1995.09.24 / Robin Spaandonk /  Re: Lost neutron mass and electron capture
     
Originally-From: rvanspaa@netspace.net.au (Robin van Spaandonk)
Newsgroups: sci.physics.fusion
Subject: Re: Lost neutron mass and electron capture
Date: Sun, 24 Sep 1995 09:56:53 GMT
Organization: Improving

In article <hheffner-2009951632200001@204.57.193.67>, Horace Heffner
wrote :

>In article <hheffner-2009950540030001@204.57.193.69>,
>hheffner@matsu.ak.net (Horace Heffner) wrote:
>
>> In article <MkLjIBK00iV9E5i7d8@andrew.cmu.edu>, Paul Karol
>> <pk03+@andrew.cmu.edu> wrote:
>> 
>> > Excerpts from netnews.sci.physics.fusion: 19-Sep-95 Lost neutron mass
>> > and elect.. by Robin van Spaandonk@nets 
>> > > In fact if you do the sums, it turns out that the electron needs to
>> > > approach the proton to within a distance of 1.841 F in order that the
>> > > energy delivered is equal to that required to make up the .782 MeV
>> > > difference in mass between H and a neutron. Now it seems to me that
>> > > 1.841 F is a very reasonable distance for electron capture. In other
>> > 
>> > Assuming total energy is constant, how does an electron approach a
>> > nucleus, changing the potential energy, and therefore its kinetic
>> > energy, and therefore its velocity, and not radiate energy as required
>> > classically for an accelerating charge?
>> > 
>> > PJK
>> 
>> According to my own personal copy of the CRC handbook, 29th edition, not
>> the unreliable 71st edition I borrow from the library, the masses are
>> shown on page 2589 as follows:
>> 
>> p=    1.67248 E-28       (CRC)
>> e=    0.00091066 E-24    (CRC)
>> -----------------------
>> H=    1.67339 E-24       (CRC)
>> 
>> Robin is definitely on to something here!  This is proof positive that
>> relativity, not just QM, is all just a bunch of malarkey.  Therefore we
>> would not expect the accelerating electron to radiate, would we?  Or would
>> we?   8^)
>
>I apologize for being flippant. I thought this post was a humorous
>reference to some other posts, but I was very wrong. I need to remember to
>get at least 4 hours sleep and to cut out the humour.
>
>I am not trained in QM at all, but since it is involved in any reasonable
>answer I'll risk messing this up so others might be motivated to correct
>it.
>
>It seems to me that the main reason the orbital electron is not attracted
>to the nucleus is that the integral of it's charge probability density,
>i.e. it's mean charge location, is already at the nucleus. Any move of the
>"orbital" or "particle electron" toward the nucleus requires energy,
>because the mean location moves away and thus work is required. Any

If this were true, then the 3s shell would also have its charge center
at the nucleus, and by your own reasoning energy would be needed to
move the electron "closer to the nucleus", i.e. from 3s to 1s. We all
know this is preciesly the opposite of what happens.
 
>acceleration toward the nucleus requires that a non-orbital (free)
>electron approach the nucleus. 
>
>Various interference experiments indicate that slow moving electrons (or
>any particle) have a wave like nature, or a "de-localiztion". The slower a
>particle moves the greater the wavelength of that de-localization.
>Conversely the faster a particle moves, the shorter the wavelength.  This
>is called the de Broglie wavelength. For a particle with momentum p,
>wavelength L, we can calculate
> 
>      L=h/p, where h is Plank's constant.
>
>For electrons in orbit, the orbit is (already) of a size determined by the
>deBroglie wavelength. For electrons approaching a proton (or any positive
>charge), the classical force must be reduced, as it is for orbital
>electrons, to compensate for the portion of the electron's charge
>probability density on the opposite side of the proton, and vice versa.
>One mechanism that reduces the deBroglie wavelength sufficiently that the
>coulomb force operates at maximum attractive force for the duration of the
>approach to the nucleus, is for the electron to acquire adequate momentum
>before the collision interaction, i.e. before approaching within the
>deBroglie wavelength.  For complete stable atoms this potential energy of
>collapse does not exist, so would not be expected to show up in the mass.
>
>Regards,
>
>Horace
>
>-- 
>Horace Heffner 907-746-0820    <hheffner@matsu.ak.net>
>PO Box 325 Palmer, AK 99645
>

