<p>A bit of wild speculation that I'll need to check.</p>

<p>Photodisintegration produces simpler particles. These simpler particles are able to "neutronise" (i.e. undergo inverse beta decay/electron-capture) with electrons, because the energy threshold for neutronisation is <em>much</em> lower for free protons and alpha particles than it is for say oxygen or iron nuclei. This process removes free electrons from a core that is supported by electron degeneracy pressure and it collapses.</p>

<p>Pair production converts photon energy into electrons and positrons. The net effect will be to take kinetic energy out of the gas and turn it into rest mass. This will lower the pressure, but an increase in density/temperature can restore it.</p>

<p>I think the process of neutronisation/electron-capture is the key difference. The threshold electron energy for neutronising a free proton is 1.3 MeV, but is about 11 MeV for a proton inside an oxygen nucleus, so photodisintegration is required to trigger collapse (or the Fermi energy of the electrons reaching the neutronisation threshold).</p>

CC BY-SA 4.0
2020-06-02T13:44:39.730Z
Have you had a chance to check this yet? My *guess* is that the difference is partly because pair production can easily become a runaway process, as described in https://en.wikipedia.org/wiki/Pair-instability_supernova#Pair-instability whereas with photodisintegration some of the energy lost to photodisintegration is recovered in further fusion reactions.
PM 2Ring
CC BY-SA 4.0
2020-05-03T11:23:40.970Z
2020-05-03T11:23:40.970Z
ProfRob
<p>Its a question of energies, time-scales, and peak temperatures. As you say at the lowest stellar masses pair production causes a star to merely contract slightly, it can continue to maintain equilibrium and continues with a slightly hotter core. As the star mass increases this contraction becomes more energetic, the core heats further and reaches the point where the star can then ignite oxygen explosively. That oxygen burning then provides the energy to generate pulsations/shocks which remove some mass from the star. Even more massive stars have effectively a single large pulsation (pair instability supernovae) which unbinds the entire star. Now what happens when the star mass goes even higher? It still collapses due to pair production, ignites oxygen, and tries to unbind it self. But now the collapse is so violent, the core gets hot enough that photo-disintergrations can occur. Suddenly all that energy from oxygen burning isn't going into unbinding the star but unbinding nuclei. There is now no longer enough energy left to unbind the star and it collapses. </p>

CC BY-SA 4.0
2020-06-02T19:04:36.380Z
2020-06-02T19:04:36.380Z
Rob
<p>Stars of at least 100 solar masses or so can reach core temperatures (and, thus, core photon energies) great enough that <a href="https://en.wikipedia.org/wiki/Pair_production" rel="noreferrer">pair production</a> (where a very-high-energy photon strikes another particle, transforming the photon into a matched particle-antiparticle pair - usually an electron-positron pair, though other types are possible - and causing the other particle to recoil slightly) starts to occur in earnest.  As this is an endothermic (energy-absorbing) process, it reduces the temperature and pressure within the star’s core, causing the star to start to collapse under its own weight.</p>

<p>One of three things can now happen:</p>

<ul>
<li>In high-metallicity stars, and in low-metallicity stars up to ~130 M<sub>S</sub>, the resultant increase in core temperature and pressure halts the collapse before it can do anything irrevocable, and the star heats up and expands back out again, blowing off significant mass in the process.  These pulsations will continue until either the star becomes too small and cold for much pair production or it explodes for some other reason.</li>
<li>In low-metallicity stars from ~130 to ~250 M<sub>S</sub>, the collapse compresses and heats the interior of the star quickly and vigorously enough that the resulting increase in core reaction rate and energy release is sufficient to <a href="https://en.wikipedia.org/wiki/Pair-instability_supernova" rel="noreferrer">unbind the entire star</a>.</li>
<li>In low-metallicity stars exceeding ~250 M<sub>S</sub>, the collapse is so rapid, and the rise in core temperature so great, that an increasingly-large fraction of the photons produced have energies high enough to cause <a href="https://en.wikipedia.org/wiki/Photodisintegration" rel="noreferrer">photodisintegration</a> (where an extremely-high-energy photon is absorbed by an atomic nucleus, causing the nucleus to break apart into two or more smaller pieces), robbing the star’s core of energy and causing the entire star to collapse directly into a black hole.</li>
</ul>

<p>Both pair production and photodisintegration are endothermic (energy-absorbing) processes, and, thus, tend to cause the star to collapse.  However, their eventual effects on the collapsing star are different; pair production causes only a partial collapse, followed by runaway thermonuclear fusion and a supernova explosion, while photodisintegration causes a complete collapse into a black hole, with nothing escaping.</p>

<p>Why does core energy loss resulting from pair production result in a partial collapse, runaway fusion, and a supernova, while core energy loss resulting from photodisintegration results in a complete collapse to a black hole?  Why don’t these two endothermic processes either both result in a partial collapse with resulting explosive fusion, or both result in total collapse to a black hole?</p>

CC BY-SA 4.0
2020-05-06T11:03:23.627Z
Because of the mass difference? The strength of the gravitational force in a star with >250 solar masses just overwhelms any other physical process. Is it not as simple as that?
adrianmcmenamin
<supernova>
Why do pair production and photodisintegration have such different effects in collapsing supermassive stars?
Vikki
# Why do pair production and photodisintegration have such different effects in collapsing supermassive stars?

