Tritium.

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Steve Lajoie

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Jun 1, 2000, 3:00:00 AM6/1/00
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Data at:
http://www.nde.lanl.gov/cf/tritweb.htm
Note, please, that this is a Los Alamos National Lab site.

We left off here with someone saying it was a hot fusion
result and not a cold fusion result. It was said that the
reason why it only happens with metal hydrides is because
the metal hydride lattice holds the target atoms stationary
for the hot deuterons to hit.

This was debunked, because:

1) Two hot deuterons in the plasma should have more energy
to overcome the coulomb repulsion than one moving deuteron
and one stationary one. Thus, non hydride metals should also
produce some tritium, but they don't.

2) The state of the palladium has a strong correlation to the
amount of tritium produced. The annealed Pd can hold as much
deuterium as the work hardened Pd, but the work hardened Pd
will tend to let deuterium gather at lattice defects. Since
work hardened Pd produced three times as much tritium as the
annealed Pd, it is strongly correlated with lattice defects
and cannot thus tritium production is due to the close proximity
of the deuterons and not hot fusion.

The response to this latter argument was a great deal of
anger and frustration. No real rebuttal was offered.

One of my arguments that this was hot fusion because of
diffusion was successfully shot down, I admit. Which doesn't
negate the validity of arguments 1 and 2.

Thus, cold fusion.

T L Clarke

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Jun 1, 2000, 3:00:00 AM6/1/00
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Steve Lajoie wrote:


> We left off here with someone saying it was a hot fusion
> result and not a cold fusion result. It was said that the
> reason why it only happens with metal hydrides is because
> the metal hydride lattice holds the target atoms stationary
> for the hot deuterons to hit.
>
> This was debunked, because:
>
> 1) Two hot deuterons in the plasma should have more energy
> to overcome the coulomb repulsion than one moving deuteron
> and one stationary one. Thus, non hydride metals should also
> produce some tritium, but they don't.
>

You neglect the influence of density. In a deuterated metal the
density of target deutrons will be much greater than in the rarified
plasma that maintains the discharge. Yes, there will be tritium
produced in the plasma alone, but it will occur at a much, much
lower rate than with deuterated metal.

I also think the little cones observed to form on the surface of the
Pd may serve to focus the electric field and enhance fusion.

> 2) The state of the palladium has a strong correlation to the
> amount of tritium produced. The annealed Pd can hold as much
> deuterium as the work hardened Pd,

Is this true? I don't think the electrolytic work supports this.

> but the work hardened Pd
> will tend to let deuterium gather at lattice defects. Since
> work hardened Pd produced three times as much tritium as the
> annealed Pd, it is strongly correlated with lattice defects
> and cannot thus tritium production is due to the close proximity
> of the deuterons and not hot fusion.

Even assuming that annealed and hardened Pd hold the same amount
of D, other explanations of the difference in fusion are available.
The lattice defects in the hardened Pd may help channel D+ ions
into the Pd enhance collision rates with D in the Pd lattice and
produce more T.

> The response to this latter argument was a great deal of
> anger and frustration. No real rebuttal was offered.

Consider it rebutted.

> One of my arguments that this was hot fusion because of
> diffusion was successfully shot down, I admit. Which doesn't
> negate the validity of arguments 1 and 2.
>
> Thus, cold fusion.

Still unproven.

Now if the experimenters could reliably get the T without 2000 volts of
electric potential .... then I might start to think CF is the
explanation.

Tom Clarke

Stephen Lajoie

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Jun 2, 2000, 3:00:00 AM6/2/00
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In article <3936CB2B...@ist.ucf.edu>,

T L Clarke <tcl...@ist.ucf.edu> wrote:
>Steve Lajoie wrote:
>
>
>> We left off here with someone saying it was a hot fusion
>> result and not a cold fusion result. It was said that the
>> reason why it only happens with metal hydrides is because
>> the metal hydride lattice holds the target atoms stationary
>> for the hot deuterons to hit.
>>
>> This was debunked, because:
>>
>> 1) Two hot deuterons in the plasma should have more energy
>> to overcome the coulomb repulsion than one moving deuteron
>> and one stationary one. Thus, non hydride metals should also
>> produce some tritium, but they don't.
>>
>
>You neglect the influence of density. In a deuterated metal the
>density of target deutrons will be much greater than in the rarified
>plasma that maintains the discharge. Yes, there will be tritium
>produced in the plasma alone, but it will occur at a much, much
>lower rate than with deuterated metal.

It is true that the density inside the Pd will be about 900 times as
as great as it is in the surrounding plasma, maybe even greater.

But it will be for a very short depth into the metal. Is the mean
free path of the plasmated deuterium 900 times longer? I don't know,
but it probably is longer in the gas than in the metal.

>I also think the little cones observed to form on the surface of the
>Pd may serve to focus the electric field and enhance fusion.

I think this was discussed in the paper.

>> 2) The state of the palladium has a strong correlation to the
>> amount of tritium produced. The annealed Pd can hold as much
>> deuterium as the work hardened Pd,
>
>Is this true? I don't think the electrolytic work supports this.

Sandia national labs keeps a data base on metal hydride properties.
I saw nothing that indicated that the metal state affected the ability of
Pd to form hydrides. It doesn't appear to be a factor. If you're asking for
a study that shows that it isn't, I don't have it, no.

>> but the work hardened Pd
>> will tend to let deuterium gather at lattice defects. Since
>> work hardened Pd produced three times as much tritium as the
>> annealed Pd, it is strongly correlated with lattice defects
>> and cannot thus tritium production is due to the close proximity
>> of the deuterons and not hot fusion.
>
>Even assuming that annealed and hardened Pd hold the same amount
>of D, other explanations of the difference in fusion are available.
>The lattice defects in the hardened Pd may help channel D+ ions
>into the Pd enhance collision rates with D in the Pd lattice and
>produce more T.

How would it help channel D+ ions into the Pd?


>> The response to this latter argument was a great deal of
>> anger and frustration. No real rebuttal was offered.
>
>Consider it rebutted.

You have pointed out that:
1) I haven't pointed out a study that shows that annealed Pd can absorb as
much hydrogen as work hardened Pd. This is true. There is a conspicuous
absence of it in the literature, thus it doesn't appear to be a factor.
That's all I have to go on. A weakness in my argument, but not a serious
one. If there was such a dependence, it would be mentioned along with CO
and surface area effects.

2) Hot fusion would be a large number of deuterons in a Pd lattice that is
exposed to hot deuterons. The probability of hot fusion would be the same
if the deuterons were clumped together or spread out, as long as there was
nearly the same number of deuterons per area. From (1) I'm saying that the
deuterons per area is about the same for both annealed and work hardened
Pd. Experiment shows that the tritium increased 3 fold in work hardened
over annealed Pd.

I really don't think that there is three times as much deuterium in the
work hardened over the annealed Pd. Such a strong dependence would have
been noted already in the literature.

That would leave the lattice defects as a factor. I can easily see there
being 3x as many lattice defects in work hardened Pd than in annealed Pd.

That would seem to say that the fusion only happens in the lattice
defects. Either something unknown about the defect is causing hot fusion
to occur there; or cold fusion is occurring there. Since there are
experiments that indicate that no hot deuterons are required, this seems
to support cold fusion.


>> One of my arguments that this was hot fusion because of
>> diffusion was successfully shot down, I admit. Which doesn't
>> negate the validity of arguments 1 and 2.
>>
>> Thus, cold fusion.
>
>Still unproven.


No, you are unconvinced. Slightly different.

>Now if the experimenters could reliably get the T without 2000 volts of
>electric potential .... then I might start to think CF is the
>explanation.

2000 volts isn't enough to cause hot fusion. This plasma isn't very hot.

>Tom Clarke
>
>

T L Clarke

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Jun 2, 2000, 3:00:00 AM6/2/00
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T L Clarke wrote:

I hit the wrong button and this was sent early.

> Stephen Lajoie wrote:

Snip most of earlier reply to avoid duplication.

> > 2000 volts isn't enough to cause hot fusion. This plasma isn't very hot.
>

> Shall we restart the discussion about the role of tunneling in host fusion?
> 2000 electron volts corresponds to 22,000,000 degrees more or less
> (someone posted the exact conversion factor, but I've forgotten it).
>
> I found this table at the nuclear weapons faq site
>

> http://www.fas.org/nuke/hew/Nwfaq/Nfaq4-4.html

> Reaction Cross Sections (cm^2)
> T (KeV) D/T D/D D/He-3
> 1.0 5.5x10^-21 1.5x10^-22 3 x10^-26
> 2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
> 5.0 1.3x10^-17 1.8x10^-19 6.7x10^-21
> 6.0 2.6x10^-17 2.3x10^-19 3.3x10^-20
> 7.0 4.1x10^-17 3.5x10^-19 5.3x10^-20
> 8.0 6.0x10^-17 5.0x10^-19 8.0x10^-20
> 9.0 8.2x10^-17 6.7x10^-19 1.3x10^-19
> 10.0 1.1x10^-16 1.2x10^-18 2.3x10^-19
> 15.0 2.6x10^-16 1.9x10^-18 1.3x10^-18
> 20.0 4.2x10^-16 5.2x10^-18 3.8x10^-18
> 30.0 6.6x10^-16 6.3x10^-18 1.0x10^-17
> 40.0 7.9x10^-16 1.0x10^-17 2.3x10^-17
> 50.0 8.7x10^-16 2.1x10^-17 5.4x10^-17
>

The cross section is 3 or 4 orders of magnitude down from
that at very high energies, but it is not zero. The LANL
experiments took several days to accumulate tritium
so their reaction was not very vigorous. Fusion at 2 KeV
is to me quite plausible as an explanation for their data.

Tom Clarke


T L Clarke

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Jun 2, 2000, 3:00:00 AM6/2/00
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Stephen Lajoie wrote:

> In article <3936CB2B...@ist.ucf.edu>,
> T L Clarke <tcl...@ist.ucf.edu> wrote:

> >Even assuming that annealed and hardened Pd hold the same amount
> >of D, other explanations of the difference in fusion are available.
> >The lattice defects in the hardened Pd may help channel D+ ions
> >into the Pd enhance collision rates with D in the Pd lattice and
> >produce more T.
>
> How would it help channel D+ ions into the Pd?
>

My thinking is that since it is work hardned it contains more defects,
that is planes and regions in which the crystal structure is perturbed.
This planes might serve as conduits along which the D+ ions could
penetrate to greater dept into the Pd.
You mentioned that such defects/dislocations etc might serve to
concentrate D. Thus as the D+ ions are channeled down along
dislocations they might encounter enhanced D concentrations.

> >> The response to this latter argument was a great deal of
> >> anger and frustration. No real rebuttal was offered.
>
> >Consider it rebutted.
>
> You have pointed out that:
> 1) I haven't pointed out a study that shows that annealed Pd can absorb as
> much hydrogen as work hardened Pd. This is true. There is a conspicuous
> absence of it in the literature, thus it doesn't appear to be a factor.
> That's all I have to go on. A weakness in my argument, but not a serious
> one. If there was such a dependence, it would be mentioned along with CO
> and surface area effects.

It occured to me that the authors mentiond annealing in _air_. Heating in
an oxidizing atmosphere could perhaps change the surface characteristics
etc of the Pd in addition to the internal structure.

> 2) Hot fusion would be a large number of deuterons in a Pd lattice that is
> exposed to hot deuterons. The probability of hot fusion would be the same
> if the deuterons were clumped together or spread out, as long as there was
> nearly the same number of deuterons per area. From (1) I'm saying that the
> deuterons per area is about the same for both annealed and work hardened
> Pd. Experiment shows that the tritium increased 3 fold in work hardened
> over annealed Pd.

> I really don't think that there is three times as much deuterium in the
> work hardened over the annealed Pd. Such a strong dependence would have
> been noted already in the literature.
>

OK

> That would leave the lattice defects as a factor. I can easily see there
> being 3x as many lattice defects in work hardened Pd than in annealed Pd.
>
> That would seem to say that the fusion only happens in the lattice
> defects. Either something unknown about the defect is causing hot fusion
> to occur there; or cold fusion is occurring there. Since there are
> experiments that indicate that no hot deuterons are required, this seems
> to support cold fusion.

I already addressed how the defects might affect D+ interation with D
within the Pd lattice.

> >> One of my arguments that this was hot fusion because of
> >> diffusion was successfully shot down, I admit. Which doesn't
> >> negate the validity of arguments 1 and 2.
> >>
> >> Thus, cold fusion.
> >
> >Still unproven.
>
> No, you are unconvinced. Slightly different.

Not very much. You think it is proven, I don't. We both agree, I hope,
on the theory of gravity which is proven.

> >Now if the experimenters could reliably get the T without 2000 volts of
> >electric potential .... then I might start to think CF is the
> >explanation.

> 2000 volts isn't enough to cause hot fusion. This plasma isn't very hot.

Shall we restart the discussion about the role of tunneling in host fusion.


2000 electron volts corresponds to 22,000,000 degrees more or less
(someone posted the exact conversion factor, but I've forgotten it).

