Fusion and stars
🎯What you need to be able to do
- Explain nuclear fusion and why it requires extreme temperatures.
- Describe the equilibrium that keeps a main-sequence star stable.
- Use luminosity and apparent brightness.
- Interpret the Hertzsprung–Russell diagram and the mass–luminosity relationship.
- Outline stellar evolution and the possible end states.
📚The physics
Fusion needs heat because nuclei repel. Two positively charged nuclei must be brought within about \( 10^{-15} \) m for the strong nuclear force to take over, and to get that close they must overcome the electric repulsion — the Coulomb barrier. Only at temperatures of order \( 10^{7} \) K do nuclei move fast enough. This is why fusion happens in stellar cores and is so difficult to sustain on Earth.
In the Sun the proton–proton chain converts four hydrogen nuclei into one helium-4 nucleus, releasing about 26 MeV. Referring back to E.4: helium-4 sits higher on the binding energy per nucleon curve than hydrogen, so the reaction runs downhill in energy terms.
Stellar equilibrium is the balance a main-sequence star holds for most of its life: gravity pulling inward against radiation and gas pressure from fusion pushing outward. The balance is self-correcting. If the core cools, gravity wins slightly, the core compresses, the temperature rises and fusion speeds up, pushing back out. A star is a thermostat, and this is why a star’s life is stable rather than explosive.
Luminosity and apparent brightness are different quantities and the distinction is examined. Luminosity \(L\) is the total power the star radiates, from Stefan’s law \( L = \sigma A T^{4} \). Apparent brightness \(b\) is the power we receive per square metre:
which falls off as the inverse square of distance. A dim-looking star may be an intrinsically bright one that is very far away — the reason apparent brightness alone tells you almost nothing about a star.
The Hertzsprung–Russell diagram plots luminosity against surface temperature, with temperature increasing to the left, which catches people out. Most stars lie on a diagonal band, the main sequence, running from hot bright stars at the top left to cool dim ones at the bottom right. Off that band sit the red giants (cool but luminous, so they must be enormous) and the white dwarfs (hot but faint, so they must be tiny). Reading a star’s size off its position is exactly the reasoning \( L = \sigma A T^{4} \) supports.
The mass–luminosity relationship for main-sequence stars is roughly \( L \propto M^{3.5} \). That steep power has a consequence worth stating: a star ten times the Sun’s mass is around three thousand times more luminous, so it burns through its fuel far faster and lives a far shorter life. Massive stars die young.
Stellar evolution. When core hydrogen runs out, the core contracts and heats while the outer layers expand and cool — a red giant. What follows depends on mass. A star like the Sun sheds its outer layers as a planetary nebula and leaves a white dwarf, supported by electron degeneracy pressure, but only if its remnant mass is below the Chandrasekhar limit of about 1.4 solar masses. Above that the collapse continues to a neutron star, and beyond roughly 3 solar masses of remnant, to a black hole. Fusion in massive stars builds elements only up to iron, because iron sits at the peak of the binding energy curve and fusing beyond it absorbs energy rather than releasing it — which is why the core collapse happens at all.
✏️Worked example
(a) Luminosity. Surface area \( A = 4\pi r^{2} = 4\pi \times (1.8 \times 10^{9})^{2} = 4.07 \times 10^{19} \) m\(^{2}\). Then \( L = \sigma A T^{4} = 5.67 \times 10^{-8} \times 4.07 \times 10^{19} \times 9600^{4} \). Since \( 9600^{4} = 8.50 \times 10^{15} \), \( L = 1.96 \times 10^{28} \) W.
(b) Compared with the Sun. \( 1.96 \times 10^{28}/3.85 \times 10^{26} = \) about 51 times the Sun’s luminosity.
(c) Apparent brightness at 25 light-years. \( d = 25 \times 9.46 \times 10^{15} = 2.37 \times 10^{17} \) m. Then \( b = L/(4\pi d^{2}) = 1.96 \times 10^{28}/(4\pi \times 5.60 \times 10^{34}) = 2.8 \times 10^{-8} \) W m\(^{-2}\).
(d) Where does it sit on the HR diagram? Hot and about fifty times more luminous than the Sun — upper left, on the main sequence. These are roughly the figures for Sirius A.
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. A star has radius \( 7.0 \times 10^{8} \) m and surface temperature 5800 K. Find its luminosity. Take \( \sigma = 5.67 \times 10^{-8} \) W m\(^{-2}\) K\(^{-4}\).
2. Find the apparent brightness of that star at a distance of \( 1.5 \times 10^{11} \) m (1 AU).
3. A main-sequence star has five times the mass of the Sun. Estimate its luminosity relative to the Sun, and comment on its lifetime.
4. Find the wavelength at which a 5800 K star radiates most strongly, using \( \lambda_{\max}T = 2.9 \times 10^{-3} \) m K.
5. Explain why fusion in a massive star builds elements only as far as iron.
6. State what the Chandrasekhar limit is and what it determines.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.
- HyperPhysics — stellar structure, fusion and the HR diagram
- NASA — stellar evolution and the life cycles of stars