Something must produce those electrons and that something (Alpha. beta} produces EMF energy at a well defined gamma level.
Bright mode release of "*photons*" from SPPs when they decay...before an SPP BEC becomes active. On Fri, Mar 11, 2016 at 5:05 PM, Bob Cook <frobertc...@hotmail.com> wrote: > 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 >>> >> >
