Tritium.
Steve Lajoie
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
> 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
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
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
> 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
:
: Thus, cold fusion.
ROFL. Not even as good as wrong.
Harry C.
Barry Kearns
"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
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
>>
>
>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
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
>>> 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
>
>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
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
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
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
>
> 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
" 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
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
>>This is T. Clarke replying from a different account
>> >>> 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
[...]
> 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
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
> 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
>
>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
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
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
>
> 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
>
> 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
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
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
>
> Fred McGalliard wrote:
> >
> > Steve Lajoie wrote:
> > ...
> > > is not going to reach some sort of thermal equilibrium with the
> > > plasma at the surface of the wire.
> >
> > 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.
>
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
> 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
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
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
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
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
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
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
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
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
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
> 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
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
>
>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
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
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
>> 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
[...]
> 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
: 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
: 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
: 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
<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
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
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
>
> ... (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?)
You must have missed my earlier post"
> 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
> 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.
As I said to Steve, this is a very complicated system. I appreciate
you ideas as to how the 2 KeV can become input to fusion.
> ...In my opinion, this discussion
> should be put on ice until the experimenters supply us with some data
> of that sort.
You are probably correct in this.
Tom Clarke
Kirk Shanahan
d...@kemi.aau.dk says...
Well, I don't think I really qualify as an expert in the hydride field.
I've only been at it for 5 years now. The real experts have been going
for 30+ years. But, I haven't heard of what is being proposed here, that
10K cycles causes Pd to stop absorbing H2. It doesn't make sense to me,
but 10K cycles is larger than usual studies. (I would also appreciate
a reference to this.)
Cycling virgin Pd with H2 does induce dislocations as a means to relieve
the stress induced by the lattice expansion Pd undergoes upon hydriding.
But this process is usually finished with the first 10 cycles or so
(depending on how the Pd was made and treated). After that it just
keeps on going, just like the Energizer bunny.
There is a proposal out there that there is also some vacancy creation
that occurs during hydriding (Y. Fukai I believe), but the jury is
still out on that one as I recall.
Dieter is also correct in that defects are trap sites for H, but all this
tends to do is make it harder to remove all the H from the Pd. It doesn't
alter the net overall capacity, just the working capacity at temperature,
and it is only a noticeable minor component of the hydride.
At least that's what I know today. I'm still learning, it's a big field,
so if you have refs to contradict me, please let me know. Thanks.
I really haven't been following this tread too closely until recently.
Why are we talking about 10000 cycles?
Also, on the 'defects cause increased fusion' comment, I think it depends
who you listen to. The general process of loading Pd to high levels is
reasonably know to create internal voids and cracks that may reach the
surface if born close enough to it. In electrochemical cells, this makes
it easier to lose H, and thus harder to hold the high H/Pd ratios needed
fo 'CF'. So, in that point of view, defects hurt fusion.
Arthur Carlson
are already averaged over a Maxwellian distribution, i.e., they are a
function of temperature, not energy. An approximate formula for the
cross section as a function of energy (beam-target) can be found at
http://www.rzg.mpg.de/~dpc/nrl/44.html with the coefficients on
.../45.html.
A5 + [ (A4-A3*E)^2 + 1 ]^-1 * A2
sigma = --------------------------------
E * [ exp(A1*E^-0.5) - 1 ]
for D+D->T+p:
A1 = 46.097
A2 = 372
A3 = 4.36e-4
A4 = 1.220
A5 = 0
for D+D->He3+n:
A1 = 47.88
A2 = 482
A3 = 3.08e-4
A4 = 1.177
A5 = 0
with sigma in barns (10^-24 cm^2) and E in keV.
There are more accurate formulas available, especially for low
energies, but this is the one I can lay my hands on most quickly. For
1<E<10, we can ignore A3 and the -1 in the denominator, and A5 is zero
anyway:
[ A4^2 + 1 ]^-1 * A2 175
sigma = -------------------- = ------------------
E * exp(A1*E^-0.5) E * exp(47*E^-0.5)
= 6.78e-19 at 1 keV
= 3.22e-13 at 2 keV
= 2.72e-09 at 4 keV
If my guess of 1e-4 of the flux having an energy of 4 keV is even
vaguely related to reality, then it looks like the decrease in flux
might be compensated by the increase in cross section, but probably
not greatly over compensated. I would propose, for the purposes of
the present discussion, that the full flux at 2 keV could be used for
back-of-the-envelope calculations of the the hot fusion hypothesis,
good within one or two orders of magnitude. Now we would need a guess
at the ion density. The velocity will be near the sound speed, say
1e4 m/s. Finally, we need a guess at the target density. How about a
monolayer to start with, under the assumption that the reduction in
cross section with energy will make collisions within the bulk of the
metal less effective? It looks like we have all the pieces except the
density. Do they mention a neutral pressure? Then we could assume
that n_e/n_neutral is something between 0.01 and 1. Of course, we
will also have to translate Curies of tritium per minute into a fusion
rate for comparison. Right, and we need an area. Who will take the
ball while I get back to work?
Art Carlson
Steve Lajoie
Dieter Britz wrote:
>
> 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;
You seem to be laboring under a misunderstanding.
I never said they were. I said defects have been found to
cause increased fusion and decreased absorption, and that
most of the hydrogen absorbed into a metal hydride occupies
interstitial lattice 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.
I think it because the people who are interested in metal
hydrides for hydrogen storage (not cold fusion) say it is
an experimental fact and a problem with metal hydrides as
a storage device. That's why I think it.
> 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.
I give up.
Steve Lajoie
Arthur Carlson wrote:
>
> 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.
Wouldn't such reflections from the surface also be present in
the non-hydride metals which showed no tritium production? I
don't see why they wouldn't.
> 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.
I shouldn't have implied that it was an unambiguous, knock down
proof, maybe, if that's what I said. There are no proofs. You have
to go with the most likely answer. I see the apparent continuation
of fusion in figure four (after the plasma was cut off due to the
wire folding over). The tritium concentration in the gas stopped
increasing due to the CO contaminating the surface. When the system
was pumped up, tritium concentration shot upwards to a level that
is consistent with continued tritium production after the plasma
was shut off.
I suppose that you could say it was just releasing excess trapped
tritium in the metal produced while the plasma was on, but it just
doesn't look that way.
> 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.
>
Steve Lajoie
Richard Schultz wrote:
>
> 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?
:-) Either you don't understand the paragraph above,
or you are just mean spirited.
I know you can read.
> : 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.
Last I checked, electrical potential is measured in Volts, and
electrical energy in Joules, there being a variable called charge
that distinguished them. 2 keV can be directly converted into
Joules, but not Volts.
In this case, 2 thousand Volts is applied across a rarefied
gas, producing a local potential difference across the gas atoms
that their electrons could fall through that exceeded their
ionization energy and become conductive. That doesn't make the
ion 2 keV. It is far closer to their ionization energy at break
down, when the plasma first forms.
Now, Ohmic heating of plasmas to 2 keV is not unheard of, but
I would be very surprised if it could be done so easily, and
I find it highly unlikely because the plasma is in contact with
a metallic solid heat sink.
There are archival machines that keep usenet post a very, very long
time, and are open to anyone that cares to search on them. Are you
sure that you want this type of post to represent you to them? I'm
not questioning your physics error, but rather your rudeness. It looks
for all the world that mocking people makes you happy. Is that how
you want people to think of you? Just asking. I really don't care
what people think of you.
Dieter Britz
For once, I include all of the foregoing. Steve, there you go again. I
thought there was a real discussion going on without any rancour, and
you "give up". You did not answer my question, which not even you can
interpret as a personal attack: Who claims this? You write that "the
people ... say..." - who are these people and where do they say it,
i.e. can you provide references? I gave you reasons why I doubt this,
and Kirk Shanahan, who has worked with metal hydrides also doubts it,
but you repeat your assertion without any backup. This is not science,
mate, this is personal jostling. So come across with that backup, or
back down.
Steve Lajoie
Roland Smith wrote:
>
> On Tue, 6 Jun 2000 17:51:30 GMT, Steve Lajoie <laj...@eskimo.com>
> wrote:
>
> Steve, your claim was that _you_ could tell the temperature of the
> plasma by looking at the visible emission ...... by eye :).
