How long does a positron last in the Sun's core? Kicking off of this question, I have a short follow-up. The fusion of hydrogen into helium in the Sun's core requires the emission of two positrons per final helium nucleus, because we start off with four protons and end up with two protons and two neutrons confined inside the helium product, and the process therefore needs to get rid of two units of positive charge.
These positrons, of course, are quickly annihilated by two electrons from the environment, squaring the final balance in the books - we take four proton-electron pairs and produce a helium atom with only two electrons.
My question is what, exactly, "quickly" means in this situation. The process is probably instantaneous with respect to most physically relevant phenomena (including, for the purposes of the previous question, transport phenomena), but it will have some nonzero timescale (zeptoseconds? nanoseconds? hours?). So: what is the mean lifetime of one of these positrons? Does it depend on conditions such as the ambient temperature and pressure or the positron's energy? Does it change if we go from the Sun to stars of other masses or in other stages of development?
This question feels to me like it is pretty elementary given sufficient knowledge of solar physics, and that it must have been answered in the 1960s at the latest. Heuristic arguments are therefore reasonable as long as they justify their hypothesis, but I'm ideally looking for an answer explicitly rooted in solid solar physics.
 A: The elementary approach would be to consider the mean free path. This is given by the number of particles per volume $n~=~N/V$ times the cross section
$$
x~=~\frac{1}{\sqrt{2}n\sigma}.
$$
The $\sqrt{2}$ comes from the Boltzmann distribution. The natural gas law $pV~=~NkT$ and $\sigma~=~2\pi r^2$ lets this be
$$
x~=~\frac{kT}{\sqrt{2}\pi pr^2}.
$$
The temperature in the sun is $p~=~1.6\times 10^7K$, the classical radius of the electron is $r~=~2.8\times 10^{-15}m$ and the pressure at the center of the sun is estimated from the average density, $1.4g/cm^3$ and its radius $6.96\times 10^{5}km$ as $p~=~9.7\times 10^{10}g/cm^2$ or $9.7\times 10^{11}Pa$. In the above formula that gives a surprising $6.5m$ for the mean free path. This might be as far as a positron travels before scattering off of either a nucleon or another electron or positron. If you assume that the positron is moving at a significant percentage of the speed of light this means it will be on this path for only $2.2\times 10^{-8}s$. If about half of these scattering events are with an electron the lifetime of a positron is then about half this number.
A: Interesting question.  In condensed matter at room temperature, there seem to be three components to the positron lifetime: fastest is direct annihilation on free electrons (more important in metals than in insulators); then annihilation of para-positronium to two photons; then annihilation of ortho-positronium to three photons.  My feeling is that you ought to be able to estimate the the direct annihilation lifetime from the number density of electrons in the Sun's core.  If positronium has a hydrogen-like excitation spectrum, though, it may be that positronium atoms are unbound in the Sun's core due to the high ambient temperature and only the free-capture lifetime matters.
(This is more of a comment than an answer, but it got too long for the comment box.)
