dirac ended that by putting forth his notion of an "ocean of
extraordinary electrons" which filled all of space, and by making accurate
predictions (e.g., the existence of the positron) based on that view. Of
course, he denied that his theory was an ether theory. Not even he dared
to use such a loaded term. But he did take on, and overturn, the
associated taboo against imagining that a substance filled all of "space."
Because of him, it became OK once again, as it had been in the 19th
century, to imagine that the vacuum was not empty. What had been a taboo
against a certain kind of thought was, by him, reduced to a taboo about
the use of a word. That, in my view, was a significant accomplishment.

In summary, you are right to say that nobody believes in "Dirac's ocean,"
if by that you mean that few physicists today would use those words.
However, every person today who has thought deeply about the problems of
electromagnetism recognizes that "the vacuum" is a complex substance with
myriad properties. Dirac, by removing a taboo, made that possible. Even
you, I dare say, would not claim that space is "empty" in the sense in
which the term was used in the early years of this century! --Mitchell
Jones}***      
> 
> And even worse, from Mitchell's viewpoint, is that even if
> you wanted to believe in the existence of the Dirac sea it
> wouldn't serve as a classical ether.  After all, the Dirac
> equation _is_ a relativistic theory.

***{You fail to understand that the curve fitted mathematics associated
with "relativity theory" can be interpreted classically. One need not, for
example, treat Einstein's t' as indicating time distortion. An accelerated
clock may simply run slower than a clock that is not accelerated. Why?
*Perhaps because the pressure exerted by particles of "Dirac's ocean"
(i.e., "the ether") moving between the atoms of the clock causes it to run
slowly!* Moreover, similar alternative interpretations are available for
all aspects of the mathematics used by Einstein. Just as there is nothing
in the mathematics of "quantum mechanics" that requires the "Copenhagen
interpretation," so there is nothing in the mathematics of "relativity"
that requires Einstein's interpretation. (The math, as I have said
repeatedly, doesn't give a hoot in hell how it is "interpreted.") Bottom
line: since Einstein's mathematics can be interpreted classically, the
fact that Dirac incorporated it into his own theories does not imply that
his "ocean" cannot serve as a classical ether. --Mitchell Jones}*** 
> 
> 
> Martin Gelfand
> Department of Physics,
> Colorado State University

===========================================================
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------------------------------
1995.09.23 /  Politics@usa.c /  Proposed State & Federal Regulations for the INTERNET!
     
Originally-From: Politics@usa.com
Newsgroups: sci.physics.fusion
Subject: Proposed State & Federal Regulations for the INTERNET!
Date: 23 Sep 1995 07:31:58 GMT
Organization: Wisconsin Internet Services

  My name is Scott Glasrud, and I am running for the New Mexico State Senate
during the 1996 elections. One of the reasons I have chosen to run is to combat 
the 
proposed state and federal regulations of the Internet.  As you know, the 
Internet 
was never designed to be regulated!  It was designed to allow communications in
the event of anuclear war or a major catastrophe. I OPPOSE REGULATION, and if 
elected
will fight to preserve your constitutional rights. HOWEVER, I NEED YOUR HELP!

I am asking each person who reeives this message to send $5.00 to the 
Scott Glasrud Campaign Committee.  If we pull together, we CAN protect our first
amendment rights!  HELP ME show the politicians the POWER behind this
important NETWORK.  Please send contributions to:

                                    The Scott Glasrud Campaign Committee
                                    11024 Montgomery Blvd. NE, Suite 179
                                         Albuquerque, New Mexico  87111



Thank you!

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1995.09.22 / MARSHALL DUDLEY /  Re: Pathological Skepticism!
     
