I wrote:
I have not yet looked closely at Holmlid's results, but I don't write them
> off. I'm keeping a distinction in my mind between his experimental
> observations and his theoretical speculations.
>
I have now had a chance to look more closely at Leif Holmlid's 2013 paper,
"Direct
I suppose a real nuclear physicist will correct me on a lot of my assumptions.
Date: Tue, 20 Oct 2015 04:16:37 -0400
Subject: Re: [Vo]:MMDD Muon Mediated Deuteron Disintegration
From: janap...@gmail.com
To: vortex-l@eskimo.com
I believe that what you imagine is what is happening in one o
On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke
wrote:
If any was produced we would need to balance this against those the energy
> required for pion production.
>
The amount of energy needed to create a free pion is large; the rest mass
for a pion is ~ 135 MeV.
I believe that what you imagine is what is happening in one of the many
cases involving SPP extreme magnetic projections and entanglement,
In case it is interesting I have found a couple of interesting papers on pionic
deuterium.
http://www.slac.stanford.edu/econf/C070910/PDF/290.pdf
-
> From: stephen_coo...@hotmail.com
> To: vortex-l@eskimo.com
> Subject: RE: [Vo]:MMDD Muon Mediated Deuteron Disintegration
> Date: Tue, 20 Oct 2015 16:16:09 +0200
>
>
> 'The amount of energy needed to create a free pion is large; the rest mass
> for a pion is ~ 135 MeV'
&g
On Tue, Oct 20, 2015 at 9:16 AM, Stephen Cooke
wrote:
Very true this is also true for the muon which has a rest mass for a pion
> is ~ 106 MeV.
This is one of the reasons I don't think muons are involved either.
Another reason -- if muons were being generated, you'd
say he could well have another explanation.
From: eric.wal...@gmail.com
Date: Tue, 20 Oct 2015 08:47:08 -0500
Subject: Re: [Vo]:MMDD Muon Mediated Deuteron Disintegration
To: vortex-l@eskimo.com
On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke <stephen_coo...@hotmail.com>
wrote:
with
> sufficient energy to interact with another Deuterium nucleus to form a
> diproton that decays to 2 high energy protons, or it decays to a muon of
> characteristic energy that is later detected.
>
> I suppose a real nuclear physicist will correct me on a lot of my
> assu
Oct 2015 08:47:08 -0500
Subject: Re: [Vo]:MMDD Muon Mediated Deuteron Disintegration
To: vortex-l@eskimo.com
On Tue, Oct 20, 2015 at 2:21 AM, Stephen Cooke <stephen_coo...@hotmail.com>
wrote:
If any was produced we would need to balance this against those the energy
required fo
This is definitely a crazy train of thought on my part but it's got me
wondering: Bearing in mind Holmlids detection of muons and possibility they
come from decay of positive or negative pions along with the fact that they are
generated in ultra dense deuterium. Is It possible that the nuclei
The lifetime of pions is a factor of 100 times shorter than the muon, which is
itself short.
How pions can be detected at all, with so short a lifetime is a question worth
asking.
There is a good possibility that they are inferred – from finding muons, which
can be detected.
I guess there is a chance that they saw X-ray spectra from Pionic Deuterium as
well as inferring from muons of specific energy. It will be interesting to see
what he says on Thursday, I hope someone can ask these kind of questions.
Sent from my iPad
> On 19 okt. 2015, at 18:18, Jones Beene
m. So I hope someone with more knowledge can clarify and knock
some holes in what i just said.
Thanks
Stephen
From: jone...@pacbell.net
To: vortex-l@eskimo.com
Subject: RE: [Vo]:MMDD Muon Mediated Deuteron Disintegration
Date: Mon, 12 Oct 2015 10:15:30 -0700
RE: [Vo]:MMDD Mu
From: Stephen Cooke
* Could there be characteristic photon emission from transitions in
muon shell levels similar to those from electrons and at what frequencies
these occur. Could these be observed experimentally? If characteristic
radiation can be seen from muon energy level transitions
MMPD Muon Mediated Deuteron Disintegration
The work of Leif Holmlid and others has opened up the possibility of
understanding what appears to be a new kind of nuclear reaction - a limited
type of chain reaction which is not fusion nor fission. The result of this
reaction is the complete
Jones,
While this is interesting speculation, I have come to assess probabilities
of of heretofore unknown reactions based inversely on binding energy. If
we look at molecular binding energy, it is less than atomic binding.
Nuclear binding energy is greater than atomic binding. Sub-nucleon
Correction
*
* In this reaction of relatively cold deuterons, gamma emission cannot
proceed, and fusion to deuterium is suppressed in favor of complete
disintegration of protons and neutrons into quarks.
. should read: "gamma emission cannot proceed, and fusion of deuterium to
helium
Bob,
Good analysis. Subnuclear binding energy is significantly higher than nuclear
binding energy, in general. As we know, it takes many terawatts to show
evidence of the Higgs boson. After that, we must bootstrap power into energy to
make this happen.
Fortunately – terawatt pulses are
On Mon, Oct 12, 2015 at 12:15 PM, Jones Beene wrote:
> According to the HRM, the process develops in
> three steps: a photon (in this case, the 24 MeV internalized photon) knocks
> a quark from the nucleon; the struck quark rescatters off a quark from
> another nucleon; then
It takes many TeV to show the Higgs - not TeraWatts. A 1.0 TeV particle
only has 1.6e-7 joules of kinetic energy, but that energy is in a single
particle and can create a TeV event in a collision.
A TW laser will not really help. The individual laser photons are only eV
level, and a TW laser
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