5.5FieldsStretch

Black Body Radiation, Wien’s Law and Stefan’s Law

A star’s colour is a direct thermometer reading. Stars radiate almost exactly like idealised "black bodies", and two remarkably simple laws let astronomers work out both a star’s surface temperature and, from that, its physical size — all without ever leaving Earth.

40 min Video by For the Love of Physics Module 5.5 Astrophysics and cosmology

Blackbody Radiation - Experimental Observations: Stefan Law, Wiens Displacement LawWatch the full walkthrough before the notes below.

Open on YouTube

What you'll be able to do

Describe a black body and its emission spectrum
Use Wien’s displacement law
Use the Stefan-Boltzmann law
Combine both laws to estimate a star’s temperature and radius

Black body radiation

A $black body$ is a perfect absorber and emitter of electromagnetic radiation — it absorbs all radiation falling on it, and its emission spectrum depends only on its temperature, not on the material it is made of. Stars are excellent approximations of black bodies. A black body’s emission spectrum is continuous, with intensity peaking at a particular wavelength that shifts depending on temperature.

Tip — A hotter black body doesn’t just radiate more overall — its emission spectrum shifts to shorter (bluer) wavelengths at the peak, which is exactly why very hot stars look blue-white and cooler stars look orange-red.

Wien’s displacement law

Wien’s law states that the wavelength at which a black body’s emission peaks is inversely proportional to its absolute temperature.

λ_{m a x} T = constant = 2.9 \times 1 0^{- 3} m K

Wien’s displacement law.

1Rearrange

λ_{m a x} T = 2.9 \times 1 0^{- 3}

T = \frac{2.9 \times 1 0 ^{- 3}}{λ _{m a x}}

T = \frac{2.9 \times 1 0 ^{- 3}}{480 \times 1 0 ^{- 9}}

Answer

T \approx 6040

The Stefan-Boltzmann law

The Stefan-Boltzmann law states that the total power radiated per unit surface area of a black body is proportional to the fourth power of its absolute temperature. For a spherical star of radius $r$ , total luminosity follows by multiplying by its surface area.

L = σ A T^{4} = 4 π r^{2} σ T^{4}

σ = Stefan-Boltzmann constant (5.67 × 10⁻⁸ W m⁻² K⁻⁴).

1Use

L = 4 π r^{2} σ T^{4}

L = 4 π \times (5.0 \times 1 0^{8})^{2} \times 5.67 \times 1 0^{- 8} \times 600 0^{4}

= 4 π \times 2.5 \times 1 0^{17} \times 5.67 \times 1 0^{- 8} \times 1.296 \times 1 0^{15}

Answer

L \approx 2.31 \times 1 0^{26}

Tip — Combining Wien’s law (find T from peak wavelength) and Stefan’s law (find radius from L and T) is exactly how astronomers estimate the physical sizes of stars they can never directly measure with a ruler.

Equation recap

λ_{m a x} T = 2.9 \times 1 0^{- 3} m K

Wien’s displacement law.

L = 4 π r^{2} σ T^{4}

Stefan-Boltzmann law (luminosity of a spherical star).

Common mistakes to avoid

Assuming a hotter star simply looks "brighter" with no colour change.

Wien’s law means a hotter star’s peak emission shifts to shorter (bluer) wavelengths — hot stars look blue-white, cool stars look red.

Forgetting to raise temperature to the FOURTH power in the Stefan-Boltzmann law.

L ∝ T⁴ — a modest increase in temperature produces a very large increase in radiated power.

Using the star’s surface area formula for a flat disc instead of a sphere.

Use the full spherical surface area, 4πr², since a star radiates in all directions from its whole surface.

Key takeaways

A black body’s emission spectrum depends only on temperature; peak wavelength follows Wien’s law: λmaxT = 2.9×10⁻³ m K.
Stefan-Boltzmann law: L = 4πr²σT⁴, giving a star’s luminosity from its radius and temperature.
Combining Wien’s and Stefan’s laws lets astronomers estimate both a star’s temperature and its radius.

Test yourself

Ready to lock in Black Body Radiation, Wien’s Law and Stefan’s Law? Pick a mode and earn XP & Dobloons.

Black Body Radiation, Wien’s Law and Stefan’s Law

40 min Video by For the Love of Physics Module 5.5 Astrophysics and cosmology