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Death of a High-mass star

Danger

There is likely to be inaccurate information here. That and the evolution of high-mass stars is very poorly understood.

High-mass stars evolve much faster than their low-mass counterparts. The more massive a star, the faster it consumes fuel and the shorter its main-sequence lifespan becomes. For example, the sun has a lifespan of around 10 billion years, but a \(5M_\odot\) B-type star only lasts 100 million years. A \(10M_\odot\) O-type star will die in only 20 million years. Changes happen much faster in general because high-mass stars' stronger gravity generates more heat, which speeds up all phases of stellar evolution.

Leaving the main sequence

High-mass stars will run out of hydrogen in their cores and leave the main sequence far sooner than the sun. At first it is similar like the sun, an inert helium core surrounded by a hydrogen burning shell.

ZAMS = Zero Age Main Sequence

The above image shows Helium-burning evolutionary tracks, following on from the hydrogen-burning phase shown in above figure. The cyan (red if light mode) lines show the hydrogen-burning phase, and the orange (blue if light mode) lines show the helium-burning phase.

In stars with \(M\leq2.5M_\bigodot\), helium is ignited in a degenerate core, and the star moves back down from the top of its red-giant track to the red-giant clump halfway up the track (hollow circle).

In stars with \(2.5M_\bigodot \leq M \leq 9M_\bigodot\), the core is not degenerate when helium ignites, and these stars move blueward (i.e. to the left) in the H–R diagram, but move redward again (i.e. to the right) once core-helium burning is replaced by shell-helium burning. This path is called a blue loop.

In stars with \(M>9M_\bigodot\), helium ignites when the star is still close to the main sequence and does not interrupt its redward evolution.

No Helium Flash?

As covered earlier

White dwarf or Supernova?

The "common" separating line between high-mass stars and low-mass stars is around \(8M_\odot\). Stars below this mass limit never reach the 600 million K required to fuse carbon nuclei, so it becomes a white dwarf made of Carbon and Oxygen (or maybe Neon and Oxygen).

A high mass star, however, can fuse not just Hydrogen and Helium, but Carbon, Neon, and even heavier elements

Helium Burning

In high-mass stars, evolution is so fast that the star doesn't even reach the red giant region before it starts fusing Helium.

As each element is burned to depletion at the center, the core contracts and heats up, and fusion starts again. A new inner core forms, contracts again, heats again, and so on. The star’s evolutionary track continues smoothly across the supergiant region of the H–R diagram, seemingly unaffected by each new phase of burning. The star’s radius increases as its surface temperature drops, so the star swells to become a red supergiant.

As the star fuses heavier and heavier elements,

  1. the total number of nuclei decreases, which means there is less nuclei to fuse, reducing the amount of time spent on the stage.
  2. the number of protons in the nucleus increases, and as a result the repulsion between nuclei increases, requiring more energy to fuse nuclei.
  3. heavier nuclei are more stable, so fusing them releases less energy than lighter nuclei, so the star has to go through more nuclei faster.

Put all three together and each stage gets much shorter than the last.

Carbon Burning

fun fact

Betelgeuse is estimated to be at this stage right now.

Carbon Burning essentially allows to carbon nuclei to form a few products such as:

  1. \(^{12}\text{C} + \!^{12}\text{C} \rightarrow \!^{24}\text{Mg}\)

    \(^{12}\text{C} + \!^{12}\text{C} \rightarrow \!^{23}\text{Mg} + \!^1\text{n}\)

  2. \(^{12}\text{C} + \!^{12}\text{C} \rightarrow \!^{23}\text{Na} + \!^1\text{H}\)

  3. \(^{12}\text{C} + \!^{12}\text{C} \rightarrow \!^{20}\text{Ne} + \!^4\text{He}\)
  4. \(^{12}\text{C} + \!^{12}\text{C} \rightarrow \!^{16}\text{O} + 2\,^4\text{He}\)

This builds up an Oxygen-Neon-Magnesium core. Carbon Burning is inefficient, not just because of nucleide stability but also because a large fraction of energy is radiated away as neutrinoes, that do not provided any radiation pressure to support the gravity of the star.

Intermediate-mass Stars

For stars whose mass lie within \(4\) to \(8M_\odot\), this is about all they will do. After approximately 1000 years of carbon burning, the inert core consists of O/Ne/Mg. The core contracts and heats up, and creates unstable shell burning. This leads to large thermal pulses that destabilize the envelope of the star.

