How long do stars live, and how do we know?
March 10, 2026
One cluster.
One birth event.
Why do the highest-mass stars vanish first?
A \(10\,M_\odot\) star has about \(10\times\) the Sun’s fuel but about \(3000\times\) the Sun’s luminosity.
Its lifetime should be:
Target idea: lifetime = available energy / burn rate.
\[ \text{turnoff in HR diagram} \;\longrightarrow\; \text{stellar lifetime model} \;\longrightarrow\; \text{cluster age} \]
Observe → model → infer
Every Star Is a Clock
Mass sets the fuel supply.
Mass sets the burn rate.
\[ \tau \sim \frac{\text{fuel}}{\text{burn rate}} \]
More mass → more fuel
More mass → MUCH higher luminosity
Burn rate wins.
The accessible fuel supply rises with mass, but the luminosity rises much faster.
Ratio method
Given: \(M = 2\,M_\odot\), \(L = 10\,L_\odot\)
Want: \(\dfrac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}}\)
\[ \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx \left(\frac{2\,\cancel{M_\odot}}{1\,\cancel{M_\odot}}\right) \left(\frac{10\,\cancel{L_\odot}}{1\,\cancel{L_\odot}}\right)^{-1} \]
\[ = 2 \times 10^{-1} = 0.2 \]
This star gets only 20% of the Sun’s main-sequence lifetime.
\[ \tau_\text{dyn} \sim \frac{1}{\sqrt{G\bar{\rho}}} \tag{1}\]
\[ \bar{\rho} = \frac{3M}{4\pi R^3} \]
smaller radius at fixed mass → larger density
larger density → shorter dynamical time
\[ \frac{\tau_{\text{dyn}}}{\tau_{\text{dyn},\odot}} = \left(\frac{\bar{\rho}}{\bar{\rho}_\odot}\right)^{-1/2} = \left(\frac{M}{M_\odot}\right)^{-1/2} \left(\frac{R}{R_\odot}\right)^{3/2} \]
\(G\) cancels, so the ratio keeps only the physics that changes from star to star.
\[ \tau_{\text{dyn},\odot} \approx 3 \times 10^3\,\text{s} \]
50 minutes
Structural imbalances get corrected very quickly.
\[ E_{\text{grav}} \sim \frac{GM^2}{R} \]
More mass → deeper gravitational well
Larger radius → weaker binding
This is the energy gravity can release if the star contracts.
\[ L = \frac{\Delta E}{\Delta t} \]
Luminosity is power.
It tells us how fast the star is losing energy.
\[ \text{timescale} \sim \frac{\text{available energy}}{\text{energy loss rate}} \]
\[ \tau_\text{KH} \sim \frac{E_{\text{grav}}}{L} \]
\[ \tau_\text{KH} \sim \frac{GM^2}{RL} \]
Gravity-only lifetime = gravitational reservoir / luminosity.
\[ \tau_{\text{KH},\odot} \approx 3 \times 10^7\,\text{yr} \]
30 million years
Gravity alone cannot power the Sun for billions of years.
\[ \frac{\tau_{\text{KH}}}{\tau_{\text{KH},\odot}} = \left(\frac{M}{M_\odot}\right)^2 \left(\frac{R}{R_\odot}\right)^{-1} \left(\frac{L}{L_\odot}\right)^{-1} \]
Keep \(M\) and \(R\) fixed, but double \(L\).
The Kelvin-Helmholtz timescale becomes:
Kelvin’s model gave the Sun only tens of Myr.
Geology needed much more time.
Biology needed much more time.
The model was missing a deeper energy source.
Why Kelvin sounded reasonable
Why Kelvin failed
\[ \Delta E = \Delta m\,c^2 \]
Fusion taps rest-mass energy, not just thermal or chemical energy.
\[ \tau_\text{nuc} \sim \frac{\varepsilon f_M M c^2}{L} \tag{2}\]
Now the Sun can last billions of years.
\[ \begin{aligned} \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx\;& \left(\frac{\varepsilon}{\varepsilon_\odot}\right) \left(\frac{f_M}{f_{M,\odot}}\right) \left(\frac{M}{M_\odot}\right) \\ &\times \left(\frac{L}{L_\odot}\right)^{-1} \end{aligned} \]
\(\varepsilon\) is similar for H \(\rightarrow\) He fusion, but \(f_M\) can vary.
\[ \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx \left(\frac{M}{M_\odot}\right) \left(\frac{L}{L_\odot}\right)^{-1} \]
Fuel in the numerator
Burn rate in the denominator
The core competition is still fuel / burn rate.
\[ \tau_{\text{nuc},\odot} \approx 10^{10}\,\text{yr} \]
10 billion years
That is why fusion solves Kelvin’s age problem.
Given: \(M = 0.2\,M_\odot\), \(L \approx 0.008\,L_\odot\)
Given: \(f_M \approx 10\,f_{M,\odot}\) for a fully convective star
Want: \(\dfrac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}}\)
Set up the ratios first so the algebra has somewhere to go.
\[ \begin{aligned} \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} &\approx \left(\frac{\cancel{\varepsilon}}{\cancel{\varepsilon_\odot}}\right) \left(\frac{10\,\cancel{f_{M,\odot}}}{1\,\cancel{f_{M,\odot}}}\right) \left(\frac{0.2\,\cancel{M_\odot}}{1\,\cancel{M_\odot}}\right) \left(\frac{0.008\,\cancel{L_\odot}}{1\,\cancel{L_\odot}}\right)^{-1} \\ &= 10 \times 0.2 \times \frac{1}{0.008} \approx 250 \end{aligned} \]
250 \(\tau_\odot\)
Low-mass stars can outlive the current universe by orders of magnitude.
\[ \tau \sim \frac{\text{fuel}}{\text{burn rate}} \]
\[ \tau \sim \frac{M}{L} \]
\[ L \propto M^{3.5} \]
\[ \tau \propto M^{-2.5} \]
\[ \frac{\tau_{\text{MS}}}{\tau_{\text{MS},\odot}} \approx \left(\frac{M}{M_\odot}\right)^{-2.5} \]
More mass → more fuel
More mass → MUCH higher luminosity
Extra fuel does not keep up.
The hierarchy remains huge even as the thermal and nuclear clocks shrink for massive stars.
\[ \tau_{\text{dyn}} \ll \tau_{\text{KH}} \ll \tau_{\text{nuc}} \]
The most massive star still on the main sequence sets the cluster age.
A cluster still contains a blue main-sequence star of about \(10\,M_\odot\).
Target idea: younger than about \(30\,\text{Myr}\).
If the Sun would collapse in about 50 minutes without support,
what is actually holding it up?
Next time: hydrostatic equilibrium

ASTR 201 • Module 3, Lecture 1