Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Dec 11, 2015 at 3:17 PM,  wrote:

The implication being that as particles go slower, their De Broglie
> wavelength increases, thus so does the distance at which the force reversal
> applies. This looks a lot like the increase in cross section for slower
> neutrons.


Indeed.  That is interesting.  The charge requirement seems like an
unnecessary addition, but I like the general direction.

The intuition for me is that the nuclear force might present a broadband
spectrum of sorts, and that this spectrum can have resonances at different
frequencies.  In high-energy particle physics, we see the resonances at
high frequencies.  Perhaps at much lower frequencies (analogous to but not
the same as radio waves), you'd also find resonances.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Wed, 9 Dec 2015 21:50:31 -0600:
Hi,
[snip]
>On Wed, Dec 9, 2015 at 9:00 PM,  wrote:
>
>The nuclear force is very short range.
>
>
>Here is where I'm inclined to part with conventional wisdom.  Consider that
>1 barn is the approximate area of a medium-sized nucleus presented to an
>oncoming neutron, that nuclei such as 135Xe have neutron-capture cross
>sections of 1e6 barns, and that with a neutron the Coulomb interaction is
>not involved.  It seems to me that the nuclear force must be working at
>longer distance than the usual 1 fm that is mentioned.  Perhaps a more
>nuanced analysis would show that it works on very fast nuclei at short
>distances and on slower-moving nuclei at longer distances.
>
>Eric

Interesting that you mention this. It brings to mind the theory of CC (kicked
off the list), where the force between like charged particles reverses when they
are separated by less than their De Broglie wavelength in the center of mass
frame. The implication being that as particles go slower, their De Broglie
wavelength increases, thus so does the distance at which the force reversal
applies. This looks a lot like the increase in cross section for slower
neutrons. (Assuming that neutrons comprise charged particles.)

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  David Roberson's message of Wed, 9 Dec 2015 22:06:19 -0500:
Hi,
[snip]
>Guys, what would you expect to happen if the identity of individual nucleons 
>is lost once they enter the nucleus?  Since each is supposed to be constructed 
>from 3 quarks, it may be logical to assume that nearby nucleons behave as one 
>greater one composed of 6 or more quarks.  How would one prove that each 
>proton and neutron keeps its identity separate?
>
>Dave
As far as the arguments I presented regarding both Coulomb & nuclear force, it
wouldn't make any difference, even if it were "soup". The total number of
sub-particles (quarks if you prefer), would remain unchanged, as would their
charge, and I had already assumed a more or less random distribution.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 10:45 PM, Jones Beene  wrote:

*From:* Eric Walker
>
>
>
> The nuclear force is very short range.
>
>
>
> Ø  Here is where I'm inclined to part with conventional wisdom.
>  [...snip...]
>
This is arguable not true.
>

Interesting points.

Also, it occurred to me that the de Broglie wavelength of a thermal neutron
is approx. 100 pm, which is on the order of the distance between lattice
sites in a metal.  This large wavelength is behind the diffraction patterns
one sees in detectors for neutron scattering.  Perhaps the neutron
wavefunction readily overlaps with that of the 135Xe nucleus as it passes
by, and that for hand-wavy reasons the strong force need not have a longer
range than 1 fm.  (Can there be a similar resonance for alphas?)

Eric

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

From: Eric Walker 

 

The nuclear force is very short range.

 

Ø  Here is where I'm inclined to part with conventional wisdom.  Consider that 
1 barn is the approximate area of a medium-sized nucleus presented to an 
oncoming neutron, that nuclei such as 135Xe have neutron-capture cross sections 
of 1e6 barns, and that with a neutron the Coulomb interaction is not involved.  

 

Eric – 

 

This is arguable not true. A degree of Coulomb interaction can be involved at 
close range with neutrons due to the spatial geometry of charge distribution.

 

Don’t forget the neutron has a magnetic moment, and therefore has at least a 
near-field or segmented charge. Luis Alvarez discovered the neutron's magnetic 
moment many moons ago, and there is no doubt about this detail.

 

Now consider at the implications of having magnetic moment and zero net charge. 
For a particle to have an intrinsic magnetic moment, it must have both spin and 
electric charge at some level. The neutron has half spin, but no net charge. 
Now– place the emphasis on the “net” in net charge…

 

… so that when we consider that the neutron is composed of three charged quarks 
 “no net charge” is a relative statement. Consequently, the smaller negative 
down quark charge (of two down quarks) is technically balanced by the larger up 
charge of one quark – but there is spatial imbalance at femtometers geometry of 
this charge due to the location of charge carriers vis-à-vis the center of 
mass. This is a charge imbalance at close range. 

 

A similar oddity is seen when a deuteron approaches another deuteron – there is 
a bit of geometric shielding of net positive repulsion which is provided by the 
neutrons of either nucleon, which only appears at close range. 

 

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 9:00 PM,  wrote:

The nuclear force is very short range.


Here is where I'm inclined to part with conventional wisdom.  Consider that
1 barn is the approximate area of a medium-sized nucleus presented to an
oncoming neutron, that nuclei such as 135Xe have neutron-capture cross
sections of 1e6 barns, and that with a neutron the Coulomb interaction is
not involved.  It seems to me that the nuclear force must be working at
longer distance than the usual 1 fm that is mentioned.  Perhaps a more
nuanced analysis would show that it works on very fast nuclei at short
distances and on slower-moving nuclei at longer distances.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

I realize I am asking strange questions, but sometimes it pays off to consider 
alternate ideas.

For example, is 12 a magic number when it comes to the number of quarks in one 
grouping?  Since alpha particles are emitted during some radioactive decays but 
no D's, perhaps 12 is magic while 6 is not strongly bound as an individual 
group.   The 6 quark combination of proton and neutron must be more firmly 
connected to the remainder of the nucleus than to each other.

We typically think of an alpha particle as composed of 4 units, when it might 
make sense to consider it 12 quarks of the best possible combination.  The 
binding energy of helium is unique when compared to other elements.  Just 
speculation. :-)

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Wed, Dec 9, 2015 10:21 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Wed, Dec 9, 2015 at 9:06 PM, David Roberson  wrote:


Guys, what would you expect to happen if the identity of individual nucleons is 
lost once they enter the nucleus?


There is an assumption that the identity is lost, in a sense. Through meson 
exchange, neutrons are thought to regularly become protons and vice versa, very 
rapidly, and it might be considered experimentally impossible to keep track of 
any one nucleon over time.


Since each is supposed to be constructed from 3 quarks, it may be logical to 
assume that nearby nucleons behave as one greater one composed of 6 or more 
quarks.  How would one prove that each proton and neutron keeps its identity 
separate?





I think the current understanding is that by using probes with sufficient 
energy to "see" individual nucleons rather than the nucleus as a whole (e.g., 
using electrons accelerated to very high energies), the data are consistent 
with nucleons containing three valence quarks.  Beyond this I don't know much 
more.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 9:06 PM, David Roberson  wrote:

Guys, what would you expect to happen if the identity of individual
> nucleons is lost once they enter the nucleus?


There is an assumption that the identity is lost, in a sense. Through meson
exchange, neutrons are thought to regularly become protons and vice versa,
very rapidly, and it might be considered experimentally impossible to keep
track of any one nucleon over time.

Since each is supposed to be constructed from 3 quarks, it may be logical
> to assume that nearby nucleons behave as one greater one composed of 6 or
> more quarks.  How would one prove that each proton and neutron keeps its
> identity separate?
>

I think the current understanding is that by using probes with sufficient
energy to "see" individual nucleons rather than the nucleus as a whole
(e.g., using electrons accelerated to very high energies), the data are
consistent with nucleons containing three valence quarks.  Beyond this I
don't know much more.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Guys, what would you expect to happen if the identity of individual nucleons is 
lost once they enter the nucleus?  Since each is supposed to be constructed 
from 3 quarks, it may be logical to assume that nearby nucleons behave as one 
greater one composed of 6 or more quarks.  How would one prove that each proton 
and neutron keeps its identity separate?

Dave

 

 

 

-Original Message-
From: mixent 
To: vortex-l 
Sent: Wed, Dec 9, 2015 10:00 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?

In reply to  Eric Walker's message of Wed, 9 Dec 2015 20:49:09 -0600:
Hi,
[snip]
>Understood.  I only wanted to get agreement on what Krane's understanding
>is.  I think Krane's understanding is the mainstream position. This is not
>necessarily the correct one, but it's good to know what it is if one is
>going to take a position against it.
>
>Given the extreme subtleties of the experimental data in this particular
>field and the success of practitioners in untangling a number of details,
>I'm personally inclined to go with Krane's understanding as a first pass,
>but this doesn't mean everyone should.
>
>Eric

I just thought of something else, which might better align with Krane's
explanation.

The nuclear force is very short range. In an elongated nucleus the curvature of
the surface at the poles is greater than the curvature at the equator. That
means that particles at the poles have on average fewer neighbors than those at
the equator, so the nuclear binding force they feel is also weaker. This
combined with the increased Coulomb repulsion makes the barrier thinner
(actually, it makes it lower, but the effect is the same).

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Wed, 9 Dec 2015 20:49:09 -0600:
Hi,
[snip]
>Understood.  I only wanted to get agreement on what Krane's understanding
>is.  I think Krane's understanding is the mainstream position. This is not
>necessarily the correct one, but it's good to know what it is if one is
>going to take a position against it.
>
>Given the extreme subtleties of the experimental data in this particular
>field and the success of practitioners in untangling a number of details,
>I'm personally inclined to go with Krane's understanding as a first pass,
>but this doesn't mean everyone should.
>
>Eric

I just thought of something else, which might better align with Krane's
explanation.

The nuclear force is very short range. In an elongated nucleus the curvature of
the surface at the poles is greater than the curvature at the equator. That
means that particles at the poles have on average fewer neighbors than those at
the equator, so the nuclear binding force they feel is also weaker. This
combined with the increased Coulomb repulsion makes the barrier thinner
(actually, it makes it lower, but the effect is the same).

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 8:26 PM,  wrote:

1) I'm curious as to how they know that tunneling from the poles is more
> likely,
> given that they can't actually see what's going on. (Perhaps the anisotropy
> shows up in experiments done in a strong magnetic field?)
>

Yes -- this type of experiment requires first aligning the magnetic dipole
or electric quadrupole moments of the alpha-emitting nuclei in a magnetic
field or a crystalline electric field gradient. In order to keep the spins
aligned they must keep the atoms below 0.01 K.

2) If the nucleus is oblong because it is spinning, then particles at the
> poles
> will be subject to a stronger centrifugal force.
>

I get the impression they were looking at deformed nuclei, whose ground
states are oblong.

3) Note that he only suggests the barrier is thinner as a *possible*
> explanation, implying possibly that no one knows for sure.
>

True.


> 4) A positively charged particle at a pole will feel the repulsive force of
> *all* the other positively charged nucleons in the nucleus, pushing in the
> same
> direction. A particle at the equator, will feel approximately equal
> numbers of
> particles pushing from the one pole as from the other, so the repulsive
> forces
> tend to counteract one another.
> At the equator there will be some component of the repulsive forces that is
> perpendicular to the long axis, i.e. pointing out of the equator, but that
> will
> be small compared to the absolute magnitude of the force. (sine of a small
> angle).
>

Yes, I imagine this is possible.

5) In short, I doubt the validity of the proffered explanation.
>

Understood.  I only wanted to get agreement on what Krane's understanding
is.  I think Krane's understanding is the mainstream position. This is not
necessarily the correct one, but it's good to know what it is if one is
going to take a position against it.

Given the extreme subtleties of the experimental data in this particular
field and the success of practitioners in untangling a number of details,
I'm personally inclined to go with Krane's understanding as a first pass,
but this doesn't mean everyone should.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Wed, 9 Dec 2015 15:11:42 -0600:
Hi,
[snip]
>On Wed, Dec 9, 2015 at 2:39 PM,  wrote:
>
>IOW the barrier does work in two directions (due to the two forces at
>> work), but
>> is never named accordingly. So I suspect that it's just the naming
>> convention
>> that is confusing you.
>>
>
>When calculating the tunneling probability of an alpha particle, the width
>of the Coulomb barrier is an important term in the calculation.  The
>tunneling rate ranges from fractions of a second to billions of years as a
>result of small changes in this term. The Coulomb barrier width extends far
>beyond the reach of the nuclear force, which is on the order of ~ 1 fm.  If
>the nuclear force were the only thing keeping the alpha particle within the
>potential well of the nucleus, one would expect any geometrical
>considerations in the calculation to extend beyond the nucleus itself on
>the order of ~ 1 fm, which is far from the case in this calculation.
>
>I think we went over this ground earlier, and I'm having trouble finding
>your reply.  Of course, you could very well be correct, and I will keep an
>open mind.

Yes we did. :) When the width of the barrier is calculated, it is based upon the
energy remaining to the alpha particle once it has escaped. The lower that
energy, the longer is the half life.
Note however that if a particle once having escaped only has a little energy
left, then it apparently didn't have much to start with. If it didn't have much
to start with, then it's not surprising that it found it difficult to overcome
the nuclear barrier, hence the long half life.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Wed, 9 Dec 2015 16:06:32 -0600:
Hi,
[snip]
>In oblong nuclei, there is an angular dependency on the alpha tunneling 
>probability.  Alpha particles are more likely to tunnel out of poles of such 
>nuclei rather than at the circumference.  Krane writes, "Since many 
>alpha-emitting nuclei are deformed, these angular distribution measurements 
>can also help us to answer another question: if we assume a stable prolate 
>(elongated) nucleus, will more alpha's be emitted from the poles or from the 
>equator?  Figure 8.9 suggests a possible answer to this question: at the 
>larger radius of the poles, the alpha particle feels a weaker Coulomb 
>potential and therefore must penetrate a thinner and lower barrier."
>I read this to mean that Krane believes that it is the Coulomb potential and 
>not the nuclear potential that is thinner and therefore easier to traverse at 
>the poles.  Do you disagree?

1) I'm curious as to how they know that tunneling from the poles is more likely,
given that they can't actually see what's going on. (Perhaps the anisotropy
shows up in experiments done in a strong magnetic field?)

2) If the nucleus is oblong because it is spinning, then particles at the poles
will be subject to a stronger centrifugal force.

3) Note that he only suggests the barrier is thinner as a *possible*
explanation, implying possibly that no one knows for sure.

4) A positively charged particle at a pole will feel the repulsive force of
*all* the other positively charged nucleons in the nucleus, pushing in the same
direction. A particle at the equator, will feel approximately equal numbers of
particles pushing from the one pole as from the other, so the repulsive forces
tend to counteract one another. 
At the equator there will be some component of the repulsive forces that is
perpendicular to the long axis, i.e. pointing out of the equator, but that will
be small compared to the absolute magnitude of the force. (sine of a small
angle).


5) In short, I doubt the validity of the proffered explanation.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 2:39 PM,  wrote:

So I suspect that it's just the naming convention that is confusing you.
>

In oblong nuclei, there is an angular dependency on the alpha tunneling
probability.  Alpha particles are more likely to tunnel out of poles of
such nuclei rather than at the circumference.  Krane writes, "Since many
alpha-emitting nuclei are deformed, these angular distribution measurements
can also help us to answer another question: if we assume a stable prolate
(elongated) nucleus, will more alpha's be emitted from the poles or from
the equator?  Figure 8.9 suggests a possible answer to this question: at
the larger radius of the poles, the alpha particle feels a weaker Coulomb
potential and therefore must penetrate a thinner and lower barrier."

I read this to mean that Krane believes that it is the Coulomb potential
and not the nuclear potential that is thinner and therefore easier to
traverse at the poles.  Do you disagree?

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 9, 2015 at 2:39 PM,  wrote:

IOW the barrier does work in two directions (due to the two forces at
> work), but
> is never named accordingly. So I suspect that it's just the naming
> convention
> that is confusing you.
>

When calculating the tunneling probability of an alpha particle, the width
of the Coulomb barrier is an important term in the calculation.  The
tunneling rate ranges from fractions of a second to billions of years as a
result of small changes in this term. The Coulomb barrier width extends far
beyond the reach of the nuclear force, which is on the order of ~ 1 fm.  If
the nuclear force were the only thing keeping the alpha particle within the
potential well of the nucleus, one would expect any geometrical
considerations in the calculation to extend beyond the nucleus itself on
the order of ~ 1 fm, which is far from the case in this calculation.

