On Jan 25, 2013, at 1:31 PM, Axil Axil wrote:
Quantum mechanics lives in the realm of the wave. The electron will
exert it influence on the positive charge nucleus in bits and pieces.
Alex, you are using the wave model and I'm using the particle model.
Both are accepted by science and are useful. However, it is best to
stick to one or the other in a discussion. Otherwise, the discussion
gets too confusing to be useful.
Take a look at this to give your imagination a brake:
http://en.wikipedia.org/wiki/Thomas%E2%80%93Fermi_screening
The Thomas-Fermi formula is a more general potential than the
Coulomb's law.
Yes, screening occurs. The question is, "Is this process alone
sufficient to create LENR at over 10^11 times/sec and how does it
allow the resulting energy be dissipated? Please answer this question.
For the nonlinear Thomas-Fermi formula, solving these simultaneously
can be difficult, and usually there is no analytical solution.
However, the linearized formula has a simple solution:
R= (Q/r)((e)exp(-kr))
With k=0 (no screening), this becomes the familiar Coulomb's law.
The infuence of about 2000 electrons near the site of fusion will
lower the coulomb barrier.
No material has 2000 electrons at any nucleus where they must be
located to lower the barrier.
Ed
On Fri, Jan 25, 2013 at 3:01 PM, David Roberson <[email protected]>
wrote:
That is an interesting complication Axil. There is no doubt that
the electrons can act as a screen of the electric field to an
extent. Once, I tried to get a handle upon the magnitude of this
effect from a simple mental model point of view and a few things
seemed to show up. The COE and COM like to make it difficult to
visualize. I placed an electron between two protons and realized
that as long as the electron was in the middle, there was no Coulomb
barrier to counter since the negative charge exerted a slightly
larger pull than the opposite positive charge repelled as the
combination gets smaller.
This model leads to an interesting idea. If the electron could be
judiciously placed precisely between the protons, there would be no
net force acting upon it. If we then allow the protons to slowly
come together, there would be no net energy imparted upon the
electron as the system shrinks. Each proton would actually be
drawn towards the other one and a small amount of energy would be
imparted upon each. This is due to the fact that the electron
charge is closer to the proton charge than is the other positive
repelling charge.
This process could be continued until something gives. A net amount
of energy is given to the protons as they head towards each other.
The electron is merely kept in the center without expending any
energy.
Now, if the electron squirts out of the line at right angles to the
axis between the protons, then it must be given energy equal to the
amount of Coulomb energy that it helped overcome as the protons came
towards each other. This would be expected if the electron were to
escape the vicinity. The protons would then possess the same amount
of energy that they would have obtained had they not had the
electron to help.
If an electron could be coaxed into this behavior and remain between
the proton pair until the group merges, then fusion would be
common. Since this is not true, one must assume that the electron
diverts at some point. Perhaps a gamma ray comes along to set it
free, but more likely, quantum mechanics intervenes and the electron
begins some form of orbital motion around one or both protons.
Unless the orbit that it settles within allows for the release of
extremely high energy, then the protons are not close enough to
fuse. I suspect that a process of this general nature might lower
the net Coulomb barrier to a degree, but I have no idea how much.
I began to think of a multiple electron case, but grew weary as my
mind wasted away.
Dave
-----Original Message-----
From: Axil Axil <[email protected]>
To: vortex-l <[email protected]>
Sent: Fri, Jan 25, 2013 2:21 pm
Subject: Re: [Vo]:Chemonuclear Transitions
For one, it is not possible for an alpha with that total energy to
be released.
I would like to introduce a complicating factor: electron screening..
Both the cross section of alpha decay and nuclear fusion can be
significantly reduced by electron screening.
In fact I believe that the helium 4 seen in cold fusion experiments
are many times derived from enhanced alpha emissions from high Z
elements rather than fusion of hydrogen.
In the presence of an electron cloud, the consideration of the
coulomb barrier potential must be replaced by the Tomas Fermi
potential to account for electron screening.
Furthermore In astrophysics, cross sections of low energy fusion
events can increase by a factor of one million based on the extent
of electron screening around the fusion site. In fact, it is
impossible to experimentally produce correct stellar fusion reaction
cross sections because both theory and experiment is not able to
explain astrophysical fusion based observations due to the electron
screening problem.
Astrophysics uses the Trojan horse approximation to get around this
electron screening conundrum.
Cheers: Axil
On Fri, Jan 25, 2013 at 1:17 PM, David Roberson <[email protected]>
wrote:
Sometimes the emails do get crossed up with the number of
responses. In this particular case I think that my input helped to
clarify the problem to many others who may be following this
discussion. My choice of observation locations proves that there
are two bodies or body equivalents that must exit the reaction. Now
it is plain for all to see that it is not possible for an alpha
particle to be the only result since I have demonstrated that the
conservation of momentum would be violated it this were to happen.