Regards,

Robin van Spaandonk <rvanspaa@netspace.net.au>
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Man is the creature that comes into this world knowing everything,
Learns all his life,
And leaves knowing nothing.
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*

cudkeys:
cuddy24 cudenrvanspaa cudfnRobin cudlnSpaandonk cudmo9 cudqt3 cudszL cudyr1995 
------------------------------
1995.09.24 / Robin Spaandonk /  Re: Lost neutron mass and electron capture
     
Originally-From: rvanspaa@netspace.net.au (Robin van Spaandonk)
Newsgroups: sci.physics.fusion
Subject: Re: Lost neutron mass and electron capture
Date: Sun, 24 Sep 1995 09:56:58 GMT
Organization: Improving

In article <43qo47$3br@newsbf02.news.aol.com>, MRichar353 wrote :

>hheffner@matsu.ak.net (Horace Heffner) wrote:
>
>>I am not trained in QM at all, but since it is involved in any reasonable
>>answer I'll risk messing this up so others might be motivated to correct
>>it.
>>
>>It seems to me that the main reason the orbital electron is not attracted
>>to the nucleus is that the integral of it's charge probability density,
>>i.e. it's mean charge location, is already at the nucleus. Any move of
>the
>>"orbital" or "particle electron" toward the nucleus requires energy,
>>because the mean location moves away and thus work is required. Any
>>acceleration toward the nucleus requires that a non-orbital (free)
>>electron approach the nucleus. 
>>
>>Various interference experiments indicate that slow moving electrons (or
>>any particle) have a wave like nature, or a "de-localiztion". The slower
>a
>>particle moves the greater the wavelength of that de-localization.
>>Conversely the faster a particle moves, the shorter the wavelength.  This
>>is called the de Broglie wavelength. For a particle with momentum p,
>>wavelength L, we can calculate
>> 
>>      L=h/p, where h is Plank's constant.
>>
>>For electrons in orbit, the orbit is (already) of a size determined by
>the
>>deBroglie wavelength. For electrons approaching a proton (or any positive
>>charge), the classical force must be reduced, as it is for orbital
>>electrons, to compensate for the portion of the electron's charge
>>probability density on the opposite side of the proton, and vice versa.
>>One mechanism that reduces the deBroglie wavelength sufficiently that the
>>coulomb force operates at maximum attractive force for the duration of
>the
>>approach to the nucleus, is for the electron to acquire adequate momentum
>>before the collision interaction, i.e. before approaching within the
>>deBroglie wavelength.  For complete stable atoms this potential energy of
>>collapse does not exist, so would not be expected to show up in the mass.
>>
>>
>
>It's actually much simpler than this. The wave function of the electron in
>a hydrogen atom is calculated by solving the schrodinger equation which
>incorporates the coulomb potential. The result is, for the ground state, a
>function which behaves as exp(-r/a) where r is the proton-electron
>separation and "a" is the "Bohr radius". This means that the probability
>that the electron is at zero separation from the proton is maximum, with
>the probability decreasing with increasing separation.

The radial probability density for the electron in the ground state of
the hydrogen atom, actually has its maximum at the Bohr radius, not at
the nucleus, as you appear to be saying here (i.e. your function has a
value of 1 when r = 0).

>
>In QM it is not necessary for the electron to "fall in" to the proton, it
>just has a non-zero probability to "be there", with the probability given
>above for the case of the electron being in the ground state.
>
>The question brought up earlier in this thread concerning the classical
>electron radiating as it fell into the proton was intended to show the
>original poster that the classical model cannot be used at all for the
>hydrogen atom, it is thoroughly inconsistent.
>
>Also, as I posted earlier in this thread, there is no mystery with the
>masses, energies, etc. The total energy is conserved. If you must speak
>classically, the kinetic energy of the electron would increase as it
>approached the proton, but it's (negative) potential energy increases (in
>absolute value) as well, so the total energy remains constant. Since that
>total energy is less than the mass of a single neutron, electron capture
>in hydrogen does not occur.

