The discovery of a black-hole pair with "forbidden" masses has gotten me trying to understand pair-instability supernovae. A well-crafted sentence from a recent paper gives the explanation

Population III stars above $65 M_\text{sun}$ encounter the pair instability after central carbon burning, when thermal energy creates $e^+e^-$ pairs rather than maintaining pressure support against collapse. The cores of these stars subsequently contract, triggering explosive thermonuclear burning of O and Si.

I interpret this as follows. At low temperature, the electromagnetic species in a heavy star's core are nuclei, electrons, and photons, which are in thermal equilibrium with each other. Any positrons that happen along (from e.g. weak interactions) are rapidly annihilated via

$$ e^+e^-\to\gamma\gamma.$$

However, as the temperature increases, the high-energy tail of the photon energy spectrum begins to contain a nonnegligible population with enough energy to allow the inverse process

$$ \gamma\gamma \to e^+e^-.$$

Once the pair-creation process turns on, we have a new population of particles participating in the electromagnetic thermal equilibrium. The new degree of freedom increases the heat capacity of the star's interior, and heat flows into the newly-expanded lepton sector. Most of this heat comes from the missing highest-energy photons, whose absence softens the radiation pressure; with less radiation pressure the core is allowed to contract.

My question is about the "runaway" nature of this instability. Is this a process that must run away, so that the core of the star will reach arbitrarily high temperature unless a new nuclear reaction pathway (such as O/Si burning, above) becomes available?

It seems at first like there ought to be a part of the configuration space where the core contains a secular population of positrons --- that is, where $\gamma\gamma \longleftrightarrow e^+e^-$ reaches a dynamic equilibrium, and radiation pressure recovers enough to support the more complicated core at this higher temperature. That first guess of mine is bolstered by some sources which refer to the drop in radiation pressure as "temporary." But I would think a possible late stage of some stellar evolution were a star with a stable positron core, I would have heard about it already; what I'm reading suggests that any star which develops the pair-creation instability is destroyed by it. Is this a process that must run away, or is this a process that does run away except in some cases that are unphysical for other reasons? And if it's a process that must run away, is the instability due to the chemistry of the core (so that, say, a He core and and O core would behave in some fundamentally different way), or would it behave in basically the same way regardless of the star's composition?

  • 2
    $\begingroup$ I've learned while writing this question that the ADS abstract service includes full text for lots of old papers that it didn't used to. Next on my reading list are Fraley, 1968 and Rakavy and Shaviv, 1967, the latter of which says "This [pair creation] instability seems to be quite independent of the nature of the nuclear reactions, the mode of energy losses, convection, and probably also composition." $\endgroup$
    – rob
    Commented Sep 3, 2020 at 18:39
  • $\begingroup$ I think the pair production is $\gamma\rightarrow e^+e^-$ in the field of the abundant nuclei. $\endgroup$
    – JEB
    Commented Sep 3, 2020 at 23:55
  • $\begingroup$ Pair creation from nuclei (or equivalently, real photons scattering from virtual photons) happens even at zero temperature, if you have a non-thermal source of photons above threshold. But until you have a secular population of positrons, that's just a roundabout method of $\gamma\to\gamma\gamma$ downscattering which couples the temperatures of the radiation and matter sectors. I thought that considering $\gamma\to e^+e^-$ would lower the temperature of the phase change, but I guessed that it doesn't affect whether the positron-rich phase can be stable, so I left it for a footnote/comment. $\endgroup$
    – rob
    Commented Sep 4, 2020 at 4:13

1 Answer 1


A star supported by radiation pressure is on the cusp of instability. A radiation pressure dominated star has an adiabatic index close to 4/3 - i.e. $P \propto \rho^{4/3}$.

Removing energy density from the photon gas and turning it into the rest mass of electrons and positrons softens the equation of state and the star must contract.

Hydrostatic equilibrium demands $dP/dR = -\rho g$. Looking just at proportionalities and assuming $P \propto \rho^\alpha$ and $\rho \propto M/R^3$, then the LHS of hydrostatic equilibrium is proportional to $M^\alpha R^{-3\alpha-1}$ and the RHS is proportional to $M^2 R^{-5}$.

Now the mass of the star is fixed, so whether equilibrium can be recovered at a smaller radius requires that $R^{-3\alpha -1}$ grows faster than $R^{-5}$, i.e. that $$ -3\alpha - 1 > -5$$ $$ \alpha > 4/3$$

But $\alpha = 4/3$ at best and because when the star contracts, the interior temperatures goes up, and an increasing fraction of the photon gas energy density is converted to pressure-free rest mass, then $\alpha$ stays below 4/3 unless some additional energy source, like nuclear burning, boosts the energy density.

In some circumstances (lower mass cores) this is enough to temporarily halt and reverse the contraction and you get a pulsational instability. e.g. Woosley et al. (2017) https://iopscience.iop.org/article/10.3847/1538-4357/836/2/244 These stars can evolve towards stable silicon burning prior to a supernova. But for higher mass stars and higher interior temperatures, strong neutrino cooling just hastens the rapid collapse and ultimate thermonuclear destruction of the star before an iron core is reached.


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