This star is known as a Cepheid Variable (Although they can form from higher-mass stars).

Eventually they eject their envelopes like our sun, giving us a planetary nebula and leaving a hot and dense O/Ne/Mg core behind (white dwarf)

High-mass Stars

In high-mass stars, carbon burning goes so fast that the envelope does not have time to repond, and we do not see much signs of carbon burning.

Neon Burning

For high-mass stars, the O/Ne/Mg core contracts and heats up (again) until it reaches around 1.5 billion K, with a density of approximately \(10^9 kg/m^3\) Neon Burning will make Oxygen and Magnesium nuclei, which creates an inert O/Mg core. Neon Burning is incredibly inefficient, with a large amount of energy radiated away as neutrinoes, causing reaction rates to go even higher. At such high temperatures, high-energy photons can disintegrate a fraction of Neon Nuclei, causing there to be even less Neon nuclei to support the star's gravity.

Warning

This stage lasts (approximately) a few years only.

Oxygen Burning

The O/Mg core contracts and heats up (yet again) until it reaches somewhere around 2 billion K with a density of very approximately \(10^{12} kg/m^3\) (and it's still a gas somehow)

Wait why is it Neon Burning before Oxygen?

Even though Oxygen-16 is lighter than Neon, Oxygen is magical (doubly-magic) and hence extremely stable. And this Neon is less stable and fuses at lower temperatures.

Neutrino losses from Oxygen burning are also significant, such that oxygen must burn at temperatures higher than a billion kelvins in order to maintain a radiation pressure strong enough to support the star against gravity.

Warning

This stage lasts (very approximately) a year only.

Oxygen can fuse together to form Silicon, Phosphorus, Sulfur, Chlorine, Argon, Potassium and Calcium. (Mostly Silicon and Sulfur) This creates an inert Silicon (among other things) core.

Silicon Burning (guys I swear this is the last one)

The Silicon core contracts and heats up (come on not again) until it reaches somewhere around 3 Billion Kelvin and Silicon burning starts.

Photodisintegration (where light breaks apart nuclei) removes like half the available silicon nuclei and decomposes it into a soup of Helium, protons and neutrons. The remaining Silicon nuclei fuse with the disintegrated Helium until it reaches Nickel-56 (process known as the alpha ladder or alpha process), which decays into Iron-56. And no more fusion can happen from here due to Iron being at the peak of the nuclear binding energy chart.

But isn't Nickel-62 most stable nucleus, not Iron-56?

Yes, in fact \(^{56}\text{Fe} + \!^4\text{He} \rightarrow \,^{60}\text{Ni}\) is exothermic.

But photodisintegration is also a factor here, and it turns out that the rate of photodisintegration is faster than the alpha process for that step.

Hence \(^{56}\text{Ni}\) is produced more than \(^{62}\text{Ni}\)

This builds up an inert Nickel/Iron core. Around which has a Si-burning shell, around it has a O-burning shell, around it has a Ne-burning shell, around it has a C-burning shell, around it has a He-burning shell, and around it has a H-burning shell, like an onion.

Warning

This process lasts like a day

The iron core reaches a mass of around \(1\) to \(2M_\odot\), and contracts and heat up (for the last time).

Iron core collapse

Here is where things go badly for the star. There are photons that are so high energy that they disintegrate the nuclei (photodisintegration). However, the only things that were supporting up the star was photons via radiation pressure, so the star gets robbed of its energy.

Free protons and electrons fuse into a neutron and a neutrino, however the neutrino does not contribute to radiation pressure, so this takes energy away from the core.

Eventually radiation pressure caves to gravity and the core collapses. Catastrophically.

Supernova(?)

This core collapse causes a rapid increase in temperature and density of the core. What happens next depends on the collapsing core. It will either form a neutron star for lower-mass stars and a black hole for higher-mass stars.

The earth-sized iron core (6000km) shrinks to about the width of Singapore (30~50km) in one second, at about \(\frac{1}{4}\) the speed of light. Its density is similar to an atomic nucleus. Neutron degeneracy pressure tries to hold up the core.

depending on the core mass it will become a supernova (neutron star/black hole) or go straight to a black hole (direct collapse)