I found this table at the nuclear weapons faq site

Reaction Cross Sections (cm^2)


T (KeV) D/T D/D D/He-3
1.0 5.5x10^-21 1.5x10^-22 3 x10^-26
2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
5.0 1.3x10^-17 1.8x10^-19 6.7x10^-21
6.0 2.6x10^-17 2.3x10^-19 3.3x10^-20
7.0 4.1x10^-17 3.5x10^-19 5.3x10^-20
8.0 6.0x10^-17 5.0x10^-19 8.0x10^-20
9.0 8.2x10^-17 6.7x10^-19 1.3x10^-19
10.0 1.1x10^-16 1.2x10^-18 2.3x10^-19
15.0 2.6x10^-16 1.9x10^-18 1.3x10^-18
20.0 4.2x10^-16 5.2x10^-18 3.8x10^-18
30.0 6.6x10^-16 6.3x10^-18 1.0x10^-17
40.0 7.9x10^-16 1.0x10^-17 2.3x10^-17
50.0 8.7x10^-16 2.1x10^-17 5.4x10^-17

>
> >Tom Clarke
> >
> >


Harry H Conover

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Jun 2, 2000, 3:00:00 AM6/2/00
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Steve Lajoie (laj...@eskimo.com) wrote:
:
: Thus, cold fusion.

ROFL. Not even as good as wrong.

Harry C.

Barry Kearns

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Jun 2, 2000, 3:00:00 AM6/2/00
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"T L Clarke" <tcl...@ist.ucf.edu> wrote in message
news:39381474...@ist.ucf.edu...

> Stephen Lajoie wrote:
> > >Still unproven.
> >
> > No, you are unconvinced. Slightly different.
>
> Not very much. You think it is proven, I don't. We both agree, I hope,
> on the theory of gravity which is proven.

[Off-topic warning]

I don't. :)

I'm assuming, of course, that you're referring to Newton's law of universal
gravitation, where every object in the universe attracts every other object
with a force directed along the line of centers for the two objects -- that is
proportional to the product of their masses and inversely proportional to the
square of the distance between them?

That theory? I think this is one that gets a "right answer" in local cases,
but for the wrong reason. But as I mentioned, a discussion of it would be
rather rather far off-topic.... ;)

--
Barry Kearns
bke...@frii.com

Stephen Lajoie

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Jun 5, 2000, 3:00:00 AM6/5/00
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In article <3938157F...@ist.ucf.edu>,
T L Clarke <tcl...@ist.ucf.edu> wrote:

>T L Clarke wrote:
>
>> Reaction Cross Sections (cm^2)
>> T (KeV) D/T D/D D/He-3
>> 1.0 5.5x10^-21 1.5x10^-22 3 x10^-26
>> 2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
>> 5.0 1.3x10^-17 1.8x10^-19 6.7x10^-21
>> 6.0 2.6x10^-17 2.3x10^-19 3.3x10^-20
>> 7.0 4.1x10^-17 3.5x10^-19 5.3x10^-20
>> 8.0 6.0x10^-17 5.0x10^-19 8.0x10^-20
>> 9.0 8.2x10^-17 6.7x10^-19 1.3x10^-19
>> 10.0 1.1x10^-16 1.2x10^-18 2.3x10^-19
>> 15.0 2.6x10^-16 1.9x10^-18 1.3x10^-18
>> 20.0 4.2x10^-16 5.2x10^-18 3.8x10^-18
>> 30.0 6.6x10^-16 6.3x10^-18 1.0x10^-17
>> 40.0 7.9x10^-16 1.0x10^-17 2.3x10^-17
>> 50.0 8.7x10^-16 2.1x10^-17 5.4x10^-17
>>
>
>The cross section is 3 or 4 orders of magnitude down from
>that at very high energies, but it is not zero. The LANL
>experiments took several days to accumulate tritium
>so their reaction was not very vigorous. Fusion at 2 KeV
>is to me quite plausible as an explanation for their data.
>
>Tom Clarke

An eV is the energy that a single charge will have after
falling through a potential of 1 volt.

I am not sure what the mean free path is in this plasma, but it probably
isn't going to fall through the whole 2000 volt potential without
collision and thus, radiation and energy loss. A 2 keV plasma would have
vaporized the Pd. At the very least, the Pd has cooled off the plasma.

Still, this ignores the issue that if the fusion effect was caused by
deuterium in high densities in the Pd being exposed to higher energy
deuterium, it would only be a function of Pd loading and not lattice
defects.

Dennis Towne

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Jun 5, 2000, 3:00:00 AM6/5/00
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His point is that significant fusion occurs, even at potentials much
less than 2000 eV.

Additionally, a 2 kev plasma need not necessarily vaporize anything, at
least not immediately. If the plasma density is low, as it was in this
experiment, the temperature of the pd need not exceed its melting
point. See also references for the farnsworth fusor, which regularly
use many-kev plasmas with solid electrodes. The electrodes last for
quite a while, though eventually they erode beyond the point of
usefulness.

> Still, this ignores the issue that if the fusion effect was caused by
> deuterium in high densities in the Pd being exposed to higher energy
> deuterium, it would only be a function of Pd loading and not lattice
> defects.

Pd loading is dependent on lattice defects. What is the problem here?

-dennis towne

GRADinc

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Jun 5, 2000, 3:00:00 AM6/5/00
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This is T. Clarke replying from a different account.>In article


>>> Reaction Cross Sections (cm^2)
>>> T (KeV) D/T D/D D/He-3

>>> ...
>>> 2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
>>> ...

>>The cross section is 3 or 4 orders of magnitude down from
>>that at very high energies, but it is not zero. The LANL
>>experiments took several days to accumulate tritium
>>so their reaction was not very vigorous. Fusion at 2 KeV
>>is to me quite plausible as an explanation for their data.

>An eV is the energy that a single charge will have after

>falling through a potential of 1 volt.

Yes indeed.

>I am not sure what the mean free path is in this plasma, but it probably
>isn't going to fall through the whole 2000 volt potential without
>collision and thus, radiation and energy loss.

Yes I realize that. That is why I think the little cones that
form on the Pd may be so significant. They would locally
concentrate the field at the tips of the cones so that an
ion could obtain a substantial fraction of the full 2 keV
as it fell toward the cone before being scattered in
a collision.

>A 2 keV plasma would have
>vaporized the Pd. At the very least, the Pd has cooled off the plasma.

A 2 keV plasma does not vaporize the Pd. The experiment
proves this. What matters
is the heat delivered to the Pd, not the temperature.
If the plasma is not very dense, as this one is, the heating
rate will not cause substantial vaporization. There is,
however, sputtering going on as the experimenters note
so the effect of the plasma on the Pd is significant.

>Still, this ignores the issue that if the fusion effect was caused by
>deuterium in high densities in the Pd being exposed to higher energy
>deuterium, it would only be a function of Pd loading and not lattice
>defects.

You snipped my explanation of how I think lattice
defects might serve to provide a "channel" down which
the ions could penetrate into the Pd encountering more
D which, as you have pointed out, may concentrate
along defects.

Tom Clarke


Kirk Shanahan

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Jun 5, 2000, 3:00:00 AM6/5/00
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In article <3936BA20...@eskimo.com>, laj...@eskimo.com says...

>
>Data at:
>http://www.nde.lanl.gov/cf/tritweb.htm
>Note, please, that this is a Los Alamos National Lab site.

Oooohh, sends shivers down my spine....Wait was that a 'call to
authority' I just heard...

{snip}

> No real rebuttal was offered.
>

That is such a common occurrance isn't it?

>Thus, cold fusion.

Or it might be plasma cleaning....

---
Kirk L. Shanahan {{My opinions...noone else's}}


Steve Lajoie

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Jun 5, 2000, 3:00:00 AM6/5/00
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Dennis Towne wrote:
>
> Stephen Lajoie wrote:

[snip]

> His point is that significant fusion occurs, even at potentials much
> less than 2000 eV.

My point was that a plasma created with a 2000 volt potential
is not a 2000 keV plasma.



> Additionally, a 2 kev plasma need not necessarily vaporize anything, at
> least not immediately. If the plasma density is low, as it was in this
> experiment, the temperature of the pd need not exceed its melting
> point.

If a 2 keV plasma comes in touch with a 1 eV piece of Palladium,
the temperature of the plasma decreases by a great deal due to
it's lower heat capacity. 2 keV plasmas have are around 20 million
degrees in temperature.

> See also references for the farnsworth fusor, which regularly
> use many-kev plasmas with solid electrodes. The electrodes last for
> quite a while, though eventually they erode beyond the point of
> usefulness.

The solid electrodes do not come in contact with the plasma
in the Farnsworth fusor. See Patent 3,386,886. This is also
evidenced in photographs of working Farnsworth fusors where
the plasma is well within the first solid electrode.

It's a law of thermodynamics that if a 20 million degree
plasma contracts a >600 C metal, the plasma will cool
off and the metal will heat up, until both plasma and
metal have the same temperature.


> > Still, this ignores the issue that if the fusion effect was caused by
> > deuterium in high densities in the Pd being exposed to higher energy
> > deuterium, it would only be a function of Pd loading and not lattice
> > defects.
>

> Pd loading is dependent on lattice defects. What is the problem here?

The problem is that Pd loading has a NEGATIVE dependent on
lattice defects. In alpha phase Pd, the hydrogen randomly occupies
interstitial lattice locations. That's >inside< the lattice,
not in defects.

Pd can usefully absorb and desorb hydrogen for about 10,000
cycles. The defects created by the repeated cycles make it
absorb less and less Pd. The same defects enable it to produce
more and more fusion.

Steve Lajoie

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Jun 5, 2000, 3:00:00 AM6/5/00
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GRADinc wrote:
>
> This is T. Clarke replying from a different account.>In article
>
> >>> Reaction Cross Sections (cm^2)
> >>> T (KeV) D/T D/D D/He-3
> >>> ...
> >>> 2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
> >>> ...
>
> >>The cross section is 3 or 4 orders of magnitude down from
> >>that at very high energies, but it is not zero. The LANL
> >>experiments took several days to accumulate tritium
> >>so their reaction was not very vigorous. Fusion at 2 KeV
> >>is to me quite plausible as an explanation for their data.
>
> >An eV is the energy that a single charge will have after
> >falling through a potential of 1 volt.
>
> Yes indeed.
>
> >I am not sure what the mean free path is in this plasma, but it probably
> >isn't going to fall through the whole 2000 volt potential without
> >collision and thus, radiation and energy loss.
>
> Yes I realize that. That is why I think the little cones that
> form on the Pd may be so significant. They would locally
> concentrate the field at the tips of the cones so that an
> ion could obtain a substantial fraction of the full 2 keV
> as it fell toward the cone before being scattered in
> a collision.

Please explain how this works with your next statement about
how the lattice defects channel the deuterium.

> >A 2 keV plasma would have
> >vaporized the Pd. At the very least, the Pd has cooled off the plasma.
>
> A 2 keV plasma does not vaporize the Pd. The experiment
> proves this.

The experiment shows that the plasma does not vaporize the
Pd. It has not been shown that the plasma has a temperature
of 2 keV.

> What matters
> is the heat delivered to the Pd, not the temperature.

The temperature of two objects in thermal contact must
reach an equilibrium at the point of where the temperature
is the same. Thermal gradients can develop according thermal
conductivities.

What I'm trying to say is, if the plasma contacts the metal,
you either melt the metal or have a cooler plasma. In this
case, the metal didn't melt so the plasma is cooler.

> If the plasma is not very dense, as this one is, the heating
> rate will not cause substantial vaporization.

But the cooling rate of the plasma will be very great.

The assumption that this is anywhere near a 22,000,000 K
plasma is weak.

No, I don't have temperature data.

> There is,
> however, sputtering going on as the experimenters note
> so the effect of the plasma on the Pd is significant.
>

> >Still, this ignores the issue that if the fusion effect was caused by
> >deuterium in high densities in the Pd being exposed to higher energy
> >deuterium, it would only be a function of Pd loading and not lattice
> >defects.
>

> You snipped my explanation of how I think lattice
> defects might serve to provide a "channel" down which
> the ions could penetrate into the Pd encountering more
> D which, as you have pointed out, may concentrate
> along defects.

Yes, I did.

So, it is your argument that hot fusion only occurs
at the defects, and that defects attracts deuterons,
and that no significant fusion occurs at non defective
lattice points.

If hot fusion, I don't see why a lone deuteron in the Pd
lattice cannot fuse and groups of deuterons at a defect
can. It appears that NO fusion occurs at the interstitial
lattice points, yet your argument doesn't explain this.

Why wouldn't the hot deuterons fuse with the millions of
interstitial lattice deuterons? Why would it be exclusively
fuse with deuterons at defects?

The cross section would be the same.

If it has to do with close proximity, yes, deuterons are
closer at defects than they are at lattice points, as a
prior argument has shown. Cold fusion explains why the
fusion is defect dependent. Hot fusion doesn't.

> Tom Clarke

Steve Lajoie

unread,
Jun 5, 2000, 3:00:00 AM6/5/00
to

Kirk Shanahan wrote:
>
> In article <3936BA20...@eskimo.com>, laj...@eskimo.com says...
> >
> >Data at:
> >http://www.nde.lanl.gov/cf/tritweb.htm
> >Note, please, that this is a Los Alamos National Lab site.
>
> Oooohh, sends shivers down my spine....Wait was that a 'call to
> authority' I just heard...

It could be considered an appeal to authority.

Other independent researchers with private funding have been slammed
for being self serving, however, and it was my intent to point out
that this was a supervised government lab and the self serving, or
"fraud" argument, was therefore moot.


> {snip}
>
> > No real rebuttal was offered.
> >
>
> That is such a common occurrance isn't it?

All too often, I'm afraid. Many people feel that ridicule and
sarcasm is sufficient and that the actual physics can be ignored.
Especially in cases where the physics doesn't support them.



> >Thus, cold fusion.
>
> Or it might be plasma cleaning....

That was ruled out.

Dennis Towne

unread,
Jun 5, 2000, 3:00:00 AM6/5/00
to
Steve Lajoie wrote:
>
> GRADinc wrote:

[snip]

> > >A 2 keV plasma would have
> > >vaporized the Pd. At the very least, the Pd has cooled off the plasma.
> >
> > A 2 keV plasma does not vaporize the Pd. The experiment
> > proves this.
>
> The experiment shows that the plasma does not vaporize the
> Pd. It has not been shown that the plasma has a temperature
> of 2 keV.

