1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.