You (the general "you", not the personal "I") can tell the
difference between a hot, 2 keV plasma and a barely ionized
plasma fairly easily, which is what I was getting at.
The barely ionized plasma is going to be colored. In this case,
the plasma is pink to light blue.
Note that the plasma in the old pixie tubes was reddish, and not
very hot at all.
Now, a 2 keV plasma will be detectable by eye. It would be white
hot and your eyeballs will dribble out because of the UV will
burn your corneas off.
Steve Lajoie
Kirk Shanahan wrote:
>
> In article <Pine.OSF.4.21.000607...@kemi.aau.dk>,
> d...@kemi.aau.dk says...
> >
> >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.
> >
> >-- Dieter Britz alias d...@kemi.aau.dk; http://www.kemi.aau.dk/~db
> >*** Echelon, bomb, sneakers, GRU: swamp the snoops with trivia! ***
> >
>
> Well, I don't think I really qualify as an expert in the hydride field.
> I've only been at it for 5 years now. The real experts have been going
> for 30+ years. But, I haven't heard of what is being proposed here, that
> 10K cycles causes Pd to stop absorbing H2. It doesn't make sense to me,
> but 10K cycles is larger than usual studies. (I would also appreciate
> a reference to this.)
>
> Cycling virgin Pd with H2 does induce dislocations as a means to relieve
> the stress induced by the lattice expansion Pd undergoes upon hydriding.
> But this process is usually finished with the first 10 cycles or so
> (depending on how the Pd was made and treated). After that it just
> keeps on going, just like the Energizer bunny.
>
> There is a proposal out there that there is also some vacancy creation
> that occurs during hydriding (Y. Fukai I believe), but the jury is
> still out on that one as I recall.
>
> Dieter is also correct in that defects are trap sites for H, but all this
> tends to do is make it harder to remove all the H from the Pd.
Humm. That would explain why when George loaded the Pd with
hydrogen first and then with deuterium, he got no fusion at
all.
> It doesn't
> alter the net overall capacity, just the working capacity at temperature,
> and it is only a noticeable minor component of the hydride.
>
> At least that's what I know today. I'm still learning, it's a big field,
> so if you have refs to contradict me, please let me know. Thanks.
>
> I really haven't been following this tread too closely until recently.
> Why are we talking about 10000 cycles?
'Cuz.
http://www.psc.edu/science/Wolf/Wolf-ext.html
"Hydrogen is loaded into the metal-hydride," explains Wolf,
"and then desorbed back off, through as many as 10,000 cycles.
You keep absorbing and desorbing hydrogen, or charging and
discharging, and like a sponge getting wet and drying out
over and over again, it expands and contracts. Over time
the material begins to break down."
Wolf: http://www.psc.edu/science/Wolf/Wolf-bio.html
I know, I'm getting my info off the net again and it's not
The Physical Review...
> Also, on the 'defects cause increased fusion' comment, I think it depends
> who you listen to.
If you don't think there's fusion going on at all, then it is a moot
point. But Claytor, Jackson and Tuggel
http://www.nde.lanl.gov/cf/tritweb.htm
aren't saying that it's cold fusion, only that the defects in Pd are
associated with tritium production. They found 3x the tritium from work
hardened Pd than from annealed Pd.
> The general process of loading Pd to high levels is
> reasonably know to create internal voids and cracks that may reach the
> surface if born close enough to it. In electrochemical cells, this makes
> it easier to lose H, and thus harder to hold the high H/Pd ratios needed
> fo 'CF'. So, in that point of view, defects hurt fusion.
But in a gas loading system, where the gas or is kept at the equilibrium
temperature and pressure, it wouldn't.
T L Clarke
> Note that the plasma in the old pixie tubes was reddish, and not
> very hot at all.
Probably not very hot but the plasma in the old nixie tubes
originated from neon gas. Same technology still used in
neon indicator lights. Argon regulator tubes glow blue as
do mercury vapor tubes.
> Now, a 2 keV plasma will be detectable by eye. It would be white
> hot and your eyeballs will dribble out because of the UV will
> burn your corneas off.
Rather an exageration since the plasma could only be viewed through
a transparent envelope that would absorb most of the UV. The
intensity would depend on the optical density.
What would you eyeball the temperature of this plasma as?
http://fus.x0r.com/images/fusor_on2.JPG
Tom Clarke
Steve Lajoie
GRADinc wrote:
>
> 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.
It is sort of like carbon steel. The area around the
carbon is a harder point than the surrounding areas,
thus they form visible grains when polished. The defect
makes it harder.
> >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.
http://www.nde.lanl.gov/cf/tritweb.htm
I think it's figure 3 that you refer to. You are correct.
Score one for the cone theory. :-)
> >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. "
You're right. I was wrong about tritium production being
immediate.
> 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.
I understood figure 2 to illustrate "background" vs.
"foreground" runs and the credibility of the data, they
having said that similar studies were dismissed as producing
tritium amounts too small to be considered significant.
They are demonstrating that the data is significant compared
to background. OF course, it could all be due to matrix
effect. My whole life has been matrix effect.
> >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.
You're right, I'm wrong. Cones first, and then the
tritium.
2 A/s would mean that the cones would be about 14 micrometers
high when tritium started being produced. Beyond 14 micrometers,
tritium production seems unaffected. Could be a 'threshold'
height for tritium production, couldn't it?
Or it could be that it takes that long to load the Pd
and get sufficient concentrations of deuterium for cold
fusion to start, and the cones are just a coincidence.
I can't prove it either way at this point, or give good
reason why I think cold fusion without pointing to other
experiments.
> > 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 misread the graph. But the rate is constant after the
cones are formed, the 14 micrometer height (20 hours at
2 A/s) seems to be the threshold and higher cones (perhaps
they are not higher? Perhaps at that point the tips of the
cones burn off?) the tritium production rate becomes constant.
I'm confused. I thought I saw it that way too!
> 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.
Okay. I'll buy that.
> >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.
I don't see where they said that the pressure was increased with
deuterium gas. Was it deuterium, or CO2? I don't know.
Like you, I think it was deuterium, but I couldn't prove
that to you.
> >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.
Okay. I see why.
> ...
> >The test would be, does fusion occur without the hot
> >deuterium, and it does.
>
> And I don't think it does.
It does in George's, Li's, Case's, Arata's, and McKubre's
vessels. I don't see why it wouldn't do it in this experiment.
Given their results, you'd expect tritium (and helium)
production in this plasma experiment.
> >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.
Yes, you're right, but you're going to have to go pretty
far out on the tail of a 13 eV distribution to get to
fusible energies, I think.
> >> 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.
They found that annealed Pd produces less tritium than work
hardened. Strain relieved but not annealed Pd produced 1/3
the tritium.
> >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.
Too bad we don't have any data showing that the more
defect there are the more cones there are. I think
that would be the case, and it would support your
cone idea.
> >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.
If one deuterium is behind the other, you have to show
forward scattering. That DOES happen, but not often.
Hey, maybe a hot deuterium entering your defect with
deuterium in it is like an energetic que ball hitting
racked billiard balls? You get them ALL bouncing off
each other and the Pd walls and some fuse.
Or, they just gather there, form a bose einstein condensate,
and occasionally a there's a fusion.
> >> 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.
Very few. But then, we're just hand waving and you're
going to wave back "and there are very few tritons being
produced", so that isn't going to work for me, is it?
> Tom Clarke
Roland Smith
wrote:
>Roland Smith wrote:
>>
>> On Tue, 6 Jun 2000 17:51:30 GMT, Steve Lajoie <laj...@eskimo.com>
>> wrote:
>>
>> Steve, your claim was that _you_ could tell the temperature of the
>> plasma by looking at the visible emission ...... by eye :).
>
>You (the general "you", not the personal "I") can tell the
>difference between a hot, 2 keV plasma and a barely ionized
>plasma fairly easily, which is what I was getting at.
The ionisation potential of hydrogen is 13.6 eV. A 13 keV D2 plasma
is therefore NOT barely ionised. Just about everything is fully
stripped. Note that in plasmas you generally excite transitions of
energies up to about 3xTe so anything up to 40 eV in this case so if
there were transitions well byond 13 keV (which there are not here)
they would light up as well..
>The barely ionized plasma is going to be colored. In this case,
>the plasma is pink to light blue.
