According to introduction level textbooks $\beta^+$ decay, that is conversion of protons into neutrons only occurs in atomic nuclei.

$$p \xrightarrow{\beta^+} n + e^+ + \nu_e$$

I understand that it takes energy that free protons don't have. But what about inside stellar cores? Is there a similar, perhaps slightly more complex reaction with the same outcome, going on in there? If so, what would be the ratio, compared to better known fusion reactions?

I know there are $\beta^+$ reactions going on in bigger nuclei in the stars as part of the PP and CNO cycles. I'm asking about free protons bumping around.

  • $\begingroup$ Consider how a proton being in a stellar interior would significantly change the situation inside the proton. Note there is a large difference between a proton being in a plasma as a proton, and a proton being in a nucleus (for example, the strong and weak forces are not a factor in the first, but definitely are in the second). $\endgroup$
    – Jon Custer
    Jul 31 '19 at 15:16
  • $\begingroup$ I don't know anything about weak and strong forces except they happen in nuclei, I'm a chemist :) I'm asking if there is maybe some other similar reaction, like high energy collision between two protons or a proton and something else, that results in a positron emission. $\endgroup$
    – Milo Bem
    Jul 31 '19 at 16:23
  • 1
    $\begingroup$ physics.stackexchange.com/q/488207 might be worth a look. $\endgroup$
    – Jon Custer
    Jul 31 '19 at 17:18

"Neutronisation" or inverse beta decay is endothermic, in the sense that the neutron mass is greater than the proton by 1.29 MeV, so even neglecting the energy of any lepton number-balancing neutrino emission, you need to find kinetic energy from somewhere and convert it into rest mass.

The reaction you suggest also creates a positron, which needs even more proton kinetic energy. A slightly easier way is $$ p + e \rightarrow n + \nu_e$$ but the combined kinetic energy of the proton and electron would still need to exceed 0.785 MeV.

Temperatures in the core of a star are of order 10-100 million K, which corresponds to particle kinetic energies of only 1-10 keV.

If you work out what fraction of the tail of a Maxwell-Boltzmann distribution has energies of hundreds of times the mean energy, it is vanishingly small (as in you wouldn't expect any particle to have that kinetic energy in the whole star).

Where this reaction can happen is in the core of a collapsing massive star. Here the temperatures approach $10^{11}$ K ($\sim 10$ MeV) and free protons and electrons can combine to form a... neutron star.

The reactions can also happen in "cold", but very dense degenerate matter. For example at the centres of very massive white dwarfs, although it would likely occur for protons in nuclei in that case. Neutronisation (or inverse beta decay as it is more normally known in this context) can happen because the degenerate electrons can have much higher kinetic energies (by factors of 1000 or more) than in a perfect gas of a similar temperature. Their kinetic energies increase at higher densities and can be sufficiently high to overcome the energy threshold for neutronisation. This may lead to instability as free electrons, that provide the majority of the pressure support, are removed from the gas.

  • $\begingroup$ Thanks, that's pretty much what I was looking for. Could you estimate how rare this reaction happens in a typical star like the Sun? like, once a million years, more, less? $\endgroup$
    – Milo Bem
    Aug 19 '19 at 9:50

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.