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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.

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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.

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  • $\begingroup$ Can you provide more information on why $\sigma \sim r_\mathrm{cl}^2$ is a good approximation to the annihilation cross-section? $\endgroup$
    – Emilio Pisanty
    Commented Aug 21, 2016 at 19:24
  • $\begingroup$ I would say that for energy in the classical to semi-relativistic domain that is would work. The temperatures in the sun are $10^7K$, which corresponds to energy $E~\sim~10^2$ to $10^3eV$. This is far lower than the mass energy of the electron/positron, and corresponds to a $\gamma~\sim~1.002$ or velocity $v~\sim~.1c$. $\endgroup$
    – Lawrence B. Crowell
    Commented Aug 21, 2016 at 22:11
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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.)

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  • $\begingroup$ I suspect the unbound channel dominates, in which case we need the mean free path, the fraction of object in the plasma that are electrons, and a cross-section for annihilation at the right energies. The first two are pretty easy (density and pressure for one and hydrogen-to-helium ration should do for the second), but the third means finding some actual data. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 21, 2016 at 17:07
  • $\begingroup$ Yeah, my naive guess would be on the unbound channel as @dmckee - this is a highly ionized plasma, right? I would also be very wary of extending room-temperature results to that environment - surely there is a large population of electrons at energies comparable to the positron's, so the rest-frame energy is all over (some region of) the map, possibly changing the cross section? In any case, I mostly thought this would've been worked out by solar physics a good time ago (but maybe that's wrong, for hopefully interesting reasons?). $\endgroup$
    – Emilio Pisanty
    Commented Aug 21, 2016 at 17:13
  • $\begingroup$ @EmilioPisanty Thermal energy in the Sun's core is $kT \approx 1\rm\,keV$. Assuming the positrons have β-decay-like energies, that's hundreds of keV. So at least initially the electrons are "slow". An alternative way to phrase your question would be to ask whether the positrons thermalize before annihilating. $\endgroup$
    – rob ♦
    Commented Aug 21, 2016 at 22:16

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