>Note that the plasma in the old pixie tubes was reddish, and not
>very hot at all.
>
>Now, a 2 keV plasma will be detectable by eye. It would be white
>hot and your eyeballs will dribble out because of the UV will
>burn your corneas off.
You forgot the bit about temperature gradients and boundary layers.
It also rather depends on the windows you use. The last few
multi-keV plasmas I took a look at left me able to see well enough to
type, although a few CCD cameras have some dead spots in now :).
Roland
Dennis Towne
>
> 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.
Actually no, it was directed at you, not at anyone else. It says little
about my intelligence. It does, however, convey appropriately what I
think of yours.
[snip]
> 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?
Apparently, the reason I can't tell you where you're wrong is because
you're a 'complete fucking idiot'. It's not my problem that you are too
stupid to understand anything. Quite frankly, its not worth my time to
try to tell you anything - again, you just don't seem to be smart enough
to understand. Once again, not my problem.
[snip]
> 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.
Think whatever you like. No-one that I'm aware of holds your opinion in
any high regard.
As a side note, I find your statement quite funny - you've described
your own posts far better than my own.
[snip]
> Isn't this fellow worth kill filing?
Damn straight I am. After all, I don't agree with your conclusions, and
I don't believe any of the things you've presented are valid evidence.
Of course I should be killfiled. The fact that I think you're a
bonehead shouldn't even need to come to light - simply disagreeing with
you is more than sufficient.
It's too bad you're so stupid however. Just a little bit more
brainpower and you might even be able to figure out how to use a real
kill file instead of just ignoring what you don't like.
-dennis towne
Steve Lajoie
Dieter Britz wrote:
>
> On Wed, 7 Jun 2000, Steve Lajoie wrote:
>
> >
> >
> > Dieter Britz wrote:
> > >
> > > On Tue, 6 Jun 2000, Steve Lajoie wrote:
[snip]
> > I think it because the people who are interested in metal
> > hydrides for hydrogen storage (not cold fusion) say it is
> > an experimental fact and a problem with metal hydrides as
> > a storage device. That's why I think it.
> >
> > > 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.
> >
> > I give up.
>
> For once, I include all of the foregoing. Steve, there you go again. I
> thought there was a real discussion going on without any rancour, and
> you "give up".
No rancor, I give up because you found me out, Deiter.
I'm just guessing at everything, and trying to jerk your chain.
You found me out. What's the purpose of going on now that you've
found me out. All those websites I've pointed to are just websites
me and my co-conspirators have created to try and fool you. But
you've realized we're just making it all up.
So, what's the point in denying it now? I just give up.
> You did not answer my question, which not even you can
> interpret as a personal attack: Who claims this?
But you're absolutely right! I'm just guessing! That wasn't
a personal attack of yours, it was an observation! I'm a liar,
making things up. I've been found out. Just because some guy
named Wolf says so on a website is no reason at all for me to
say this. After all, a website isn't The Physical Review, so I
must be full of shit, right? A liar who's just making things up,
and that's a fact, not a personal attack. I shouldn't be offended,
right?
Never mind what I've read. I get no credit at all for that. As
some have pointed out, textbooks are not the Physical Review
either.
> You write that "the
> people ... say..." - who are these people and where do they say it,
> i.e. can you provide references?
Not from the Physical Review, or any other peer reviewed journal
that Jim Carr likes. So, fuck me and the horse I road in on. All
my sources are like that.
> I gave you reasons why I doubt this,
> and Kirk Shanahan, who has worked with metal hydrides also doubts it,
Well, there you go. You doubt it so I MUST be wrong. It's always
been that way and it always will be that way. You're the authority
and I'm the crackpot that needs to be moderated away. Then this would be
just a happy little group where people pop up and say "cold fusion
doesn't exist". Damn, we don't need assholes like me making an actual
issue of it, even if that is what the newsgroup is for.
I mean, if you think something, then it's up to me to come up with
documented proof, notarized and signed by a judge with my first born
as guarantee, agreed to by three Nobel prize winners in that field,
that it's different or >I'M< personal jostling. You don't need to do
ANYTHING to support your position. I'm presumed to be making it up
until PROVEN otherwise, and then it has to be so damned airtight
a whale couldn't get a fart out of it.
> but you repeat your assertion without any backup. This is not science,
> mate, this is personal jostling. So come across with that backup, or
> back down.
I did back down. I said I give up. What the is your problem
now? YOU MUST BE ABSOLUTELY RIGHT. No point in my carrying on
this charade anymore, is there? Isn't giving up backing down?
I hope my confession here is good enough for you. If not, let
me know and I'll admit to even more lies. If you want, I'll
admit that I shot JFK and whatever else you want. I've had it
with you always being on my ass. Yes, I hacked into that Los
Alamos site and posted that thing on tritium. Pretty good, huh?
Had most of you people going. And I hacked into all those other
websites, too. Wasn't it a good story?
This newsgroup sucks.
Dennis Towne
>
> Dennis Towne wrote:
> >
[snip]
> > > 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.
And what is that obvious error?
> 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.
From what part of your ass did you pull this gem? Please explain.
> 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.
Two plates in vacuum, one positive, one negative. 2KV potential between
them.
Introduce small quantity of deuterium gas and a few free electrons near
the positive plate.
Electrons ionize the gas, deuterons accelerate to negative plate.
Negative plate sees an impinging plasma with a temperature of just less
than 2KeV.
> 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.
Break down voltage has nothing to do with this experiment. All that
matters is what potential the average deuteron falls through on its way
to the target electrode. If the gas is sufficiently rarified that
collisions are rare, you easily get plasma temperatures at the target in
the KeV range.
-dennis towne
Oh yeah, for your remarks above, you definitely deserve this one:
Steve, you're an idiot.
Dennis Towne
>
> 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?
Hrm, nope - I don't think that's what he was saying. Looks to me like
he doesn't feel that there's any need to call you an idiot. I have to
agree, but I do it anyway because I find it entertaining. In all
honesty, your actions scream 'idiot' far louder than even my tactless
posts - and believe me, I was trying pretty hard.
> You really believe it convinces other people? Even about
> physics?
Was it supposed to? It's not like you are any better. Your posts, when
not insults or lacking in content, are almost invariably wrong. You'd
do well to learn some physics before trying to discuss it. Might want
to get a new brain first tho - that one you got right now seems down
right crusty.
> 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.
No, I'd say this is definitely not my normal mode of exchange. Usually,
I have rather solid discussions with people, with very little if any
name calling or insults. In your case however, I fail to see the
point. The polite approach to dealing with you is obviously not
working, so let's see how the tactless approach fares.
-dennis towne
Dennis Towne
>
> Roland Smith wrote:
> >
> > wrote:
> >
> > Now I wouldnt normally do this ..... but it's just too good an
> > opertunity to miss this time ...............
> >
> > [snip]
> >
> > >(22 million Kelvins in temperature...) from a 2 kV spark
> > >less, given the pink and blue colors showing the different
> > >excitation levels of the deuterium's electrons.
> >
> >
> > 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
> you can't see a 2 keV plasma.... Yet, is visible. They
> have reported it as so.
So you're saying that you can tell the difference between a 13 eV plasma
and a 2 KeV plasma by looking at them with your naked eye?
(Hint: you're an idiot, steve.)
-dennis towne
Steve Lajoie
Dennis Towne wrote:
>
> Steve Lajoie wrote:
> >
> > 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.
>
> Actually no, it was directed at you, not at anyone else. It says little
> about my intelligence. It does, however, convey appropriately what I
> think of yours.
You think so? All the more pathetic.
[snip rest of Towne's profanities and anti-social remarks.]
I am absolutely amazed that no one besides me protests the
nature of your posts. It appears they approve of your behavior.
Steve Lajoie
Dennis Towne wrote:
> (Hint: you're an idiot, steve.)
Someone once said words to the effect of "Evil men do their
work because good men don't stand up to them."
I don't see much denunciation of this kind of behavior,
and it is getting far too common. Is there so little interest
in cold fusion that people cannot even speak out against such
rants?
If people spoke up against such post as Towne's and Schultz's,
they wouldn't make them. They are motivated by a need to seem
intelligent, and if people pointed out how unintelligent these
post paint them, I am sure they would restrain their behaviors.
