# How does the $p+p + e^-\rightarrow D + \gamma\gamma+ \nu$ reaction proceed in the sun?

In this video on nuclear fusion the explanation of the reaction $$p+p\rightarrow D + \nu + \gamma\gamma$$ is that the two protons collide, form a weakly bound state, and then sometimes decay into the right hand side via inverse beta decay. Call that $$p + p + e^- \rightarrow pp + e^-\rightarrow D + e^+ + e^- + \nu \rightarrow D + \gamma\gamma + \nu.$$

Another way for this process to occur is for the inverse beta decay to happen first, that is \begin{align} p + p + e^- &\rightarrow p + n + \nu + \gamma\gamma &&\rightarrow D + \nu + \gamma\gamma,\text{ or} \\ p + p + e^- &\rightarrow p + n + e^+ + \nu + e^- &&\rightarrow D + \nu + \gamma\gamma. \end{align}

In the core of the sun which chain contributes more, overall, to the deuterium production? I know that the first reaction needs the inverse beta decay to happen during a very short window while the di-proton exists. The latter is less constrained by that, but is hampered by the virtual neutron. So, I'd be especially interested to see how the numbers shake out.

• Why do you say "virtual neutron"? Inverse beta decay can be $p + e \rightarrow n + \nu_e$. – Rob Jeffries Aug 9 at 15:29
• @RobJeffries Right! The neutron is still virtual (it's not on an external leg). Thanks for catching that. – Sean E. Lake Aug 10 at 5:36
• Inverse beta decay produces real neutrons. I still don't understand your use of "virtual". – Rob Jeffries Aug 10 at 11:41
• Are you sure that that should be an electron (instead of a positron) in the title? – Emilio Pisanty Aug 10 at 12:56
• @RobJeffries Any particle without any external legs on the overall Feynman diagram is virtual, right? The overall Feynman diagram starts with two protons and an electron, and ends with a deuteron, neutrino, and some photons. Everything else exists for a very short time, and is, therefore, virtual. Strictly speaking, inverse beta decay is a process with many Feynman diagrams. I'm actually only referring to the lowest order diagram for that decay as a sub-diagram of this process. – Sean E. Lake Aug 10 at 18:55

Inverse beta decay, in the form $$p + e \rightarrow n + \nu_e$$ (or some variant thereof) requires the kinetic energy of the proton and electron to at least make up the mass difference $$(m_n - m_p - m_e)c^2 = 0.785$$ MeV.
If I have my use of Wolfram Integrator correct, then the fraction of particles in the tail of the M-B distribution with $$E/k_BT > 603$$ is $$\sim 10^{-261}$$ so neutronisation of free protons just doesn't happen.