Originally-From: mdudley@brbbs.brbbs.com (MARSHALL DUDLEY)
Newsgroups: sci.physics.fusion
Subject: Re: Pathological Skepticism!
Date: Fri, 22 Sep 1995 16:20 -0500 (EST)

"Alan M. Dunsmuir" <Alan@moonrake.demon.co.uk> writes:
 
-> Hydrogen - created by the solar-powered hydrolysis of sea water - will
                                           ^^^^^^^^^^
Huh?  I show hydrolysis to mean a chemical reaction in which a substance reacts
with water to produce another substance, such as starch & water to make
glucose.  I am not sure if you are misusing a term here, or if there is another
definition of which I am unaware.
 
Please explain exactly what solar-powered hydrolysis is.  If you are talking
about a process that converts water to hydrogen and oxygen directly from
sunlight I would be very interested in knowing how to do that.  If you can cite
any papers that would even be better.
                                                                Marshall
 
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------------------------------
1995.09.23 / Alan M /  Re: Pathological Skepticism!
     
Originally-From: "Alan M. Dunsmuir" <Alan@moonrake.demon.co.uk>
Newsgroups: sci.physics.fusion
Subject: Re: Pathological Skepticism!
Date: Sat, 23 Sep 1995 07:48:26 +0100
Organization: Home

In article: <USE2PCB336859156@brbbs.brbbs.com>  mdudley@brbbs.brbbs.com 
(MARSHALL DUDLEY) writes:
> Huh?  I show hydrolysis to mean a chemical reaction in which a 
substance reacts
> with water to produce another substance, such as starch & water to 
make
> glucose.  I am not sure if you are misusing a term here, or if there 
is another
> definition of which I am unaware.
>
Sorry - finger/brain trouble. For 'hydrolysis' read 'electrolysis'. 
-- 
Alan M. Dunsmuir [@ his wits end]     (Can't even quote poetry right)

         I am his Highness' dog at Kew
         Pray tell me sir, whose dog are you?
			      [Alexander Pope]

PGP Public Key available on request.


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------------------------------
1995.09.23 / Alan M /  Re: Testing the Mitchell Jones Hypothesis
     
Originally-From: "Alan M. Dunsmuir" <Alan@moonrake.demon.co.uk>
Newsgroups: sci.physics.fusion
Subject: Re: Testing the Mitchell Jones Hypothesis
Date: Sat, 23 Sep 1995 07:50:01 +0100
Organization: Home

Mitchell Jones seems to have developed paralysis of the fingers 
regarding his 'slower than light' photons. Whatever can this mean?
-- 
Alan M. Dunsmuir [@ his wits end]     (Can't even quote poetry right)

         I am his Highness' dog at Kew
         Pray tell me sir, whose dog are you?
			      [Alexander Pope]

PGP Public Key available on request.


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------------------------------
1995.09.24 / Bruce Simpson /  Re: French nuclear test agenda
     
Originally-From: bruce@faxmail.co.nz (Bruce Simpson)
Newsgroups: sci.physics.fusion
Subject: Re: French nuclear test agenda
Date: Sun, 24 Sep 1995 06:17:49 GMT
Organization: FaxMail Technologies

Mario Pain <pain@drfc.cad.cea.fr> wrote:

>bruce@faxmail.co.nz (Bruce Simpson) wrote:

>>But France has repeatedly claimed that there is *NO* risk associated
>>with these tests.  If this is indeed the case then it matters not
>>whether the devices are tested in downtown Paris or anywhere else for
>>that matter.  You can't claim total safety on one hand and "reduced
>>risk" on the other.
>>
> I do not think France has ever claimed there was NO risk associated 
>with this tests. Such a claim will of course be absurd, as there is
>no human act which has "zero risk". What is claimed is that this risk
>is "in the noise", that is, the probability of an accident happening is
>negligible when compared with any other possible accident. To be schematic,
>a polynesian has more chance of being eaten by an alligator when getting
>out of bed than to be affected by this tests.

But that is a risk that the polynesian is well aware of and accepts as
part of his daily life.

French nuclear testing in the South Pacific is somewhat akin to New
Zealand placing a nuclear device 100Km off the French coast and then
connecting the trigger mechanism to a huge "wheel of fortune" device
which gave a one in ten million chance of detonation.  This is a
*very* low risk - but would it be acceptable?  The French would surely
say "why not do this in your own country?" or perhaps "why endanger
the lives of Frenchmen when there is no benefit to us associated with
the risk?".  This is all we are saying.

