Thanks Dave, I welcome the opportunity. Please forgive my brief style
and frequent typos. This results from slow typing skill and an
assumption that much of what I might say is already known by the
reader, requiring only a hint to reach the answer. Also, I do not
encourage discussion about detail or arguments about basic ideas. We
all know that a lot is missing in our understanding of Nature, but I
do not have the time to address any of these interesting issues except
LENR. My policy is fight only one war at a time.:-)
Ed
On Jan 25, 2013, at 11:38 AM, David Roberson wrote:
Thanks Ed, I think we are pretty much in agreement at this time. I
tend to view processes from the "other side" which sometimes can
simplify understanding of complex events and that is why I
commented. Perhaps I got a bit too riled at the suggestion that my
post was a total waste of time!
I greatly honor your contributions to and knowledge of this
important field and I look forward to receiving additional guidance
from your inputs to vortex. We all appreciate the opportunity to
converse with you when you join us.
Dave
-----Original Message-----
From: Edmund Storms <[email protected]>
To: vortex-l <[email protected]>
Cc: Edmund Storms <[email protected]>
Sent: Fri, Jan 25, 2013 1:27 pm
Subject: Re: [Vo]:Chemonuclear Transitions
On Jan 25, 2013, at 11:17 AM, David Roberson 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.
I agree
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.
Yes, Dave that is the basic conclusion that results from the law of
conservation of momentum. Thanks for making this clearer.
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.
Your point was not a waste. However, everyone should read every
message before replying.
Ed
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]