This statement is fine...


> > What matters
> > is the heat delivered to the Pd, not the temperature.
>
> The temperature of two objects in thermal contact must
> reach an equilibrium at the point of where the temperature
> is the same. Thermal gradients can develop according thermal
> conductivities.
>
> What I'm trying to say is, if the plasma contacts the metal,
> you either melt the metal or have a cooler plasma. In this
> case, the metal didn't melt so the plasma is cooler.

This is a gross oversimplification, and there is another mechanism
here: the plasma heats the metal, and the metal conducts/radiates away
the heat fast enough that it does not vaporize or ever reach equilibrium
temperature with the plasma. Because of the very small contact area
with the plasma, this is unsuprising.


> > If the plasma is not very dense, as this one is, the heating
> > rate will not cause substantial vaporization.
>
> But the cooling rate of the plasma will be very great.
>
> The assumption that this is anywhere near a 22,000,000 K
> plasma is weak.
>
> No, I don't have temperature data.

Regardless of assumptions made, if the plasma is a plasma, then it is
above the melting temperature of the palladium. Point being that the
palladium need not be in thermal equilibrium with the plasma.

Perhaps you should get some temperature data.

-dennis towne

Steve Lajoie

unread,
Jun 5, 2000, 3:00:00 AM6/5/00
to

" After 20 hours, palladium was visibly sputtered onto the plate. The
sputter
rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"

That tells me that the plasma is has heated some of the Pd to
vapor.

The entire wire is engulfed in plasma. Again, from the web site:

" In operation, the plasma is adjusted so that it envelopes the whole
wire and
contacts the plate at a small spot."

Tritium is found in the wire, not the plate, where there was only
a small spot of plasma touching it.

"In order to resolve whether the tritium was originating in the plate or
wire,
they were separately heated after plasma 4. The wire released about 12.4
nCi
of tritium while the plate had no measurable (< 0.3 nCi) release."

I don't see how a wire completely surrounded in plasma for 300 hours
is not going to reach some sort of thermal equilibrium with the
plasma at the surface of the wire.


> > > If the plasma is not very dense, as this one is, the heating
> > > rate will not cause substantial vaporization.
> >
> > But the cooling rate of the plasma will be very great.
> >
> > The assumption that this is anywhere near a 22,000,000 K
> > plasma is weak.
> >
> > No, I don't have temperature data.
>
> Regardless of assumptions made, if the plasma is a plasma, then it is
> above the melting temperature of the palladium. Point being that the
> palladium need not be in thermal equilibrium with the plasma.

It seems to be a thermodynamic law that they are.

Your argument would be more properly stated that there is a large
thermal gradient in the plasma from near the surface (gas) to the
plasma, which is what actually is going on.


> Perhaps you should get some temperature data.

Really? You are just happy to assume that you have a 22 million
Kelvin plasma? Never mind that it is in contact with a solid
palladium wire.

Hydrogen ionizes at what? Around 13 eV? That's only 150,000 K,
isn't it? Not much fusion going on there.

> -dennis towne

Fred McGalliard

unread,
Jun 5, 2000, 3:00:00 AM6/5/00
to Steve Lajoie

Steve Lajoie wrote:
...


> I don't see how a wire completely surrounded in plasma for 300 hours
> is not going to reach some sort of thermal equilibrium with the
> plasma at the surface of the wire.

I thought the point is that if the wire were in fact at equilibrium with
the plasma, even with a plasma at only 1-2 ev, the wire would be way
over it's melting point. You have to appreciate just how low the density
is in most plasmas. The ones that are slightly dense are used in
welding, where the purpose is to melt the material.

GRADinc

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
Steve Lajoie laj...@eskimo.com
>GRADinc wrote:
>>This is T. Clarke replying from a different account
ditto re the source.

>> >>> Reaction Cross Sections (cm^2)
>> >>> T (KeV) D/T D/D D/He-3
>> >>> ...
>> >>> 2.0 2.6x10^-19 5.4x10^-21 1.4x10^-23
>> >>> ...

>> >An eV is the energy that a single charge will have after


>> >falling through a potential of 1 volt.

>> Yes indeed.

>> >I am not sure what the mean free path is in this plasma, but it probably
>> >isn't going to fall through the whole 2000 volt potential without
>> >collision and thus, radiation and energy loss.

>> Yes I realize that. That is why I think the little cones that
>> form on the Pd may be so significant. They would locally
>> concentrate the field at the tips of the cones so that an
>> ion could obtain a substantial fraction of the full 2 keV
>> as it fell toward the cone before being scattered in
>> a collision.

>Please explain how this works with your next statement about
>how the lattice defects channel the deuterium.

The lattice defects extend into the little cones.
This is a very complicated experimental system,
many things can be happening both singly and in
combination.
Perhaps the little cones grow from the lattice defects
in the as-shipped (un-annealed) Pd.

>> >A 2 keV plasma would have
>> >vaporized the Pd. At the very least, the Pd has cooled off the plasma.

>> A 2 keV plasma does not vaporize the Pd. The experiment
>> proves this.

>The experiment shows that the plasma does not vaporize the
>Pd. It has not been shown that the plasma has a temperature
>of 2 keV.

Yes. The temperature of the plasma is unknown.
It probably varies all over the place in the region of
the electrodes.

>> What matters
>> is the heat delivered to the Pd, not the temperature.

>The temperature of two objects in thermal contact must
>reach an equilibrium at the point of where the temperature
>is the same. Thermal gradients can develop according thermal
>conductivities.

I'm not so sure this applies to gas/plasma-solid interfaces.

>What I'm trying to say is, if the plasma contacts the metal,
>you either melt the metal or have a cooler plasma. In this
>case, the metal didn't melt so the plasma is cooler.

But even if this statement is true - see above doubt -
what is important is the speed with which an individual
ion impacts the Pd. The plasma would go from high temp
to metal temp in a distance about equal to the mean
free path. So at least some ions (don't forget the cones,
though) would impact at 2 KeV in this case. Some
might be less, but it only takes some to do fusion.
In fact it would be the impact of these ions that would
heat the metal.

>> If the plasma is not very dense, as this one is, the heating
>> rate will not cause substantial vaporization.

>But the cooling rate of the plasma will be very great.

Within a mean free path. There is some dimensionless
number - Knudsen number? - that governs flow and
such in rarified gases and plasmas.

>The assumption that this is anywhere near a 22,000,000 K
>plasma is weak.

The fact is there is an electric potential of 2 KeV available
to drive the ions into the Pd. The plasma may not be
at the corresponding temperature, I rather doubt if it is,
but mechanisms exist for accelerating individual ions
to KeV energies and into the Pd.

>No, I don't have temperature data.

>> There is,


>> however, sputtering going on as the experimenters note
>> so the effect of the plasma on the Pd is significant.

>> >Still, this ignores the issue that if the fusion effect was caused by
>> >deuterium in high densities in the Pd being exposed to higher energy
>> >deuterium, it would only be a function of Pd loading and not lattice
>> >defects.

>> You snipped my explanation of how I think lattice
>> defects might serve to provide a "channel" down which
>> the ions could penetrate into the Pd encountering more
>> D which, as you have pointed out, may concentrate
>> along defects.

>Yes, I did.

Well its back now.

>So, it is your argument that hot fusion only occurs
>at the defects, and that defects attracts deuterons,
>and that no significant fusion occurs at non defective
>lattice points.

Did I say only? I was merely suggesting a way that
defects could in part determin the 3:1 difference
between non-annealed and annealed. I didn't think
of Towne's explanation of simply appealing to increased
D loading.
So no that is not my argument. I am suggesting that
it is possible that the rate of fusion may increase when
defects are present.

>If hot fusion, I don't see why a lone deuteron in the Pd
>lattice cannot fuse and groups of deuterons at a defect
>can. It appears that NO fusion occurs at the interstitial
>lattice points, yet your argument doesn't explain this.

Its a matter of cross sections and probabilities.
If the D+ ion can travel further into the lattice it has
a higher chance of encountering a D atom in the lattice
and fusing. If deuterium concentration is higher at
defects this may enhance the probabiliteis as well.

>Why wouldn't the hot deuterons fuse with the millions of
>interstitial lattice deuterons? Why would it be exclusively
>fuse with deuterons at defects?

The do fuse with the millions. But unless they hit
a defect the D+ doesn't go very far and doesn't get as
many chances as a D+ "channeling" down a defect.
Annealed Pd did produce tritium so there is no question
of exclusivity.

>The cross section would be the same.

D+ to D, yes. But if the D+ scatters of a Pd and looses
its energy, then the cross section for subsequent D+/D
encounters goes way down. In a deffect Pd scattering
might be minimized.

>If it has to do with close proximity, yes, deuterons are
>closer at defects than they are at lattice points, as a
>prior argument has shown. Cold fusion explains why the
>fusion is defect dependent. Hot fusion doesn't.

Yes proximity would be important as I pointed out above.
The effect is much the same for hot LANL experiment
2 KeV as for the putative cold fusion results.

Tom Clarke


Dieter Britz

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
On Mon, 5 Jun 2000, Steve Lajoie wrote:

[...]


> If a 2 keV plasma comes in touch with a 1 eV piece of Palladium,
> the temperature of the plasma decreases by a great deal due to
> it's lower heat capacity. 2 keV plasmas have are around 20 million
> degrees in temperature.

Just wondering: you believe that fusion takes place, and that the main
reaction pathway is to 4He. This produces about 24 MeV of energy. So,
if you think that the odd 2 keV particle hitting the Pd will evaporate
the metal, how about that whopping 24 MeV?

> Pd can usefully absorb and desorb hydrogen for about 10,000
> cycles. The defects created by the repeated cycles make it
> absorb less and less Pd. The same defects enable it to produce
> more and more fusion.

Again, asking, "really wanting to know", as I think Eeyore said: I
take it you believe in the conservation of mass (of Pd), so how come
the bits it has broken up into absorb less hydrogen than the solid
initial bit? (Actually, you write that it absorbs less and less Pd,
but I think you mean hydrogen). Are you saying that Pd absorbs less
hydrogen at its surface?

-- Dieter Britz alias d...@kemi.aau.dk; http://www.kemi.aau.dk/~db
*** Echelon, bomb, sneakers, GRU: swamp the snoops with trivia! ***


Roland Smith

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
wrote:

[snip]

>" After 20 hours, palladium was visibly sputtered onto the plate. The
>sputter
>rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
>
>That tells me that the plasma is has heated some of the Pd to
>vapor.

And that tells me that Steve doesn't know a whole lot of plasma
physics. Perhaps someone should tell him about spallation and e-beam
sputtering etc. You can knock out atoms from a solid surface layer
_without_ heating it substantially this way. I _would_ tell the poor
little lamb myself but I got kill-filed for talking physics at him :).

>I don't see how a wire completely surrounded in plasma for 300 hours
>is not going to reach some sort of thermal equilibrium with the
>plasma at the surface of the wire.

Well, there is this thing called _thermal transport_ that moves energy
around and sets up thermal gradients. These can be _extremely_ large
when a plasma is bounded by a cold, sold surface. Note that the
fluorescent bulb above your desk contains a plasma at a very
substantially higher temperature than the rest of your office :).

>> > > If the plasma is not very dense, as this one is, the heating
>> > > rate will not cause substantial vaporization.
>> >
>> > But the cooling rate of the plasma will be very great.
>> >

>> > The assumption that this is anywhere near a 22,000,000 K
>> > plasma is weak.
>> >

>> > No, I don't have temperature data.

And there is the big problem. Steve wants us to believe him because
he is RIGHT, but doesnt have any data to support his position. Ho
Hum.

>> Regardless of assumptions made, if the plasma is a plasma, then it is
>> above the melting temperature of the palladium. Point being that the
>> palladium need not be in thermal equilibrium with the plasma.
>
>It seems to be a thermodynamic law that they are.

Gradients

Gradients

Gradients.

[Snip]

All the best, Roland

T L Clarke

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
Steve Lajoie wrote:

> Tritium is found in the wire, not the plate, where there was only
> a small spot of plasma touching it.

This is to be expected. On electrode (the wire) will have polarity to
accelerate D+ ions, the other (the plate) to accelerate electrons
and negative ions. Accelerated D+ ions would be expected to
cause fusion, but not e-. I would expect D- or other negative ions
to be rare.

Tom Clarke

Kirk Shanahan

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
In article <393C0013...@eskimo.com>, laj...@eskimo.com says...

>
>Kirk Shanahan wrote:
>>
>> In article <3936BA20...@eskimo.com>, laj...@eskimo.com says...
>> >
>> >Data at:
>> >http://www.nde.lanl.gov/cf/tritweb.htm
>> >Note, please, that this is a Los Alamos National Lab site.
>>
>> Oooohh, sends shivers down my spine....Wait was that a 'call to
>> authority' I just heard...
>
>It could be considered an appeal to authority.
>
>Other independent researchers with private funding have been slammed
>for being self serving, however, and it was my intent to point out
>that this was a supervised government lab and the self serving, or
>"fraud" argument, was therefore moot.
>

I'm glad to hear that. Since I am paid out of the same DOE funds I
obviously inherit the same consideration right? I really enjoy being
infallible and sacrosanct by default...:-) Oh wait, I only work for
a 'Technology Center', not a 'National Lab', so I must only walk in
puddles instead of on water...

