Turn a star’s temperature up just a little, and the energy it pours out doesn’t rise a little — it rises by a lot. The Stefan-Boltzmann law shows exactly why temperature has such an outsized effect on radiated power.
📘 What you need to know
The total power radiated by a perfect black body depends on its absolute temperature and its surface area
The Stefan-Boltzmann law is written as P = σAT⁴
σ is the Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴
Radiated power depends on T⁴ — a small rise in temperature causes a much larger rise in radiated power
For stars, which can be approximated as black bodies, the total power emitted is just their luminosity, L
The Stefan-Boltzmann Law
The total power radiated by a perfect black body depends on two things: its surface area, and its absolute temperature raised to the fourth power. This relationship is known as the Stefan-Boltzmann law.
Stefan-Boltzmann lawP = σAT⁴
Where P is the total power emitted across all wavelengths in watts, σ is the Stefan-Boltzmann constant, A is surface area in m², and T is absolute temperature in kelvin.
Applying Stefan-Boltzmann to Stars
Stars can be treated as good approximations of black bodies, since almost all radiation incident on them is absorbed. For a star, the total power emitted across all wavelengths, P, is exactly its luminosity, L — the total power output of the star. Since stars are (approximately) spherical, their surface area is A = 4πr², where r is the star’s radius.
Stefan-Boltzmann law for a starL = 4πr²σT⁴
A star radiates power outward in all directions — its total output depends on both its surface area and the fourth power of its temperature
🧭 Recipe: Using the Stefan-Boltzmann Law for a Star
Identify what you’re solving for — luminosity, radius, or temperature
Write down L = 4πr²σT⁴ and rearrange if needed for the unknown quantity
Make sure temperature is in kelvin before raising it to the fourth power
Substitute and calculate, keeping careful track of powers of ten
Quick recap: P = σAT⁴ in general; L = 4πr²σT⁴ for a star. Radiated power scales with the fourth power of temperature — small temperature changes have a big effect.
WE 1
A star has a radius of 3.2 × 10⁸ m and a surface temperature of 6200 K. Calculate its luminosity.
Step 1 — Write the equation
L = 4πr²σT⁴
Step 2 — Substitute the known valuesL = 4π × (3.2 × 10⁸)² × (5.67 × 10⁻⁸) × 6200⁴≈ 1.08 × 10²⁶ W
WE 2
A star has a luminosity of 3.2 × 10²⁴ W and a surface temperature of 4500 K. Calculate its radius in solar radii, given the Sun’s radius is 6.96 × 10⁸ m.
Step 1 — Rearrange the equation for radiusr = √[L ÷ (4πσT⁴)]Step 2 — Substitute the known valuesr = √[(3.2 × 10²⁴) ÷ (4π × 5.67 × 10⁻⁸ × 4500⁴)]Step 3 — Calculate and compare to the Sun’s radiusr ≈ 1.05 × 10⁸ m → r ÷ R☉ = (1.05 × 10⁸) ÷ (6.96 × 10⁸)≈ 0.150 R☉
💡 Top tips
Always convert temperature to kelvin before raising it to the fourth power
Use the star’s radius, not its diameter — a common source of factor-of-two errors
Because power scales with T⁴, doubling a star’s temperature multiplies its luminosity by sixteen, not two
Stars are only approximated as black bodies — the model is a very good one, but not perfectly exact
⚠ Common mistakes
Forgetting to raise temperature to the fourth power, rather than squaring or cubing it
Substituting diameter into the equation where radius is required
Dropping the factor of 4π when converting between the general P = σAT⁴ and the stellar version L = 4πr²σT⁴
Mixing up the Stefan-Boltzmann constant, σ, with other constants that use similar-looking symbols
Up next: Wien’s Displacement Law — where we look at how a star’s colour reveals its surface temperature.
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