And I get greatly disappointed when usually moral, upright
and thinking people like Deiter start to fall into the
same behavior, apparently wanting to go in with the crowd.
In his book about his opinions, Albert Einstein tells of a
time when he was asked to speak out against anti-semitics
in Paris. He declined, his reason being that a Jew defending
Jews against the hate mongers would not be as convincing as
Christians denouncing the hate mongers.
I don't feel it's right that I should have to defend myself
from the profanity and verbal garbage of Dennis Towne, and
the others.
I have to assume that the readers of sci.physics.fusion find
this Towne spew and hate garbage intellectually stimulating.
So be it.
No wonder all the pro-CF people don't post here. No one is
willing to life so much as a finger to support civility and
intellectual freedom.
And anyone who says different is a god damned liar.
Kirk Shanahan
>
>Kirk Shanahan wrote:
{snip}
>> Dieter is also correct in that defects are trap sites for H, but all this
>> tends to do is make it harder to remove all the H from the Pd.
>
>Humm. That would explain why when George loaded the Pd with
>hydrogen first and then with deuterium, he got no fusion at
>all.
>
Yes, but in normal material the heel content is very, very small (<.03 H/Pd
units I would guess). However, I agree that it is posible such a
contamination might affect the 'CF' process, especially since we can't
really specify what that is.
{snip}
>> I really haven't been following this tread too closely until recently.
>> Why are we talking about 10000 cycles?
>
>'Cuz.
>http://www.psc.edu/science/Wolf/Wolf-ext.html
>
>"Hydrogen is loaded into the metal-hydride," explains Wolf,
>"and then desorbed back off, through as many as 10,000 cycles.
>You keep absorbing and desorbing hydrogen, or charging and
>discharging, and like a sponge getting wet and drying out
>over and over again, it expands and contracts. Over time
>the material begins to break down."
>
>Wolf: http://www.psc.edu/science/Wolf/Wolf-bio.html
>
I'm laughing man, we've come full cycle here. Earlier on the page you
quote, WSRC is referred to. WSRC is Westinghouse Savannah River Company.
The R&D arm of the company is the Savannah River Technology Center, which
is where I work. And which is where Ralph Wolf, Khalid Mansour, and Myung
Lee all used to work (Ralph and Myung directly with me, and I knew Khalid from
back when I was doing chemical process simulations, my office was across the
hall from his). All coauthors of the study you're referring to.
Unfortunately all are gone now, Ralph and Khalid work elsewhere, Myung
passed away in December.
Ralph was the computer chemist, Myung was the senior and very bright hydride
chemist, and Khalid was a Cray computer specialist.
But! I could check this out and what you quote is correct as far as it
goes. Pd will decrepitate with cycling, and 10,000 cycles is about right
because Pd is a 'good' actor on the front. Some other alloys we work with
like LaNi4.24Al0.75 will do this within 10-50 cycles. The key point
however is that the hydrogen absorbtion capacity and characteristics are not
noticeably affected by this. The main impact is on the engineering
properties of the materials. You have to design with the decrepitation
in mind.
I should have remenbered this, as that's why palladium purifiers are not
made of pure palladium, but of a palladium-silver alloy instead. Pd/Ag
alloys operated at 400C or above don't breakdown this way. (Why I'm not
sure, more homework I have to do.)
>I know, I'm getting my info off the net again and it's not
>The Physical Review...
>
Nothing wrong with the 'net per se. You got a process description, and
not data tables, etc. But this work isn't controversial and
revolutionary. I'm still going to fish around a bit for the original
work in this case.
>
>> Also, on the 'defects cause increased fusion' comment, I think it depends
>> who you listen to.
>
>If you don't think there's fusion going on at all, then it is a moot
>point. But Claytor, Jackson and Tuggel
>http://www.nde.lanl.gov/cf/tritweb.htm
>aren't saying that it's cold fusion, only that the defects in Pd are
>associated with tritium production. They found 3x the tritium from work
>hardened Pd than from annealed Pd.
>
You're right, until I believe that plasma cleaning-type effects are not
causing the observations, it's moot. But if it was fusion, then
observations like that would be pertinent.
>> The general process of loading Pd to high levels is
>> reasonably know to create internal voids and cracks that may reach the
>> surface if born close enough to it. In electrochemical cells, this makes
>> it easier to lose H, and thus harder to hold the high H/Pd ratios needed
>> fo 'CF'. So, in that point of view, defects hurt fusion.
>
>But in a gas loading system, where the gas or is kept at the equilibrium
>temperature and pressure, it wouldn't.
Usually gas loading systems can't reach the high pressures necessary to
load Pd to the level where CF is supposed to occur (>.85 H/Pd, typically >
several thousand atmospheres are needed). Ararta'a apparatus achieves this
by using electrochemistry to establish chemical potential equivalent to
this high P regime. A plasma may do the same. There is some work by
Flanagan et. al. that suggests a plasma can also get to very high loadings
like H/Pd=1. However, I believe that was done with a sulfur contaminant
on the Pd surace that appeared to hinder the recombination of H to form
desorbable H2. Thus the plasma injects H, and it can't get out. If you
don't have something to hinder H-H recombination, I doubt you can reach
high loadings easily.
Dennis Towne
Or perhaps they find my posts to be factual. I'll be the first to say
they are tactless and offensive - but even a tactless and offensive post
can be correct.
-dennis towne
T L Clarke
> and me posting from GRADinc wrote:
reach some consensus?
.........
> > ...
> > >The test would be, does fusion occur without the hot
> > >deuterium, and it does.
> >
> > And I don't think it does.
>
> It does in George's, Li's, Case's, Arata's, and McKubre's
> vessels. I don't see why it wouldn't do it in this experiment.
Since I don't accept those others as proving CF, especially George's
because of the sketchy description, I don't expect CF to happen in
the LANL experiment either.
> > >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.
>
> Yes, you're right, but you're going to have to go pretty
> far out on the tail of a 13 eV distribution to get to
> fusible energies, I think.
You assume that the high energy ions come from the tail of
a maxwellian distribution, not from directly acceleration by
voltage gradient at a cone or across the cathode "dark space".
> ...
> > >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.
>
> If one deuterium is behind the other, you have to show
> forward scattering. That DOES happen, but not often.
Not really. The scattering in question is not atom off atom,
but nuclei off nuclei. The nuclei are much smaller than the atoms
so it is like shooting a rocket into the asteroid belt. Lots of
object, but they are small and far apart.
> Hey, maybe a hot deuterium entering your defect with
> deuterium in it is like an energetic que ball hitting
> racked billiard balls? You get them ALL bouncing off
> each other and the Pd walls and some fuse.
No. I think most of the D+ ions would miss the nuclei (Pd or D)
and would slowly loose their energy in ~13 ev collisions with the
marshmallow cloud of electrons around the hard nuclei.
> Or, they just gather there, form a bose einstein condensate,
> and occasionally a there's a fusion.
Way too much energy for a BEC.
> > >> 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.
>
> Very few. But then, we're just hand waving and you're
> going to wave back "and there are very few tritons being
> produced", so that isn't going to work for me, is it?
No. To me this experiment proves that enough
are being produced.
Tom Clarke
Kirk Shanahan
>
>I understood figure 2 to illustrate "background" vs.
>"foreground" runs and the credibility of the data, they
>having said that similar studies were dismissed as producing
>tritium amounts too small to be considered significant.
>They are demonstrating that the data is significant compared
>to background. OF course, it could all be due to matrix
>effect. My whole life has been matrix effect.
>
Thought you'd sneak that by me didn't you....
The matrix effect here might have been in the T detection system.
Femtotechs are a type of ion chamber, and ion chambers are sensitive
to _any_ change in gas composition that alters the ion flow in the
detection chamber.
Water is an especially bad actor. I'm trying to recall the mechanism.
I think it involves the ease of ionizing water vs. H2 or D2, but I'll
have to check. All I know is that if our building humidity controls
go out for awhile, we get kicked out of the labs because of false
positives on our T detectors.
So, if a little water blew down the line, you'd get a "T" signal. Claytor
knew this though, and he cross-checked with LSC studies, thereby
refuting the "the signal is water" matrix effect. If it wasn't for
that, I'd not even accept he had tritium.