Yes, I'm being sarcastic. I don't feel where you have been or are
currently is anything other than the most general indication of
qualifications. In any given specific case, anyone can be wrong or
can have missed something. Therefore, I don't believe your above
comment has merit. The reasons will be different, but the possibility
of self-serving and even fraud is as valid at a National Lab as
anywhere else. That being said, I am not a conspiracy theorist, and I
rarely give those kind of arguments any weight anyway.

I prefer to appeal to the logic and the body of data supporting the
logic.

>
>> {snip}
>>
>> > No real rebuttal was offered.
>> >
>>
>> That is such a common occurrance isn't it?
>
>All too often, I'm afraid. Many people feel that ridicule and
>sarcasm is sufficient and that the actual physics can be ignored.
>Especially in cases where the physics doesn't support them.
>

Add the word 'chemistry' into that and I'll agree, from both sides of
the fence.

>> >Thus, cold fusion.
>>
>> Or it might be plasma cleaning....
>
>That was ruled out.

Not to my satisfaction at least.

Stephen Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
In article <Pine.OSF.4.21.000606...@kemi.aau.dk>,

Dieter Britz <d...@kemi.aau.dk> wrote:
>On Mon, 5 Jun 2000, Steve Lajoie wrote:
>
>[...]
>> If a 2 keV plasma comes in touch with a 1 eV piece of Palladium,
>> the temperature of the plasma decreases by a great deal due to
>> it's lower heat capacity. 2 keV plasmas have are around 20 million
>> degrees in temperature.
>
>Just wondering: you believe that fusion takes place, and that the main
>reaction pathway is to 4He. This produces about 24 MeV of energy. So,
>if you think that the odd 2 keV particle hitting the Pd will evaporate
>the metal, how about that whopping 24 MeV?

There was no 24 MeV anything measured by George. That energy appears to be
distributed in smaller amounts.

>> Pd can usefully absorb and desorb hydrogen for about 10,000
>> cycles. The defects created by the repeated cycles make it
>> absorb less and less Pd. The same defects enable it to produce
>> more and more fusion.
>
>Again, asking, "really wanting to know", as I think Eeyore said: I
>take it you believe in the conservation of mass (of Pd), so how come
>the bits it has broken up into absorb less hydrogen than the solid
>initial bit? (Actually, you write that it absorbs less and less Pd,
>but I think you mean hydrogen). Are you saying that Pd absorbs less
>hydrogen at its surface?

How do you get "bits broken up" out of what I said?


Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Fred McGalliard wrote:
>
> Steve Lajoie wrote:
> ...


> > I don't see how a wire completely surrounded in plasma for 300 hours
> > is not going to reach some sort of thermal equilibrium with the
> > plasma at the surface of the wire.
>

> I thought the point is that if the wire were in fact at equilibrium with
> the plasma, even with a plasma at only 1-2 ev, the wire would be way
> over it's melting point. You have to appreciate just how low the density
> is in most plasmas. The ones that are slightly dense are used in
> welding, where the purpose is to melt the material.

I can see now that this argument of mine is going to be
unconvincing, and for good reason.

One reasonable way to make my point would be to
set up the equations, and solve them.

I don't care to get that involved.

Dennis Towne

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
Roland Smith wrote:
>
> On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
> wrote:
>
> [snip]
>
> >" After 20 hours, palladium was visibly sputtered onto the plate. The
> >sputter
> >rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
> >
> >That tells me that the plasma is has heated some of the Pd to
> >vapor.
>
> And that tells me that Steve doesn't know a whole lot of plasma
> physics. Perhaps someone should tell him about spallation and e-beam
> sputtering etc. You can knock out atoms from a solid surface layer
> _without_ heating it substantially this way. I _would_ tell the poor
> little lamb myself but I got kill-filed for talking physics at him :).

I wouldn't worry too much about it roland, I'm sure he's seeing your
posts. You see, I don't think he really has a kill file - or if he
does, he obviously isn't smart enough to figure out how to use it. See,
he's been responding to my posts recently - which indicates to me that
he's just holding the names of people he doesn't like in his head and
ignoring the posts manually. After a while of course, that list of
people falls out of his brain and he starts responding again.

My advice? At the end of every post, add a line like 'Steve, you're an
idiot'. That way you make sure to stay on the 'hot' part of the list.

-dennis towne

Dennis Towne

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
Steve Lajoie wrote:
>
> Dennis Towne wrote:
> >
> > Steve Lajoie wrote:

[snip]

> > > What I'm trying to say is, if the plasma contacts the metal,
> > > you either melt the metal or have a cooler plasma. In this
> > > case, the metal didn't melt so the plasma is cooler.
> >

> > This is a gross oversimplification, and there is another mechanism
> > here: the plasma heats the metal, and the metal conducts/radiates away
> > the heat fast enough that it does not vaporize or ever reach equilibrium
> > temperature with the plasma. Because of the very small contact area
> > with the plasma, this is unsuprising.
>

> " After 20 hours, palladium was visibly sputtered onto the plate. The
> sputter
> rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
>
> That tells me that the plasma is has heated some of the Pd to
> vapor.
>

> The entire wire is engulfed in plasma. Again, from the web site:
>
> " In operation, the plasma is adjusted so that it envelopes the whole
> wire and
> contacts the plate at a small spot."
>

> Tritium is found in the wire, not the plate, where there was only
> a small spot of plasma touching it.
>

> "In order to resolve whether the tritium was originating in the plate or
> wire,
> they were separately heated after plasma 4. The wire released about 12.4
> nCi
> of tritium while the plate had no measurable (< 0.3 nCi) release."
>

> I don't see how a wire completely surrounded in plasma for 300 hours
> is not going to reach some sort of thermal equilibrium with the
> plasma at the surface of the wire.

Jesus christ steve, you're a complete fucking idiot. I can't even bring
myself to continue talking to you - it's just not worth the time. You
just don't understand. Put me back in that mythical kill file of
yours. You know, the one that loses all its entries at the end of every
month.

-dennis towne

Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Roland Smith wrote:
>
> On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
> wrote:
>
> [snip]
>

> >" After 20 hours, palladium was visibly sputtered onto the plate. The
> >sputter
> >rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
> >
> >That tells me that the plasma is has heated some of the Pd to
> >vapor.
>

> And that tells me that Steve doesn't know a whole lot of plasma
> physics.

And this tells me that you argue with sarcasm and ridicule and
not physics.

> Perhaps someone should tell him about spallation and e-beam
> sputtering etc. You can knock out atoms from a solid surface layer
> _without_ heating it substantially this way. I _would_ tell the poor
> little lamb myself but I got kill-filed for talking physics at him :).

A damnable lie, and you know it. You were killfiled for making
antisocial, rude and low content post like this.


> >I don't see how a wire completely surrounded in plasma for 300 hours
> >is not going to reach some sort of thermal equilibrium with the
> >plasma at the surface of the wire.
>

> Well, there is this thing called _thermal transport_ that moves energy
> around and sets up thermal gradients. These can be _extremely_ large
> when a plasma is bounded by a cold, sold surface. Note that the
> fluorescent bulb above your desk contains a plasma at a very
> substantially higher temperature than the rest of your office :).

You clipped my references to thermal gradients, and ignore
the fact that I said "at the surface of the wire". That
can only be motivated by your intent to deceive and ridicule
something I didn't even say. Pretty low, in my book. Why do
you stoop to such despicable behaviors?

> >> > > If the plasma is not very dense, as this one is, the heating
> >> > > rate will not cause substantial vaporization.
> >> >
> >> > But the cooling rate of the plasma will be very great.
> >> >
> >> > The assumption that this is anywhere near a 22,000,000 K
> >> > plasma is weak.
> >> >
> >> > No, I don't have temperature data.
>
> And there is the big problem. Steve wants us to believe him because
> he is RIGHT, but doesnt have any data to support his position. Ho
> Hum.

Let me see.... The statement I am arguing against is that
you don't get a 2 keV plasma (22 million Kelvins) by sparking
an 2000 volt electric arc.

Now, without any data at all, you are DEFENDING this view,
so that you can explain the tritium production without resorting
to cold fusion explanations. Mind you, you ironically heap
gobs of ridicule about how much >I< don't know about plasma
physics.

Nope. I don't have the temperature data. Nor do you.
What's pathetic is that because I don't have the data
you think you can get away with any preposterous assumption
you want.

Tell you what. I don't care to go down this path anymore.
Go publish your great discovery about how you can EASILY
make a 1 MeV deuterium plasma by sparking an arc across
two electrodes with a million volts. At that temperature,
you're going to make break even easily.

When you get your great discovery published in the Physical
Review, I shall stand back in awe of your greatness in solving
the hot fusion problem. Or maybe I'll laugh when they tell
you the obvious.


> >> Regardless of assumptions made, if the plasma is a plasma, then it is
> >> above the melting temperature of the palladium. Point being that the
> >> palladium need not be in thermal equilibrium with the plasma.
> >
> >It seems to be a thermodynamic law that they are.
>
> Gradients
>
> Gradients
>
> Gradients.
>
> [Snip]

That about sums your arguments, a lot of snipping and then
lies, the snip being necessary to keep your lies from being
blatantly obvious.

Here is what you snipped that I said from that same post:

"Your argument would be more properly stated that there is a large
thermal gradient in the plasma from near the surface (gas) to the
plasma, which is what actually is going on."

You're being intentionally deceptive in order to ridicule me for
not knowing things I clearly explained.

You owe me an apology.

> All the best, Roland

Look, here's a much better rebuttal to what I said that doesn't
require lies, snips, ridicule and bullshit.

Steve:

You have demonstrated the fact that there is a heat energy loss
from the plasma to the palladium. Surely, you are correct that
the energy to sputter the Pd in the plasma's container is heat
that came from the plasma. However, that has nothing to do with
the steady state temperature of the system, and you've not shown
it to be less than the 22 million Kelvin assumed in the previous
post. Since energy is constantly being added to the plasma, you
cannot show that the plasma is cold simply because it is losing
heat to the Pd due to thermal contact.

^
|--- This argument blows my prior argument out of the water.
Some people have beat around the bush about it and I was
thinking about what they said this morning, and this is
what they were getting at.

The ONLY way to show what the temperature of the plasma
is, is to either measure it or set up the equations and solve
it. Since it involves a lot of values I don't have handy, I'm
not going to bother.

I will point out that if it was only a matter of sparking a
million Volts across some deuterium gas to make a 1 MeV deuterium
plasma, that hot fusion would be a no brainer. There is a big
misconception between what 2 kV and 2 keV means.

Steve Lajoie

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Jun 6, 2000, 3:00:00 AM6/6/00
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T L Clarke wrote:


>
> Steve Lajoie wrote:
>
> > Tritium is found in the wire, not the plate, where there was only
> > a small spot of plasma touching it.
>

> This is to be expected. On electrode (the wire) will have polarity to
> accelerate D+ ions, the other (the plate) to accelerate electrons
> and negative ions. Accelerated D+ ions would be expected to
> cause fusion, but not e-. I would expect D- or other negative ions
> to be rare.


I agree completely.

Dennis Towne

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Jun 6, 2000, 3:00:00 AM6/6/00
to
Steve Lajoie wrote:
>
> Fred McGalliard wrote:
> >
> > Steve Lajoie wrote:
> > ...
> > > I don't see how a wire completely surrounded in plasma for 300 hours
> > > is not going to reach some sort of thermal equilibrium with the
> > > plasma at the surface of the wire.
> >
> > I thought the point is that if the wire were in fact at equilibrium with
> > the plasma, even with a plasma at only 1-2 ev, the wire would be way
> > over it's melting point. You have to appreciate just how low the density
> > is in most plasmas. The ones that are slightly dense are used in
> > welding, where the purpose is to melt the material.
>
> I can see now that this argument of mine is going to be
> unconvincing, and for good reason.
>
> One reasonable way to make my point would be to
> set up the equations, and solve them.
>
> I don't care to get that involved.

In other words, you don't know what the hell you are talking about, and
if you tried to set up the equations you know you'd either do it wrong
or invalidate your own viewpoint. I'm sure neither of those would be
acceptable to you. Best to just ignore any further posts on the topic.

-dennis towne

Dieter Britz

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Jun 6, 2000, 3:00:00 AM6/6/00
to
On 6 Jun 2000, Stephen Lajoie wrote:

> In article <Pine.OSF.4.21.000606...@kemi.aau.dk>,
> Dieter Britz <d...@kemi.aau.dk> wrote:
> >On Mon, 5 Jun 2000, Steve Lajoie wrote:
> >
> >[...]
> >> If a 2 keV plasma comes in touch with a 1 eV piece of Palladium,
> >> the temperature of the plasma decreases by a great deal due to
> >> it's lower heat capacity. 2 keV plasmas have are around 20 million
> >> degrees in temperature.
> >
> >Just wondering: you believe that fusion takes place, and that the main
> >reaction pathway is to 4He. This produces about 24 MeV of energy. So,
> >if you think that the odd 2 keV particle hitting the Pd will evaporate
> >the metal, how about that whopping 24 MeV?
>
> There was no 24 MeV anything measured by George. That energy appears to be
> distributed in smaller amounts.

Now, Steve, you yourself have, not long ago, calculated the energy
exchanged for mass, when two D's form one 4He. That is 24 MeV or so,
and, broken up or not, remains that much. In fact, it is a point of
contention, as you have to be aware, just how it does get distributed -
with massive release of radiation, or - as the CNF lot maintains - in
some kind of benign Moessbauer-type process, ending in only heat.

