In fusion terminology, what does it imply by taking orientational average of the fusion cross section when one of the nuclei has some static deformation/orientation? How is this average taken?

  • $\begingroup$ Presumably all orientations are equally probable, so the average would just be $\int d\Omega \sigma(\Omega)$. $\endgroup$ Feb 16, 2017 at 5:44
  • $\begingroup$ I meant weighted average over nuclear orientation. $\endgroup$
    – Ana
    Feb 16, 2017 at 6:31
  • $\begingroup$ @ZeroTheHero What is it strong magnetic field? It non-QM analogy, I would think it changes the level distribution of the orientation. $\endgroup$
    – peterh
    Feb 16, 2017 at 6:34

1 Answer 1


It means that you're accounting for the fact that the spins of the nuclei in the fusion interaction aren't (or are, perhaps) oriented in a particular way. For instance, consider the D-T fusion reaction $$ \rm ^2H + {}^3H \to {}^4He + n + 18\,MeV $$ The deuteron $\rm^2H$ (or d) has nuclear spin $\hbar$, while the triton $\rm^3H$ (or t) has nuclear spin $\frac12\hbar$. But the final state, $\rm^4He+n$, has spin $\frac12\hbar$. If the deuteron and the triton happen to interact with their spins aligned, the minimum angular momentum in the system is $\frac32\hbar$, and the final state can't occur unless there's some orbital angular momentum to it. Most of the fusion will occur from collisions where the d and t spins happen to be anti-aligned with each other. Neglecting to account the times when the d and t are aligned parallel, and the fusion coefficient is much smaller, would overestimate the cross section by a factor of $\frac23$, or thereabouts.


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