>My whole life has been matrix effect.
So sorry...
{snip}
Steve Lajoie
Kirk Shanahan wrote:
>
[snip]
Perhaps you could correct the web site so that people like
me are not mislead and make a fool of themselves.
Steve Lajoie
T L Clarke wrote:
>
> Steve Lajoie wrote:
>
> > and me posting from GRADinc wrote:
>
> reach some consensus?
How so? You disagreed with everything I said.
Barry Kearns
news:393E9CE5...@eskimo.com...
>
> Dennis Towne wrote:
>
> > (Hint: you're an idiot, steve.)
>
> Someone once said words to the effect of "Evil men do their
> work because good men don't stand up to them."
>
> I don't see much denunciation of this kind of behavior,
> and it is getting far too common.
And, of course, you're one of the prime offenders, so mewling about the
injustice of it is bound to fall upon deaf ears. Hypocrisy has a way of doing
that, you know.
> Is there so little interest
> in cold fusion that people cannot even speak out against such
> rants?
Of course they can "speak out"... there is no underlying censorship taking
place here. That does *not*, however, mean that no one else is going to offer
*their* comments and opinions about what's said. That's part of the deal...
if you don't get censored, others who disagree with you don't either.
> If people spoke up against such post as Towne's and Schultz's,
> they wouldn't make them.
Oh really? Do you think people speaking up about *your* posts has somehow
stopped *you* from making them? Why, then, would you suppose that it would
apply to others?
Or is it because you think you're on the side of the angels, and the opponents
are (in your words) "evil men"?
> They are motivated by a need to seem
> intelligent, and if people pointed out how unintelligent these
> post paint them, I am sure they would restrain their behaviors.
It certainly doesn't seem to have slowed you down in the slightest, despite
the rather large contingent of posts pointing out how unintelligent your posts
paint you...
> I don't feel it's right that I should have to defend myself
> from the profanity and verbal garbage of Dennis Towne, and
> the others.
You don't have to "defend yourself" against these things. Since you don't,
it's illogical (IMO) to discuss whether it's "right" that you have to.
If you think it's verbal garbage, then either convey that impression... or
don't. There's no "have to" here.
They are as free to offer "profanity and verbal garbage" as you are... and
since you've dished out quite a heaping helping of that yourself, I fail to
see the moal justification in bemoaning it in others.
Matthew 7:5 "Thou hypocrite, first cast out the beam out of thine own eye; and
then shalt thou see clearly to cast out the mote out of thy brother's eye."
> I have to assume that the readers of sci.physics.fusion find
> this Towne spew and hate garbage intellectually stimulating.
>
> So be it.
As usual, you've found a way to draw an inaccurate conclusion from the
evidence at hand.
It is more likely, in my estimation, that the readers find the reactions to be
understandable enough that they are willing to overlook the lapse. We are,
after all, human.... and few can accurately claim to have the patience of
Job... and that seems to be what is required when dealing with your
persistent obtuseness.
> No wonder all the pro-CF people don't post here.
False. I am a pro-CF person, and I post here... therefore the proposition
that all of them don't is falsified.
> No one is
> willing to life so much as a finger to support civility and
> intellectual freedom.
Balderdash. Mitchell Jones has attempted on many occassions to actively
support and advocate both. And he's not the only one, of course... though
only one is needed to show the flaw in your claim.
> And anyone who says different is a god damned liar.
Not only intellectually vacuous.. this is also profane, and false on its face.
Since I say different, I must assume that I am then a "god damned liar".
While I personally don't care whether any so-called god cares to damn me
(being decidedly irreligious)... I do care when I'm called a liar.
Please cease with posting tripe such as this... especially since you are
decrying the behavior in others.
--
Barry Kearns
bke...@frii.com
David Gaskill
wrote:
>I don't feel it's right that I should have to defend myself
>from the profanity and verbal garbage of Dennis Towne, and
>the others.
>
>I have to assume that the readers of sci.physics.fusion find
>this Towne spew and hate garbage intellectually stimulating.
>
>So be it.
>
>No wonder all the pro-CF people don't post here. No one is
>willing to life so much as a finger to support civility and
>intellectual freedom.
OK, I will.
I must confess that I have become increasingly concerned of late about
the personal attacks on you.
I have read this group since its inception and have always found it
stimulating. If it wasn't for people like you who are prepared to
respond to the overwhelming preponderance of sceptics, (of which I am
one) the group would have died long ago. I was delighted a-month or
two ago when one of those who over the years has contributed hugely to
the enjoyment I get from reading the group made an all too brief
reappearance before retreating to his Jedite lair. The group would be
much the poorer if you did the same thing.
Even though I think you have a unique ability to enrage others who
post to this group this is really is no excuse for schoolboy
name-calling and profanity..
I admire your willingness to admit that you are wrong when you believe
this to be the case and I marvel at the sheer volume of your postings.
(How you find time to earn a living as well I find difficult to
imagine).
I would hate to see this group die and I would appeal to all of those
who post to refrain from personal abuse however infuriating you find
Steve's posts to be. I would suggest that it is much more productive
to attack his ideas, however provocative the expression of them, than
it is to attack him personally. If he disappears into the night,
(taking his killfile with him), he will be impossible to replace.
David
Dieter Britz
[...]
> But! I could check this out and what you quote is correct as far as it
> goes. Pd will decrepitate with cycling, and 10,000 cycles is about right
> because Pd is a 'good' actor on the front. Some other alloys we work with
> like LaNi4.24Al0.75 will do this within 10-50 cycles. The key point
> however is that the hydrogen absorbtion capacity and characteristics are not
> noticeably affected by this. The main impact is on the engineering
> properties of the materials. You have to design with the decrepitation
> in mind.
And also, if broken-up Pd does not absorb hydrogen so well, why do
some CNF workers use just that (sponge, powder)?
> Usually gas loading systems can't reach the high pressures necessary to
> load Pd to the level where CF is supposed to occur (>.85 H/Pd, typically >
> several thousand atmospheres are needed). Ararta'a apparatus achieves this
> by using electrochemistry to establish chemical potential equivalent to
> this high P regime. A plasma may do the same. There is some work by
Arata uses electrochemical potential to drive deuterium through the
wall of that Pd bottle, creating pressurised D2 inside it, to charge
up the Pd black in the bottle. But he doesn't get more than 800 atm
inside the bottle (limited by a safety valve), whereas charging
directly into a Pd cathode can get you up to 10^4 atm real pressure or
so (enormous fugacity, 10^{26}, but not quite so enormous real
pressure).
T L Clarke
I guess I'll have to pull up the post and quote:
You said:
>You're right. I was wrong about tritium production being
immediate.
...
>You're right, I'm wrong. Cones first, and then the
tritium.
...
>I misread the graph.
...
>Okay. I'll buy that.
...
>> I disagree.[this from T. Clarke]
>Okay. I see why.
...
Now I insert the remainder of my post as I'm curious about your
response:
.........
> > ...
> > >The test would be, does fusion occur without the hot
> > >deuterium, and it does.
> >
> > And I don't think it does.
>
> It does in George's, Li's, Case's, Arata's, and McKubre's
> vessels. I don't see why it wouldn't do it in this experiment.
Since I don't accept those others as proving CF, especially George's
because of the sketchy description, I don't expect CF to happen in
the LANL experiment either.
> > >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.
>
> Yes, you're right, but you're going to have to go pretty
> far out on the tail of a 13 eV distribution to get to
> fusible energies, I think.
You assume that the high energy ions come from the tail of
a maxwellian distribution, not from directly acceleration by
voltage gradient at a cone or across the cathode "dark space".
> ...
> > >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.
>
> If one deuterium is behind the other, you have to show
> forward scattering. That DOES happen, but not often.
Not really. The scattering in question is not atom off atom,
but nuclei off nuclei. The nuclei are much smaller than the atoms
so it is like shooting a rocket into the asteroid belt. Lots of
object, but they are small and far apart.
> Hey, maybe a hot deuterium entering your defect with
> deuterium in it is like an energetic que ball hitting
> racked billiard balls? You get them ALL bouncing off
> each other and the Pd walls and some fuse.
No. I think most of the D+ ions would miss the nuclei (Pd or D)
and would slowly loose their energy in ~13 ev collisions with the
marshmallow cloud of electrons around the hard nuclei.