With the nuclear testing at Mururoa the benefit is only to France but
the risk is shared by many other countries.  I do not accept that
France has any right to endanger the lives of myself or any of my
friends and familly simply to ensure its own sovereignty.

> That does not mean of course that France should not take any measure
>which will reduce both the probability of a nuclear accident, and the
>potential effect of it. In that sense, the Mururoa site is the best of
>the available ones.

*best* for who?  Certainly not the inhabitants of the South Pacific.
France's attitude is not only arrogant but selfish.


*----[Fixed-price software development over the net ]----*
|     bsimpson@iprolink.co.nz or bruce@faxmail.co.nz     |
*--[C/C++, Win, OS/2, POSIX, device-drivers, fax, comms]-*

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------------------------------
1995.09.23 /  beefjerk /  cmsg cancel <440d5e$a91@newsie.wis.com>
     
Originally-From: Politics@usa.com (beefjerk)
Newsgroups: sci.physics.fusion
Subject: cmsg cancel <440d5e$a91@newsie.wis.com>
Date: Sat, 23 Sep 1995 05:40:43 GMT
Organization: Wisconsin Internet Services


Cancelled-This post is inappropriate for this group.

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------------------------------
1995.09.23 / John Logajan /  Re: Seeking explanation for observed events (revisited)
     
Originally-From: jlogajan@skypoint.com (John Logajan)
Newsgroups: sci.physics.fusion
Subject: Re: Seeking explanation for observed events (revisited)
Date: 23 Sep 1995 14:02:40 GMT
Organization: SkyPoint Communications, Inc.

Marcus Leech (mleech@bnr.ca) wrote:
: >   After a few experiments, we concluded that it
: >   had to do with the compression of the packed-granular snow at the site.
: >   Every time the snow made a 'crunching' noise, we observed broad-spectrum
: >   RF noise on the spectrum analyser.
: > 
: > Is this actually related to the compression of the snow? And if so, how?
: >   I can concieve of both a small-scale piezoelectric mechanism and
: >   a static-discharge mechanism.  On the day we did this, it was very
: >   cold (-15C), and dry.


Airborne snow, of course, carries static charge -- we often see lightning
strikes at a nearby TV antenna tower in the dead of winter just after
a snowfall starts.  I've also seen a snow fall generate high static
charge in a small antenna I had mounted on the roof.  The coaxial cable
coming into the house would "twitch" from the changing high potential
and sparks would periodically jump from the connector tip to the connector
ground sheath.  I ended up installing a permanent bleeder resistor to
prevent possible damage to my radio equipment.

I don't know if snow packed on the ground would be as effective at isolating
charge.

Another mechanism to consider is the material of the boot contacting and
seperating from the snow.  Just last week I was holding a 0-scope probe
when I got up off my cloth chair -- that created a voltage spike.  Then
I noticed that stomping my feet caused a voltage spike.  Lifting my foot
off the carpet caused a polarity spike, and placing my foot down caused
the opposite polarity spike.


--
 - John Logajan -- jlogajan@skypoint.com  --  612-633-0345 -
 - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA -
 -   WWW URL = http://www.skypoint.com/members/jlogajan    -
cudkeys:
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1995.09.23 /  ZoltanCCC /  Re: I have a fusion reactor in my garage!!!!!!!
     
Originally-From: zoltanccc@aol.com (ZoltanCCC)
Newsgroups: sci.physics.fusion
Subject: Re: I have a fusion reactor in my garage!!!!!!!
Date: 23 Sep 1995 11:43:02 -0400
Organization: America Online, Inc. (1-800-827-6364)

I would appretiate it if you could email me a description of your reactor.
If you want to use regular mail I provide my address:

Zoltan Szakaly
11517 Dellmont dr
Tujunga CA 91042
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1995.09.23 /  ZoltanCCC /  Re: I have a fusion reactor in my garage!!!!!!!
     
Originally-From: zoltanccc@aol.com (ZoltanCCC)
Newsgroups: sci.physics.fusion
Subject: Re: I have a fusion reactor in my garage!!!!!!!
Date: 23 Sep 1995 11:43:36 -0400
Organization: America Online, Inc. (1-800-827-6364)

We would all appretiate a brief description here in this newsgroup.

Zoltan Szakaly
cudkeys:
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1995.09.23 / Mitchell Jones /  Re: Testing the Mitchell Jones Hypothesis
     
Originally-From: 21cenlogic@i-link.net (Mitchell Jones)
Newsgroups: sci.physics.fusion
Subject: Re: Testing the Mitchell Jones Hypothesis
Date: Sat, 23 Sep 1995 10:50:29 -0500
Organization: 21st Century Logic

In article <402216017wnr@moonrake.demon.co.uk>, Alan@moonrake.demon.co.uk wrote:

> Mitchell Jones seems to have developed paralysis of the fingers 
> regarding his 'slower than light' photons. Whatever can this mean?
> -- 
> Alan M. Dunsmuir [@ his wits end]     (Can't even quote poetry right)
> 
>          I am his Highness' dog at Kew
>          Pray tell me sir, whose dog are you?
>                               [Alexander Pope]
> 
> PGP Public Key available on request.

Alan, you are a glutton for punishment. The fact is that I had already
replied to your post before you made it, in a reply to Bill Rowe which you
either didn't read or else conveniently forgot. Frankly, re-posting
material for the benefit of people who didn't read it when I posted it the
first time gets a low priority with me. I prefer to respond to new
comments. However, for your benefit, the relevant excerpts from my reply
to Bill Rowe follow:

 -------------------------------------------
"At this point, we have to ask ourselves how a gamma photon gets spat out
as the isotope shifted nucleus drops to ground state. The question is, do
we visualize a series of transitional states in which the photon
accelerates to the speed of light, or not? By the "quantum mechanical"
mode of thinking, there is not merely no problem in imagining an entity
leaping from zero velocity to the speed of light without passing through
all of the intervening transitional states, but such images are positively
to be encouraged! However, as I have noted elsewhere, such imagined gaps
in motion violate the principle of continuity. However brief the velocity
rise time may be, the principle of continuity requires that the photon
pass through all the intervening states of motion. Period. The fact that
we have not measured such states does not mean they do not exist. Indeed,
they must exist because the principle that requires their existence (the
principle of continuity) supports the entire structure of human knowledge.
To question its validity self-reduces to an absurdity.

By the way, please do not assume that I am alone in visualizing such
transitional states. Even physics gurus routinely do it. Unlike me,
however, they are careful to kiss the behinds of the powers that be by
pretending to believe in the sacred mumbo-jumbo of "quantum mechanics."
For example--and it is one among many--check out the various "dressing
transformations" that have been applied to the original Hamiltonian to
curve fit it to newly emerging experimental data. If memory serves, Pauli
applied such a transformation to explain the Biot-Savart law. His
interpretation was that the magnetic field which accompanies a moving
charged particle consists of a cloud of "bound photons" which move along
with it. (Yes, Virginia, photons don't always move at the speed of light!)"
 -------------------------------------------
[End of quoted material]

So, Alan, are you still laughing? If so, then I must note that you are not
merely laughing at me, but also at Wolfgang Pauli! Pretty funny, huh? 

--Mitchell Jones

===========================================================
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1995.09.23 / A Plutonium /  Re: First experimental predictions of Hydrogen Atom Systems
     
Originally-From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Re: First experimental predictions of Hydrogen Atom Systems
Date: 23 Sep 1995 23:27:02 GMT
Organization: Plutonium College

In article <441g71$842@dartvax.dartmouth.edu>
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:

>   Physicists for the past 40 years have not been truthful to
> themselves. They should realize that when a particle exists for such a
> fleeting time as 10^-6 sec that such a particle is really not on "par"
> to the existence of say the proton or electron etc.
>    To put it on terms that all readers can understand. If we were to
> give a comprehensive science study of Homo sapiens, should we spend any
> time whatsoever or some little time on the fact that a worm lives in
> the eyelashes of humans, considering that we are really talking about
> Homo sapiens and not an organism that lives on Homo sapiens? In the
> same vain, is the muon a particle on par with the electron, or is it a
> energy exchange, that is, the muon is merely an electron with stored up
> extra energy and not a true particle. 

  Here is a better analogy of the present day particle physics zoo.
Suppose particle physics was the start of the classification of life, a
la Linneas.

  How would present day particle physicists have done a job on the
starting out of the classification of life and models thereof.  Well we
have the Plant Kingdom and the Animal Kingdom and we have things like
the archopods and mammalia and invertibrates etc.  That was just a
passing trespass of the biology classification so some biologist do not
holler at me for it. BTW I need a complete biology classification
scheme so if someone has it please post.

  But getting to the analogy of transporting today's particle zoo
physicists back to the time of Linneas and ask them to work up a
classification. Then I suspect what they would include is bread loaves
in grocery stores because some particle zoo physicist noticed that
people eat bread and so concluded that bread was alive. And so the
classification of life by our present day particle zoo physicists would
have been a far larger scheme to include every imaginable brand of
bread and lollipops and suckers would have their space I am sure.
cudkeys:
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------------------------------
1995.09.22 / James Panetta /  Re: Testing the Mitchell Jones Hypothesis
     
Originally-From: panetta@finch.SLAC.Stanford.EDU (James H. Panetta)
Newsgroups: sci.physics.fusion
Subject: Re: Testing the Mitchell Jones Hypothesis
Date: Fri, 22 Sep 1995 02:56:00 GMT
Organization: Stanford Linear Accelerator Center

Hi All,

Mitchell sent me a copy of the original protoneutron post.  I replied
to him via email, and he has encouraged me to post it to sci.physics.fusion.

/**************start included message*********************/

Hi Mitchell,
> You mentioned that you had not seen the original post. It follows. It
> should answer most of the questions in your recent post. The one about
> photons below lightspeed is addressed in a reply to Bill Rowe, already
> posted. If you have further questions, feel free to ask.

Thanks for the prompt reply.  I have a few comments on some of your
physics.

> ***{To repeat: I do not see a need for an alternative model. A voltage
> gradient is all that is required to attract an H+ ion into the unit cells
> of the cathode. It doesn't have to be "forced" in: the openings in the
> triangular faces of the unit cells are large enough for it to move in
> *very* easily--as easily as a fly into an amphitheater, in fact. Remember:
> an H+ is simply a proton--i.e., the nucleus of a hydrogen atom. The
> standard formula, curve-fitted to the experimental data, for computing the
> radius of a nucleus is simply r =3D (1.37E-13)A^(1/3), where r is the radius
> in cm and A is the mass number. For H+ the mass number (number of protons
> plus number of neutrons) is 1, so r =3D 1.37E-13 cm =3D .0000137 A.  Once
> inside the cathode, such an H+ ion will drift toward the electron source
> until it acquires an orbital electron. At that point, it will become an
> electrically neutral H atom, with an effective radius as per the innermost
> Bohr orbit (.53 =C5 or slightly larger). This means, roughly speaking, that
> its radius will increase by four orders of magnitude when it captures an
> electron, and it will become trapped in the unit cell which it is
> occupying at that time. --Mitchell Jones}***

You really can't use the proton radius to determine the size of a
H+ ion, you should determine the radius at which the coulomb force
is powerful enough to remove an electron from a neighboring atom.
This distance is probably not less than half of the Bohr radius.

