# Pion production in D-D fusion

1. Jan 6, 2014

### gildomar

1. The problem statement, all variables and given/known data

"Assuming that the incident deuteron has sufficient energy, why is the reaction $d+d \rightarrow He^{4}_{2}+\pi^{0}$ not allowed?"

2. Relevant equations

3. The attempt at a solution

I've gone through the various conservation laws that apply in a strong reaction (energy, charge, baryon number, lepton number, spin, isospin, isospin z-component, strangeness, charge conjugation, time reversal), and as far as I can tell, they're conserved (though it's a little tricky checking parity conservation). And since there's two products, momentum should be able to be conserved as well. And I know that there's nothing intrinsically wrong with just the left side, since deuteron-deuteron fusion is a viable process, but that had a nucleus and a nucleon as products, not a nucleus and a meson. Unless I messed up checking one of the conserved quantities, the closest that I can figure is that emitting a nucleon instead of a meson lowers the Q-value of the process, and is thus more favored. Or that the secondary product from the reaction (other than the nucleus) has to have directly participated in the reaction in order to properly carry off the remaining energy/momentum instead of having been part of the field around the reaction (hence why a nucleon is emitted).

2. Jan 6, 2014

### Staff: Mentor

Apart from parity (where I don't know it), I see nothing that would prevent the process. I guess it is unlikely as you need significant and nearly equal momentum transfer for two nuclei at the same time, but the question does not ask about the probability.