# Why do later fusion stages in a star last shorter?

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?

• 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. Dec 26 '19 at 10:43