It all seems so simple, when you only look at half of the picture.

The point I am trying to make is that it is not true that the
potential energy of the electron proton pair is zero when they are
infinitely far apart, and thus representable by zero mass.
You may wish to define it so, that doesn't make it true. In other
words, you are not free to choose your zero point of potential energy
where you choose. If you don't believe this then consider the
following thought experiment.
1) Take a 6 foot long piece of elastic.
2) Tie one end around your head.
3) Tie the other end around a tennis ball.
4) Ask a friend to take the tennis ball and back up a few steps, until

the elastic is stretched to the limit.
5) Arbitrarily assign a total energy of zero to this state of the
system.
6) Ask your friend to let go of the tennis ball.
7) You have nothing to worry about....,the potential energy of the
system is zero...you just defined it that way! :->

>
>Mark Richardson
PS If you now have a black eye, don't blame me. I told you it was a
thought experiment.
Regards,

Robin van Spaandonk <rvanspaa@netspace.net.au>
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Man is the creature that comes into this world knowing everything,
Learns all his life,
And leaves knowing nothing.
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*

cudkeys:
cuddy24 cudenrvanspaa cudfnRobin cudlnSpaandonk cudmo9 cudqt3 cudszL cudyr1995 
------------------------------
1995.09.24 / Horace Heffner /  Re: Testing the Mitchell Jones Hypothesis
     
Originally-From: hheffner@matsu.ak.net (Horace Heffner)
Newsgroups: sci.physics.fusion
Subject: Re: Testing the Mitchell Jones Hypothesis
Date: Sun, 24 Sep 1995 22:04:44 -0900
Organization: none

In article <21cenlogic-2309951050290001@austin-2-15.i-link.net>,
21cenlogic@i-link.net (Mitchell Jones) wrote:

> 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
> 
> ===========================================================

I just don't get it. If the rest mass of the photon is zero, acceleration
would be infinite. If the rest mass is not zero, the photon must end up
with infinite mass and never reach light speed. Are you also discarding
relativity?

Regards,                          <hheffner@matsu.ak.net>
                                  PO Box 325 Palmer, AK 99645
Horace Heffner                    907-746-0820

cudkeys:
cuddy24 cudenhheffner cudfnHorace cudlnHeffner cudmo9 cudqt3 cudszM cudyr1995 
------------------------------
1995.09.24 / 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: Sun, 24 Sep 1995 08:38:45 +0100
Organization: Home

In article: <21cenlogic-2309951050290001@austin-2-15.i-link.net>  
21cenlogic@i-link.net (Mitchell Jones) writes:
> 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:

No need to repost. I'd already read that drivel. It doesn't reply to my 
point at all, which is simply, as every undergraduate student *ought* to 
know when he starts, and ceratvinly *does* know by the end of his first 
semester, photons do and can only travel at c. They are emitted at c, 
they are absorbed at c. Before emission and after absorption they are 
not photons.

As I said, with a zero mass and a finite (non-zero) momentum, a photon 
*can* only exist if it is travelling at c.

Indeed, *all* observable wave-forms travel only at the form's 
propagating speed. When you drop a stone into a pool, the outgoing 
ripples don't accelerate up gradually from a propagating speed of zero 
to the terminal speed for outgoing ripples. It happens instantaneously.

(Now the *amplitude* of the weater ripples - that *is* something which 
builds up.) I suggest a basic undergraduate course in physics, 
Mitchell, before next attempting to post here any variants of your 
'Theory of Everything'.
-- 
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.


cudkeys:
cuddy24 cudenAlan cudfnAlan cudlnM cudmo9 cudqt3 cudszM cudyr1995 
------------------------------
1995.09.20 / Matthew Kennel /  Re: Making He3 and related questions
     
Originally-From: mbk@caffeine.engr.utk.edu (Matthew Kennel)
Newsgroups: sci.space.tech,sci.physics,sci.physics.fusion
Subject: Re: Making He3 and related questions
Date: 20 Sep 1995 19:45:54 GMT
Organization: Oak Ridge National Laboratory and Univ. Tenn.