Before my mental example, it was just a statement that was difficult
to defend. Now we can more readily understand the type of reaction
that must take place in this form of fusion. For one, it is not
possible for an alpha with that total energy to be released. If we
could get a measure of the energy of the alphas that actually are
emitted, then that information can be directly used to calculate the
transferred momentum and energy which is received by the matrix.
Now, I have shown that some reactionary force is required through
which the energy and momentum is transferred to the system. This is
an important observation in my opinion.
It is good that the members of vortex-l can discuss issues of this
nature since much is not known about the reactions that take place.
Sometimes a small spark of incite at the correct moment will lead to
added knowledge. Perhaps others now will realize that what I have
written here is educational. The next time, they might use my ideal
observation location or something of a similar nature to understand
other physics problems. Had I written a paper, it is likely that I
would have overlooked this particular tidbit of knowledge and left
out a major issue that should have been considered.
So, I suggest that we continue to engage in similar discussions
within vortex and enlarge our knowledge base since no one person is
required to be the holder of all that is important. Knowledge is
always advancing as more minds are engaged.
I vote for open discussion within vortex. And, my post was not a
waste of anybodies time. Proof of this assertion will be from this
point forth since most of those engaged in the current discussion
will now understand the issue of energy and momentum requirements.
Dave
-----Original Message-----
From: Edmund Storms <[email protected]>
To: vortex-l <[email protected]>
Cc: Edmund Storms <[email protected]>
Sent: Fri, Jan 25, 2013 12:12 pm
Subject: Re: [Vo]:Chemonuclear Transitions
The problem with such exchanges is that the messages to different
people cross so that I have to explain the same thing several times,
which is a waste of time. That is why I write papers so that
everyone can study the same explanation.
On Jan 25, 2013, at 9:51 AM, David Roberson wrote:
Ed, I am confused by your statement that cold fusion is a 2-body to
1 body reaction. I see two reaction components unless I am missing
something. One is the alpha particle and the other appears in the
form of mass released as energy into the surrounding structure.
The energy release must result from emission of something. Normally
in hot fusion, the release results from emission of a strong gamma
when He4 forms. This gamma is not present when He4 forms during cold
fusion. Why not? The mechanism of energy transfer is obviously not
conventional, yet it must be consistent with the law of conservation
of momentum. I try to solve this problem in my theory. Most people
ignore the issue.
Ed
Every observer must see that the laws of physics apply to what he
sees. My favorite point is to be located precisely between the two
protons as they head toward each other with exactly the same
energy. In this location an observer sees that a finite amount of
kinetic energy is measured for the two particles and that there is
exactly zero momentum for the equal velocity pair. When they
collide together, there is no motion required for the resulting
alpha particle until it releases the excess energy. When that
energy is finally emitted in some form, then a reaction force would
result in relative motion of the alpha particle. In this manner,
both conservation of energy as well as conservation of momentum is
shown.
In my experience, when these laws are seen by any one observer,
then they are true for all of the others. Do you see a hole in
this argument? How are the laws true for others but not for the
one ideally located?
Dave
-----Original Message-----
From: Edmund Storms <[email protected]>
To: vortex-l <[email protected]>
Cc: Edmund Storms <[email protected]>
Sent: Fri, Jan 25, 2013 10:38 am
Subject: Re: [Vo]:Chemonuclear Transitions
The human mind is able to imagine endless possibilities. In order
to make any progress, a triage must be done by eliminating the
ideas that are so improbable or so illogical that they have very
little chance of being correct. That is what I'm attempting to do.
In any case, several basic rules MUST be considered. Hot fusion is
a conventional 2 body-2 body reaction as is required to carry away
the energy and momentum. Cold fusion is a 2-body to 1 body reaction
that violates this condition. That violation MUST be acknowledged
and explained.
People are not free to imaginary any thing. Certain rules are known
to apply. These rules are so basic that they MUST not be ignored.
Ed Storms
On Jan 25, 2013, at 8:22 AM, Daniel Rocha wrote:
d+d=n+He3 and d+d=t+p
What about d+d+...+d=? We don't know. This is what many many
particle models ends up being. Theyare hot fusion. The only
difference it is that there are many, more than 2>, incoming
nuclei to fuse. You cannot do that in experiments using colliders,
it is too unlikely. So, you cannot say that cold fusion is any
different than hot fusion that easily.
2013/1/25 Edmund Storms <[email protected]>
Yes, people try to explain LENR using the behavior described in
the paper.
--
Daniel Rocha - RJ
[email protected]
