RE: [Vo]: Bremsstrahlung experimental note

[Russ George]

It’s right there for all to see hidden behind the Cheshire Cat’s grin!

 

From: David Roberson [mailto:dlrober...@aol.com] 
Sent: Monday, March 14, 2016 11:07 AM
To: vortex-l@eskimo.com
Subject: Re: [Vo]: Bremsstrahlung experimental note

 

Can you point out the location of the mouse in Rossi's patent?

Dave

 

 

 

-Original Message-
From: Axil Axil <janap...@gmail.com <mailto:janap...@gmail.com> >
To: vortex-l <vortex-l@eskimo.com <mailto:vortex-l@eskimo.com> >
Sent: Sun, Mar 13, 2016 3:34 pm
Subject: Re: [Vo]: Bremsstrahlung experimental note

Something is getting out of the LENR reactor. The mouse is stimulating the cat 
in Rossi's reactor clustering scheme. The some emission of the mouse is 
producing the LENR reaction inside the Cat type reactor.  

 

That emission only exits the Mouse when the power to the heater coils of the 
Mouse is turned off so that the emission is a some sort of charged particle.

 

On Sun, Mar 13, 2016 at 3:26 PM, Bob Higgins <rj.bob.higg...@gmail.com 
<mailto:rj.bob.higg...@gmail.com> > wrote:

Muons with less than about 4MeV are not going to escape the reactor.  Cosmic 
muons are average 2GeV.  No magnetic field that I could generate is going to 
significantly deflect either of these.

 

On Sun, Mar 13, 2016 at 11:39 AM, Axil Axil <janap...@gmail.com 
<mailto:janap...@gmail.com> > wrote:

@Bob 

 

Use a magnetic shield to divert muons and other charged particles.

 I describe it here

 

https://www.lenr-forum.com/forum/index.php/Thread/2862-A-Simple-LENR-Magnetic-Radiation-Shield/?postID=15183#post15183

 



 

Re: [Vo]: Bremsstrahlung experimental note

[David Roberson]

Can you point out the location of the mouse in Rossi's patent?

Dave

 

 

 

-Original Message-
From: Axil Axil <janap...@gmail.com>
To: vortex-l <vortex-l@eskimo.com>
Sent: Sun, Mar 13, 2016 3:34 pm
Subject: Re: [Vo]: Bremsstrahlung experimental note



Something is getting out of the LENR reactor. The mouse is stimulating the cat 
in Rossi's reactor clustering scheme. The some emission of the mouse is 
producing the LENR reaction inside the Cat type reactor. 


That emission only exits the Mouse when the power to the heater coils of the 
Mouse is turned off so that the emission is a some sort of charged particle.



On Sun, Mar 13, 2016 at 3:26 PM, Bob Higgins <rj.bob.higg...@gmail.com> wrote:

Muons with less than about 4MeV are not going to escape the reactor.  Cosmic 
muons are average 2GeV.  No magnetic field that I could generate is going to 
significantly deflect either of these.



On Sun, Mar 13, 2016 at 11:39 AM, Axil Axil <janap...@gmail.com> wrote:

@Bob


Use a magnetic shield to divert muons and other charged particles.
 I describe it here



https://www.lenr-forum.com/forum/index.php/Thread/2862-A-Simple-LENR-Magnetic-Radiation-Shield/?postID=15183#post15183

Re: [Vo]: Bremsstrahlung experimental note

[Axil Axil]

Something is getting out of the LENR reactor. The mouse is stimulating the
cat in Rossi's reactor clustering scheme. The some emission of the mouse is
producing the LENR reaction inside the Cat type reactor.

That emission only exits the Mouse when the power to the heater coils of
the Mouse is turned off so that the emission is a some sort of charged
particle.

On Sun, Mar 13, 2016 at 3:26 PM, Bob Higgins 
wrote:

> Muons with less than about 4MeV are not going to escape the reactor.
> Cosmic muons are average 2GeV.  No magnetic field that I could generate is
> going to significantly deflect either of these.
>
> On Sun, Mar 13, 2016 at 11:39 AM, Axil Axil  wrote:
>
>> @Bob
>>
>> Use a magnetic shield to divert muons and other charged particles.
>>  I describe it here
>>
>>
>> https://www.lenr-forum.com/forum/index.php/Thread/2862-A-Simple-LENR-Magnetic-Radiation-Shield/?postID=15183#post15183
>>
>> [image: Inline image 1]
>>
>

Re: [Vo]: Bremsstrahlung experimental note

[Bob Higgins]

Muons with less than about 4MeV are not going to escape the reactor.
Cosmic muons are average 2GeV.  No magnetic field that I could generate is
going to significantly deflect either of these.

