I Classical description of fusion

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    Classical Fusion
Aidan Davis
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Fusion is, in most cases (stars, etc.), considered probabilistic. The Gamow-Sommerfeld factor is used to calculate the probability that two colliding nuclei will undergo fusion, considering the fact that the particles have a chance of fusing by quantum tunneling. However, one can calculate an energy (and corresponding speed) that a particle must have to overcome the coloumb barrier and approach within a set distance of the target nuclei, using only classical mechanics. This energy is 1.44 MeV•Z(1)•Z(2)/D, where D is the set distance in fermis, aka 10^-15 meters. If a particle is incoming onto a target nucleus enough energy to come close enough to fuse, even by classical mechanics, is fusion guaranteed or is there still a probabilistic nature to it?
 
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You cannot guarantee fusion. In addition, often this energy is sufficient to get various other reactions.
 
mfb said:
You cannot guarantee fusion. In addition, often this energy is sufficient to get various other reactions.
Ah, pair production of electrons and positrons occurs at these high energies. So fusion would be optimized best by using the Gamow factor https://en.m.wikipedia.org/wiki/Gamow_factor (Ok, wiki isn't the best source but it gives the equations straightforwardly. Here's a derivation of sorts: http://www.astro.princeton.edu/~gk/A403/fusion.pdf)
to calculate the probability of fusion f fusion at a certain energy and then treating that as an optimization problem to get the lowest energy per fusion. I worked it this way, and came to the conclusion that if optimized in this way, D-D fusion, for example, peaks at 246.6 KeV, and all combinations peak at a probability of 13.535% (e^-2). The peak energy is proportional to the product of the charges of the two nuclei and the square root of their reduced mass.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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