> ***{At this point, I would like to toss out what I call the protoneutron
> theory of "cold fusion." If this theory is correct, then "cold fusion"
> ain't fusion!
>
> Note that the drift of the H+ ions into the cathode does not require a
> free path through unoccupied unit cells. In an occupied unit cell, the
> occupant is a neutral H atom. Thus there is no voltage gradient barring an
> H+ ion from drifting into an occupied unit cell, and there is no voltage
> gradient preventing an outward drifting electron from passing into an
> occupied unit cell, either. This means it is purely a matter of
> statistical probability, regarding the question of whether a drifting H+
> meets an electron in an occupied unit cell. If it does, it will attempt to
> "pop" into a neutral hydrogen atom. Unfortunately, because the unit cell
> is already occupied, when the second H atom attempts to form it will not
> have sufficient room for its electron at even its innermost Bohr orbit.
> Result: the electron will spiral down toward the nucleus, where it will
> linger at grazing altitude in a particle form which we may term a
> *protoneutron.*

First problem, electrons do not *spiral* in to the nucleus.

>                 Why doesn't it become an actual neutron? The answer: in
> the reaction p + e --> n, the mass of the proton is 1.67239E-24 grams
> while that of the electron is .00091E-24 grams.

Second problem, p+e-->n violates lepton number conservation.  Where's
the neutrino?

Third problem, electron capture is known to proceed via W exchange.
This process is well known, and calculable.  The theory fits the known
data.  What prediction would your protoneutron theory give for
sodium-22 EC?  Iron-55?

Fourth problem, if protoneutrons exist, they will cause an effect in
the endpoint energy of tritium beta decay.  The endpoint curve is
extremely well known due to the search for massive neutrinos.
What is this effect?

> I would add that I am perfectly aware of the conflict between the
> protoneutron theory and "quantum mechanics." This is a classical
> mechanical theory, and classical mechanics is based on the principle of
> continuity--i.e., the principle that no entity may come into existence out
> of nothing or vanish into nothing. The principle of continuity indicates
> that an entity can arrive at, or leave, a location in one way only: by
> successively occupying each position in a continuous spatial pathway to,
> or away from, that location. The implication is that motion is continuous,
> not a series of quantized "jumps."

Then explain the photoelectric effect.

>                                    This means, for example, that an
> electron moving from one stable orbit to another does not "jump" in the
> quantum mechanical sense: it follows a continuous spatial pathway from the
> former orbit to the latter, and it exists just as surely during those
> instants when it is between the "preferred" orbits, as when it is in one
> of them. In the case presently under discussion, the principle of
> continuity means that an electron exists and follows a continuous spatial
> pathway when it is transiting from the lower Bohr orbit to a position in
> the nucleus. It does not merely exist "in" the nucleus and "in" the
> innermost Bohr orbit, but also in between. The protoneutron, in short, is
> a classical mechanical phenomenon. It violates the most fundamental
> precept of "quantum mechanics" and, if it exists, it constitutes one more
> piece of evidence, in a pile that is already mountainous, indicating that
> motion in the microcosm is not "quantized" and, hence, that "quantum
> mechanics" is wrong.

What evidence is there that quantum mechanics is wrong?  All evidence I
have seen and *performed* experiments to examine has shown that quantum
mechanics fits the data.

> Needless to say, I present the protoneutron theory as a hypothesis subject
> to verification or falsification. If it proves to be indefensible, then I
> will abandon it. On that, you have my absolute guarantee. --Mitchell
> Jones}***

I think that protoneutrons are indefensible, but that's the opinion of
a particle physicist at Caltech and you may take it how you wish.

Thanks again for responding quickly.

