After the Main Sequence

Section 2 of 8

Core Hydrogen Exhaustion

Part 1: Core Hydrogen Exhaustion Starts the Evolutionary Track

In old star clusters, we do not see stars vanish from the main sequence and disappear. We see them peel away onto the subgiant branch and then ascend the red giant branch. That means the end of core hydrogen burning does not shut the star off immediately. It launches a new phase.

On the main sequence, a low-mass star is in hydrostatic equilibrium (pressure gradients balance gravity) and thermal equilibrium (luminosity leaving the surface is replenished by nuclear burning in the core). Once core hydrogen is exhausted, the core can no longer replenish the energy it loses. The correct starting point is the stellar virial theorem:

Use the virial theorem as a quasi-static, ideal-gas argument for the contracting core. We are not claiming that every detail of the post-main-sequence interior can be read off from this one equation alone.

Let $K_{\text{th}}$ be the thermal kinetic energy of the gas (in erg) and $U_{\text{grav}}$ the gravitational potential energy (in erg, and negative). The total energy of the core is $E = K_{\text{th}} + U_{\text{grav}}$ . Now use the virial theorem step by step:

2K_{\text{th}} + U_{\text{grav}} = 0 \quad\Longrightarrow\quad U_{\text{grav}} = -2K_{\text{th}}.

Substitute into the total-energy expression:

E = K_{\text{th}} + U_{\text{grav}} = K_{\text{th}} - 2K_{\text{th}} = -K_{\text{th}},

so we also have

E = -K_{\text{th}} = \frac{U_{\text{grav}}}{2}.

This line is the key to the whole section. For a self-gravitating system, losing total energy makes the bound state deeper. If the core radiates energy away, then $\Delta E < 0$ . Because $E = U_{\text{grav}}/2$ , making $E$ more negative means $\Delta U_{\text{grav}} < 0$ : the potential well deepens, so the core must contract. Because $E = -K_{\text{th}}$ , making $E$ more negative means $\Delta K_{\text{th}} > 0$ : contraction raises the thermal kinetic energy. For an ideal gas,

K_{\text{th}} = \frac{3}{2}N k_B T,

so with $N$ fixed, higher $K_{\text{th}}$ means higher $T$ . This is the negative-heat-capacity behavior of self-gravitating objects: the core loses energy, contracts, and gets hotter.

After central hydrogen is exhausted, the core is mostly helium ash left behind by earlier fusion. Because the temperature is still too low for helium fusion, this helium core is inert: it contributes mass and gravity, but not new nuclear power. At the outer edge of that contracting core, a hydrogen-rich layer that was previously too cool for fusion gets compressed and heated. The shell that ignites is the layer just outside the inert helium core, because it is the first hydrogen-rich region compressed strongly by the contracting core. The center itself cannot resume hydrogen burning because its hydrogen fuel has already been exhausted.

How much inert core can a star carry? — the Schönberg–Chandrasekhar limit

There is a threshold hiding in the contraction. An inert, isothermal helium core can hold up the overlying envelope with ordinary ideal-gas pressure only up to a critical fraction of the star’s mass — the Schönberg–Chandrasekhar limit, about $\sim 10\%$ of the stellar mass. While the helium core is below that fraction, contraction is slow. Once the growing core exceeds $\sim 10\%$ , ideal-gas pressure can no longer support the envelope’s weight, and the core contracts rapidly on a thermal (Kelvin–Helmholtz) timescale — accelerating the climb up the red giant branch. (A core is reprieved only when it becomes degenerate, whose pressure needs no temperature to hold the line.) It is the same kind of statement we keep meeting: a balance — envelope weight against the maximum pressure an isothermal core can supply — solved for a critical mass.

The turnoff and subgiant observations imply that core hydrogen exhaustion does not make the star simply switch off and fade at roughly fixed structure. Instead, gravity regains control of the center, forcing contraction of the inert helium core and heating of the surrounding shell. That is why the star leaves the main sequence and enters subgiant evolution rather than disappearing from the HR diagram.

Problem

A stellar core loses energy, so $\Delta E < 0$ . Using $E = -K_{\text{th}} = U_{\text{grav}}/2$ , decide whether the core temperature rises or falls. Write your answer as a chain of algebraic statements and a one-sentence physical interpretation.