On Sun, Mar 13, 2016 at 11:39 AM, Axil Axil  wrote:

> @Bob
>
> Use a magnetic shield to divert muons and other charged particles.
>  I describe it here
>
>
> https://www.lenr-forum.com/forum/index.php/Thread/2862-A-Simple-LENR-Magnetic-Radiation-Shield/?postID=15183#post15183
>
> [image: Inline image 1]
>

Re: [Vo]: Bremsstrahlung experimental note

[Axil Axil]

@Bob

Use a magnetic shield to divert muons and other charged particles.
 I describe it here

https://www.lenr-forum.com/forum/index.php/Thread/2862-A-Simple-LENR-Magnetic-Radiation-Shield/?postID=15183#post15183

[image: Inline image 1]


On Sun, Mar 13, 2016 at 10:56 AM, Bob Higgins 
wrote:

> Muonic decay in the reactor is an interesting prospect that I would like
> to read more about.  However, I don't think the muons, electrons, or
> protons are going to escape the reactor in any large number due to the
> mass/cm^2 they would have to traverse to get out.  Muons are no more likely
> to penetrate the reactor walls than electrons or protons of the same
> energy.  The reason that muons are an issue with the lead in the
> scintillator shield is that the cosmogenic muons have a typical energy of
> 2GeV - probably 1000x that of what could be created inside the reactor.
> The penetration is directly related to the energy of the muon.
>
> Certainly some in-the-cave vs. out-of-the-cave measurements are in order,
> but can't easily be done while the experiment is running.
>
> On Fri, Mar 11, 2016 at 7:45 AM, Jones Beene  wrote:
>
>> Bob,
>>
>>
>>
>> There is simply too little nickel. If looking for bremsstrahlung, and in
>> the absence of gamma - a possible source of high speed electrons would be
>> muon decay.
>>
>>
>>
>> At least this would be true in a situation like the glow-tube, where
>> dense hydrogen would be expected to form.
>>
>>
>>
>> If the counts are higher inside the lead cave, compared to outside
>> (bare), it is very likely that the source is muonic from the reactor, not
>> cosmic - and the target is lead.
>>
>>
>>
>> *From:* Bob Higgins
>>
>>
>>
>> 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
>>
>
>

RE: [Vo]: Bremsstrahlung experimental note

[Jones Beene]

From: Bob Higgins 

 

Ø  Muonic decay in the reactor is an interesting prospect that I would like to 
read more about. However, I don't think the muons, electrons, or protons are 
going to escape the reactor in any large number due to the mass/cm^2 they would 
have to traverse to get out. Muons are no more likely to penetrate the reactor 
walls than electrons or protons of the same energy.  

Holmlid is communicative, speaks English, and answers his email, so it would be 
wise to pose the questions directly to him. I am pretty certain that he 
believes that almost all muons escape the reactor.

As I understand the situation, your logical error above is to assume that the 
muons formed in the reactor are not relativistic but are of comparable energy 
to fusion. 

This is not the interpretation of Holmlid which I have. The muons from UDD 
disintegration are much more energetic than protons from fusion. The reason the 
reactor does not heat up, commensurate with their energy is because almost all 
of them escape. 

The “decay halo” of fast electrons from muon decay could be a 
probability-sphere which is a hundred meters in diameter around the reactor. At 
any place within this zone, the counts are fairly low, but not inverse square – 
and the falsifiability comes from finding the precise zone where counts are 
much higher, which is far removed from the reactor - corresponding to the peak 
of decay activity.

 

 

 

 

 

 

The reason that muons are an issue with the lead in the scintillator shield is 
that the cosmogenic muons have a typical energy of 2GeV - probably 1000x that 
of what could be created inside the reactor.  The penetration is directly 
related to the energy of the muon.

Certainly some in-the-cave vs. out-of-the-cave measurements are in order, but 
can't easily be done while the experiment is running.

 

On Fri, Mar 11, 2016 at 7:45 AM, Jones Beene  wrote:

Bob,

 

There is simply too little nickel. If looking for bremsstrahlung, and in the 
absence of gamma - a possible source of high speed electrons would be muon 
decay. 

 

At least this would be true in a situation like the glow-tube, where dense 
hydrogen would be expected to form.

 

If the counts are higher inside the lead cave, compared to outside (bare), it 
is very likely that the source is muonic from the reactor, not cosmic - and the 
target is lead.

