# AHL Stellar processes Having established the characteristics and evolution of different types of star, here we will dig a little deeper. Under what circumstances will a nebula collapse? What energy changes are happening during fusion beyond the main sequence? What happens to make a star die? And what does all of this have to do with gold ingots?

Key Concepts

Jeans criterion

Stars form when a giant molecular cloud collapses. Clouds collapse due to gravitational attraction but this is opposed by the random movement of the particles, so a cloud of given mass will only collapse if the temperature is low and the mass of the particles is high.

We can calculate whether a cloud will collapse based on whether its mass exceeds the Jeans criterion:

$$M_J={3\over 2}{kTR\over Gm}$$

• $$M_J$$ is the Jeans criterion, above which an interstellar cloud may collapse (kg)
• $$k$$ is the Boltzmann constant (1.38 × 10-23 JK-1)
• $$T$$ is absolute temperature (K)
• $$R$$ is the radius of the cloud (m)
• $$G$$ is the universal gravitational constant (6.67 × 10−11 m3 kg−1 s−2)
• $$m$$ is the average molecular mass (kg)

The Jeans criterion varies with temperature, radius and composition.

Nuclear fusion

Once a giant molecular cloud has collapsed, it forms a protostar. Nucleosynthesis (the production of new elements by combining nuclei) will commence and the star will enter the main sequence if the temperature is sufficient for hydrogren nuclei to overcome electrostatic repulsion due to like charges and fuse to form helium. Fusion releases binding energy as the strong nuclear force combines the hydrogren nuclei, meaning that a radiation pressure acts outwards, balancing against the inward force of gravity.

As the hydrogen starts to run out, the rate of fusion decreases, the radiation pressure is reduced and the star starts to collapse. This can be modelled in Algodoo.

It is stars with the highest mass that spend least time in the main sequence. This is a little contrary to common sense - shouldn't stars with the highest availability of hydrogren nuclei need longer for hydrogen fusion to complete? Instead, let's consider all relevant relationships in turn.

The mass–luminosity relation shows that luminosity is proportional to mass raised to a high power:

$$L\propto M^{3.5}$$

• $$L$$ is the luminosity of the star (W, or any consistent unit of power due to the proportional relationship)
• $$M$$ is the mass of the star (kg, or any consistent unit of mass due to the proportional relationship)
• $$3.5$$ is the exponent for main sequence stars

Assuming that luminosity is constant over the lifetime of the main sequence and knowing that luminosity is equivalent to power, we see that luminosity is inversely proportional to time:

$$L\propto {E\over t_{MS}}$$

• $$E$$ is the total binding energy released by fusion (J)
• $$t_{MS}$$ is the lifetime of the main sequence (s)

Considering that binding energy is released due to a mass defect:

$$E\propto M$$

• $$M$$ is the mass available for fusion

Therefore, combining these relationships, we see that the lifetime of the main sequence actually decreases with the mass of the star raised to a power:

$$t_{MS}\propto{1\over M^{2.5}}$$

Essentials

Nucleosynthesis off the main sequence

When a star has fused all available hydrogren, it ceases to be in its main sequence. However, fusion may yet continue depending on the temperature of the core after the collapse due to decreasing radiation pressure.

As the core gets hotter and more dense in the helium collapse, the helium nuclei may begin to fuse to beryllium. This requires a higher temperature than for hydrogren fusion, as more energy is required to overcome the electrostatic repulsion between the nuclei. When the rate of fusion decreases once more, the process of star contraction and increased temperature repeats.

In large stars with a core of $$M>4 M_\text{sun}$$, cycles of nuclear fusion continue until iron is formed. This is because iron nuclei have the maximum binding energy per nucleon and so further fusion would require more kinetic energy than is released as binding energy. The temperatures required are not sustained.

How do we know all of this? Observations of stellar spectra show the existence of different elements in stars. Nuclear fusion theory explains these.

So where do the heavier elements (like gold!) come from? One mechanism is beta decay. In radioactivity we learned that a surplus of neutrons in a nucleus (e.g. due to absorption of excess neutrons) leads to instability. Energy is released when a neutron is changed into a proton and an electron, and the retaining of the proton means that a new element is formed.

Absorption of neutrons can happen in one of the following ways:

• Rapid (r) processes in which there is not time to decay between each addition. Resulting nuclei are neutron-rich.
• Slow (s) processes in which each neutron decays before the next is added. Resulting nuclei are proton-rich. This is how almost all of the heavy elements are created.

Type Ia and II supernovae

The other route for fusion of larger nuclei occurs if the core becomes so dense that electrons combine with protons to form neutrons. The ensuing sudden collapse of the core leads to an explosion known as a supernova in which there is a rapid increase in temperature and luminosity, as shown in the simulation.

There are two types of supernovae:

• Type Ia - an exploding white dwarf that contains all the elements synthesized in the core. Their reliable light curve means that Type Ia supernovae are used as standard candles; their luminosity is known during the decay of nickel and cobalt to iron and so their distance from an observer can be determined. • Type II (as shown above) - composed mainly of hydrogen since this is what makes up the outer layers of the star. The remaining core is a neutron star.

Test Yourself

Use flashcards to practise your recall.

Use quizzes to practise application of theory.