An additional reply from Ron, below. Eric
"Fusion doesn't have a rate as a function of distance, it's a coherent tunneling effect in deuterons and it depends on the whole wavefunction shape not on the classical 'distance between nucleons'. "You can only calculate it using a two-body -> resonance -> final product effective (nonrelativisitc) field theory. This requires the two-body -> resonance coupling constant and the resonance-> final product coupling constant, neither of which do we know for each of the ~5 relevant alpha-particule resonances, and also the dependence on an external strong electric field (provided by the nucleus). With good estimates for these coupling constants, you can simulate the process. "Muon catalyzed fusion as an estimate is a good idea, I looked at muon catalyzed fusion, but it's not an easy estimator, because the deuterons are stationary, so the process is different from colliding deuterons, and the muon is far away, and has a small charge, so it doesn't introduce a strong electric field." On Tue, Mar 5, 2013 at 11:32 PM, <[email protected]> wrote: > In reply to Eric Walker's message of Tue, 5 Mar 2013 21:43:21 -0800: > Hi, > [snip] > >A further reply from Ron, below. > > > >Eric > > > > > >"The stopping distance is around 600 atoms according to Spaandonk, and I > >don't dispute the theoretical value, but I don't know if this is accurate, > >it can be larger or smaller, because the calculations are theoretical, and > >this energy regime is comparable to the energy levels involved, so > >experiment is the best way. > > True, note however that Deuterons can only penetrate to near the K shell > while > they still have most of their energy, so the useful range is likely to be > less, > perhaps much less. > > > > >"The reason that the deuterons have unnatural concentration near the > >nucleus is precisely because of the repulsion--- they turn around at a > >distance of 100 fermis from the nucleus. This is because their kinetic > >energy is equal to the potential energy of an electron-hole in a K-shell, > >so that you get stopping at exactly the K-shell radius, give or take a > >factor of 2. > > > >"The turn-around region is where the wavefunction is most concentrated, > >because semi-classical wavefunctions go as 1 over the square root of the > >speed. This enhancement is like a beam-focusing device. > > Thanks, understood. > > > > >"Robin is right that there are two parts to test--- the fusion and the > >Auger transfer. The Auger transfer one knows how to test, but this is > >likely to work, but the details of the banding are still mysterious. The > >other part is the fusion rates. This is confounding to me. > > I have a Mathcad file that makes an estimate of fusion rates based on > separation > distance if Ron is interested. Note however that it assumes a constant > distance, > so will overestimate the rate. However he may be able to adapt it to suit > his > needs. > [snip] > >"Other than that, I'm stuck, because I can't think of any circumstance > >where you would see 3-body events other than in cold fusion. So the actual > >rate needs to be attacked theoretically, and it's annoyingly hard. > > I used muon catalyzed fusion as a base. That should at least provide a > ballpark > estimate, then he can guess whether the D+D->He4 rate would be higher or > lower. > My guess would be that because the limiting factor is essentially the > tunneling > time, the rate would be about the same (once the 2 D's are together, they > are > going to react, one way or another. Only the branching ratio is determined > by > other factors such as conservation of momentum), and when it comes to > energy > production, the actual resulting reaction is not really so important > anyway. > (There is plenty of D to last a very long time. :) > [snip] > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > >
