The Stefan–Boltzmann Law

Spec mapping: OCR H556 Module 5.5 — Astrophysics and Cosmology (the Stefan–Boltzmann law as the total power radiated by a black body per unit area: $P/A = \sigma T^{4}$P/A=σT4; the spherical-star form $L = 4\pi r^{2}\sigma T^{4}$L=4πr2σT4; the Stefan constant $\sigma = 5.67 \times 10^{-8}$σ=5.67×10−8 W m$^{-2}$−2 K$^{-4}$−4; rearrangement to solve for any of $L$L, $r$r, $T$T given the other two). Refer to the official OCR H556 specification document for exact wording.

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:

1. The total area under the curve — representing the total power radiated per unit area — grows rapidly with temperature (Stefan–Boltzmann law, this lesson).
2. The wavelength at which the curve peaks shifts towards shorter wavelengths as temperature rises (Wien's displacement law, Lesson 3).
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:

$$P/A = \sigma T^{4}$$P/A=σT4

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

$$\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$$σ=5.67×10−8Wm−2K−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

$$L = 4 \pi r^{2} \sigma T^{4}$$L=4πr2σT4

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.

$$L = 4 \pi r^{2} \sigma T^{4}$$L=4πr2σT4

Compute each factor:

$$\begin{aligned} r^{2} = (6.96 \times 10^{8})^{2} = 4.84 \times 10^{17} m^{2} \\ T^{4} = (5800)^{4} = 1.13 \times 10^{15} K^{4} \end{aligned}$$r2=(6.96×108)2=4.84×1017m2T4=(5800)4=1.13×1015K4​

Substitute:

$$\begin{aligned} L = 4 \pi \times (4.84 \times 10^{17}) \times (5.67 \times 10^{-8}) \times (1.13 \times 10^{15}) \\ = 4 \pi \times 4.84 \times 10^{17} \times 5.67 \times 10^{-8} \times 1.13 \times 10^{15} \\ = 4 \pi \times 3.10 \times 10^{25} \\ = 3.90 \times 10^{26} W \end{aligned}$$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=3.90×1026W​

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:

$$r^{2} = L / (4 \pi \sigma T^{4})$$r2=L/(4πσT4)

Compute step-by-step:

$$\begin{aligned} T^{4} = (3500)^{4} = 1.50 \times 10^{14} K^{4} \\ 4 \pi \sigma T^{4} = 4 \pi \times (5.67 \times 10^{-8}) \times (1.50 \times 10^{14}) \\ = 4 \pi \times 8.51 \times 10^{6} \\ = 1.07 \times 10^{8} W m^{-2} \\ r^{2} = (3.83 \times 10^{31}) / (1.07 \times 10^{8}) \\ = 3.58 \times 10^{23} m^{2} \\ r = 5.98 \times 10^{11} m \end{aligned}$$T4=(3500)4=1.50×1014K44πσT4=4π×(5.67×10−8)×(1.50×1014)=4π×8.51×106=1.07×108Wm−2r2=(3.83×1031)/(1.07×108)=3.58×1023m2r=5.98×1011m​

This is about 860 solar radii. Betelgeuse is so large that if it replaced the Sun, it would engulf Mercury, Venus, Earth and Mars, extending nearly to the orbit of Jupiter. Despite being more than 500 times cooler per unit area than a hotter star, its sheer size — with surface area \approx 7 × 10⁵ times that of the Sun — makes it one of the brightest stars in the night sky.

This example illustrates why we cannot say "hot = bright" without qualification. Betelgeuse is cooler than the Sun but vastly more luminous, because it is vastly larger. The interplay between r and T in L = 4πr²σT⁴ is what gives the Hertzsprung–Russell diagram (Lesson 5) its characteristic structure.

---

Worked Example 3: Temperature From Luminosity And Radius

A white dwarf has radius r = 7.0 × 10⁶ m (about the size of the Earth) and luminosity L = 4.0 × 10²³ W (roughly 10⁻³ L_☉). Calculate its surface temperature.

