In reply to Jones Beene's message of Fri, 12 May 2006 08:50:50 -0700: Hi, [snip] >> Negative muons orbit at the Bohr radius (BR) * electron mass / >> muon mass. Hence hydrinohydride should try to do the same, > >Not necessarily, Robin. There is some evidence that leptons >experience inertia (and gravity) differently than fermions. Mills >even buys into that one. The point being that this is not an >apples-to-apples comparison as the Bohr radius applies >specifically to leptons.
This is an excellent point. The only thing IMO that determines an electron orbit is the fact that it orbits at radii where it's De Broglie wave is self reinforcing. However for a self-contained and stable entity such as hydrinohydride this isn't necessary at all, so that any radius ought to be possible. OTOH since it is a charged particle orbiting in a magnetic field (K nucleus has a magnetic moment), it should radiate and spiral into the nucleus, or at least as close as it's own radius will allow. > >> Either way, it is larger than the distance at which it should >> orbit. This leaves several possible scenarios:- > >> 1) It sits snug against the nucleus at it's own radius. > >This can probably be ruled out because of the strong force... That depends on the distance, and the charge on the nucleus. Both are parameters in the exponent of "e" which determines the half life, so even small changes in either, can result in huge changes to the half life of the fusion reaction. [snip] >Perhaps the shrinkage is either n=1/16 or 1/17 ... in the range of >the maximum enthalpy. IOW maximum enthalpy instead of maximum Also true. At n=1/16, the radius of the hydrino is either BR/16 = 3300 F (according to Mills), or BR/16^2 = 206 F (according to me). For Potassium, I get a K shell electron radius of either 2785 F (ab initio) or 7583 F (based on X-ray absorption). I must be doing something wrong! :) >shrinkage. Perhaps these are also the only ones which can easily >escape - in the solar corona - that is: the hydrinos which are >favored to be expelled in a solar-genesis situation, over geologic >time. Lets see - at 1/16 the radius of the hydrino is a factor of >32,000 times reduced over the Bohr radius. Not sure where you get 32000 from. >Of course there are >many other problems with this whole scenario - to wit: > >Mills has apparently collected gram-sized supplies of tightly >bound K-Hy- (images on his web site) and we can assume from that >situation that he has proven in so doing that the excess charge is >nullified by the strong covalence (ionic --> covalent bonding), >meaning that 20 total electrons remain in the compound; but can >that be "confused" (in "inertness") with the net 18 electrons of >argon? i.e. instead of 20 of a normal hydride. The scenario we are >getting at would be easier to pull-off starting with chlorine. For >this to work out chemically with potassium, the 20 electrons must >"look like" the 18 of Argon, no? That might be the expected >outcome of an extremely tight bonding scenario... Think of the Hy- as a negative proton. It adds the mass of a proton, but reduces the central charge by 1. Hence K39 becomes Ar40. As far as the electrons are concerned, they think they are orbiting an Argon nucleus, and will arrange themselves accordingly. > >...plus it provides another route to falsifiability - in that in a >normal tank of Argon, where there are three stable isotopes: 36, >38, and 40 and with 40Ar accounting for 99.6% of the total - >THEN - of that 99.6, the "special-K" component should have an >extra mass of 2 electrons, compared with the "real" argon, no? Well no. It should weigh in at about 8.5 MeV more than real Argon, due to the "free" proton. (When protons fuse with other nuclei there is a mass/energy conversion. Since this hasn't taken place in special-K, the excess mass is still present). >Even with a ppm population of this heavier "argon" of special-K, >there should be a "blip" on the mass-spec chart. Yes, unless of course *all* Ar40 is actually special-K, in which case there wouldn't be any difference between the currently measured value of Ar40 and K+H. However there is a difference, 8.5 MeV. From which we may conclude that certainly not all Ar40 is special K (and probably none of it ;). > >We must also assume that the Hy- shrinkage in those samples Mills >has collected is near the minimum (i.e they are much larger in >radius - one or two steps of shrinkage) than the tiny size needed >to "gasify" the atom into "special-K" which would be the Argon >substitute. Yes. > >Bizarro! The real problem in the whole scenario seems to be to get >a molecule to bond so tightly that it "looks like" a monatomic >noble gas, correct? It is no wonder that NO mainstream scientist >would touch this line of reasoning "with a ten-foot pole" Actually there may be another example. Some Deuterium may actually be "Hy- + proton" nucleus with an electron circling this "nucleus" just as though it were a true Deuterium nucleus. This "molecular" nucleus would split at relatively low energies (keV iso MeV), which might give rise to some of the Phillips-Oppenheimer reactions. (Though there isn't any real neutron present, so it would have to be reactions where a proton is absorbed into the other nucleus, or else a proton absorbtion combined with a near instantaneous electron capture reaction enhanced by the close proximity of the shrunken electron.) One might then wonder why the faux deuterium nucleus hadn't already fused into a real deuterium nucleus? This might be explained by the greater nergy release available from fusion with another nucleus compared to deuterium formation (perhaps nuclear reactions tend to happen more readily, the larger the energy involved). IOW the half-life of deuterium formation could be long enough to ensure that there is always a population of faux deuterium. [snip] >However, that little "blip" mentioned above - the putative one on >the mass-spec charts, previously ignored or dismissed as a "relic" >of the instrumentation - yes ! ... that would be enough. But is it >there? The $64000 question! :) > >> BTW for the Auger scenario to play out, it would seem to me that >> one would have to put as much energy into the compound atom to >> dislodge the hydrino hydride as was initially released when it >> entered. > >Here again, this is a logical assumption based on what happens >with leptons. No, it's really just based on conservation of energy. >The assumption might NOT be valid for species whose >charge to mass ratio is massively different. Not to mention, the >Hy- may not become totally dislodged itself - but due to its much >lower mass (compared to the K nucleus). If it is severely jostled >about, under circumstance which go beyond normal acceleration into >"jerk" then we could have the Auger cascade with ZPE stepping-in >later to restore some normalcy to the situation. Assuming there is a ZPE. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