(Asked by: username_0 on 2020-05-02)

Stars of at least 100 solar masses or so can reach core temperatures (and, thus, core photon energies) great enough that [pair production](https://en.wikipedia.org/wiki/Pair_production) (where a very-high-energy photon strikes another particle, transforming the photon into a matched particle-antiparticle pair - usually an electron-positron pair, though other types are possible - and causing the other particle to recoil slightly) starts to occur in earnest. As this is an endothermic (energy-absorbing) process, it reduces the temperature and pressure within the star’s core, causing the star to start to collapse under its own weight.

One of three things can now happen:

* In high-metallicity stars, and in low-metallicity stars up to ~130 MS, the resultant increase in core temperature and pressure halts the collapse before it can do anything irrevocable, and the star heats up and expands back out again, blowing off significant mass in the process. These pulsations will continue until either the star becomes too small and cold for much pair production or it explodes for some other reason.
* In low-metallicity stars from ~130 to ~250 MS, the collapse compresses and heats the interior of the star quickly and vigorously enough that the resulting increase in core reaction rate and energy release is sufficient to [unbind the entire star](https://en.wikipedia.org/wiki/Pair-instability_supernova).
* In low-metallicity stars exceeding ~250 MS, the collapse is so rapid, and the rise in core temperature so great, that an increasingly-large fraction of the photons produced have energies high enough to cause [photodisintegration](https://en.wikipedia.org/wiki/Photodisintegration) (where an extremely-high-energy photon is absorbed by an atomic nucleus, causing the nucleus to break apart into two or more smaller pieces), robbing the star’s core of energy and causing the entire star to collapse directly into a black hole.

Both pair production and photodisintegration are endothermic (energy-absorbing) processes, and, thus, tend to cause the star to collapse. However, their eventual effects on the collapsing star are different; pair production causes only a partial collapse, followed by runaway thermonuclear fusion and a supernova explosion, while photodisintegration causes a complete collapse into a black hole, with nothing escaping.

Why does core energy loss resulting from pair production result in a partial collapse, runaway fusion, and a supernova, while core energy loss resulting from photodisintegration results in a complete collapse to a black hole? Why don’t these two endothermic processes either both result in a partial collapse with resulting explosive fusion, or both result in total collapse to a black hole?

**Tags:** supernova

> **Comments:**
> **username_1**: Because of the mass difference? The strength of the gravitational force in a star with >250 solar masses just overwhelms any other physical process. Is it not as simple as that?  

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## 2 answers

### Answer by username_2 on 2020-05-03. Score: 1

A bit of wild speculation that I'll need to check.

Photodisintegration produces simpler particles. These simpler particles are able to "neutronise" (i.e. undergo inverse beta decay/electron-capture) with electrons, because the energy threshold for neutronisation is *much* lower for free protons and alpha particles than it is for say oxygen or iron nuclei. This process removes free electrons from a core that is supported by electron degeneracy pressure and it collapses.

Pair production converts photon energy into electrons and positrons. The net effect will be to take kinetic energy out of the gas and turn it into rest mass. This will lower the pressure, but an increase in density/temperature can restore it.

I think the process of neutronisation/electron-capture is the key difference. The threshold electron energy for neutronising a free proton is 1.3 MeV, but is about 11 MeV for a proton inside an oxygen nucleus, so photodisintegration is required to trigger collapse (or the Fermi energy of the electrons reaching the neutronisation threshold).

> **Comments:**
> **username_3**: Have you had a chance to check this yet? My *guess* is that the difference is partly because pair production can easily become a runaway process, as described in https://en.wikipedia.org/wiki/Pair-instability_supernova#Pair-instability whereas with photodisintegration some of the energy lost to photodisintegration is recovered in further fusion reactions.  

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### Answer by username_4 on 2020-06-02. Score: 0

Its a question of energies, time-scales, and peak temperatures. As you say at the lowest stellar masses pair production causes a star to merely contract slightly, it can continue to maintain equilibrium and continues with a slightly hotter core. As the star mass increases this contraction becomes more energetic, the core heats further and reaches the point where the star can then ignite oxygen explosively. That oxygen burning then provides the energy to generate pulsations/shocks which remove some mass from the star. Even more massive stars have effectively a single large pulsation (pair instability supernovae) which unbinds the entire star. Now what happens when the star mass goes even higher? It still collapses due to pair production, ignites oxygen, and tries to unbind it self. But now the collapse is so violent, the core gets hot enough that photo-disintergrations can occur. Suddenly all that energy from oxygen burning isn't going into unbinding the star but unbinding nuclei. There is now no longer enough energy left to unbind the star and it collapses.


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