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