I think we went over this ground earlier, and I'm having trouble finding
your reply.  Of course, you could very well be correct, and I will keep an
open mind.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Tue, 8 Dec 2015 20:00:24 -0600:
Hi,
[snip]
>About the matter of the Coulomb barrier -- I like your and Dave's argument 
>that the Coulomb barrier should be expected to work in one direction (and this 
>would also seem to be implied by the shell theorem).  But Krane on three or so 
>occasions has written things that imply that the Coulomb barrier works in two 
>directions, suggesting that it's not just a misinterpretation on my part.  One 
>possibility here is that a side effect of decreasing the Coulomb barrier 
>surrounding the nucleus is that this somehow alters the nuclear potential in a 
>way that makes it seem as though the Coulomb barrier works in two directions.

What is called the "Coulomb barrier" is the result of two opposing forces. The
repulsive Coulomb force (for positively charged particles), and the attractive
nuclear force. Most references make no distinction, but refer simply to the
Coulomb barrier, even when IMO they should be referring to the "Nuclear force
barrier", which I have never even seen mentioned. 
IOW the barrier does work in two directions (due to the two forces at work), but
is never named accordingly. So I suspect that it's just the naming convention
that is confusing you.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Tue, 8 Dec 2015 20:00:24 -0600:
Hi,
[snip]
>That would also create daughters with much higher energies.  A nice thing 
>about the lower energies involved in alpha captures is that the daughters end 
>up having ~ 300 keV/nucleon, which is not that much.  Also, 2 D's and 4 H's 
>require 3-body and 5-body reactions, which seem improbable to me.

If they are bound together in shrunken molecules it is only a two body reaction.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

I wrote:

About the matter of the Coulomb barrier -- I like your and Dave's argument
> that the Coulomb barrier should be expected to work in one direction (and
> this would also seem to be implied by the shell theorem).  But Krane on
> three or so occasions has written things that imply that the Coulomb
> barrier works in two directions, suggesting that it's not just a
> misinterpretation on my part.
>

I think the difficulties come down to the fact that the Coulomb barrier not
only provides a force against which incoming charged particles must
counteract on approaching the nucleus (a classical effect). It also serves
as a Faraday cage of sorts that prevents charged particles, which can be
thought of as a type of RF, from "leaking" through it (a nonclassical
effect).  To the extent that that Faraday cage is weakened by surplus
electron charge, RF (charged particles) can leak (tunnel) through it in
either direction.

In this understanding, the Coulomb barrier works in one direction as a
center of Coulomb charge and in two directions as a Faraday cage.  The
analogy of a Faraday cage is meant to be illustrative and not taken too
literally.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Tue, Dec 8, 2015 at 6:18 PM,  wrote:

Note that with alpha fusion some extra energy is available, so I suppose
> that in
> theory that means you could start a little lower, however that would also
> not be
> spontaneous fission, but rather triggered fission.
>

Agreed. The reference to spontaneous fission in the appendix was an
interesting detail, but the fragmentation would be induced, presumably, by
the additional energy in the compound nucleus resulting from the capture of
an alpha particle.

This is not to say that spontaneous fission might not also be a possibility
if something is increasing the tunneling probability for fission.

If you add a neutron to U238 it doesn't fission, despite the addition of 4.8
> MeV. So unless the alpha fusion releases quite a bit more than this, I
> don't
> think you are going to get much fission for lower isotopes.
>

Agreed.  Added to this picture is the assumption that something is
simultaneously increasing the rate of alpha decay, the alpha capture cross
section and the probability of fragmentation.

Note that addition of 2 D's or 4 H's yields some 20+MeV more than addition
> of an
> alpha, making such a combination much more likely to be able to trigger
> fission.
>

That would also create daughters with much higher energies.  A nice thing
about the lower energies involved in alpha captures is that the daughters
end up having ~ 300 keV/nucleon, which is not that much.  Also, 2 D's and 4
H's require 3-body and 5-body reactions, which seem improbable to me.

About the matter of the Coulomb barrier -- I like your and Dave's argument
that the Coulomb barrier should be expected to work in one direction (and
this would also seem to be implied by the shell theorem).  But Krane on
three or so occasions has written things that imply that the Coulomb
barrier works in two directions, suggesting that it's not just a
misinterpretation on my part.  One possibility here is that a side effect
of decreasing the Coulomb barrier surrounding the nucleus is that this
somehow alters the nuclear potential in a way that makes it seem as though
the Coulomb barrier works in two directions.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Tue, 8 Dec 2015 16:18:31 -0600:
Hi,

Note that with alpha fusion some extra energy is available, so I suppose that in
theory that means you could start a little lower, however that would also not be
spontaneous fission, but rather triggered fission.

If you add a neutron to U238 it doesn't fission, despite the addition of 4.8
MeV. So unless the alpha fusion releases quite a bit more than this, I don't
think you are going to get much fission for lower isotopes.
(Neutron addition to U235 yields 6.5 MeV, which apparently is enough, at least
for this somewhat unstable isotope.)

Note that addition of 2 D's or 4 H's yields some 20+MeV more than addition of an
alpha, making such a combination much more likely to be able to trigger fission.

>I had in mind atomic mass (i.e., nuclides in the neighborhood of
>zirconium).  I got this tidbit from Wikipedia (second paragraph):
>
>https://en.wikipedia.org/wiki/Spontaneous_fission
>
>I have now added a reference to this page in the paper.  Wikipedia shows
>"(SF)" in some cases for isotopes in this range, e.g., in the table on the
>righthand side for niobium:
>
>https://en.wikipedia.org/wiki/Niobium
>
>I think this means, "theoretically there could be spontaneous fission, but
>it hasn't been observed."  I've also been assuming that this is also the
>approximate threshold for the start of fragmentation reactions.
>
>Eric
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

I had in mind atomic mass (i.e., nuclides in the neighborhood of
zirconium).  I got this tidbit from Wikipedia (second paragraph):

https://en.wikipedia.org/wiki/Spontaneous_fission

I have now added a reference to this page in the paper.  Wikipedia shows
"(SF)" in some cases for isotopes in this range, e.g., in the table on the
righthand side for niobium:

https://en.wikipedia.org/wiki/Niobium

I think this means, "theoretically there could be spontaneous fission, but
it hasn't been observed."  I've also been assuming that this is also the
approximate threshold for the start of fragmentation reactions.

Eric


On Tue, Dec 8, 2015 at 3:53 PM,  wrote:

> In reply to  Eric Walker's message of Mon, 7 Dec 2015 22:20:18 -0600:
> Hi Eric,
>
> In the sentence "Spontaneous fission becomes energetically possible at
> atomic
> masses greater than 92."
> Do mean "masses" or "numbers". IOW are you talking about Uranium and up, or
> upward of about Zirconium?
>
> >For those who have had the patience to follow the argument this far, here
> >is a paper that goes into further detail on the general idea:
> >
> >https://drive.google.com/open?id=0BzKtdce19-wySFBVLXJET3k2TlU
> >
> >There should be a "download" link at the top that will allow it to be
> >downloaded as a PDF.  I am particularly happy with the appendix.
> >
> >Eric
> [snip]
> Regards,
>
> Robin van Spaandonk
>
> http://rvanspaa.freehostia.com/project.html
>
>

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Mon, 7 Dec 2015 22:20:18 -0600:
Hi Eric,

In the sentence "Spontaneous fission becomes energetically possible at atomic
masses greater than 92." 
Do mean "masses" or "numbers". IOW are you talking about Uranium and up, or
upward of about Zirconium?

>For those who have had the patience to follow the argument this far, here
>is a paper that goes into further detail on the general idea:
>
>https://drive.google.com/open?id=0BzKtdce19-wySFBVLXJET3k2TlU
>
>There should be a "download" link at the top that will allow it to be
>downloaded as a PDF.  I am particularly happy with the appendix.
>
>Eric
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

For those who have had the patience to follow the argument this far, here
is a paper that goes into further detail on the general idea:

https://drive.google.com/open?id=0BzKtdce19-wySFBVLXJET3k2TlU

There should be a "download" link at the top that will allow it to be
downloaded as a PDF.  I am particularly happy with the appendix.

Eric


On Fri, Dec 4, 2015 at 6:12 PM, Eric Walker  wrote:

> On Fri, Dec 4, 2015 at 3:07 PM,  wrote:
>
> I guess there are some exceptions. :)
>>
>
> For exothermic 4He + Pd reactions with 1-3 daughters which also occur in
> nature, I get a count of 269 reactions.  If one removes the limitation on
> unstable daughters, I get a count of 4556 reactions.
>
> 4He + 102Pd => 26Mg + 80Kr + 8602 keV
> 4He + 106Pd => 22Ne + 88Sr + 8464 keV
> 4He + 104Pd => 4He + 30Si + 74Ge + 8460 keV
>
> (Some of the reactions spit a 4He back out.)
>
> However try this for the reactions that Iwamura reported.
>>
>
> To get from cesium to praseodymium, we have:
>
> 4He + 133Cs => gamma + 137La + 1495 keV
> 4He + 137La => gamma + 141Pr + 1300 keV
>
> To get from strontium to molybdenum, we have:
>
> 4He + 88Sr => gamma + 92Zr + 2963 keV
> 4He + 92Zr => gamma + 96Mo + 2759 keV
>
> The case of barium to samarium does appear to be endothermic; e.g.,
>
> 4He + 137Ba => gamma + 141Ce + 139 keV
> 4He + 141Ce => gamma + 145Nd + -1579 keV
> 4He + 145Nd => gamma + 149Sm + -1872 keV
>
> Iwamura and the other authors had trouble confirming samarium when they
> started with 137Ba, although they thought it was probable.  It makes sense
> that this chain would have been attenuated due to the endothermic steps.
> But if the result was in fact real, it also suggests that the alphas are
> travelling with ~ 3-9 MeV of energy. :)
>
> The actual chains would have been more complex in the case of barium
> because some of the intermediate steps are unstable.  But a more complete
> analysis shows that several paths that diverge will eventually reconverge
> on samarium.
>
> Eric
>
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Mon, Dec 7, 2015 at 8:57 AM, Bob Higgins 
wrote:

OTOH, if even one electron was placed in a deep Dirac level, would it
> enhance the possibility of electron capture reactions?  How would we even
> know if happened?  Chemically, it would behave like the electron captured
> element even without the capture.
>

Interesting questions and points. Perhaps electron capture would proceed
quickly if there is overlap with the nucleus. In Mills's theory,
specifically, this would not happen, because the orbits are said to be thin
spheres rather than volumes, and so there would never be overlap, unless
one takes the position that Mills's theory is a steady-state one and that
something else might happen when there's a transition of some kind.

I do not yet see a compelling need to resort to below-ground-state electron
levels, myself, although I appreciate other people taking this route.  In
the context of LENR such theories have been proposed to explain evidence
that I think can be accounted for through other means.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Bob Higgins]

"Shrunken" in this context is about deep Dirac electron orbitals or Mills
orbitals.  Yes, this should be possible with atoms having more than 1
electron in the neutral state, but to be of any fusion use, the orbitals of
all electrons must be simultaneously shrunk.  This shrunken state should be
easier to form and detect in a hydrogen atom with only 1 electron for the
neutral atom, and yet we don't have any confirmation that it does happen
(there are many thing about nature that we haven't learned how to detect).

If a helium had only one of its 2 electrons in a shrunken orbital, the net
charge would be +1 when the outer electron orbital was crossed.  The outer
electron orbital would probably be different with one of the electrons
shrunk to a deep Dirac level.  This would cause the He to look like a heavy
hydrogen chemically.

OTOH, if even one electron was placed in a deep Dirac level, would it
enhance the possibility of electron capture reactions?  How would we even
know if happened?  Chemically, it would behave like the electron captured
element even without the capture.

On Mon, Dec 7, 2015 at 3:56 AM, Roarty, Francis X  wrote:

> Eric,   it seems likely that any gas atoms that can migrate into the
> required nano geometry should also shrink… so helium should do fine but I
> don’t know about metal atoms which might need plasma temperatures and that
> would seem to threaten the standard skeletal catalyst or nano powders,
> however there is the Mills mechanism of self catilization that would
> present a possible avenue for this.
>
> Fran
>
>
>
> *From:* Eric Walker [mailto:eric.wal...@gmail.com]
> *Sent:* Sunday, December 06, 2015 4:36 PM
> *To:* vortex-l@eskimo.com
> *Subject:* EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?
>
>
>
> On Sun, Dec 6, 2015 at 3:00 PM,  wrote:
>
>
>
> BTW, If "shrunken Helium" should exist (along the same lines as Mills'
> shrunken
> Hydrogen), then it might provide a simple means of overcoming the Coulomb
> barrier.
>
>
>
> If there is shrunken hydrogen, it seems likely that there would also be
> shrunken helium (and shrunken zinc, lead, uranium, etc.).
>
>
>
> Eric
>
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Roarty, Francis X]

Eric,   it seems likely that any gas atoms that can migrate into the 
required nano geometry should also shrink… so helium should do fine but I don’t 
know about metal atoms which might need plasma temperatures and that would seem 
to threaten the standard skeletal catalyst or nano powders, however there is 
the Mills mechanism of self catilization that would present a possible avenue 
for this.
Fran

From: Eric Walker [mailto:eric.wal...@gmail.com]
Sent: Sunday, December 06, 2015 4:36 PM
To: vortex-l@eskimo.com
Subject: EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?

On Sun, Dec 6, 2015 at 3:00 PM, mailto:mix...@bigpond.com>> 
wrote:

BTW, If "shrunken Helium" should exist (along the same lines as Mills' shrunken
Hydrogen), then it might provide a simple means of overcoming the Coulomb
barrier.

If there is shrunken hydrogen, it seems likely that there would also be 
shrunken helium (and shrunken zinc, lead, uranium, etc.).

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sun, Dec 6, 2015 at 3:00 PM,  wrote:

BTW, If "shrunken Helium" should exist (along the same lines as Mills'
> shrunken
> Hydrogen), then it might provide a simple means of overcoming the Coulomb
> barrier.
>

If there is shrunken hydrogen, it seems likely that there would also be
shrunken helium (and shrunken zinc, lead, uranium, etc.).

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Sun, 6 Dec 2015 08:11:00 -0600:
Hi,

BTW, If "shrunken Helium" should exist (along the same lines as Mills' shrunken
Hydrogen), then it might provide a simple means of overcoming the Coulomb
barrier.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Sun, 6 Dec 2015 08:11:00 -0600:
Hi,
[snip]
>Do you have a sense of a lower bound?  My impression so far has been that it 
>would be hard to fuse a 4He with anything lighter than beryllium, say, even if 
>energetically it is possible.  

The production of Carbon in stars is based upon Helium fusion. Note however that
A=5 is a very bad number for nuclear isotopes. They tend to be very unstable
with extremely short half lives.

>Perhaps the 4He would simply break apart the light nucleus upon impact, for 
>example.  But this is just speculation on my part.

That probably depends on how much kinetic energy the He4 has, and how much is
needed to break apart the target nucleus. 
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Dec 5, 2015 at 2:33 PM,  wrote:

In short it should
> fuse with almost anything stable, provided that the Coulomb barrier can be
> overcome.
>

Do you have a sense of a lower bound?  My impression so far has been that
it would be hard to fuse a 4He with anything lighter than beryllium, say,
even if energetically it is possible.  Perhaps the 4He would simply break
apart the light nucleus upon impact, for example.  But this is just
speculation on my part.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Fri, 4 Dec 2015 18:12:56 -0600:
Hi Eric,
[snip]

Ok, I had that completely wrong. I forgot that He4 itself doesn't have as high a
binding energy per nucleon as most of the other elements. In short it should
fuse with almost anything stable, provided that the Coulomb barrier can be
overcome.