Rearrange:

$$\begin{aligned} T^{4} = L / (4 \pi r^{2} \sigma ) \\ = (4.0 \times 10^{23}) / (4 \pi \times (7.0 \times 10^{6})^{2} \times 5.67 \times 10^{-8}) \end{aligned}$$T4=L/(4πr2σ)=(4.0×1023)/(4π×(7.0×106)2×5.67×10−8)​

Compute the denominator:

$$\begin{aligned} r^{2} = 4.9 \times 10^{13} m^{2} \\ 4 \pi r^{2} \sigma = 4 \pi \times 4.9 \times 10^{13} \times 5.67 \times 10^{-8} \\ = 3.49 \times 10^{7} W K^{-4} \end{aligned}$$r2=4.9×1013m24πr2σ=4π×4.9×1013×5.67×10−8=3.49×107WK−4​

So:

$$\begin{aligned} T^{4} = (4.0 \times 10^{23}) / (3.49 \times 10^{7}) \\ = 1.15 \times 10^{16} K^{4} \\ T = (1.15 \times 10^{16})^(1/4) \\ = 1.04 \times 10^{4} K \end{aligned}$$T4=(4.0×1023)/(3.49×107)=1.15×1016K4T=(1.15×1016)(1/4)=1.04×104K​

So the white dwarf has a surface temperature of about 10 000 K, even hotter than the Sun. Despite being intrinsically faint (because it is so small), each square metre of its surface radiates more power than the Sun — consistent with T⁴ dominating when T > T_☉.

---

Units And Conversions

Always keep SI units:

Quantity	SI Unit	Common Non-SI Units (convert before use)
L	W	solar luminosities L_☉
r	m	solar radii R_☉, kilometres
T	K	celsius (add 273.15)
σ	W m⁻² K⁻⁴	(always this)

Common errors:

• Using celsius. T = 27°C gives T⁴ = 5.3 × 10⁵. Correct: T = 300 K gives T⁴ = 8.1 × 10⁹. Four orders of magnitude difference.
• Using solar radii. r = 100 R_☉ is not a number in metres; you must convert.
• Mixing luminosity and flux. L is in W, F is in W m⁻². The Stefan–Boltzmann law gives P/A = σT⁴ (flux at the star's surface) or L = 4πr²σT⁴ (total power). Do not confuse them with the observed flux at Earth, F_obs = L/(4πd²).

---

Why T⁴? (Background)

You will not be asked to derive the Stefan–Boltzmann law in an OCR exam, but you may find it interesting to see why T⁴ appears. The argument comes from thermodynamics and kinetic theory:

• The energy density of blackbody radiation u turns out to be u \propto T⁴, from integrating the Planck distribution.
• The power radiated per unit area is u c / 4 — the factor of c/4 comes from averaging over the angles at which photons leave the surface.

Putting these together:

$$P/A = (u c)/4 ∝ T^{4}$$P/A=(uc)/4∝T4

The proportionality constant works out to $\sigma = (2\pi^{5} k_{B}^{4})/(15 h^{3} c^{2})$σ=(2π5kB4​)/(15h3c2), which in SI units evaluates to $5.67 \times 10^{-8}$5.67×10−8 W m$^{-2}$−2 K$^{-4}$−4 — the value you use every day. The fourth power is ultimately a consequence of integrating over all wavelengths and all directions. None of this is examinable, but it shows that the "mysterious" $T^{4}$T4 is not an empirical accident: it falls out of straightforward thermodynamics.

---

A Visual Anchor: Two Black-Body Spectra At Different Temperatures

This picture summarises the two great laws of stellar radiation. Stefan–Boltzmann (this lesson) controls the area under the curve — total radiated flux per unit area. Wien's displacement law (Lesson 3) controls the position of the peak. Together they let us turn a measured spectrum into a star's temperature and radius.

---

Synoptic Links

• Inverse-square law (Lesson 1): Stefan–Boltzmann gives $L = 4\pi r^{2}\sigma T^{4}$L=4πr2σT4 at the star; the inverse-square law $F = L/(4\pi d^{2})$F=L/(4πd2) gives the flux at Earth. Combining them lets us extract $r$r from $T$T (spectroscopy) and $F$F (photometry) once $d$d is known.
• Thermal physics (Module 5.1): $T^{4}$T4 here is the kelvin temperature of the radiating surface. The same kelvin scale governs the ideal-gas law $pV = nRT$pV=nRT and Boltzmann statistics $\propto e^{-E/k_{B}T}$∝e−E/kB​T — celsius would break all three.
• Quantum physics — Planck distribution (Module 6.5 / undergraduate): $\sigma T^{4}$σT4 is the integral of the Planck spectral radiance $B(\lambda,T)$B(λ,T) over all wavelengths and solid angles. The $T^{4}$T4 scaling is a direct consequence of the photon-energy quantisation $E = hf$E=hf you met in the photoelectric-effect topic.

---

Specimen question modelled on the OCR H556 paper format