The Stefan–Boltzmann Law

A star's luminosity (Lesson 1) is not set by some arbitrary engineering constraint; it is determined by the physics of thermal radiation from a hot body. Every object above absolute zero emits electromagnetic radiation, and the rate at which it does so depends on only two things: its surface area and its temperature. For a star, this gives us a beautifully simple law that relates luminosity, radius and temperature.

That law is the Stefan–Boltzmann law, and it is the single most important equation in OCR Module 5.5. Learning it, deriving its consequences, and using it to compute stellar properties will occupy most of this and the next two lessons.

This lesson introduces the law, its physical meaning, and how to apply it to real stars. In Lesson 3 we shall complement it with Wien's displacement law, and in Lesson 4 we shall combine the two to extract the full set of stellar properties from observations.

Black Bodies And Thermal Radiation

Any object at a non-zero absolute temperature radiates electromagnetic energy. A cup of tea glows in the infrared; a heated poker glows red, then yellow, then white-hot as you raise its temperature. The shape of the spectrum and the total radiated power both depend on temperature in a precise way.

A black body is an idealisation: an object that absorbs all incident electromagnetic radiation at every wavelength. Because it does not reflect, its own thermal emission is particularly simple — it depends only on temperature. Stars are not perfect black bodies, but they are close enough that the idealisation gives extremely good predictions for their temperatures and luminosities.

The black body spectrum — intensity as a function of wavelength — has a characteristic shape known as the Planck curve:

graph LR
    T1[Low T<br/>cool<br/>peak IR] --> S1((spectrum<br/>shifted<br/>right))
    T2[High T<br/>hot<br/>peak UV] --> S2((spectrum<br/>shifted<br/>left))
    S1 -. Wien's law .- S2

Key features of the Planck curve:

The total area under the curve — representing the total power radiated per unit area — grows rapidly with temperature (Stefan–Boltzmann law, this lesson).
The wavelength at which the curve peaks shifts towards shorter wavelengths as temperature rises (Wien's displacement law, Lesson 3).
At every wavelength, hotter black bodies emit more than cooler ones: the Planck curves do not cross.

OCR does not ask you to derive the Planck curve or use it directly, but you should know its shape qualitatively — this comes up in multiple choice questions on distinguishing stars.

Statement Of The Stefan–Boltzmann Law

The Stefan–Boltzmann law states that the total power radiated per unit area from the surface of a black body is proportional to the fourth power of its absolute temperature:

P/A = σ T⁴

where σ is the Stefan constant (also called the Stefan–Boltzmann constant):

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

This value appears on the OCR data sheet. Commit it to memory to the extent that you recognise it on sight; you do not need to memorise it exactly.

For a spherical star of radius r, the total surface area is 4πr², and the luminosity (total power) is therefore

L = 4π r² σ T⁴

This is the form of the law you will use most often. Know it by heart; it is specification point 5.5.1(c).

Exam Tip: The Stefan–Boltzmann law is dimensional — every term must be in SI units. Temperature is in kelvin, not celsius. Radius is in metres, not kilometres or solar radii. Luminosity is in watts, not solar luminosities.

The Power Of The Fourth Power

The T⁴ dependence is what makes the law so striking. Doubling the temperature of a star multiplies its radiated power per unit area by 2⁴ = 16. Tripling the temperature multiplies it by 81. This is why a small change in surface temperature can correspond to an enormous change in luminosity.

graph LR
    T1[T = 3000 K<br/>red dwarf] --> L1[P/A ≈ 4.6 × 10⁶<br/>W/m²]
    T2[T = 6000 K<br/>Sun-like] --> L2[P/A ≈ 7.3 × 10⁷<br/>W/m²]
    T3[T = 12000 K<br/>hot B star] --> L3[P/A ≈ 1.2 × 10⁹<br/>W/m²]

Going from 3000 K to 6000 K — doubling the temperature — raises the radiated power per unit area by a factor of sixteen. A blue-hot B star at 12 000 K radiates 256 times more per square metre than a red dwarf at 3000 K. The blackbody T⁴ law is why colour and temperature dominate stellar physics.

Worked Example 1: Luminosity Of The Sun

The Sun has radius R_☉ = 6.96 × 10⁸ m and surface temperature T_☉ = 5800 K. Predict its luminosity using the Stefan–Boltzmann law.

L = 4π r² σ T⁴

Compute each factor:

r² = (6.96 × 10⁸)² = 4.84 × 10¹⁷ m²
T⁴ = (5800)⁴ = 1.13 × 10¹⁵ K⁴

Substitute:

L = 4π × (4.84 × 10¹⁷) × (5.67 × 10⁻⁸) × (1.13 × 10¹⁵)
  = 4π × 4.84 × 10¹⁷ × 5.67 × 10⁻⁸ × 1.13 × 10¹⁵
  = 4π × 3.10 × 10²⁵
  = 3.90 × 10²⁶ W

The tabulated value is L_☉ = 3.83 × 10²⁶ W, so our prediction agrees within 2%. The small discrepancy is because the Sun is not a perfect black body — its spectrum is slightly distorted by absorption features from the photosphere — and because we rounded T to two significant figures.

This calculation is a standard OCR exam question, with slightly different values, and it confirms that a huge astronomical quantity (the Sun's power output) can be derived from only its radius and surface temperature.

Worked Example 2: Radius Of A Supergiant

The red supergiant Betelgeuse has luminosity L ≈ 10⁵ L_☉ = 3.83 × 10³¹ W and surface temperature T ≈ 3500 K. Find its radius.

Rearrange the Stefan–Boltzmann law:

r² = L / (4π σ T⁴)

Compute step-by-step:

T⁴ = (3500)⁴ = 1.50 × 10¹⁴ K⁴
4π σ T⁴ = 4π × (5.67 × 10⁻⁸) × (1.50 × 10¹⁴)
        = 4π × 8.51 × 10⁶
        = 1.07 × 10⁸ W m⁻²

Stefan–Boltzmann Law