>On Fri, Dec 4, 2015 at 3:07 PM,  wrote:
>
>I guess there are some exceptions. :)
>>
>
>For exothermic 4He + Pd reactions with 1-3 daughters which also occur in
>nature, I get a count of 269 reactions.  If one removes the limitation on
>unstable daughters, I get a count of 4556 reactions.
>
>4He + 102Pd => 26Mg + 80Kr + 8602 keV
>4He + 106Pd => 22Ne + 88Sr + 8464 keV
>4He + 104Pd => 4He + 30Si + 74Ge + 8460 keV
>
>(Some of the reactions spit a 4He back out.)
>
>However try this for the reactions that Iwamura reported.
>>
>
>To get from cesium to praseodymium, we have:
>
>4He + 133Cs => gamma + 137La + 1495 keV
>4He + 137La => gamma + 141Pr + 1300 keV
>
>To get from strontium to molybdenum, we have:
>
>4He + 88Sr => gamma + 92Zr + 2963 keV
>4He + 92Zr => gamma + 96Mo + 2759 keV
>
>The case of barium to samarium does appear to be endothermic; e.g.,
>
>4He + 137Ba => gamma + 141Ce + 139 keV
>4He + 141Ce => gamma + 145Nd + -1579 keV
>4He + 145Nd => gamma + 149Sm + -1872 keV
>
>Iwamura and the other authors had trouble confirming samarium when they
>started with 137Ba, although they thought it was probable.  It makes sense
>that this chain would have been attenuated due to the endothermic steps.
>But if the result was in fact real, it also suggests that the alphas are
>travelling with ~ 3-9 MeV of energy. :)
>
>The actual chains would have been more complex in the case of barium
>because some of the intermediate steps are unstable.  But a more complete
>analysis shows that several paths that diverge will eventually reconverge
>on samarium.
>
>Eric
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Dec 4, 2015 at 3:07 PM,  wrote:

I guess there are some exceptions. :)
>

For exothermic 4He + Pd reactions with 1-3 daughters which also occur in
nature, I get a count of 269 reactions.  If one removes the limitation on
unstable daughters, I get a count of 4556 reactions.

4He + 102Pd => 26Mg + 80Kr + 8602 keV
4He + 106Pd => 22Ne + 88Sr + 8464 keV
4He + 104Pd => 4He + 30Si + 74Ge + 8460 keV

(Some of the reactions spit a 4He back out.)

However try this for the reactions that Iwamura reported.
>

To get from cesium to praseodymium, we have:

4He + 133Cs => gamma + 137La + 1495 keV
4He + 137La => gamma + 141Pr + 1300 keV

To get from strontium to molybdenum, we have:

4He + 88Sr => gamma + 92Zr + 2963 keV
4He + 92Zr => gamma + 96Mo + 2759 keV

The case of barium to samarium does appear to be endothermic; e.g.,

4He + 137Ba => gamma + 141Ce + 139 keV
4He + 141Ce => gamma + 145Nd + -1579 keV
4He + 145Nd => gamma + 149Sm + -1872 keV

Iwamura and the other authors had trouble confirming samarium when they
started with 137Ba, although they thought it was probable.  It makes sense
that this chain would have been attenuated due to the endothermic steps.
But if the result was in fact real, it also suggests that the alphas are
travelling with ~ 3-9 MeV of energy. :)

The actual chains would have been more complex in the case of barium
because some of the intermediate steps are unstable.  But a more complete
analysis shows that several paths that diverge will eventually reconverge
on samarium.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Thu, 3 Dec 2015 23:41:26 -0600:
Hi,
[snip]
>On Thu, Dec 3, 2015 at 10:39 PM,  wrote:
>
>For elements heavier than Fe/Ni, alpha capture is endothermic, which implies
>> that it could only happen if fast alphas are available.
>>
>
>Here is what my script is telling me about that:
>
>4He + 110Pd => gamma + 114Cd + 4108 keV
>4He + 108Pd => gamma + 112Cd + 3476 keV
>4He + 107Pd => gamma + 111Cd + 3306 keV
>4He + 106Pd => gamma + 110Cd + 2866 keV
>4He + 104Pd => gamma + 108Cd + 2283 keV
>4He + 102Pd => gamma + 106Cd + 1626 keV
>
>Am I doing something wrong?

Yes, you are surprising me! ;)

I guess there are some exceptions. :)

However try this for the reactions that Iwamura reported.

>Furthermore, only a tiny fraction of fast alphas would actually undergo a
>> reaction.
>
>
>This kind of thing leads me to wonder about an enhanced alpha capture cross
>section.
>
>Eric
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Thu, Dec 3, 2015 at 10:39 PM,  wrote:

For elements heavier than Fe/Ni, alpha capture is endothermic, which implies
> that it could only happen if fast alphas are available.
>

Here is what my script is telling me about that:

4He + 110Pd => gamma + 114Cd + 4108 keV
4He + 108Pd => gamma + 112Cd + 3476 keV
4He + 107Pd => gamma + 111Cd + 3306 keV
4He + 106Pd => gamma + 110Cd + 2866 keV
4He + 104Pd => gamma + 108Cd + 2283 keV
4He + 102Pd => gamma + 106Cd + 1626 keV

Am I doing something wrong?

Furthermore, only a tiny fraction of fast alphas would actually undergo a
> reaction.


This kind of thing leads me to wonder about an enhanced alpha capture cross
section.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  David Roberson's message of Wed, 2 Dec 2015 11:24:13 -0500:
Hi,
[snip]
>Your description of the field fluctuations occurring due to random processes 
>taking place does seem logical.   What would you expect to observe if a 
>nucleus that typically emits alphas is placed within a strong electric field?  
>For example, placing some of these ions within the field located between the 
>plates of a high voltage capacitor?  One might expect that type of arrangement 
>to have an effect upon the alpha energy and decay rate.  In this structure the 
>field could be adjusted to quantify the functional relationships.

I like this idea. IIRC the experiments that lead to the Barker patent also
detected a larger effect when the radioisotope was placed near (in/on) the
surface of the VdG generator. I think that means that the field would be
asymmetric.

If your capacitor idea pans out, it might lead to a useful device that could
produce energy on demand, if the isotope and voltages are chosen carefully.

What one would aim for is as large a shift in decay rates as possible, with
minimal expenditure of applied energy.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Tue, 1 Dec 2015 16:11:27 -0600:
Hi,
[snip]
>I came upon the suggestion because it provides a nice explanation for
>results like those of Iwamura as being successive alpha captures.  In his
>case, only two or three captures are needed to get from strontium to
>molybdenum, cesium to praseodymium, or barium to samarium.  Successive
>alpha captures seem like a reasonable hypothesis if there are lots of alpha
>particles being emitted under induced decay.

For elements heavier than Fe/Ni, alpha capture is endothermic, which implies
that it could only happen if fast alphas are available. Furthermore, only a tiny
fraction of fast alphas would actually undergo a reaction.

IMO a more reasonable reaction involves the addition of a small cluster of H/D
atoms. Such a reaction would be exothermic, and may well lead to the
transmutation results found.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Wed, Dec 2, 2015 at 10:24 AM, David Roberson  wrote:

Robin, are you aware of any direct correlation between the energy emitted
> by a particle and its decay rate?


This is a well-established finding.  Alpha decays in nature are between 4
and 9 MeV (approx.).  The more energy, the shorter the decay, and the less
energy, the longer the decay.  The decay rates are logarithmic in the
energy, going from 10e-3 s at 9 MeV to 4.4e17 s at 4 MeV for thorium
isotopes.  If you have a copy of Krane's "Introductory Nuclear Physics" (I
highly recommend it), this phenomenon is discussed starting on p. 251 (I'm
having trouble finding a comparable discussion on the Internet).

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Robin, I agree that the field strength originating from the central tiny charge 
would be the same as without the presence of the external ring once the alpha 
breaches that ring.   The main idea is that the alpha would only require an 
incoming amount of energy associated with passage from that outer ring to the 
center charge.  This would of course be a smaller energy requirement than 
without the ring being in place.

The reduction in alpha incoming energy is inversely proportional to the square 
of the radius of the outer ring.   That implies that the central charge will 
dominate the total required energy to a very large extent unless the radius of 
the outer spherical ring of negative charge is small compared to that of the 
central charge.   For this reason I also would not expect the energy reduction 
to be very large, but it should be present.

You are correct in suggesting that a symmetrical sphere of negative charge 
surrounding a nucleus would not impact the field within that nucleus.  And, one 
would not expect the alpha to be encouraged to leave a nucleus for that reason. 
 On the other hand, one leaving should enjoy less net energy than an alpha 
ejected from a bare nucleus once it has left the near region.  This appears to 
be an interesting concept.  I suppose that it implies that since the alpha was 
once part of the nucleus, it shares a portion of the energy loss that occurs 
when the external electrons shed orbital energy as the atom reaches ground 
state by electromagnetic radiation.

Robin, are you aware of any direct correlation between the energy emitted by a 
particle and its decay rate?  The spherical symmetrical shell structure should 
likely result in a slightly lower energy alpha emission by the process outlined 
above.  Even though the electric field within the nucleus would remain the 
same, the energy of the released alpha would be less.  At the moment it is not 
clear as to how to handle the energy associated with the electrons that are 
released from the atom at the same time as the alpha.

Your description of the field fluctuations occurring due to random processes 
taking place does seem logical.   What would you expect to observe if a nucleus 
that typically emits alphas is placed within a strong electric field?  For 
example, placing some of these ions within the field located between the plates 
of a high voltage capacitor?  One might expect that type of arrangement to have 
an effect upon the alpha energy and decay rate.  In this structure the field 
could be adjusted to quantify the functional relationships.

Dave

 

 

-Original Message-
From: mixent 
To: vortex-l 
Sent: Tue, Dec 1, 2015 4:34 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?

In reply to  David Roberson's message of Mon, 30 Nov 2015 18:10:02 -0500:
Hi,
[snip]

Dave, I like your analysis. However it implies that if the field were
spherically symmetrical alpha decay would not be enhanced since the nucleus
would not feel the external electrons at all. Since increasing the electron
density is known to slightly increase the alpha decay rate, one can only draw
the conclusion that the field is not (always) spherically symmetrical. In fact
given the high mobility of electrons, I would expect there to be very high
frequency and essentially random fluctuations in the local electron density at
any given point. A momentary peak would temporarily enhance the chances of alpha
decay. Furthermore a specific arrangement of atoms in a molecule or lattice may
well create an asymmetric field. Such fields may play a role in catalysis. (I'm
thinking especially of active sites in enzymes here.)


>I was thinking along the same lines as you Eric.  If you take a positive 
>charge of tiny size and surround it with an equal amount of symmetrically 
>distributed negative charge the structure is overall electrically neutral when 
>viewed at a distance.   An alpha approaching from the outside would not 
>encounter any force until it passes through the negative electrical spherical 
>shell.
>
>Once the alpha passes through the negative charge shell it encounters a 
>portion of the original positive field that is the same as previously observed 
>without the negative charge shell present.  In effect the alpha has avoided 
>the energy required to breach the negative shell distance from the central 
>charge.  The negative field is balanced out within the region from its surface 
>all the way to the central charge due to its symmetrical structure. 
>
>Dave
>
> 
>
> 
>
> 
>
>-Original Message-
>From: Eric Walker 
>To: vortex-l 
>Sent: Mon, Nov 30, 2015 4:03 pm
>Subject: Re: [Vo]: How many atoms to make condensed matter?
>
>
>
>
>On Mon, Nov 30, 2015 at 2:41 PM,   wrote:
>
>
>
>No, I'm saying it does both. When the alpha particle is far away it enhances 
>it,
>
>but wh

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Tue, Dec 1, 2015 at 3:33 PM,  wrote:

Furthermore a specific arrangement of atoms in a molecule or lattice may
> well create an asymmetric field.


If what is needed is a field that is not spherical, that is also an
interesting idea.  Perhaps hydrogen and deuterium accomplish this, e.g., as
in this photo, which shows charge density diagrams with 0, 1, and 2
hydrogen atoms in the middle of a palladium lattice:

http://i.imgur.com/3uKjFpi.png

The electron charge density for electrons orbiting the hydrogen atoms at
the center of the images are suggested by the Tsyganov and others to be
forced into a p-orbital, although I don't know how likely this is.  This
image is from p. 103 of this issue of JMCNS:
http://www.iscmns.org/CMNS/JCMNS-Vol17.pdf.

In this case, the electron orbitals pass by the sides of the palladium
atoms, creating a non-uniform charge distribution.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Tue, Dec 1, 2015 at 3:36 PM,  wrote:

The emphasis I think should be placed on the word "opposite". IOW one might
> expect the result to be "opposite" as well.
>

True.  I suspect it will be hard to get much further insight into the
matter at the level of physical argumentation.

I came upon the suggestion because it provides a nice explanation for
results like those of Iwamura as being successive alpha captures.  In his
case, only two or three captures are needed to get from strontium to
molybdenum, cesium to praseodymium, or barium to samarium.  Successive
alpha captures seem like a reasonable hypothesis if there are lots of alpha
particles being emitted under induced decay.

In addition, there's something in the LENR transmutation experiments that
seems to be fragmenting heavier nuclei such as palladium.  It would not be
surprising to me if this kind of thing is the result of a botched alpha
capture.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Mon, 30 Nov 2015 15:19:03 -0600:
Hi Eric,

This is just a restatement of your original position with no further supporting
evidence or logic.

The emphasis I think should be placed on the word "opposite". IOW one might
expect the result to be "opposite" as well.


>On Mon, Nov 30, 2015 at 2:41 PM,  wrote:
>
>No, I'm saying it does both. When the alpha particle is far away it
>> enhances it,
>> but when it get close to a target nucleus it works against it. I'm not
>> sure what
>> the net result would be.
>>
>
>More to the point, it might be possible to think of alpha decay and alpha
>capture as being opposite directions of the same process. So if there's
>something that catalyzes the "leaking" of an alpha particle out of heavier
>nuclides, the same thing might enhance the ability of lighter nuclides to
>capture them if it would be energetically favorable.
>
>Eric
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  David Roberson's message of Mon, 30 Nov 2015 18:10:02 -0500:
Hi,
[snip]

Dave, I like your analysis. However it implies that if the field were
spherically symmetrical alpha decay would not be enhanced since the nucleus
would not feel the external electrons at all. Since increasing the electron
density is known to slightly increase the alpha decay rate, one can only draw
the conclusion that the field is not (always) spherically symmetrical. In fact
given the high mobility of electrons, I would expect there to be very high
frequency and essentially random fluctuations in the local electron density at
any given point. A momentary peak would temporarily enhance the chances of alpha
decay. Furthermore a specific arrangement of atoms in a molecule or lattice may
well create an asymmetric field. Such fields may play a role in catalysis. (I'm
thinking especially of active sites in enzymes here.)


>I was thinking along the same lines as you Eric.  If you take a positive 
>charge of tiny size and surround it with an equal amount of symmetrically 
>distributed negative charge the structure is overall electrically neutral when 
>viewed at a distance.   An alpha approaching from the outside would not 
>encounter any force until it passes through the negative electrical spherical 
>shell.
>
>Once the alpha passes through the negative charge shell it encounters a 
>portion of the original positive field that is the same as previously observed 
>without the negative charge shell present.  In effect the alpha has avoided 
>the energy required to breach the negative shell distance from the central 
>charge.  The negative field is balanced out within the region from its surface 
>all the way to the central charge due to its symmetrical structure. 
>
>Dave
>
> 
>
> 
>
> 
>
>-Original Message-
>From: Eric Walker 
>To: vortex-l 
>Sent: Mon, Nov 30, 2015 4:03 pm
>Subject: Re: [Vo]: How many atoms to make condensed matter?
>
>
>
>
>On Mon, Nov 30, 2015 at 2:41 PM,   wrote:
>
>
>
>No, I'm saying it does both. When the alpha particle is far away it enhances 
>it,
>
>but when it get close to a target nucleus it works against it. I'm not sure 
>what
>the net result would be.
>
>
>
>
>If the volume of the surplus negative charge is spherical about the 
>positively-charged nucleus, the shell theorem implies that one can neglect any 
>negative charge that lies on the far side of the alpha particle from the 
>nucleus.  (It is probably not spherical, whatever it is, unless that s-orbital 
>thing is what's going on.)
>
>
>Also, I'm going to guess that we have to be careful not to treat the negative 
>and positive charge separately; i.e., what is seen by the alpha particle is 
>the result of their overlap.  So in this understanding, if the field of the 
>nucleus is overwhelmingly positive, the negative charge is experienced by the 
>alpha particle to be a little less positive charge.
>
>
>Eric
>
>
>
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

I was thinking along the same lines as you Eric.  If you take a positive charge 
of tiny size and surround it with an equal amount of symmetrically distributed 
negative charge the structure is overall electrically neutral when viewed at a 
distance.   An alpha approaching from the outside would not encounter any force 
until it passes through the negative electrical spherical shell.

Once the alpha passes through the negative charge shell it encounters a portion 
of the original positive field that is the same as previously observed without 
the negative charge shell present.  In effect the alpha has avoided the energy 
required to breach the negative shell distance from the central charge.  The 
negative field is balanced out within the region from its surface all the way 
to the central charge due to its symmetrical structure. 