Ian Gay (gay@sfu.ca) wrote:
: In article <43k16f$h4@hopper.kenyon.edu>,
:    Timothy Sullivan <Sullivan@kenyon.edu> wrote:
: >He-3 is a decay product of tritium and tritium is used in hydrogen 
: bombs. 
: >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?) and say 'please call'.

It's probably a DOE controlled material.
cudkeys:
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------------------------------
1995.09.22 / Joshua Levy /  Re: Jed's definition of CF (at last!)
     
Originally-From: Joshua Levy <joshua@intrinsa.com>
Newsgroups: sci.physics.fusion
Subject: Re: Jed's definition of CF (at last!)
Date: 22 Sep 1995 17:17:19 GMT
Organization: Intrinsa Corporation


Joshua Levy <joshua@intrinsa.com> wrote: 
>>By this definition, the Griggs device (and the similar Russian device) is not
>>CF (no helium).  Actually, P&F are not CF either, since they did not produce
>>helium either.

jedrothwell@delphi.com replied:
>P&F
>definitely are doing CF, and their cells do produce helium as far as I know.
>Have they told you anything different? Can you point to any experimental
>evidence to back up your claims? Ha, ha ha ha! Just kidding, of course you
>can't. 

You have it backwards: I can point at P&F 1989 paper, and later retractions, to
show that they did NOT produce helium.  I don't remember any helium claims in
their "heat after death" paper, do you?  What papers can you point at which 
show P&F produced helium?

On the one had you want P&F to be doing CF work, on the other hand you want
to require helium production, which hey have never been able to show. Which 
is it?  Is the real Jed CF definition "You must product helium, except for
P&F who do cold fusion even if they are not producing helium.  Everyone
else, however, must produce helium."  

And then he adds the usual Jed ad hominisms....
>You are a "CF skeptic." That means you never read anything, you never
>know what you are talking about, and you just make up pretend facts as you go
>along. Right?

You claim that P&F product heliumn "as far as you know", and previously you talk
about getting information from scientific papers.  So which scientific paper
shows that P&F get helium?  One good citation will do!

Joshua Levy <joshua@intrinsa.com>

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------------------------------
1995.09.24 /  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: 24 Sep 1995 13:18:35 -0400
Organization: America Online, Inc. (1-800-827-6364)

Its all right Mark, I was not offended. I am seeking the truth and your
comment helped me understand what's going on. The reality of elementary
particle processes is complex and it takes a while to understand. The
explanation I posted regarding the field disappearing and its energy
increasing the impulse of the electron is totally consistent with quantum
mechanics. I am a QM fan and would not propose anything that is not
consistent with it. Classical visualization helps our understanding.

Zoltan Szakaly
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------------------------------
1995.09.24 / Horace Heffner /  Re: Is Griggs Experiment Hot Water Simplicity Incarnate?
     
Originally-From: hheffner@matsu.ak.net (Horace Heffner)
Newsgroups: sci.physics.fusion
Subject: Re: Is Griggs Experiment Hot Water Simplicity Incarnate?
Date: Sun, 24 Sep 1995 13:35:37 -0900
Organization: none

As a newbie, I may have missed something in the long history of threads on
the Griggs device tests. However, it seems like a couple of variables
might have been overlooked, so I'll post this anyway for the benefit of
others who may have missed the discussion also. Below is quoted a set of
sample data points and calculations provided by Mitchell Jones in an
earlier post of his. In reference to this calculation, it appears the
density and specific heat (heat content) of the water is not addressed. 

Specific heat is important because a specific heat for aqueous inorganic
solutions can be less than .85 that of distilled water. Water with a
specific heat of .85 will only require .85 the energy per gram that
distilled water takes  to heat to the same temperature, thus the
calculation of heat energy produced below could be off by 1/.85x100 -100 =
17 percent. Specific heat could be depressed by hard water, water treated
with chlorine, chlorine compounds, or sulfates.

Density plays an identical role with respect to potential error.
Measurements would be affected by any density loss immediately prior to
the meter. This can be caused by outgassing due to pressure loss just
prior to the meter. Very small bubbles at the meter will be measured as
water volume. These bubbles could be chlorine, air, or CO2. The effect
would be significant if the water came from a carbonaceous (limestone)
aquifer, or was treated with aeriation or chlorination. In addition,
sulfate treatment, now newly mandated in many regions, produces dissolved
gasses.

