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Large stars collapse and if they are large enough form black holes. But the likelihood reduces with metallicity.

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What mechanism facilitates this? I believe it has something to do with opacity and Eddiington luminosity but haven't been able to find a source that pulls the information together in a conceptual manner.

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Metals have many electrons and many energy transitions available, therefore they are more likely to interact with photons (they increase the opacity). As a consequence, a star with higher metallicity will experience stronger winds, driven by the metal lines, that lead to a higher mass loss rate.

the mass of a star as a function of time This plot from Mapelli (2018) shows the mass of a $90 M_\odot$ star as a function of time, for different metallicities. The higher $Z$, the more dramatic is mass loss.

This of course influences the mass of the black holes that can be produced. If at the end of its life, a star has retained only a small fraction of its initial mass, it will form a black hole much smaller than if it had not lost mass. And if the mass loss has been so dramatic that the mass of the compact remnant is below the Tolman-Oppenheimer-Volkoff limit, the result might be a neutron star instead of a black hole.

This is the main reason why the likelihood of forming a black hole decreases with metallicity. High metallicity stars can loose enough mass during their life, that they are more likely to form a neutron star.

Useful reference: Heger et al. (2003)

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  • $\begingroup$ So just to make this 100% clear in my mind. A stars matter is like a sail to the photons, metallicity is like plugging any potential holes in that sail to increase it's capture of the wind? $\endgroup$
    – TheJeran
    Commented Sep 12, 2022 at 7:31
  • $\begingroup$ @TheJeran Yes, it is a good analogy $\endgroup$
    – Prallax
    Commented Sep 12, 2022 at 14:46

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