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Mon, Nov 30, 2015 4:03 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Mon, Nov 30, 2015 at 2:41 PM,   wrote:



No, I'm saying it does both. When the alpha particle is far away it enhances it,

but when it get close to a target nucleus it works against it. I'm not sure what
the net result would be.




If the volume of the surplus negative charge is spherical about the 
positively-charged nucleus, the shell theorem implies that one can neglect any 
negative charge that lies on the far side of the alpha particle from the 
nucleus.  (It is probably not spherical, whatever it is, unless that s-orbital 
thing is what's going on.)


Also, I'm going to guess that we have to be careful not to treat the negative 
and positive charge separately; i.e., what is seen by the alpha particle is the 
result of their overlap.  So in this understanding, if the field of the nucleus 
is overwhelmingly positive, the negative charge is experienced by the alpha 
particle to be a little less positive charge.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Mon, Nov 30, 2015 at 2:41 PM,  wrote:

No, I'm saying it does both. When the alpha particle is far away it
> enhances it,
> but when it get close to a target nucleus it works against it. I'm not
> sure what
> the net result would be.
>

More to the point, it might be possible to think of alpha decay and alpha
capture as being opposite directions of the same process. So if there's
something that catalyzes the "leaking" of an alpha particle out of heavier
nuclides, the same thing might enhance the ability of lighter nuclides to
capture them if it would be energetically favorable.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Mon, Nov 30, 2015 at 2:41 PM,  wrote:

No, I'm saying it does both. When the alpha particle is far away it
> enhances it,
> but when it get close to a target nucleus it works against it. I'm not
> sure what
> the net result would be.
>

If the volume of the surplus negative charge is spherical about the
positively-charged nucleus, the shell theorem implies that one can neglect
any negative charge that lies on the far side of the alpha particle from
the nucleus.  (It is probably not spherical, whatever it is, unless that
s-orbital thing is what's going on.)

Also, I'm going to guess that we have to be careful not to treat the
negative and positive charge separately; i.e., what is seen by the alpha
particle is the result of their overlap.  So in this understanding, if the
field of the nucleus is overwhelmingly positive, the negative charge is
experienced by the alpha particle to be a little less positive charge.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Sun, 29 Nov 2015 21:34:17 -0600:
Hi,
[snip]
>On Sun, Nov 29, 2015 at 7:09 PM,  wrote:
>
>
>> As the particle gets
>> closer to the target, the screening electrons get fewer, and the effect
>> eventually reverses, with there being more behind the particle than in
>> front of
>> it.
>>
>
>Just so I am clear on what you're arguing -- you're arguing that additional
>electron charge permeating the Coulomb barrier surrounding the nucleus
>would not increase the alpha-capture cross section; if anything it would
>decrease it. If I have understood your position, it seems counterintuitive
>to me.
>
>Eric

No, I'm saying it does both. When the alpha particle is far away it enhances it,
but when it get close to a target nucleus it works against it. I'm not sure what
the net result would be.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sun, Nov 29, 2015 at 7:09 PM,  wrote:


> As the particle gets
> closer to the target, the screening electrons get fewer, and the effect
> eventually reverses, with there being more behind the particle than in
> front of
> it.
>

Just so I am clear on what you're arguing -- you're arguing that additional
electron charge permeating the Coulomb barrier surrounding the nucleus
would not increase the alpha-capture cross section; if anything it would
decrease it. If I have understood your position, it seems counterintuitive
to me.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Sat, 28 Nov 2015 16:38:41 -0600:
Hi,
[snip]
>It would increase the alpha-capture cross section by screening the Coulomb 
>repulsion, allowing a more direct approach by the incident prompt alpha 
>particle. In this thinking, tunnelling across the barrier into a nucleus for 
>which it would be energetically favorable to capture an alpha particle (i.e., 
>many lighter ones) becomes more likely if there is less Coulomb repulsion to 
>deflect the alpha out of the way on its approach. I don't think the approach 
>even has to be fully on-center.

What the effect of electrons would be depends on whether they are between the
particle & it's target or behind the particle.
At normal atomic distances they can provide some screening. As the particle gets
closer to the target, the screening electrons get fewer, and the effect
eventually reverses, with there being more behind the particle than in front of
it.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 28, 2015 at 2:44 PM,  wrote:

>At any rate, in this case, we
> >seem to have an example of a single environmental factor that would
> >increase the rate alpha emission and increase the alpha capture cross
> >section.
>
> ...and exactly how would it do the latter?


It would increase the alpha-capture cross section by screening the Coulomb
repulsion, allowing a more direct approach by the incident prompt alpha
particle. In this thinking, tunnelling across the barrier into a nucleus
for which it would be energetically favorable to capture an alpha particle
(i.e., many lighter ones) becomes more likely if there is less Coulomb
repulsion to deflect the alpha out of the way on its approach. I don't
think the approach even has to be fully on-center.

Note that simple fusion is not necessarily what is needed; a compound
nucleus that decays briefly after the capture is also good for producing
heat.

I would think that it would actually
> slow down fusion, since it would effectively reduce the energy of the
> (positively charged) source particle.


By a small amount, I would imagine.  An alpha particle emitted from a
typical alpha-emitter is going to have 4-9 MeV.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Sat, 28 Nov 2015 09:34:55 -0600:
Hi,
[snip]
>On Sat, Nov 28, 2015 at 1:47 AM,  wrote:
>
>Yes, I know, but the presence of more negative charge close to the nucleus
>> increases the energy of the positively charged alpha particle because, not
>> only
>> is it leaving the positively charged nucleus behind, but it now also has
>> extra
>> negatively charged particles ahead of it. In short the "voltage drop"
>> increases.
>>
>
>Btw, it occurred to me that you're the one who originally brought that
>patent to my attention.

Indeed. However it was Jones (IIRC) who first brought it to my attention years
ago .

>
>I think the negative charge would be somewhat isotropic, so I don't see how
>it would set up a voltage that would help to pull the alpha out of the
>nucleus, 

"isotropic" is scale dependent in this case. On the scale of humans it would be.
On the atomic scale, it is all external to the nucleus.

>but you might be right of course. At any rate, in this case, we
>seem to have an example of a single environmental factor that would
>increase the rate alpha emission and increase the alpha capture cross
>section.

...and exactly how would it do the latter? I would think that it would actually
slow down fusion, since it would effectively reduce the energy of the
(positively charged) source particle. (It would effectively make the hill that
needed to be climbed even higher, by lowering the base from which it had to
start.)
However an excess external *positive* charge might make fusion easierwhich
reminds me of the you-tube video on Coulombic explosions that someone recently
posted. If all the electrons suddenly desert a surface to bind with something in
solution, then the surface is left locally with an excess positive charge, so
any hydrogen already in the material might have an easier time fusing???
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 28, 2015 at 1:47 AM,  wrote:

Yes, I know, but the presence of more negative charge close to the nucleus
> increases the energy of the positively charged alpha particle because, not
> only
> is it leaving the positively charged nucleus behind, but it now also has
> extra
> negatively charged particles ahead of it. In short the "voltage drop"
> increases.
>

Btw, it occurred to me that you're the one who originally brought that
patent to my attention.

I think the negative charge would be somewhat isotropic, so I don't see how
it would set up a voltage that would help to pull the alpha out of the
nucleus, but you might be right of course. At any rate, in this case, we
seem to have an example of a single environmental factor that would
increase the rate alpha emission and increase the alpha capture cross
section.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Fri, 27 Nov 2015 21:28:12 -0600:
Hi Eric,
[snip]
>On Fri, Nov 27, 2015 at 7:29 PM,  wrote:
>
>Note that the Coulomb barrier width is a function of the particle's energy.
>
>
>Not only do we have a simple theoretical argument that the Coulomb barrier
>width can be lessened by electron screening, precisely as one might imagine
>it could, somewhat independent of the departing alpha-particle's energy,
>there is also evidence that this has been accomplished in practice.

Yes, I know, but the presence of more negative charge close to the nucleus
increases the energy of the positively charged alpha particle because, not only
is it leaving the positively charged nucleus behind, but it now also has extra
negatively charged particles ahead of it. In short the "voltage drop" increases.

>
>Following is one example:
>
>http://www.google.com/patents/US5076971
>
>Eric
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Nov 27, 2015 at 7:29 PM,  wrote:

Note that the Coulomb barrier width is a function of the particle's energy.


Not only do we have a simple theoretical argument that the Coulomb barrier
width can be lessened by electron screening, precisely as one might imagine
it could, somewhat independent of the departing alpha-particle's energy,
there is also evidence that this has been accomplished in practice.

Following is one example:

http://www.google.com/patents/US5076971

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Fri, 27 Nov 2015 18:21:42 -0600:
Hi,


[snip]
>This may be true. But the calculation of the tunneling probability of an alpha 
>particle to escape a nucleus depends upon the width of the Coulomb barrier.  
>The narrower the width, the more likely tunneling is to occur.  Here we're 
>talking about the electromagnetic rather than the nuclear force.  I.e., it 
>seems that somehow the Coulomb barrier works in both directions rather than 
>just from an approach from the outside.

Note that the Coulomb barrier width is a function of the particle's energy. The
more energetic the particle is, the higher up the barrier the intersection point
will be, and hence the thinner the barrier. So saying that the barrier is
thinner is just another way of saying that the particle energy is higher.
However it is hardly surprising that more energetic particles have a greater
chance of escaping the nuclear force.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Nov 27, 2015 at 6:10 PM,  wrote:


> For alpha emission it is actually a nuclear force barrier, since the
> Coulomb
> force actually helps in escaping, rather than hindering.
>

This may be true. But the calculation of the tunneling probability of an
alpha particle to escape a nucleus depends upon the width of the Coulomb
barrier.  The narrower the width, the more likely tunneling is to occur.
Here we're talking about the electromagnetic rather than the nuclear
force.  I.e., it seems that somehow the Coulomb barrier works in both
directions rather than just from an approach from the outside.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Fri, 27 Nov 2015 18:00:46 -0600:
Hi,
[snip]
>This suggests, then, that the same process that is leading to alpha decay 
>(e.g., suppression of the Coulomb barrier) might also be increasing the cross 
>section for alpha capture.

The Coulomb barrier is actually two different barriers depending on which
direction you are going in. For fusion it really is a Coulomb barrier, i.e. you
have to overcome the Coulomb force in order to enter the nucleus.

For alpha emission it is actually a nuclear force barrier, since the Coulomb
force actually helps in escaping, rather than hindering.

Ergo, something which helps alpha emission is unlikely to help fusion.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Nov 27, 2015 at 5:44 PM,  wrote:

In order to overcome the repulsion, they need to strike another nucleus at
> high
> energy. The need for high energy implies that they must get lucky, and hit
> another nucleus before they lose too much energy to ionization.
> Even if they do collide, there is a good chance that the result will just
> be an
> elastic collision rather than a nuclear event.
>

This suggests, then, that the same process that is leading to alpha decay
(e.g., suppression of the Coulomb barrier) might also be increasing the
cross section for alpha capture.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[mixent]

In reply to  Eric Walker's message of Fri, 20 Nov 2015 01:18:19 -0600:
Hi,
[snip]
>The energy is produced as the lighter elements undergo a series of transitions 
>under successive alpha captures.

Even fast alphas produces only a trivial amount of alpha captures (e.g.
1/1). This is primarily due to three things.

1) The alphas rapidly lose energy while ionizing other atoms.
2) They are repelled by the nuclei of other atoms.
3) Nuclei are small relative to the size of atoms.

In order to overcome the repulsion, they need to strike another nucleus at high
energy. The need for high energy implies that they must get lucky, and hit
another nucleus before they lose too much energy to ionization.
Even if they do collide, there is a good chance that the result will just be an
elastic collision rather than a nuclear event.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

The Papp engine and the E-Cat just work at different optical frequencies.
The E-Cat uses light in the infrared and the Papp engine in the XUV range.
The Papp system, like in the Holmlid experiment which is also a XUV based
system, there is a shock wave produced that reaches a significant fraction
of the light speed. This produces the pressure pulse seen in the Papp
engine that pushes the piston.

Noble gases work mostly  in the XUV light range. No noble metals are
required.

On Fri, Nov 20, 2015 at 10:36 AM, Eric Walker  wrote:

> On Fri, Nov 20, 2015 at 9:16 AM, David Roberson 
> wrote:
>
> Good point.  As long as it takes a very tiny amount of precious metal the
>> cost could be contained.   It would be much better to use one of the high
>> mass elements that is lower cost if you have any choice.
>>
>
> I suspect there's a choice here, and that something heavy but less
> expensive could be used.  In my reading of the transmutation
> experiments, induced fission appears to be a competing channel to that of
> alpha decay.  Palladium seems to offer an example of the latter.  If
> whatever is increasing the decay rate of alpha and beta emitters is also
> inducing fission, then palladium would simply break apart (e.g., into Fe,
> Cr, Ca, Si, Ti, etc.).  If this is true, although palladium would provide
> plenty of energy density as a result, but it might not be a good alpha
> particle source for the nickel fuel stock in the E-Cat.
>
> Eric
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Nov 20, 2015 at 9:16 AM, David Roberson  wrote:

Good point.  As long as it takes a very tiny amount of precious metal the
> cost could be contained.   It would be much better to use one of the high
> mass elements that is lower cost if you have any choice.
>

I suspect there's a choice here, and that something heavy but less
expensive could be used.  In my reading of the transmutation
experiments, induced fission appears to be a competing channel to that of
alpha decay.  Palladium seems to offer an example of the latter.  If
whatever is increasing the decay rate of alpha and beta emitters is also
inducing fission, then palladium would simply break apart (e.g., into Fe,
Cr, Ca, Si, Ti, etc.).  If this is true, although palladium would provide
plenty of energy density as a result, but it might not be a good alpha
particle source for the nickel fuel stock in the E-Cat.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Fri, Nov 20, 2015 at 9:12 AM, David Roberson  wrote:

I find it difficult to compare the Papp engine with a Rossi ECAT.  They are
> very different in structure and behavior.


They are definitely different in operation, but I would not be surprised to
learn that they work under a similar principle.  Just to speculate, in the
case of the E-Cat, alpha decay might be induced in some alpha emitter, and
alpha capture on the part of the fuel stock is occurring as a result.  In
the case of the Papp engine, alpha decay might be induced in an alpha
emitter and the prompt alphas are used to ionize a gas and drive a piston.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Good point.  As long as it takes a very tiny amount of precious metal the cost 
could be contained.   It would be much better to use one of the high mass 
elements that is lower cost if you have any choice.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Fri, Nov 20, 2015 2:19 am
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Thu, Nov 19, 2015 at 10:21 PM, David Roberson  wrote:


The counts for elements of that m value appear quite small when compared to the 
other elements.  Also, why on earth would anyone use such an expensive element 
if a dirt cheap one can substitute?  My suspicion is that this is a dead end 
idea if production costs are taken into consideration.




I can only guess at what role they play.  My current working hypothesis is that 
the heavier elements serve as alpha emitters in Rossi's case. You probably 
would only need a small fraction of the total fuel. The nickel and iron are the 
feedstock. The energy is produced as the lighter elements undergo a series of 
transitions under successive alpha captures.


If this is true, it could explain the difficulties people are having getting 
nickel to produce excess heat in the garage experiments.  It might not hurt to 
them again with an alpha emitter added in in the range of those elements at 
m>100 in the chart and see what happens.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

I find it difficult to compare the Papp engine with a Rossi ECAT.  They are 
very different in structure and behavior.

In the Papp there is obvious motion of the piston which could generate 
electrical voltage by normal processes.  The ECAT has no moving parts that I am 
aware of.

This is not to say that the ECAT does not generate electrical power, only that 
if it does the process is likely different than the one occurring within the 
Papp.  I still require convincing that the Papp engine is capable of performing 
as advertised.

Dave

 

 

 

-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Fri, Nov 20, 2015 12:25 am
Subject: Re: [Vo]: How many atoms to make condensed matter?



Electrical production has been done in the Papp engine. I go back to the Papp 
engine that was self powered by recycling overunity electrons produced by the 
LENR reaction. 


Look at starting at 21:10 of the video below. Watch the motor driven by the 
feedback current gathered by the thorium filled buckets. 


https://www.youtube.com/watch?v=_zWJNyoFgJM




On Fri, Nov 20, 2015 at 12:08 AM, David Roberson  wrote:

First it must be determined that nickel based systems can not generate 
electricity in the same general manner.   So far we have not actually seen 
proof that either metal works all that well.   I remain skeptical until more 
evidence becomes available in support of this concept.