 

From: Bob Higgins 

 

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

 

Re: [Vo]: Bremsstrahlung experimental note

[Bob Higgins]

Muonic decay in the reactor is an interesting prospect that I would like to
read more about.  However, I don't think the muons, electrons, or protons
are going to escape the reactor in any large number due to the mass/cm^2
they would have to traverse to get out.  Muons are no more likely to
penetrate the reactor walls than electrons or protons of the same energy.
The reason that muons are an issue with the lead in the scintillator shield
is that the cosmogenic muons have a typical energy of 2GeV - probably 1000x
that of what could be created inside the reactor.  The penetration is
directly related to the energy of the muon.

Certainly some in-the-cave vs. out-of-the-cave measurements are in order,
but can't easily be done while the experiment is running.

On Fri, Mar 11, 2016 at 7:45 AM, Jones Beene  wrote:

> Bob,
>
>
>
> There is simply too little nickel. If looking for bremsstrahlung, and in
> the absence of gamma - a possible source of high speed electrons would be
> muon decay.
>
>
>
> At least this would be true in a situation like the glow-tube, where dense
> hydrogen would be expected to form.
>
>
>
> If the counts are higher inside the lead cave, compared to outside (bare),
> it is very likely that the source is muonic from the reactor, not cosmic -
> and the target is lead.
>
>
>
> *From:* Bob Higgins
>
>
>
> 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
>

Re: [Vo]: Bremsstrahlung experimental note

[Axil Axil]

Why does the burst last for just a second even when excess heat is
produced after the radiation burst?

On Fri, Mar 11, 2016 at 4:25 PM,   wrote:
> In reply to  Bob Cook's message of Fri, 11 Mar 2016 09:34:55 -0800:
> Hi,
> [snip]
>>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.
>
> Shielding is based primarily on the electrons of an atom being ionized.
> Bremsstrahlung is created when a fast particle interacts with a nucleus.
> Most fast electrons impinging on solid matter will create ionization, i.e. 
> they
> get stopped by other electrons. AFAIK Only about 1% get through to the nucleus
> and create Bremsstrahlung. I think that both nuclear charge and number of 
> nuclei
> per unit volume would be important for Bremsstrahlung production. Mass of a
> nucleus not so much, because even a single proton is already about 2000 times
> more massive than an electron. When it comes to collisions, it makes little
> difference whether the nucleus is light or heavy. In short any nucleus is
> effectively an "immovable object" as far as an electron is concerned.
>
> BTW if MeV level electrons are stopped by Aluminium foil, then the can would
> have to be very thin not to stop them.
>
> Has anyone considered the possibility that some (little) bremsstrahlung might 
> be
> caused by fast protons impacting on heavier nuclei?
>
> Regards,
>
> Robin van Spaandonk
>
> http://rvanspaa.freehostia.com/project.html
>

Re: [Vo]: Bremsstrahlung experimental note

[mixent]

In reply to  Bob Cook's message of Fri, 11 Mar 2016 09:34:55 -0800:
Hi,
[snip]
>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. 

Shielding is based primarily on the electrons of an atom being ionized.
Bremsstrahlung is created when a fast particle interacts with a nucleus.
Most fast electrons impinging on solid matter will create ionization, i.e. they
get stopped by other electrons. AFAIK Only about 1% get through to the nucleus
and create Bremsstrahlung. I think that both nuclear charge and number of nuclei
per unit volume would be important for Bremsstrahlung production. Mass of a
nucleus not so much, because even a single proton is already about 2000 times
more massive than an electron. When it comes to collisions, it makes little
difference whether the nucleus is light or heavy. In short any nucleus is
effectively an "immovable object" as far as an electron is concerned.

BTW if MeV level electrons are stopped by Aluminium foil, then the can would
have to be very thin not to stop them.

Has anyone considered the possibility that some (little) bremsstrahlung might be
caused by fast protons impacting on heavier nuclei?

Regards,

Robin van Spaandonk

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

Re: [Vo]: Bremsstrahlung experimental note

[Axil Axil]

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  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  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 

Re: [Vo]: Bremsstrahlung experimental note

[Axil Axil]

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  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

Re: [Vo]: Bremsstrahlung experimental note

[Bob Cook]

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

RE: [Vo]: Bremsstrahlung experimental note

[Jones Beene]

Bob,

 

There is simply too little nickel. If looking for bremsstrahlung, and in the 
absence of gamma - a possible source of high speed electrons would be muon 
decay. 

 

At least this would be true in a situation like the glow-tube, where dense 
hydrogen would be expected to form.

 

If the counts are higher inside the lead cave, compared to outside (bare), it 
is very likely that the source is muonic from the reactor, not cosmic - and the 
target is lead.

 

From: Bob Higgins 

 

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