The combined effect of specific heat and density depression could
potentially account for the observed results. To compensate, water samples
of a known specific volume could be taken before and after the experiment,
and calorimitry performed using a resistive heating element to measure the
amount of time it takes to raise the temperature of that particular volume
from the observed inlet temperature to the observed outlet temperature.
(Note that the heat content and density of water varies with temperature
and pressure.) 

Another option would be to run the experiment with degassed deionized
water, but this could change the actual observed effect. 

From an earlier posting:

>Originally-From: 21cenlogic@i-link.net (Mitchell Jones)
>Newsgroups: sci.physics.fusion
>Subject: Re: Silly nonsense from Robert Heeter
>Date: Tue, 01 Aug 1995 11:19:55 -0500
>Organization: 21st Century Logic
>
[snip]
>Let me be very specific: in the hot water runs, tap water flows into the
>rotor housing of the Hydrosonic Pump and then out again through the other
>side, and thence goes down the drain. The temperature of the input water
>flow is measured by six calibrated thermocouples, and an average taken.
>Likewise, the outflow temperature is measured by six calibrated
>thermocouples and an average taken. For example, during one sample
>interval the average incoming water temperature was 80.39 degrees F., and
>the average outgoing water temperature was 148.2 degrees F. Subtracting
>the former from the latter, we get an average change in temperature of
>67.86 degrees F. During the test, the flow rate of the water held steady
>at 5.382 gallons/min. Thus the heat added to the water by the device in an
>hour was 5.382 times 60 times 8.3 times 67.86, which gives 181,881 BTU9s
>per hour. (Note: gallons per minute times 60 gives gallons per hour, and
>gallons per hour times 8.3 gives pounds per hour. Number of pounds per
>hour times the Farenheit temperature change per pound gives BTU9s per
>hour.)
>
>Do you have any problems with any of this? Do you see any possibility of a
>major error here?
>
>Moving on: input energy was calculated using (a) torque measured on the
>rotor shaft by a dynamometer: 1161.8 inch-pounds; and (b) shaft rotation
>rate: 3561 rpm. Calculated power comes out to about 65.63 HP, which
>equates to a rate of 167,084 BTU9s per hour. [Note: The calculation makes
>use of the fact that power in ft-lbs/sec equals torque in ft-lbs times
>angular speed in radians per sec. First, convert to appropriate units:
>1161.8 inch-pounds = 96.8 ft-lbs; and 3561 rpm times 2</60 = 372.9 radians
>per sec. (Since each rpm equals 1 revolution or 360 degrees, it also
>equals 2< radians.) Power thus is 96.8 times 372.9 = 36097 ft-lbs/sec.
>Since 1 HP is 550 ft-lbs/sec, we then divide by 550 to get 65.63 HP as our
>input power reading, which equates to 167,084 BTU9s per hour.] In
>addition, input energy was measured by means of a power meter, yielding a
>rate of about 48.9 KW, which is roughly the same as 65.63 HP.
>
>Do you see any major errors looming up in any of this?
>
>If you are still with me, then I assume you agree that the COP
>(coefficient of performance) is 181,881 divided by 167,084, or 1.089,
>(which, in percentage terms, is 108.9%). Note further that this device is
>producing considerably more heat than that imparted to the water: heat
>transferred from the pump housing and from the electric motor housing to
>the external environment is not being taken into account by this
>calculation. (According to Jed, the pump housing is *very* hot, and
>radiates heat as if it were an electric space heater.) What this means,
>speaking conservatively, is that you would need to find an error on the
>order of 20% to invalidate these results. My question for you, therefore,
>is this: how can it be that this device is not performing over unity,
>given the facts of the situation as just described? If you can, please be
>specific: tell me where the error, or errors, may lie. And remember, the
>device produces these sorts of results over and over, for hours and hours.
>Some other data frames (taken at one minute intervals) gave the following
>COPs:
>
>1.089
>1.095
>1.09
>1.103
>1.097
>1.098
>1.10
>1.091
>1.105
>1.109
>
>Where is the possibility of a major error, Barry? As I noted above, before
>I actually focused my mind on these data and worked through the
>calculations, I had doubts very similar to yours.  But, afterwards, I
>discovered that my doubts were gone. The reason: it is one thing to, upon
>discovering that your wallet isn't in your pocket, theorize that it may be
>on the kitchen table, and quite another to hold to that theory after
>checking the kitchen table and discovering that the wallet isn't there! By
>the same token, it is OK to hypothesize that an experimental result may be
>due to errors before checking out the design. But it makes no sense to
>hold to that hypothesis after checking out the design and finding that the
>suspected possibilities for error aren't there! --Mitchell Jones}***
>
[snip]