I wonder what fraction of the energy release can be directly converted into 
electricity?   Unless it is a large proportion some form of heat powered 
generator is going to be required anyway.  The overall system will have to be 
optimized.

Dave

 

 

 

-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Thu, Nov 19, 2015 11:28 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?



The cost is cheap if the reactor can produce electricity directly.


On Thu, Nov 19, 2015 at 11:21 PM, David Roberson  wrote:

The counts for elements of that m value appear quite small when compared to the 
other elements.  Also, why on earth would anyone use such an expensive element 
if a dirt cheap one can substitute?  My suspicion is that this is a dead end 
idea if production costs are taken into consideration.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Thu, Nov 19, 2015 10:27 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?





On Thu, Nov 19, 2015 at 7:34 PM, David Roberson  wrote:


Rossi has never mentioned palladium use within his reactors Axil.  That is your 
thought as far as I am aware.




I don't know whether Rossi is now using or has used palladium in the past. But 
one detail in the Lugano report that only became apparent to me much later was 
that there appears to have been a number of heavy elements in one of the fuel 
assays in Appendix 3 (see the lower graph, outlined in red).


http://i.imgur.com/7RQon11.png



Some of these heavier masses were likely to have been compound ions. But it's 
seems unlikely that 100 percent of them were.  Note that palladium overlaps 
with m=105.  Even if palladium were not present, I would not be surprised if 
one or more heavier elements were.


I do not know why this was the only instance in which a graph for m>100 was 
shown in the appendix.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Thu, Nov 19, 2015 at 10:21 PM, David Roberson  wrote:

The counts for elements of that m value appear quite small when compared to
> the other elements.  Also, why on earth would anyone use such an expensive
> element if a dirt cheap one can substitute?  My suspicion is that this is a
> dead end idea if production costs are taken into consideration.
>

I can only guess at what role they play.  My current working hypothesis is
that the heavier elements serve as alpha emitters in Rossi's case. You
probably would only need a small fraction of the total fuel. The nickel and
iron are the feedstock. The energy is produced as the lighter elements
undergo a series of transitions under successive alpha captures.

If this is true, it could explain the difficulties people are having
getting nickel to produce excess heat in the garage experiments.  It might
not hurt to them again with an alpha emitter added in in the range of those
elements at m>100 in the chart and see what happens.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

Electrical production has been done in the Papp engine. I go back to the
Papp engine that was self powered by recycling overunity electrons produced
by the LENR reaction.

Look at starting at 21:10 of the video below. Watch the motor driven by the
feedback current gathered by the thorium filled buckets.

https://www.youtube.com/watch?v=_zWJNyoFgJM

On Fri, Nov 20, 2015 at 12:08 AM, David Roberson  wrote:

> First it must be determined that nickel based systems can not generate
> electricity in the same general manner.   So far we have not actually seen
> proof that either metal works all that well.   I remain skeptical until
> more evidence becomes available in support of this concept.
>
> I wonder what fraction of the energy release can be directly converted
> into electricity?   Unless it is a large proportion some form of heat
> powered generator is going to be required anyway.  The overall system will
> have to be optimized.
>
> Dave
>
>
>
> -Original Message-
> From: Axil Axil 
> To: vortex-l 
> Sent: Thu, Nov 19, 2015 11:28 pm
> Subject: Re: [Vo]: How many atoms to make condensed matter?
>
> The cost is cheap if the reactor can produce electricity directly.
>
> On Thu, Nov 19, 2015 at 11:21 PM, David Roberson 
> wrote:
>
>> The counts for elements of that m value appear quite small when compared
>> to the other elements.  Also, why on earth would anyone use such an
>> expensive element if a dirt cheap one can substitute?  My suspicion is that
>> this is a dead end idea if production costs are taken into consideration.
>>
>> Dave
>>
>>
>>
>> -----Original Message-
>> From: Eric Walker 
>> To: vortex-l 
>> Sent: Thu, Nov 19, 2015 10:27 pm
>> Subject: Re: [Vo]: How many atoms to make condensed matter?
>>
>> On Thu, Nov 19, 2015 at 7:34 PM, David Roberson 
>> wrote:
>>
>> Rossi has never mentioned palladium use within his reactors Axil.  That
>>> is your thought as far as I am aware.
>>>
>>
>> I don't know whether Rossi is now using or has used palladium in the
>> past. But one detail in the Lugano report that only became apparent to me
>> much later was that there appears to have been a number of heavy elements
>> in one of the fuel assays in Appendix 3 (see the lower graph, outlined in
>> red).
>>
>> http://i.imgur.com/7RQon11.png
>>
>> Some of these heavier masses were likely to have been compound ions. But
>> it's seems unlikely that 100 percent of them were.  Note that palladium
>> overlaps with m=105.  Even if palladium were not present, I would not be
>> surprised if one or more heavier elements were.
>>
>> I do not know why this was the only instance in which a graph for m>100
>> was shown in the appendix.
>>
>> Eric
>>
>>
>>
>

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

First it must be determined that nickel based systems can not generate 
electricity in the same general manner.   So far we have not actually seen 
proof that either metal works all that well.   I remain skeptical until more 
evidence becomes available in support of this concept.

I wonder what fraction of the energy release can be directly converted into 
electricity?   Unless it is a large proportion some form of heat powered 
generator is going to be required anyway.  The overall system will have to be 
optimized.

Dave

 

 

 

-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Thu, Nov 19, 2015 11:28 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?



The cost is cheap if the reactor can produce electricity directly.


On Thu, Nov 19, 2015 at 11:21 PM, David Roberson  wrote:

The counts for elements of that m value appear quite small when compared to the 
other elements.  Also, why on earth would anyone use such an expensive element 
if a dirt cheap one can substitute?  My suspicion is that this is a dead end 
idea if production costs are taken into consideration.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Thu, Nov 19, 2015 10:27 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?





On Thu, Nov 19, 2015 at 7:34 PM, David Roberson  wrote:


Rossi has never mentioned palladium use within his reactors Axil.  That is your 
thought as far as I am aware.




I don't know whether Rossi is now using or has used palladium in the past. But 
one detail in the Lugano report that only became apparent to me much later was 
that there appears to have been a number of heavy elements in one of the fuel 
assays in Appendix 3 (see the lower graph, outlined in red).


http://i.imgur.com/7RQon11.png



Some of these heavier masses were likely to have been compound ions. But it's 
seems unlikely that 100 percent of them were.  Note that palladium overlaps 
with m=105.  Even if palladium were not present, I would not be surprised if 
one or more heavier elements were.


I do not know why this was the only instance in which a graph for m>100 was 
shown in the appendix.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

The cost is cheap if the reactor can produce electricity directly.

On Thu, Nov 19, 2015 at 11:21 PM, David Roberson  wrote:

> The counts for elements of that m value appear quite small when compared
> to the other elements.  Also, why on earth would anyone use such an
> expensive element if a dirt cheap one can substitute?  My suspicion is that
> this is a dead end idea if production costs are taken into consideration.
>
> Dave
>
>
>
> -Original Message-
> From: Eric Walker 
> To: vortex-l 
> Sent: Thu, Nov 19, 2015 10:27 pm
> Subject: Re: [Vo]: How many atoms to make condensed matter?
>
> On Thu, Nov 19, 2015 at 7:34 PM, David Roberson 
> wrote:
>
> Rossi has never mentioned palladium use within his reactors Axil.  That
>> is your thought as far as I am aware.
>>
>
> I don't know whether Rossi is now using or has used palladium in the past.
> But one detail in the Lugano report that only became apparent to me much
> later was that there appears to have been a number of heavy elements in one
> of the fuel assays in Appendix 3 (see the lower graph, outlined in red).
>
> http://i.imgur.com/7RQon11.png
>
> Some of these heavier masses were likely to have been compound ions. But
> it's seems unlikely that 100 percent of them were.  Note that palladium
> overlaps with m=105.  Even if palladium were not present, I would not be
> surprised if one or more heavier elements were.
>
> I do not know why this was the only instance in which a graph for m>100
> was shown in the appendix.
>
> Eric
>
>
>

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

The counts for elements of that m value appear quite small when compared to the 
other elements.  Also, why on earth would anyone use such an expensive element 
if a dirt cheap one can substitute?  My suspicion is that this is a dead end 
idea if production costs are taken into consideration.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Thu, Nov 19, 2015 10:27 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Thu, Nov 19, 2015 at 7:34 PM, David Roberson  wrote:


Rossi has never mentioned palladium use within his reactors Axil.  That is your 
thought as far as I am aware.




I don't know whether Rossi is now using or has used palladium in the past. But 
one detail in the Lugano report that only became apparent to me much later was 
that there appears to have been a number of heavy elements in one of the fuel 
assays in Appendix 3 (see the lower graph, outlined in red).


http://i.imgur.com/7RQon11.png



Some of these heavier masses were likely to have been compound ions. But it's 
seems unlikely that 100 percent of them were.  Note that palladium overlaps 
with m=105.  Even if palladium were not present, I would not be surprised if 
one or more heavier elements were.


I do not know why this was the only instance in which a graph for m>100 was 
shown in the appendix.


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Thu, Nov 19, 2015 at 7:34 PM, David Roberson  wrote:

Rossi has never mentioned palladium use within his reactors Axil.  That is
> your thought as far as I am aware.
>

I don't know whether Rossi is now using or has used palladium in the past.
But one detail in the Lugano report that only became apparent to me much
later was that there appears to have been a number of heavy elements in one
of the fuel assays in Appendix 3 (see the lower graph, outlined in red).

http://i.imgur.com/7RQon11.png

Some of these heavier masses were likely to have been compound ions. But
it's seems unlikely that 100 percent of them were.  Note that palladium
overlaps with m=105.  Even if palladium were not present, I would not be
surprised if one or more heavier elements were.

I do not know why this was the only instance in which a graph for m>100 was
shown in the appendix.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

This is a deduction based on Rossi's patent. The E-Cat X runs at high
temperatures higher than the Hot Cat. Rossi also has had problems with
structural failures in the E-Cat X because of high temperature operation.
It is doubtful that nickel can handle long term high temperature reactor
operations.

Rossi also says that the E-Cat X is different from the Hot Cat and that the
E-Cat X has a different effect than the low temperature reactors.

On Thu, Nov 19, 2015 at 8:34 PM, David Roberson  wrote:

> Rossi has never mentioned palladium use within his reactors Axil.  That
> is your thought as far as I am aware.
>
> Dave
>
>
>
> -Original Message-
> From: Axil Axil 
> To: vortex-l 
> Sent: Thu, Nov 19, 2015 2:20 pm
> Subject: Re: [Vo]: How many atoms to make condensed matter?
>
> In the Co-deposition video, the narrator says that the reaction sets in
> withing minutes for the start of Co-deposition even though the current is
> at very low power levels maintained at Co-deposition initialization.
>
> During Co-deposition initiation no hydrogen is deposited until three
> atomic levels of palladium are transferred to the copper substrate.
>
> This means that the palladium chloride envelope is the active LENR factor
> and not the hydrogen deposited on the electrode.
>
> As Rossi has done on the E-Cat X, palladium powder used in a pure hydrogen
> envelope WITHOUT oxygen is effective as a LENR reaction mechanism.
>
> On Thu, Nov 19, 2015 at 6:08 AM, Roarty, Francis X <
> francis.x.roa...@lmco.com> wrote:
>
>> Axil, Jones,
>> Good insights and dot connecting, would it apply to Patterson
>> beads submerged in water with a lithium sulfate (Li2SO4) electrolyte
>> solution? (Li2SO4)  vs palladium chloride absorption of UV?  His claim
>> that it would neutralize radiation without emitting harmful radiation is
>> consistent with more recent Japanese patents for remediation.
>> Fran
>>
>> *From:* Axil Axil [mailto:janap...@gmail.com]
>> *Sent:* Wednesday, November 18, 2015 2:15 PM
>> *To:* vortex-l 
>> *Subject:* EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?
>>
>> As posted before, water absorbs UV light about 100,000,000 times better
>> than infrared light. This makes it a poor partner with any noble metal at
>> producing polaritons at UV frequencies.
>>
>>
>> However, when chlorine is added to the palladium solution to form  palladium
>> chloride in the electrolyte, the absorption of UV light is greatly reduced.
>> This favors polariton formation using UV light.
>>
>> See:
>>
>>
>> http://www.researchgate.net/publication/229233040_Speciation_of_aqueous_palladium(II)_chloride_solutions_using_optical_spectroscopies
>> adium(
>>
>>
>> On Wed, Nov 18, 2015 at 1:33 PM, Jones Beene  wrote:
>>
>> On further examination, Ken may be on to an important insight here which
>> is relevant to LENR. Here is another reference with more detail.
>>
>> “Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”
>> YANG, et al.
>>
>> When read in the context of the recently mentioned Szpak interview, where
>> we see the highly credible report of 3 out of 10 meltdown events, using
>> only plain water but with palladium chloride in the electrolyte – this
>> makes me think that the magnetic field facilitated single atom palladium to
>> first densify and accumulate, and then after 3-4 days to react in bulk.
>>
>> *From:* Ken Deboer
>>
>> … most recently JM Thomas (Nature 17 Sept 2015) showed that single
>> atoms, of Pd especially, make better catalysts than nanoparticles. super
>> catalysts, in fact.
>>
>>
>> Now that you mention it – if you look back at Pd-D cold fusion, one of
>> the most effective techniques is “co-dep” or co-deposition.
>>
>> In co-dep, palladium chloride is in the electrolyte, which means
>> essentially that individual ions of palladium are present.
>>
>>
>>
>>
>>
>>
>
>

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Rossi has never mentioned palladium use within his reactors Axil.  That is your 
thought as far as I am aware.

Dave

 

 

 

-Original Message-
From: Axil Axil 
To: vortex-l 
Sent: Thu, Nov 19, 2015 2:20 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?



In the Co-deposition video, the narrator says that the reaction sets in withing 
minutes for the start of Co-deposition even though the current is at very low 
power levels maintained at Co-deposition initialization.  


During Co-deposition initiation no hydrogen is deposited until three atomic 
levels of palladium are transferred to the copper substrate. 


This means that the palladium chloride envelope is the active LENR factor and 
not the hydrogen deposited on the electrode.


As Rossi has done on the E-Cat X, palladium powder used in a pure hydrogen 
envelope WITHOUT oxygen is effective as a LENR reaction mechanism.



On Thu, Nov 19, 2015 at 6:08 AM, Roarty, Francis X  
wrote:


Axil, Jones,
Good insights and dot connecting, would it apply toPatterson beads 
submerged in water with a lithium sulfate (Li2SO4) electrolyte solution? 
(Li2SO4)  vs palladium chloride absorption of UV?  His claim that it would 
neutralize radiation without emitting harmful radiation is consistent with more 
recent Japanese patents for remediation. 
Fran
 
From: Axil Axil [mailto:janap...@gmail.com]
Sent: Wednesday, November 18, 2015 2:15 PM
To: vortex-l 
Subject: EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?

 

As posted before, water absorbs UV light about 100,000,000 times better than 
infrared light. This makes it a poor partner with any noble metal at producing 
polaritons at UV frequencies. 

 

 

However, when chlorine is added to the palladium solution to form  palladium 
chloride in the electrolyte, the absorption of UV light is greatly reduced. 
This favors polariton formation using UV light.

 

See:

 

http://www.researchgate.net/publication/229233040_Speciation_of_aqueous_palladium(II)_chloride_solutions_using_optical_spectroscopiesadium(

 


 

On Wed, Nov 18, 2015 at 1:33 PM, Jones Beene  wrote:


On further examination, Ken may be on to an important insight here which is 
relevant to LENR. Here is another reference with more detail.
 
“Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”
YANG, et al.
 
When read in the context of the recently mentioned Szpak interview, where we 
see the highly credible report of 3 out of 10 meltdown events, using only plain 
water but with palladium chloride in the electrolyte – this makes me think that 
the magnetic field facilitated single atom palladium to first densify and 
accumulate, and then after 3-4 days to react in bulk.

 
From: Ken Deboer
 

… most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms, of Pd 
especially, make better catalysts than nanoparticles. super catalysts, in fact.
 
 
Now that you mention it – if you look back at Pd-D cold fusion, one of the most 
effective techniques is “co-dep” or co-deposition.
 
In co-dep, palladium chloride is in the electrolyte, which means essentially 
that individual ions of palladium are present.
 