> Or, they just gather there, form a bose einstein condensate,
> and occasionally a there's a fusion.
Way too much energy for a BEC.
> > >> 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.
>
> Very few. But then, we're just hand waving and you're
> going to wave back "and there are very few tritons being
> produced", so that isn't going to work for me, is it?
No. To me this experiment proves that enough
Kirk Shanahan
>
>Perhaps you could correct the web site so that people like
>me are not mislead and make a fool of themselves.
Unfortnately the Web Site belongs to the Pittsburgh Supercomputing Center,
and is not influenced by me. The work you read was conducted on their
computers and was posted as a technical information package about their
(the PSC) work, as a way to convince others to do work there as well. In
other words, they are doing PR with their Web site, like most people do to
one extent or the other. So, since this was a done deal several years ago,
and since I have no interaction with them, I doubt my suggestion would be
very effective.
It seems to me though, that you are being too hard on yourself with your
one-liner above. I think it is apparent that one could read into the Web
page what you did, and it is only the more extensive experience well that
I was able to tap that clarified the point. It seems to me that that is
the point of 'scientific communication', of which Usenet is one example.
So, we have just done what we were all trying to do here on one point in
the discussion. At least that's one of the reasons I post here.
The field of CF is in dire need of this kind of point-by-point discussions.
What has always disappointed me about the field however is that the
_primary_ researchers are so unwilling to participate in this kind of
give-and-take. In the end, it diminishes their credibility.
I ran across an interesting quote this week that I think applies:
There is no place for dogma in science. The scientist
is free to ask any question, to doubt any assertion,
to seek any evidence, to correct any error.
-- J. Robert Oppenheimer (1904-1967)
---
Kirk L. Shanahan {{My opinions...noone else's}}
(As a heads up, I start a long vacation soon, so my number of posts will
again decrease.)
Kirk Shanahan
says...
>
>
{snip}
>
>Arata uses electrochemical potential to drive deuterium through the
>wall of that Pd bottle, creating pressurised D2 inside it, to charge
>up the Pd black in the bottle. But he doesn't get more than 800 atm
>inside the bottle (limited by a safety valve), whereas charging
>directly into a Pd cathode can get you up to 10^4 atm real pressure or
>so (enormous fugacity, 10^{26}, but not quite so enormous real
>pressure).
>
Actually, Arata and Zhang's report in J. of High Temp. Soc. says that they just
couldn't measure any higher than that. They explicitly state on page 47 that
they feel the P reached far over 1000 atm.
Dennis Towne
>
> Dennis Towne wrote:
>
> > (Hint: you're an idiot, steve.)
>
> Someone once said words to the effect of "Evil men do their
> work because good men don't stand up to them."
[snip]
> In his book about his opinions, Albert Einstein tells of a
> time when he was asked to speak out against anti-semitics
> in Paris. He declined, his reason being that a Jew defending
> Jews against the hate mongers would not be as convincing as
> Christians denouncing the hate mongers.
It's not a true usenet flame unless there's a reference to Hitler and
the Nazis. Way to go, Steve! Glad you are here to keep the discussion
on track.
-dennis towne
Mitchell Jones
<laj...@eskimo.com> wrote:
> Dieter Britz wrote:
> >
> > On Wed, 7 Jun 2000, Steve Lajoie wrote:
> >
> > >
> > >
> > > Dieter Britz wrote:
> > > >
> > > > On Tue, 6 Jun 2000, Steve Lajoie wrote:
>
> [snip]
>
> > > I think it because the people who are interested in metal
> > > hydrides for hydrogen storage (not cold fusion) say it is
> > > an experimental fact and a problem with metal hydrides as
> > > a storage device. That's why I think it.
> > >
> > > > 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.
***{Note: the first comment (three leading arrows) came *after* Dieter's
question (four leading arrows). It is therefore misleading to quote it
this way, thereby suggesting that Dieter was responding to it when he
asked if you were guessing. Here (between the lines of asterisks) is the
actual context in which Dieter's question appeared:
*******************************
> 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.
*******************************
As you can see, there was nothing untoward in Dieter's question: it was
perfectly reasonable, given the *actual* context (as opposed to the
context falsely implied by you, above).
Having said that, I must add that Dieter was out of line to say "I suggest
you look up some basic crystallography," since that sounds like a veiled
accusation of ignorance. It would have been more civil to delete the first
sentence and, after the second one, give a reference to an elementary
textbook on crystallography.
By the way, if that is what triggered your lengthy rant (see below), it is
what you should have focused on, if you wanted to make any headway with
your argument. (A person who feels he has been pricked by a thorn needs to
identify where it is, in order to pull it out.)
--MJ}***
> > >
> > > I give up.
***{Thereby implying that Dieter was asking a stupid or an insulting
question? Why? The material to which he was responding was snipped out by
you, and there is nothing intrinsically stupid about his question, so
anyone who reads your "I give up" and the lengthy diatribe that follows is
guaranteed to not have the slightest idea what triggered it. (Indeed, it
is apparent from the above you don't even know what triggered it.)
--MJ}***
> >
> > For once, I include all of the foregoing. Steve, there you go again. I
> > thought there was a real discussion going on without any rancour, and
> > you "give up".
>
> No rancor, I give up because you found me out, Deiter.
>
> I'm just guessing at everything, and trying to jerk your chain.
>
> You found me out. What's the purpose of going on now that you've
> found me out. All those websites I've pointed to are just websites
> me and my co-conspirators have created to try and fool you. But
> you've realized we're just making it all up.
>
> So, what's the point in denying it now? I just give up.
>
> > You did not answer my question, which not even you can
> > interpret as a personal attack: Who claims this?
***{Steve, before you respond to an irritant, you need to identify it. The
only legitimate gripe that you can level at Dieter's statements has to do
with his insinuation that you are ignorant. There was nothing wrong with
his question per se. Thus I have yet another suggestion, in furtherance of
the self-improvement of Steve Lajoie: calm down before you compose a post,
and never send it off until you have re-read it while in a tranquil state
of mind. --MJ}***
[rest of misguided rant deleted]
>
> This newsgroup sucks.
***{Sometimes it does, but you can deal with the barbs, if you first
identify *exactly* what the barb consists of, and how it works. --MJ}***
=====================================================
Mitchell Jones
<cga...@ibm.net > wrote:
***{I share your hope that Steve will stay with us, but I cannot,
realistically, expect that to happen unless he learns to deal with the
barbs that distress him so much. As his misguided, hugely sarcastic rant
directed at Dieter demonstrates, when he is pricked he becomes too
enraged, hurt, or whatever, to think clearly, and wastes most of his ammo
shooting at the wrong target. Until he ceases to do that, his experiences
in this group will continue to be unpleasant. --MJ}***
=====================================================
Bruce Scott TOK
Arthur Carlson <car...@ipp.mpg.de> wrote:
[...]
>quantitative input. Who is willing to look up the actual energy
>dependence of the fusion cross section?)
Clayton, D, Principles of Stellar Evolution and Nucleosynthesis,
McGraw-Hill (1968)
Both the physics and the numbers are there. But this is p-p fusion, or
the pieces of the CNO cycle. For strong interactions the math is
different, but Clayton treats that, too. The formulae are not simple,
and stellar modelling usually uses fits to functions like a power series
in some temperature function, times an exponential whose argument is
a negative constant times T to the minus 1/3 or 2/3 power.
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
Dieter Britz
Caught out again in sloppy reading! You are right. I read the figure
legend on Fig. 5, where the open type bottle is described as having a
pressure gauge, "up to about 900 atmosphere which is limited by only
pressure gage" and took that to mean a relief valve of some kind
being present. Sorry, and thanks for pointing this out.
Mitchell Jones
[snip]
>
> Pd loading is dependent on lattice defects. What is the problem here?