>
> >> Pd can usefully absorb and desorb hydrogen for about 10,000
> >> cycles. The defects created by the repeated cycles make it
> >> absorb less and less Pd. The same defects enable it to produce
> >> more and more fusion.
> >
> >Again, asking, "really wanting to know", as I think Eeyore said: I
> >take it you believe in the conservation of mass (of Pd), so how come
> >the bits it has broken up into absorb less hydrogen than the solid
> >initial bit? (Actually, you write that it absorbs less and less Pd,
> >but I think you mean hydrogen). Are you saying that Pd absorbs less
> >hydrogen at its surface?
>
> How do you get "bits broken up" out of what I said?

Well, these defects cause it to become a matrix of isolated crystals,
from the initial largely crystalline matrix. In other words, bits
separated from each other by cracks, voids, defects, all accommodating
a bit of D2. In fact, this is thought (by some propopnents of CNF, I
might add) to lead to greater, not less, hydrogen absorption. How do
you see all this?

Steve Lajoie

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Jun 6, 2000, 3:00:00 AM6/6/00
to

I was considering this when I was driving in today. This
is possible, but I'd only have a hand waving argument to
support it.

> This is a very complicated experimental system,
> many things can be happening both singly and in
> combination.
> Perhaps the little cones grow from the lattice defects
> in the as-shipped (un-annealed) Pd.


If the cones had to do with fusion, then the rate of
tritium production would increase as the palladium was
sputtered off the wire and the cones formed. As the
Pd was sputtered off at a constant rate, our tritium
production would start from zero and increase as some
function as the cones formed.

Tritium production, however, is almost immediate, as
shown on a number of graphs. And the rate doesn't change.

So, we have tritium production at a fix rate before
the cones formed, and at the same fixed rate after
the cones have formed. I would conclude that the
formation of cones doesn't affect the rate and the
cones have nothing to do with tritium production.


> >> >A 2 keV plasma would have
> >> >vaporized the Pd. At the very least, the Pd has cooled off the plasma.
>
> >> A 2 keV plasma does not vaporize the Pd. The experiment
> >> proves this.
>
> >The experiment shows that the plasma does not vaporize the
> >Pd. It has not been shown that the plasma has a temperature
> >of 2 keV.
>
> Yes. The temperature of the plasma is unknown.
> It probably varies all over the place in the region of
> the electrodes.

I mostly agree.



> >> What matters
> >> is the heat delivered to the Pd, not the temperature.
>
> >The temperature of two objects in thermal contact must
> >reach an equilibrium at the point of where the temperature
> >is the same. Thermal gradients can develop according thermal
> >conductivities.
>
> I'm not so sure this applies to gas/plasma-solid interfaces.

It does, but all I have shown is that the plasma is losing
heat to the Pd, which doesn't show that the plasma is cool
because there's energy being added.


> >What I'm trying to say is, if the plasma contacts the metal,
> >you either melt the metal or have a cooler plasma. In this
> >case, the metal didn't melt so the plasma is cooler.
>
> But even if this statement is true - see above doubt -
> what is important is the speed with which an individual
> ion impacts the Pd. The plasma would go from high temp
> to metal temp in a distance about equal to the mean
> free path. So at least some ions (don't forget the cones,
> though) would impact at 2 KeV in this case.

I think that the assumption that you get a 2 keV plasma
(22 million Kelvins in temperature...) from a 2 kV spark
is incorrect. I think the plasma is more like 13 eV or
less, given the pink and blue colors showing the different
excitation levels of the deuterium's electrons.

> Some
> might be less, but it only takes some to do fusion.
> In fact it would be the impact of these ions that would
> heat the metal.

It's too cold. If I thought it was 2 keV, I might agree.



> >> If the plasma is not very dense, as this one is, the heating
> >> rate will not cause substantial vaporization.
>
> >But the cooling rate of the plasma will be very great.
>
> Within a mean free path. There is some dimensionless
> number - Knudsen number? - that governs flow and
> such in rarified gases and plasmas.



> >The assumption that this is anywhere near a 22,000,000 K
> >plasma is weak.
>
> The fact is there is an electric potential of 2 KeV available
> to drive the ions into the Pd.

You are confusing the 2 kV potential with the temperature
of the deuterium plasma, which is sometimes expressed in
eV. There is barely enough energy to ionize the deuterium,
which is more like 13 eV.

> The plasma may not be
> at the corresponding temperature, I rather doubt if it is,
> but mechanisms exist for accelerating individual ions
> to KeV energies and into the Pd.
>
> >No, I don't have temperature data.
>
> >> There is,
> >> however, sputtering going on as the experimenters note
> >> so the effect of the plasma on the Pd is significant.
>
> >> >Still, this ignores the issue that if the fusion effect was caused by
> >> >deuterium in high densities in the Pd being exposed to higher energy
> >> >deuterium, it would only be a function of Pd loading and not lattice
> >> >defects.
>
> >> You snipped my explanation of how I think lattice
> >> defects might serve to provide a "channel" down which
> >> the ions could penetrate into the Pd encountering more
> >> D which, as you have pointed out, may concentrate
> >> along defects.
>
> >Yes, I did.
>
> Well its back now.
>
> >So, it is your argument that hot fusion only occurs
> >at the defects, and that defects attracts deuterons,
> >and that no significant fusion occurs at non defective
> >lattice points.
>
> Did I say only?

No, I added that because it would make more sense.
compared to the number of interstitial lattice sites,
the defect locations are very few.

> I was merely suggesting a way that
> defects could in part determin the 3:1 difference
> between non-annealed and annealed. I didn't think
> of Towne's explanation of simply appealing to increased
> D loading.

Yes, but Towne was wrong because the more defects there
are, the less loading there is, not more. He knew there
was an effect but didn't know or neglected to mention
that the correlation was negative.

The more times you load deuterium into Pd, the more
fusion you get because it creates more defects but
you load less and less deuterium into the Pd because
of those defects. Pd can be effectively loaded about
10,000 times.

> So no that is not my argument. I am suggesting that
> it is possible that the rate of fusion may increase when
> defects are present.

So is mine!

But looking at figure 4 of the graph, I note that
there is tritium production at the drift rate until
CO is added, which poison's the surface but the plasma
kept it clean. When the wire melted and bent away from
the plate, (plasma off) tritium in the surrounding gas
reached a constant. (No tritium production OR no tritium
released from the wire.) Then they pumped up the D2
pressure to 600 torr and the more tritium appeared.

What produced that additional tritium?! The plasma
was off, but the amount is as if the reaction continued
for ~ 25 hours at the same rate.

If you're producing tritium at the same rate without
the plasma, then plasma cannot be causing the fusion.

The plasma is LOADING the Pd. Once loaded, the tritium
must be forming from some other process, which can only
be cold fusion.


> >If hot fusion, I don't see why a lone deuteron in the Pd
> >lattice cannot fuse and groups of deuterons at a defect
> >can. It appears that NO fusion occurs at the interstitial
> >lattice points, yet your argument doesn't explain this.
>
> Its a matter of cross sections and probabilities.

Yes.

> If the D+ ion can travel further into the lattice it has
> a higher chance of encountering a D atom in the lattice
> and fusing. If deuterium concentration is higher at
> defects this may enhance the probabiliteis as well.
>
> >Why wouldn't the hot deuterons fuse with the millions of
> >interstitial lattice deuterons? Why would it be exclusively
> >fuse with deuterons at defects?
>
> The do fuse with the millions. But unless they hit
> a defect the D+ doesn't go very far and doesn't get as
> many chances as a D+ "channeling" down a defect.

I suppose channeling can be considered less weird a discovery
than cold fusion, to some. :-)

The test would be, does fusion occur without the hot
deuterium, and it does.

And I would point out that at 13 eV, where the mean energy
of the deuterons will be, you are going to get a very,
very small cross section.

> Annealed Pd did produce tritium so there is no question
> of exclusivity.

I don't understand this last statement. Even annealed Pd
has defects, it just has fewer of them.


> >The cross section would be the same.
>
> D+ to D, yes. But if the D+ scatters of a Pd and looses
> its energy, then the cross section for subsequent D+/D
> encounters goes way down. In a deffect Pd scattering
> might be minimized.

Actually, D+ scattering off Pd would have more energy
than D+ scattering off a deuteron.

So I don't see what good bunching them together would
do in hot fusion. Your target area is decreased a bit.



> >If it has to do with close proximity, yes, deuterons are
> >closer at defects than they are at lattice points, as a
> >prior argument has shown. Cold fusion explains why the
> >fusion is defect dependent. Hot fusion doesn't.
>
> Yes proximity would be important as I pointed out above.
> The effect is much the same for hot LANL experiment
> 2 KeV as for the putative cold fusion results.

But it's not at 2 keV.

> Tom Clarke

Steve Lajoie

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Jun 6, 2000, 3:00:00 AM6/6/00
to

Kirk Shanahan wrote:
>
> In article <393C0013...@eskimo.com>, laj...@eskimo.com says...
> >
> >Kirk Shanahan wrote:
> >>
> >> In article <3936BA20...@eskimo.com>, laj...@eskimo.com says...
> >> >
> >> >Data at:
> >> >http://www.nde.lanl.gov/cf/tritweb.htm
> >> >Note, please, that this is a Los Alamos National Lab site.
> >>
> >> Oooohh, sends shivers down my spine....Wait was that a 'call to
> >> authority' I just heard...
> >
> >It could be considered an appeal to authority.
> >
> >Other independent researchers with private funding have been slammed
> >for being self serving, however, and it was my intent to point out
> >that this was a supervised government lab and the self serving, or
> >"fraud" argument, was therefore moot.
> >
>
> I'm glad to hear that. Since I am paid out of the same DOE funds I
> obviously inherit the same consideration right? I really enjoy being
> infallible and sacrosanct by default...:-) Oh wait, I only work for
> a 'Technology Center', not a 'National Lab', so I must only walk in
> puddles instead of on water...
>
> Yes, I'm being sarcastic. I don't feel where you have been or are
> currently is anything other than the most general indication of
> qualifications.

I agree. But it matters to some people, and they feel that research
that is not government funded lacks credibility.

> In any given specific case, anyone can be wrong or
> can have missed something. Therefore, I don't believe your above
> comment has merit. The reasons will be different, but the possibility
> of self-serving and even fraud is as valid at a National Lab as
> anywhere else. That being said, I am not a conspiracy theorist, and I
> rarely give those kind of arguments any weight anyway.
>
> I prefer to appeal to the logic and the body of data supporting the
> logic.

Great. Read George's helium experiment in cold fusion times.

Steve Lajoie

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Jun 6, 2000, 3:00:00 AM6/6/00
to

Dennis Towne wrote:
>
> Roland Smith wrote:
> >
> > On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
> > wrote:
> >
> > [snip]
> >
> > >" After 20 hours, palladium was visibly sputtered onto the plate. The
> > >sputter
> > >rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
> > >
> > >That tells me that the plasma is has heated some of the Pd to
> > >vapor.
> >
> > And that tells me that Steve doesn't know a whole lot of plasma

> > physics. Perhaps someone should tell him about spallation and e-beam


> > sputtering etc. You can knock out atoms from a solid surface layer
> > _without_ heating it substantially this way. I _would_ tell the poor
> > little lamb myself but I got kill-filed for talking physics at him :).
>

> I wouldn't worry too much about it roland, I'm sure he's seeing your
> posts. You see, I don't think he really has a kill file - or if he
> does, he obviously isn't smart enough to figure out how to use it. See,
> he's been responding to my posts recently - which indicates to me that
> he's just holding the names of people he doesn't like in his head and
> ignoring the posts manually. After a while of course, that list of
> people falls out of his brain and he starts responding again.

Physics, please.


> My advice? At the end of every post, add a line like 'Steve, you're an
> idiot'. That way you make sure to stay on the 'hot' part of the list.

I am sure that this post has enhanced your credibility with seriously
interested people a great deal, Townes.

> -dennis towne

Steve Lajoie

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Jun 6, 2000, 3:00:00 AM6/6/00
to

Dennis Towne wrote:

[snip]


> Jesus christ steve, you're a complete fucking idiot. I can't even bring
> myself to continue talking to you - it's just not worth the time. You
> just don't understand. Put me back in that mythical kill file of
> yours. You know, the one that loses all its entries at the end of every
> month.

Another example of a Dennis Towne post. Mr. Towne believes post like
this prove to others how intelligent he is and how inferior my thinking
is.

Where I come from, if you cannot tell someone why they are wrong, you
have no business claiming you're right.

I have had to back off a couple of times recently because I:
1) Had a weak argument for diffusion.
2) Had a really bad argument (that the plasma was losing
heat energy thus temperature was lower) to explain why
2 kV does not produce a 2 keV plasma when the plasma is
in contact with metal.

Bad arguments. I wish I hadn't made them.

But here, with Dennis Towne, what do you have?

If I am "a complete fucking idiot", then why can't Dennis Towne
tell me where I'm wrong? He says he is so smart, but where is
his explanation? The irony is that I am not even wrong!
I said:

" I don't see how a wire completely surrounded in plasma for 300 hours
is not going to reach some sort of thermal equilibrium with the
plasma at the surface of the wire."

Anyone with an electric range knows this is true. You turn on the
element, it gets hot, and after some time it reaches a constant
temperature with the bottom of the pan. Heat flows at a constant
rate from the element through the pan to the food, and stays
constant (within reason, you can boil off all the water in the
pan, for example...)

For this obvious physical truth, Dennis Towne calls me "a complete
fucking idiot" and says >I< don't understand?

I don't know what others think of this fellow. My thoughts
are that this is a crude and anti-social outburst, it completely
lacks scientific content and maturity, and speaks very, very
ill of him.

If you find it a convincing argument against a steady state
heat flow like Dennis Towne does...

Well, I his type of argument is what you're use to and you
find it convincing, I find that tragic.

Isn't this fellow worth kill filing?