 

 




 

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Thu, Nov 19, 2015 at 1:20 PM, Axil Axil  wrote:

This means that the palladium chloride envelope is the active LENR factor
> and not the hydrogen deposited on the electrode.


This sounds likely to me, although hydrogen may help out.

As Rossi has done on the E-Cat X, palladium powder used in a pure hydrogen
> envelope WITHOUT oxygen is effective as a LENR reaction mechanism.


As you may have already pointed out, oxygen is a big absorber of UV.

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

On Thu, Nov 19, 2015 at 6:08 AM, Roarty, Francis X <
francis.x.roa...@lmco.com> wrote:

> Axil, Jones,
>
> Good insights and dot connecting, would it apply to Patterson
> beads submerged in water with a lithium sulfate (Li2SO4) electrolyte
> solution? (Li2SO4)  vs palladium chloride absorption of UV?  His claim
> that it would neutralize radiation without emitting harmful radiation is
> consistent with more recent Japanese patents for remediation.
>
>
>
There must be a match between the surface of the particle, the envelope
that covers the particle and the operating temperature of the reaction. The
surface of the particle must contain light in a highly reflective optical
box reflecting light at low dispersion to produce polaritons.

If the wavelength of the reaction is in the infrared, if the surface of the
   Patterson beads is nickel(reflects infrared light well), and if the
envelope of (Li2SO4) does not absorb infrared, then the LENR environment is
well matched to optimally generate polaritons.

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

In the Co-deposition video, the narrator says that the reaction sets in
withing minutes for the start of Co-deposition even though the current is
at very low power levels maintained at Co-deposition initialization.

During Co-deposition initiation no hydrogen is deposited until three atomic
levels of palladium are transferred to the copper substrate.

This means that the palladium chloride envelope is the active LENR factor
and not the hydrogen deposited on the electrode.

As Rossi has done on the E-Cat X, palladium powder used in a pure hydrogen
envelope WITHOUT oxygen is effective as a LENR reaction mechanism.

On Thu, Nov 19, 2015 at 6:08 AM, Roarty, Francis X <
francis.x.roa...@lmco.com> wrote:

> Axil, Jones,
>
> Good insights and dot connecting, would it apply to Patterson
> beads submerged in water with a lithium sulfate (Li2SO4) electrolyte
> solution? (Li2SO4)  vs palladium chloride absorption of UV?  His claim
> that it would neutralize radiation without emitting harmful radiation is
> consistent with more recent Japanese patents for remediation.
>
> Fran
>
>
>
> *From:* Axil Axil [mailto:janap...@gmail.com]
> *Sent:* Wednesday, November 18, 2015 2:15 PM
> *To:* vortex-l 
> *Subject:* EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?
>
>
>
> As posted before, water absorbs UV light about 100,000,000 times better
> than infrared light. This makes it a poor partner with any noble metal at
> producing polaritons at UV frequencies.
>
>
>
>
>
> However, when chlorine is added to the palladium solution to form  palladium
> chloride in the electrolyte, the absorption of UV light is greatly reduced.
> This favors polariton formation using UV light.
>
>
>
> See:
>
>
>
>
> http://www.researchgate.net/publication/229233040_Speciation_of_aqueous_palladium(II)_chloride_solutions_using_optical_spectroscopies
> adium(
>
>
>
>
>
> On Wed, Nov 18, 2015 at 1:33 PM, Jones Beene  wrote:
>
> On further examination, Ken may be on to an important insight here which
> is relevant to LENR. Here is another reference with more detail.
>
>
>
> “Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”
>
> YANG, et al.
>
>
>
> When read in the context of the recently mentioned Szpak interview, where
> we see the highly credible report of 3 out of 10 meltdown events, using
> only plain water but with palladium chloride in the electrolyte – this
> makes me think that the magnetic field facilitated single atom palladium to
> first densify and accumulate, and then after 3-4 days to react in bulk.
>
>
>
> *From:* Ken Deboer
>
>
>
> … most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms,
> of Pd especially, make better catalysts than nanoparticles. super
> catalysts, in fact.
>
>
>
>
>
> Now that you mention it – if you look back at Pd-D cold fusion, one of the
> most effective techniques is “co-dep” or co-deposition.
>
>
>
> In co-dep, palladium chloride is in the electrolyte, which means
> essentially that individual ions of palladium are present.
>
>
>
>
>
>
>
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Roarty, Francis X]

Axil, Jones,
Good insights and dot connecting, would it apply to Patterson beads 
submerged in water with a lithium sulfate (Li2SO4) electrolyte solution? 
(Li2SO4)  vs palladium chloride absorption of UV?  His claim that it would 
neutralize radiation without emitting harmful radiation is consistent with more 
recent Japanese patents for remediation.
Fran

From: Axil Axil [mailto:janap...@gmail.com]
Sent: Wednesday, November 18, 2015 2:15 PM
To: vortex-l 
Subject: EXTERNAL: Re: [Vo]: How many atoms to make condensed matter?

As posted before, water absorbs UV light about 100,000,000 times better than 
infrared light. This makes it a poor partner with any noble metal at producing 
polaritons at UV frequencies.


However, when chlorine is added to the palladium solution to form  palladium 
chloride in the electrolyte, the absorption of UV light is greatly reduced. 
This favors polariton formation using UV light.

See:

http://www.researchgate.net/publication/229233040_Speciation_of_aqueous_palladium(II)_chloride_solutions_using_optical_spectroscopiesadium(


On Wed, Nov 18, 2015 at 1:33 PM, Jones Beene 
mailto:jone...@pacbell.net>> wrote:
On further examination, Ken may be on to an important insight here which is 
relevant to LENR. Here is another reference with more detail.

“Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”
YANG, et al.

When read in the context of the recently mentioned Szpak interview, where we 
see the highly credible report of 3 out of 10 meltdown events, using only plain 
water but with palladium chloride in the electrolyte – this makes me think that 
the magnetic field facilitated single atom palladium to first densify and 
accumulate, and then after 3-4 days to react in bulk.

From: Ken Deboer

… most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms, of Pd 
especially, make better catalysts than nanoparticles. super catalysts, in fact.


Now that you mention it – if you look back at Pd-D cold fusion, one of the most 
effective techniques is “co-dep” or co-deposition.

In co-dep, palladium chloride is in the electrolyte, which means essentially 
that individual ions of palladium are present.

Re: [Vo]: How many atoms to make condensed matter?

[Axil Axil]

As posted before, water absorbs UV light about 100,000,000 times better
than infrared light. This makes it a poor partner with any noble metal at
producing polaritons at UV frequencies.


However, when chlorine is added to the palladium solution to form  palladium
chloride in the electrolyte, the absorption of UV light is greatly reduced.
This favors polariton formation using UV light.

See:

http://www.researchgate.net/publication/229233040_Speciation_of_aqueous_palladium(II)_chloride_solutions_using_optical_spectroscopies
adium(


On Wed, Nov 18, 2015 at 1:33 PM, Jones Beene  wrote:

> On further examination, Ken may be on to an important insight here which
> is relevant to LENR. Here is another reference with more detail.
>
>
>
> “Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”
>
> YANG, et al.
>
>
>
> When read in the context of the recently mentioned Szpak interview, where
> we see the highly credible report of 3 out of 10 meltdown events, using
> only plain water but with palladium chloride in the electrolyte – this
> makes me think that the magnetic field facilitated single atom palladium to
> first densify and accumulate, and then after 3-4 days to react in bulk.
>
>
>
> *From:* Ken Deboer
>
>
>
> … most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms,
> of Pd especially, make better catalysts than nanoparticles. super
> catalysts, in fact.
>
>
>
>
>
> Now that you mention it – if you look back at Pd-D cold fusion, one of the
> most effective techniques is “co-dep” or co-deposition.
>
>
>
> In co-dep, palladium chloride is in the electrolyte, which means
> essentially that individual ions of palladium are present.
>
>
>
>
>
>
>

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

On further examination, Ken may be on to an important insight here which is 
relevant to LENR. Here is another reference with more detail.

 

“Single-Atom Catalysts: A New Frontier in Heterogeneous Catalysis”

YANG, et al.

 

When read in the context of the recently mentioned Szpak interview, where we 
see the highly credible report of 3 out of 10 meltdown events, using only plain 
water but with palladium chloride in the electrolyte – this makes me think that 
the magnetic field facilitated single atom palladium to first densify and 
accumulate, and then after 3-4 days to react in bulk.

 

From: Ken Deboer 

 

… most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms, of Pd 
especially, make better catalysts than nanoparticles. super catalysts, in fact.

 

 

Now that you mention it – if you look back at Pd-D cold fusion, one of the most 
effective techniques is “co-dep” or co-deposition.

 

In co-dep, palladium chloride is in the electrolyte, which means essentially 
that individual ions of palladium are present.

 

 

 

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

From: Ken Deboer 

 

… most recently JM Thomas (Nature 17 Sept 2015) showed that single atoms, of Pd 
especially, make better catalysts than nanoparticles. super catalysts, in fact.

 

 

Now that you mention it – if you look back at Pd-D cold fusion, one of the most 
effective techniques is “co-dep” or co-deposition.

 

In co-dep, palladium chloride is in the electrolyte, which means essentially 
that individual ions of palladium are present.

 

 

 

Re: [Vo]: How many atoms to make condensed matter?

[Ken Deboer]

Question:  Not sure if it has been discussed before, but could it be that
nanoparticulate fuel arrangements are not the ideal?  Many workers, most
recently JM Thomas (Nature 17 Sept 2015) showed that single atoms, of Pd
especially, make better catalysts than nanoparticles. super catalysts, in
fact.

On Sun, Nov 15, 2015 at 5:42 PM, Jones Beene  wrote:

> Yet another interesting possibility for anomalous energy, showing up in
> nature but heretofore unappreciated - which arguably fits into a version
> of the Holmlid effect is in biology. If Holmlid is correct that iron-oxide
> catalyst along with an alkali (potassium) and a source of light, can
> create energetic particles, we can look at biology in a new light, so to
> speak.
>
> Disregarding Kervran, there are two biological energy anomalies in
> particular which are really rusty, so to speak. One is the Monarch
> butterfly and the other is the class of plants called epiphytes. Both of
> these are strong energy anomalies of a biological sort with iron oxide
> and light as contributory.
>
> Epiphytes are non-parasitic plants that grow on trees, deriving nutrients
> from air and rain, not from the host. Spanish Moss is the classic example,
> growing on oak trees for physical support, but having no roots, it
> generally does not negatively affect the oak. It has been shown that Spanish
> Moss, Tillandsia, grows faster on electrical cables than on trees and has
> 17% Fe2O3, iron oxide in its ashes, when burned, as well as potassium. This
> plant captures sunlight and possibly induced electrical current as a
> supplement (or alternative) to photosynthesis. It is a fast grower, not
> as fast as Kudzu but in times past, tens of millions of pounds was
> harvested annually in the USA for such things as the Model-T Ford (seat
> padding). I love these old PopSci references:
>
>
> *https://books.google.com/books?id=eiYDMBAJ&pg=PA32&dq=Popular+Science+1932+plane&hl=en&ei=uwhRTaffEsq9tgf9vIG6CQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEkQ6AEwCTgy#v=onepage&q&f=true*
> 
>
> ___
>
> Another interesting possibility for anomalous heat due to the Holmlid
> effect (nucleon disintegration) is the planet Jupiter.
>
> Jupiter has a core temperature estimated to be 36,000 K (64,300 °F)
> despite the cold surface - but its large gravitational force is far from
> being able to trigger nuclear reactions. It is a factor of 75 too low to
> trigger fusion. It also has an iron core, the surface of which is probably
> catalytic and where heat is generated via proton disintegration. Jupiter
> could be the ideal mass range for finding hydrogen planets with hot cores -
> where the Holmlid effect predominates. There really are few other choices
> that make as much sense.
>
> The present consensus for Jupiter’s internal heating is that the great
> mass and compressibility is making energy available from gravitational
> contraction. But that explanation is close to brain-dead, IMO since there
> is no net gain from gravitation oscillation around an static average value.
> And there is no indication of permanent shrinkage.
>
> However, theoretical models do indicate that if Jupiter had even slightly
> more mass than it does at present, it would shrink significantly, meaning
> that it is in a metastable state already which is probably exacerbated due
> to the internal heating. So to the extent that internal heat offsets
> gravity, then yes gravity heating is arguable, but not without the internal
> heat. Thus we cannot attribute this core temperature to gravity at all,
> except as the transfer medium/mechanism.
>
> *From:* Bob Higgins
>
> Ø   Can you say what evidence the natural state should exhibit if
> such a sub-nuclear shuffle were as "less difficult" as you describe?  Are
> there natural occurrences that can be looked for that could validate such a
> proposition?
>
> Indeed – such a radical shift would have dramatic, even Universal
> repercussions (turtles all the way down) .
>
> The obvious first place to look is our sun. Do we really understand the
> solar hydrogen fusion cycle?  My opinion is that we could have it partly
> wrong, especially the basic P+P reaction- which is statistically difficult
> to reconcile. Here is the way the mainstream looks at it:
>
> *https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction*
> 
>
> But many observers are struck by the mechanics of the solar fusion cycle
> being absolutely dependent on a rare beta decay in the diproton. Can that
> really happen during the short lifetime of the species? Despite what you
> may think, this critical detail has never been observed, and is merely an
> educated guess. It is a guess which is based mo

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

Yet another interesting possibility for anomalous energy, showing up in nature 
but heretofore unappreciated - which arguably fits into a version of the 
Holmlid effect is in biology. If Holmlid is correct that iron-oxide catalyst 
along with an alkali (potassium) and a source of light, can create energetic 
particles, we can look at biology in a new light, so to speak.

Disregarding Kervran, there are two biological energy anomalies in particular 
which are really rusty, so to speak. One is the Monarch butterfly and the other 
is the class of plants called epiphytes. Both of these are strong energy 
anomalies of a biological sort with iron oxide and light as contributory. 

Epiphytes are non-parasitic plants that grow on trees, deriving nutrients from 
air and rain, not from the host. Spanish Moss is the classic example, growing 
on oak trees for physical support, but having no roots, it generally does not 
negatively affect the oak. It has been shown that Spanish Moss, Tillandsia, 
grows faster on electrical cables than on trees and has 17% Fe2O3, iron oxide 
in its ashes, when burned, as well as potassium. This plant captures sunlight 
and possibly induced electrical current as a supplement (or alternative) to 
photosynthesis. It is a fast grower, not as fast as Kudzu but in times past, 
tens of millions of pounds was harvested annually in the USA for such things as 
the Model-T Ford (seat padding). I love these old PopSci references:

https://books.google.com/books?id=eiYDMBAJ&pg=PA32&dq=Popular+Science+1932+plane&hl=en&ei=uwhRTaffEsq9tgf9vIG6CQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEkQ6AEwCTgy#v=onepage&q&f=true
___
Another interesting possibility for anomalous heat due to the Holmlid effect 
(nucleon disintegration) is the planet Jupiter.

Jupiter has a core temperature estimated to be 36,000 K (64,300 °F) despite the 
cold surface - but its large gravitational force is far from being able to 
trigger nuclear reactions. It is a factor of 75 too low to trigger fusion. It 
also has an iron core, the surface of which is probably catalytic and where 
heat is generated via proton disintegration. Jupiter could be the ideal mass 
range for finding hydrogen planets with hot cores - where the Holmlid effect 
predominates. There really are few other choices that make as much sense.

The present consensus for Jupiter’s internal heating is that the great mass and 
compressibility is making energy available from gravitational contraction. But 
that explanation is close to brain-dead, IMO since there is no net gain from 
gravitation oscillation around an static average value. And there is no 
indication of permanent shrinkage.

However, theoretical models do indicate that if Jupiter had even slightly more 
mass than it does at present, it would shrink significantly, meaning that it is 
in a metastable state already which is probably exacerbated due to the internal 
heating. So to the extent that internal heat offsets gravity, then yes gravity 
heating is arguable, but not without the internal heat. Thus we cannot 
attribute this core temperature to gravity at all, except as the transfer 
medium/mechanism. 

From: Bob Higgins 

*   Can you say what evidence the natural state should exhibit if such a 
sub-nuclear shuffle were as "less difficult" as you describe?  Are there 
natural occurrences that can be looked for that could validate such a 
proposition? 

Indeed – such a radical shift would have dramatic, even Universal repercussions 
(turtles all the way down) .

