[Vo]:Re: Bremsstrahlung experimental note

[Bob Cook]

Axil--

Bremsstrahlung radiation is due to inelastic scattering of electrons as they 
pass through matter.  There are no resonances.  The radiations occurs as a 
result of an electron changing direction as a result of the electric field 
it is passing through.  This change in direction (acceleration) saps energy 
from the kinetic energy of the free electron and distributes that energy as 
electromagnetic radiation equivalent to the loss of kinetic energy of the 
electron.   The spectrum is random photons because the distance and charge 
of particles being encountered by an energetic electron is random.  Thus the 
forces on the electron, whether due to other lattice electrons or positive 
charges in the lattice are random in magnitude.

Landau distributions of the energy of photons do not apply to free electrons 
unless they are at relativistic velocities and have an effective mass like a 
proton, pion, alpha or other heavy particle.

What do you consider is the likely mechanism producing the  "Landau 
distribution" you suggest?  Specifically, what particles are involved in the 
generation of the spectrum?

Bob Cook

-----Original Message----- 
From: Axil Axil
Sent: Friday, March 11, 2016 10:19 AM
To: vortex-l
Subject: Re: [Vo]: Bremsstrahlung experimental note

The seconds long MFMP X-ray burst is smooth and demonstrates no
resonance energy peaks caused by the interaction of electrons with
matter. The MFMP burst is strictly a release of photons in a random
energy distribution.

A Landau distribution is what we are seeing in the MFMP radiation
plot. It is the release of energy by particles based on a random
release process. This is seen when a particle gives up its kinetic
energy to a thin film as the particles interact randomly with the
matter in the thin film.

If SPPs are releasing their energy based on a random timeframe and/or
based on a random accumulation amount, a Landau distribution of energy
release will be seen.

You might see a Landau distribution if there is a random mixing of
both low energy photons (infrared) and high energy photons (gamma's
from the nucleus);

Such mixing is produced by Fano resonance, where an SPPs are being fed
by both infrared photon pumping and nuclear based gamma photon
absorption.



On Fri, Mar 11, 2016 at 1:05 PM, Axil Axil <janap...@gmail.com> wrote:
Electrons may have nothing to do with the x-ray radiation.

The radiation could be produced by photon based quasiparticles.

The LENR reaction might start with Surface Plasmon Polaritons
initiated nuclear reactions and then after thermalization, the decay
of those SPPs. When the SPPs decay, they release their energy content
as photons of varng energies,

After a second or two, a Bose condensate of these SPPs form and the
energy of the photons are released as hawking radiation which is
thermal.

The radiation seen only lasts for a second.

In LENR we get either high energy radiation (x-rays) or heat; not
both. This is based on the temperature of the reactor. A cold reactor
produces X-Rays because of weak SPP pumping..

The SPP absorbs nuclear binding energy and stores it in a whispering
gallery wave (WGW) in a dark mode. The energy is stored inside the WGW
until the WGW goes to a bright mode when the SPP decays. This
conversion from dark mode to bright mode happens in a random
distribution.

When the temperature is raised over a thermal conversion limit, a BEC
is formed where the stored nuclear binding energy is released from the
SPP BEC as hawking radiation which is thermal.

On Fri, Mar 11, 2016 at 12:34 PM, Bob Cook <frobertc...@hotmail.com> 
wrote:
The effectiveness of the SS can at stopping any high energy electrons 
that
cause Bremsstrahlung would depend upon the thickness of the can (or 
alumina)
and the energy of the incident electrons.  I think the loss of energy per
scattering event is proportional to Z ^2 for the nucleus that is doing 
the
scattering.  Al at Z=13 and with  Fe at Z=26 the intensity of the
Bremsstrahlung signal would be about a factor of 4 different.  The mean
length of the path of an electron is a good parameter to know for any 
given
substance (basically its density) vs the incident energy of the electron.
Shielding engineering curves provide this information I believe.   Iron
being significantly more dense than Al2O3 would be much better at slowing
electrons and thus producing Bremsstrahlung IMHO.

At high electron energies the change of direction of the electron going
through SS can would be less than for a low energy electron.  For slow
electrons scattering can significantly change the direction of an 
incident
electron such that all Bremsstrahlung would be emitted from the material
that stopped the electron.

I think with a SS can present in the system vs no can and only Alumina
stopping the electrons, one would expect to see a more intense signal at
high energy  compared to the spectrum from the Alumina reactor chamber. 
The
absorption of the EM Bremsstrahlung by the respective media would also 
have
to be considered.  Neither Alumina nor SS may transmit some of the
Bremsstrahlung spectrum very well.  Thus the effective shielding of the 
EM
radiation considering a distributed source would have to be evaluated for
the resulting high energy EM and the signal intensity corrected 
accordingly.
The cut off at the high energy spectrum will be a useful value to know to
understand the maximum energy of the electron source.  This may provide
information about the reaction producing the electrons.   The change of 
the
intensity of the Bremsstrahlung signal as a function of the magnetic 
field
would also provide information as to whether or not the lattice 
orientation
of the nano fuel was important.   One might expect that the electrons 
being
produced by the respective LENR reaction would produced in some preferred
direction.

Bob Cook
From: Bob Higgins
Sent: Friday, March 11, 2016 6:09 AM
To: vortex-l@eskimo.com
Subject: [Vo]: Bremsstrahlung experimental note

I don't know if other Vorts thought of this already... but I had a minor
epiphany regarding the radiation that MFMP measured in GS5.2.  We 
identified
this radiation tentatively as bremsstrahlung.  This has certain
implications.  Bremsstrahlung requires that the high speed electrons 
impact
on a high atomic mass element so as to be accelerated/decelerated quickly 
to
produce the radiation.  It could be that the stainless steel can that
contained the fuel was an important component in seeing the 
bremsstrahlung.
Without the can, there would still be the Ni for the electrons to hit, 
but
the Ni is covered with light atomic mass Li.  If the electrons were to
strike alumina (no fuel can present), I don't think there would be nearly 
as
much bremsstrahlung because alumina is comprised of light elements.

Thus, the stainless steel can for the fuel may be an important component 
for
seeing the bremsstrahlung.

Bob Higgins