  --Jim Panetta


-- 
--
My opinions are mine...not SLAC's...not Caltech's...not DOE's...mine.
(except by random, unforseeable coincidences)
panetta@cithex.caltech.edu   panetta@slacvm.slac.stanford.edu
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------------------------------
1995.09.23 /   /  Re: The electron capture theory of cold fusion
     
Originally-From: mrichar353@aol.com (MRichar353)
Newsgroups: sci.physics.fusion
Subject: Re: The electron capture theory of cold fusion
Date: 23 Sep 1995 21:37:10 -0400
Organization: America Online, Inc. (1-800-827-6364)

zoltanccc@aol.com (ZoltanCCC) wrote:

>In article <4409ec$nlq@newsbf02.news.aol.com>, mrichar353@aol.com
>(MRichar353) writes:
>
>>
>>Oh, come on. I have refuted Robin's idea conclusively, at least for
>people
>>who know physics at an undergraduate level. What you and he are talking
>>about violates energy conservation, it's as simple as that.
>>
>>
>
>Mark, unfortunately your half cocked reference to the conservation of
>energy did not cut it with me. Up to this point nobody has described the
>situation as the way I see it now. Here it follows:
>
  [snip of an attempt to classically describe electron capture, followed
by the statement that the previously-proposed method won't work]

Any classical description of the process is perforce intrinsically
inconsistent, and this is probably the source of the confusion in your
theory. I'm glad you were still able to derive the fact that energy can't
come from nowhere. The nice thing about conservation laws is that they are
independent of (although consistent with) the underlying details.

Sorry about my tone in my previous post, I was in a foul mood. Although
you are a "creative thinker" you are not a crank, as you are able to
profit from constructive criticism.

Mark Richardson
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1995.09.23 /   /  Re: Mark Richardson, Mr. Dupree, and Mr. Esch
     
Originally-From: mrichar353@aol.com (MRichar353)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Re: Mark Richardson, Mr. Dupree, and Mr. Esch
Date: 23 Sep 1995 21:44:17 -0400
Organization: America Online, Inc. (1-800-827-6364)

Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote:

>In article <43tc0g$q4s@newsbf02.news.aol.com>
>mrichar353@aol.com (MRichar353) writes:
>
>> 
>> God how idiotic!
>> 
>> Superposition has absolutely nothing to do with building up a
>> multi-particle system from individual particles. It is the principle
that
>> a linear combination of two wave functions is another wave function.
These
>> two (or more) wave functions represent *alternatives*; they cannot both
be
>> true at the same time (at least when a measurement occurs).
>> 
>> Then again, considering the source, what should I expect?
>> 
>> I *must* get a newsreader with a killfile capability!
>> 
>> Mark Richardson
>
>  I was updating the go-to-hell file which I keep and saw that Mr.Craig
>Dupree and Mr. Patrick P. E. Esch were on it. Reconsider.
>  PU, PLuto revert both Craig Dupree and Patrick P. E. Esch back to
>point of neutrality, away from hell. Atom
>
>  As to Mark Richardson who I told to go to hell, his spite and hatred
>still spews forth in greater amounts, I say go to hell, Styx this time.

That was not hatred, Archie, it was frustration mixed with exasperation,
tucked in the middle of which was some constructive criticism.

If you want to talk about superposition you might as well know what the
word means. However you don't seem to want to respond to the constructive
part.

Mark Richardson
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1995.09.24 / Jacques Distler /  Re: All atoms are Hydrogen Atom Systems -> Superposition Principle
     
Originally-From: distler@golem.ph.utexas.edu (Jacques Distler)
Newsgroups: alt.sci.physics.plutonium,sci.physics,sci.physics.electromag
sci.physics.fusion,sci.physics.accelerators,sci.physics.particle
Subject: Re: All atoms are Hydrogen Atom Systems -> Superposition Principle
Date: Sun, 24 Sep 1995 00:05:30 -0600
Organization: Physics Department, University of Texas at Austin

In article <4424aa$jku@dartvax.dartmouth.edu>,
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote:

>
>  Someone please mind posting a representative list of elementary
>particles, just the most important members of each family showing the
>MEV, all in MEV. Plus indicate the lifetime of the particle all in
>seconds. Then what the particle decays into. Please post with a name of
>the particle not the Greek symbol.  Thanks

Hey, so you'd actually like to learn some physics? Very commendable. 

I'll do better than give you a little table. The entire Particle Data Book
is online:

   http://www-pdg.lbl.gov/rpp/book/contents.html

Check it out. More than enough information for a 100 of your posts.

Jacques
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