The obvious first place to look is our sun. Do we really understand the solar 
hydrogen fusion cycle?  My opinion is that we could have it partly wrong, 
especially the basic P+P reaction- which is statistically difficult to 
reconcile. Here is the way the mainstream looks at it:
https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction

But many observers are struck by the mechanics of the solar fusion cycle being 
absolutely dependent on a rare beta decay in the diproton. Can that really 
happen during the short lifetime of the species? Despite what you may think, 
this critical detail has never been observed, and is merely an educated guess. 
It is a guess which is based mostly on lack of another viable mechanism. 

If Holmlid is shown to be correct – then on our sun, we should find that 
nucleon disintegration could be happening instead of, or in addition to, the 
fusion of protons. Of course, there would be some of both, since muons catalyze 
fusion and we know that helium is formed. The proportions could be close to 
even, however. 

The precise details are impossible to frame without more information, but if 
Holmlid is replicated, you will see solar cosmologists in a desperate scramble 
to cover their proverbial trailing edges.

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Bob, I suspect that the resistance that is most important is contained within 
the volume of the permanent magnet in this instance.  Eddy currents induced 
there would cause the magnet to heat up effectively converting the 
gravitational potential energy into thermal energy.

Do you have reason to believe that the external field originating from the 
current flow within the superconductor would not penetrate the permanent 
magnet?  If present, I also suspect that this induced lossy current flowing 
within the permanent magnetic would itself interact with the rest of the fields 
and cause a reduction in net lift force.

Eric has proposed some interesting thought experiments associated with this 
subject.   Hopefully we can come to a reasonable consensus of some sort.

Dave

 

-Original Message-
From: Bob Higgins 
To: vortex-l 
Sent: Sat, Nov 14, 2015 5:57 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?



When you introduce the magnet to the presence of the superconductor, currents 
are induced in the superconductor that cause the magnetic field to exactly 
cancel at the surface of the superconductor such that there is no penetration 
of the magnetic field into the superconductor.  However, this induction of a 
new current is of non-zero frequency, and hence these non-zero frequency 
current components will experience a finite resistance.



On Sat, Nov 14, 2015 at 3:47 PM, Eric Walker  wrote:


On Sat, Nov 14, 2015 at 4:26 PM, David Roberson  wrote:



So, I would expect to see the falling velocity of the magnet to become less and 
less as the conductor used for the pipe become less resistive.  But, the 
geometry is also going to enter into the equation.


Let's assume a geometry favorable to the suspension of the heavy permanent 
magnet for indefinite duration.


If we assume that the pipe diameter is sufficiently small then it would make 
sense to assume that the magnet would be suspended at zero velocity if placed 
within a super conducting pipe.


It's clear that with sufficient current in the superconductor, the heavy magnet 
will just sit there, with the force of gravity counterbalanced by the magnetic 
field that is induced.  My question is whether the system is a stable one, or 
whether there's a gradual decrease in the current as a function of time.


In one scenario, the heavy magnet is suspended in the superconductor 
indefinitely.  In another scenario it is suspended there for a long time, until 
the current drops below a critical threshold, at which point it will move down 
a little, and so on, until it falls through.


Just so I understand your argument -- is your understanding that the force of 
gravity does not indirectly counteract and do work against the current 
circulating in the superconductor as the heavy magnet is suspended there?


Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Interesting final question Eric.  If in fact the heavy magnetic remains 
suspended then yes.  On the other hand there is a mechanism that might result 
in the magnet being able to move slowly downwards through the tube.  The 
complete system includes an additional loss factor associated with eddy 
currents flowing within the magnet body.  The counter current flowing within 
the super conductor causes it to produce a magnetic field that penetrates the 
body of the permanent magnet.

If we assume that the magnet moves slowly within the tubing, then a current 
will be induced within the magnet itself that can convert gravitational 
potential energy of the magnet into heat.  If this mechanism applies, which 
seems likely, then energy would be conserved as is required by the laws of 
physics.   In the earlier thought experiment I was assuming an ideal model of a 
permanent magnet that has zero volume conductivity.   A real world magnetic 
should slowly fall within a super conducting tube as the magnet converts its 
gravitational potential energy into internal heat energy.

An interesting experiment would be to actually place a permanent magnet within 
a super conducting tube and see how it responds.  If the magnet has zero volume 
conductivity, it will remain frozen within the small tube.  On the other hand, 
if the magnet has current induced within its resistive body by the field 
originating within the super conducting tube, it will fall at some constant 
rate.

Eric, we are conducting a thought experiment during this discussion instead of 
actually performing a real experiment.  Unless all of the possible secondary 
effects are taken into consideration it is quite likely that the concepts need 
to be adjusted as real data is examined.

When considering the above arguments, I do not agree with your last statement 
unless the heavy magnetic actually remains suspended and stationary within the 
superconductor.   My thoughts are that this static arrangement is not going to 
be observed under real life circumstances.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Sat, Nov 14, 2015 5:47 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Sat, Nov 14, 2015 at 4:26 PM, David Roberson  wrote:



So, I would expect to see the falling velocity of the magnet to become less and 
less as the conductor used for the pipe become less resistive.  But, the 
geometry is also going to enter into the equation.


Let's assume a geometry favorable to the suspension of the heavy permanent 
magnet for indefinite duration.


If we assume that the pipe diameter is sufficiently small then it would make 
sense to assume that the magnet would be suspended at zero velocity if placed 
within a super conducting pipe.


It's clear that with sufficient current in the superconductor, the heavy magnet 
will just sit there, with the force of gravity counterbalanced by the magnetic 
field that is induced.  My question is whether the system is a stable one, or 
whether there's a gradual decrease in the current as a function of time.


In one scenario, the heavy magnet is suspended in the superconductor 
indefinitely.  In another scenario it is suspended there for a long time, until 
the current drops below a critical threshold, at which point it will move down 
a little, and so on, until it falls through.


Just so I understand your argument -- is your understanding that the force of 
gravity does not indirectly counteract and do work against the current 
circulating in the superconductor as the heavy magnet is suspended there?


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 14, 2015 at 4:57 PM, Bob Higgins 
wrote:

When you introduce the magnet to the presence of the superconductor,
> currents are induced in the superconductor that cause the magnetic field to
> exactly cancel at the surface of the superconductor such that there is no
> penetration of the magnetic field into the superconductor.  However, this
> induction of a new current is of non-zero frequency, and hence these
> non-zero frequency current components will experience a finite resistance.
>

This sounds as though the magnet will fall through eventually, due to the
finite resistance brought about by the non-zero frequency of the current.
I assume this component is radiated away as heat, even in a superconductor.

Is the understanding that at the atomic level this part of the analogy
breaks down, and you would not see the same phenomenon?

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Bob Higgins]

When you introduce the magnet to the presence of the superconductor,
currents are induced in the superconductor that cause the magnetic field to
exactly cancel at the surface of the superconductor such that there is no
penetration of the magnetic field into the superconductor.  However, this
induction of a new current is of non-zero frequency, and hence these
non-zero frequency current components will experience a finite resistance.

On Sat, Nov 14, 2015 at 3:47 PM, Eric Walker  wrote:

> On Sat, Nov 14, 2015 at 4:26 PM, David Roberson 
> wrote:
>
> So, I would expect to see the falling velocity of the magnet to become
>> less and less as the conductor used for the pipe become less resistive.
>> But, the geometry is also going to enter into the equation.
>
>
> Let's assume a geometry favorable to the suspension of the heavy permanent
> magnet for indefinite duration.
>
> If we assume that the pipe diameter is sufficiently small then it would
>> make sense to assume that the magnet would be suspended at zero velocity if
>> placed within a super conducting pipe.
>
>
> It's clear that with sufficient current in the superconductor, the heavy
> magnet will just sit there, with the force of gravity counterbalanced by
> the magnetic field that is induced.  My question is whether the system is a
> stable one, or whether there's a gradual decrease in the current as a
> function of time.
>
> In one scenario, the heavy magnet is suspended in the superconductor
> indefinitely.  In another scenario it is suspended there for a long time,
> until the current drops below a critical threshold, at which point it will
> move down a little, and so on, until it falls through.
>
> Just so I understand your argument -- is your understanding that the force
> of gravity does not indirectly counteract and do work against the current
> circulating in the superconductor as the heavy magnet is suspended there?
>
> Eric
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 14, 2015 at 4:26 PM, David Roberson  wrote:

So, I would expect to see the falling velocity of the magnet to become less
> and less as the conductor used for the pipe become less resistive.  But,
> the geometry is also going to enter into the equation.


Let's assume a geometry favorable to the suspension of the heavy permanent
magnet for indefinite duration.

If we assume that the pipe diameter is sufficiently small then it would
> make sense to assume that the magnet would be suspended at zero velocity if
> placed within a super conducting pipe.


It's clear that with sufficient current in the superconductor, the heavy
magnet will just sit there, with the force of gravity counterbalanced by
the magnetic field that is induced.  My question is whether the system is a
stable one, or whether there's a gradual decrease in the current as a
function of time.

In one scenario, the heavy magnet is suspended in the superconductor
indefinitely.  In another scenario it is suspended there for a long time,
until the current drops below a critical threshold, at which point it will
move down a little, and so on, until it falls through.

Just so I understand your argument -- is your understanding that the force
of gravity does not indirectly counteract and do work against the current
circulating in the superconductor as the heavy magnet is suspended there?

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

I like those types of videos!  In the pipe versus magnet example, the falling 
magnet causes a current to be induced within the copper pipe.  The magnitude of 
the current is proportional to the velocity of the moving magnet and is 
inversely proportional to the resistance of the pipe material.

I suppose you can think of this simple system as a type of motor where the 
generated force is directed against the motion of the magnet and proportional 
to its velocity.  In that case the downward velocity would reach a value where 
the motor force is balanced against the force of gravity.  The current induced 
within the metal is going to depend upon how rapidly the magnet falls hence a 
better conductor would maintain a greater current at the same velocity.

So, I would expect to see the falling velocity of the magnet to become less and 
less as the conductor used for the pipe become less resistive.  But, the 
geometry is also going to enter into the equation.  This is because a very wide 
opening for the pipe would not allow the material to capture a significant 
amount of the magnetic field and therefore offer only a small force as the 
magnet passes through.  This scenario would be interesting to model.

If we assume that the pipe diameter is sufficiently small then it would make 
sense to assume that the magnet would be suspended at zero velocity if placed 
within a super conducting pipe.  I believe you are thinking of the inverse case 
for the effect of resistance which would apply to an insulator.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Sat, Nov 14, 2015 1:24 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:


The loss in the current carrying magnet is due to series resistance and if that 
resistance is eliminated it would not require any additional power once the 
current is set up.




Consider this video:


https://www.youtube.com/watch?v=uh0bbW6S3BY



Here the falling of a permanent magnet is slowed down as it falls through a 
copper pipe.  As it falls, it is inducing a current in the pipe.  My uninformed 
impression is that the only thing that is keeping it from falling faster is the 
resistance in the pipe to the flow of the current.  If we replace the copper 
pipe with a superconductor, the magnet could then fall through as fast as 
gravity will accelerate it, will it not?  If there is a current already set up 
in the superconductor, it would counteract the falling of the magnet, but this 
would be a matter of degree and not a permanent situation, or am I mistaken?


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:

The loss in the current carrying magnet is due to series resistance and if
> that resistance is eliminated it would not require any additional power
> once the current is set up.
>

Consider this video:

https://www.youtube.com/watch?v=uh0bbW6S3BY

Here the falling of a permanent magnet is slowed down as it falls through a
copper pipe.  As it falls, it is inducing a current in the pipe.  My
uninformed impression is that the only thing that is keeping it from
falling faster is the resistance in the pipe to the flow of the current.
If we replace the copper pipe with a superconductor, the magnet could then
fall through as fast as gravity will accelerate it, will it not?  If there
is a current already set up in the superconductor, it would counteract the
falling of the magnet, but this would be a matter of degree and not a
permanent situation, or am I mistaken?

Eric

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

_
Another interesting possibility for anomalous heat due to the Holmlid effect 
(nucleon disintegration) is the planet Jupiter.

Jupiter has a core temperature estimated to be 36,000 K (64,300 °F) despite the 
cold surface - but its large gravitational force is far from being able to 
trigger nuclear reactions. It is a factor of 75 too low to trigger fusion. It 
also has an iron core, the surface of which is probably catalytic and where 
heat is generated via proton disintegration. Jupiter could be the ideal mass 
range for finding hydrogen planets with hot cores - where the Holmlid effect 
predominates. There really are few other choices that make as much sense.

The present consensus for Jupiter’s internal heating is that the great mass and 
compressibility is making energy available from gravitational contraction. But 
that explanation is close to brain-dead, IMO since there is no net gain from 
gravitation oscillation around an static average value. And there is no 
indication of permanent shrinkage.

However, theoretical models do indicate that if Jupiter had even slightly more 
mass than it does at present, it would shrink significantly, meaning that it is 
in a metastable state already which is probably exacerbated due to the internal 
heating. So to the extent that internal heat offsets gravity, then yes gravity 
heating is arguable, but not without the internal heat. Thus we cannot 
attribute this core temperature to gravity at all, except as the transfer 
medium/mechanism. 

From: Bob Higgins 

*   Can you say what evidence the natural state should exhibit if such a 
sub-nuclear shuffle were as "less difficult" as you describe?  Are there 
natural occurrences that can be looked for that could validate such a 
proposition? 

Indeed – such a radical shift would have dramatic, even Universal repercussions 
(turtles all the way down) .

The obvious first place to look is our sun. Do we really understand the solar 
hydrogen fusion cycle?  My opinion is that we could have it partly wrong, 
especially the basic P+P reaction- which is statistically difficult to 
reconcile. Here is the way the mainstream looks at it:
https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction

But many observers are struck by the mechanics of the solar fusion cycle being 
absolutely dependent on a rare beta decay in the diproton. Can that really 
happen during the short lifetime of the species? Despite what you may think, 
this critical detail has never been observed, and is merely an educated guess. 
It is a guess which is based mostly on lack of another viable mechanism. 

If Holmlid is shown to be correct – then on our sun, we should find that 
nucleon disintegration could be happening instead of, or in addition to, the 
fusion of protons. Of course, there would be some of both, since muons catalyze 
fusion and we know that helium is formed. The proportions could be close to 
even, however. 

The precise details are impossible to frame without more information, but if 
Holmlid is replicated, you will see solar cosmologists in a desperate scramble 
to cover their proverbial trailing edges.

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

I do not think the domains will change.  An example is the ability of an atomic 
clock to keep extremely accurate time.  This would not be true unless the 
orbitals maintained constant energy levels.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Sat, Nov 14, 2015 2:03 am
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:


I consider electrons in orbits as being equivalent to a superconductor current 
since the orbits do not collapse with time.  No power is radiated by an 
electron orbital and hence no work is required to keep it in the proper 
location.




Another way to come at the question I just raised is this -- even though the 
electrons may be superconducting in their orbits, is there something the force 
of a weight that is held up might do to gradually decohere the magnetic domains?


Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

It will remain held in place forever unless the magnet heat up.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Sat, Nov 14, 2015 1:53 am
Subject: Re: [Vo]: How many atoms to make condensed matter?




On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:


The loss in the current carrying magnet is due to series resistance and if that 
resistance is eliminated it would not require any additional power once the 
current is set up.



I was thinking about that.  But let's make the example extreme.  Suppose you 
have a superconducting magnet set up in an industrial grade structure, and you 
run a current through sufficient to keep a one-hundred ton block of iron ten 
meters off the ground.  Will the solenoid hold up the block indefinitely, or 
will some process of loss (other than through resistance) cause the electrons 
to gradually radiate away the input energy until the block eventually falls?  
(We can neglect for the moment the energy needed to cool the current loop in 
order to keep it superconducting.)


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Bob Higgins]

I think the answer to your question about gradual decoherence of the
magnetic domains might actually be the opposite.  Remember your old
horseshoe magnets?  They were always stored with a "keeper" so as to keep
the magnetic field strong.  I think what would happen over time is that the
magnet will align itself for minimum energy (due to thermal dither), and
minimum energy when there is a magnetic material (reluctor) on the magnet
means gradual alignment of the domains to include the reluctor as part of
the minimum energy solution.

In the case of a superconductor, don't forget that a superconductor is only
zero resistance for DC (F=0).  True DC has been held constant from -inf to
+inf for all time.  At some point in a superconducting magnet, current must
be added once the superconducting state is achieved.  This may typically be
a ramp in current, but it boggles me somewhat to understand how a true 0
frequency component is introduced into the superconductor.  While the
storage of that current does not require work, it requires a great deal of
work to load that current into the superconductor.

