Stellar Evolution Explorer

Move through the main sequence and see how mass changes a star's brightness, lifetime, and final remnant.

Worked example — Stellar Evolution Explorer

Luminosity	1 L☉
Surface temperature	5,778 K
Radius	1 R☉
Main-sequence lifetime	10 Gyr
Spectral class	G
Likely remnant	White dwarf
Calculation	1 M☉
Star mass (M☉)	1 M☉

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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the brightest stars are the shortest-lived 🖖

Two readouts move from one slider, in opposite directions. This tool sets luminosity by L = M⁴ and main-sequence lifetime by τ = 10/M³ Gyr — which is exactly fuel divided by burn rate, since τ = 10 · M/L. So extra mass defeats itself: click Rigel at 21 M☉ and the panel reads 194,481 L☉ but only 1.08 million years, against the Sun's 1 L☉ and 10 billion. Twenty-one times the mass buys 194,000 times the light and costs 9,000 times the life.

why stars fall on a single line 🖖

Plot stars by temperature and brightness and they don't scatter at random — most lie along one diagonal band, the main sequence. The reason is that a single property, mass, fixes both a star's temperature and its luminosity together, so the two climb hand in hand. Slide the mass here and watch the star travel along that band: heavier means hotter, bluer and brighter, while lighter means cooler, redder and dimmer.

the diagram has a nearly empty stripe 🖖

Between the main sequence and the cool red-giant branch sits a sparsely populated region, the Hertzsprung gap. Stars don't avoid it — they cross it, but so quickly (in a few hundred thousand years, a Kelvin-Helmholtz thermal timescale) that catching one mid-crossing is rare. So the near-empty band is really a speed trap: its emptiness measures how briefly a star lingers there before ballooning into a red giant.

Problem solved in full

Luminosity, temperature, radius and lifetime of a 21 M☉ star 6 steps
A 21 M☉ star. Get its luminosity, temperature, radius and lifetime from the three power laws, then check whether the three can all be true at once. This is the Rigel state.
1. Luminosity is the steepest of the three relations, and it is why massive stars are so unforgiving. Twenty-one times the mass is nearly two hundred thousand times the light.
  
  $L = M^{4} = 2 1^{4} = 194481 \approx 1.94 \times 1 0^{5} L_{⊙}$
2. Surface temperature climbs far more gently — a factor of five for a factor of twenty-one in mass — which is why the main sequence is a long diagonal on the H–R diagram rather than a vertical line.
  
  $T_{eff} = 5778 M^{0.54} = 5778 \times 2 1^{0.54} = 5778 \times 5.176 = 29907 K$
3. Radius and lifetime come from the remaining two fits. A millionth of the Sun's lifetime is the whole point of the tool: this star will die before the Earth's continents finish rearranging.
  
  $R = M^{0.8} = 2 1^{0.8} = 11.42 R_{⊙}, τ = \frac{10}{M ^{3}} = \frac{10}{9261} = 1.08 \times 1 0^{- 3} Gyr = 1.08 Myr$
4. The lifetime law is the one thing here that is not a fit. Fuel available scales as the mass; fuel burnt per second scales as the luminosity; divide.
  
  $\frac{τ}{τ _{⊙}} = M^{- 3}, \frac{L}{L _{⊙}} = M^{4} ⟹ \frac{fuel used}{fuel available} \propto \frac{M ^{4} \cdot M ^{- 3}}{M} = 1$
5. Now the check. Stefan–Boltzmann is not optional — a star's luminosity is its area times σT⁴ — so the radius and temperature exponents already determine the luminosity exponent.
  
  $L = 4 π R^{2} σ T^{4} ⟹ \frac{L}{L _{⊙}} = M^{1.6} (M^{0.54})^{4} = M^{3.76} = 9.37 \times 1 0^{4}$
6. Take the ratio of the two answers for L. It is a clean power of the mass, which tells you the disagreement is structural rather than arithmetic.
  
  $\frac{M ^{4}}{M ^{3.76}} = M^{0.24} = 2 1^{0.24} = 2.08$
Answer

They cannot — the panel's own numbers disagree with Stefan–Boltzmann by a factor of 2.08. That is not sloppiness, it is what chaining fitted exponents costs. L = M⁴, R = M^0.8 and T = 5778 M^0.54 are three separate empirical fits to three separate observations, and Stefan's law demands the third follow from the other two: 4 must equal 2(0.8) + 4(0.54) = 3.76. It does not, and the leftover M^0.24 is exactly the 2.08 you find at 21 M☉. Each fit is fine to 10% over the mass range it was drawn for; multiply three of them and the errors are exponents, not percentages. The physics that survives all of it is the lifetime: fuel goes as M and burn rate as M⁴, so τ ∝ M⁻³, and this star gets a millionth of the Sun's ten billion years.

$21 M_{⊙} : L = 1.94 \times 1 0^{5} L_{⊙}, T = 29907 K, R = 11.42 R_{⊙}, τ = 1.08 Myr$

References (2)

The mass–luminosity relation the tool approximates as L = M⁴: Z. Eker et al., "Interrelated main-sequence mass–luminosity, mass–radius, and mass–effective temperature relations." Monthly Notices of the Royal Astronomical Society 479(4), 5491–5511, 2018.
Measured masses and radii the relation is fitted to: G. Torres, J. Andersen and A. Giménez, "Accurate masses and radii of normal stars: modern results and applications." The Astronomy and Astrophysics Review 18(1–2), 67–126, 2009.

By the 42Math team · Updated 2026-07-05

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Sun (1 M☉) - Sun-like star: about 10 Gyr main sequence, ends as a white dwarf
Rigel (21 M☉) - Massive blue supergiant: short main-sequence lifetime, likely core-collapse remnant
Proxima (0.12 M☉) - Tiny red dwarf: extremely long lifetime, far beyond the current age of the universe
Sirius (2.1 M☉) - A-type main-sequence star: brighter and shorter-lived than the Sun