Roland Smith

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Jun 6, 2000, 3:00:00 AM6/6/00
to
On Tue, 06 Jun 2000 08:50:17 -0500, Dennis Towne <so...@xirr.com>
wrote:

>Roland Smith wrote:
>>
>> On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
>> wrote:
>>
>> [snip]
>>
>> >" After 20 hours, palladium was visibly sputtered onto the plate. The
>> >sputter
>> >rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
>> >
>> >That tells me that the plasma is has heated some of the Pd to
>> >vapor.
>>
>> And that tells me that Steve doesn't know a whole lot of plasma
>> physics. Perhaps someone should tell him about spallation and e-beam
>> sputtering etc. You can knock out atoms from a solid surface layer
>> _without_ heating it substantially this way. I _would_ tell the poor
>> little lamb myself but I got kill-filed for talking physics at him :).
>
>I wouldn't worry too much about it roland, I'm sure he's seeing your
>posts. You see, I don't think he really has a kill file - or if he
>does, he obviously isn't smart enough to figure out how to use it. See,
>he's been responding to my posts recently - which indicates to me that
>he's just holding the names of people he doesn't like in his head and
>ignoring the posts manually. After a while of course, that list of
>people falls out of his brain and he starts responding again.

Well lets just say that Steves "leaky" kill file and my position in or
out of it is not causing me too much loss of sleep right now. Oh and
in a strange sort of masochistic way I kind of enjoy the occasional
Steve post ...... well let me qualify that. I have just marked 318
finals exams scripts so just about _anything_ other than marking
amuses me greatly right now :).

>My advice? At the end of every post, add a line like 'Steve, you're an
>idiot'. That way you make sure to stay on the 'hot' part of the list.
>

>-dennis towne

Oh no, I couldn't do that. That would be taking away his thunder. I
think he does it so much better in his own words and in his own
special way :).

All the best, Roland

Roland Smith

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Jun 6, 2000, 3:00:00 AM6/6/00
to
On Tue, 6 Jun 2000 14:06:33 GMT, Steve Lajoie <laj...@eskimo.com>
wrote:

Hay Steve ....... What's this then ? I thought you had me safely kill
filed. Not telling "porkies" were you.

>Roland Smith wrote:
>>
>> On Mon, 5 Jun 2000 22:10:11 GMT, Steve Lajoie <laj...@eskimo.com>
>> wrote:
>>
>> [snip]
>>
>> >" After 20 hours, palladium was visibly sputtered onto the plate. The
>> >sputter
>> >rate at 300 torr, 3.5 A, was about ~2 Angstroms/s"
>> >
>> >That tells me that the plasma is has heated some of the Pd to
>> >vapor.

>> And that tells me that Steve doesn't know a whole lot of plasma
>> physics.

Steve ..... repeat after me very carefully word you wrote above

SPUTTERING !

Not melting

Not boiling

Not vaporisation

SPUTTERING !

>And this tells me that you argue with sarcasm and ridicule and
>not physics.

Well lets see Steve. When I see one of your "post from the hip and
never mind the physics" threads on a subject I know a bit about it
does bring out the sarky side in me. I _used_ to keep that in check,
but though "what the hell" after you started making up non-existant
experimental details and then had the cheek to call people
creationists !

>> Perhaps someone should tell him about spallation and e-beam
>> sputtering etc. You can knock out atoms from a solid surface layer
>> _without_ heating it substantially this way. I _would_ tell the poor
>> little lamb myself but I got kill-filed for talking physics at him :).
>
>A damnable lie, and you know it. You were killfiled for making
>antisocial, rude and low content post like this.

If I was kill filed what are you doing replying to the post ?

And no Steve, the reason I got kill filed was that I asked you some
questions on the physics that you didn't or couldn't answer and I
pressed you on the issue. I note that you very carefully avoided the
main point of the argument by the way. Would you dain to answer the
bit about e-beam sputtering above ?



>> >I don't see how a wire completely surrounded in plasma for 300 hours
>> >is not going to reach some sort of thermal equilibrium with the
>> >plasma at the surface of the wire.
>>
>> Well, there is this thing called _thermal transport_ that moves energy
>> around and sets up thermal gradients. These can be _extremely_ large
>> when a plasma is bounded by a cold, sold surface. Note that the
>> fluorescent bulb above your desk contains a plasma at a very
>> substantially higher temperature than the rest of your office :).
>
>You clipped my references to thermal gradients, and ignore
>the fact that I said "at the surface of the wire". That
>can only be motivated by your intent to deceive and ridicule
>something I didn't even say. Pretty low, in my book. Why do
>you stoop to such despicable behaviors?

Ok Steve why not have another go then. Do you mean gradients in the
Pd or in the plasma. The two have rather different thermal
conductivites. Does the surface of the wire get above 3240 K in your
opinion ?

>> >> > > If the plasma is not very dense, as this one is, the heating
>> >> > > rate will not cause substantial vaporization.
>> >> >
>> >> > But the cooling rate of the plasma will be very great.
>> >> >
>> >> > The assumption that this is anywhere near a 22,000,000 K
>> >> > plasma is weak.
>> >> >
>> >> > No, I don't have temperature data.
>>
>> And there is the big problem. Steve wants us to believe him because
>> he is RIGHT, but doesnt have any data to support his position. Ho
>> Hum.
>
>Let me see.... The statement I am arguing against is that
>you don't get a 2 keV plasma (22 million Kelvins) by sparking
>an 2000 volt electric arc.

I don't believe I claimed that you did get such a plasma. However it
rather depends on the electron mean free path and local formation of
current fillaments and shocks that can make conditions highly non
uniform. Plasmas with discharges running through them are _not_ nice
isotropic blobs neatly in LTE.

>Now, without any data at all, you are DEFENDING this view,
>so that you can explain the tritium production without resorting
>to cold fusion explanations. Mind you, you ironically heap
>gobs of ridicule about how much >I< don't know about plasma
>physics.

Nope, I most definitely did _non_ say the plasma must be at 2keV.
What I said is that your claim that the outside of the wire _must_
have been raised to above the vaporisation point of Pd was probably
wrong for several reasons.

>Nope. I don't have the temperature data. Nor do you.
>What's pathetic is that because I don't have the data
>you think you can get away with any preposterous assumption
>you want.

I rather think that is your methodology Steve. Want to provide the
link to "Magic Boron Neutron Shields" (TM) that you didn't have data
for but made up in a previous thread ?

>Tell you what. I don't care to go down this path anymore.
>Go publish your great discovery about how you can EASILY
>make a 1 MeV deuterium plasma by sparking an arc across
>two electrodes with a million volts. At that temperature,
>you're going to make break even easily.

Such things are called Z-pinches and plasma-foci, they are non
equilibrium, produce local acceleration of particles to high energies,
will produce bucket loads of neutrons every shot and are nowhere near
break even. Not my field but the guys down the corridor have a big
one and think its rather cool :). For fusion aficionados see :-

http://www.ssc.ntu.edu.sg:8000/ckplee/Adrianpaper1/speed.htm

a "little" 14KeV machine ...... or

http://www.pp.ph.ic.ac.uk/~magpie/carb_results.htm

a "big" MeV, MA Machine. Note that the X-ray temperatures in some of
this work is _significantly_ higher (5 MeV) than the drive voltage (2
MeV) as a result of local ion beam formation.

>When you get your great discovery published in the Physical
>Review, I shall stand back in awe of your greatness in solving
>the hot fusion problem. Or maybe I'll laugh when they tell
>you the obvious.

Actually my last fusion PRL used a somewhat different method. I like
to think it was rather more elegant :). Phys. Rev. Lett. Vol.84,
No.12, pp.2634-2637 (2000).

>> >> Regardless of assumptions made, if the plasma is a plasma, then it is
>> >> above the melting temperature of the palladium. Point being that the
>> >> palladium need not be in thermal equilibrium with the plasma.
>> >
>> >It seems to be a thermodynamic law that they are.
>>
>> Gradients
>>
>> Gradients
>>
>> Gradients.
>>
>> [Snip]
>
>That about sums your arguments, a lot of snipping and then
>lies, the snip being necessary to keep your lies from being
>blatantly obvious.
>
>Here is what you snipped that I said from that same post:
>
>"Your argument would be more properly stated that there is a large
>thermal gradient in the plasma from near the surface (gas) to the
>plasma, which is what actually is going on."
>
>You're being intentionally deceptive in order to ridicule me for
>not knowing things I clearly explained.
>
>You owe me an apology.

Tell you what Steve, you appologise to the professional physicist you
called a creationist and I will look into it :).



>> All the best, Roland
>
>Look, here's a much better rebuttal to what I said that doesn't
>require lies, snips, ridicule and bullshit.
>
>Steve:
>
>You have demonstrated the fact that there is a heat energy loss
>from the plasma to the palladium. Surely, you are correct that
>the energy to sputter the Pd in the plasma's container is heat
>that came from the plasma. However, that has nothing to do with
>the steady state temperature of the system, and you've not shown
>it to be less than the 22 million Kelvin assumed in the previous
>post. Since energy is constantly being added to the plasma, you
>cannot show that the plasma is cold simply because it is losing
>heat to the Pd due to thermal contact.

Ok, so you found someone more polite than I am :). I used to post
that way as well. Most of the time and with most people I still do.

> |--- This argument blows my prior argument out of the water.
> Some people have beat around the bush about it and I was
> thinking about what they said this morning, and this is
> what they were getting at.
>
> The ONLY way to show what the temperature of the plasma
>is, is to either measure it or set up the equations and solve
>it. Since it involves a lot of values I don't have handy, I'm
>not going to bother.
>
> I will point out that if it was only a matter of sparking a
>million Volts across some deuterium gas to make a 1 MeV deuterium
>plasma, that hot fusion would be a no brainer.

It _is_ a no brainer. Just not an EFICIENT no brainer :). See the
links above.

>There is a big
>misconception between what 2 kV and 2 keV means.

The first link above is to a rig that runs at 14 KeV and makes 10E8
neutrons a shot. Not too far away from 2 KeV is it now.

Comments on the physics Steve ?

Roland

Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Dieter Britz wrote:
>
> On 6 Jun 2000, Stephen Lajoie wrote:
>
> > In article <Pine.OSF.4.21.000606...@kemi.aau.dk>,
> > Dieter Britz <d...@kemi.aau.dk> wrote:
> > >On Mon, 5 Jun 2000, Steve Lajoie wrote:
> > >
> > >[...]
> > >> If a 2 keV plasma comes in touch with a 1 eV piece of Palladium,
> > >> the temperature of the plasma decreases by a great deal due to
> > >> it's lower heat capacity. 2 keV plasmas have are around 20 million
> > >> degrees in temperature.
> > >
> > >Just wondering: you believe that fusion takes place, and that the main
> > >reaction pathway is to 4He. This produces about 24 MeV of energy. So,
> > >if you think that the odd 2 keV particle hitting the Pd will evaporate
> > >the metal, how about that whopping 24 MeV?
> >
> > There was no 24 MeV anything measured by George. That energy appears to be
> > distributed in smaller amounts.
>
> Now, Steve, you yourself have, not long ago, calculated the energy
> exchanged for mass, when two D's form one 4He. That is 24 MeV or so,
> and, broken up or not, remains that much. In fact, it is a point of
> contention, as you have to be aware, just how it does get distributed -
> with massive release of radiation, or - as the CNF lot maintains - in
> some kind of benign Moessbauer-type process, ending in only heat.

I'm puzzled here. Yes, I can see how 2 keV deuterium (or even
13 eV) plasma can knock off Pd atoms from the lattice. But George
didn't find any ionizing radiation other than from the alpha particles
emitted by cold fusion. The vessels just got hot. 24 MeV spread out
over the lattice isn't going to (necessarily) heat it to vapor.
Yes, it would add heat.

> > >> Pd can usefully absorb and desorb hydrogen for about 10,000
> > >> cycles. The defects created by the repeated cycles make it
> > >> absorb less and less Pd. The same defects enable it to produce
> > >> more and more fusion.
> > >
> > >Again, asking, "really wanting to know", as I think Eeyore said: I
> > >take it you believe in the conservation of mass (of Pd), so how come
> > >the bits it has broken up into absorb less hydrogen than the solid
> > >initial bit? (Actually, you write that it absorbs less and less Pd,
> > >but I think you mean hydrogen). Are you saying that Pd absorbs less
> > >hydrogen at its surface?
> >
> > How do you get "bits broken up" out of what I said?
>
> Well, these defects cause it to become a matrix of isolated crystals,
> from the initial largely crystalline matrix. In other words, bits
> separated from each other by cracks, voids, defects, all accommodating
> a bit of D2. In fact, this is thought (by some propopnents of CNF, I
> might add) to lead to greater, not less, hydrogen absorption. How do
> you see all this?

Even annealed metal has isolated crystal grains, they have just
been allowed to grow larger.

Hydrogen is absorbed into Pd and mostly it occupies the interstitial
lattice points. Some of those "points" are defect locations.

Defects have been found to cause increased fusion, and decreased
absorption. As I've said, Pd can be cycled with hydrogen about
10,000 times, then the defects caused by the repeated hydrogen
cycling causes it to lose it's ability to store hydrogen.

Roland Smith

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
On Tue, 6 Jun 2000 15:08:18 GMT, Steve Lajoie <laj...@eskimo.com>
wrote:

Now I wouldnt normally do this ..... but it's just too good an
opertunity to miss this time ...............

[snip]

>I think that the assumption that you get a 2 keV plasma
>(22 million Kelvins in temperature...) from a 2 kV spark
>is incorrect. I think the plasma is more like 13 eV or
>less, given the pink and blue colors showing the different
>excitation levels of the deuterium's electrons.

