Repeated many times in previous posts and except in part here as follows: I have referenced papers here to show how the confinement of electrons actually SPPs on the surface of gold nanoparticles: a nanoplasmonic mechanism can change the half-life of U232 from 69 years to 6 microseconds. It also causes thorium to fission.
See references: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=1&cad=rja&sqi=2&ved=0CC4QFjAA&url=http%3A%2F%2Farxiv.org%2Fpdf%2F1112.6276&ei=nI6UUeG1Fq-N0QGypIAg&usg=AFQjCNFB59F1wkDv-NzeYg5TpnyZV1kpKQ&sig2=fhdWJ_enNKlLA4HboFBTUA&bvm=bv.46471029,d.dmQ Nothing happens when there is only a laser used with NO nanoparticles, Using this nanoparticle method, I wonder if increased radioactive decay could be detected if simply caused initiated by a bright light source or a milliwatt laser pointer when that light energy is cataluzed to magnetic energy. On Tue, Aug 19, 2014 at 11:52 PM, Eric Walker <[email protected]> wrote: > On Tue, Aug 19, 2014 at 12:06 PM, Axil Axil <[email protected]> wrote: > > >> http://egooutpeters.blogspot.ro/2014/08/fundamental-causation-mechanisms-of-lenr.html >> >> What is the issues with this line of thinking as a source of muons? >> > > I am out of my element in this topic, but I will offer some feedback > nonetheless. First, I'm infinitely skeptical that any kind of fusion will > occur with virtual mesons, some of which decay to muons with "mostly" > virtual energy. For anything interesting to happen, I'm assuming you will > need real mesons and real muons. > > I understand that mesons can lead to nuclear reactions on their own. But > for the sake of thinking things through, we can ask how many muons would be > needed for 1 Watt power production (if only muons were catalyzing nuclear > reactions). Consider that a typical nickel proton capture reaction will > yield ~ 5 MeV. That means 1 Joule * s^-1 = 6.24e12 MeV * s^-1 = 1.25e12 > proton captures * s^-1. Using your number, a muon can catalyze 150 > reactions. Assuming this is the right order of magnitude not only for d+t > muon catalyzed fusion but also for proton capture in nickel, I think over > time that would average out to around 1.25e12 captures * s^-1 / (150 > captures * muon^-1) = 8.32e9 muons per second which would need to be > produced by the magnetic field. The muons will come about as a result of > pion decays, for which we will need 8.32e9 negative pions per second. > > The energy needed to produce a negative pion is ~ 140 MeV. Your > challenge, then, would seem to be to work out how strong a magnetic field > is needed to generate 8.32e9 pions per second along the Boltzmann tail > (assuming a Boltzmann distribution). Even if the energy needed for the > pion production is found in the long tail, I'm guessing the average energy > of the distribution will still be considerable at this rate of production. > I'm also skeptical that human beings have ever even created a magnetic > field that is strong enough to simply will negative pions from out of the > vasty deep. > > (If anyone spots a mistake in any of these calculations, please call it > out.) > > Note that a negative muon reacts with a proton to create a neutral pion > and a neutron. Note also that a proton capture in nickel is likely to > cause short-lived radioisotopes and energetic states in the daughter nuclei > which will need to decay somehow. This is likely to happen through beta > and beta plus decay, and there's likely to be annihilation photons. So if > this is what is going on it would seem to be inconsistent with your > assumption early in the article about radioactive byproducts: "The fact > that no radioactive isotopes are found in the ash of the cold fusion > reaction is unequivocal proof ...". > > Eric > >