I think something similar can be said about the magnet and the ring in the
photo.  Just like the weight on the table is not doing work even though it
has potential energy, something put that potential energy into the system
by doing work in the first place.

On Sat, Nov 14, 2015 at 12:02 AM, Eric Walker  wrote:

> On Sat, Nov 14, 2015 at 12:25 AM, David Roberson 
> wrote:
>
> I consider electrons in orbits as being equivalent to a superconductor
>> current since the orbits do not collapse with time.  No power is radiated
>> by an electron orbital and hence no work is required to keep it in the
>> proper location.
>>
>
> Another way to come at the question I just raised is this -- even though
> the electrons may be superconducting in their orbits, is there something
> the force of a weight that is held up might do to gradually decohere the
> magnetic domains?
>
> Eric
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:

I consider electrons in orbits as being equivalent to a superconductor
> current since the orbits do not collapse with time.  No power is radiated
> by an electron orbital and hence no work is required to keep it in the
> proper location.
>

Another way to come at the question I just raised is this -- even though
the electrons may be superconducting in their orbits, is there something
the force of a weight that is held up might do to gradually decohere the
magnetic domains?

Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

On Sat, Nov 14, 2015 at 12:25 AM, David Roberson  wrote:

The loss in the current carrying magnet is due to series resistance and if
> that resistance is eliminated it would not require any additional power
> once the current is set up.
>

I was thinking about that.  But let's make the example extreme.  Suppose
you have a superconducting magnet set up in an industrial grade structure,
and you run a current through sufficient to keep a one-hundred ton block of
iron ten meters off the ground.  Will the solenoid hold up the block
indefinitely, or will some process of loss (other than through resistance)
cause the electrons to gradually radiate away the input energy until the
block eventually falls?  (We can neglect for the moment the energy needed
to cool the current loop in order to keep it superconducting.)

Eric

Re: [Vo]: How many atoms to make condensed matter?

[David Roberson]

Eric, replace the lossy magnet by a superconducting magnet and you get the same 
result without requiring any additional work to be done.  The loss in the 
current carrying magnet is due to series resistance and if that resistance is 
eliminated it would not require any additional power once the current is set up.

I consider electrons in orbits as being equivalent to a superconductor current 
since the orbits do not collapse with time.  No power is radiated by an 
electron orbital and hence no work is required to keep it in the proper 
location.

Dave

 

 

 

-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Fri, Nov 13, 2015 9:21 pm
Subject: Re: [Vo]: How many atoms to make condensed matter?



Some of this thread has gotten to some of the basics relating to magnetism, 
which is a bit of a mystery to me.  There's the dynamic magnetism that arises 
through a moving current.  And there's the static magnetism that is created 
through the formation of magnetic domains in a ferromagnetic material, in which 
the spins of the atoms are aligned in one or another direction.  At a high 
level, these concepts make sense to me.


What I don't fully understand is how conservation of energy applies in the case 
of the system in this photo:


http://i.imgur.com/YzC8KlI.jpg



Here we have a strong permanent magnet and a keyring.  They are configured in 
an arrangement that, without the influence of the permanent magnetism, would be 
unstable against the force of gravity.  But the magnetism of the magnet keeps 
the two components together in the assembly against gravity.


A common explanation for this kind of thing will be something to the effect 
that no work is being done in this system because there is no movement.  But I 
think that oversimplifies the mystery of it.  We can suspect that work is in 
fact being done at the atomic level if in our minds we replace the permanent 
magnet with a magnet formed from a current carrying wire wrapped around a piece 
of metal.  We can set up a magnetic field in this system by keeping current 
flowing through the wire, and we must keep the current flowing in order to 
continue to have the field.  We could do that by turning a crank on a small 
hand generator or burning petroleum to power an electrical generator.  With the 
permanent magnet, one suspects that there must be something comparable going on 
as well.


My question is -- what is it that seems to be adding energy to the system in 
order to keep the permanent magnetic field in place, analogous to the motor 
with the crank or the electrical generator?  What is the fuel in this system 
that does the work?


Eric

Re: [Vo]: How many atoms to make condensed matter?

[Eric Walker]

Some of this thread has gotten to some of the basics relating to magnetism,
which is a bit of a mystery to me.  There's the dynamic magnetism that
arises through a moving current.  And there's the static magnetism that is
created through the formation of magnetic domains in a ferromagnetic
material, in which the spins of the atoms are aligned in one or another
direction.  At a high level, these concepts make sense to me.

What I don't fully understand is how conservation of energy applies in the
case of the system in this photo:

http://i.imgur.com/YzC8KlI.jpg

Here we have a strong permanent magnet and a keyring.  They are configured
in an arrangement that, without the influence of the permanent magnetism,
would be unstable against the force of gravity.  But the magnetism of the
magnet keeps the two components together in the assembly against gravity.

A common explanation for this kind of thing will be something to the effect
that no work is being done in this system because there is no movement.
But I think that oversimplifies the mystery of it.  We can suspect that
work is in fact being done at the atomic level if in our minds we replace
the permanent magnet with a magnet formed from a current carrying wire
wrapped around a piece of metal.  We can set up a magnetic field in this
system by keeping current flowing through the wire, and we must keep the
current flowing in order to continue to have the field.  We could do that
by turning a crank on a small hand generator or burning petroleum to power
an electrical generator.  With the permanent magnet, one suspects that
there must be something comparable going on as well.

My question is -- what is it that seems to be adding energy to the system
in order to keep the permanent magnetic field in place, analogous to the
motor with the crank or the electrical generator?  What is the fuel in this
system that does the work?

Eric

RE: [Vo]: How many atoms to make condensed matter?

[Hoyt A. Stearns Jr.]

As Professor KVK Nehru has elucidated,  you're right, the sun is powered by 
decays of the heavy elements

in the center (  That seems so self evident, it's hard to imagine where 
this other stuff came from,

as if the neutrino experiments ad nauseum wouldn't have invalidated that theory 
long ago,

but you know, if the theory and experiment disagree, the theory wins :-) .

)



Glimpses Into the Structure of the Sun ( KVK,Nehru ) 
<https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB0QFjAAahUKEwj58dy86I3JAhWM7iYKHZ9fDGQ&url=http%3A%2F%2Freciprocalsystem.org%2FPDFa%2FGlimpses%2520into%2520the%2520Structure%2520of%2520the%2520Sun%2520(KVK%2C%2520Nehru).pdf&usg=AFQjCN>



Hoyt Stearns

Scottsdale, Arizona US







From: Jones Beene [mailto:jone...@pacbell.net]
Sent: Friday, November 13, 2015 9:08 AM
To: vortex-l@eskimo.com
Subject: RE: [Vo]: How many atoms to make condensed matter?



From: Bob Higgins

Ø   Can you say what evidence the natural state should exhibit if such a 
sub-nuclear shuffle were as "less difficult" as you describe?  Are there 
natural occurrences that can be looked for that could validate such a 
proposition?

Indeed – such a radical shift would have dramatic, even Universal repercussions 
(turtles all the way down) .

The obvious first place to look is our sun. Do we really understand the solar 
hydrogen fusion cycle?  My opinion is that we could have it partly wrong, 
especially the basic P+P reaction- which is statistically difficult to 
reconcile. Here is the way the mainstream looks at it:

 <https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction> 
https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction

But many observers are struck by the mechanics of the solar fusion cycle being 
absolutely dependent on a rare beta decay in the diproton. Can that really 
happen during the short lifetime of the species? Despite what you may think, 
this critical detail has never been observed, and is merely an educated guess. 
It is a guess which is based mostly on lack of another viable mechanism.

If Holmlid is shown to be correct – then on our sun, we should find that 
nucleon disintegration could be happening instead of, or in addition to, the 
fusion of protons. Of course, there would be some of both, since muons catalyze 
fusion and we know that helium is formed. The proportions could be close to 
even, however.

The precise details are impossible to frame without more information, but if 
Holmlid is replicated, you will see solar cosmologists in a desperate scramble 
to cover their proverbial trailing edges.



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RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

From: Bob Higgins 

*   Can you say what evidence the natural state should exhibit if such a 
sub-nuclear shuffle were as "less difficult" as you describe?  Are there 
natural occurrences that can be looked for that could validate such a 
proposition? 

Indeed – such a radical shift would have dramatic, even Universal repercussions 
(turtles all the way down) .

The obvious first place to look is our sun. Do we really understand the solar 
hydrogen fusion cycle?  My opinion is that we could have it partly wrong, 
especially the basic P+P reaction- which is statistically difficult to 
reconcile. Here is the way the mainstream looks at it:
https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction

But many observers are struck by the mechanics of the solar fusion cycle being 
absolutely dependent on a rare beta decay in the diproton. Can that really 
happen during the short lifetime of the species? Despite what you may think, 
this critical detail has never been observed, and is merely an educated guess. 
It is a guess which is based mostly on lack of another viable mechanism. 

If Holmlid is shown to be correct – then on our sun, we should find that 
nucleon disintegration could be happening instead of, or in addition to, the 
fusion of protons. Of course, there would be some of both, since muons catalyze 
fusion and we know that helium is formed. The proportions could be close to 
even, however. 

The precise details are impossible to frame without more information, but if 
Holmlid is replicated, you will see solar cosmologists in a desperate scramble 
to cover their proverbial trailing edges.

Re: [Vo]: How many atoms to make condensed matter?

[Bob Higgins]

It is an interesting speculation.  Nature is a truly immense experiment,
particularly when considering the number of atoms present.  Immense nuclear
trials are constantly happening all around us.  If this type of sub-nuclear
shuffle were happening with the "less difficulty" that you describe, it
seems that the effects of that would be far more evident in what we see
around us.  Can you say what evidence the natural state should exhibit if
such a sub-nuclear shuffle were as "less difficult" as you describe?  Are
there natural occurrences that can be looked for that could validate such a
proposition?

On Fri, Nov 13, 2015 at 8:07 AM, Jones Beene  wrote:

> Side note for the aestheticists amongst us: Isn’t it likely that a
> hexagonal geometry of pico-snowflakes is a generic form which is reflected
> in structures all the way down to dense hydrogen?  It’s no coincidence
> that iron oxide as catalyst, takes on the classic hexagonal
> nanostructure, and this geometry becomes reflected in a most stable form
> of dense hydrogen (LiH6 and others)... and then reflected the other way,
> in snowflakes or gems and rocks.
>
> *http://www.sciencedirect.com/science/article/pii/S0370269304006860*
> 
>
> Of further interest is a hint at how LENR could be explained… if, that is
> … it is found (via Holmlid) that nucleons are actually less difficult to
> disintegrate than they are to fuse (via UDD symmetry-breaking where the
> baryon octet resists becoming a nucleon sextet):
>
>
> *http://www.treehugger.com/corporate-responsibility/nobel-prize-in-physics-for-beautiful-symmetry-breaking-discoveries.html*
> 
>
> We’ve talked about D'Arcy Thompson here before. He was the Scottish
> biologist and mathematician remembered for “On Growth and Form” (1917)
> possibly the most timeless book of science since “Principia”. Peter
> Medawar, Nobel Laureate, calls it “the finest work of literature in all
> the annals of science that have been recorded in the English tongue.”
>
> Even without the full picture (or the necessary proof) there can be a
> hidden realization which almost jumps out at you, when enough pieces of
> the jigsaw puzzle are fitted together …
>
>

Re: [Vo]: How many atoms to make condensed matter?

[Bob Higgins]

I think you are asking the correct questions.  As I have come to read more
about the RM and think about their behavior, I have come to respect
Winterberg's concept to a greater degree.

The RM snowflakes have a high magnetic moment due to their large flat
orbitals, and apparently the atoms in the RM snowflake align with all of
their spins parallel to create a giant magnetic moment for the flake.  It
made me think of the neodymium iron disk magnets I have.  These magnets
really want to stack in a column.  And, in a column, the magnetic moments
add such that the column has a greater magnetic field running along its
axis.

In a previous comment I posited the possibility that if you had such a
stack of atoms, could the ultra-dense state switch between D(1) and the
D(-1)=D(0) state once the columns of flakes had formed?  This is still a
pretty big leap of faith because the atomic spacing of the D(1) flake is
150pm and the proposed D(-1)=D(0) state is 2.3pm (soley based upon Coulomb
Explosion particle velocity measurements).

Yes, it seems that Winterberg's columns of flakes could be detectable in
rotational spectra - but at a much lower frequency than the already low
frequency (50 MHz) for the individual RM flakes.  This may show as a 1-30
MHz resonance in a shielded cavity containing such stacked flakes, with a
resonance frequency dependent on the number of stacked flakes.  However,
while the individual RM flakes may remain scattered like a gas in a
container (they are very low density), the same may not be true for the
stacks of flakes.  With their high magnetic moment and much higher mass /
surface area, these stacks may want to collect on the walls - particularly
if the walls are a magnetic material (like Ni or iron).  Note that my stack
of disk magnets really wants to stay stuck together and attached to my tool
box.

It is interesting to consider that inside a container with Ni particles
that these stacks of RM flakes may want to attach themselves to the Ni
surface.  I think there is less chance of this being a factor in a Rossi
hotCat/Parkhomov style reactor because the Ni particles are completely
coated with LiH.  However, in Rossi's original eCat, H2 gas was admitted to
Ni particles mixed with some type of Fe catalyst.  Could that catalyst have
been an Fischer-Tropsch Fe2O3 based catalyst that caused formation of RM
flakes that stacked and attached to the surface of his Ni powder particles
like my disk magnet stack on my toolbox?

On Fri, Nov 13, 2015 at 5:30 AM, Stephen Cooke 
wrote:

> Its very hard to see how a single flake can transform between a planar
> atomic crystal state and ultra dense linear paired vortex. But perhaps
> there is a mechanism based on energetic and state conservation effects.
>
> Assuming the effect is more classical and simple however could the switch
> between planar and UDD form be explained by first having a stack of flakes
> in the form of a nano wire?:
>
> Do we know how the Winterberg stack of Rydberg matter flakes forms. Does
> he have a theory for this? Is it just a consequence of the planar nature of
> the Rydberg matter it self or is there a kind of dipole magnetic effect
> between the flakes that can cause the flakes to align and stack in this way
> to form a Rydberg nanowire?
>
> I know I'm being very speculative here, but I wonder if a stack of Rydberg
> matter flakes (h(1) or (d1)) each made up with of magnetically aligned
> atoms, between the flakes, could under the right stimulation (such as a
> strong magnetic field or SPP) switch to a bunch of columns of Ultra dense
> matter (h(0) or (d0)) with each pair of atoms in the column coming from
> adjacent flakes. For example if each flake had 50 or so atoms could a stack
> of them switch to form 50 or maybe 25 Ultra dense vortexes.
>
> Perhaps this is too speculative I'm sure its possible to come up with any
> number of ideas. I suppose we would first need evidence of the Winterberg
> stack occurring before speculating on these lines.
>
> Would a Winterberg type stack have any observable signature such as
> emission spectra etc?
>
>

RE: [Vo]: How many atoms to make condensed matter?

[Jones Beene]

Side note for the aestheticists amongst us: Isn't it likely that a hexagonal
geometry of pico-snowflakes is a generic form which is reflected in
structures all the way down to dense hydrogen?  It's no coincidence that
iron oxide as catalyst, takes on the classic hexagonal nanostructure, and
this geometry becomes reflected in a most stable form of dense hydrogen
(LiH6 and others)... and then reflected the other way, in snowflakes or gems
and rocks. 
http://www.sciencedirect.com/science/article/pii/S0370269304006860

Of further interest is a hint at how LENR could be explained. if, that is .
it is found (via Holmlid) that nucleons are actually less difficult to
disintegrate than they are to fuse (via UDD symmetry-breaking where the
baryon octet resists becoming a nucleon sextet):
http://www.treehugger.com/corporate-responsibility/nobel-prize-in-physics-fo
r-beautiful-symmetry-breaking-discoveries.html

We've talked about D'Arcy Thompson here before. He was the Scottish
biologist and mathematician remembered for "On Growth and Form" (1917)
possibly the most timeless book of science since "Principia". Peter Medawar,
Nobel Laureate, calls it "the finest work of literature in all the annals of
science that have been recorded in the English tongue."

Even without the full picture (or the necessary proof) there can be a hidden
realization which almost jumps out at you, when enough pieces of the jigsaw
puzzle are fitted together .