Regards,                          <hheffner@matsu.ak.net>
                                  PO Box 325 Palmer, AK 99645
Horace Heffner                    907-746-0820

cudkeys:
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------------------------------
1995.09.24 / Jim Carr /  Re: Multineutron systems
     
Originally-From: jac@ds8.scri.fsu.edu (Jim Carr)
Newsgroups: sci.physics.fusion
Subject: Re: Multineutron systems
Date: 24 Sep 1995 21:15:13 -0400
Organization: Supercomputer Computations Research Institute

In article <43svov$kos@newsbf02.news.aol.com> 
zoltanccc@aol.com (ZoltanCCC) writes:
>
>Unfortunately the formula does not work very well under 20 nucleons so it
>is useless in our quest for knowledge about deuterium electron capture.

The semi-empirical mass formula is wholly unnecessary for the two-body 
system where you can solve the equations with potentials constructed 
to reproduce the scattering data. 

>I suspect that electron capture in deuterium will result in an n-n system
>that is marginally stable under thermal conditions. Perhaps there is a few
>electron volts of binding energy, enough to keep the two together until
>they collide with another d.

There are two reasons to doubt this: 

1) People have looked very hard for the di-neutron for many years. 
   They have not found any indication of it.  There are many many 
   references in the A .lt. 4 literature reviews. 

2) There would be an analog state in deuterium that would be either 
   bound or slightly unbound.  Again, the energy of these resonances 
   are known (for decades) and they do not favor a di-neutron.  

-- 
 James A. Carr   <jac@scri.fsu.edu>     |  What a long strange trip it's 
    http://www.scri.fsu.edu/~jac/       |  been.        
 Supercomputer Computations Res. Inst.  |              Jerry Garcia
 Florida State, Tallahassee FL 32306    |                1942-1995 
cudkeys:
cuddy24 cudenjac cudfnJim cudlnCarr cudmo9 cudqt3 cudszM cudyr1995 
------------------------------
1995.09.25 / 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: Mon, 25 Sep 1995 02:32:06 GMT
Organization: Stanford Linear Accelerator Center

In article <21cenlogic-2209951952380001@austin-2-6.i-link.net>,
Mitchell Jones <21cenlogic@i-link.net> wrote:
>In article <DFADHC.947@unixhub.SLAC.Stanford.EDU>,
>panetta@finch.SLAC.Stanford.EDU (James H. Panetta) wrote:
>>
>> 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.
>
>***{The implication of your suggestion seems to be that H+ ions cannot
>exist either in the electrolyte or in the palladium electrode, because
>each will tear an electron away from the outer shell of the first atom it
>approaches, thereby becoming a neutral hydrogen atom. In fact, that is not
>what happens. The reason is that the other atoms also have a grip on those
>electrons, and in the ensuing tugs of war, they have the advantage. I
>could tell you why that is the case, but it would require a lengthy
>digression. Suffice it to say that it is a fact: ions exist. Since your
>suggestion implies that they do not, it must be wrong. --Mitchell
>Jones}*** 

No.  I did not say this at all.  I said that you cannot treat the 
H+ ion as a proton in vacuo.  You must account for the coulomb field
and it's interaction with neighboring atoms.  Also, remember that the
palladium electrode contains excess electrons, so these are available
to the H+ ion as well as valence electrons from Pd.

