Why does the formation of deuterium involve the release of an antielectron rather than electron capture

Question

In chapter 2 of Particle Physics Brick by Brick, Still describes the proton-proton chain which involves a proton changing to a neutron at 2 different stages, once through decay and once by electron capture.

In the first stages of the Proton-Proton chain, two protons collide one of which can decay into a neutron with the release of an antielectron and an electron neutrino, to create Deuterium.

Later in the chain, Berylium-7 changes a proton to a neutron with the release of an electron neutrino, but this time through the process of electron capture.

My question is why does the proton in Deuterium decay and the proton in Berryllium-7 doesn't?

Answer 1

Electron capture on beryllium is a two-body process.

Electron capture in proton-proton fusion to deuterium is a three-body process. All of the matrix elements (jargon for, roughly, reaction probabilities) are the same in the two-body $\rm pp\to d e^+\nu$ as in the three-body $\rm pe^-p\to d\nu$, but the latter requires more particles to be in the same place at the same time. This so-called "PEP reaction" apparently accounts for about 1/400th of deuterium production events in the sun's core.

The two-body process $\rm pe^-\to n\nu$ requires more energy than the three-body pep reaction; the sun isn't hot enough for that one.

Anyone who has ever scheduled a meeting can explain why three-body processes are less common than two-body processes.