When most of the hydrogen in the core of a massive star has fused to form helium, the next fusion stages (helium, carbon, neon, oxygen, ...) produce less and less energy in a single fusion reaction. As a consequence, the later stages last shorter and shorter. This is often explained by saying that the power output must remain constant.

This explanation seems to assume that the star (or it's core) must remain in some form of equilibrium. Why equilibrium? Why can't the total power output decrease because of the lower energy from a single fusion reaction?

My guess: if the core's luminosity decreases, the core will contract and become hotter and denser, which accelerates the fusion reactions in the core. Is this correct?

  • $\begingroup$ In my opinion, none of the answers given so far gets to the actual question: why does the total luminosity of the star not decrease as the energy production efficiency drops. To stress this point: when I did stellar models, I calculated stars of same mass (20 solar masses) with different sizes of He Cores, e.g. 4 and 2 solar masses (at the beginning of He Burning). •Both• models exhibited almost the same luminosity (within 10%)! It seems like the outer layers of a star govern it’s luminosity while the core has to adopt to provide it. $\endgroup$ Dec 26, 2019 at 10:43

2 Answers 2


In general, the minimum luminosity a star has is whilst it is on the main sequence, "burning" hydrogen in its core. Subsequent changes are driven by changes in the core composition and opacity. Fewer electrons per mass unit for heavier species leads to lower radiative opacities and higher luminosities; fewer particles per mass unit means higher temperatures are required to generate a given pressure; whilst greater Coulomb repulsion between heavier elements means that higher temperatures are required to supply these luminosities.

The timescale for a burning phase will be roughly given by the number of nuclei available to react, multiplied by the binding energy released in the fusion processes, divided by the luminosity.

All of these factors conspire to shorten the lifetime of subsequent nuclear burning phases: there are fewer reactants of greater atomic mass in a core that stays at roughly the same mass; each reaction produces less energy as the change in binding energy per nucleon is maximised in turning hydrogen to helium, but is an order of magnitude lower (and getting smaller) for subsequent burning phases; the luminosity gradually increases as the star becomes more evolved.

  • $\begingroup$ Your answer helps. I suppose "Fewer electrons per mass unit for heavier species leads to lower radiative opacities and higher luminosities" means that the number of electrons in the core decreases, so the contribution from electron scattering to the opacity decreases, so more photons leave the core: higher core luminosity. If this is correct, why does the number of electrons per unit mass decrease? My guess: because the positrons from the CNO-cycle (or pp-chains) annihilate with electrons. During He-burning and beyond the number of electrons wouldn't change much since alpha captures dominate. $\endgroup$
    – gamma1954
    Dec 25, 2019 at 19:24
  • $\begingroup$ @gamma1954 Because turning hydrogen into helium and heavier elements means turning protons into neutrons, by adding electrons. You are correct that this is not as important beyond He. However then, the number of mass units per particle increases. I will add something. $\endgroup$
    – ProfRob
    Dec 25, 2019 at 22:34

You're absolutely correct. When the fusion reactions fail to create a hydrostatic equilibrium, the star will contract. This can do a few things. If a shell around the core becomes hot enough because of the increased pressure, then it can begin fusing some elements (typically hydrogen, even in massive stars as this part of the star couldn't fuse hydrogen during the main sequence). The core can also begin fusing at a higher rate because additional pressure means greater collision rates and therefore a higher fusion rate. I would also suspect that the decreasing time period for each reaction also relies on two other aspects: the lower binding energy of progressive elements, and also lower concentrations. Progressive elements have higher binding energies, but this progression has a decreasing rate (until Iron, at which point the entire trend goes negative). This means that each successive stage of fusion produces less net energy than the last (in general. Star fusion can be a little more complex, but this is mostly true). Because of this, the star will collapse to a higher density, and therefore a higher collision rate until it reaches an equilibrium, and so burn through each progressive fuel more quickly.


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