<Protoneutron stuff deleted>

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

>***{Here, you are merely quoting standard "quantum mechanical" dogma. In
>case you haven't read my comments on this subject in other posts, let me
>state for the record that I bear an active contempt for "quantum
>mechanics."

You then bear an "active contempt" for a theory which fits the known
data.

>            In my view, the "preferred" orbits--i.e., the Bohr-Sommerfeld
>orbits--are not the only orbits. The classical, transitional orbits also
>exist, and are sometimes occupied. I view them as wildly unstable, and
>normally occupied only so fleetingly that, with present instruments, such
>occupancy cannot be detected.

Wrong.  The standard orbits can be easily detected by a common spectrograph.
If these are normally occupied only fleetingly, then show a reasonable
explanation for the Balmer, Lyman, and Paschen series of spectroscopic
lines.  The *very* well known frequencies of these are well defined with
the use of standard undergraduate level quantum mechanics.  If these
orbits weren't occupied, then explain the 21cm line.

>                              However, under exceptional circumstances
>such as the ones we are discussing, where insufficient space is available
>for the Bohr-Sommerfeld orbits, 

I still have no idea why you are talking about "insufficient space" for
the electron.  By this argument there is "insufficient space" in the
solar core.  Densities there are much higher than the densities of 
hydrogen in a palladium electrode.  By your arguments, the entire
sun should have converted by now.

>for the Bohr-Sommerfeld orbits, and insufficient energy is available for
>the nucleus (the proton) to absorb the electron, I believe that the
>electron is essentially forced into the classical orbits. Result: it
>spirals down to what I call "grazing altitude" above the nucleus,
>radiating away its excess energy as it goes, and forms a protoneutron. 

Is this radiation continuous?  If so, why has it never been seen in
observations of astrophysical hydrogen clouds?

>As a matter of curiosity, what do you think happens to an electron under
>these circumstances? Remember: it lacks the energy needed to escape from
>the grip of the proton; it lacks the space needed to utilize the
>Bohr-Sommerfeld orbits; and it lacks the energy needed to merge with the
>proton and form a neutron. So what does it do? --Mitchell Jones}*** 

Get this straight:  Nobody really talks about the Bohr-Sommerfeld theory
anymore.  It doesn't fit the data.  Quantum Mechanics does.  Use
your theory to derive standard observables like the spectrographic
lines listed above, or toss it.  

>> >                 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?
>
>***{I have noted in other posts that I consider it a waste of time to
>mention the neutrino. It has nothing to do with the points presently at
>issue. --Mitchell Jones}*** 

It has something to do with the points in the issue.  Unless you can
account for *all* known facts in your theory, the theory is wrong.

>> 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?
>
>***{You misunderstand. I am not proposing a mathematical procedure for
>calculating anything about electron captures. I am putting forward the
>hypothesis that when a hydrogen ion acquires an electron under certain
>very specific conditions (i.e., where insufficient space is available to
>form a Bohr-Sommerfeld orbit, and insufficient energy is available to
>permit neutron formation) a wildly unstable type of particle which I call
>a protoneutron can come into existence and accumulate.  It is my conjecture
>that such particles have extraordinarily large gamma absorption cross
>sections, and that their existence can be used to explain the "cold
>fusion" results. If you suspect that there is more to my position than
>meets the eye, you are correct: I do, in fact, have very specific reasons
>for expecting this particle to have the properties that I attribute to it.
>However, no purpose would be served by getting into those reasons here.
>The hypothesis is both clear and testable, and that is all that matters.
>If the tests that I have proposed support the existence of protoneutrons,
>then you "quantum mechanics" guys will simply have to tweak your
>mathematics again, like you have done 10,000 times already, to make room
>for it. Afterwards, you can claim that you predicted it all along, as you
>have also done 10,000 times already. ("Quantum mechanics" gurus are like
>stock market prognosticators: they predict the past to 12 or 15 digits of
>accuracy, but the future is quite another matter!)    

What tests have you proposed?  What are the possible sources of error?
What controls do you propose?  What mathematics have we tweaked?

