In reply to Horace Heffner's message of Tue, 12 Apr 2011 21:05:45 -0800: Hi, [snip] > The feasibility of such states indicates that >when QM computation means are available, they will likely show such >states to be Rydberg like, with quantum fuzziness unimportant to the >state. >
Gee Horace, this sounds a lot like my model. :) (See http://rvanspaa.freehostia.com/theory-paper.html , in particular the section on HUP). > >ENERGY TRAPPING > >Assume the kinetic energy of a specific electron in the deflated >state is momentarily around 1 MeV, thus the potential energy in that >case is about -1 MeV. Upon fusion with Ni, the kinetic energy in that >case remains initially at 1 MeV, but, due to the suddenly present 28 >extra Ni nucleus charges, the potential energy is reduced to -29 Mev, >and the net potential plus kinetic energy is reduced by 28 MeV. The >potential energy gained by the proton approach to the nucleus is >actually present, but for exogenous thermodynamic purposes lost due >to the interaction of the strong force. This loss of potential >energy does not prevent electron capture of the now energetically >trapped electron, if capture occurs very fast, because that electron >initially has the kinetic energy necessary. It isn't clear why kinetic energy is "necessary". Is it necessary that the energy be kinetic in nature, or just that a specific quantity of energy be available, and if the latter then how much? I am left wondering (literally) where the energy comes from during normal beta decay for the electron to escape the nucleus. The only conclusion I can come to it that conversion of a neutron to a proton results in 28 MeV more, of nuclear binding energy, than we usually calculate as the beta decay energy, which additional 28 MeV is then carried away by the electron, and mostly lost during it's uphill battle to escape the Coulomb attraction of the nucleus. This in turn would imply that adding a "fresh" proton to the nucleus adds an additional 28 MeV of binding energy, and since the proton didn't need to overcome the Coulomb barrier on the way in, that additional 28 MeV is available to the electron to escape. (Sorry, I just don't buy the ZPE extraction part of your theory). > > >NUCLEAR HEAT > >I think it is also true that nuclear heat may prolong the ability of >the electron to both radiate and be captured. The nuclear heat to >which I refer is provided by zero point energy, in a manner as >described here: > >http://mtaonline.net/~hheffner/NuclearZPEtapping.pdf > >in which I estimate the nuclear temperature for Ni to be 1.02 MeV. >The energetically trapped electron thereby provides a method of >capturing zero point energy. > > >UNCERTAINTY PRESSURE > >The electron only expands its orbital outside the nucleus if a weak >reaction does not quickly follow the strong reaction. This orbital >expansion is driven by zero point energy, uncertainty energy. The >proximity of the electron to the deflated state hydrogen nucleus, and >its kinetic energy, prior to tunneling into a heavy nucleus, are for >practical purposes random variables. However, I think you can impose reasonable limits on the distance between proton and electron in the deflated state. If the distance is too great, then the electron won't shield the proton adequately, and tunneling won't happen in a reasonable time (i.e. a short enough time to be useful for LENR). You can probably use the ordinary QM equation of the H atom to determine the percentage of time that the electron is within an adequately small radius (though this is a bit deceptive as tunneling is possible at all distances, so the net probability requires integration of the probabilities at each distance). >The resulting associated values >post tunneling are thus also random variables. The energy balance for >individual LENR reactions are therefore also random variables. >Energy does not appear to be conserved, because vacuum energy >transactions are involved. Time of electron near the composite fused >nucleus is a random variable, and one which, along with the other >random variables, affects the branching ratios. > >It seems clear that (1) the electron which catalyzes the hydrogen >entry to the nucleus is necessarily trapped there in many cases, >because there is not enough energy from the strong force reaction to >expel it, (2) some heavy LENR reactions clearly do not involve weak >reactions, and (3) the trapped electron can not remain in the >nucleus, otherwise it would be indistinguishable chemically from the >product of a follow-on weak reaction, and the mass of the nucleus >would remain about 0.8 MeV too high. The mass wouldn't be 0.8 MeV too high, because that mass was lost to the creation of the kinetic energy in the first place. Note your own words: "The deflated state is a degenerate state of the hydrogen within its molecular or lattice environment, and has the near ground state binding energy of that state." ...IOW there is no extra 0.8 MeV. (The combined *rest* mass of the electron and proton in the deflated state has to be 0.8 MeV less than that of an H atom, because 0.8 MeV of rest mass has been converted into kinetic energy of the electron, leaving the total relativistic mass constant and equal to the rest mass of the Hydrogen atom - which BTW is still 782 keV shy of the energy required to create a free neutron - which energy deficit is precisely why free neutrons decay into protons and electrons). >Experimental results thus >indicate the trapped electron has an exit. The energy for that exit >clearly can be provided by zero point energy, "uncertainty pressure", >if you will, expanding the electron wavefunction, and its de Broglie >wavelength. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