[snip]

Would you care to give us a description of what a 2 keV deuterium
plasma _would_ look like and how you would tell the difference between
it and a 13 eV plasma given that the human eyes cut of range is
something less than 4 eV.

Hay ...... pass me those X-ray specs will you :).

Roland

Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Dennis Towne wrote:
>
> Steve Lajoie wrote:
> >
> > Fred McGalliard wrote:
> > >
> > > Steve Lajoie wrote:
> > > ...

> > > > I don't see how a wire completely surrounded in plasma for 300 hours
> > > > is not going to reach some sort of thermal equilibrium with the
> > > > plasma at the surface of the wire.
> > >

> > > I thought the point is that if the wire were in fact at equilibrium with
> > > the plasma, even with a plasma at only 1-2 ev, the wire would be way
> > > over it's melting point. You have to appreciate just how low the density
> > > is in most plasmas. The ones that are slightly dense are used in
> > > welding, where the purpose is to melt the material.
> >
> > I can see now that this argument of mine is going to be
> > unconvincing, and for good reason.
> >
> > One reasonable way to make my point would be to
> > set up the equations, and solve them.
> >
> > I don't care to get that involved.
>
> In other words, you don't know what the hell you are talking about, and
> if you tried to set up the equations you know you'd either do it wrong
> or invalidate your own viewpoint. I'm sure neither of those would be
> acceptable to you. Best to just ignore any further posts on the topic.

Nope. I don't know the thermal conductivity of a plasma of
unknown temperature, and there's an obvious error in the
paper that makes it difficult to know how much energy is
going into the plasma.

It is rather obvious that the plasma is only near 13 eV or
so and not 2 keV, and that I don't even need to make that
argument anyway.

Please feel free to explain how you think you can get a plasma
temperature of 22 million Kelvins with a 2 kV, 5 Amp arc.

When I was in school, you had to consider the density of the
gas and the ionization energy of the gas to calculate the
break down voltage. It wasn't a matter of 2 kV creates
a 2 keV in temperature gas. 2 kV is a potential, 2 keV is
an energy; a very significant difference.

T L Clarke

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to Steve Lajoie
Steve Lajoie wrote:

> It is rather obvious that the plasma is only near 13 eV or
> so and not 2 keV, and that I don't even need to make that
> argument anyway.

I would tend to think the plasma or conducting gas - can it
be said to be fully a plasma? - is only at several (tens) of thousands
of degrees not millions.

> Please feel free to explain how you think you can get a plasma
> temperature of 22 million Kelvins with a 2 kV, 5 Amp arc.

You don't need a plasma temperature that high. You only need
D+ ions hitting the cathode with energies of the order of 2 KeV.

This discussion reminded me of the "cathode dark space" and I found
the following at http://www.cerac.com/pubs/cmn/cmn6_3.htm

"In the case of plasma sputtering, the target, which is the source
material, is made the cathode, and the chamber walls or some other
electrode is the anode. A voltage is developed across these
electrodes; a discharge plasma is developed which generates
electrons and ions and imparts kinetic energy to the ionized working
gas. Ar+ ions bombard the target freeing surface material. The
interactions between electrodes and ionized species and electrons is
complicated, and the variety of sputtering configurations existent
emphasize specific aspects of the plasma physics that is involved.
... In all forms of plasma
sputtering, a virtual electrode is created at the boundary between
the plasma and a volume known as the Crook's dark space, where
electronic and ionic interactions are absent. Ar+ ions are extracted
from the plasma and accelerated across the dark space to impinge
on the target. ..."

In a similar way one would expect the D+ ions to accelerate across
the dark space and hit the Pd. The energy of the ions depends on
the voltage gap across the dark space which depends on stuff we don't
know, the conductivity of the plasma and thus voltage drop prior
to the dark space at 5 amps etc.

> When I was in school, you had to consider the density of the
> gas and the ionization energy of the gas to calculate the
> break down voltage.

The D gas is broken down.

> It wasn't a matter of 2 kV creates
> a 2 keV in temperature gas. 2 kV is a potential, 2 keV is
> an energy; a very significant difference.

Yes. But we are arguing about whether 2 kV can accelerate a
charge-one ion to 2 keV or some significant fraction of the
2 keV such as 1.5 keV (or even 1 keV). Energetically this
is possible.

Tom Clarke


Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Roland Smith wrote:
>

> Well lets just say that Steves "leaky" kill file and my position in or
> out of it is not causing me too much loss of sleep right now. Oh and
> in a strange sort of masochistic way I kind of enjoy the occasional
> Steve post ...... well let me qualify that. I have just marked 318
> finals exams scripts so just about _anything_ other than marking
> amuses me greatly right now :).
>
> >My advice? At the end of every post, add a line like 'Steve, you're an
> >idiot'. That way you make sure to stay on the 'hot' part of the list.
> >
> >-dennis towne
>
> Oh no, I couldn't do that. That would be taking away his thunder. I
> think he does it so much better in his own words and in his own
> special way :).

So, do you really believe that this reflects well of your
knowledge and character, and badly upon mine?

You really believe it convinces other people? Even about
physics?

Is it the type of exchange that you are use to? Is this
normal for you?

Just asking. You and Mr. Towne seem so at home with this
exchange. I'm curious from an anthropological point of
view. I'm pretty sure what your answers would be. It
just amazes me.

Kirk Shanahan

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to
In article <393D14A9...@eskimo.com>, laj...@eskimo.com says...
>
>Kirk Shanahan wrote:
>>
{snip}

>> I prefer to appeal to the logic and the body of data supporting the
>> logic.
>
>Great. Read George's helium experiment in cold fusion times.
>
>

I have, in great detail. I have also contatced some of the other principals
involved. I also discussed the He analytical chemistry, or lack thereof,
several months ago. I remain unconvinced...

I understand some new data has been presented at ICCF8.

{snip}

Steve Lajoie

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to

Roland Smith wrote:
>
> On Tue, 6 Jun 2000 15:08:18 GMT, Steve Lajoie <laj...@eskimo.com>
> wrote:
>
> Now I wouldnt normally do this ..... but it's just too good an
> opertunity to miss this time ...............
>
> [snip]
>

> >I think that the assumption that you get a 2 keV plasma
> >(22 million Kelvins in temperature...) from a 2 kV spark
> >is incorrect. I think the plasma is more like 13 eV or
> >less, given the pink and blue colors showing the different
> >excitation levels of the deuterium's electrons.
>

> [snip]
>
> Would you care to give us a description of what a 2 keV deuterium
> plasma _would_ look like and how you would tell the difference between
> it and a 13 eV plasma given that the human eyes cut of range is
> something less than 4 eV.
>
> Hay ...... pass me those X-ray specs will you :).

It doesn't bother you that your (erroneous) argument
bites its own tail? If you can't see a 13 eV plasma,
you can't see a 2 keV plasma.... Yet, is visible. They
have reported it as so.

But in this case, we mean 13 eV and 2 keV as a temperature,
and the radiation given off as photons is distributed over
a wide range and is not monochromatic. There are plenty
of photons in the visible range for the eye to see.

Fred McGalliard

unread,
Jun 6, 2000, 3:00:00 AM6/6/00
to lo...@www.spec.ph.ic.ac.uk

Roland Smith wrote:
...


> The first link above is to a rig that runs at 14 KeV and makes 10E8
> neutrons a shot. Not too far away from 2 KeV is it now.


I recall hearing recently of a variation on the zeta pinch that I
thought was very interesting. They used a cylindrical collection of
exploding wires to produce an X-Ray compression wave to heat and
compress a preheated Z arc before it could twist itself up too much.
Sort of like an electronic version of the fission fusion bomb. Neat
idea, but rather too one shot for my liking.

GRADinc

unread,
Jun 7, 2000, 3:00:00 AM6/7/00
to
Steve Lajoie

>> The lattice defects extend into the little cones.

>I was considering this when I was driving in today. This
>is possible, but I'd only have a hand waving argument to
>support it.

Yes, hand waving on my part as well.
Without detailed modeling or measurements on
can only wave one's hands plausibly.

>If the cones had to do with fusion, then the rate of
>tritium production would increase as the palladium was
>sputtered off the wire and the cones formed. As the
>Pd was sputtered off at a constant rate, our tritium
>production would start from zero and increase as some
>function as the cones formed.

I believe that is what the experimenters found.
I don't have the web site handy from this account,
but I think they remarked that tritium production
started after 20 hours into a run. The formation
of cones taking many hours would account for this.

>Tritium production, however, is almost immediate, as
>shown on a number of graphs. And the rate doesn't change.

Well that prompts me to go to the effort to close one
window and open another and double click and ....
that wasn't so hard.
Here is the quote I mis-remembered:
"After a few hours of plasma operation the voltage-current
stabilized, presumably due to the formation of small cones
(10-20 microns high) all over the surface of the wire. After

20 hours, palladium was visibly sputtered onto the plate. "

But looking at Figure 2, to me all the runs are
rather irregular in rate. The green starts slowly
after a few hours, makes a jump in rate at 40 hours,
then plateaus around 75 hours, starts up again
with some wiggles at an intermediate rate until the end.

The black trace doesn't do much until 20 hours, then
is more or less constant with maybe a slight increase
at 50, then a increase in rate at 70 followed by some
irregularities up to the jump at 110 ours that is due
to addition of CO2(?)

The brown trace doesn't do much for the first 70
hours, the ramps up to 130 hours where it plateaus
until the (CO2) jump.

The purple trace is constant until about 40 hours
and irregularly approaches a plateay at about
150 hours.

>So, we have tritium production at a fix rate before
>the cones formed

No. Combining the author's remarks with the
data in Figure 2 it seems to be little if any
production until cones form.

> and at the same fixed rate after
>the cones have formed. I would conclude that the
>formation of cones doesn't affect the rate and the
>cones have nothing to do with tritium production.

I don't see how you can say the rate is the same
after cone formation.

>I think that the assumption that you get a 2 keV plasma
>(22 million Kelvins in temperature...) from a 2 kV spark
>is incorrect.

That is not my assumption. All I am assuming is
that you get some ions at singificant energy,
say 1-2 keV, from the experimental apparatus.
The majority of the plasma - away from the little
cones - may be at a fairly moderate temperature.
The concept of temperature may not even apply
to such localized occurences of energetic ions
as the plasma may not be in equilibrium.

> I think the plasma is more like 13 eV or
>less, given the pink and blue colors showing the different
>excitation levels of the deuterium's electrons.

Could be, but this does not rule out the occurrence
of high energy fast ions locally near the cones
[or in the Crook's dark layer - although I am informed
via an e-mail that the dark layer does not work in such
a way as to produce energetic ions, but I'm not absolutely
convinced that something like that might not be going on]

>> Some
>> might be less, but it only takes some to do fusion.
>> In fact it would be the impact of these ions that would
>> heat the metal.

>It's too cold. If I thought it was 2 keV, I might agree.

Some, I used the word "some". If only some are
at high energy then the average temperature can
be low.

>> The fact is there is an electric potential of 2 KeV available
>> to drive the ions into the Pd.

>You are confusing the 2 kV potential with the temperature
>of the deuterium plasma,

I am not confusing this. I am only recognizing
the availability of a 2 kV potential for accelerating
ions in this apparatus. We are debating whether
there is some way to locally - I favor at cones -
accelerate ions to near 2KeV energy.
The average temperature of the bulk of the plasma
is not important. What matters is the maximum
energy at which D+ ions hit the deuterated Pd.

>There is barely enough energy to ionize the deuterium,
>which is more like 13 eV.

2000 volts at 5 amps is 10 kW. Quite a lot of power
applied to not very much gas. I know it is irrelevant
to look at it this way, but it may help intuit how
there is more than enough energy by orders of
magnitude to ionize the deuterium.
....

>> >So, it is your argument that hot fusion only occurs
>> >at the defects, and that defects attracts deuterons,
>> >and that no significant fusion occurs at non defective
>> >lattice points.

>> Did I say only?

>No, I added that because it would make more sense.
>compared to the number of interstitial lattice sites,
>the defect locations are very few.

It does not make more sense. Consider the "only"
removed.

>>I didn't think
>> of Towne's explanation of simply appealing to increased
>> D loading.

>Yes, but Towne was wrong because the more defects there
>are, the less loading there is, not more.

Doesn't matter, cones and defects explain the data.

>> So no that is not my argument. I am suggesting that
>> it is possible that the rate of fusion may increase when
>> defects are present.

>So is mine!

Well we are both trying to explain the same data,
so of course we tend to agree in this regard.

>But looking at figure 4 of the graph, I note that
>there is tritium production at the drift rate until
>CO is added, which poison's the surface but the plasma
>kept it clean.

But they say the pressure of D decreases more than
can be accounted for by reaction with CO2 implying
absorbtion of additional D in the Pd.
As they note:
"However, in the presence of a reactive energetic plasma
the surface is cleaned of these materials and deuterium is
allowed to disassociate on the surface and diffuse in.
When the plasma ceases, the surface poison reabsorbs
inhibiting deuterium from recombining on the surface. "

>When the wire melted and bent away from
>the plate, (plasma off) tritium in the surrounding gas
>reached a constant. (No tritium production OR no tritium
>released from the wire.) Then they pumped up the D2
>pressure to 600 torr and the more tritium appeared.

I don't read the graph that way. It looks to me like
the increas in T is coincident with the line going to
600 torr. The increase in tritium can be attributed
to trace T in the D gas that is introduced. They say the
D has 90 pCi/l of tritium. The increase is about 200 pCi/l,
so this is a plausible explanation.

>What produced that additional tritium?! The plasma
>was off, but the amount is as if the reaction continued
>for ~ 25 hours at the same rate.

It looks to me like it came in with the D at the
concentration of 90 pCi/l of tritium.