>As for your questions about sodium and iron, my reaction is that it would
>be *enormously* difficult to apply the protoneutron scenario to anything
>other than hydrogen or its isotopes. Think about it: to apply the scenario
>to Fe-55, you would have to somehow strip 26 electrons away, place the
>resulting nucleus into circumstances where none of the Bohr-Sommerfeld
>orbits were possible, and then introduce a single free electron. The hope
>would be that the electron would spiral down and orbit the nucleus at
>grazing altitude, where it would become... what? You couldn't very well
>call the result a protoneutron, because it would contain 26 protons and 29
>neutrons. I suppose we could make a stretch and call it a "compound
>protoneutron," but doing so, in my view, would be a large leap indeed.
>Moreover, leaving nomenclature aside, how would you create such a thing? I
>frankly have no idea, but if you can somehow manage it, I am willing to
>compliment your technical wizardry in advance! --Mitchell Jones}*** 

One cannot create something that does not exist.  Besides it's *easy* to
strip the electrons off Fe-55.  Nuclear physicists do it all the time.

>> 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?
>
>***{Protoneutrons do not exist under the circumstances where tritium beta
>decay has been measured, and they have nothing whatsoever to do with those
>measurements.

Why not?  Given your explanation of how the electrons behave, there must 
be some part of a tritium sample where there is "not enough room" for
the standard orbits.  Therefore there is an effect from your model.  But
there is no effect seen in the data.  Therefore your model is *incorrect*.

>> > 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.
>
>***{There is nothing about the photoelectric effect that violates
>classical mechanics. Suppose you drill a small hole in the surface of a
>billiard table, drop a strand of 10 pound test fishing line down into it,
>pour in a drop of epoxy, and then, 6 inches away, attach the other end to
>a billiard ball by the same method. If you then shoot ten cue balls of
>steadily increasing mass at it, all with the same velocity, you will
>create a situation strictly analogous to the photoelectric effect. If you
>pick the right velocity then there will be a cutoff mass where, for cue
>balls lighter than that, total energy will be insufficient to break the
>fishing line and the object ball will remain bound. For heavier balls, on
>the other hand, the string will break and the object ball will fly free.
>Since you cannot very well deny that classical mechanics explains this
>situation, how can you deny that it also explains the photoelectric
>effect?

But why is it *always the same frequency* of light that induces the
effect for each element?  By your arguments above, the electrons don't
really sit in the standard energy levels.

>And now, a more important point: you have strongly implied that you
>endorse the standard "quantum mechanical" interpretation--to wit: that
>motion in the microcosm consists of a series of quantized "jumps" or
>"leaps" in which the entity disappears from one spatial position and
>magically reappears at the next one, without passing through the gap
>between the two positions.

This is not what quantum mechanics says.  It says that particles in
a system can only exist in certain discrete *states*.  These
states are *not* positions.  Given a hydrogen atom, an electron in
the state n=1 has the maximum probability of being at
the Bohr radius.  In the state n=2, l=0, the electron has the
maximum probability of being at twice the Bohr radius.  THIS 
DOES NOT MEAN that an electron in the state n=1 cannot be found
outside the Bohr radius.  Please read and *understand* something
like Eisberg and Resnick (an elementary quantum text) before
you talk about things which you don't understand.

>> 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.
>
>***{You misunderstand. It is necessary to sharply distinguish between the
>curve fitted mathematical constructs of physics, and the visual models by
>which the mathematics is "interpreted." I do not deny that the curve
>fitted mathematics fits the experimental results. (That would be a
>contradiction in terms!) What I do deny is the validity of the "quantum
>mechanical"--i.e., the "Copenhagen"--interpretation of those mathematical
>formulations. However much they may wish it were so, the fact is that the
>advocates of "quantum mechanics" have no more right to claim the curve
>fitted mathematics than does anybody else. Mathematical constructs have no
>viewpoint about physical theory, and do not give a hoot in hell how they
>are "interpreted." The issue here is not about the math, but about the
>"interpretations"--i.e., about whether the "quantum mechanical" attack on
>the principle of continuity can be sustained or not. --Mitchell Jones}*** 

Classical physics is just as much a construct as quantum mechanics.
Should we throw that out too?

  --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
cudkeys:
cuddy25 cudenpanetta cudfnJames cudlnPanetta cudmo9 cudqt3 cudszL cudyr1995 
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