>
> -dennis towne
***{Hi Dennis. I just noticed this thread. Regarding the claim that Pd
loading is dependent on lattice defects, I am not sure exactly what you
mean. Are you merely saying that a larger surface area is exposed if there
are lots of lattice defects--i.e., irregularities in the crystal
structure--which in turn *speeds up* loading? In other words, are you
talking about the *rate* of loading? Or are you saying that loading
*requires* lattice defects--in other words, that it is only within the
defects that loading occurs? If the latter, then I must disagree: once a
deuterium atom has been ionized, you have a bare nucleus, which is easily
small enough to pass into the crystal lattice of the Pd even if there are
no defects; and, if the positively charged deuterium nucleus meets an
electron while inside the lattice, it will become a neutral deuterium atom
at the location where the meeting occurs. Since the ground state radius of
D is just right for a snug fit inside one of the octagonal unit cells of
the Pd, and is too large for the D atom to exit from such a cell (the
openings in the face of the octagon are less than the radius of the
deuterium atom), the result is that the D remains wedged in the
cell--i.e., the cell is loaded. This occurs even if there are *zero*
defects in the lattice. --MJ}***
=====================================================
T L Clarke
Mitchell Jones wrote:
> Since the ground state radius of
> D is just right for a snug fit inside one of the octagonal unit cells of
> the Pd, and is too large for the D atom to exit from such a cell (the
> openings in the face of the octagon are less than the radius of the
> deuterium atom), the result is that the D remains wedged in the
> cell--i.e., the cell is loaded. This occurs even if there are *zero*
> defects in the lattice. --MJ}***
> =====================================================
Is that the way it happens? I can't find the numerical values quickly, but I
believe
the Pd lattice expands as the hydrogen is absorbed.
See http://www.psc.edu/science/Wolf/Wolf.html
which has the caption:
Onset of the beta phase in palladium hydride at 300 degrees Kelvin. This phase
change occurs as the concentration of hydrogen atoms (yellow) in the palladium
(purple) increases. At early stages (the alpha phase), hydrogen atoms randomly
populate small interstices in the lattice structure. At a critical point, the
lattice
expands, allowing hydrogen to cluster at higher density, as visualized here.
This
image shows the lattice from the (001) direction.
Tom Clarke
Dieter Britz
>
>
> Mitchell Jones wrote:
>
> > Since the ground state radius of
> > D is just right for a snug fit inside one of the octagonal unit cells of
> > the Pd, and is too large for the D atom to exit from such a cell (the
> > openings in the face of the octagon are less than the radius of the
> > deuterium atom), the result is that the D remains wedged in the
> > cell--i.e., the cell is loaded. This occurs even if there are *zero*
> > defects in the lattice. --MJ}***
> > =====================================================
>
> Is that the way it happens? I can't find the numerical values quickly, but I
> believe
> the Pd lattice expands as the hydrogen is absorbed.
It does, by about 16%, if I remember rightly. But Mitch's point (which
I also made a week or two ago) is that the hydrogen likes it inside
the Pd, and defects will not affect the extent of loading - although,
as he writes, if they are extensive, they might speed up the loading
process. This is not unimportant, which is why some use Pd black or
Pd sponge, etc. There are, however, those who reckon that a big solid
lump is the way to get fusion, the chunkier the better.
Mitchell Jones
<tcl...@ist.ucf.edu> wrote:
> Mitchell Jones wrote:
>
> > Since the ground state radius of
> > D is just right for a snug fit inside one of the octagonal unit cells of
> > the Pd, and is too large for the D atom to exit from such a cell (the
> > openings in the face of the octagon are less than the radius of the
> > deuterium atom), the result is that the D remains wedged in the
> > cell--i.e., the cell is loaded. This occurs even if there are *zero*
> > defects in the lattice. --MJ}***
> > =====================================================
>
> Is that the way it happens? I can't find the numerical values quickly
***{This was all worked out in exhaustive mathematical detail on this
group, years ago. I will send you a copy of the calculations via private
e-mail, if you would like. --MJ}***
, but I
> believe the Pd lattice expands as the hydrogen is absorbed.
***{Yes, the Pd lattice expands as the hydrogen is absorbed, but that is
to be expected: when one of the octagonal unit cells is loaded, the
electron shell of the D is right up against the outer shells of the 6 Pd's
that make up the octagon. Since the electrons repel one another, the
lattice expands a bit, as each additional D is loaded. --MJ}***
>
> See http://www.psc.edu/science/Wolf/Wolf.html
> which has the caption:
>
> Onset of the beta phase in palladium hydride at 300 degrees Kelvin. This phase
> change occurs as the concentration of hydrogen atoms (yellow) in the palladium
> (purple) increases. At early stages (the alpha phase), hydrogen atoms
randomly
> populate small interstices in the lattice structure. At a critical
point, the lattice
> expands, allowing hydrogen to cluster at higher density, as visualized here.
> This image shows the lattice from the (001) direction.
***{I looked at the site. It had the character of advertisement, not
science. No details were given about the algorithms they are using in
their simulation program, or about the physical assumptions they are
making. They are, of course, correct in thinking that the lattice expands
a bit as the hydrogen is absorbed. --MJ}***
>
> Tom Clarke
=====================================================
Mitchell Jones
>
> 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.
***{I'll take a stab at this. First, the 2,000 volt arc will give an
energy of 2 keV to any deuteron that shoots across the plasma unimpeded,
which, in the case of a rarified plasma, will be most of them. Let us,
therefore, calculate the temperature of an ideal monoatomic gas for which
the average particle has an energy of 2 keV and a mole weighs of 2 grams.
The internal energy per mole (U) of such a gas is the sum of the
rotational and translational kinetic energy of the elements of which it is
comprised, and, since it is presumed to be monoatomic, collisions will be
perfectly elastic, and the rotational component can be neglected. Under
such conditions the average specific heat, C, will be such that C =
(3/2)(R/M), where R is the universal gas constant and M is the molecular
weight of the gas. Thus we can multiply both sides by the absolute
temperature, T, obtaining: TC = (3/2)(R/M)(T). Substituting U for TC, we
have U = (3/2)(R/M)(T). Solving for T, we have: T = 2U/3(R/M). Taking the
average kinetic energy of translation to be 2 keV, a mole of monoatomic
deuterium gas will consist of 2 grams, and will contain 6.02x10^23 atoms
(Avogadro's number). Thus U = (6.02x10^23)(2)(1.6x10^-12) = 1.93x10^12
ergs. Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
2U/3(R/M) = 2(1.93x10^12)/[3(8.317x10^7)/2] = 30,967 degrees K.
Of course, the above very rough calculations neglect pressure energy,
which will be due not merely to the internal energy per mole, but also to
the mutual repulsion of the particles as a result of stripping off some of
their outer electrons. Thus if we assume that outer electrons are
*completely removed* from every deuterium atom, then that mutual repulsion
would become *stupendous*. Result: to contain them all within the same
volume as before might very well result in a temperature rise to 22
million degrees. However, in my view the attempt to argue on such grounds
that the effective temperature in this sort of plasma is 22 million
degrees would be absurd. The reason: only a tiny fraction of the deuterium
atoms actually have their electrons completely stripped away. For the
others--the vast majority--"ionization" merely means that their electrons
have been boosted up into the n = 2 through n = 5 orbits. Thus most of the
atoms in such a plasma remain electrically neutral, albeit with expanded
outer shells, and the enormous pressures and temperatures that would be
required to contain them (at constant volume) if they were completely
bereft of electrons simply do not apply.
Of course, since I am a latecomer to this thread, I really have no idea
how the 22 million degree figure was obtained. Was a detailed calculation
ever posted? If so, what were the assumptions on which that result was
based?
--Mitchell Jones}***
=====================================================
Dieter Britz
> >Steve Lajoie wrote:
> >
> > 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.
>
> ***{I'll take a stab at this. First, the 2,000 volt arc will give an
> energy of 2 keV to any deuteron that shoots across the plasma unimpeded,
> which, in the case of a rarified plasma, will be most of them. Let us,
> therefore, calculate the temperature of an ideal monoatomic gas for which
> the average particle has an energy of 2 keV and a mole weighs of 2 grams.
>
> The internal energy per mole (U) of such a gas is the sum of the
> rotational and translational kinetic energy of the elements of which it is
> comprised, and, since it is presumed to be monoatomic, collisions will be
> perfectly elastic, and the rotational component can be neglected. Under
> such conditions the average specific heat, C, will be such that C =
> (3/2)(R/M), where R is the universal gas constant and M is the molecular
> weight of the gas. Thus we can multiply both sides by the absolute
> temperature, T, obtaining: TC = (3/2)(R/M)(T). Substituting U for TC, we
> have U = (3/2)(R/M)(T). Solving for T, we have: T = 2U/3(R/M). Taking the
> average kinetic energy of translation to be 2 keV, a mole of monoatomic
> deuterium gas will consist of 2 grams, and will contain 6.02x10^23 atoms
> (Avogadro's number). Thus U = (6.02x10^23)(2)(1.6x10^-12) = 1.93x10^12
> ergs. Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
> 2U/3(R/M) = 2(1.93x10^12)/[3(8.317x10^7)/2] = 30,967 degrees K.