>If you're producing tritium at the same rate without
>the plasma, then plasma cannot be causing the fusion.

Yes. But there are alternatives to fusion without
the 2kV electrical input.

>The plasma is LOADING the Pd. Once loaded, the tritium
>must be forming from some other process, which can only
>be cold fusion.

I disagree.
...


>The test would be, does fusion occur without the hot
>deuterium, and it does.

And I don't think it does.

>And I would point out that at 13 eV, where the mean energy
>of the deuterons will be, you are going to get a very,
>very small cross section.

And I point out that it is the maximum energy, not
the mean energy that is significant.

>> Annealed Pd did produce tritium so there is no question
>> of exclusivity.

>I don't understand this last statement. Even annealed Pd
>has defects, it just has fewer of them.

OK. So there is not data concerning fusion, hot
or cold in deuterated Pd without defects.

>Actually, D+ scattering off Pd would have more energy
>than D+ scattering off a deuteron.

Yes, center of mass and all that.

>So I don't see what good bunching them together would
>do in hot fusion.

Only the D+/D collisions can cause fusion.
D+/Pd collisions no fusion, but less energy
loss. This rapidly becomes handwaving without
math. D+/Pd could have great scattering angles.

Well perhaps I should drop considerations of defects
and just stick to my favorite suspect the cones.

>Your target area is decreased a bit.

But one final last remark, my hand waving intuition
says this might be counterbalanced by the increased
concentration of D in a defect.

>> The effect is much the same for hot LANL experiment
>> 2 KeV as for the putative cold fusion results.

>But it's not at 2 keV.

But _some_ of them are.

Tom Clarke


Dieter Britz

unread,
Jun 7, 2000, 3:00:00 AM6/7/00
to
On Tue, 6 Jun 2000, Steve Lajoie wrote:

[...]


> Hydrogen is absorbed into Pd and mostly it occupies the interstitial
> lattice points. Some of those "points" are defect locations.
>
> Defects have been found to cause increased fusion, and decreased
> absorption. As I've said, Pd can be cycled with hydrogen about
> 10,000 times, then the defects caused by the repeated hydrogen
> cycling causes it to lose it's ability to store hydrogen.

I suggest you look up some basic crystallography. Interstitial sites
are not defect sites; the word interstitial implies a region of intact
crystal structure without defects. I am not saying that there can't be
hydrogen at defect sites. But you still have not explained WHY you
think Pd absorbs less hydrogen after that many cycles, i.e. after it
has got a lot of defects. Are you, btw, guessing at this, or is it an
observation made by somebody? Who? I wonder what the experts on metal
hydrides would say about this.

Richard Schultz

unread,
Jun 7, 2000, 3:00:00 AM6/7/00
to
Steve Lajoie (laj...@eskimo.com) wrote:

: Where I come from, if you cannot tell someone why they are wrong, you


: have no business claiming you're right.

For some reason, no posts for s.p.f. appeared on our server from some time
in mid-May until today. So I don't know the absolute number, but I
know that the number of times he corrected you was >1. People get
frustrated when they explain things to you and you refuse to accept the
explanation because of its source.

: But here, with Dennis Towne, what do you have?

: If I am "a complete fucking idiot", then why can't Dennis Towne
: tell me where I'm wrong? He says he is so smart, but where is
: his explanation? The irony is that I am not even wrong!

The irony is that you are so wrong that it boggles the mind.

: I said:

: " I don't see how a wire completely surrounded in plasma for 300 hours


: is not going to reach some sort of thermal equilibrium with the

: plasma at the surface of the wire."

: Anyone with an electric range knows this is true. You turn on the

: element, it gets hot, and after some time it reaches a constant
: temperature with the bottom of the pan. Heat flows at a constant
: rate from the element through the pan to the food, and stays
: constant (within reason, you can boil off all the water in the
: pan, for example...)

You cook in a plasma? And I thought that microwaving was a fast
way of cooking!

: For this obvious physical truth, Dennis Towne calls me "a complete


: fucking idiot" and says >I< don't understand?

It's not an obvious physical truth. It reveals that you know neither
what a gradient is, nor the difference between temperature and heat.

-----
Richard Schultz sch...@mail.biu.ac.il
Department of Chemistry tel: 972-3-531-8065
Bar-Ilan University, Ramat-Gan, Israel fax: 972-3-535-1250
-----
". . .Mr Schutz [sic] acts like a functional electro-terrorist who
impeads [sic] scientific communications with his too oft-silliness."
-- Mitchell Swartz, sci.physics.fusion article <EEI1o...@world.std.com>

Richard Schultz

unread,
Jun 7, 2000, 3:00:00 AM6/7/00
to
Dieter Britz (d...@kemi.aau.dk) wrote:

: Now, Steve, you yourself have, not long ago, calculated the energy


: exchanged for mass, when two D's form one 4He. That is 24 MeV or so,
: and, broken up or not, remains that much. In fact, it is a point of
: contention, as you have to be aware, just how it does get distributed -
: with massive release of radiation, or - as the CNF lot maintains - in
: some kind of benign Moessbauer-type process, ending in only heat.

Just to clear up a possible misconception: the process may be benign,
but it can't be Moessbauer. I explained this in great detail to
Mitchell Swartz, on more than one occasion. Not that that stopped him
from continuing to invoke the Moessbauer Effect.

Richard Schultz

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Jun 7, 2000, 3:00:00 AM6/7/00
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Steve Lajoie (laj...@eskimo.com) wrote:

: Nope. I don't know the thermal conductivity of a plasma of


: unknown temperature, and there's an obvious error in the

: paper that makes it difficult to know how much energy is
: going into the plasma.

Do you even know if the plasma *has* a temperature?

: 2 kV is a potential, 2 keV is an energy; a very significant difference.

I think that you should consider rephrasing that sentence. After all,
you wouldn't want people to think that you were dumb or something.

Roland Smith

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Jun 7, 2000, 3:00:00 AM6/7/00
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On Tue, 6 Jun 2000 19:27:01 GMT, Fred McGalliard
<frederick.b...@boeing.com> wrote:

>
>
>Roland Smith wrote:
>...


>> The first link above is to a rig that runs at 14 KeV and makes 10E8
>> neutrons a shot. Not too far away from 2 KeV is it now.
>

>I recall hearing recently of a variation on the zeta pinch that I
>thought was very interesting. They used a cylindrical collection of
>exploding wires to produce an X-Ray compression wave to heat and
>compress a preheated Z arc before it could twist itself up too much.
>Sort of like an electronic version of the fission fusion bomb.

I remember hearing about some of the first Sandia wire array shots at
an APS meeting in around 1996 and work has been progressing in a
number of labs since. The amount of x-ray power you can get from
these devices is simply stunning. Something like 300 TW (10E12 W) for
several 100 ns if memory serves me correctly. The "wall plug"
eficiency is also extremely high. However the temperature of the
radiation drive is a little too cool for eficient compression of a
fuel pellet. They get something like a 100 eV radiation temperature
and would really like >350 eV. There is quite a lot of work going on
to study the wire array collaps physics. The Sandia machine is hugely
powerful but actually quite hard to get diagnostics on I believe. We
have a "smaller" machine here (its still bloody big) that is running a
lot of probing experiments to look at related physics.

Take a look at :- http://www.pp.ph.ic.ac.uk/~magpie/ for a few recent
results.

>Neat idea, but rather too one shot for my liking.

Well yes and no :). The Sandia crew were thinking of building a
"spare" load section that could be slotted in while the other was
being refurbished which would double the shot rate. Putting that
amount of current through anything tends to make a bit of a mess :).
I have seen some ideas for high rep rat versions with liquid metal
contacts etc that could be pulsed more rapidly. Dont think anyone is
building one yet though.

All the best, Roland


Roland Smith

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Jun 7, 2000, 3:00:00 AM6/7/00
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On Tue, 6 Jun 2000 17:51:30 GMT, Steve Lajoie <laj...@eskimo.com>
wrote:

>
>
>Roland Smith wrote:
>>
>> On Tue, 6 Jun 2000 15:08:18 GMT, Steve Lajoie <laj...@eskimo.com>
>> wrote:
>>
>> Now I wouldnt normally do this ..... but it's just too good an
>> opertunity to miss this time ...............
>>
>> [snip]
>>

>> >I think that the assumption that you get a 2 keV plasma
>> >(22 million Kelvins in temperature...) from a 2 kV spark
>> >is incorrect. I think the plasma is more like 13 eV or
>> >less, given the pink and blue colors showing the different
>> >excitation levels of the deuterium's electrons.
>>

>> [snip]
>>
>> Would you care to give us a description of what a 2 keV deuterium
>> plasma _would_ look like and how you would tell the difference between
>> it and a 13 eV plasma given that the human eyes cut of range is
>> something less than 4 eV.
>>
>> Hay ...... pass me those X-ray specs will you :).
>
>It doesn't bother you that your (erroneous) argument
>bites its own tail?

Nope, it doesn't ...... see below.

>If you can't see a 13 eV plasma,
>you can't see a 2 keV plasma.... Yet, is visible. They
>have reported it as so.

That was my point :).

Steve, your claim was that _you_ could tell the temperature of the
plasma by looking at the visible emission ...... by eye :). I have no
doubt that there is a visible plasma if you strike a 2 kV arc in
deuterium, I strongly contest what you claim to be able to say about
the plasma temperature from that.

Now to get back to the physics, there is this thing called the
Boltzmann distribution. If a plasma has a mean temp of 13 eV then all
the visible transitions can be excited. If a plasma has a mean temp
of 2 keV then all the visible transitions can be excited to.

Note that not all of the plasma must be (or in this case can be) at
that mean temperature as there is no magnetic confinement and the
plasma touches the cold walls of the container. Ergo there are
temperature gradients and plenty of recombination, bound-bound
transitions and bremstrahlung going on to produce visible light in
either case.

You don't tell the difference by looking at visible light, you go and
do a continuum slope measurement in the XUV. And just to make life
interesting, you do it space resolved to make sure you are not looking
at a "cold" boundary layer instead of the hot bits. Oh, and if you
want to do it well you do some interferometry as well to get the
electron density and the density profile. If you _really_ want to
nail it you do it all again short pulse time framed to look for local
hot spots and instabilities.

>But in this case, we mean 13 eV and 2 keV as a temperature,
>and the radiation given off as photons is distributed over
>a wide range and is not monochromatic. There are plenty
>of photons in the visible range for the eye to see.

Yes I PERFECTLY understand that, there will of course be bucket loads
of visible light in either case.

Now back to the original question, could you please tell us how you
can tell the difference between a 13 eV deuterium plasma and a 2 keV
plasma by looking at a visible wavelength image. In both cases ALL
the available visible transitions can be excited.

Roland ....... who just happens to have made multi eV and multi keV
deuterium plasmas from time to time :).


Arthur Carlson

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Jun 7, 2000, 3:00:00 AM6/7/00
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Excuse me, Tom, but I think you are violating the rule against
introducing valid physics in this thread!

Anytime there are charged particles around kilovolts of potential, one
must consider the possibility that they are tapping this energy and
thermalizing it. I would certainly investigate this possibility
thoroughly before postulating novel nuclear physics. That said, I
agree with Steve Lajoie (who says it's "rather obvious") and Tom
Clarke (who "tend[s] to think") that the plasma temperature is likely
to be closer to a few eV than a few keV.

Tom then refocuses the discussion on the fact that the fundamental
issue is the energy of the ions, not their temperature. He emphasizes
the "stuff we don't know", but I am willing to state with some degree
of certainty my feeling that most of the voltage drop will be across
the cathode sheath, so that most of the ions reaching the surface will
have an energy nearly equal to the applied potential times their
charge.

The real difficulty for the hot fusion crowd, to which I still belong,
is that the reaction rate of 2 keV monoenergetic ions is a heck of a
lot lower than that for an ion temperature of 2 keV. Most of the
reactions in a thermal plasma at temperatures up to several tens of
keV occur in the tail of the Maxwell distribution. This is true with
a vengeance at energies below a few keV. (In a serious scientific
discussion, we would never have gotten this far without some
quantitative input. Who is willing to look up the actual energy
dependence of the fusion cross section?)

However, there are processes other than thermalization which can raise
the energy of a fraction of the ions. For example, the incoming ions
will be neutralized upon contact with the surface. Most will be
absorbed, but several per cent will be reflected from the surface with
nearly their initial energy. If one of these makes a head-on
collision with an incoming ion, the energy available for the reaction
will be equivalent to an 8 keV ion colliding with a stationary
deuteron. Alternatively, the outgoing neutral may be ionized and
backscattered from an impurity ion, so that it has an energy near 4
keV after passing a second time through the sheath. Such processes
may only affect a small fraction of the flux (~10e-4?), but they may
dominate the fusion rate due to the very sensitive dependence on
energy.

It should be obvious by now that this experiment, at the least, is not
the unambiguous, knock-down proof of cold fusion that Steve Lajoie
originally suggested it was. There are classes of explanations which
can be ruled in or out unambiguously by relatively simple
measurements. The hot-fusion hypothesis, e.g., by a neutron
measurement. The isotopic enrichment hypothesis by experiments with
varying initial ratios of H:D:T. In my opinion, this discussion
should be put on ice until the experimenters supply us with some data
of that sort.

--
To study, to finish, to publish. -- Benjamin Franklin

Dr. Arthur Carlson
Max Planck Institute for Plasma Physics
Garching, Germany
car...@ipp.mpg.de
http://www.ipp.mpg.de/~Arthur.Carlson/home.html

As usual, if I am caught or killed, the Institute
will disavow any knowledge of my actions.

T L Clarke

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Jun 7, 2000, 3:00:00 AM6/7/00