I think you made a mistake there somewhere along the line; that figure is
certainly too low. Using the more direct E = 3/2 kT and converting the
2 keV to Joules, I get about 15 * 10^6 K, not that far from the 22 million
someone quoted. I seem to remember once being told that 1 eV is about
equiivalent to 10000 K, so this figures roughly.
GRADinc
>> > Since the ground state radius of
>> > D is just right for a snug fit inside one of the octagonal unit cells of
>> > the Pd,
.....
>> Is that the way it happens? I can't find the numerical values quickly
....
>***{Yes, the Pd lattice expands as the hydrogen is absorbed, but that is
to be expected:
That answers my concern.
From your discussion of "snug fit"
I thought you were implying that there
was no expansion.
Tom Clarke
John Park
> On Wed, 21 Jun 2000, Mitchell Jones wrote:
>
>> >Steve Lajoie wrote:
>> >
>> > 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.
>>
>> ***{I'll take a stab at this. First, the 2,000 volt arc will give an
>> energy of 2 keV to any deuteron that shoots across the plasma unimpeded,
>> which, in the case of a rarified plasma, will be most of them. Let us,
>> therefore, calculate the temperature of an ideal monoatomic gas for which
>> the average particle has an energy of 2 keV and a mole weighs of 2 grams.
>>
>> The internal energy per mole (U) of such a gas is the sum of the
>> rotational and translational kinetic energy of the elements of which it is
>> comprised, and, since it is presumed to be monoatomic, collisions will be
>> perfectly elastic, and the rotational component can be neglected. Under
>> such conditions the average specific heat, C, will be such that C =
>> (3/2)(R/M), where R is the universal gas constant and M is the molecular
>> weight of the gas. Thus we can multiply both sides by the absolute
>> temperature, T, obtaining: TC = (3/2)(R/M)(T). Substituting U for TC, we
>> have U = (3/2)(R/M)(T). Solving for T, we have: T = 2U/3(R/M). Taking the
>> average kinetic energy of translation to be 2 keV, a mole of monoatomic
>> deuterium gas will consist of 2 grams, and will contain 6.02x10^23 atoms
>> (Avogadro's number). Thus U = (6.02x10^23)(2)(1.6x10^-12) = 1.93x10^12
This looks like 2 eV, not 2 keV, in ergs.
>> ergs. Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
>> 2U/3(R/M) = 2(1.93x10^12)/[3(8.317x10^7)/2] = 30,967 degrees K.
or 31 million K with the correction.
Mitchell Jones
Britz <d...@kemi.aau.dk> wrote:
> On Wed, 21 Jun 2000, Mitchell Jones wrote:
>
> > >Steve Lajoie wrote:
> > >
> > > 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.
> >
> > ***{I'll take a stab at this. First, the 2,000 volt arc will give an
> > energy of 2 keV to any deuteron that shoots across the plasma unimpeded,
> > which, in the case of a rarified plasma, will be most of them. Let us,
> > therefore, calculate the temperature of an ideal monoatomic gas for which
> > the average particle has an energy of 2 keV and a mole weighs of 2 grams.
> >
> > The internal energy per mole (U) of such a gas is the sum of the
> > rotational and translational kinetic energy of the elements of which it is
> > comprised, and, since it is presumed to be monoatomic, collisions will be
> > perfectly elastic, and the rotational component can be neglected. Under
> > such conditions the average specific heat, C, will be such that C =
> > (3/2)(R/M), where R is the universal gas constant and M is the molecular
> > weight of the gas. Thus we can multiply both sides by the absolute
> > temperature, T, obtaining: TC = (3/2)(R/M)(T). Substituting U for TC, we
> > have U = (3/2)(R/M)(T). Solving for T, we have: T = 2U/3(R/M). Taking the
> > average kinetic energy of translation to be 2 keV, a mole of monoatomic
> > deuterium gas will consist of 2 grams, and will contain 6.02x10^23 atoms
> > (Avogadro's number). Thus U = (6.02x10^23)(2)(1.6x10^-12) = 1.93x10^12
> > ergs. Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
> > 2U/3(R/M) = 2(1.93x10^12)/[3(8.317x10^7)/2] = 30,967 degrees K.
>
> I think you made a mistake there somewhere along the line
***{Damn, Dieter, can't I turn my attention away from this group for a
couple of weeks without being shot down? What's the matter with you guys,
anyway? :-) --MJ}***
; that figure is
> certainly too low. Using the more direct E = 3/2 kT and converting the
> 2 keV to Joules, I get about 15 * 10^6 K, not that far from the 22 million
> someone quoted. I seem to remember once being told that 1 eV is about
> equiivalent to 10000 K, so this figures roughly.
***{Yes, your ballpark figure does, indeed, demonstrate that I went astray
somewhere, and John Park found the actual error. (See the next post.)
Thanks. --MJ}***
>
> -- Dieter Britz alias d...@kemi.aau.dk; http://www.kemi.aau.dk/~db
> *** Echelon, bomb, sneakers, GRU: swamp the snoops with trivia! ***
=====================================================
Mitchell Jones
(John Park) wrote:
> Dieter Britz (d...@kemi.aau.dk) writes:
> > On Wed, 21 Jun 2000, Mitchell Jones wrote:
> >
> >> >Steve Lajoie wrote:
> >> >
> >> > 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.
> >>
> >> ***{I'll take a stab at this. First, the 2,000 volt arc will give an
> >> energy of 2 keV to any deuteron that shoots across the plasma unimpeded,
> >> which, in the case of a rarified plasma, will be most of them. Let us,
> >> therefore, calculate the temperature of an ideal monoatomic gas for which
> >> the average particle has an energy of 2 keV and a mole weighs of 2 grams.
> >>
> >> The internal energy per mole (U) of such a gas is the sum of the
> >> rotational and translational kinetic energy of the elements of which it is
> >> comprised, and, since it is presumed to be monoatomic, collisions will be
> >> perfectly elastic, and the rotational component can be neglected. Under
> >> such conditions the average specific heat, C, will be such that C =
> >> (3/2)(R/M), where R is the universal gas constant and M is the molecular
> >> weight of the gas. Thus we can multiply both sides by the absolute
> >> temperature, T, obtaining: TC = (3/2)(R/M)(T). Substituting U for TC, we
> >> have U = (3/2)(R/M)(T). Solving for T, we have: T = 2U/3(R/M). Taking the
> >> average kinetic energy of translation to be 2 keV, a mole of monoatomic
> >> deuterium gas will consist of 2 grams, and will contain 6.02x10^23 atoms
> >> (Avogadro's number). Thus U = (6.02x10^23)(2)(1.6x10^-12) = 1.93x10^12
> ^^^^^^^^^^^^^^
>
> This looks like 2 eV, not 2 keV, in ergs.
>
>
> >> ergs. Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
> >> 2U/3(R/M) = 2(1.93x10^12)/[3(8.317x10^7)/2] = 30,967 degrees K.
>
> or 31 million K with the correction.
***{Woops! Absolutely correct! Thanks. The calculation should have been as
follows:
U = (3/2)(R/M)(T) = (6.02x10^23)(2000)(1.6x10^-12) = 1.93x10^15 ergs.
Since R = 8.317x10^7 ergs per deg. K per mole, it follows that T =
2U/3(R/M) = 2(1.93x10^15)/[3(8.317x10^7)/2] = 30,967,000 degrees K.
--Mitchell Jones}***
>
> >
> > I think you made a mistake there somewhere along the line; that figure is
> > certainly too low. Using the more direct E = 3/2 kT and converting the
> > 2 keV to Joules, I get about 15 * 10^6 K, not that far from the 22 million
> > someone quoted. I seem to remember once being told that 1 eV is about
> > equiivalent to 10000 K, so this figures roughly.
> >